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Form of Rienmann Tensor isotrpic & homogenous metric quick Question

  1. Dec 18, 2014 #1
    Context: Deriving the maximally symmetric- isotropic and homogenous- spatial metric

    I've seen a fair few sources state that the Rienamm tensor associated with the metric should take the form:

    * ##R_{abcd}=K(g_{ac}g_{bd}-g_{ad}g_{bc})##

    The arguing being that a maximally symmetric space has constant curvature and so Riemann Tensor can only depend upon the metric and the k which specifies the curvature. This combined with respecting the anti-symmetries of the Riemann tensor leads to form *.

    My question:

    I assume it should be obvious, since most sources simply state it without any justifcation, but, I don't understand how the requirement of constant curvature forces the Riemann Tensor to only depend on the metric ?

    Here's what I do know:

    The Riemann tensor arises from quantifying the extent to which the second covariant deriviatives fail to commute. It has a associate Ricci vector and scalar. And, the Ricci scalar is the curvature scalar of the manifold. I know that taking traces we can 'reduce' the Riemann tensor to get the Ricci scalar.

    Actually, I may partly have an idea why: to get the Ricci scalar from the Riemann tensor (in the form of 4 lower indices) we need to multiply by 2 metrics with raised indicies, so if there are only metrics on the RHS of *, we get traces of the metric with itself - so just 3 in space - and so we have got R=constant as desired.

    But this idea doesn't seem complete to me - it doesnt say why we can't include other tensors - would this be along the right lines anyway or not?

    Also, I'm not looking for a mathematical proof, I have found a couple of sources for these, using killing vectors, I'm just looking for a logical explanation.

    Thanks very much for your help !
     
  2. jcsd
  3. Dec 18, 2014 #2
    You could in principle use other tensors, it's just that you have to use "proper" tensors, and that form is the only one that has the correct (anti)symmetry properties. A "lighter" proof can be found by going to a locally flat coordinate system, and noting that in it, your Riemann tensor should be Lorentz-invariant, and as such formed of Lorentz-invariant tensors, which are the Levi-Civita tensor, Kronecker delta and the metric itself. If you try form the most general quantity that has the same index structure as R with these and apply the (anti)symmetry relations, you'll get the (*) you posted (of course, it has also has a proportionality constant that's a Lorentz scalar, but that can be determined independently).
     
  4. Dec 21, 2014 #3
    Thanks for your reply. I've never really done something like this before, so I'm unsure what my starting expression should be. i'm thinking a term with 2metrics, and cross-terms which give the correct number of indicies. Looking at the Levi-Civita tensor though, it has an upper indice, so would I be better starting with ##R^{a}_{bcd}## as a pose to ##R_{abcd}##. Also should each term be multiplied by a constant?
     
  5. Dec 21, 2014 #4
    You should be fine with starting with all lower indices (after all, that's what you want to end up with), as you can always use the metric to lower and raise the indices of the L–C tensor. And yes, your initial expression should include the most general possible linear combination of these objects, so each comes with an independent (Lorentz) scalar coefficient.

    Other than that they are (a priori) only restricted by the correct amount & position of indices – so, in addition to the 2-metric terms, you in principle need a L–C-term, cross-terms with the the L–C tensor and two metrics (contracting one metric tensor with the L–C tensor – these can be easily eliminated) and so on.

    Since you say that you haven't done something like this before: you might want to use the symmetries as much and as soon as possible. In addition to the Riemann tensor identities, the metric is symmetric and L–C tensor is antisymmetric. For example, terms of the form ##A_1\eta_{\mu\nu}\eta_{\rho\sigma}## and ##A_2\eta_{\nu\mu}\eta_{\sigma\rho}##, which in principle should both be included separately, are not independent since the metric is symmetric, and can be combined to ##A\eta_{\mu\nu}\eta_{\rho\sigma}## (and, as it happens, this term does not contribute, since it doesn't respect the ##R_{abcd}=-R_{bacd}## and ##R_{abcd}=-R_{abdc}## antisymmetries).
     
