Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Covariant Derivative and metric tensor

  1. Jun 28, 2012 #1
    Hi all,

    I am wondering if it is possible to derive the definition of a Christoffel symbols using the Covariant Derivative of the Metric Tensor. If yes, can I get a step-by-step solution?

    Thanks!

    Joe W.
     
  2. jcsd
  3. Jun 28, 2012 #2

    WannabeNewton

    User Avatar
    Science Advisor

    You know [itex]\triangledown _{\mu }g_{\alpha \beta } = 0[/itex] so permute the indexes, [itex]\triangledown _{\mu }g_{\alpha \beta } = 0, \triangledown _{\alpha }g_{\mu \beta } = 0,\triangledown _{\beta }g_{\alpha \mu } = 0 [/itex]. Then just write out each one, [itex]\triangledown _{\mu }g_{\alpha \beta } = \partial _{\mu }g_{\alpha \beta } - \Gamma ^{\sigma }_{\alpha \mu }g_{\sigma \beta } - \Gamma ^{\sigma }_{\beta \mu }g_{\alpha \sigma }[/itex] and similarly for the other two. Subtract the expressions for [itex]\triangledown _{\mu }g_{\alpha \beta }[/itex] and [itex]\triangledown _{\alpha }g_{\mu \beta } [/itex] and add the result to the expression for [itex]\triangledown _{\beta }g_{\alpha \mu }[/itex] and after using the symmetry of the christoffel symbols on the lower two indexes you get [itex]\partial _{\mu }g_{\alpha \beta } + \partial _{\beta }g_{\alpha \mu } - \partial _{\alpha}g_{\mu \beta } -2\Gamma ^{\sigma }_{\mu \beta }g_{\alpha \sigma } = 0 [/itex]. So you just solve for the Christoffel symbols (multiply both sides by [itex]g^{\alpha \sigma } [/itex] to just get [itex] \Gamma ^{\sigma }_{\mu \beta }\delta ^{\sigma}_{\sigma } = \Gamma ^{\sigma }_{\mu \beta }[/itex]) and you have your usual expression for the Christoffel symbols in terms of the metric.
     
  4. Jun 28, 2012 #3
    Thanks for reply.

    But I would like to have Christofell symbols in terms of the metric to be pluged in this equation. I mean, prove that covariant derivative of the metric tensor is zero by using metric tensors for Gammas in the equation.
     
    Last edited: Jun 28, 2012
  5. Jun 28, 2012 #4

    haushofer

    User Avatar
    Science Advisor

    Well, plug the Christoffel symbol (the ( ) indicate symmetrization of the indices with weight one)
    [tex]
    \Gamma^{\lambda}_{\rho\mu} = \frac{1}{2}g^{\lambda\sigma} \Bigl( 2\partial_{(\rho}g_{\mu)\sigma} - \partial_{\sigma}g_{\rho\mu} \Bigr)
    [/tex]
    into the definition of the covariant derivative of the metric and write it out.
    [tex]
    \nabla_{\rho}g_{\mu\nu} = \partial_{\rho}g_{\mu\nu} - 2 \Gamma^{\lambda}_{\rho(\mu}g_{\nu)\lambda} \\
    = \partial_{\rho}g_{\mu\nu} - \partial_{\rho}g_{\mu\nu} - \partial_{(\mu}g_{\nu)\rho} +
    \partial_{(\nu}g_{\mu)\rho} = 0
    [/tex]
    This is really a textbook question, so it would help if you point out what your precise problem is.
     
  6. Jun 28, 2012 #5
    Thank you!

    It is not from textbook; it was from Susskind Lecture 5 on GR. He asked students to prove what you just did.

    Also, what is the solution of WannaBeNewton's method? (post above) It only shows that Gammas are symmetric but it doesn't get the metric definition.
     
  7. Jun 28, 2012 #6
    \Gamma ^{\sigma }_{\mu \beta }\delta ^{\sigma}_{\sigma } = \Gamma ^{\sigma }_{\mu \beta }

    What I should do next?
     
  8. Jun 28, 2012 #7

    haushofer

    User Avatar
    Science Advisor

    The Christoffel symbols I mentioned in my post. Just do the calculation as he shows you.
     
