• #### I found one spherically symmetric solution of Einstein's vacuum field e

From Hannu Poropudas@21:1/5 to All on Sun Oct 15 01:35:20 2023
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland

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• From JanPB@21:1/5 to Hannu Poropudas on Mon Oct 16 17:21:41 2023
On Sunday, October 15, 2023 at 1:35:22 AM UTC-7, Hannu Poropudas wrote:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland

There is only one solution to the spherically symmetric vacuum equations
(see e.g. Hawking & Ellis).

So you either made a mistake somewhere or your solution is the same
as Schwarzschild's, only written differently.

--
Jan

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• From Ross Finlayson@21:1/5 to JanPB on Tue Oct 17 18:39:13 2023
On Monday, October 16, 2023 at 5:21:43 PM UTC-7, JanPB wrote:
On Sunday, October 15, 2023 at 1:35:22 AM UTC-7, Hannu Poropudas wrote:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
There is only one solution to the spherically symmetric vacuum equations (see e.g. Hawking & Ellis).

So you either made a mistake somewhere or your solution is the same
as Schwarzschild's, only written differently.

--
Jan

Ha, spherically linearly, spherically rotationally, symmetric, ....

"Only one", ....

It's usually Schwarzschild and Chandrasekhar on the inside,
and Kruszkeles and Kerr, or so, on the outside, one stationary black hole, spherical.

Outside the forces are like inverse cube outside but inside linear,
the inverse square singularity, cube wall, here that the inertial
system is usualy considred to be inside and stationary, also,
the stationary, non-rotating well of a gravitational black hole.

Mathematically, ....

It seems you are to post it to "arxiv" then have others include it.

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• From patdolan@21:1/5 to Hannu Poropudas on Tue Oct 17 21:41:51 2023
On Sunday, October 15, 2023 at 1:35:22 AM UTC-7, Hannu Poropudas wrote:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland

I congratulate you, H. Poropudas. Is it Doctor Poropudas by chance? In any event, I would like to request that you apply your new metric to a) a photon grazing the surface of the sun, and b) the apogee of planet Mercury. If you decide to do this, then

PS--What have you heard about a new gadanken experiment that is reputed to have originated in the Northwestern region of the United States which appears to be taking the world of relativity by storm. I believe they call it the Big Ben Paradox.

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• From Hannu Poropudas@21:1/5 to All on Thu Oct 19 00:22:40 2023
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland

I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here. Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi. Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

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• From JanPB@21:1/5 to Hannu Poropudas on Thu Oct 19 11:41:06 2023
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi. Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan

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• From Hannu Poropudas@21:1/5 to All on Thu Oct 19 23:54:10 2023
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan

Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu

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• From Ross Finlayson@21:1/5 to Hannu Poropudas on Sat Oct 21 11:02:49 2023
On Thursday, October 19, 2023 at 11:54:12 PM UTC-7, Hannu Poropudas wrote:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu

It's either "more symmetries" or "less symmetries", or same.

You mention integration constants and so we call that "quantities",
when quantities aren't atomic algebraically to their symbolic notation. (Analytically.)

The differential quantities, here is for an ideal, "the field equations
of a singularity", "the orbit equations of a singularity", as above
this "cube wall" reflects geometrically, the large and small,
inside/outside, that the horizon is local everywhere, makes
for not necessarily centralized moments, why outside it is
inverse cube and inside flat, or that ideally that's zero instead
of infinity, in or out of a singularity.

Then it's figured that the inertial _systems_, are in their potential
and actual, for example having "singularities that vanish"
vis-a-vis "singularities that as infinite ideally always grow
while growing less than their growing Scharzschild radius",
within which matter is so dense as singular, affecting inverse cube.

It's theories of stellar pulsation. Accretion, this and that,
it's like, when twenty or more years ago,
science announced "science now makes black holes on Earth",
that, there was "of course by definition that would eat the Earth",
what "definitions" result, singularity theories in ideals,
that "maintaining the inertial center" of a black hole, is arbitrary.

Here it's about inverse cube or weight to power, "gravity",
then in coordinates "the dynamics of the singularity, in
effect". Then these are linear broadly and orthogonal
directly, "cube wall" and "oncoming cube" and "incoming cube",
inverse cube and flat.

Good luck Hannu, here then it's as about "constants and quantities", differential.

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• From mitchrae3323@gmail.com@21:1/5 to Ross Finlayson on Sun Oct 22 12:00:17 2023
On Saturday, October 21, 2023 at 11:02:52 AM UTC-7, Ross Finlayson wrote:
On Thursday, October 19, 2023 at 11:54:12 PM UTC-7, Hannu Poropudas wrote:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
It's either "more symmetries" or "less symmetries", or same.

You mention integration constants and so we call that "quantities",
when quantities aren't atomic algebraically to their symbolic notation. (Analytically.)

The differential quantities, here is for an ideal, "the field equations
of a singularity", "the orbit equations of a singularity", as above
this "cube wall" reflects geometrically, the large and small, inside/outside, that the horizon is local everywhere, makes
for not necessarily centralized moments, why outside it is
inverse cube and inside flat, or that ideally that's zero instead
of infinity, in or out of a singularity.

Then it's figured that the inertial _systems_, are in their potential
and actual, for example having "singularities that vanish"
vis-a-vis "singularities that as infinite ideally always grow
while growing less than their growing Scharzschild radius",
within which matter is so dense as singular, affecting inverse cube.

It's theories of stellar pulsation. Accretion, this and that,
it's like, when twenty or more years ago,
science announced "science now makes black holes on Earth",
that, there was "of course by definition that would eat the Earth",
what "definitions" result, singularity theories in ideals,
that "maintaining the inertial center" of a black hole, is arbitrary.

Here it's about inverse cube or weight to power, "gravity",
then in coordinates "the dynamics of the singularity, in
effect". Then these are linear broadly and orthogonal
directly, "cube wall" and "oncoming cube" and "incoming cube",
inverse cube and flat.

Good luck Hannu, here then it's as about "constants and quantities", differential.

How is a vacuum round?
The closed universe is flat geometry and in the hypersphere boundary

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Hannu Poropudas@21:1/5 to All on Tue Oct 24 01:56:47 2023
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu

I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published) to calculate two integration constants K1 and K2 of Euler-Largange equations (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Hannu Poropudas@21:1/5 to All on Wed Oct 25 02:01:53 2023
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published) to calculate two integration constants K1 and K2 of Euler-Largange equations (NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas

I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.

I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Hannu Poropudas@21:1/5 to All on Wed Oct 25 04:37:33 2023
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas

CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From JanPB@21:1/5 to Hannu Poropudas on Wed Oct 25 14:53:02 2023
On Wednesday, October 25, 2023 at 2:01:55 AM UTC-7, Hannu Poropudas wrote:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas

Again: all vacuum spherically symmetric solutions are locally
isometric to the Schwarzschild metric.

Period.

--
Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From The Starmaker@21:1/5 to JanPB on Wed Oct 25 23:17:05 2023
JanPB wrote:

and he just wasted more of your time.

--
to think the unthinkable, mention the unmentionable, say the unsayable,
and challenge the unchallengeable.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Hannu Poropudas@21:1/5 to All on Thu Oct 26 01:04:38 2023
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu

I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Hannu Poropudas@21:1/5 to All on Fri Oct 27 00:46:53 2023
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge. This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas

In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From mitchrae3323@gmail.com@21:1/5 to Hannu Poropudas on Fri Oct 27 12:27:59 2023
On Friday, October 27, 2023 at 12:46:55 AM UTC-7, Hannu Poropudas wrote:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P) rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge. This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas

Gravity has a center and is a closed curve.
It is not an open parabola metric..

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Hannu Poropudas@21:1/5 to All on Sat Oct 28 01:49:20 2023
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P) rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge. This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas

Error estimations of these two series approximations (function - (series approx function), not integrated here):

For REIF(r) : Error is approximately = 0 when 5.2*10^11 <r<7.2*10^11,
Max negative error about -2.15 near 2.72*10^11,
Max negative error about -0.2 near 8.307*10^11.

For IMIF(r): Error is approximately = 0 when 5.4*10^1<r<7.0*10^11,
Max negative error about -1.09 near 2.72*10^11,
Max negative error about -0.1 near 8.307*10^11.

These REIF(r) and IMIF(r) seems to me to be TWO BRANCHES of proper time t ?

Best Regards,
Hannu Poropudas

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Ross Finlayson@21:1/5 to Hannu Poropudas on Sat Oct 28 11:53:14 2023
On Saturday, October 28, 2023 at 1:49:22 AM UTC-7, Hannu Poropudas wrote:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P) rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge. This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
Error estimations of these two series approximations (function - (series approx function), not integrated here):

For REIF(r) : Error is approximately = 0 when 5.2*10^11 <r<7.2*10^11,
Max negative error about -2.15 near 2.72*10^11,
Max negative error about -0.2 near 8.307*10^11.

For IMIF(r): Error is approximately = 0 when 5.4*10^1<r<7.0*10^11,
Max negative error about -1.09 near 2.72*10^11,
Max negative error about -0.1 near 8.307*10^11.

These REIF(r) and IMIF(r) seems to me to be TWO BRANCHES of proper time t ?

Best Regards,
Hannu Poropudas

If you apply what to happen so they meet at rest again, what would that be?

Galilean, it'd be Galilean, invariant.

Here "invariant" is also introduction to "singularity theory" as "multiplicity theory".
Then it becomes a continuum mechanics, while having a proper time everywhere.

If it's time invariant add back what bands bring them back together.

It's for "multiplicity theory" and its closures to "singularity theory".

"Complex analysis" is a usual sort of "principled singularity theory".
Just like roots it's got a principled branch, what with respect to phase, singularities would map in singularity-free systems, or principled singluarities ,
"coordinate free", coordinates free, then with alternations in asymptotes as extremes, regimes, multiplicities. (Principled singularities and principled multiplicities.)

In this way whatever would be restitutive, to extremes and regimes,
potential is all "imaginary", virtual, while real, fleeting or light-like.

In its own closures which is complete also, flux out to the edges or
static potential, makes that in continuity theories, above conservation,
the otherwise dimensionaless fundamental terms, all are each others'
regimes, extremes, also.

--- SoupGate-Win32 v1.05
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• From patdolan@21:1/5 to Hannu Poropudas on Sat Oct 28 13:02:56 2023
On Saturday, October 28, 2023 at 1:49:22 AM UTC-7, Hannu Poropudas wrote:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P) rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge. This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
Error estimations of these two series approximations (function - (series approx function), not integrated here):

For REIF(r) : Error is approximately = 0 when 5.2*10^11 <r<7.2*10^11,
Max negative error about -2.15 near 2.72*10^11,
Max negative error about -0.2 near 8.307*10^11.

For IMIF(r): Error is approximately = 0 when 5.4*10^1<r<7.0*10^11,
Max negative error about -1.09 near 2.72*10^11,
Max negative error about -0.1 near 8.307*10^11.

These REIF(r) and IMIF(r) seems to me to be TWO BRANCHES of proper time t ?

Best Regards,
Hannu Poropudas

Hannu Poropudas, I have a challenge for you. I want you to use the new theory of gravitation, aka General Relativity, and your new spherical solution, to explicitly calculate the normal force of a 1.0 kg brick resting on my kitchen tabletop at an
altitude of 100m above sea level in Seattle Warshington. Neither Tom Roberts nor Legion nor Gary Harnagel, nor Dono, nor Jan(s), nor Prokary, nor Pyth, nor Dirk, nor Paul B. Anderson, nor Athel will even attempt such an impossibly complex computation.
Will YOU, Hannu Poropudas? If not, why not?

--- SoupGate-Win32 v1.05
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• From Ross Finlayson@21:1/5 to patdolan on Sat Oct 28 13:19:26 2023
On Saturday, October 28, 2023 at 1:02:58 PM UTC-7, patdolan wrote:
On Saturday, October 28, 2023 at 1:49:22 AM UTC-7, Hannu Poropudas wrote:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
Error estimations of these two series approximations (function - (series approx function), not integrated here):

For REIF(r) : Error is approximately = 0 when 5.2*10^11 <r<7.2*10^11,
Max negative error about -2.15 near 2.72*10^11,
Max negative error about -0.2 near 8.307*10^11.

For IMIF(r): Error is approximately = 0 when 5.4*10^1<r<7.0*10^11,
Max negative error about -1.09 near 2.72*10^11,
Max negative error about -0.1 near 8.307*10^11.

These REIF(r) and IMIF(r) seems to me to be TWO BRANCHES of proper time t ?

Best Regards,
Hannu Poropudas
Hannu Poropudas, I have a challenge for you. I want you to use the new theory of gravitation, aka General Relativity, and your new spherical solution, to explicitly calculate the normal force of a 1.0 kg brick resting on my kitchen tabletop at an
altitude of 100m above sea level in Seattle Warshington. Neither Tom Roberts nor Legion nor Gary Harnagel, nor Dono, nor Jan(s), nor Prokary, nor Pyth, nor Dirk, nor Paul B. Anderson, nor Athel will even attempt such an impossibly complex computation.
Will YOU, Hannu Poropudas? If not, why not?

