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Ricci Curvature Tensor

  1. Mar 18, 2014 #1
    I've been studying the Einstein field equations. I learned that the Ricci curvature tensor was expressed as the following commutator:

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

    I know that these covariant derivatives are being applied to some vector(s).

    What I don't know however, is whether or not both covariant derivatives are being applied to the same vector and in the same frame of reference. Is this the case or not? Also, does the evaluation of both covariant derivatives need to use the same summation variables or can each covariant derivative involve different summation terms? If you do have to use the same summation terms for each covariant derivative, do all indices have to switch places when you do the other covariant derivative (ie. If one covariant derivative involves ∂V[itex]\mu[/itex]/∂y[itex]\nu[/itex], then does the other have to have ∂V[itex]\nu[/itex]/∂y[itex]\mu[/itex])? Sorry if the notation is off or if the equation looks weird. This is my first time making a thread on this forum (as I am new here) so it is my first time typing an equation like this.
     
  2. jcsd
  3. Mar 19, 2014 #2

    Bill_K

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    You're thinking of the Riemann tensor.

    The covariant derivative is a covariant version of the gradient operator, and in curved space two successive derivatives don't commute. The commutator includes a term for each tensor index of the quantity being differentiated. For example for Sμν one would have:

    (∇μν - ∇νμ) Sαβ = Sακ Rκβμν + Sλβ Rαλμν

    An index can appear at most twice in any term. An index that appears twice is called a dummy index and is summed over. For a pair of dummy indices you can always substitute any unused letter.
     
    Last edited: Mar 19, 2014
  4. Mar 19, 2014 #3
    Thank you for your correction. The man in the video that I watched said that the Riemann tensor could be regarded as the Ricci tensor for the purposes of the Einstein field equations.

    Here is a link to the video (I copied the URL of the video at the time that I wanted you to start from). If you don't mind, could you please watch to 1:36:11? This is only about one minute beyond where I started you off.

    https://www.youtube.com/watch?feature=player_embedded&v=pES_tNZJm3Q#t=5699

    As you can see, the man had evaluated the commutator that I presented before you. He called the solution the Riemann tensor and said that it could function as the Ricci tensor. My point of confusion came in due to the fact that he did a lot of abbreviations. Basically, what I want to know is this: Do all of those Cristoffel symbols need to have the same super scripts? I know the subscripts [itex]\mu[/itex] and [itex]\nu[/itex] need to be in very specific and fixed places. However, do the dummy variables need to be the same for each cristoffel symbol (in perhaps a rearranged order) or correlate to each other in any way? Also, if the Riemann tensor and the Ricci curvature tensor are two different tensors, then how does the Riemann tensor satisfy the purposes of the field equations?
     
  5. Mar 19, 2014 #4

    Bill_K

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    This guy is probably well-intentioned, but my impression is that giving half the detail and leaving the other half out is bound to result in confusion. When he writes just Γμ for the Christoffel symbol, then a natural question to ask is what the other two indices are supposed to be. I think you need to learn it the right way first, and then afterwards you can start making notational abbreviations like this.

    When he writes ∂μ + Γμ, he means this to act on something, a vector Vμ maybe. And here I've intentionally used the same symbol μ twice on both ∂μ and Vμ in order to point out that you must often rename indices to step around the summation convention. ∂μ Vμ would imply that μ is being summed over, so you should rewrite it as perhaps ∂μ Vσ.

    Likewise, the covariant derivative ∇μ Vσ written out will contain a term perhaps like Γσμλ Vλ, but if you differentiate V twice, getting two Christoffel symbols in the same term, you must be careful to change one pair of λ's to something else - anything else that's not already in use. Never have more than two of the same index in the same term.

    The Riemann tensor is the geometrical quantity that describes in full the curvature. It can be split into two parts with different physical meanings. One half, the Ricci tensor, represents matter and appears in Einstein's Equations. The other half, the Weyl tensor, represents the gravitational field, and is nonzero even in places outside the source where there is no matter present.
     
  6. Mar 19, 2014 #5

    pervect

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    What a confusing lecture! I think you'll need a better source of info to learn even a little bit about GR, I can' t imagine any meaningful way in which the Riemann tensor could be equated with the Ricci tensor for instance.

    To try to give you a little insight, though: (I have no idea what you know, I'll try to cover the basics, which may either be overkill, or more likely too terse to be really understandable).

