Prove that metric tensor is covariant constant

[lerus]

TL;DR Summary

How to prove that metric tensor is covariant constant

I'm reading "Problem Book In Relativity and Gravitation".
In this book there is a problem
7.5 Show that metric tensor is covariant constant.
To prove it, authors suggest to use formulae for covariant derivative:
Aαβ;γ=Aαβ,γ−AσαΓβγσ−AσβΓαγσ
after that they write this formulae for tensor g and after that it is easy to prove that
gαβ;γ=0
I think I found another way to to prove it and I like it much more :).
It is always possible to find a coordinate system such that for some given point in the manifold, not only is the coordinate basis orthonormal but additionally all first-order partial derivatives of the metric components vanish.
It means that in this coordinate system:
gαβ,γ=0
and all
Γαγσ=0
It means that it this coordinate system
gαβ;γ=0
But this is a tensor equation it means it will be correct in any coordinate system.
Is this approach correct?

 

[PeterDonis]

Is this approach correct?

No, because it only proves that the first derivatives of the metric are zero at one point. But what needs to be proved is that the metric is covariantly constant everywhere.

 

[Orodruin]

First of all, please use LaTeX. Particularly index equations are unreadable as posted.

To the point: It is not always the case that the connection is metric compatible. It depends on the connection and metric. However, it is the case for the Levi-Civita connection, which is the unique torsion free and metric compatible connection. As such, you will need to use the particular form of the connection coefficients (Christoffel symbols in the case of the Levi-Civita connection) to prove metric compatibility.

 

[lerus]

First of all, please use LaTeX. Particularly index equations are unreadable as posted.

To the point: It is not always the case that the connection is metric compatible. It depends on the connection and metric. However, it is the case for the Levi-Civita connection, which is the unique torsion free and metric compatible connection. As such, you will need to use the particular form of the connection coefficients (Christoffel symbols in the case of the Levi-Civita connection) to prove metric compatibility.

Thank you for your answer, I actually used LaTeX and in preview everything looked OK, but when I posted, something happened. Sorry...

Did I understand correctly that that this formulae
##A_{\alpha \beta ; \gamma} = A_{\alpha \beta , \gamma} - A_{\sigma \alpha }\Gamma^{\sigma}_{\beta \gamma} - A_{\sigma \beta }\Gamma^{\sigma}_{\alpha \gamma}##
is correct only for Levi-Civita connection?

 

[lerus]

No, because it only proves that the first derivatives of the metric are zero at one point. But what needs to be proved is that the metric is covariantly constant everywhere.

Thank you for your answer.
We can repeat this for any point, isn't it enough?

 

[martinbn]

The covariant derivative is defined, in this book, by the formula you wrote, the one with the Gammas. The Gammas are defined by the usual coordinate formula using the metric and its partial derivatives. I think they expect you to write the formula for the covariant derivative of the metric. Then substitute the Gammas and check that everything cancels out.

 

[haushofer]

Thank you for your answer.
We can repeat this for any point, isn't it enough?

If you prove that the first derivative of f(x)=x^2 vanishes in x=0, does this mean it vanishes in every point?

You can change your grid such that the point formarly known as x=0 is moved to x=b such that now f'(x=b) becomes 0, but this is only in this particular point again.

 

[Orodruin]

Thank you for your answer.
We can repeat this for any point, isn't it enough?

Yes it is enough. However, you need to use also the argument that the Christoffel symbols themselves are a linear combination of first derivatives of the metric.

If you prove that the first derivative of f(x)=x^2 vanishes in x=0, does this mean it vanishes in every point?

You can change your grid such that the point formarly known as x=0 is moved to x=b such that now f'(x=b) becomes 0, but this is only in this particular point again.

This is not the point. The point is that regardless of the point you choose, you can find a coordinate system such that the metric in that point is orthonormal with vanishing first derivatives. Since ##\nabla_a g_{bc} = 0## is independent of the coordinate system, it is true regardless of the coordinates at that point. But since this can be done for every point ##\nabla_a g_{bc} = 0## everywhere.

The crucial part is that the argument can be repeated for any point you choose. However, as noted above, it relies on also noting that each term in the Christoffel symbols contain a first derivative of the metric.

The statement can also be proven without any reference to a coordinate system.

 

[lerus]

Yes it is enough. However, you need to use also the argument that the Christoffel symbols themselves are a linear combination of first derivatives of the metric.


This is not the point. The point is that regardless of the point you choose, you can find a coordinate system such that the metric in that point is orthonormal with vanishing first derivatives. Since ##\nabla_a g_{bc} = 0## is independent of the coordinate system, it is true regardless of the coordinates at that point. But since this can be done for every point ##\nabla_a g_{bc} = 0## everywhere.

The crucial part is that the argument can be repeated for any point you choose. However, as noted above, it relies on also noting that each term in the Christoffel symbols contain a first derivative of the metric.

The statement can also be proven without any reference to a coordinate system.

Thank you.
You described better than I could describe myself.
And, yes , you are right I still need this formulae for covariant derivative
$$A_{\alpha\beta;\gamma} = A_{\alpha\beta,\gamma} - A_{\alpha\sigma}\Gamma^{\sigma\beta\gamma} - A_{\beta\sigma}\Gamma^{\sigma\alpha\gamma}$$
I also need the argument that Christoffel symbols themselves are a linear combination of first derivatives of the metric.
It means that my approach is not shorter than the approach that is suggested in the book.
I have another question, probably it is better to start new thread, but still let's try here. The above mentioned formulae is correct only for metric compatible connection. Is it logically correct that we use this formulae prove that
##g_{ab;c}=0## ?
Thank you

 

[Orodruin]

Thank you.
You described better than I could describe myself.
And, yes , you are right I still need this formulae for covariant derivative
$$A_{\alpha\beta;\gamma} = A_{\alpha\beta,\gamma} - A_{\alpha\sigma}\Gamma^{\sigma\beta\gamma} - A_{\beta\sigma}\Gamma^{\sigma\alpha\gamma}$$

With the latter two indices of the connection coefficients lowered. Otherwise it doesn’t make sense.

I also need the argument that Christoffel symbols themselves are a linear combination of first derivatives of the metric.

The Christoffel symbols represent the connection coefficients of the Levi-Civita connection. The formula for expressing the derivative of the metric is universal, but the gammas could technically be the connection coefficients of some other connection.

It means that my approach is not shorter than the approach that is suggested in the book.
I have another question, probably it is better to start new thread, but still let's try here. The above mentioned formulae is correct only for metric compatible connection.

Which formula? The expression for the derivative of the metric is general, but the gammas could be the coefficients of a non-metric compatible connection. They will then not take the Levi-Civita form, but the expression will look the same.

Is it logically correct that we use this formulae prove that
##g_{ab;c}=0## ?
Thank you

 

[lerus]

With the latter two indices of the connection coefficients lowered. Otherwise it doesn’t make sense.

The Christoffel symbols represent the connection coefficients of the Levi-Civita connection. The formula for expressing the derivative of the metric is universal, but the gammas could technically be the connection coefficients of some other connection.

Which formula? The expression for the derivative of the metric is general, but the gammas could be the coefficients of a non-metric compatible connection. They will then not take the Levi-Civita form, but the expression will look the same.

I was talking about formula for gammas. But I need to think a bit more before asking questions.
Thank you.

 

[Orodruin]

I was talking about formula for gammas.

In GR, the connection used is the Levi-Civita connection, where the connection coefficients take that particular form. For a general connection, they can be anything.