• #### 8 pi G over c^3 or c^4 or what?

From rockbrentwood@gmail.com@21:1/5 to All on Wed Aug 12 08:50:47 2020
I see, for the coupling coefficient for Einstein's field equations, different things in the literature even though one choice appears natural and obvious.

The one seen in numerous Wikipedia articles and on a wall in Leiden uses c^4. I have a problem with that, and I'll explain what's wrong with it.

The Einstein-Hilbert action is, up to a proportionality, the integral of R root(|g|) d^4 x, where g is the determinant of the metric, and R the curvature scalar associated with the metric; the volume element being d^4 x = dt ^ dx ^ dy ^ dz when using
local coordinates with t for time, (x,y,z) for space. The proportionality is inverse to the gravitational coupling, k.

Let M, L, T denote the dimensions respectively of mass, length and time; so [x] = [y] = [z] = L, [t] = T, [d^4 x] = L^3 T and [action] = ML^2/T. Assume for the line element [g_{mn} dx^m dx^n] = A, where x^0 = t, x^1 = x, x^2 = y, x^3 = z, where A is left
to be determined.

Denoting the dimensions [x^m] = [m] for brevity (with [0] = T, [1] = [2] = [3] = L), then for the metric, [g_{mn}] = A/[mn] and inverse metric [g^{mn}] = [mn]/A. Thus, [g] = A^4/[0123]^2 and [root(|g|) d^4 x] = A^2.

For the connection coefficient [Gamma_{mnr}] = A/[mnr], and [Gamma^m_{nr}] = [m]/[nr], since Gamma^m_{nr} = Gamma_{snr} g^{ms} is raised with the inverse metric. (We're using the summation convention here and below).

Thus [R^s_{rmn}] = [s]/[rmn] and [R_{mn}] = 1/[mn], where R_{mn} = R^s_{msn}. Finally, [R] = 1/A, since R = g^{mn} R_{mn} is contracted with the inverse metric.

Combining, we get [integral R root(|g|) d^4 x] = A. So, in order for this to produce an action, the proportionality has to be ML^2/AT and the coupling coefficient (which is inverse to this) is [k] = AT/ML^2.

As an aside, Einstein's equation with the cosmological coefficient Lambda is: G_{mn} + Lambda g_{mn} = k T_{mn}, where T_{mn} is the stress tensor. Since G_{mn} = R_{mn} - 1/2 g_{mn} R, then [G_{mn}] = 1/[mn]. Consequently, for the cosmological
coefficient, we must have [Lambda] = 1/A. Normally, it is taken to have dimensions [Lambda] = 1/L^2, which requires A = L^2.

For the Minkowski metric [eta_{mn}] = A/[mn], we have [eta_{00}] = A/T^2. Combining this with the Newton gravitational coefficient [G] = L^3/MT^2, and light speed [c] = L/T, this yields [eta_{00} 8 pi G/c^5] = AT/ML^2 = [k]. And we assume that |eta_{00}|
is a power of c. Therefore,

k = +/- eta_{00} 8 pi G/c^5.

For k = +/- 8 pi G/c^n, if n = 3, 4 or 5, then we have:

(A, [Lambda], |eta_{00}|) = L^2, 1/L^2, c^2, for n = 3;
(A, [Lambda], |eta_{00}|) = LT, 1/LT, c, for n = 4; or
(A, [Lambda], |eta_{00}|) = T^2, 1/T^2, 1, for n = 5.

Nobody, that I am aware of, ever takes the line element with dimensions A = LT. So k = +/- 8 pi G/c^4 makes no sense at all. The natural choices are +/- 8 pi G/c^3 with A = L^2 and the line element being regarded as a measure for spatial distance; or +/-
8 pi G/c^5 with A = T^2 and the line element being regarded as a measure of proper time.

Of these two, the only one consistent with the condition [Lambda] = 1/L^2 is A = L^2, |eta_{00}| = c^2 and k = +/- 8 pi G/c^3.

