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