General Relativity - Tensor Identities

[Tangent87]

I am doing this question on page 75 here (http://www.maths.cam.ac.uk/undergrad/pastpapers/2006/Part_2/list_II.pdf ).

Here is my solution so far:

\nabla_a\nabla_b\phi=\partial_a\partial_b\phi=\partial_b\partial_a\phi=\nabla_b\nabla_a\phi

Then for the energy-momentum tensor I just wrote down this from my notes:

T_{ab}=\frac{c^4}{8{\pi}G}(R_{ab}-\frac{1}{2}Rg_{ab}+{\Lambda}g_{ab}) where R=g^{cd}R_{cd}.

Now, I believe this is the contracted Bianchi identity: \nabla^{b}(R_{ab}-\frac{1}{2}g_{ab}R)=0 which using their expression for R_{ab} gives:

\nabla^{b}(\partial_a\phi\partial_b\phi)-\frac{1}{2}g_{ab}\nabla^{b}(g^{cd}R_{cd})=0<br /> <br /> \Rightarrow \nabla^{b}(\partial_a\phi\partial_b\phi)=\frac{1}{2}g_{ab}g^{cd}\nabla^{b}(\partial_c\phi\partial_d\phi).

Now I'm stuck cos I don't see what's going to cancel to give me their (*), do I have to use the energy-momentum tensor somehow?

Thanks for any help.

 

[Phrak]

It might help if you identified the page number.

 

[Tangent87]

I said page 75 didn't I?

 

[dextercioby]

I am doing this question on page 75 here (http://www.maths.cam.ac.uk/undergrad/pastpapers/2006/Part_2/list_II.pdf ).

Here is my solution so far:

\nabla_a\nabla_b\phi=\partial_a\partial_b\phi=\partial_b\partial_a\phi=\nabla_b\nabla_a\phi

It's wrong. The double covariant derivative, under the assumptions of GR, is not equal to the double normal derivative, not even for scalars. \nabla_a\nabla_b\phi = \nabla_a\left(\nabla_b\phi\right) = \nabla_a T_{b} = .... As for the sequel, in the context of the exercise, I'm pretty sure that the cosmological constant is 0.

 

[Tangent87]

It's wrong. The double covariant derivative, under the assumptions of GR, is not equal to the double normal derivative, not even for scalars. \nabla_a\nabla_b\phi = \nabla_a\left(\nabla_b\phi\right) = \nabla_a T_{b} = .... As for the sequel, in the context of the exercise, I'm pretty sure that the cosmological constant is 0.

Ah yes thank you that makes sense now. However it's getting equation (*) and the last bit that's really bothering me, do you have any ideas for those?

 

[latentcorpse]

It's wrong. The double covariant derivative, under the assumptions of GR, is not equal to the double normal derivative, not even for scalars. \nabla_a\nabla_b\phi = \nabla_a\left(\nabla_b\phi\right) = \nabla_a T_{b} = .... As for the sequel, in the context of the exercise, I'm pretty sure that the cosmological constant is 0.

Isn't the double covariant derivative equal to the double partial derivative for scalars when we are working witha torsion free connection such as the levi civita connection in this case though?

 

[dextercioby]

Isn't the double covariant derivative equal to the double partial derivative for scalars when we are working witha torsion free connection such as the levi civita connection in this case though?

No, why would it be ? Just perform the calculations and convince yourself. The connection will appear.

 

[Tangent87]

No, why would it be ? Just perform the calculations and convince yourself. The connection will appear.

I'm confused now because in my notes it definitely says that \nabla_a\phi=\partial_a\phi so let's just call \partial_a\phi=\psi some other scalar then do the same thing again \nabla_b\psi=\partial_b\psi. So what's wrong with the way I originally did it?

 

[dextercioby]

Well, \partial_{a}\phi = \psi_{a} is the right equation. WRT general coordinate changes, both terms of the equality I wrote are covectors, while your equation is not balanced. You cannot equate a covector with a scalar. The indices wouldn't match.

 

[Tangent87]

Well, \partial_{a}\phi = \psi_{a} is the right equation. WRT general coordinate changes, both terms of the equality I wrote are covectors, while your equation is not balanced. You cannot equate a covector with a scalar. The indices wouldn't match.