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• From Savin Beniwal@21:1/5 to rockbr...@gmail.com on Thu Aug 13 17:51:51 2020
On Wednesday, 12 August 2020 14:20:49 UTC+5:30, rockbr...@gmail.com wrote:
I see, for the coupling coefficient for Einstein's field equations,
different things in the literature even though one choice appears
natural and obvious.

[[Mod. note -- Quoted text trimmed. -J.T.]

Nobody, that I am aware of, ever takes the line element with
dimensions A = LT. So k = +/- 8 pi G/c^4 makes no sense at all. The
natural choices are +/- 8 pi G/c^3 with A = L^2 and the line element
being regarded as a measure for spatial distance; or +/- 8 pi G/c^5
with A = T^2 and the line element being regarded as a measure of
proper time.

Of these two, the only one consistent with the condition
[Lambda] = 1/L^2 is A = L^2, |eta_{00}| = c^2 and k = +/- 8 pi G/c^3.

First of all, as per my knowledge This dimensional analysis
does not depend on the system of coordinates neither on the metric tensor as g_mn has no dimension.

So the Einstein field equation is G_mn+lambda*g_mn=kT_mn. And we
have to calculte the dimension of k. As G_mn is already defined as
R_mn-1/2R. The scalar curvature (R) is a contraction of the Ricci
tensor (R_mn). The units do not alter through a contraction. The
Ricci tensor is therefore a contraction of the Riemann tensor and
all terms (R, R_mn) are the dimension of lenght (IInd derivative
of field).

[[Mod. note -- Line-wrapping adjusted so equations aren't broken
across lines. Alas, there were garbled 8-bit characters whose meaning
wasn't clear; I've replaced them with "???". -J.T.]

And hence, [G_mn]=1/L^2; [Lambda]=1/L^2, and [kT_mn]=1/L^2.
T_mn is energy density of dimension ML^2T^(???)/L^3, which
implies
[k]=(L^3/(MT^(-2)L^2))*(1/L^2)-->[k]=T^2/(ML) ...............(1)

On the other hand, [G]=L^3/(MT^2) and [c]=L/T---->[G/c^4]=(L^3/(MT^2))/(T^4/L^4)
---->[G/c^4]=T^2/(ML) .............(2)

From Equ (1) and (2), we can conclude that only k=8*pi*G/c^4 consistant
with the Einstein Field equation even with the Einstein-Hilbert Action.

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• From rockbrentwood@gmail.com@21:1/5 to Savin Beniwal on Thu Aug 20 09:04:34 2020
On Thursday, August 13, 2020 at 7:51:55 PM UTC-5, Savin Beniwal wrote:
T_mn is energy density

No it's not. And that's where your mistake ultimately comes from. T_mn is *not* even a tensor density at all, but a tensor. The correct statement is that it is #T_mn = root(|g|) T_mn is the density; and it is this which has the various densities (mass/
energy density, pressure) as its components; not T_mn.

Second, the densities you're thinking of (those that are associated with the transport laws for mass, momentum and energy) come out of the (1,1) tensor density, not the (2,0) tensor density. It is #T^0_0 which is mass-energy density, not #T_00. The
native form of that tensor density (before any metric is applied to raise or lower indexes) is (1,1), not (2,0); this corresponding to the tensor that gives the (N-1)-current associated with the local diffeomorphism given by the vector field X = X^m @_m:

T_X = boundary-integral (#T^m_n X^n) (@_m _| d^N x)

where @_m = @/@x^m and _| is the contraction operator and d^N x the N-form

d^N x = dx^0 ^ dx^1 ^ ... ^ dx^{N-1}

So, let's go back over this in further detail ... and generalize it to N dimensions ... because the factor 8 pi is also wrong, except for N = 4, and that also needs to be brought up. Oftentimes the literature glibly keeps 8 pi in when generalizing to
higher dimensions, forgetting about where it came from - people like Mansouri have pointed out the error and provided the correct analysis, which I'll replicate below

On Thursday, August 13, 2020 at 7:51:55 PM UTC-5, Savin Beniwal wrote:
First of all, as per my knowledge This dimensional analysis
does not depend on the system of coordinates neither on the metric tensor as g_mn has no dimension.