Paddy what you do is go down to a flea market and find an old scale according to
spring pendulum or the off-weight, toward a balance scale, and maybe a known fixed mass weight for a balance, then also one of those new densitometers of the sorts of bathroom scales, or whatever's considered dead-weight, then you yourself can confirm small differences among the two configurations of experiment,
then explain how clocks drifted last year back around to being on time.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From patdolan@21:1/5 to Ross Finlayson on Sat Oct 28 19:33:21 2023
On Saturday, October 28, 2023 at 1:19:28 PM UTC-7, Ross Finlayson wrote:
On Saturday, October 28, 2023 at 1:02:58 PM UTC-7, patdolan wrote:
On Saturday, October 28, 2023 at 1:49:22 AM UTC-7, Hannu Poropudas wrote:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations", which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles? How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
Error estimations of these two series approximations (function - (series approx function), not integrated here):

For REIF(r) : Error is approximately = 0 when 5.2*10^11 <r<7.2*10^11, Max negative error about -2.15 near 2.72*10^11,
Max negative error about -0.2 near 8.307*10^11.

For IMIF(r): Error is approximately = 0 when 5.4*10^1<r<7.0*10^11,
Max negative error about -1.09 near 2.72*10^11,
Max negative error about -0.1 near 8.307*10^11.

These REIF(r) and IMIF(r) seems to me to be TWO BRANCHES of proper time t ?

Best Regards,
Hannu Poropudas
Hannu Poropudas, I have a challenge for you. I want you to use the new theory of gravitation, aka General Relativity, and your new spherical solution, to explicitly calculate the normal force of a 1.0 kg brick resting on my kitchen tabletop at an
altitude of 100m above sea level in Seattle Warshington. Neither Tom Roberts nor Legion nor Gary Harnagel, nor Dono, nor Jan(s), nor Prokary, nor Pyth, nor Dirk, nor Paul B. Anderson, nor Athel will even attempt such an impossibly complex computation.
Will YOU, Hannu Poropudas? If not, why not?
Paddy what you do is go down to a flea market and find an old scale according to
spring pendulum or the off-weight, toward a balance scale, and maybe a known fixed mass weight for a balance, then also one of those new densitometers of the sorts of bathroom scales, or whatever's considered dead-weight, then you yourself can confirm small differences among the two configurations of experiment,
then explain how clocks drifted last year back around to being on time.

Ross, yes, that's how you do it. And it is easy to write out the equation(s) representing that experiment.

I now request that any of the aforementioned write out the equations of the new gravitation, aka general relativity, so that we may check the results of the new theory against the old. I confidently reiterate that to do so is impossibly complex. And I
give a my proof the fact that not a single denizen of this forum can or will do it; nor have they even deigned to respond why they won't do it. Hannu Poropudas and the rest just look at me as though I am crazy for even asking. Even though Hannu is in
possession of a brand new metric contrived for this exact purpose.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From mitchrae3323@gmail.com@21:1/5 to patdolan on Sun Oct 29 11:47:23 2023
On Saturday, October 28, 2023 at 7:33:23 PM UTC-7, patdolan wrote:
On Saturday, October 28, 2023 at 1:19:28 PM UTC-7, Ross Finlayson wrote:
On Saturday, October 28, 2023 at 1:02:58 PM UTC-7, patdolan wrote:
On Saturday, October 28, 2023 at 1:49:22 AM UTC-7, Hannu Poropudas wrote:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations", which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models. Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles? How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree. >IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
Error estimations of these two series approximations (function - (series approx function), not integrated here):

For REIF(r) : Error is approximately = 0 when 5.2*10^11 <r<7.2*10^11, Max negative error about -2.15 near 2.72*10^11,
Max negative error about -0.2 near 8.307*10^11.

For IMIF(r): Error is approximately = 0 when 5.4*10^1<r<7.0*10^11,
Max negative error about -1.09 near 2.72*10^11,
Max negative error about -0.1 near 8.307*10^11.

These REIF(r) and IMIF(r) seems to me to be TWO BRANCHES of proper time t ?

Best Regards,
Hannu Poropudas
Hannu Poropudas, I have a challenge for you. I want you to use the new theory of gravitation, aka General Relativity, and your new spherical solution, to explicitly calculate the normal force of a 1.0 kg brick resting on my kitchen tabletop at an
altitude of 100m above sea level in Seattle Warshington. Neither Tom Roberts nor Legion nor Gary Harnagel, nor Dono, nor Jan(s), nor Prokary, nor Pyth, nor Dirk, nor Paul B. Anderson, nor Athel will even attempt such an impossibly complex computation.
Will YOU, Hannu Poropudas? If not, why not?
Paddy what you do is go down to a flea market and find an old scale according to
spring pendulum or the off-weight, toward a balance scale, and maybe a known
fixed mass weight for a balance, then also one of those new densitometers of
the sorts of bathroom scales, or whatever's considered dead-weight, then you
yourself can confirm small differences among the two configurations of experiment,
then explain how clocks drifted last year back around to being on time.
Ross, yes, that's how you do it. And it is easy to write out the equation(s) representing that experiment.

I now request that any of the aforementioned write out the equations of the new gravitation, aka general relativity, so that we may check the results of the new theory against the old. I confidently reiterate that to do so is impossibly complex. And I
give a my proof the fact that not a single denizen of this forum can or will do it; nor have they even deigned to respond why they won't do it. Hannu Poropudas and the rest just look at me as though I am crazy for even asking. Even though Hannu is in
possession of a brand new metric contrived for this exact purpose.

Gravity field is everywhere round. And has a center.
What is the geometry of a detached gravity wave?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Hannu Poropudas@21:1/5 to All on Mon Oct 30 00:59:20 2023
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P) rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge. This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.

+,- sign for REIF(r)

REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.

+,- sign for IMIF(r)

IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)

plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)

plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Hannu Poropudas@21:1/5 to All on Mon Oct 30 03:35:03 2023
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P) rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge. This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas

I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From patdolan@21:1/5 to Hannu Poropudas on Mon Oct 30 10:23:55 2023
On Monday, October 30, 2023 at 3:35:05 AM UTC-7, Hannu Poropudas wrote:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas

Hannu, is this how you treat the discoverer of the Big Ben Paradox? Complete silence? You will not be portrayed in a very generous way when I write my autobiography.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Hannu Poropudas@21:1/5 to All on Mon Oct 30 23:31:21 2023
maanantai 30. lokakuuta 2023 klo 19.23.57 UTC+2 patdolan kirjoitti:
On Monday, October 30, 2023 at 3:35:05 AM UTC-7, Hannu Poropudas wrote:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations", which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles? How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
Hannu, is this how you treat the discoverer of the Big Ben Paradox? Complete silence? You will not be portrayed in a very generous way when I write my autobiography.

This is question is NOT subject of my posting chain.

I noticed that in the stack exchage of physics was written about "Big Ben Paradox" question:

SR is not a theory of gravity.

Hannu

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Hannu Poropudas@21:1/5 to All on Tue Oct 31 01:40:20 2023
maanantai 30. lokakuuta 2023 klo 12.35.05 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P) r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas

I'am sorry about error in 30.10.2023 posting of mine.

Here is CORRECTED 30.10.2023 posting of mine

# CORRECTED. Two branches of coordinate time T series approx.31.10.2023 H.P.

# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 7.5*10^(-13)
# error estimation max abs negative side about -1.3*10^(-12)

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 4.8*10^(-13)
# error estimation max abs negative side about -2.4*10^(-13)

Best Regards,
Hannu Poropudas

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• From Hannu Poropudas@21:1/5 to All on Tue Oct 31 04:41:53 2023
tiistai 31. lokakuuta 2023 klo 10.40.22 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 12.35.05 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations", which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles? How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
I'am sorry about error in 30.10.2023 posting of mine.

Here is CORRECTED 30.10.2023 posting of mine

# CORRECTED. Two branches of coordinate time T series approx.31.10.2023 H.P. ># This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part) >REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 7.5*10^(-13)
# error estimation max abs negative side about -1.3*10^(-12)
# +,- formula (Primitive function, Imaginary part) >IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 4.8*10^(-13)
# error estimation max abs negative side about -2.4*10^(-13)

Best Regards,
Hannu Poropudas

ONE NOTE about one "little strange" function used in my maple calculations

csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)

Hannu

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• From patdolan@21:1/5 to Hannu Poropudas on Tue Oct 31 10:49:39 2023
On Monday, October 30, 2023 at 11:31:23 PM UTC-7, Hannu Poropudas wrote:
maanantai 30. lokakuuta 2023 klo 19.23.57 UTC+2 patdolan kirjoitti:
On Monday, October 30, 2023 at 3:35:05 AM UTC-7, Hannu Poropudas wrote:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations", which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models. Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles? How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
Hannu, is this how you treat the discoverer of the Big Ben Paradox? Complete silence? You will not be portrayed in a very generous way when I write my autobiography.
This is question is NOT subject of my posting chain.

I noticed that in the stack exchage of physics was written about "Big Ben Paradox" question:

SR is not a theory of gravity.

Hannu
I accept this answer, Hannu. You have been fully restored to my good graces.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Ross Finlayson@21:1/5 to Hannu Poropudas on Thu Nov 2 01:55:37 2023
On Tuesday, October 31, 2023 at 4:41:56 AM UTC-7, Hannu Poropudas wrote:
tiistai 31. lokakuuta 2023 klo 10.40.22 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 12.35.05 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations", which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models. Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles? How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
I'am sorry about error in 30.10.2023 posting of mine.

Here is CORRECTED 30.10.2023 posting of mine

# CORRECTED. Two branches of coordinate time T series approx.31.10.2023 H.P.
# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part) >REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 7.5*10^(-13)
# error estimation max abs negative side about -1.3*10^(-12)
# +,- formula (Primitive function, Imaginary part) >IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 4.8*10^(-13)
# error estimation max abs negative side about -2.4*10^(-13)

Best Regards,
Hannu Poropudas
ONE NOTE about one "little strange" function used in my maple calculations

csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)

Hannu

You mentioned the sign term reflecting that earlier you wrote from your derivation,
that part of it was as under-defined or de facto, a compensating term.

It seems what you are integrating is power terms resolving, why the halfs either
way have a "pseudo" product, what is a law that results why in your terms, they add
up, if you haven't explained "why" it's legal those terms wouldn't resolve, radiating usually.

It seems those would be waves falling so would result "why", is because, they are under the area terms, the absorption or radiation, how then those have to add up, to make the estimate, which as you note appears accurate.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Hannu Poropudas@21:1/5 to All on Thu Nov 2 02:41:55 2023
torstai 2. marraskuuta 2023 klo 10.55.39 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, October 31, 2023 at 4:41:56 AM UTC-7, Hannu Poropudas wrote:
tiistai 31. lokakuuta 2023 klo 10.40.22 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 12.35.05 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations", which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models. Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance. Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time ># Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^
7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^
7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
I'am sorry about error in 30.10.2023 posting of mine.

Here is CORRECTED 30.10.2023 posting of mine

# CORRECTED. Two branches of coordinate time T series approx.31.10.2023 H.P.
# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part) >REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 7.5*10^(-13)
# error estimation max abs negative side about -1.3*10^(-12)
# +,- formula (Primitive function, Imaginary part) >IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 4.8*10^(-13)
# error estimation max abs negative side about -2.4*10^(-13)

Best Regards,
Hannu Poropudas
ONE NOTE about one "little strange" function used in my maple calculations

csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)

Hannu
You mentioned the sign term reflecting that earlier you wrote from your derivation,
that part of it was as under-defined or de facto, a compensating term.

It seems what you are integrating is power terms resolving, why the halfs either
way have a "pseudo" product, what is a law that results why in your terms, they add
up, if you haven't explained "why" it's legal those terms wouldn't resolve, radiating usually.

It seems those would be waves falling so would result "why", is because, they
are under the area terms, the absorption or radiation, how then those have to
add up, to make the estimate, which as you note appears accurate.

1. I put more clearly my incomplete error estimation procedure here:

Solution was actually +,- Int(function(r),r) and only +function(r) was used
in error estimation, -function(r) was not used in error estimation for my convenience,
so error estimation is was not complete in this sense, but it gives correctly order of the error?

Error estimation was made only of (+) branch, of function(r) and (-) branch of function(r) was not used ,
so error estimation was not complete in that sense, but I made so for my convenience not
to make too long posting about error estimation.

function(r) - (series approx of function(r)),

function(r) is NOT integrated here due it is too complicated to do that.

(+) branch of the function(r) only used in error estimation (compare proper time case,
similar way was done in this case in error estimation) .

2. It should be remembered here that all mathematics in this special example was done with
complex numbers inside r=2*MG event horizon.