I've worked it through now how you suggested and see what you mean. The connection stays the same because it's symmetric and we use \partial_aT_b=\partial_bT_a.

 

[dextercioby]

\partial_aT_b=\partial_bT_a.

This is generally NOT true, unless, for example, a=b.

Did you get the (*) part ? You are advised to use the Bianchi identity for the Einstein tensor. Do you know this identity ?

 

[Tangent87]

This is generally NOT true, unless, for example, a=b.

Did you get the (*) part ? You are advised to use the Bianchi identity for the Einstein tensor. Do you know this identity ?

No, but it's true in the case where T_a=\nabla_a\phi=\partial_a\phi right?

I did not the get (*), see my original post at the top for what I've done on it so far (my Bianchi identity may be wrong, can you check for me?)

Thanks.

 

[dextercioby]

Well, I like using Greek indices, so could you follow in my footsteps ?

\nabla^{\mu} R_{\mu\nu} = \frac{1}{2}g_{\mu\nu}g^{\alpha\beta}\nabla^{\mu}R_{\alpha\beta}

using the abovementioned identity and the metric compatibility of the connection. Now in what I wrote above just plug

R_{\mu\nu}=\left(\nabla_{\mu}\phi\right)\left(\nabla_{\nu}\phi\right)

and reshuffle and relabel indices.

 

[latentcorpse]

Well, I like using Greek indices, so could you follow in my footsteps ?

\nabla^{\mu} R_{\mu\nu} = \frac{1}{2}g_{\mu\nu}g^{\alpha\beta}\nabla^{\mu}R_{\alpha\beta}

using the abovementioned identity and the metric compatibility of the connection. Now in what I wrote above just plug

R_{\mu\nu}=\left(\nabla_{\mu}\phi\right)\left(\nabla_{\nu}\phi\right)

and reshuffle and relabel indices.

Well I know this isn't my thread but I'm interested in doing the question for practice anyway. I find, after using the product rule that

\nabla^\mu \nabla_\mu \phi \nabla_\nu \phi + \nabla_\mu \phi \nabla^\mu \nabla_\nu \phi= \frac{1}{2} \nabla_v \nabla^\mu \phi \nabla_\mu \phi + \frac{1}{2} \nabla^\mu \phi \nabla_\nu \nabla_\mu \phi

Now I don't see how we can simplify this?

 

[dextercioby]

Well I know this isn't my thread but I'm interested in doing the question for practice anyway. I find, after using the product rule that

\nabla^\mu \nabla_\mu \phi \nabla_\nu \phi + \nabla_\mu \phi \nabla^\mu \nabla_\nu \phi= \frac{1}{2} \nabla_v \nabla^\mu \phi \nabla_\mu \phi + \frac{1}{2} \nabla^\mu \phi \nabla_\nu \nabla_\mu \phi

Now I don't see how we can simplify this?

The RHS of the equality you wrote contains 2 identical terms. Summing them you'll find one of the terms in the LHS. Then the desired conclusion follows easily.

 

[Tangent87]

Well I know this isn't my thread but I'm interested in doing the question for practice anyway. I find, after using the product rule that

\nabla^\mu \nabla_\mu \phi \nabla_\nu \phi + \nabla_\mu \phi \nabla^\mu \nabla_\nu \phi= \frac{1}{2} \nabla_v \nabla^\mu \phi \nabla_\mu \phi + \frac{1}{2} \nabla^\mu \phi \nabla_\nu \nabla_\mu \phi

Now I don't see how we can simplify this?

Wait, where did the g's go?

 

[latentcorpse]

Wait, where did the g's go?

Because g is covariantly conserved, we can move g in and out of covariant derivatives at will, can we not?

 

[latentcorpse]

The RHS of the equality you wrote contains 2 identical terms. Summing them you'll find one of the terms in the LHS. Then the desired conclusion follows easily.

I can't see how those two terms are equal? OR how they cancel one of the terms on the LHS?

 

[Tangent87]

Because g is covariantly conserved, we can move g in and out of covariant derivatives at will, can we not?

But there are no g in your latest identity either in or out of the covariant derivatives. We have two g in the RHS of what we start with: <br /> \nabla^{\mu} R_{\mu\nu} = \frac{1}{2}g_{\mu\nu}g^{\alpha\beta}\nabla^{\mu}R_ {\alpha\beta} <br /> how are these going to disappear? I vaguely recall from somewhere that we can use the g to raise/lower indices, is that what you do with them? Does anyone know the exact specifics of how that works?