Everything has dimensions - including even things that "don't". Actually, it is *my* analysis, not yours, which is independent of what choice of dimensions one ascribes to the coordinates, and covers all possibilities in a single universal framework -
which also happens to subsume yours. Yours is heavily-dependent on the particular choices (already accounted for my mine) and so sheds no additional light, but only obscures the light already shed.

Your choice corresponds to [g_mn] = 1, "no dimension" meaning "dimensionless" and [x^m] = 1 = [x^n]. That's a special case of [g_mn dx^m dx^n] = A, [x^m] = [m] and [x^n] = [n] corresponding to the selection A = 1, [m] = 1 = [n].

But your assumption is wrong: coordinates *do* have dimensions: space-like coordinates are generally taken with dimension L, time-like with T. But it doesn't matter, since my analysis doesn't depend on how they're set, though yours did. (Indeed, that's
part of what's proven by the analysis - so it doesn't need to be assumed up front, as you erroneously thought it had to be).

(It also contradicts what you later said below, which amounts to the choice [0] = [1] = ... = [N-1] = L and A = L^2).

In general, as per the analysis the dimensions of [g_mn] are A/[mn], while those of the inverse metric are [g^mn] = [mn]/A (which - once again - includes your analysis as a special case and subsumes it).

The coefficient kappa is given first and foremost by the action principle via

S = integral 1/(2 kappa) R root(|g|) d^N x

analyses involving the stress tensor are not actually relevant at all and have no say on the final outcome. Quite the opposite, it's the other way around: the stress tensor too comes out of the action principle as follows.

The right hand side of Einstein's equations are determined from the action principle by variation with respect to g^{mn}.

#T_{mn} = -2 @(#L)/@g^{mn};
@ = partial derivative operator

where #L and #T are tensor DENSITIES, #L being the Lagrangian density, normally written L root(|g|).

The units for #T are given directly in these terms:

ML^2/T = [S] = [g^{mn}] [#T_{mn}] [d^N x] = [mn]/A [#T_{mn}] [01...(N-1)]

hence

[#T_{mn}] = A ML^2/(T [01...(N-1)] [mn])

With the normal choices A = L^2, [0] = T, [1] = [2] = [3] = L, this works out to [#T_{mn}] = ML/T^2 1/[mn] - which yield the expected result for the entire tensor density [#T_{mn} dx^m (x) dx^n] = ML/T^2.

(x) denote the tensor product operator (circle with x inside).

When the indices are put back in native form, this yields

[#T^m_n] = [g^{mr}] [#T_{nr}] = [mr]/A A ML^2/(T[01...(N-1)][nr])
or
[#T^m_n] = ML^2/(T[01...(N-1)]) [m]/[n]

For the energy density (m=0=n) and pressure (m=n=1,...,(N-1)), this produces the result ML^{3-N}/T^2 or for N = 3, M/(LT^2), as expected. This is the same dimensions as for the entire tensor density:

[#T^m_n @_m (x) dx^n] = ML^2/(T[01...(N-1)]).

The scalar curvature (R) is a contraction of the Ricci tensor

with [R] = 1/A, as I already showed; with [R_mn] = 1/[mn].

The contract changes the dimension, except in the case where [0] = [1] = ... = [N-1] = root(A). Your "doesn't change the dimension" is the special case corresponding to A = 1 (or later, below, to A = L^2), so is already subsumed.

all terms (R, R_mn) are the dimension of lenght

which contradicts what you earlier stated that all the quantities have "no dimensions" (i.e. are dimensionless). What you're now saying here - *this* statement - corresponds to the choice:

A = L^2, [0] = [1] = ... = [N-1] = L.

With *that* choice

[#T_{mn}] = ML^{2-N}/T = [#T^m_n]

while with your previous choice (A = 1, [0] = [1] = ... = [N-1] = 1):

[#T_{mn}] = ML^2/T = [#T^m_n].

And hence, [G_mn]=1/L^2; [Lambda]=1/L^2, and [kT_mn]=1/L^2.

which is the special case of the analysis I provided:

[G_mn] = 1/[mn] = [kappa T_mn].

corresponding to your choice [m] = L = [n].