3. I only tried to point out here that how complex mathematics can be used to make calculations in
this special example case which I have selected to calculate completely.

who understand astrophysics better than me, if this was sensible at all ?

Best Regards,
Hannu Poropudas

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• From Hannu Poropudas@21:1/5 to All on Fri Nov 3 03:01:47 2023
torstai 2. marraskuuta 2023 klo 11.41.57 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 10.55.39 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, October 31, 2023 at 4:41:56 AM UTC-7, Hannu Poropudas wrote:
tiistai 31. lokakuuta 2023 klo 10.40.22 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 12.35.05 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models. Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better, if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023 ># REMARK: My letter convenience t=proper time T=coordinate time ># Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.
6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.
6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
I'am sorry about error in 30.10.2023 posting of mine.

Here is CORRECTED 30.10.2023 posting of mine

# CORRECTED. Two branches of coordinate time T series approx.31.10.2023 H.P.
# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here
#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part) >REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 7.5*10^(-13)
# error estimation max abs negative side about -1.3*10^(-12)
# +,- formula (Primitive function, Imaginary part) >IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 4.8*10^(-13)
# error estimation max abs negative side about -2.4*10^(-13)

Best Regards,
Hannu Poropudas
ONE NOTE about one "little strange" function used in my maple calculations

csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)

Hannu
You mentioned the sign term reflecting that earlier you wrote from your derivation,
that part of it was as under-defined or de facto, a compensating term.

It seems what you are integrating is power terms resolving, why the halfs either
way have a "pseudo" product, what is a law that results why in your terms, they add
up, if you haven't explained "why" it's legal those terms wouldn't resolve,

It seems those would be waves falling so would result "why", is because, they
are under the area terms, the absorption or radiation, how then those have to
add up, to make the estimate, which as you note appears accurate.
1. I put more clearly my incomplete error estimation procedure here:

Solution was actually +,- Int(function(r),r) and only +function(r) was used in error estimation, -function(r) was not used in error estimation for my convenience,
so error estimation is was not complete in this sense, but it gives correctly order of the error?

Error estimation was made only of (+) branch, of function(r) and (-) branch of function(r) was not used ,
so error estimation was not complete in that sense, but I made so for my convenience not
to make too long posting about error estimation.

function(r) - (series approx of function(r)),

function(r) is NOT integrated here due it is too complicated to do that.

(+) branch of the function(r) only used in error estimation (compare proper time case,
similar way was done in this case in error estimation) .

2. It should be remembered here that all mathematics in this special example was done with
complex numbers inside r=2*MG event horizon.

3. I only tried to point out here that how complex mathematics can be used to make calculations in
this special example case which I have selected to calculate completely.

who understand astrophysics better than me, if this was sensible at all ?

Best Regards,
Hannu Poropudas

SUMMARY:

# I. SUMMARY: formulae in complex calcs.(Maple9) H.P. 03.11.2023

# Spherically symmetric metric used, satisfies Einstein vac. eqs.
# m=M*G=MG, G=gravitational const.in c.g.s units, c=1 units calc.
# (conversion to c.g.s units: time -> time/c, length -> length.)
# S2-star around SgrA* black hole numerical values used
# aphelion and perihelion numerical values used
# (my earlier sci.physics.relativity posts).

# Only one example was chosen, Euler-Lagrange eqs. constants
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;

# S2-star around SgrA* black hole numerical values used
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

##############################
# Proper time t case
# +,- formula >#m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# +,- formula for proper time t-t0. Here Primitive function. >#Int(m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

#############
# Series approx. up to 7 degree Real part of proper time
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Real part of proper time t

#REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7

# From error curve
# Error estimation: about = 0 between 5.2e11<r<7.2e11
# Error estimation: max negative about -2.15 near 2.72e11
# Error estimation: max negative about -0.2 near 8.307e11

# Series approx. up to 7 degree Imaginary part of proper time
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Imaginary part of proper time

#IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7

# From error curve
# Error estimation: about = 0 between 5.4e11<r<7.0e11
# Error estimation: max negative about -1.09 near 2.72e11
# Error estimation: max negative about -0.1 near 8.307e11

# +,- formula
#plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIF(r),r=2.720522631*10^11..8.306841627*10^11);
# +,- formula
#plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIF(r),r=2.720522631*10^11..8.306841627*10^11);
#############

#########################################################
# Coordinate time T case
# +,- formula. >#(K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# coordinate time T and proper time t relation
#dT = dt*K2/(1-2*m/r), equation is from Euler-Lagrange eqs.

#+,- formula for coordinate time T-T0. Here Primitive function. >#Int((K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

############################
# Series approx. up to 7 degree
# Coordinate time T, Primitive function, Real part
# +,- formula (Primitive function, Real part) >#REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-0.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r
# From error curve
# Error estimation: max positive side about 7.5e-13
# Error estimation: max abs negative side about -1.3e-12

# +,- formula (Primitive function, Imaginary part)
# Coordinate time T, Primitive function, Imaginary part >#IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;
# From error curve
# Error estimation max positive side about 4.8e-13
# Error estimation max abs negative side about -2.4e-13

#plot(REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >############################

###############################################################
# II. SUMMARY: Earlier made complex calculations H.P. 3.11.2023

# First analytic solution. Primitive function.
#a2<=r<=a1
#a1 := 0.8306841627e12;
#a2 := 0.2720522631e12;

# +,- sign for integral, Primitive function, calculated form
# 0<=P<=Pi/2

#phi := P->-0.8328841065*I*((-2500000000+1348688373*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.7344898564)/((2500000000+1348688373*sin(P)^4-3848688373*sin(P)^2)^(1/2)*cos(P));

#r := P->-0.2259895064e24/(0.5586318996e12*sin(P)^2-0.8306841627e12);

#plot([Im(phi(P)),r(P),P=0..Pi/2]);
#plot([-Im(phi(P)),r(P),P=0..Pi/2]);
#plot([Im(phi(P)),r(P),P=0..Pi]);

###############################
# Second analytic solution. Primitive function.
#a4<=rr<=a3
#a3 := 0;
#a4 := -0.1103327381e13;

# 0<=P<=Pi/2, Primitive function, calculated form

phiphi := -0.8328841065*I*((-0.1000000000e11+4605246509*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.6786196659)/((0.1000000000e11+4605246509*sin(P)^4-0.1460524651e11*sin(P)^2)^(1/2)*cos(P));

rr := P->(0.9165165817e24*sin(P)^2-0.9165165817e24)/(0.1103327381e13*sin(P)^2+0.8306841627e12);

plot([Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([-Im(phiphi(P)),rr(P),P=0..Pi/2]);
plot([Im(phiphi(P)),rr(P),P=0..Pi]);

Best Regards,
Hannu Poropudas

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• From Hannu Poropudas@21:1/5 to All on Tue Nov 14 03:33:29 2023
perjantai 3. marraskuuta 2023 klo 12.01.50 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 11.41.57 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 10.55.39 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, October 31, 2023 at 4:41:56 AM UTC-7, Hannu Poropudas wrote:
tiistai 31. lokakuuta 2023 klo 10.40.22 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 12.35.05 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like "Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11,
2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better, if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics. Texas Symposium on Relativistic Astrophysics, Geneva 2015. 33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023 ># REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.
6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.
6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
I'am sorry about error in 30.10.2023 posting of mine.

Here is CORRECTED 30.10.2023 posting of mine

# CORRECTED. Two branches of coordinate time T series approx.31.10.2023 H.P.
# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part) >REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 7.5*10^(-13)
# error estimation max abs negative side about -1.3*10^(-12)
# +,- formula (Primitive function, Imaginary part) >IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 4.8*10^(-13)
# error estimation max abs negative side about -2.4*10^(-13)

Best Regards,
Hannu Poropudas
ONE NOTE about one "little strange" function used in my maple calculations

csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)

Hannu
You mentioned the sign term reflecting that earlier you wrote from your derivation,
that part of it was as under-defined or de facto, a compensating term.

It seems what you are integrating is power terms resolving, why the halfs either
way have a "pseudo" product, what is a law that results why in your terms, they add
up, if you haven't explained "why" it's legal those terms wouldn't resolve,

It seems those would be waves falling so would result "why", is because, they
are under the area terms, the absorption or radiation, how then those have to
add up, to make the estimate, which as you note appears accurate.
1. I put more clearly my incomplete error estimation procedure here:

Solution was actually +,- Int(function(r),r) and only +function(r) was used
in error estimation, -function(r) was not used in error estimation for my convenience,
so error estimation is was not complete in this sense, but it gives correctly order of the error?

Error estimation was made only of (+) branch, of function(r) and (-) branch of function(r) was not used ,
so error estimation was not complete in that sense, but I made so for my convenience not
to make too long posting about error estimation.

function(r) - (series approx of function(r)),

function(r) is NOT integrated here due it is too complicated to do that.

(+) branch of the function(r) only used in error estimation (compare proper time case,
similar way was done in this case in error estimation) .

2. It should be remembered here that all mathematics in this special example was done with
complex numbers inside r=2*MG event horizon.

3. I only tried to point out here that how complex mathematics can be used to make calculations in
this special example case which I have selected to calculate completely.

who understand astrophysics better than me, if this was sensible at all ?

Best Regards,
Hannu Poropudas
SUMMARY:

# I. SUMMARY: formulae in complex calcs.(Maple9) H.P. 03.11.2023

# Spherically symmetric metric used, satisfies Einstein vac. eqs.
# m=M*G=MG, G=gravitational const.in c.g.s units, c=1 units calc.
# (conversion to c.g.s units: time -> time/c, length -> length.)
# S2-star around SgrA* black hole numerical values used
# aphelion and perihelion numerical values used
# (my earlier sci.physics.relativity posts).

# Only one example was chosen, Euler-Lagrange eqs. constants
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
# S2-star around SgrA* black hole numerical values used
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
##############################
# Proper time t case
# +,- formula >#m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# +,- formula for proper time t-t0. Here Primitive function. >#Int(m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

#############
# Series approx. up to 7 degree Real part of proper time
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Real part of proper time t

#REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7

# From error curve
# Error estimation: about = 0 between 5.2e11<r<7.2e11
# Error estimation: max negative about -2.15 near 2.72e11
# Error estimation: max negative about -0.2 near 8.307e11

# Series approx. up to 7 degree Imaginary part of proper time
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Imaginary part of proper time

#IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7

# From error curve
# Error estimation: about = 0 between 5.4e11<r<7.0e11
# Error estimation: max negative about -1.09 near 2.72e11
# Error estimation: max negative about -0.1 near 8.307e11

# +,- formula
#plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIF(r),r=2.720522631*10^11..8.306841627*10^11);
# +,- formula
#plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIF(r),r=2.720522631*10^11..8.306841627*10^11);
#############

#########################################################
# Coordinate time T case
# +,- formula. >#(K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# coordinate time T and proper time t relation
#dT = dt*K2/(1-2*m/r), equation is from Euler-Lagrange eqs.

#+,- formula for coordinate time T-T0. Here Primitive function. >#Int((K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

############################
# Series approx. up to 7 degree
# Coordinate time T, Primitive function, Real part
# +,- formula (Primitive function, Real part) >#REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-0.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r
# From error curve
# Error estimation: max positive side about 7.5e-13
# Error estimation: max abs negative side about -1.3e-12
# +,- formula (Primitive function, Imaginary part)
# Coordinate time T, Primitive function, Imaginary part >#IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;
# From error curve
# Error estimation max positive side about 4.8e-13
# Error estimation max abs negative side about -2.4e-13

#plot(REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >############################

###############################################################
# II. SUMMARY: Earlier made complex calculations H.P. 3.11.2023

# First analytic solution. Primitive function.
#a2<=r<=a1
#a1 := 0.8306841627e12;
#a2 := 0.2720522631e12;

# +,- sign for integral, Primitive function, calculated form
# 0<=P<=Pi/2

#phi := P->-0.8328841065*I*((-2500000000+1348688373*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.7344898564)/((2500000000+1348688373*sin(P)^4-3848688373*sin(P)^2)^(1/2)*cos(P));

#r := P->-0.2259895064e24/(0.5586318996e12*sin(P)^2-0.8306841627e12);

#plot([Im(phi(P)),r(P),P=0..Pi/2]);
#plot([-Im(phi(P)),r(P),P=0..Pi/2]);
#plot([Im(phi(P)),r(P),P=0..Pi]);

###############################
# Second analytic solution. Primitive function.
#a4<=rr<=a3
#a3 := 0;
#a4 := -0.1103327381e13;

# 0<=P<=Pi/2, Primitive function, calculated form

phiphi := -0.8328841065*I*((-0.1000000000e11+4605246509*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.6786196659)/((0.1000000000e11+4605246509*sin(P)^4-0.1460524651e11*sin(P)^2)^(1/2)*cos(P));

rr := P->(0.9165165817e24*sin(P)^2-0.9165165817e24)/(0.1103327381e13*sin(P)^2+0.8306841627e12);

plot([Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([-Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([Im(phiphi(P)),rr(P),P=0..Pi]);

Best Regards,
Hannu Poropudas

I forget to mention that when you replace m by -m (or other words to say m->MG -> -MG, in c=1 units calc., with c.g.s units),
then this metrics ALSO satisfies Einstein's vacuum spherically symmetric field equations.

