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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EThere is only one solution to the spherically symmetric vacuum equations (see e.g. Hawking & Ellis).
90900 Kiiminki / Oulu
Finland
So you either made a mistake somewhere or your solution is the same
as Schwarzschild's, only written differently.
--
Jan
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9E
90900 Kiiminki / Oulu
Finland
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--
Jan
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuIt'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.
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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
Stop wasting your time.
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
Hannu
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,
Hannu Poropudas
# 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);
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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,Error estimations of these two series approximations (function - (series approx function), not integrated here):
Hannu Poropudas
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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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,Error estimations of these two series approximations (function - (series approx function), not integrated here):
Hannu Poropudas
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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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);
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.Best Regards,Error estimations of these two series approximations (function - (series approx function), not integrated here):
Hannu Poropudas
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, 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
Hannu Poropudas
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
Hannu Poropudas
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);
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.Best Regards,Error estimations of these two series approximations (function - (series approx function), not integrated here):
Hannu Poropudas
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, 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
Hannu Poropudas
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.
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
Hannu Poropudas
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);
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.Best Regards,Error estimations of these two series approximations (function - (series approx function), not integrated here):
Hannu Poropudas
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, 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
Hannu Poropudas
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 inPaddy what you do is go down to a flea market and find an old scale according toRoss, yes, that's how you do it. And it is easy to write out the equation(s) representing that experiment.
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.
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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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
+,- sign for REIF(r)#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;
+,- sign for IMIF(r)#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;
+,- 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
# 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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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
+,- sign for REIF(r)#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;
+,- sign for IMIF(r)#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;
+,- 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);
I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
Hannu Poropudas
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
+,- sign for REIF(r)#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;
+,- sign for IMIF(r)#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;
+,- 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);
I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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, 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.
Hannu Poropudas
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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
+,- sign for REIF(r)#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;
+,- sign for IMIF(r)#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;
+,- 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);
I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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
# 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)
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
Hannu Poropudas
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
+,- sign for REIF(r)#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;
+,- sign for IMIF(r)#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;
+,- 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);
I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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,I'am sorry about error in 30.10.2023 posting of mine.
Hannu Poropudas
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
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
Hannu Poropudas
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
+,- sign for REIF(r)#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;
+,- sign for IMIF(r)#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;
+,- 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);
I accept this answer, Hannu. You have been fully restored to my good graces.I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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
This is question is NOT subject of my posting chain.Best Regards,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.
Hannu Poropudas
I noticed that in the stack exchage of physics was written about "Big Ben Paradox" question:
SR is not a theory of gravity.
Hannu
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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
Hannu Poropudas
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
+,- sign for REIF(r)#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;
+,- sign for IMIF(r)#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;
+,- 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);
I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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,I'am sorry about error in 30.10.2023 posting of mine.
Hannu Poropudas
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,ONE NOTE about one "little strange" function used in my maple calculations
Hannu Poropudas
csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)
Hannu
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.
7;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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
Hannu Poropudas
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
+,- sign for REIF(r)#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;+,- sign for IMIF(r)#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)^
#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);
I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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,I'am sorry about error in 30.10.2023 posting of mine.
Hannu Poropudas
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,ONE NOTE about one "little strange" function used in my maple calculations
Hannu Poropudas
csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)
HannuYou 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.
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.
6292090968e12)^7;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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
Hannu Poropudas
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
+,- sign for REIF(r)#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;+,- sign for IMIF(r)#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.
#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);
I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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,I'am sorry about error in 30.10.2023 posting of mine.
Hannu Poropudas
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,ONE NOTE about one "little strange" function used in my maple calculations
Hannu Poropudas
csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)
HannuYou 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, they1. I put more clearly my incomplete error estimation procedure here:
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.
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.
4. I have made here NO physical interpretations about this special example, I leave to make them to those
who understand astrophysics better than me, if this was sensible at all ?
Best Regards,
Hannu Poropudas
# 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]);
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.
6292090968e12)^7;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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
Hannu Poropudas
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
+,- sign for REIF(r)#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;+,- sign for IMIF(r)#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.
#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);
I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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,I'am sorry about error in 30.10.2023 posting of mine.
Hannu Poropudas
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,ONE NOTE about one "little strange" function used in my maple calculations
Hannu Poropudas
csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)
HannuYou 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, they1. I put more clearly my incomplete error estimation procedure here:
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.
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.
4. I have made here NO physical interpretations about this special example, I leave to make them to those
who understand astrophysics better than me, if this was sensible at all ?
Best Regards,SUMMARY:
Hannu Poropudas
# 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
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.
6292090968e12)^7;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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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
+,- sign for REIF(r)#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;+,- sign for IMIF(r)#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.
#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);
I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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,I'am sorry about error in 30.10.2023 posting of mine.
Hannu Poropudas
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,ONE NOTE about one "little strange" function used in my maple calculations
Hannu Poropudas
csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)
HannuYou 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, they1. I put more clearly my incomplete error estimation procedure here:
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.
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.