    ##\nabla_a## is what's called a covariant derivative. It's a generalization of the hopefully more familiar partial derivaitive ##\partial_{\mu}## which is also written as ##\frac{\partial}{\partial \mu}##.

    You can find the exact rules that a derivative operator must satisfy online in reference like Caroll's lecture notes on GR, they basically have to satisfy the chain rule and several other more basic notions that partial derivative operators do. You'll also find a rather terse discussion that singles out a specific preferred partial derivative operator, one that is torsion free and metric compatible, which definese a specific derivative operator given a manifold with a metric that we use in GR. So in general derivative operators aren't unique, but in GR we single out one that is from the metric.

    Next you need to know or learn about tensor notation:

    You'll need to know that in tensor notation ##x^{\mu}## actually represents in 4-d space 4 numbers, where ##\mu## is allowed to take on the values 0,1,2,3. This makes it a vector, specifically a tangent vector.

    ##x_{\mu}## is the dual of ##x^{\mu}##. It is also a vector, but it's not a tangent vector, it's a cotangent vector. Also, usually when people say "vector" with no qualification, they mean tangent vectors, not cotangent vectors, but I will (perhaps confusingly) call them both vectors here for lack of a better term.

    A tensor with no indicies is a scalar quanity, a tensor with one index is a vector or rank-1 tensor, a tensor with 2 indices is a rank 2 tensor, etc.

    Now you can see that ##\partial_{\mu}## when it operates on a scalar quantity f (one with no indices) it generates a rank-1 tensor (a quantity with one index).

    ##\nabla_{\mu}## does the same, but it can be meaningfully applied to tensor quantities, and not just scalars. When you do so, it generates a tensor of one higher rank than its argument.

    The heart of the original definition of the Riemann tensor that given an arbitrary cotangent vector ##\omega_c## defined on some curved manifold, the operator ##\left( \nabla_a \nabla_b - \nabla_b \nabla_a \right)## depends only on the value of ##\omega## at a point, defining another tensor, the Riemann tensor.

    Specifically

    ##\left( \nabla_a \nabla_b - \nabla_b \nabla_a \right) \omega_c = R_{abc}{}^d \omega_d##

    where ##R_{abc}{}^d## is the Riemann.

    To interpret this in any meaningful sense requires more notions from differential geoemtry, like "parallel transport", that I don't have the time to get into. This gets tied in with what the connection and Christoffel symbols are, which I'm not going to try to explain here. (Usually you learn about connections and Christoffels first, before trying to tackle the Riemann).

    Basically this isn't something you'll learn from watching a short videotaped lecture, or a short post, it's something that's going to need some serious study to learn.
     
  7. Mar 20, 2014 #6
    Thank you very much. Now I am getting somewhere. There's only one more thing that I must ask. Let's say that I apply ∇[itex]\mu[/itex] to your example vector V[itex]\sigma[/itex] and use your example dummy variable [itex]\lambda[/itex]. Does this mean that ∇[itex]\nu[/itex] is being applied to the same vector V[itex]\sigma[/itex] since both covariant derivatives are inside of the commutator? Does this also mean that I need to use the same [itex]\lambda[/itex] as my dummy variable for ∇[itex]\nu[/itex]?
     
  8. Mar 20, 2014 #7

    Bill_K

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    In ∇νμ it's applied to the first derivative of the vector. Then you do it all over again with μ and ν reversed, and subtract the two results.

    No, you better not!

    OK, the first derivative is

    μVσ = ∂μVσ + Γσμλ Vλ

    Now we take the second derivative:

    νμVσ = ∇ν(∂μVσ + ΓσμλVλ) = ∂ν(∂μVσ + ΓσμλVλ) + (...correction terms..)

    The quantity in parentheses has two indices, so there will be two correction terms, one for each index:

    Γσν?(∂μV? + Γ?μλVλ)

    and

    Γ?νμ(∂?Vσ + ΓσVλ)

    Do you see the problem? Every place there's a question mark, we were going to use a λ. But we can't use λ again because there can be at most two like indices in a term. So we have to think of another letter, say κ. The correction terms can be written

    Γσνκ(∂μVκ + ΓκμλVλ)

    and

    Γκνμ(∂κVσ + ΓσκλVλ)
     
  9. Mar 20, 2014 #8
    Thanks! I totally get it now.
     
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