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In the case of 5d Kaluza (Klein) GR with NO charge and NO gauge field we expect 5d to reduce to 4d GR exactly. So this should be a very simple useful sanity check.s the side and corner terms of Einstein, Ricci and Energy tensors are zero, then R would be the same in 4d or 5d.

This simple example demonstrates an inconsistency or misunderstanding.

Take a small `test' particle oscillating within a thin tube through the centreof a large spherical `source' mass of uniform density rho.

Apparently it does not behave the same in 5d as in 4d?

OR more simply, taking a = b = 5 from eqn(1e) suggests R = 0 rather than the expected rho:

[tex] \:\: 0 = T_{55} = G_{55} = R_{55} - g_{55} R/2 = 0 - 1 R/2 \;\;\;\;\;\;eqn(0e) \:\:[/tex]

Detail section:

In D dimensions Where D=4,5; i,j=1-3; a,b=0-3(5).

Mass of spherical source of radius [itex] \: r_0 = (r_0)^3 4\pi\rho/3;\; g_{00} = -1 + 8\pi G \rho r^2 /3;\; 0 < r < r_0 \: [/itex]

[itex] \: g_{55}=1;\; g^{ab}(g_{ab}, R_{ab}, T_{ab}) = (D, R, T);\; R_{55} = - F^{ab}F_{ab}\epsilon_0/4 = 0; \; 8\pi G = 1\: [/itex]

here saves space (even if G differs between 4d and 5d, it does not solve the problem).

Also using Einstein's index summation convention and c = 1. For weak field in 4d or 5d:

[itex] \:T_{ab} = \rho u_a u_b,\; u_0 \approx -1 = u_a u^a,\; u_j \approx 0,\; u_5 = 0,\; T_{00} \approx \rho = -T \: [/itex]

[tex] \:\: T_{ab} = G_{ab} = R_{ab} - g_{ab} R/2 \;\;\;\;\;\;eqn(1e) \:\: [/tex]

[itex] \: T = g^{ab} G_{ab} = R - D R/2 = -(D-2)R/2 \: [/itex]

[itex] \: -R/2 = T/(D-2) \: [/itex]

[itex] \: R_{ab} = T_{ab} - g_{ab}T/(D-2) \: [/itex]

So for 4d:

[itex] \: R_{00} = \rho(1 - 1/2) = \rho/2 \: [/itex]

[itex] \: R_{11} = \rho(0 + 1/2) = \rho/2 = R_{22} = R_{33} \;\to\; R = (-1 + 3)\rho/2 = \rho \: [/itex] as wanted (corresponding to Newton)

[tex] \:\: R_{00} = \rho/2 \; (=4\pi G \rho) = R_{11} \;\;\;\;\;\;eqn(1r) \:\: [/tex]

But then it all goes wrong for 5d:

[itex] \: R_{00} = \rho(1 - 1/3) = \rho 2/3 \: [/itex]

[itex] \: R_{11} = \rho(0 + 1/3) = \rho/3 \: [/itex]

[tex] \:\: R_{00} = \rho 2/3 = 2R_{11} \neq R_{11} \;\;\;\;\;\;eqn(2r) \:\: [/tex]

The factor 1/2 in the Einstein Tensor should be independent of number of dimensions.

General Theory of Relativity - Dirac: p23-24 (Chap 13-14) suggest [tex] \: [G_{ab} = R_{ab} - g_{ab} R/2 = T_{ab}]_{;c} \: [/tex]

is to do with Bianchi relations, not on D. The weak field case is at p45-47 (Chap 25). Dirac and Misner-Thorne-Wheeler texts do not discuss 5d.

Most information on Kaluza Klein (quoted on Wiki and normal searches) set the energy tensor term to zero as the test particle is outside the source

- so 5d GR reduces to [itex] \: R_{ab} = 0 \: [/itex], avoiding the issue of R in solving the EoM for metric elements within a source.

Possible answer - I don't like it for reasons given - but anyway just try:

[tex] \:\: T_{ab} = G_{ab} = R_{ab} - g_{ab} R/(D-2) \;\;\;\;\;\;eqn(2e) \:\: [/tex]

[itex] \: T = g^{ab} G_{ab} = R - D R/(D-2) = -2R/(D-2) \: [/itex]

[itex] \: -R/(D-2) = T/2 \: [/itex]

[itex] \: R_{ab} = T_{ab} - g_{ab}T/2 \;\: [/itex] So D = 5,6, generates wanted eqn(1r) but not eqn(1e).

Question summary:

Explain the corner term of Einstein's (Kaluza's) equation (0e). are corner terms suspect anyway,

or is R = 0 also needed within the source (as well as outside the source)?

Which Einstein equation is correct eqn(1e) or eqn(2e)? How can eqn(2e) be correct anyway?

If eqn(1e) is correct, then please explain how to convert the wrong Ricci elements in eqn(2r) to the wanted eqn(1r)?

I still get very lost with the abstract formalism of differential geometry, but believe this simple test can be explained in terms of the old calculus used here and by Dirac.

How can such a simple sanity check go so horribly wrong? Thank you.

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# Contracting Einstein Tensor in 4d works, have problem in 5d

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