  9. Jun 28, 2012 #8

    haushofer

    User Avatar
    Science Advisor

    Let me elaborate. Normally in GR you state that

    1) The metric is covariantly constant
    2) The connection is symmetric in its lower indices

    The interesting thing for you to do then is the calculation WannabeNewton mentions, using (2) that the connection is symmetric. That really allows you to solve for the connection explicitly in terms of the metric and its derivatives. If you don't get it you can consult e.g. Carroll's notes on GR. The calculation I showed you is then just a simple check.

    -edit WannabeNewton makes index mistakes; just check the calculation in Carroll's notes.
     
  10. Jun 28, 2012 #9
    I didn't understand the last instruction "So you just solve for the Christoffel symbols (multiply both sides by gασ to just get Γσμβδσσ=Γσμβ) and you have your usual expression for the Christoffel symbols in terms of the metric."
     
  11. Jun 28, 2012 #10

    WannabeNewton

    User Avatar
    Science Advisor

    Hmm...what index mistake? You just get [itex]\Gamma ^{\sigma }_{\mu \beta } = \frac{1}{2}g^{\alpha \sigma }(\partial _{\mu }g_{\alpha \beta } + \partial _{\beta }g_{\alpha \mu } - \partial _{\alpha }g_{\mu \beta })[/itex] and that's the usual expression.
     
  12. Jun 28, 2012 #11

    haushofer

    User Avatar
    Science Advisor

    Your "multiply both sides by [itex]g^{\alpha\sigma}[/itex]", while you already contract over the sigma index. It should read e.g. "multiply both sides by [itex]g^{\alpha\rho}[/itex]".
     
    Last edited: Jun 28, 2012
  13. Jun 28, 2012 #12

    haushofer

    User Avatar
    Science Advisor

    Like I said, do the calculation yourself with your own indices. You only have to use that

    [tex]
    g^{\mu\nu}g_{\nu\rho} = \delta^{\mu}_{\rho}
    [/tex]

    Have you read Carroll's notes? The calculation is there stated very explicitly, and it doesn't get any more explicit than that I'm afraid. Just do the calculation yourself following the steps.

    I have the feeling that you don't get the point of this calculation in the first place. What you want in GR is to write the connection in terms of the metric, such that it doesn't introduce new degrees of freedom. The geometry is then uniquely determined by the metric!

    One way to achieve this is to put the covariant derivative of the metric to zero and to put the torsion to zero. There are pictures which make clear what this means for the geometry in every textbook on GR. Now, metric compatibility gives you

    [tex]
    D*\frac{1}{2}D(D+1)
    [/tex]

    conditions (why?). The connection has also this number of independent components (why?), and it appears only algebraically multiplied by a metric (which is invertable) in the metric compatibility condition. This means you can solve for it! The index permutation mentioned earlier and in Carroll notes gives you the solution for the connection.

    The Riemann tensor then only depends on the metric, and that is the statement that the geometry is uniquely defined by the metric.
     
    Last edited: Jun 28, 2012
  14. Jun 28, 2012 #13

    WannabeNewton

    User Avatar
    Science Advisor

    Oh I didn't see that. My bad.
     
  15. Jun 28, 2012 #14

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I think your main question has been answered. But if I may suggest, you can and should wrap your latex inside <tex> and </tex} tags, where you replace < by [ and > by ], so that it becomes formatted in a more readable way. i.e.

    [tex]\Gamma ^{\sigma }_{\mu \beta }\delta ^{\sigma}_{\sigma } = \Gamma ^{\sigma }_{\mu \beta } [/tex]

    (If this still doesn't make sense, try quoting the message, you should see how it's done).

    Better yet is

    \Gamma ^{\sigma }{}_{\mu \beta }\delta ^{\sigma}{}_{\sigma } = \Gamma ^{\sigma }{}_{\mu \beta }

    which due to the added {} comes out as

    [tex]
    \Gamma ^{\sigma }{}_{\mu \beta }\delta ^{\sigma}{}_{\sigma } = \Gamma ^{\sigma }{}_{\mu \beta }
    [/tex]
     
  16. Jun 29, 2012 #15
    I understood the explanation in S. Caroll's notes! Thank you all!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Covariant Derivative and metric tensor
Loading...