Variational methods on Lagrangians (D'Inverno Chapter 11)

[TerryW]

Oh dear! I thought I had cracked chapter 11 and had done all the problems. However, when I came to write up the answers I realized my answer to Problem 11.7 didn't work. I thought I had a simple answer to (i) but then used the same process for (ii) and ended up with +R^ab. My approach to (i) was to simply take the equation for L_g = (-g)^1/2 R

So L_g =(-g)^1/2g^abR_ab

So L_G = gothic'g' $$\delta$$R_ab

I then just take a variation wrt the dynamic variable gothic'g' to get the desired result.

I realize that I haven't taken a Leibnitz product variation so I have ignored the term gothic'g' $$\delta$$R_ab. Is it necessary to do this and then prove that gothic'g' $$\delta$$R_ab = 0? Any thoughts on how to do this?(Anyone know how to to the gothic G for tensor weights?)

 

[Altabeh]

gothic'g' $$\delta$$R_ab. Is it necessary to do this and then prove that gothic'g' $$\delta$$R_ab = 0? Any thoughts on how to do this?

(Anyone know how to to the gothic G for tensor weights?)

You don't actually need to prove the second part because the variation of the Ricci tensor wrt any metric density of either covariant or contravariant type is automatically zero. This is obviously because the Ricci tensor lacks any weight factor $$\sqrt{-g}$$ and this is all due to it being a tensor not a tensor density. Remember that the variation of the Ricci tensor does not vanish by itself unless it is contained in a boundary (surface) integral equation as in the derivation of Einstein's field equations themselves.

And I was wondering if I could know what your 'gothic' means!

AB

 

[Altabeh]

Is it necessary to do this and then prove that gothic'g' $$\delta$$R_ab = 0? Any thoughts on how to do this?

About the part (iii) you have to note that there must be a constraint of the form

$$\frac{(-\mathfrak{g}^{cd}\delta \Gamma^{k}_{cd}+\mathfrak{g}^{ck}\delta \Gamma^{d}_{cd})_{;k}}{\delta g^{ab}}=0,$$

which is derived by the use of the Palatini equation for the variation of the Ricci tensor. I have to recall that again in a surface integral as in the Hilbert-Einstein action, the numerator of this equation automatically vanishes, leading to a no-constraint form of the answer to the part (iii).

AB

 

[TerryW]

You don't actually need to prove the second part because the variation of the Ricci tensor wrt any metric density of either covariant or contravariant type is automatically zero. This is obviously because the Ricci tensor lacks any weight factor $$\sqrt{-g}$$ and this is all due to it being a tensor not a tensor density. Remember that the variation of the Ricci tensor does not vanish by itself unless it is contained in a boundary (surface) integral equation as in the derivation of Einstein's field equations themselves.
AB

I'm still not sure why I can ignore the $$\delta$$Rab. As you say 'the variation of the Ricci tensor does not vanish by itself unless it is contained in a boundary (surface) integral equation..' When we are just taking the variation on its own, it isn't contained in a surface integral so I don't see how we can just assume it is zero. I've tried to prove that
$$\delta$$R_ab = 0 but without success.

If you do assume that $$\delta$$R_ab is zero, it enables you to reach the answers for (i), (ii) and (iii) except that my answer for (ii) comes out as +R^ab not -R^ab.

My working is as follows:

L_G = $$\sqrt{(-g)}$$ R = $$\sqrt{(-g)}$$ g_ab R^ab = $$\mathfrak{g}_{ab}$$ R^ab

$$\delta$$L_G/$$\delta\mathfrak{g}^{ab}$$ = R^ab.

And I was wondering if I could know what your 'gothic' means!

AB

When I created the post, I didn't know how to write $$\mathfrak{g}^{ab}$$, which what I meant by gothic 'g', but you used the appropriate LaTeX in your reply which I have now stored in my file of useful LaTex

About the part (iii) you have to note that there must be a constraint of the form

$$\frac{(-\mathfrak{g}^{cd}\delta \Gamma^{k}_{cd}+\mathfrak{g}^{ck}\delta \Gamma^{d}_{cd})_{;k}}{\delta g^{ab}}=0,$$

which is derived by the use of the Palatini equation for the variation of the Ricci tensor. I have to recall that again in a surface integral as in the Hilbert-Einstein action, the numerator of this equation automatically vanishes, leading to a no-constraint form of the answer to the part (iii).