Spherically symmetric metrics which satisfies also
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1+m/r)^4*r^4*(1+2*m/r)), 0, 0, 0], [0, -1/(1+m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1+m/r)^2, 0], [0, 0, 0, 1+2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1+m/r)^4*r^4*(1+2*m/r)))*dr^2-(1/(1+m/r)^2)*dtheta^2-(sin(theta)^2/(1+m/r)^2)*dphi^2+(1+2*m/r)*dt^2

(m -> M*G/c^2 , if SI-units are used, m -> MG, if c=1 units calc. are used with c.g.s) .

I think that this kind of object could only be sensible inside event horizon r = 2*MG, if it would be sensible at all?

Best Regards, Hannu Poropudas

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Ross Finlayson@21:1/5 to Ross Finlayson on Tue Nov 14 20:08:25 2023
On Tuesday, November 14, 2023 at 7:27:16 PM UTC-8, Ross Finlayson wrote:
On Tuesday, November 14, 2023 at 3:33:32 AM UTC-8, Hannu Poropudas wrote:
perjantai 3. marraskuuta 2023 klo 12.01.50 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 11.41.57 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 10.55.39 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, October 31, 2023 at 4:41:56 AM UTC-7, Hannu Poropudas wrote:
tiistai 31. lokakuuta 2023 klo 10.40.22 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 12.35.05 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like "Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11, 2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics. Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function) and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.
6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.
6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9) ># Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9) ># Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
I'am sorry about error in 30.10.2023 posting of mine.

Here is CORRECTED 30.10.2023 posting of mine

# CORRECTED. Two branches of coordinate time T series approx.31.10.2023 H.P.
# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part) >REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 7.5*10^(-13)
# error estimation max abs negative side about -1.3*10^(-12)
# +,- formula (Primitive function, Imaginary part) >IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 4.8*10^(-13)
# error estimation max abs negative side about -2.4*10^(-13)

Best Regards,
Hannu Poropudas
ONE NOTE about one "little strange" function used in my maple calculations

csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)

Hannu
You mentioned the sign term reflecting that earlier you wrote from your derivation,
that part of it was as under-defined or de facto, a compensating term.

It seems what you are integrating is power terms resolving, why the halfs either
way have a "pseudo" product, what is a law that results why in your terms, they add
up, if you haven't explained "why" it's legal those terms wouldn't resolve,

It seems those would be waves falling so would result "why", is because, they
are under the area terms, the absorption or radiation, how then those have to
add up, to make the estimate, which as you note appears accurate.
1. I put more clearly my incomplete error estimation procedure here:

Solution was actually +,- Int(function(r),r) and only +function(r) was used
in error estimation, -function(r) was not used in error estimation for my convenience,
so error estimation is was not complete in this sense, but it gives correctly order of the error?

Error estimation was made only of (+) branch, of function(r) and (-) branch of function(r) was not used ,
so error estimation was not complete in that sense, but I made so for my convenience not
to make too long posting about error estimation.

function(r) - (series approx of function(r)),

function(r) is NOT integrated here due it is too complicated to do that.

(+) branch of the function(r) only used in error estimation (compare proper time case,
similar way was done in this case in error estimation) .

2. It should be remembered here that all mathematics in this special example was done with
complex numbers inside r=2*MG event horizon.

3. I only tried to point out here that how complex mathematics can be used to make calculations in
this special example case which I have selected to calculate completely.

who understand astrophysics better than me, if this was sensible at all ?

Best Regards,
Hannu Poropudas
SUMMARY:

# I. SUMMARY: formulae in complex calcs.(Maple9) H.P. 03.11.2023

# Spherically symmetric metric used, satisfies Einstein vac. eqs.
# m=M*G=MG, G=gravitational const.in c.g.s units, c=1 units calc.
# (conversion to c.g.s units: time -> time/c, length -> length.)
# S2-star around SgrA* black hole numerical values used
# aphelion and perihelion numerical values used
# (my earlier sci.physics.relativity posts).

# Only one example was chosen, Euler-Lagrange eqs. constants
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
# S2-star around SgrA* black hole numerical values used
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
##############################
# Proper time t case
# +,- formula >#m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# +,- formula for proper time t-t0. Here Primitive function. >#Int(m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

#############
# Series approx. up to 7 degree Real part of proper time
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Real part of proper time t

#REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7

# From error curve
# Error estimation: about = 0 between 5.2e11<r<7.2e11
# Error estimation: max negative about -2.15 near 2.72e11
# Error estimation: max negative about -0.2 near 8.307e11

# Series approx. up to 7 degree Imaginary part of proper time >#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Imaginary part of proper time

#IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7

# From error curve
# Error estimation: about = 0 between 5.4e11<r<7.0e11
# Error estimation: max negative about -1.09 near 2.72e11
# Error estimation: max negative about -0.1 near 8.307e11

# +,- formula
#plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIF(r),r=2.720522631*10^11..8.306841627*10^11);
# +,- formula
#plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIF(r),r=2.720522631*10^11..8.306841627*10^11);
#############

#########################################################
# Coordinate time T case
# +,- formula. >#(K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# coordinate time T and proper time t relation
#dT = dt*K2/(1-2*m/r), equation is from Euler-Lagrange eqs.

#+,- formula for coordinate time T-T0. Here Primitive function. >#Int((K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

############################
# Series approx. up to 7 degree
# Coordinate time T, Primitive function, Real part
# +,- formula (Primitive function, Real part) >#REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-0.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r
# From error curve
# Error estimation: max positive side about 7.5e-13
# Error estimation: max abs negative side about -1.3e-12
# +,- formula (Primitive function, Imaginary part)
# Coordinate time T, Primitive function, Imaginary part >#IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;
# From error curve
# Error estimation max positive side about 4.8e-13
# Error estimation max abs negative side about -2.4e-13

#plot(REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >############################

###############################################################
# II. SUMMARY: Earlier made complex calculations H.P. 3.11.2023

# First analytic solution. Primitive function.
#a2<=r<=a1
#a1 := 0.8306841627e12;
#a2 := 0.2720522631e12;

# +,- sign for integral, Primitive function, calculated form
# 0<=P<=Pi/2

#phi := P->-0.8328841065*I*((-2500000000+1348688373*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.7344898564)/((2500000000+1348688373*sin(P)^4-3848688373*sin(P)^2)^(1/2)*cos(P));

#r := P->-0.2259895064e24/(0.5586318996e12*sin(P)^2-0.8306841627e12);

#plot([Im(phi(P)),r(P),P=0..Pi/2]); >#plot([-Im(phi(P)),r(P),P=0..Pi/2]);
#plot([Im(phi(P)),r(P),P=0..Pi]);

###############################
# Second analytic solution. Primitive function.
#a4<=rr<=a3
#a3 := 0;
#a4 := -0.1103327381e13;

# 0<=P<=Pi/2, Primitive function, calculated form

phiphi := -0.8328841065*I*((-0.1000000000e11+4605246509*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.6786196659)/((0.1000000000e11+4605246509*sin(P)^4-0.1460524651e11*sin(P)^2)^(1/2)*cos(P));

rr := P->(0.9165165817e24*sin(P)^2-0.9165165817e24)/(0.1103327381e13*sin(P)^2+0.8306841627e12);

plot([Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([-Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([Im(phiphi(P)),rr(P),P=0..Pi]);

Best Regards,
Hannu Poropudas
I forget to mention that when you replace m by -m (or other words to say m->MG -> -MG, in c=1 units calc., with c.g.s units),
then this metrics ALSO satisfies Einstein's vacuum spherically symmetric field equations.

Spherically symmetric metrics which satisfies also
Einstein's vacuum field equations.

(c=1,G=1 units)
matrix([[m^2/((1+m/r)^4*r^4*(1+2*m/r)), 0, 0, 0], [0, -1/(1+m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1+m/r)^2, 0], [0, 0, 0, 1+2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1+m/r)^4*r^4*(1+2*m/r)))*dr^2-(1/(1+m/r)^2)*dtheta^2-(sin(theta)^2/(1+m/r)^2)*dphi^2+(1+2*m/r)*dt^2

(m -> M*G/c^2 , if SI-units are used, m -> MG, if c=1 units calc. are used with c.g.s) .

I think that this kind of object could only be sensible inside event horizon r = 2*MG, if it would be sensible at all?

Best Regards, Hannu Poropudas
It's remarkable and reminds of something like Clausius or Hooke's law,
about the derivations that go into the a-diabatic, and non-a-diabatic.

It reminds of Dirichlet function, and which side has measure, or both do, and they share, half.

There are various considerations about whether it's from the point of
view of entering, the horizon, or, leaving it.

You might relate it to radiation either way. There's a usual notion that "what enters never leaves", but, there's also a notion that a black hole
can have a center-of-mass that's part of a highly-relativistic system,
that oscillates either side of the horizon.

You might relate it to orthogonality and what results time/frequency
as like the Fourier analysis, about the fine line in the middle there,
or for a "Fourier-style" analysis.

Another consideration is again about Clausius and Hooke's law,
and what goes into that the a-diabatic and non-a-diabatic both
are things, and about the derivations that get to where there's
add a "and here multiply or rather divide each by 2, ...", then figure
out that's a most usual sort of doubling-space and the continuous
and discrete: mostly about the mathematics that goes into detailing
quantum behavior, in terms usually of the derivations that result inter-operable with the classical.

So, there's to be considered various forms of formalisms for "convergence", in the formula, to keep in mind that there are various laws of large numbers,
and that not all tests for convergence, agree.

A usual treatment of "renormalization" (quantum renormalization)
is that it's "re-de-normalization", about continuous/discrete and "quantum".

I think that most people don't know much about that 1'st and 2'nd thermo
law say different things, and the way that 2'nd thermo law has been made
nice and usual in defining "entropy" because the simplest theory is
"must come down", after Clausius and up to Kelvin, has wall-papered over
a bunch of the derivations that would later need be re-derived to make
over for what are various considerations in continuous/discrete and doubling-space, after the great success of "electron physics", has that
after the "ultraviolet catastrophe", there are more ways, to write systems that make for definitions for derivations, that keep things reasonably
finite and bounded for about three different kinds of way, that most people who don't follow philosophy of science and dogmatic mathematical physics, might not've heard of, and most that do, might not've thought of, but some do.

[continued in next message]

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• From Ross Finlayson@21:1/5 to Hannu Poropudas on Tue Nov 14 19:27:13 2023
On Tuesday, November 14, 2023 at 3:33:32 AM UTC-8, Hannu Poropudas wrote:
perjantai 3. marraskuuta 2023 klo 12.01.50 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 11.41.57 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 10.55.39 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, October 31, 2023 at 4:41:56 AM UTC-7, Hannu Poropudas wrote:
tiistai 31. lokakuuta 2023 klo 10.40.22 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 12.35.05 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like "Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11, 2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11
+,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration) (1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics. Texas Symposium on Relativistic Astrophysics, Geneva 2015. 33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.
6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.
6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
I'am sorry about error in 30.10.2023 posting of mine.

Here is CORRECTED 30.10.2023 posting of mine

# CORRECTED. Two branches of coordinate time T series approx.31.10.2023 H.P.
# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part) >REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 7.5*10^(-13)
# error estimation max abs negative side about -1.3*10^(-12)
# +,- formula (Primitive function, Imaginary part) >IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 4.8*10^(-13)
# error estimation max abs negative side about -2.4*10^(-13)

Best Regards,
Hannu Poropudas
ONE NOTE about one "little strange" function used in my maple calculations

csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)

Hannu
You mentioned the sign term reflecting that earlier you wrote from your derivation,
that part of it was as under-defined or de facto, a compensating term.

It seems what you are integrating is power terms resolving, why the halfs either
way have a "pseudo" product, what is a law that results why in your terms, they add
up, if you haven't explained "why" it's legal those terms wouldn't resolve,

It seems those would be waves falling so would result "why", is because, they
are under the area terms, the absorption or radiation, how then those have to
add up, to make the estimate, which as you note appears accurate.
1. I put more clearly my incomplete error estimation procedure here:

Solution was actually +,- Int(function(r),r) and only +function(r) was used
in error estimation, -function(r) was not used in error estimation for my convenience,
so error estimation is was not complete in this sense, but it gives correctly order of the error?

Error estimation was made only of (+) branch, of function(r) and (-) branch of function(r) was not used ,
so error estimation was not complete in that sense, but I made so for my convenience not
to make too long posting about error estimation.

function(r) - (series approx of function(r)),

function(r) is NOT integrated here due it is too complicated to do that.

(+) branch of the function(r) only used in error estimation (compare proper time case,
similar way was done in this case in error estimation) .

