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Symmetry of Ricci Tensor

  1. Dec 28, 2013 #1
    Hey,

    I have been doing a few proofs and stumbled across this little problem.

    Trying to show the symmetry of the Ricci tensor by using the Riemann tensor definition

    ##R^m_{\ ikp} = \partial_k \Gamma^m_{\ ip} - \partial_p \Gamma^m_{\ ki} + \Gamma^a_{\ ip} \Gamma^m_{\ ak} - \Gamma^a_{\ ik} \Gamma^m_{\ ap}##

    Now set m = k

    ##R^m_{\ \,imp} = R_{ip} = \partial_m \Gamma^m_{\ ip} - \partial_p \Gamma^m_{\ mi} + \Gamma^a_{\ ip} \Gamma^m_{\ am} - \Gamma^a_{\ im} \Gamma^m_{\ ap}##

    Checking every term for symmetry (i <-> p)

    1. symmetric because ##\Gamma^{m}_{\ ip} = \Gamma^{m}_{\ pi}##

    2. see below

    3. symmetric because ##\Gamma^{m}_{\ ip} = \Gamma^{m}_{\ pi}##

    4. symmetric because ##\Gamma^a_{\ im} \Gamma^m_{\ ap} = \Gamma^m_{\ pa} \Gamma^a_{\ mi}## (interchanging a & m, and just rotate) and ##\Gamma^{a}_{\ im} = \Gamma^{a}_{\ mi}##

    Now for the 2nd term

    ##\partial_p \Gamma^m_{\ mi} == \partial_i \Gamma^m_{\ mp}##

    This is a contracted Christoffel symbol...

    using ##\Gamma^k_{ij} = \frac{1}{2} g^{kl} \left(\partial_i g_{jl} + \partial_j g_{il} - \partial_l g_{ij}\right)##

    leads to

    ##\Gamma^{m}_{\ mi} = \frac{1}{2} g^{ml} \frac{\partial g_{ml}}{\partial {x^i}} ##

    Now the derivative

    ##\partial_p \Gamma^m_{\ mi} = \frac{1}{2} \left( \frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} + g^{ml} \frac{\partial g_{ml}}{\partial {x^i} \partial {x^p}}\right)##

    The 2nd term in this expression is symmetric (i <-> p), because order of partial differentiation doesn't matter.

    For the first term I am not sure though.

    Is ##\frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} == \frac{\partial g^{ml}}{\partial x^i} \frac{\partial g_{ml}}{\partial {x^p}}##?

    The inverse is something totally different, but you also contract it, this is the point where I am slightly confused.

    --------------

    I know there are other ways to "derive" the symmetry of the Ricci tensor, but I wanted to try this one :)

    --------------

    Thank you in advance!

    It's probably so obvious that I don't see it :rolleyes:

    Hope you guys had a nice Christmas ;)
     
  2. jcsd
  3. Dec 28, 2013 #2

    WannabeNewton

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    Well you're definitely torturing yourself needlessly by doing it this way :tongue2:

    If you just use the real definition of the Riemann tensor, ##2\nabla_{[a}\nabla_{b]}\xi^{c} = -R_{abd}{}{}^{c}\xi^{d}##, then it immediately follows that ##R_{ab} = g^{cd}R_{dabc} = g^{cd}R_{cbad} = R_{ba}##.

    But if you really want to do it using the coordinate-dependent definition (ick!)* of the Riemann tensor then make it easier for yourself and go to a locally inertial coordinate system at an arbitrary event ##p## in space-time so that ##\Gamma^{\mu}_{\nu\gamma}(p) = 0## hence ##R_{\mu\nu}(p) = 2\Gamma^{\gamma}_{\mu[\nu,\gamma]}(p)##. Now all you have to do is plug in for ##\Gamma^{\mu}_{\nu\gamma}(p)##.

    *Coordinate-independent ("abstract index") calculations are infinitely more elegant than coordinate-dependent calculations :smile:
     
  4. Dec 28, 2013 #3

    Bill_K

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    Here's the fact you need: δgμν = - gμα gνβ δgαβ. Thus

    ∂gmℓ/∂xp = - gma gℓb ∂gab/∂xp

    and therefore

    ∂gmℓ/∂xp ∂gmℓ/∂xi = - gma gℓb ∂gab/∂xp ∂gmℓ/∂xi

    which is symmetric.
     
  5. Dec 28, 2013 #4
    It wasn't much work, actually.

    But that still boils down to ##\frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} == \frac{\partial g^{ml}}{\partial x^i} \frac{\partial g_{ml}}{\partial {x^p}}##, doesn't it?

    Using your definition, I get the first term to be symmetric and the 2nd one is

    ##\Gamma^{\gamma}_{\ \mu \gamma , \nu}##

    which should be equal to

    ##\Gamma^{\gamma}_{\ \nu \gamma , \mu}##.

    And this still leads to

    ##\frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} == \frac{\partial g^{ml}}{\partial x^i} \frac{\partial g_{ml}}{\partial {x^p}}##.

    This is the only remaining thing I need to show and I don't know how :/


    Oh yes, I totally forgot this thing. Thank you!
     
  6. Dec 28, 2013 #5

    WannabeNewton

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    But keep in mind that if you do this in a locally inertial coordinate system like I told you to, ##\partial_{\gamma}g_{\mu\nu}(p) = 0## so you would just have the term with the second derivative of the metric and that's trivially symmetric.

    The point I'm trying to convey is: if you're going to do a coordinate-dependent calculation then you may as well pick the coordinate system that makes the calculation as simple as possible. However I personally like to do coordinate-free calculations whenever possible.
     
    Last edited: Dec 28, 2013
  7. Dec 28, 2013 #6
    Ok, done :)

    Just plug in, replace ab by ml, use symmetry of g_ml.

    Thank you both.

    Now I know what to prove next - or maybe I should finally start a serious book about this :rolleyes:


    I could do that for sure, but I haven't been that much into this, that I think about things like
    -lets use this to simplify this
    -use trick 123
    -...

    At the moment I am still doing this for fun, or when I am bored.
    I am planning to do "rigorous" tensor/GR from the beginning very soon, so that will work out I hope :)
     
  8. Dec 28, 2013 #7

    WannabeNewton

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    It's a ton of fun so you'll enjoy it without a doubt, especially if you like index gymnastics. Good luck!
     
  9. Dec 28, 2013 #8
    Thanks!

    Schutz GR, Carrol GR, Schutz MM, Cahill, Lovelock, Ohanian, ... should do the trick :)
     
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