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

  1. Oct 8, 2015 #1
    Question outline:
    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.
     
  2. jcsd
  3. Oct 8, 2015 #2
    This claim is a little bit ambiguous if you ask me.
    I take it they mean ##G_{\mu\nu} = {8\pi G}\, T_{\mu\nu}## in the D=4 resulting theory?

    If you want another view on this, check out Chris Pope's lectures on Kaluza Klein, which you can find at http://people.physics.tamu.edu/pope/.
    He uses a more suitable (in my opinion) approach where he considers Lagrangian densities and action principles.
    He uses frame fields (vielbeins) and spin connections, so this is a great opportunity to study them if you're unfamiliar with them.

    If you get through the first 4 pages of section 1.1 you should get a good idea of what happens. (ignore that it's described for arbitrary D)
     
  4. Oct 8, 2015 #3
    Thanks JorisL for your response and help.
    0. I suspect you are unhappy about me writing ## g_{55} ## = 1 rather than ## \phi^2. ## This is to simplify (save space). Think of a charged star rather than nucleus. Only I am setting charge and field to zero (so ## g_{55} ## = constant is valid) - to do sanity checks on very very basic index gymnastics in D=4 and D=5 dimensions - see 4.
    1. ## 8\pi G = 1 ## is only written here to save space (even if G differs between 4d and 5d, it does not solve the problem).
    2. will look at Pope's site and KK paper again, I downloaded his helpful ihplec.pdf in 2008 and studied the first parts. His reduction from his D+1 to D=4 dimensions is necessarily complicated by non-zero field and charge. And his paper is an example of 3.
    3. 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 ## R_{ab} = 0, ## avoiding the issue of R in solving the EoM for metric elements within a source.
    4. I still get very lost with the abstract formalism of differential geometry, but believe this very simple test has to work in the language of the old calculus used here and by Dirac himself.
    5. Kaluza's paper (refs in Wiki as is Pope) uses the EoM rather than Lagrangian - as the side terms are especially revealing.
     
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