2. It should be remembered here that all mathematics in this special example was done with
complex numbers inside r=2*MG event horizon.

3. I only tried to point out here that how complex mathematics can be used to make calculations in
this special example case which I have selected to calculate completely.

who understand astrophysics better than me, if this was sensible at all ?

Best Regards,
Hannu Poropudas
SUMMARY:

# I. SUMMARY: formulae in complex calcs.(Maple9) H.P. 03.11.2023

# Spherically symmetric metric used, satisfies Einstein vac. eqs.
# m=M*G=MG, G=gravitational const.in c.g.s units, c=1 units calc.
# (conversion to c.g.s units: time -> time/c, length -> length.)
# S2-star around SgrA* black hole numerical values used
# aphelion and perihelion numerical values used
# (my earlier sci.physics.relativity posts).

# Only one example was chosen, Euler-Lagrange eqs. constants
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
# S2-star around SgrA* black hole numerical values used
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
##############################
# Proper time t case
# +,- formula >#m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# +,- formula for proper time t-t0. Here Primitive function. >#Int(m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

#############
# Series approx. up to 7 degree Real part of proper time
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Real part of proper time t

#REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7

# From error curve
# Error estimation: about = 0 between 5.2e11<r<7.2e11
# Error estimation: max negative about -2.15 near 2.72e11
# Error estimation: max negative about -0.2 near 8.307e11

# Series approx. up to 7 degree Imaginary part of proper time >#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Imaginary part of proper time

#IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7

# From error curve
# Error estimation: about = 0 between 5.4e11<r<7.0e11
# Error estimation: max negative about -1.09 near 2.72e11
# Error estimation: max negative about -0.1 near 8.307e11

# +,- formula
#plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIF(r),r=2.720522631*10^11..8.306841627*10^11);
# +,- formula
#plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIF(r),r=2.720522631*10^11..8.306841627*10^11);
#############

#########################################################
# Coordinate time T case
# +,- formula. >#(K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# coordinate time T and proper time t relation
#dT = dt*K2/(1-2*m/r), equation is from Euler-Lagrange eqs.

#+,- formula for coordinate time T-T0. Here Primitive function. >#Int((K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

############################
# Series approx. up to 7 degree
# Coordinate time T, Primitive function, Real part
# +,- formula (Primitive function, Real part) >#REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-0.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r
# From error curve
# Error estimation: max positive side about 7.5e-13
# Error estimation: max abs negative side about -1.3e-12
# +,- formula (Primitive function, Imaginary part)
# Coordinate time T, Primitive function, Imaginary part >#IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;
# From error curve
# Error estimation max positive side about 4.8e-13
# Error estimation max abs negative side about -2.4e-13

#plot(REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >############################

###############################################################
# II. SUMMARY: Earlier made complex calculations H.P. 3.11.2023

# First analytic solution. Primitive function.
#a2<=r<=a1
#a1 := 0.8306841627e12;
#a2 := 0.2720522631e12;

# +,- sign for integral, Primitive function, calculated form
# 0<=P<=Pi/2

#phi := P->-0.8328841065*I*((-2500000000+1348688373*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.7344898564)/((2500000000+1348688373*sin(P)^4-3848688373*sin(P)^2)^(1/2)*cos(P));

#r := P->-0.2259895064e24/(0.5586318996e12*sin(P)^2-0.8306841627e12);

#plot([Im(phi(P)),r(P),P=0..Pi/2]);
#plot([-Im(phi(P)),r(P),P=0..Pi/2]);
#plot([Im(phi(P)),r(P),P=0..Pi]);

###############################
# Second analytic solution. Primitive function.
#a4<=rr<=a3
#a3 := 0;
#a4 := -0.1103327381e13;

# 0<=P<=Pi/2, Primitive function, calculated form

phiphi := -0.8328841065*I*((-0.1000000000e11+4605246509*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.6786196659)/((0.1000000000e11+4605246509*sin(P)^4-0.1460524651e11*sin(P)^2)^(1/2)*cos(P));

rr := P->(0.9165165817e24*sin(P)^2-0.9165165817e24)/(0.1103327381e13*sin(P)^2+0.8306841627e12);

plot([Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([-Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([Im(phiphi(P)),rr(P),P=0..Pi]);

Best Regards,
Hannu Poropudas
I forget to mention that when you replace m by -m (or other words to say m->MG -> -MG, in c=1 units calc., with c.g.s units),
then this metrics ALSO satisfies Einstein's vacuum spherically symmetric field equations.

Spherically symmetric metrics which satisfies also
Einstein's vacuum field equations.

(c=1,G=1 units)
matrix([[m^2/((1+m/r)^4*r^4*(1+2*m/r)), 0, 0, 0], [0, -1/(1+m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1+m/r)^2, 0], [0, 0, 0, 1+2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1+m/r)^4*r^4*(1+2*m/r)))*dr^2-(1/(1+m/r)^2)*dtheta^2-(sin(theta)^2/(1+m/r)^2)*dphi^2+(1+2*m/r)*dt^2

(m -> M*G/c^2 , if SI-units are used, m -> MG, if c=1 units calc. are used with c.g.s) .

I think that this kind of object could only be sensible inside event horizon r = 2*MG, if it would be sensible at all?

Best Regards, Hannu Poropudas

It's remarkable and reminds of something like Clausius or Hooke's law,
about the derivations that go into the a-diabatic, and non-a-diabatic.

It reminds of Dirichlet function, and which side has measure, or both do,
and they share, half.

There are various considerations about whether it's from the point of
view of entering, the horizon, or, leaving it.

You might relate it to radiation either way. There's a usual notion that "what enters never leaves", but, there's also a notion that a black hole
can have a center-of-mass that's part of a highly-relativistic system,
that oscillates either side of the horizon.

You might relate it to orthogonality and what results time/frequency
as like the Fourier analysis, about the fine line in the middle there,
or for a "Fourier-style" analysis.

Another consideration is again about Clausius and Hooke's law,
and what goes into that the a-diabatic and non-a-diabatic both
are things, and about the derivations that get to where there's
add a "and here multiply or rather divide each by 2, ...", then figure
out that's a most usual sort of doubling-space and the continuous
and discrete: mostly about the mathematics that goes into detailing
quantum behavior, in terms usually of the derivations that result inter-operable with the classical.

So, there's to be considered various forms of formalisms for "convergence",
in the formula, to keep in mind that there are various laws of large numbers, and that not all tests for convergence, agree.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Hannu Poropudas@21:1/5 to All on Fri Nov 17 01:30:36 2023
keskiviikko 15. marraskuuta 2023 klo 6.08.28 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, November 14, 2023 at 7:27:16 PM UTC-8, Ross Finlayson wrote:
On Tuesday, November 14, 2023 at 3:33:32 AM UTC-8, Hannu Poropudas wrote:
perjantai 3. marraskuuta 2023 klo 12.01.50 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 11.41.57 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 10.55.39 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, October 31, 2023 at 4:41:56 AM UTC-7, Hannu Poropudas wrote:
tiistai 31. lokakuuta 2023 klo 10.40.22 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 12.35.05 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like "Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11, 2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11 +,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration) (1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation,
if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function) and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.
6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.
6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9) ># Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9) ># Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
I'am sorry about error in 30.10.2023 posting of mine.

Here is CORRECTED 30.10.2023 posting of mine

# CORRECTED. Two branches of coordinate time T series approx.31.10.2023 H.P.
# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part) >REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 7.5*10^(-13)
# error estimation max abs negative side about -1.3*10^(-12) ># +,- formula (Primitive function, Imaginary part) >IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 4.8*10^(-13)
# error estimation max abs negative side about -2.4*10^(-13)

Best Regards,
Hannu Poropudas
ONE NOTE about one "little strange" function used in my maple calculations

csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)

Hannu
You mentioned the sign term reflecting that earlier you wrote from your derivation,
that part of it was as under-defined or de facto, a compensating term.

It seems what you are integrating is power terms resolving, why the halfs either
way have a "pseudo" product, what is a law that results why in your terms, they add
up, if you haven't explained "why" it's legal those terms wouldn't resolve,

It seems those would be waves falling so would result "why", is because, they
are under the area terms, the absorption or radiation, how then those have to
add up, to make the estimate, which as you note appears accurate.
1. I put more clearly my incomplete error estimation procedure here:

Solution was actually +,- Int(function(r),r) and only +function(r) was used
in error estimation, -function(r) was not used in error estimation for my convenience,
so error estimation is was not complete in this sense, but it gives correctly order of the error?

Error estimation was made only of (+) branch, of function(r) and (-) branch of function(r) was not used ,
so error estimation was not complete in that sense, but I made so for my convenience not
to make too long posting about error estimation.

function(r) - (series approx of function(r)),

function(r) is NOT integrated here due it is too complicated to do that.

(+) branch of the function(r) only used in error estimation (compare proper time case,
similar way was done in this case in error estimation) .

2. It should be remembered here that all mathematics in this special example was done with
complex numbers inside r=2*MG event horizon.

3. I only tried to point out here that how complex mathematics can be used to make calculations in
this special example case which I have selected to calculate completely.

who understand astrophysics better than me, if this was sensible at all ?

Best Regards,
Hannu Poropudas
SUMMARY:

# I. SUMMARY: formulae in complex calcs.(Maple9) H.P. 03.11.2023

# Spherically symmetric metric used, satisfies Einstein vac. eqs.
# m=M*G=MG, G=gravitational const.in c.g.s units, c=1 units calc.
# (conversion to c.g.s units: time -> time/c, length -> length.)
# S2-star around SgrA* black hole numerical values used
# aphelion and perihelion numerical values used
# (my earlier sci.physics.relativity posts).

# Only one example was chosen, Euler-Lagrange eqs. constants
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
# S2-star around SgrA* black hole numerical values used
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
##############################
# Proper time t case
# +,- formula >#m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# +,- formula for proper time t-t0. Here Primitive function. >#Int(m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

#############
# Series approx. up to 7 degree Real part of proper time
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Real part of proper time t

#REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^7

# From error curve
# Error estimation: about = 0 between 5.2e11<r<7.2e11
# Error estimation: max negative about -2.15 near 2.72e11
# Error estimation: max negative about -0.2 near 8.307e11

# Series approx. up to 7 degree Imaginary part of proper time >#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Imaginary part of proper time

#IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^7

# From error curve
# Error estimation: about = 0 between 5.4e11<r<7.0e11
# Error estimation: max negative about -1.09 near 2.72e11
# Error estimation: max negative about -0.1 near 8.307e11

# +,- formula
#plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIF(r),r=2.720522631*10^11..8.306841627*10^11);
# +,- formula
#plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIF(r),r=2.720522631*10^11..8.306841627*10^11); >#############

#########################################################
# Coordinate time T case
# +,- formula. >#(K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# coordinate time T and proper time t relation
#dT = dt*K2/(1-2*m/r), equation is from Euler-Lagrange eqs.

#+,- formula for coordinate time T-T0. Here Primitive function. >#Int((K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

############################
# Series approx. up to 7 degree
# Coordinate time T, Primitive function, Real part
# +,- formula (Primitive function, Real part) >#REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-0.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r
# From error curve
# Error estimation: max positive side about 7.5e-13
# Error estimation: max abs negative side about -1.3e-12
# +,- formula (Primitive function, Imaginary part)
# Coordinate time T, Primitive function, Imaginary part >#IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;
# From error curve
# Error estimation max positive side about 4.8e-13
# Error estimation max abs negative side about -2.4e-13

#plot(REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >############################

###############################################################
# II. SUMMARY: Earlier made complex calculations H.P. 3.11.2023

# First analytic solution. Primitive function.
#a2<=r<=a1
#a1 := 0.8306841627e12;
#a2 := 0.2720522631e12;

# +,- sign for integral, Primitive function, calculated form
# 0<=P<=Pi/2

#phi := P->-0.8328841065*I*((-2500000000+1348688373*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.7344898564)/((2500000000+1348688373*sin(P)^4-3848688373*sin(P)^2)^(1/2)*cos(P));

#r := P->-0.2259895064e24/(0.5586318996e12*sin(P)^2-0.8306841627e12);

#plot([Im(phi(P)),r(P),P=0..Pi/2]); >#plot([-Im(phi(P)),r(P),P=0..Pi/2]); >#plot([Im(phi(P)),r(P),P=0..Pi]);

###############################
# Second analytic solution. Primitive function.
#a4<=rr<=a3
#a3 := 0;
#a4 := -0.1103327381e13;

# 0<=P<=Pi/2, Primitive function, calculated form

phiphi := -0.8328841065*I*((-0.1000000000e11+4605246509*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.6786196659)/((0.1000000000e11+4605246509*sin(P)^4-0.1460524651e11*sin(P)^2)^(1/2)*cos(P));

rr := P->(0.9165165817e24*sin(P)^2-0.9165165817e24)/(0.1103327381e13*sin(P)^2+0.8306841627e12);

plot([Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([-Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([Im(phiphi(P)),rr(P),P=0..Pi]);

Best Regards,
Hannu Poropudas
I forget to mention that when you replace m by -m (or other words to say m->MG -> -MG, in c=1 units calc., with c.g.s units),
then this metrics ALSO satisfies Einstein's vacuum spherically symmetric field equations.

