Index Concordia

Some (trivial) consequences of metric-compatibility

Posted in Geometry, Relativity by Index Guy on July 13, 2007

If you were a bit nervous that I used two identities in the previous post without proof, well here I am to make you happy!

Let us start with a connection that is metric compatible. This means that

\nabla_{\alpha} g_{\mu \nu} = 0.

We now show that if this is true, then we also have

\nabla_{\alpha}g^{\mu \nu} = 0 and \nabla_{\alpha} \epsilon_{\mu \nu \sigma \rho} = 0.

We start with the first one. By definition, the inverse metric satisfies

 

g^{\alpha \gamma} g_{\gamma \beta} = \delta^{\alpha}_{\beta}.

If we operate with the covariant derivative on this equation, on the right-hand side we obtain zero, since the Kronecker delta is the same in every coordinate system and to top it all it is just a bunch of constants. Using the product rule we get

g^{\mu \nu}\nabla_{\alpha} g_{\nu \sigma} + g_{\nu \sigma}\nabla_{\alpha} g^{\mu \nu} = 0.

Since the connection is metric compatible, we have the first term vanishing. Unless the metric is trivially zero, we have

\nabla_{\alpha} g^{\mu \nu} = 0.

So one is in the bag. Now we consider the covariant derivative of the Levi-Civita tensor. By definition, this tensor can be written in terms of the Levi-Civita symbol (which is the same in every coordinate system)

\displaystyle\epsilon_{\mu \nu \sigma \rho} = \sqrt{\left|g\right|} \tilde{\epsilon}_{\mu \nu \sigma \rho}.

If we apply the covariant derivative to this equation, it will only operate on the determinant. By the chain rule we will get a term that involves the derivative of the determinant of the metric. This determinant will only depend on the metric entries, which means that the covariant derivative will vanish. VoilĂ !

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2 Responses

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  1. notsure said, on February 1, 2010 at 1:25 am

    The fact than the derivative doesn’t operate on the levi-civita symbol don’t seem obvious for me…

  2. Hanciong said, on February 21, 2010 at 1:08 pm

    hai. could you show me the formula for the determinant of the metric? thanx


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