T_mn is energy density of dimension ML^2T^(???)/L^3

which is wrong: it's not a density at all, #T_mn is.

In fact, I did the same analysis as you did long ago and ended up making that same mistake resulting in the same wrong conclusion with c^4 instead of c^3.

#T_mn is the density, not T_mn.

The dimensions of T_mn are actually given as follows

[#T_{mn}] = A ML^2/(T [01...(N-1)] [mn])
[root(|g|)] = A^{N/2}/([01...(N-1)])

hence

[T_mn] = A^{1-N/2} ML^2/(T [mn])

For N = 4 with the choice A = L^2, this yields

[T_mn] = M/(T[mn])

Thus

1/[mn] = [G_mn] = [kappa] [T_mn] = M/(T[mn])

[kappa] = T/M

And since [G] = L^3/(MT^2) and [c] = L/T, this shows that

[kappa] = [G/c^3], not [G/c^4].

In general

1/[mn] = [G_mn] = [kappa] A^{1-N/2} ML^2/(T [mn])

or

[kappa] = T A^{N/2-1}/(ML^2).

The N-dimensional gravitational coefficient, which we'll call G', is that which yields an inverse r^{N-2} field for a point-like source of mass M of the form

g = G' M/r^{N-2}

and so has dimensions

[G'] M/L^{N-2} = L/T^2
or
[G'] = L^{N-1}/MT^2
or
[G'/c^3] = L^{N-4}T/M

which matches [kappa] when A = L^2. Otherwise, there is a dependency on the dimension of A, which can only be removed by adding in the extra factor as I did

[eta_{00}] = A/T^2, [eta_{00}/c^2] = A/L^2

(for your choice, A = L^2 and [0] = L that would be [eta_{00}] = L^2).

Then

[G'/c^3 (|eta_{00}|/c^2)^{N/2-1}] = TA^{N/2-1}/(ML^2) = [kappa]

For the general N-dimensional case, where U is the gravitational potential for the point-like source. with [U] = (L/T)^2,

del^2 U corresponds to R_00

up to sign, for the choice A = L^2; while for perfect, fluids

rho c^2 corresponds to #T^0_0,

where rho is the mass density; for a perfect fluid with 0 pressure, it's the only non-zero component.

So, with that settled, adopt c = 1 for the following and [0] = [1] = ... = [N-1] = L, A = L^2.

For near-Minkowski metrics |g_{00}| ~ |eta_{00}| = 1, |g| ~ |eta| = 1, which we'll use below.

The Einstein equations in terms of the tensor R_mn is:

R_mn root(|g|) = kappa (#T_mn - g_mn #T^r_r/(N-2))

or since #T^r_r = #T^0_0, this yields

del^2 U = R_00 = kappa g_00 rho/root(|g|) (N-3)/(N-2)
or
del^2 U ~ kappa rho (N-3)/(N-2)

The Poisson equation for N-1 spatial dimensions is

del^2 U = -A_{N-2} G' rho

where A_{N-1} is the area of the unit N-2 sphere, which is

A_{N-2} = (N-1) pi^{(N-1)/2} / ((N-1)/2)!
or
A_{N-2} = 2 pi pi^{(N-3)/2} / ((N-3)/2)!

adopting the convention (-1/2)! = root(pi).

You can readily see that this works for N = 4:

A_2 = 2 pi root(pi)/(1/2)! = 4 pi.

Thus

kappa = +/-(N-1)(N-2)/(N-3) pi^{(N-1)/2} / ((N-1)/2)! G'

Or reinserting back in the factors of c:

kappa = +/-(N-1)(N-2)/(N-3) pi^{(N-1)/2}/((N-1)/2)! G'/c^3.

For N = 4, this produces the factor 8 pi, up to sign, as expected:

3*2/1 pi root(pi)/(3/2)! = 3*2/1 pi 4/3 = 8 pi.

