Covariant Derivative and metric tensor

[GRstudent]

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.

 

[WannabeNewton]

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.

 

[GRstudent]

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.

 

[haushofer]

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.

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.

 

[GRstudent]

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.

 

[GRstudent]

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

What I should do next?

 

[haushofer]

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)

The Christoffel symbols I mentioned in my post. Just do the calculation as he shows you.

 

[haushofer]

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.

 

[GRstudent]

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."

 

[WannabeNewton]

-edit WannabeNewton makes index mistakes; just check the calculation in Carroll's notes.

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.

 

[haushofer]

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]".

 

[haushofer]

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."

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.

 

[WannabeNewton]

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]".

Oh I didn't see that. My bad.

 

[pervect]

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

What I should do next?

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]

 

[GRstudent]

I understood the explanation in S. Caroll's notes! Thank you all!