General Relativity - Tensor Identities

[dextercioby]

I am 100% sure that $\nabla_b \phi = \partial_b \phi$ for $$\phi$$ scalar. What bigabau said was that the double covariant is not the same as the double partial but you can still use the innermost one being partial, you then get a connection term but since the connection is symmetric you can interchange a and b and you get the commutation relation for the covariants.

That's correct. About the last part, you have to compute the covariant D'Alembertian for a scalar field. It boils down to computing the covariant 4-divergence of a vector field and further to expressing

$$\Gamma^{\nu}_{~\nu\mu} = f\left(\partial_{\mu}\sqrt{\left| g\right|}\right)$$

which is a standard formula. A proof of that you can find on the internet, or, for example, in Dirac's little book.

 

[latentcorpse]

That's correct. About the last part, you have to compute the covariant D'Alembertian for a scalar field. It boils down to computing the covariant 4-divergence of a vector field and further to expressing

$$\Gamma^{\nu}_{~\nu\mu} = f\left(\partial_{\mu}\sqrt{\left| g\right|}\right)$$

which is a standard formula. A proof of that you can find on the internet, or, for example, in Dirac's little book.

Ok. I see that $\nabla_b \phi = \partial_b \phi$

But then $\nabla_a \nabla_b \phi = \nabla_a \partial_b \phi$
Why is this equal to $\nabla_b \partial_a \phi$?

Thanks!

 

[dextercioby]

Ok. I see that $\nabla_b \phi = \partial_b \phi$

But then $\nabla_a \nabla_b \phi = \nabla_a \partial_b \phi$
Why is this equal to $\nabla_b \partial_a \phi$?

Thanks!

Because $\phi$ is a scalar and the spacetime manifold has 0 torsion, then one can show that

$$\nabla_{a}\nabla_{b}\phi = \nabla_{b}\nabla_{a}\phi$$.

This equality should be left like this and not be recast in the noncovariant version you wrote, i.e. using normal derivatives acting on the $\phi$ i/o covariant ones, even though they are identical for a scalar field, but only for a scalar field.

 

[Tangent87]

Ok. I see that $\nabla_b \phi = \partial_b \phi$

But then $\nabla_a \nabla_b \phi = \nabla_a \partial_b \phi$
Why is this equal to $\nabla_b \partial_a \phi$?

Thanks!

Because $$\nabla_a \partial_b \phi=\partial_a\partial_b\phi-\Gamma_{ab}^c\partial_c\phi$$. Now do you see?

 

[latentcorpse]

Because $$\nabla_a \partial_b \phi=\partial_a\partial_b\phi-\Gamma_{ab}^c\partial_c\phi$$. Now do you see?

Ah, yes of course!
Thanks!

 

[Tangent87]

That's correct. About the last part, you have to compute the covariant D'Alembertian for a scalar field. It boils down to computing the covariant 4-divergence of a vector field and further to expressing

$$\Gamma^{\nu}_{~\nu\mu} = f\left(\partial_{\mu}\sqrt{\left| g\right|}\right)$$

which is a standard formula. A proof of that you can find on the internet, or, for example, in Dirac's little book.

So do you mean we have to compute $$\partial_a\partial^a\phi$$? Also could you possibly direct me to a proof of $$\Gamma^{\nu}_{~\nu\mu} = f\left(\partial_{\mu}\sqrt{\left| g\right|}\right)$$? I have never seen that before. Thanks.

 

[dextercioby]

I have attached a screenshot from Dirac's little book. I use it to make the following computations

$$\Box \phi = \nabla_a \nabla^a \phi = \partial_a \left(g^{ab}\partial_b \phi\right) + \Gamma^{c}_{~ca}g^{ab}\partial_b \phi = 0$$

Now plug the formula from Dirac for $\Gamma^{c}_{~ca}$ and multiply both sides of the new equality by $\sqrt{-g}$.

You should be then able to find the formula you wish to prove.

 

[Tangent87]

I have attached a screenshot from Dirac's little book. I use it to make the following computations

$$\Box \phi = \nabla_a \nabla^a \phi = \partial_a \left(g^{ab}\partial_b \phi\right) + \Gamma^{c}_{~ca}g^{ab}\partial_b \phi = 0$$

Now plug the formula from Dirac for $\Gamma^{c}_{~ca}$ and multiply both sides of the new equality by $\sqrt{-g}$.

You should be then able to find the formula you wish to prove.

Thank you for your help bigubau, I have derived the formula now but I don't understand the last two equalities of that screenshot from Dirac's little book, especially where the log(g) comes from. Could you elaborate please?

 

[Tangent87]

Ok I've worked out where the last equality comes from (it was quite easy actually, should of seen it). But I still don't understand why $$g^{ab}g_{ab,c}=g^{-1}g_{,c}$$

 

[dextercioby]

What is $g_{,c}$ equal to ? It's a formula very important in General Relativity. It's the derivative of the determinant of the metric tensor (or any square matrix in general). It's discussed in every GR book and surely on PF too, if not right in thir very thread. :)

 

[Tangent87]

What is $g_{,c}$ equal to ? It's a formula very important in General Relativity. It's the derivative of the determinant of the metric tensor (or any square matrix in general). It's discussed in every GR book and surely on PF too, if not right in thir very thread. :)

Sorry I should of explained, I know that $$g=det(g_{ab})$$ and therefore what we have to prove is that $$g^{ab}g_{ab,c}=\frac{1}{det(g_{ab})}\partial_c(det(g_{ab}))$$ but I get stuck here, I know it probably involves the inverse of metric somehow but having the partial derivative in there is confusing me.

 

[dextercioby]

In the lecture notes attached in this thread https://www.physicsforums.com/showthread.php?t=457123 (page 107)

you can find a derivation of that formula based merely on high school mathematics.

 

[Tangent87]

In the lecture notes attached in this thread https://www.physicsforums.com/showthread.php?t=457123 (page 107)

you can find a derivation of that formula based merely on high school mathematics.

Ah that's exactly what I was looking for, thank you.