4. I have made here NO physical interpretations about this special example, I leave to make them to those
who understand astrophysics better than me, if this was sensible at all ?
Best Regards,SUMMARY:
Hannu Poropudas
# 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,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),
Hannu Poropudas
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 PoropudasIt'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.
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.
6292090968e12)^7;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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also approximate proper time t integral (primitive function)
Hannu Poropudas
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
+,- sign for REIF(r)#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;+,- sign for IMIF(r)#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.
#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);
I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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,I'am sorry about error in 30.10.2023 posting of mine.
Hannu Poropudas
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,ONE NOTE about one "little strange" function used in my maple calculations
Hannu Poropudas
csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)
HannuYou 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, they1. I put more clearly my incomplete error estimation procedure here:
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.
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.
4. I have made here NO physical interpretations about this special example, I leave to make them to those
who understand astrophysics better than me, if this was sensible at all ?
Best Regards,SUMMARY:
Hannu Poropudas
# 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,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),
Hannu Poropudas
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
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.
6292090968e12)^7;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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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
+,- sign for REIF(r)#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;+,- sign for IMIF(r)#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.
#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);
I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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,I'am sorry about error in 30.10.2023 posting of mine.
Hannu Poropudas
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,ONE NOTE about one "little strange" function used in my maple calculations
Hannu Poropudas
csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)
HannuYou 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, they1. I put more clearly my incomplete error estimation procedure here:
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.
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.
4. I have made here NO physical interpretations about this special example, I leave to make them to those
who understand astrophysics better than me, if this was sensible at all ?
Best Regards,SUMMARY:
Hannu Poropudas
# 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,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),
Hannu Poropudas
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 PoropudasIt'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",Some examples reading about "Clausius Hooke non-a-diabatic Unruh".
in the formula, to keep in mind that there are various laws of large numbers,
and that not all tests for convergence, agree.
https://www.science.org/doi/10.1126/sciadv.1601646 https://journals.aps.org/pre/abstract/10.1103/PhysRevE.102.052119
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.
6292090968e12)^7;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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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
+,- sign for REIF(r)#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;+,- sign for IMIF(r)#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.
#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);
7I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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,I'am sorry about error in 30.10.2023 posting of mine.
Hannu Poropudas
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,ONE NOTE about one "little strange" function used in my maple calculations
Hannu Poropudas
csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)
HannuYou 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, they1. I put more clearly my incomplete error estimation procedure here:
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.
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.
4. I have made here NO physical interpretations about this special example, I leave to make them to those
who understand astrophysics better than me, if this was sensible at all ?
Best Regards,SUMMARY:
Hannu Poropudas
# 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)^
# 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,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),
Hannu Poropudas
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 PoropudasIt'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
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.
6292090968e12)^7;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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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
+,- sign for REIF(r)#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;+,- sign for IMIF(r)#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.
#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);
7I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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,I'am sorry about error in 30.10.2023 posting of mine.
Hannu Poropudas
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,ONE NOTE about one "little strange" function used in my maple calculations
Hannu Poropudas
csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)
HannuYou 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, they1. I put more clearly my incomplete error estimation procedure here:
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.
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.
4. I have made here NO physical interpretations about this special example, I leave to make them to those
who understand astrophysics better than me, if this was sensible at all ?
Best Regards,SUMMARY:
Hannu Poropudas
# 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)^
# 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,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),
Hannu Poropudas
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 PoropudasIt'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
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.
6292090968e12)^7;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.
Best Regrads,
Hannu Poropudas
Kolamäentie 9EI used random numbers for arbitrary example , which I don't know if it is sensible at all in this case
90900 Kiiminki / Oulu
Finland
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,Your solution is either:
Hannu Poropudas
(a) incorrect, or:
(b) isometric to Schwarzschild's.
Don't waste your time.
--Your (b) alternative seems not to be true due two separate event horizons in this metrics ?
Jan
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.
HannuI 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 PoropudasI 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,CORRECTION: It is proper time (t) integral in question, not coordinate time (T).
Hannu Poropudas
Sorry that I confused these two letters.
HannuI 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
Please take a look.
Best Regards,In order to me more mathematically complete I calculate also
Hannu Poropudas
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
+,- sign for REIF(r)#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;+,- sign for IMIF(r)#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.
#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);
7I calculated also coordinate time T series approximation up to 7 degree at r=MG,Best Regards,
Hannu Poropudas
(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,I'am sorry about error in 30.10.2023 posting of mine.
Hannu Poropudas
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,ONE NOTE about one "little strange" function used in my maple calculations
Hannu Poropudas
csgn(r-0.6292090968e12) = (r-0.6292090968e12)/abs(r-0.6292090968e12)
HannuYou 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, they1. I put more clearly my incomplete error estimation procedure here:
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.
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.
4. I have made here NO physical interpretations about this special example, I leave to make them to those
who understand astrophysics better than me, if this was sensible at all ?
Best Regards,SUMMARY:
Hannu Poropudas
# 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)^
# 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,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),
Hannu Poropudas
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 PoropudasIt'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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