# Derivative of the metric tensor

1. Mar 10, 2008

### Santiago

Could anybody help to spot the inconsistency in the following reasoning?

When calculating the normal derivative of the metric tensor I get:

$$\partial_\mu g^{\rho \sigma} = g^{\rho \lambda} g^{\sigma \gamma} \partial_\mu g_{\lambda \gamma} + 2 \partial_\mu g^{\rho \sigma},$$ (1)

which means that:

$$g^{\rho \lambda} g^{\sigma \gamma} \partial_\mu g_{\lambda \gamma} = -\partial_\mu g^{\rho \sigma}.$$ (2)

And I don't see how this could be.

That's how I get this result:

$$\partial_\mu g^{\rho \sigma} = \partial_\mu (g^{\rho \lambda} g^{\sigma \gamma} g_{\lambda \gamma}) = g^{\rho \lambda} g^{\sigma \gamma} \partial_\mu g_{\lambda \gamma} + g^{\rho \lambda} g_{\lambda \gamma} \partial_\mu g^{\sigma \gamma} + g_{\lambda \gamma} g^{\sigma \gamma} \partial_\mu g^{\rho \lambda} = g^{\rho \lambda} g^{\sigma \gamma} \partial_\mu g_{\lambda \gamma} + \delta^\rho_\gamma \partial_\mu g^{\sigma \gamma} + \delta^\sigma_\lambda \partial_\mu g^{\rho \lambda} =$$
$$= g^{\rho \lambda} g^{\sigma \gamma} \partial_\mu g_{\lambda \gamma} + 2 \partial_\mu g^{\rho \sigma}.$$ (3)

Could anybody show how to get directly the right hand side of equation (2) from the left hand side, or show where the mistake in the equation (3) is?

Thanks a lot.

2. Mar 10, 2008

### shoehorn

Unless I've misunderstood something terribly obvious, it's trivial to get what you want. We know that

$$\partial_cg^{ab} = g^{ad}g^{be}\partial_cg_{de} + 2\partial_cg^{ab}$$

Agreed? Now subtract $2\partial_cg^{ab}$ from both sides and you get

$$g^{ad}g^{be}\partial_cg_{de} = \partial_cg^{ab} - 2\partial_cg^{ab} = -\partial_cg^{ab}$$

which is what you're looking for.

3. Mar 10, 2008

### samalkhaiat

Last edited: Mar 10, 2008
4. Mar 12, 2008

### Santiago

Thank you shoehorn and samalkhaiat for your replies, it helped a lot. Especially samalkhaiat. That's what I needed.

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