Spherically symmetric metrics which satisfies also
Einstein's vacuum field equations.

(c=1,G=1 units)
matrix([[m^2/((1+m/r)^4*r^4*(1+2*m/r)), 0, 0, 0], [0, -1/(1+m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1+m/r)^2, 0], [0, 0, 0, 1+2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1+m/r)^4*r^4*(1+2*m/r)))*dr^2-(1/(1+m/r)^2)*dtheta^2-(sin(theta)^2/(1+m/r)^2)*dphi^2+(1+2*m/r)*dt^2

(m -> M*G/c^2 , if SI-units are used, m -> MG, if c=1 units calc. are used with c.g.s) .

I think that this kind of object could only be sensible inside event horizon r = 2*MG, if it would be sensible at all?

Best Regards, Hannu Poropudas
It's remarkable and reminds of something like Clausius or Hooke's law, about the derivations that go into the a-diabatic, and non-a-diabatic.

It reminds of Dirichlet function, and which side has measure, or both do, and they share, half.

There are various considerations about whether it's from the point of
view of entering, the horizon, or, leaving it.

You might relate it to radiation either way. There's a usual notion that "what enters never leaves", but, there's also a notion that a black hole can have a center-of-mass that's part of a highly-relativistic system, that oscillates either side of the horizon.

You might relate it to orthogonality and what results time/frequency
as like the Fourier analysis, about the fine line in the middle there,
or for a "Fourier-style" analysis.

Another consideration is again about Clausius and Hooke's law,
and what goes into that the a-diabatic and non-a-diabatic both
are things, and about the derivations that get to where there's
add a "and here multiply or rather divide each by 2, ...", then figure
out that's a most usual sort of doubling-space and the continuous
and discrete: mostly about the mathematics that goes into detailing quantum behavior, in terms usually of the derivations that result inter-operable with the classical.

So, there's to be considered various forms of formalisms for "convergence",
in the formula, to keep in mind that there are various laws of large numbers,
and that not all tests for convergence, agree.

[continued in next message]

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• From Ross Finlayson@21:1/5 to Hannu Poropudas on Fri Nov 17 13:51:24 2023
On Friday, November 17, 2023 at 1:30:39 AM UTC-8, Hannu Poropudas wrote:
keskiviikko 15. marraskuuta 2023 klo 6.08.28 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, November 14, 2023 at 7:27:16 PM UTC-8, Ross Finlayson wrote:
On Tuesday, November 14, 2023 at 3:33:32 AM UTC-8, Hannu Poropudas wrote:
perjantai 3. marraskuuta 2023 klo 12.01.50 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 11.41.57 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 10.55.39 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, October 31, 2023 at 4:41:56 AM UTC-7, Hannu Poropudas wrote:
tiistai 31. lokakuuta 2023 klo 10.40.22 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 12.35.05 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11, 2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11 +,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation, if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.
6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.
6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
I'am sorry about error in 30.10.2023 posting of mine.

Here is CORRECTED 30.10.2023 posting of mine

# CORRECTED. Two branches of coordinate time T series approx.31.10.2023 H.P.
# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part) >REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 7.5*10^(-13)
# error estimation max abs negative side about -1.3*10^(-12)
# +,- formula (Primitive function, Imaginary part) >IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 4.8*10^(-13)
# error estimation max abs negative side about -2.4*10^(-13)

Best Regards,
Hannu Poropudas
ONE NOTE about one "little strange" function used in my maple calculations

csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)

Hannu
You mentioned the sign term reflecting that earlier you wrote from your derivation,
that part of it was as under-defined or de facto, a compensating term.

It seems what you are integrating is power terms resolving, why the halfs either
way have a "pseudo" product, what is a law that results why in your terms, they add
up, if you haven't explained "why" it's legal those terms wouldn't resolve,

It seems those would be waves falling so would result "why", is because, they
are under the area terms, the absorption or radiation, how then those have to
add up, to make the estimate, which as you note appears accurate.
1. I put more clearly my incomplete error estimation procedure here:

Solution was actually +,- Int(function(r),r) and only +function(r) was used
in error estimation, -function(r) was not used in error estimation for my convenience,
so error estimation is was not complete in this sense, but it gives correctly order of the error?

Error estimation was made only of (+) branch, of function(r) and (-) branch of function(r) was not used ,
so error estimation was not complete in that sense, but I made so for my convenience not
to make too long posting about error estimation.

function(r) - (series approx of function(r)),

function(r) is NOT integrated here due it is too complicated to do that.

(+) branch of the function(r) only used in error estimation (compare proper time case,
similar way was done in this case in error estimation) .

2. It should be remembered here that all mathematics in this special example was done with
complex numbers inside r=2*MG event horizon.

3. I only tried to point out here that how complex mathematics can be used to make calculations in
this special example case which I have selected to calculate completely.

who understand astrophysics better than me, if this was sensible at all ?

Best Regards,
Hannu Poropudas
SUMMARY:

# I. SUMMARY: formulae in complex calcs.(Maple9) H.P. 03.11.2023

# Spherically symmetric metric used, satisfies Einstein vac. eqs. ># m=M*G=MG, G=gravitational const.in c.g.s units, c=1 units calc. ># (conversion to c.g.s units: time -> time/c, length -> length.)
# S2-star around SgrA* black hole numerical values used
# aphelion and perihelion numerical values used
# (my earlier sci.physics.relativity posts).

# Only one example was chosen, Euler-Lagrange eqs. constants >#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
# S2-star around SgrA* black hole numerical values used
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
##############################
# Proper time t case
# +,- formula >#m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# +,- formula for proper time t-t0. Here Primitive function. >#Int(m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

#############
# Series approx. up to 7 degree Real part of proper time >#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Real part of proper time t

#REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^
7

# From error curve
# Error estimation: about = 0 between 5.2e11<r<7.2e11
# Error estimation: max negative about -2.15 near 2.72e11
# Error estimation: max negative about -0.2 near 8.307e11

# Series approx. up to 7 degree Imaginary part of proper time >#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Imaginary part of proper time

#IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^
7

# From error curve
# Error estimation: about = 0 between 5.4e11<r<7.0e11
# Error estimation: max negative about -1.09 near 2.72e11
# Error estimation: max negative about -0.1 near 8.307e11

# +,- formula >#plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIF(r),r=2.720522631*10^11..8.306841627*10^11);
# +,- formula >#plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIF(r),r=2.720522631*10^11..8.306841627*10^11); >#############

#########################################################
# Coordinate time T case
# +,- formula. >#(K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# coordinate time T and proper time t relation
#dT = dt*K2/(1-2*m/r), equation is from Euler-Lagrange eqs.

#+,- formula for coordinate time T-T0. Here Primitive function. >#Int((K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

############################
# Series approx. up to 7 degree
# Coordinate time T, Primitive function, Real part
# +,- formula (Primitive function, Real part) >#REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-0.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r
# From error curve
# Error estimation: max positive side about 7.5e-13
# Error estimation: max abs negative side about -1.3e-12
# +,- formula (Primitive function, Imaginary part)
# Coordinate time T, Primitive function, Imaginary part >#IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;
# From error curve
# Error estimation max positive side about 4.8e-13
# Error estimation max abs negative side about -2.4e-13

#plot(REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >############################

###############################################################
# II. SUMMARY: Earlier made complex calculations H.P. 3.11.2023

# First analytic solution. Primitive function.
#a2<=r<=a1
#a1 := 0.8306841627e12;
#a2 := 0.2720522631e12;

# +,- sign for integral, Primitive function, calculated form
# 0<=P<=Pi/2

#phi := P->-0.8328841065*I*((-2500000000+1348688373*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.7344898564)/((2500000000+1348688373*sin(P)^4-3848688373*sin(P)^2)^(1/2)*cos(P));

#r := P->-0.2259895064e24/(0.5586318996e12*sin(P)^2-0.8306841627e12);

#plot([Im(phi(P)),r(P),P=0..Pi/2]); >#plot([-Im(phi(P)),r(P),P=0..Pi/2]); >#plot([Im(phi(P)),r(P),P=0..Pi]);

###############################
# Second analytic solution. Primitive function.
#a4<=rr<=a3
#a3 := 0;
#a4 := -0.1103327381e13;

# 0<=P<=Pi/2, Primitive function, calculated form

phiphi := -0.8328841065*I*((-0.1000000000e11+4605246509*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.6786196659)/((0.1000000000e11+4605246509*sin(P)^4-0.1460524651e11*sin(P)^2)^(1/2)*cos(P));

rr := P->(0.9165165817e24*sin(P)^2-0.9165165817e24)/(0.1103327381e13*sin(P)^2+0.8306841627e12);

plot([Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([-Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([Im(phiphi(P)),rr(P),P=0..Pi]);

Best Regards,
Hannu Poropudas
I forget to mention that when you replace m by -m (or other words to say m->MG -> -MG, in c=1 units calc., with c.g.s units),
then this metrics ALSO satisfies Einstein's vacuum spherically symmetric field equations.

Spherically symmetric metrics which satisfies also
Einstein's vacuum field equations.

(c=1,G=1 units)
matrix([[m^2/((1+m/r)^4*r^4*(1+2*m/r)), 0, 0, 0], [0, -1/(1+m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1+m/r)^2, 0], [0, 0, 0, 1+2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1+m/r)^4*r^4*(1+2*m/r)))*dr^2-(1/(1+m/r)^2)*dtheta^2-(sin(theta)^2/(1+m/r)^2)*dphi^2+(1+2*m/r)*dt^2

(m -> M*G/c^2 , if SI-units are used, m -> MG, if c=1 units calc. are used with c.g.s) .

I think that this kind of object could only be sensible inside event horizon r = 2*MG, if it would be sensible at all?

Best Regards, Hannu Poropudas
It's remarkable and reminds of something like Clausius or Hooke's law, about the derivations that go into the a-diabatic, and non-a-diabatic.

It reminds of Dirichlet function, and which side has measure, or both do,
and they share, half.

There are various considerations about whether it's from the point of view of entering, the horizon, or, leaving it.

You might relate it to radiation either way. There's a usual notion that "what enters never leaves", but, there's also a notion that a black hole can have a center-of-mass that's part of a highly-relativistic system, that oscillates either side of the horizon.

You might relate it to orthogonality and what results time/frequency

[continued in next message]

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• From Hannu Poropudas@21:1/5 to All on Tue Nov 21 00:10:51 2023
perjantai 17. marraskuuta 2023 klo 11.30.39 UTC+2 Hannu Poropudas kirjoitti:
keskiviikko 15. marraskuuta 2023 klo 6.08.28 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, November 14, 2023 at 7:27:16 PM UTC-8, Ross Finlayson wrote:
On Tuesday, November 14, 2023 at 3:33:32 AM UTC-8, Hannu Poropudas wrote:
perjantai 3. marraskuuta 2023 klo 12.01.50 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 11.41.57 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 10.55.39 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, October 31, 2023 at 4:41:56 AM UTC-7, Hannu Poropudas wrote:
tiistai 31. lokakuuta 2023 klo 10.40.22 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 12.35.05 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11, 2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11 +,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation, if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.
6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.
6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
I'am sorry about error in 30.10.2023 posting of mine.

Here is CORRECTED 30.10.2023 posting of mine

# CORRECTED. Two branches of coordinate time T series approx.31.10.2023 H.P.
# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part) >REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 7.5*10^(-13)
# error estimation max abs negative side about -1.3*10^(-12)
# +,- formula (Primitive function, Imaginary part) >IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 4.8*10^(-13)
# error estimation max abs negative side about -2.4*10^(-13)

Best Regards,
Hannu Poropudas
ONE NOTE about one "little strange" function used in my maple calculations

csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)

Hannu
You mentioned the sign term reflecting that earlier you wrote from your derivation,
that part of it was as under-defined or de facto, a compensating term.

It seems what you are integrating is power terms resolving, why the halfs either
way have a "pseudo" product, what is a law that results why in your terms, they add
up, if you haven't explained "why" it's legal those terms wouldn't resolve,

It seems those would be waves falling so would result "why", is because, they
are under the area terms, the absorption or radiation, how then those have to
add up, to make the estimate, which as you note appears accurate.
1. I put more clearly my incomplete error estimation procedure here:

Solution was actually +,- Int(function(r),r) and only +function(r) was used
in error estimation, -function(r) was not used in error estimation for my convenience,
so error estimation is was not complete in this sense, but it gives correctly order of the error?

Error estimation was made only of (+) branch, of function(r) and (-) branch of function(r) was not used ,
so error estimation was not complete in that sense, but I made so for my convenience not
to make too long posting about error estimation.

function(r) - (series approx of function(r)),

function(r) is NOT integrated here due it is too complicated to do that.

