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Question about Riemann and Ricci Curvature Tensors

  1. Jul 1, 2014 #1
    After my studies of metric tensors and Cristoffel symbols, I decided to move on to the Riemann tensor and the Ricci curvature tensor. Now I noticed that the Einstein Field Equations contain the Ricci curvature tensor (R[itex]\mu\nu[/itex]).

    Some sources say that you can derive this tensor by simply deriving the Riemann tensor using the commutator:

    [∇[itex]\nu[/itex] , ∇[itex]\mu[/itex]]

    However, it seems to me (and to some other sources) that this would derive Rab[itex]\nu[/itex][itex]\mu[/itex] which in turn could contract to R[itex]\nu[/itex][itex]\mu[/itex] rather than R[itex]\mu[/itex][itex]\nu[/itex].

    The Einstein field equations involve R[itex]\mu[/itex][itex]\nu[/itex] rather than R[itex]\nu[/itex][itex]\mu[/itex].

    If you are trying to work with Einstein's equations, then wouldn't you have to do the commutator:

    [∇[itex]\mu[/itex] , ∇[itex]\nu[/itex]]

    instead of the previous one that I mentioned and derive R[itex]\mu[/itex][itex]\nu[/itex] instead? (Especially since every other tensor in the equations involve the indicies in the order [itex]\mu[/itex][itex]\nu[/itex] instead of the other way around.)
  2. jcsd
  3. Jul 1, 2014 #2


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    The Ricci tensor is symmetric so these are the same tensor.
  4. Jul 1, 2014 #3
    So it doesn't matter which commutator I use?
  5. Jul 1, 2014 #4


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    Moreover, ##\mu## and ##\nu## are arbitrary labels on indices. You can rewrite the Einstein field equations by making the switches ##\mu\rightarrow\nu## and ##\nu\rightarrow\mu## everywhere and absolutely nothing will change. In fact, relabeling indices is a pretty standard trick to make calculations easier and less confusing.

    Also, I think you have some indices wrong in the first post, the Ricci tensor is a contraction of the 1st and 3rd indices of the Riemann, you seem to imply that it's a contraction of the 1st and 2nd.

  6. Jul 1, 2014 #5


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    I think you can always try to take all the Riemann tensor contractions, the only one that survives though is the one Matterwave gave.... for example the easiest one:

    [itex] R^{\mu}_{\mu \rho \sigma} = g^{\mu \alpha} R_{\alpha \mu \rho \sigma}= 0 [/itex]
    because you have a symmetric to [itex]\mu \alpha[/itex] contracting to an antisymmetric .... [itex] R_{\alpha \mu \rho \sigma}= - R_{\mu \alpha \rho \sigma}[/itex].
  7. Jul 1, 2014 #6
    So in short, when I take the commutator [∇[itex]\nu[/itex] , ∇[itex]\mu[/itex]] , which of the following am I actually deriving?:




    or is it none of the above? If it is none of the above, then how would I express the tensor that this commutator derives?

    My problem is not the math and the deriving itself. It is really just the notation for this particular tensor that confuses me since I have seen it expressed in so many different ways today alone.

    Also, wouldn't I have to use a contravariant metric tensor to contract the above tensor into the Ricci tensor if I am trying to work with the Einstein Field Equations?

    I ask this because the field equations start with the Ricci tensor, yet so many sources seem to imply that the Riemann tensor suffices.
  8. Jul 1, 2014 #7


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    The Ricci tensor IS just a contraction of the Riemann tensor. If you wanted, you could replace ##R_{\mu\nu}## everywhere with ##R^\tau_{\mu\tau\nu}##. They are equal.

    The correct expression in terms of the commutator depends a little bit on your sign conventions! In a common sign convention it is:


    For any vector ##V##
  9. Jul 1, 2014 #8
    So, if I want to use the expression $$R^\rho_{\sigma\mu\nu}$$ to get the Ricci tensor that appears in the field equations, then all I have to do is use that contravariant metric tensor to contract? (Or do I just leave this expression as it is?)
  10. Jul 1, 2014 #9


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    You can contract it by turning the ##\mu## into a ##\rho##.
  11. Jul 1, 2014 #10


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    All of them. The tensor is ##R##, and you can represent its components in different ways (contravariant/upper indices, covariant/lower indices, different coordinate systems) as you wish. ##R_{\alpha\beta\mu\nu}## specifies the tensor ##R## in a given coordinate system no more and no less than ##R^{\alpha}_{\beta\mu\nu}## does.

    Because Ricci is a contraction of Riemann, once you know Riemann you know Ricci. So Riemann does suffice.
  12. Jul 2, 2014 #11
    So, does this mean that if I am trying to solve the Einstein field equations, then I don't actually have to contract it? I could just leave it as the Riemann tensor and move on?
  13. Jul 2, 2014 #12


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    No.... It means that knowing the Riemann tensor means that you can calculate the Ricci Tensor....
    In the Einstein Equations, it's the Ricci tensor that enters in...
    But you can always write the Einstein Equation as:
    [itex] R_{\mu \nu} - \frac{1}{2} g_{\mu \nu}R = R^{\rho}_{\mu \rho \nu} - \frac{1}{2} g_{\mu \nu} g^{\alpha \sigma} R_{\sigma \alpha}=R^{\rho}_{\mu \rho \nu} - \frac{1}{2} g_{\mu \nu} g^{\alpha \sigma} R^{\rho}_{\sigma \rho \alpha} [/itex]

    but that's just too complicated... Most of times you go the other way (from the last to first equation), because you know the metric, which allows you to calculate the Christoffel Symbols (connection), from them you calculate the Riemann tensor, from the Riemann tensor you calculate the Ricci Tensor and from the Ricci Tensor the Ricci Scalar...
    the last two participate in Einstein's equations...