 

[latentcorpse]

But there are no g in your latest identity either in or out of the covariant derivatives. We have two g in the RHS of what we start with: <br /> \nabla^{\mu} R_{\mu\nu} = \frac{1}{2}g_{\mu\nu}g^{\alpha\beta}\nabla^{\mu}R_ {\alpha\beta} <br /> how are these going to disappear? I vaguely recall from somewhere that we can use the g to raise/lower indices, is that what you do with them? Does anyone know the exact specifics of how that works?

Yes. Exactly. So if you have

g^{\mu \nu} \nabla^\rho ( X_\mu Y_\rho) (I just made this up to show what happens)
Then because \nabla g=0, we can move g inside the covariant derivative as follows:
\nabla^\rho ( g^{\mu \nu} X_\mu y^\rho)

And g does raise/lower indices when we have contraction i.e. g^{\mu \nu} X_\mu = x^\nu

So we would get
\nabla^\rho ( X^\nu \rho )

Note that the free index in my answer (an upper \nu matches the free index in my original expression).

 

[Tangent87]

Yes. Exactly. So if you have

g^{\mu \nu} \nabla^\rho ( X_\mu Y_\rho) (I just made this up to show what happens)
Then because \nabla g=0, we can move g inside the covariant derivative as follows:
\nabla^\rho ( g^{\mu \nu} X_\mu y^\rho)

And g does raise/lower indices when we have contraction i.e. g^{\mu \nu} X_\mu = x^\nu

So we would get
\nabla^\rho ( X^\nu \rho )

Note that the free index in my answer (an upper \nu matches the free index in my original expression).

Oh ok, so you sort of cancel the contracted indices? In that case, would g_{ab}V^b=V_a? Could you do anything if you had something like g^{ab}V^b?

 

[latentcorpse]

Oh ok, so you sort of cancel the contracted indices? In that case, would g_{ab}V^b=V_a?

Correct.

Could you do anything if you had something like g^{ab}V^b?

This is considered blasphemy by most physicists. In index notation, you can NEVER have two up indices that are the same or two down indices that are the same. so what you wrote doesn't make sense and you would never come across it in the Einstein index notation stuff.

 

[Tangent87]

Correct.



This is considered blasphemy by most physicists. In index notation, you can NEVER have two up indices that are the same or two down indices that are the same. so what you wrote doesn't make sense and you would never come across it in the Einstein index notation stuff.

Haha fair enough. I think I've got the whole thing now, you just have to raise/lower the right indices on the RHS for it to work though it will still work in your case latentcorpse if it's true that \frac{1}{2} \nabla^\mu \phi \nabla_\nu \nabla_\mu \phi=\frac{1}{2} \nabla_\mu \phi (\nabla_\nu \nabla^\mu \phi)

 

[Tangent87]

In the last bit of the question, we're meant to derive the expression \partial_a(\sqrt{-g}g^{ab}\partial_b\phi)=0. I don't understand what the sqrt(-g) is all about? Is it the metric or something else, it's the fact that it has no indices that's making me think it's not the metric. Is it meant to be G the gravitational constant?

 

[latentcorpse]

Haha fair enough. I think I've got the whole thing now, you just have to raise/lower the right indices on the RHS for it to work though it will still work in your case latentcorpse if it's true that \frac{1}{2} \nabla^\mu \phi \nabla_\nu \nabla_\mu \phi=\frac{1}{2} \nabla_\mu \phi (\nabla_\nu \nabla^\mu \phi)

I'm not sure that those two things are equal. Why do you think they are? You're probably better at the whole maths thing than me so perhaps you can explain how you get from what I had to what the final answer is?


