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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