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• From lydiamariewilliamson@gmail.com@21:1/5 to rockbr...@gmail.com on Thu Aug 20 20:26:49 2020
On Thursday, August 20, 2020 at 4:04:39 AM UTC-5, rockbr...@gmail.com wrote:
On Thursday, August 13, 2020 at 7:51:55 PM UTC-5, Savin Beniwal wrote:
First of all, as per my knowledge This dimensional analysis
does not depend on the system of coordinates neither on the metric tensor as
g_mn has no dimension.

Your choice corresponds to [g_mn] = 1, "no dimension" meaning
"dimensionless" and [x^m] = 1 = [x^n]. That's a special case of [g_mn
dx^m dx^n] = A, [x^m] = [m] and [x^n] = [n] corresponding to the
selection A = 1, [m] = 1 = [n].

Though the rest of the analysis is fine, you misread what he said here
and there was no contradiction. The only thing you can infer from the
remark above is what you later inferred, namely that his statement
amounts to the convention:

A = [0]^2, [0] = [1] = ... = [N - 1].

But your assumption is wrong: coordinates *do* have dimensions: space-like coordinates are generally taken with dimension L, time-like with T.

In fact, you both missed pointing out key examples that show how/why the
more general analysis applies. For spherical space-time coordinates, x^0
= t, x^1 = r, x^2 = theta, x^3 = phi, you have [0] = T, [1] = L, [2] = 1
= [3]. Or ... you could also say that the dimensions are [2] = R = [3]
for radians; in which case you would have to go back to the tradition of stating the argument of trig functions with units; the correspondence
being that the classical versions (sin_C, cos_C) would be equated to the post-Calculus versions cos x + i sin x = exp(ix) by the conversion

In either case, the units of [g_mn] are not dimensionless! [g_22] =
[g_33] = A or A/R^2, which under the convention A = L^2 would equate to
L^2 or (L/R)^2.

You also misidentified the mistake he made, though you're correct in
pointing out that his characterization of T_mn is wrong. But that error
(under his conventions A = L^2, [0] = ... = [N-1] = L) is harmless.

The original poster's claim that T_mn has dimensions M/LT^2 is actually
what's wrong - when made under the assumption [0] = L. Its dimensions
depend on [0]; and M/LT^2 is only valid with [0] = T. Otherwise, it is
M/L^2T:

[#T_mn] = ML^2/T 1/[01...(N-1)] A/[mn]

becomes under the assumption A = L^2 and [m] = L = [n]:

[#T_mn] = ML^{2-N}/T.

so with [g_mn] = A/L^2 and [g] = A^N/[01...(N-1)]^2 and T_mn =
#T_mn/|g|^{1/2}, which yields dimensions (under the same assumptions A =
L^2, [m] = L = [n]):

[T_mn] = ML^2/T A^{1-N/2}/[mn] = ML^{2-N}/T

which reduces to M/L^2T for N = 4.

Only when [0] = T, instead of [0] = L; with [1] = [2] = ... = [N-1] = L,
A = L^2, do you get [T_mn] = M/LT^2, the unit for energy density and
pressure.

That's the mistake he actually made.

It would also help to see his last claim addressed *directly* - *in*
*terms* of the conventions he used!

The action S = integral 1/(2 kappa) R |g|^{1/2} d^N x

with the choice A = L^2, [0] = [1] = ... = [N-1] = L has units involving
these:

[S] = ML^2/T,
[g^{1/2} d^N x] = A^{N/2} = L^N,
[R] = 1/A = 1/L^2,
[G'] (your N-dimensional G) = L^{N-1}/MT^2,
[c] = L/T.

Therefore

ML^2/T = [S] = [R] [|g|^{1/2} d^N x]/[kappa]
or
ML^2/T = 1/A A^{N/2}/[kappa]
or
[kappa] = A^{N/2-1}T/ML^2 = L^{N-4}T/M
while
[G'/c^3] = L^{N-1}/MT^2 (T/L)^3 = L^{N-4}T/M.

So

kappa = (number factors) * G'/c^3

and the verdict is indeed that it's c^3, not c^4 ... even on grounds of
his assumptions.

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