(+) branch of the function(r) only used in error estimation (compare proper time case,
similar way was done in this case in error estimation) .

2. It should be remembered here that all mathematics in this special example was done with
complex numbers inside r=2*MG event horizon.

3. I only tried to point out here that how complex mathematics can be used to make calculations in
this special example case which I have selected to calculate completely.

who understand astrophysics better than me, if this was sensible at all ?

Best Regards,
Hannu Poropudas
SUMMARY:

# I. SUMMARY: formulae in complex calcs.(Maple9) H.P. 03.11.2023

# Spherically symmetric metric used, satisfies Einstein vac. eqs. ># m=M*G=MG, G=gravitational const.in c.g.s units, c=1 units calc. ># (conversion to c.g.s units: time -> time/c, length -> length.)
# S2-star around SgrA* black hole numerical values used
# aphelion and perihelion numerical values used
# (my earlier sci.physics.relativity posts).

# Only one example was chosen, Euler-Lagrange eqs. constants >#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
# S2-star around SgrA* black hole numerical values used
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
##############################
# Proper time t case
# +,- formula >#m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# +,- formula for proper time t-t0. Here Primitive function. >#Int(m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

#############
# Series approx. up to 7 degree Real part of proper time >#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Real part of proper time t

#REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^
7

# From error curve
# Error estimation: about = 0 between 5.2e11<r<7.2e11
# Error estimation: max negative about -2.15 near 2.72e11
# Error estimation: max negative about -0.2 near 8.307e11

# Series approx. up to 7 degree Imaginary part of proper time >#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Imaginary part of proper time

#IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^
7

# From error curve
# Error estimation: about = 0 between 5.4e11<r<7.0e11
# Error estimation: max negative about -1.09 near 2.72e11
# Error estimation: max negative about -0.1 near 8.307e11

# +,- formula >#plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIF(r),r=2.720522631*10^11..8.306841627*10^11);
# +,- formula >#plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIF(r),r=2.720522631*10^11..8.306841627*10^11); >#############

#########################################################
# Coordinate time T case
# +,- formula. >#(K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# coordinate time T and proper time t relation
#dT = dt*K2/(1-2*m/r), equation is from Euler-Lagrange eqs.

#+,- formula for coordinate time T-T0. Here Primitive function. >#Int((K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

############################
# Series approx. up to 7 degree
# Coordinate time T, Primitive function, Real part
# +,- formula (Primitive function, Real part) >#REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-0.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r
# From error curve
# Error estimation: max positive side about 7.5e-13
# Error estimation: max abs negative side about -1.3e-12
# +,- formula (Primitive function, Imaginary part)
# Coordinate time T, Primitive function, Imaginary part >#IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;
# From error curve
# Error estimation max positive side about 4.8e-13
# Error estimation max abs negative side about -2.4e-13

#plot(REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >############################

###############################################################
# II. SUMMARY: Earlier made complex calculations H.P. 3.11.2023

# First analytic solution. Primitive function.
#a2<=r<=a1
#a1 := 0.8306841627e12;
#a2 := 0.2720522631e12;

# +,- sign for integral, Primitive function, calculated form
# 0<=P<=Pi/2

#phi := P->-0.8328841065*I*((-2500000000+1348688373*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.7344898564)/((2500000000+1348688373*sin(P)^4-3848688373*sin(P)^2)^(1/2)*cos(P));

#r := P->-0.2259895064e24/(0.5586318996e12*sin(P)^2-0.8306841627e12);

#plot([Im(phi(P)),r(P),P=0..Pi/2]); >#plot([-Im(phi(P)),r(P),P=0..Pi/2]); >#plot([Im(phi(P)),r(P),P=0..Pi]);

###############################
# Second analytic solution. Primitive function.
#a4<=rr<=a3
#a3 := 0;
#a4 := -0.1103327381e13;

# 0<=P<=Pi/2, Primitive function, calculated form

phiphi := -0.8328841065*I*((-0.1000000000e11+4605246509*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.6786196659)/((0.1000000000e11+4605246509*sin(P)^4-0.1460524651e11*sin(P)^2)^(1/2)*cos(P));

rr := P->(0.9165165817e24*sin(P)^2-0.9165165817e24)/(0.1103327381e13*sin(P)^2+0.8306841627e12);

plot([Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([-Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([Im(phiphi(P)),rr(P),P=0..Pi]);

Best Regards,
Hannu Poropudas
I forget to mention that when you replace m by -m (or other words to say m->MG -> -MG, in c=1 units calc., with c.g.s units),
then this metrics ALSO satisfies Einstein's vacuum spherically symmetric field equations.

Spherically symmetric metrics which satisfies also
Einstein's vacuum field equations.

(c=1,G=1 units)
matrix([[m^2/((1+m/r)^4*r^4*(1+2*m/r)), 0, 0, 0], [0, -1/(1+m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1+m/r)^2, 0], [0, 0, 0, 1+2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1+m/r)^4*r^4*(1+2*m/r)))*dr^2-(1/(1+m/r)^2)*dtheta^2-(sin(theta)^2/(1+m/r)^2)*dphi^2+(1+2*m/r)*dt^2

(m -> M*G/c^2 , if SI-units are used, m -> MG, if c=1 units calc. are used with c.g.s) .

I think that this kind of object could only be sensible inside event horizon r = 2*MG, if it would be sensible at all?

Best Regards, Hannu Poropudas
It's remarkable and reminds of something like Clausius or Hooke's law, about the derivations that go into the a-diabatic, and non-a-diabatic.

It reminds of Dirichlet function, and which side has measure, or both do,
and they share, half.

There are various considerations about whether it's from the point of view of entering, the horizon, or, leaving it.

You might relate it to radiation either way. There's a usual notion that "what enters never leaves", but, there's also a notion that a black hole can have a center-of-mass that's part of a highly-relativistic system, that oscillates either side of the horizon.

You might relate it to orthogonality and what results time/frequency

[continued in next message]

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• From Hannu Poropudas@21:1/5 to All on Tue Nov 21 03:19:13 2023
perjantai 17. marraskuuta 2023 klo 11.30.39 UTC+2 Hannu Poropudas kirjoitti:
keskiviikko 15. marraskuuta 2023 klo 6.08.28 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, November 14, 2023 at 7:27:16 PM UTC-8, Ross Finlayson wrote:
On Tuesday, November 14, 2023 at 3:33:32 AM UTC-8, Hannu Poropudas wrote:
perjantai 3. marraskuuta 2023 klo 12.01.50 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 11.41.57 UTC+2 Hannu Poropudas kirjoitti:
torstai 2. marraskuuta 2023 klo 10.55.39 UTC+2 Ross Finlayson kirjoitti:
On Tuesday, October 31, 2023 at 4:41:56 AM UTC-7, Hannu Poropudas wrote:
tiistai 31. lokakuuta 2023 klo 10.40.22 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 12.35.05 UTC+2 Hannu Poropudas kirjoitti:
maanantai 30. lokakuuta 2023 klo 9.59.22 UTC+2 Hannu Poropudas kirjoitti:
perjantai 27. lokakuuta 2023 klo 10.46.55 UTC+3 Hannu Poropudas kirjoitti:
torstai 26. lokakuuta 2023 klo 11.04.40 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 14.37.35 UTC+3 Hannu Poropudas kirjoitti:
keskiviikko 25. lokakuuta 2023 klo 12.01.55 UTC+3 Hannu Poropudas kirjoitti:
tiistai 24. lokakuuta 2023 klo 11.56.49 UTC+3 Hannu Poropudas kirjoitti:
perjantai 20. lokakuuta 2023 klo 9.54.12 UTC+3 Hannu Poropudas kirjoitti:
torstai 19. lokakuuta 2023 klo 21.41.08 UTC+3 JanPB kirjoitti:
On Thursday, October 19, 2023 at 12:22:43 AM UTC-7, Hannu Poropudas wrote:
sunnuntai 15. lokakuuta 2023 klo 11.35.22 UTC+3 Hannu Poropudas kirjoitti:
Spherically symmetric metrics which satisfies
Einstein's vacuum field equations.

(c=1,G=1 units)

matrix([[m^2/((1-m/r)^4*r^4*(1-2*m/r)), 0, 0, 0], [0, -1/(1-m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1-m/r)^2, 0], [0, 0, 0, 1-2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1-m/r)^4*r^4*(1-2*m/r)))*dr^2-(1/(1-m/r)^2)*dtheta^2-(sin(theta)^2/(1-m/r)^2)*dphi^2+(1-2*m/r)*dt^2

(m -> m*G/c^2 , if SI-units are used.)

I don't know that would this solution have any astrophysical applications?

There exist a book called something like
"Exact Solutions of the Einstein Field Equations",
which have about 740 pages and
I don't know if this solution is among them?

Three singularity points of the metrics are the following:

r = 0, r = m*G/c^2 and r = 2*m*G/c^2.

I have used generalized form of eq. (9) on page 171, when I calculated this solution with my Maple 9.

Reference:
Tolman R. C., 1934.
Effect of inhomogeneity on cosmological models.
Proc. Natl. Acad. Sci. USA, 1934, Mar; 20 (30): 1679-176.

Hannu Poropudas

Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
I used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
due three integration constants from Euler-Lagrange equations does not have
same interpretations as in Schwarzschild case (energy (constant) and angular momementum (constant) etc).
I used these guess numbers from aphelion and perihelion of S2-star around SgrA* black hole which calculations I published
some time ago in this sci.physics.relativity Google Group. (I used c=1 units , c.g.s units then and I use again c=1 units and c.g.s units here)

MG = 6.292090968*10^11, 2*MG=1.258418194*10^12.
I found two parametric form analytic solutions (Both are primitive functions, 0<=P<=Pi/2):

2.720522631*10^11<=r<=8.306841627*10^11 +,- sign for integral
phi= Int(-0.8328841065*I/sqrt(1-0.5394753492*sin(P)^2),P)
r=-2.259895064*10^23/(5.586318996*10^11*sin(P)^2-8.306841627*10^11)

and

-1.103327381*10^12<=rr<=0
+,- sign for integral phiphi=Int(-0.8330796374*I/sqrt(1-0.4605246509*sin(P)^2),P)
rr= (9.165165817*10^23*sin(P)^2-9.165165817*10^23)/(1.103327381*10^12*sin(P)^2+8.306841627*10^11)

I calculated also these integrals but their formulae are too long to copy here.
Both sign can be taken into account when plotting Imaginary parts 0<=P<=Pi.
Real parts = 0 in these integrals.
How to interpret pure imaginary phi and phiphi angles?
How to interpret these Imaginary angle plots?

Best Regards,
Hannu Poropudas

(a) incorrect, or:

(b) isometric to Schwarzschild's.

--
Jan
Your (b) alternative seems not to be true due two separate event horizons in this metrics ?

Schwarzschild metric comes also correctly, but with different sign selection in metrics than what I used.

Your (a) alternative is not true due this metric satisfies Einstein's vacuum field equations,
but you are correct in point of view that it may not be physically acceptable solution
of these equations at our present orthodoxic physical knowledge.
This is indicated by imaginary unit (I=sqrt(-1)) in these example of two analytic solutions.

There exist also few other integration constants from Euler-Largrange equations,
but I have selected randomly only one couple of them in this example calculation.

Hannu
I put here those strange (NO ordinary physical interpretation) formulae of integration
constants from Euler-Largrange equations:

I mark now for convenience T = coordinate time and t = proper time.

(dphi/dt)/(1-m/r)^2 = K1 (constant of integration)
(1-2*m/r)*(dT/dt) = K2 (constant of integration)
(1-2*m/r)*(dT/dt)^2 - m^2*(dr/dt)^2 / ( (1-m/r)^4*r^4*(1-2*m/r) ) - (dphi/dt)^2 / (1-m/r)^2 = 1.

I calculated for randomly selected numerical values of S2-star aphelion and perhelion
distances (c=1 units, and c.g.s units) from my earlier calculations of analytic GR solutions
for S2-star orbit around SgrA* black hole (sci.physics.relativity published)
to calculate two integration constants K1 and K2 of Euler-Largange equations
(NO ordinary physical interpretation), (I = sqrt(-1) = imaginary unit):

K1 = +,- 0.7072727132*I,
K2 = +,- 0.5943942676 +,- 0.5943942676*I,

And I selected here randomly as an example two constants of integration
in this my two analytic solutions calculation:

K1 = - 0.7072727132*I
and
K2 = 0.5943942676 - 0.5943942676*I

This selection gave those two pure imaginary analytic solutions which I gave here earlier.
(Phi(P) is pure imaginary angle and r(P) is real distance.
Phiphi(P) is pure imaginary angle and rr(P) is real distance).

Plot ([Im(phi(P)),r(P),P=0..Pi]);
Plot ([Im(phiphi(P)),rr(P),P=0..Pi]);
gives both +, - solutions in both cases (P..Pi/2 gives only one branch and P..Pi gives both branches)

Those both plots resemble somehow pendulum orbit ?