As for the next bit, g= \text{det } g_{\mu \nu}

 

[Tangent87]

I'm not sure that those two things are equal. Why do you think they are? You're probably better at the whole maths thing than me so perhaps you can explain how you get from what I had to what the final answer is? As for the next bit, g= \text{det } g_{\mu \nu}

This is how I did it from start to finish:

\nabla^a(R_{ab}-\frac{1}{2}g_{ab}g^{cd}R_{cd})=0

\nabla^a(\nabla_a\phi\nabla_b\phi)=\frac{1}{2}g_{ab}g^{cd}\nabla^a(\nabla_c\phi\nabla_d\phi)

(\nabla^a\nabla_a\phi)\nabla_b\phi+\nabla_a\phi(\nabla^a\nabla_b\phi)=\frac{1}{2}g_{ab}g^{cd}[(\nabla^a\nabla_c\phi)\nabla_d\phi+\nabla_c\phi(\nabla^a\nabla_d\phi)]=\frac{1}{2}(\nabla_b\nabla^c\phi)\nabla_c\phi+\frac{1}{2}\nabla_c\phi(\nabla_b\nabla^c\phi)=\nabla_a\phi(\nabla^a\nabla_b\phi)

Therefore \nabla^a\nabla_a\phi=\nabla_a\nabla^a\phi=0

 

[latentcorpse]

This is how I did it from start to finish:

\nabla^a(R_{ab}-\frac{1}{2}g_{ab}g^{cd}R_{cd})=0

\nabla^a(\nabla_a\phi\nabla_b\phi)=\frac{1}{2}g_{ab}g^{cd}\nabla^a(\nabla_c\phi\nabla_d\phi)

(\nabla^a\nabla_a\phi)\nabla_b\phi+\nabla_a\phi(\nabla^a\nabla_b\phi)=\frac{1}{2}g_{ab}g^{cd}[(\nabla^a\nabla_c\phi)\nabla_d\phi+\nabla_c\phi(\nabla^a\nabla_d\phi)]=\frac{1}{2}(\nabla_b\nabla^c\phi)\nabla_c\phi+\frac{1}{2}\nabla_c\phi(\nabla_b\nabla^c\phi)=\nabla_a\phi(\nabla^a\nabla_b\phi)

Therefore \nabla^a\nabla_a\phi=\nabla_a\nabla^a\phi=0

But to get
\frac{1}{2}(\nabla_b\nabla^c\phi)\nabla_c\phi+\frac{1}{2}\nabla_c\phi(\nabla_b\nabla^c\phi)=\nabla_a\phi(\nabla^a\nabla_b\phi)

you have assumed \nabla_b \nabla^c \phi = \nabla^b \nabla_c \phi Why is this true?

And then you do the cancelation which leaves you with
(\nabla^a \nabla_a \phi) \nabla_b \phi=0
That doesn't necessarily mean \nabla^a \nabla_a \phi=0 does it?
Surely it could just as easily mean \nabla_b \phi=0, no?

 

[Tangent87]

But to get
\frac{1}{2}(\nabla_b\nabla^c\phi)\nabla_c\phi+\frac{1}{2}\nabla_c\phi(\nabla_b\nabla^c\phi)=\nabla_a\phi(\nabla^a\nabla_b\phi)

you have assumed \nabla_b \nabla^c \phi = \nabla^b \nabla_c \phi Why is this true?

And then you do the cancelation which leaves you with
(\nabla^a \nabla_a \phi) \nabla_b \phi=0
That doesn't necessarily mean \nabla^a \nabla_a \phi=0 does it?
Surely it could just as easily mean \nabla_b \phi=0, no?

\nabla_b \nabla^c \phi = \nabla^b \nabla_c \phi I assumed was true from the commutation we showed in the first part of the question. And no we can't have \nabla_b \phi=0 because it says in the question to assume \partial_b\phi\neq0 which is the same thing essentially.

 

[latentcorpse]

\nabla_b \nabla^c \phi = \nabla^b \nabla_c \phi I assumed was true from the commutation we showed in the first part of the question. And no we can't have \nabla_b \phi=0 because it says in the question to assume \partial_b\phi\neq0 which is the same thing essentially.

How did you show
\nabla_a \nabla_b \phi = \nabla_b \nabla_a \phi after bigabau said that your initial way was wrong?

And also, he said that \nabla_b \phi \neq \partial_b \phi so doesn't that mean

And no we can't have \nabla_b \phi=0 because it says in the question to assume \partial_b\phi\neq0 which is the same thing essentially.

might be wrong?

 

[Tangent87]

How did you show
\nabla_a \nabla_b \phi = \nabla_b \nabla_a \phi after bigabau said that your initial way was wrong?

And also, he said that \nabla_b \phi \neq \partial_b \phi so doesn't that mean

might be wrong?

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.

 

[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.