    Last edited: Dec 21, 2014
  6. Dec 25, 2014 #5
     
  7. Dec 25, 2014 #6
    Thanks, I'm still having trouble seeing how a term with 2 metrics and L-C tensor can have the same indice position/amount as ##R_{abcd}##, if I have something like ##g_{ab}g_{cd}T^e_{fg}## and contract with ##g^{ab}## I get ##4g_{cd}T^f_{fg}##.So whilst this means I now sum over f, I still have 5 indicies.. I can only think of similar manipulations not something that will allow me to make progress. Also kronecker-delta - should these be included in the orginal expression or arising from contractions?
     
  8. Dec 25, 2014 #7
    Wait, what's your T? By Levi-Civita tensor I mean the antisymmetric tensor density ##\varepsilon_{\mu\nu\sigma\rho}##. On an orientable manifold[*], it is a "true" tensor when multiplied by a suitable coefficient (##\sqrt{|\det(g)|}##, or ##\sqrt{-g}## in typical notation on Lorentzian manifolds), and recalling that we work on a locally flat coordinate system, the coefficient becomes unity. You can get the correct index structure with ##g_{\mu\nu}\varepsilon_{\alpha\beta\sigma\rho}g^{\alpha\beta}## and (the numerous) variations of that expression.

    You should in principle include Kronecker deltas, but since you can always just use the Kronecker delta to remove a dummy index (the upper index of a delta must be contracted with something in order to get all lower indices), they contribute in a trivial way.

    Side note: Assuming you're approaching this from a physics perspective, you should note that one usually only uses latin indices with abstract index notation on Lorentzian manifolds, which is why I've been using Greek indices when referring to objects defined in the locally flat coordinate system. Of course, if you're aware of the distinction, then do whatever you please, but I thought I'd point it out since your notation seemed somewhat unconventional to me.

    [*] I assume that the manifold in question is orientable. Again, if your question is motivated by GR, this is given. I'm not sure how well this proof/reasoning works for non-orientable manifolds – it might very well be that there are subtleties that require the use of a more sophisticated proof.
     
    Last edited: Dec 25, 2014
  9. Dec 25, 2014 #8
    Thanks. I'm not entirely sure I understand this point. So if I have something like ##A_{ac}B^{a}_d## and use a delta: ##\delta^{a}_{e}A_{ac}B^{a}_{d}=A_{ac}B_{ed}##, initially I had ##a## as a dummy index...is this along the right lines?

    And for the starting expression then something like: ##Ag_{ab}g_{cd} +B\epsilon_{abcd}g_{ab}g_{cd}+(##the remaining 4! permutations of L-C tensor , each multiplied by ##g_{ab}g_{cd}## ##) + C\epsilon_{abcd}## (##A,B,C## constants, and each permutation of L-C term with a constant too.)
     
    Last edited: Dec 25, 2014
  10. Dec 25, 2014 #9
    Well, almost. Remember that 1) contractions are always between upper and lower indices (at least when you're not just working on ##\mathbb{R}^n##) and 2) You shouldn't use the same index as a dummy index and a free index (essentially, same index can appear only twice in an expression, and if it appears twice, it must be contracted) – you have a dummy ##a## in ##\delta## and ##B##, and what seems to be a free index ##a## in ##A##.

    Using your (0,2) and (1,1) tensors, the expression should read, for example, ##A_{ac}B_{ed}=A_{ac}\delta_{e}^{f}B_{fd}##, where ##f## is not the same as the other indices.

    Not quite. You need multiple ##gg##-terms, in principle 4! to account for all permutations of the indices like I explained in post #4 (and, like I said, *some* of them can be combined with others and then possibly eliminated via symmetries – but not all). This is what makes the calculation tedious if you want to do it completely explicitly. Note that otherwise you won't be able to get the result you quote in the OP, since it only consists of ##gg##-terms. The constants are also independent for each term, so you shouldn't denote them all by A,B,C, but rather by something like ##A_1,A_2,...##;##B_1,B_2,...## and so on.
     
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