    The contraction means that when you see a symbol as:
    [itex]R^{\rho}_{\sigma \rho \alpha}= R^{0}_{\sigma 0 \alpha}+R^{1}_{\sigma 1 \alpha}+R^{2}_{\sigma 2 \alpha}+R^{3}_{\sigma 3 \alpha} = R_{\sigma \alpha} [/itex]
    In 4dimensions.... So knowing the elements [itex]R^{\mu}_{\sigma \rho \alpha}[/itex] you can calculate the ricci tensor by contracting [itex] \mu[/itex] with [itex]\rho[/itex]...so the Riemann tensor does suffice as Nugatory said...

    But you can't leave it without contraction... because at least if you accept that the Einstein Equations are:
    [itex] R_{\mu \nu} - \frac{1}{2} g_{\mu \nu}R = 8 \pi G T_{\mu \nu}[/itex]
    Then I don't see how you can leave a rank 4 tensor and add or equate it with rank 2 tensors.... it's like adding a matrix (similar to a 2 rank tensor) with a vector (similar to a 1 rank tensor)
    Last edited: Jul 2, 2014
  14. Jul 2, 2014 #13


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    I feel like we have answered the same question over 6 different replies, 6 different ways. I'm not sure what your confusion is OP, but it seems that you consider a contraction to be something that requires fundamentally new information?

    Maybe it's more clear if we re-phrase contraction as simply a sum of some components? The Ricci tensor, as we showed before, is simply the sum of 4 specific components of the Riemann tensor. It's analogous to saying 4x=x+x+x+x. In equations we can use 4x, or we can use x+x+x+x, it's just 4x is much simpler to write.
  15. Jul 2, 2014 #14
    I tried considering this option, but I came across some things that were seemingly problematic when I did so.

    As we can calculate:

    [∇[itex]\mu[/itex] , ∇[itex]\nu[/itex]] =

    [itex]\mu[/itex][itex]\Gamma[/itex]a[itex]\nu[/itex]b - ∂[itex]\nu[/itex][itex]\Gamma[/itex]a[itex]\mu[/itex]b + [itex]\Gamma[/itex]a[itex]\mu[/itex]c[itex]\Gamma[/itex]c[itex]\nu[/itex]b - [itex]\Gamma[/itex]a[itex]\nu[/itex]c[itex]\Gamma[/itex]c[itex]\mu[/itex]b

    This all equals Rab[itex]\mu[/itex][itex]\nu[/itex]

    You are telling me that I can turn this into the Ricci tensor if I turn the [itex]\mu[/itex] into an a.

    Now as we know, a Cristoffel symbol (in 3 dimensions) like this: [itex]\Gamma[/itex]a[itex]\nu[/itex]b

    would create 3 different 3 by 3 matricies. If I turn the [itex]\mu[/itex] into an a then:

    This expression (∂a[itex]\Gamma[/itex]a[itex]\nu[/itex]b) would mean that I could only differentiate elements with respect to the axis that corresponds to that particular matrix. Therefore, if I am working with spherical coordinates for example, then I can only differentiate the elements of my first matrix with respect to r, my 2nd matrix with respect to θ and my third matrix with respect to ø. I could not differentiate my first matrix with respect to θ even if it has elements that are in terms of θ.

    The expression (∂[itex]\nu[/itex][itex]\Gamma[/itex]aab) would mean that I could only differentiate elements in the first row in the first matrix, the 2nd row in the 2nd matrix, and the 3rd row in the 3rd matrix. This would leave in total 6 other rows undifferentiated.

    Essentially, turning the [itex]\mu[/itex] into an a would leave so many elements unaccounted for when I did all of my operations on the Cristoffel symbols.

    What should I do about this? Could I even legally just change the [itex]\mu[/itex] into an a without doing some kind of operation first?
  16. Jul 2, 2014 #15
    Disregard this post. I finally got it after a recent post that was posted while I was typing this post.
  17. Jul 2, 2014 #16


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    [itex] \Gamma^{a}_{\nu \beta} = \Gamma^{a}_{\nu \beta} (x^{\mu}) [/itex]
    So the derivative will act on the [itex]\Gamma[/itex]'s because of the argument [itex]x^{\mu}[/itex]....
    I don't see how you can see it as 3 different matrices which are 3x3....it's one quantity, with 27 elements (of course not all independent)...if I let my imagination to get wild I'd see it as a cubic [3D] matrix, but I don't think it makes much sense...

    Also one more thing.... the result of a contraction doesn't imply much about the previous quantity... as a simple example, if I tell you that the trace of a matrix is equal to 4, you can't possibly know the form of the matrix.... However knowing the matrix, you can always determine its trace....
    Last edited: Jul 2, 2014
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