I have NO physical interpretations of these solutions
and I think that these have NO real physical applications.

Hannu Poropudas
I investigated also question that what kind of coordinate time (T) solution would be in parametric form ?

It seems to me that this integral is too complicated to calculate analytically, but it could be so
with those above K1 and K2 (plus K3 = 0 additional integration constant in Euler-Lagrange equations)
in this above case that the coordinate time T could be two dimensional complex number ?

This also seems to support what I said above.
I have NO physical interpretations of these solutions
and I think at the moment that these have NO real physical applications.

And we should study two dimensional complex mathematics of two dimensional
coordinate time (T) in this complicated integral better,
if we try to better understand this situation, if this would be sensible at all ?

Best Regards,
Hannu Poropudas
CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Sorry that I confused these two letters.

Hannu
I found one interesting reference, which show that there
are really only few astrophysically significant exact solutions to Einstein's field equations.

Ishak, M. 2015.
Exact Solutions to Einstein's Equations in Astrophysics.
Texas Symposium on Relativistic Astrophysics, Geneva 2015.
33 pages. https://personal.utdallas.edu/~mishak/ExactSolutionsInAstrophysics_Ishak_Final.pdf

Best Regards,
Hannu Poropudas
In order to me more mathematically complete I calculate also
approximate proper time t integral (primitive function)
and plotted both real part and imaginary part of it.
I have NO interpretations of these.

# Approximate proper time t integral calculated HP 27.10.2023
# REMARK: My letter convenience t=proper time T=coordinate time
# Real part and Imaginary part plotted
#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11

#Real part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for REIF(r)
REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.
6292090968e12)^7;

#Imaginary part of primitive function t approx.
# Series approx at r = MG up to 7 degree.
+,- sign for IMIF(r)
IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.
6292090968e12)^7;

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for REIF(r)
plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11);

#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;

+,- sign for IMIF(r)
plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11);

Best Regards,
Hannu Poropudas
I calculated also coordinate time T series approximation up to 7 degree at r=MG,
(REMARK: This is preliminary calculation I have not rechecked it yet):

If my approximate calculations are correct, then it is possible to calculate more
"quantities" in this strange black hole of two event horizons space-time of mine,
if this is sensible at all?

# Two branches of coordinate time T series approx.30.10.2023 H.P.

# This coordinate time T is also complex number with two branches
# (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part)

REIG:=r->-2.851064818*ln(abs(r-0.6292090968e12))-.8288703850*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 8.3*10^(-9)
# error estimation max abs negative side about -4.2*10^(-9)
# Both max are at r=MG, other definition area error = about 0

# +,- formula (Primitive function, Imaginary part)

IMIG:=r->-1.425532409*(1-csgn(r-0.6292090968e12))*Pi+1.657740770*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 3.2*10^(-9)
# error estimation max abs negative side about -6.4*10^(-9)
# Both max are at r=MG, other definition area error = about 0

Best Regards,
Hannu Poropudas
I'am sorry about error in 30.10.2023 posting of mine.

Here is CORRECTED 30.10.2023 posting of mine

# CORRECTED. Two branches of coordinate time T series approx.31.10.2023 H.P.
# This coordinate time T is also complex number with two branches ># (real and Imaginary)

#Coordinate time T series approx. up to 7 degree
# function-(series approx function), not integrated here >#(+)branch only used error estimation (compare proper time case)

# +,- formula (Primitive function, Real part) >REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r;

# error estimation max positive side about 7.5*10^(-13)
# error estimation max abs negative side about -1.3*10^(-12)
# +,- formula (Primitive function, Imaginary part) >IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;

# error estimation max positive side about 4.8*10^(-13)
# error estimation max abs negative side about -2.4*10^(-13)

Best Regards,
Hannu Poropudas
ONE NOTE about one "little strange" function used in my maple calculations

csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)

Hannu
You mentioned the sign term reflecting that earlier you wrote from your derivation,
that part of it was as under-defined or de facto, a compensating term.

It seems what you are integrating is power terms resolving, why the halfs either
way have a "pseudo" product, what is a law that results why in your terms, they add
up, if you haven't explained "why" it's legal those terms wouldn't resolve,

It seems those would be waves falling so would result "why", is because, they
are under the area terms, the absorption or radiation, how then those have to
add up, to make the estimate, which as you note appears accurate.
1. I put more clearly my incomplete error estimation procedure here:

Solution was actually +,- Int(function(r),r) and only +function(r) was used
in error estimation, -function(r) was not used in error estimation for my convenience,
so error estimation is was not complete in this sense, but it gives correctly order of the error?

Error estimation was made only of (+) branch, of function(r) and (-) branch of function(r) was not used ,
so error estimation was not complete in that sense, but I made so for my convenience not
to make too long posting about error estimation.

function(r) - (series approx of function(r)),

function(r) is NOT integrated here due it is too complicated to do that.

(+) branch of the function(r) only used in error estimation (compare proper time case,
similar way was done in this case in error estimation) .

2. It should be remembered here that all mathematics in this special example was done with
complex numbers inside r=2*MG event horizon.

3. I only tried to point out here that how complex mathematics can be used to make calculations in
this special example case which I have selected to calculate completely.

who understand astrophysics better than me, if this was sensible at all ?

Best Regards,
Hannu Poropudas
SUMMARY:

# I. SUMMARY: formulae in complex calcs.(Maple9) H.P. 03.11.2023

# Spherically symmetric metric used, satisfies Einstein vac. eqs. ># m=M*G=MG, G=gravitational const.in c.g.s units, c=1 units calc. ># (conversion to c.g.s units: time -> time/c, length -> length.)
# S2-star around SgrA* black hole numerical values used
# aphelion and perihelion numerical values used
# (my earlier sci.physics.relativity posts).

# Only one example was chosen, Euler-Lagrange eqs. constants >#K3:=0;
#K1 := -0.7072727132*I;
#K2 := 0.5943942676-0.5943942676*I;
# S2-star around SgrA* black hole numerical values used
#m := MG;
#MG := 0.6292090968e12;
#2*MG := 0.1258418194e13;
##############################
# Proper time t case
# +,- formula >#m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# +,- formula for proper time t-t0. Here Primitive function. >#Int(m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

#############
# Series approx. up to 7 degree Real part of proper time >#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Real part of proper time t

#REIF:=r->-0.9292411964e-8*r+0.8717127610e-20*r^2+0.4446277653e-31*(r-0.6292090968e12)^3+0.2329675135e-42*(r-0.6292090968e12)^4+0.3071201158e-47*(r-0.6292090968e12)^5+0.2827253683e-58*(r-0.6292090968e12)^6+0.2065510967e-69*(r-0.6292090968e12)^
7

# From error curve
# Error estimation: about = 0 between 5.2e11<r<7.2e11
# Error estimation: max negative about -2.15 near 2.72e11
# Error estimation: max negative about -0.2 near 8.307e11

# Series approx. up to 7 degree Imaginary part of proper time >#a2<=r<=a1, definition area
#a2=2.720522631*10^11, a1=8.306841627*10^11
# +,- formula Imaginary part of proper time

#IMIF:=r->-0.2217185446e-7*r+0.3637453960e-19*r^2+0.1018586871e-30*(r-0.6292090968e12)^3+0.3376062792e-42*(r-0.6292090968e12)^4+0.4794788388e-47*(r-0.6292090968e12)^5+0.2475402813e-58*(r-0.6292090968e12)^6+0.1233132761e-69*(r-0.6292090968e12)^
7

# From error curve
# Error estimation: about = 0 between 5.4e11<r<7.0e11
# Error estimation: max negative about -1.09 near 2.72e11
# Error estimation: max negative about -0.1 near 8.307e11

# +,- formula >#plot(REIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIF(r),r=2.720522631*10^11..8.306841627*10^11);
# +,- formula >#plot(IMIF(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIF(r),r=2.720522631*10^11..8.306841627*10^11); >#############

#########################################################
# Coordinate time T case
# +,- formula. >#(K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r));

# coordinate time T and proper time t relation
#dT = dt*K2/(1-2*m/r), equation is from Euler-Lagrange eqs.

#+,- formula for coordinate time T-T0. Here Primitive function. >#Int((K2/(1-2*m/r))*m/sqrt((-K1^2*(1-m/r)^2+K2/(1-2*m/r)+K3-1)*(1-m/r)^4*r^4*(1-2*m/r)),r);

############################
# Series approx. up to 7 degree
# Coordinate time T, Primitive function, Real part
# +,- formula (Primitive function, Real part) >#REIG:=r->-1.425532409*ln(abs(r-0.6292090968e12))-0.4144351924*(1-csgn(r-0.6292090968e12))*Pi+0.3330424925e12/(r-0.6292090968e12)-0.2939012715e-11*r
# From error curve
# Error estimation: max positive side about 7.5e-13
# Error estimation: max abs negative side about -1.3e-12
# +,- formula (Primitive function, Imaginary part)
# Coordinate time T, Primitive function, Imaginary part >#IMIG:=r->-0.7127662045*(1-csgn(r-0.6292090968e12))*Pi+0.8288703849*ln(abs(r-0.6292090968e12))-0.2312741379e12/(r-0.6292090968e12)+0.1298075147e-11*r;
# From error curve
# Error estimation max positive side about 4.8e-13
# Error estimation max abs negative side about -2.4e-13

#plot(REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-REIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >#plot(-IMIG(r),r=2.720522631*10^11..8.306841627*10^11); >############################

###############################################################
# II. SUMMARY: Earlier made complex calculations H.P. 3.11.2023

# First analytic solution. Primitive function.
#a2<=r<=a1
#a1 := 0.8306841627e12;
#a2 := 0.2720522631e12;

# +,- sign for integral, Primitive function, calculated form
# 0<=P<=Pi/2

#phi := P->-0.8328841065*I*((-2500000000+1348688373*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.7344898564)/((2500000000+1348688373*sin(P)^4-3848688373*sin(P)^2)^(1/2)*cos(P));

#r := P->-0.2259895064e24/(0.5586318996e12*sin(P)^2-0.8306841627e12);

#plot([Im(phi(P)),r(P),P=0..Pi/2]); >#plot([-Im(phi(P)),r(P),P=0..Pi/2]); >#plot([Im(phi(P)),r(P),P=0..Pi]);

###############################
# Second analytic solution. Primitive function.
#a4<=rr<=a3
#a3 := 0;
#a4 := -0.1103327381e13;

# 0<=P<=Pi/2, Primitive function, calculated form

phiphi := -0.8328841065*I*((-0.1000000000e11+4605246509*sin(P)^2)*(-1+sin(P)^2))^(1/2)*(1-sin(P)^2)^(1/2)*EllipticF(sin(P), 0.6786196659)/((0.1000000000e11+4605246509*sin(P)^4-0.1460524651e11*sin(P)^2)^(1/2)*cos(P));

rr := P->(0.9165165817e24*sin(P)^2-0.9165165817e24)/(0.1103327381e13*sin(P)^2+0.8306841627e12);

plot([Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([-Im(phiphi(P)),rr(P),P=0..Pi/2]); >plot([Im(phiphi(P)),rr(P),P=0..Pi]);

Best Regards,
Hannu Poropudas
I forget to mention that when you replace m by -m (or other words to say m->MG -> -MG, in c=1 units calc., with c.g.s units),
then this metrics ALSO satisfies Einstein's vacuum spherically symmetric field equations.

Spherically symmetric metrics which satisfies also
Einstein's vacuum field equations.

(c=1,G=1 units)
matrix([[m^2/((1+m/r)^4*r^4*(1+2*m/r)), 0, 0, 0], [0, -1/(1+m/r)^2, 0, 0], [0, 0, -sin(theta)^2/(1+m/r)^2, 0], [0, 0, 0, 1+2*m/r]])])

(c=1,G=1 units)

ds^2=(m^2/((1+m/r)^4*r^4*(1+2*m/r)))*dr^2-(1/(1+m/r)^2)*dtheta^2-(sin(theta)^2/(1+m/r)^2)*dphi^2+(1+2*m/r)*dt^2

(m -> M*G/c^2 , if SI-units are used, m -> MG, if c=1 units calc. are used with c.g.s) .

I think that this kind of object could only be sensible inside event horizon r = 2*MG, if it would be sensible at all?

Best Regards, Hannu Poropudas
It's remarkable and reminds of something like Clausius or Hooke's law, about the derivations that go into the a-diabatic, and non-a-diabatic.

It reminds of Dirichlet function, and which side has measure, or both do,
and they share, half.

There are various considerations about whether it's from the point of view of entering, the horizon, or, leaving it.

You might relate it to radiation either way. There's a usual notion that "what enters never leaves", but, there's also a notion that a black hole can have a center-of-mass that's part of a highly-relativistic system, that oscillates either side of the horizon.

You might relate it to orthogonality and what results time/frequency

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