General Relativity – Why is the Covariant Derivative of the Metric Tensor Zero?

differential-geometrydifferentiationgeneral-relativitymetric-tensortensor-calculus

I've consulted several books for the explanation of why

$$\nabla _{\mu}g_{\alpha \beta} = 0,$$

and hence derive the relation between metric tensor and affine connection $\Gamma ^{\sigma}_{\mu \beta} $

$$\Gamma ^{\gamma} _{\beta \mu} = \frac{1}{2} g^{\alpha \gamma}(\partial _{\mu}g_{\alpha \beta} + \partial _{\beta} g_{\alpha \mu} – \partial _{\alpha}g_{\beta \mu}).$$

But I'm getting nowhere. May be I've to go through the concepts of manifold much deeper.

Best Answer

The connection is chosen so that the covariant derivative of the metric is zero. The vanishing covariant metric derivative is not a consequence of using "any" connection, it's a condition that allows us to choose a specific connection $\Gamma^{\sigma}_{\mu \beta}$. You could in principle have connections for which $\nabla_{\mu}g_{\alpha \beta}$ did not vanish. But we specifically want a connection for which this condition is true because we want a parallel transport operation which preserves angles and lengths.

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