AB

I didn't see any need for the additional constraint for (iii) but I did identify one for (ii). It wasn't the same as yours, mine was:

[tex]\nabla_b \sqrt{(-g)} [/tex]g^abR = 0. But this is zero anyway!
(And I just cannot get this last line of LaTeX to work!)

 

[Altabeh]

I'm still not sure why I can ignore the $$\delta$$Rab. As you say 'the variation of the Ricci tensor does not vanish by itself unless it is contained in a boundary (surface) integral equation..' When we are just taking the variation on its own, it isn't contained in a surface integral so I don't see how we can just assume it is zero. I've tried to prove that
$$\delta$$R_ab = 0 but without success.

If you do assume that $$\delta$$R_ab is zero, it enables you to reach the answers for (i), (ii) and (iii) except that my answer for (ii) comes out as +R^ab not -R^ab.

You seem to have completely misunderstood me. The variation of the Ricci tensor is by no means zero generally. I assume here that you can simply prove the part (i); so I'm going to give you a hint on how to derive (ii).

As for (ii), I assume you can simply find that

$$\delta (\mathfrak{g}^{ab}R_{ab})=\delta \mathfrak{g}^{ab}R_{ab}+\delta R_{ab}\mathfrak{g}^{ab}.$$

I gave you a clue that since the variation is with respect to a metric density, so that

$$\frac{\delta R_{ab}}{\delta \mathfrak{g}_{cd}}\mathfrak{g}^{ab}=0.$$

Having been said, this is just because the variation of a tensor with respect to a density is always zero. (Indeed, there is no weight factor within $$R_{ab}$$).

Now we are left with

$$\frac{\delta \mathfrak{L}_G}{\delta \mathfrak{g}_{cd}}=\frac{\delta \mathfrak{g}^{ab}}{\delta \mathfrak{g}_{cd}}R_{ab}.$$

Try to obtain the result

$$\delta \mathfrak{g}^{ab} R_{ab}=-\delta g_{ab}\sqrt{-g}G^{ab}.$$

(Hint: Get a hand from the equation (11.34).)

From this you are able to write

$$\delta \mathfrak{g}^{ab} R_{ab}=-\delta \mathfrak{g}_{ab}R^{ab}+\frac{1}{2}g^{ab}R\delta\mathfrak{g}_{ab}-\delta\sqrt{-g}R.$$

Now it all falls upon you to reveal the rest of the calculation.

For the part (iii), I tried to use the Palatini approach, but you seemed to fail in understanding it. It can be obtained simply in way similar to the one D'inverno shows in the equation (11.34). Just think of it and if you didn't come up with anything, I'd go into its detail, too. Remember that all three parts include in a special constraint which is relevant to the divergence of the Einstein tensor density.

AB

 

[TerryW]

Hi Altabeh

Thankyou for your reply and your patience with me!

As for (ii), I assume you can simply find that

$$\delta (\mathfrak{g}^{ab}R_{ab})=\delta \mathfrak{g}^{ab}R_{ab}+\delta R_{ab}\mathfrak{g}^{ab}.$$

I gave you a clue that since the variation is with respect to a metric density, so that

$$\frac{\delta R_{ab}}{\delta \mathfrak{g}_{cd}}\mathfrak{g}^{ab}=0.$$

Having been said, this is just because the variation of a tensor with respect to a density is always zero. (Indeed, there is no weight factor within $$R_{ab}$$).

AB

Ah! I see what you are getting at now. But I have looked back through D'Inverno and I cannot find anything that established the rather crucial result that the variation of a tensor wrt a density is always zero. Can you point me in the direction of a proof anywhere?

Try to obtain the result

$$\delta \mathfrak{g}^{ab} R_{ab}=-\delta g_{ab}\sqrt{-g}G^{ab}.$$

AB

Yes I have done this!

From this you are able to write

$$\delta \mathfrak{g}^{ab} R_{ab}=-\delta \mathfrak{g}_{ab}R^{ab}+\frac{1}{2}g^{ab}R\delta\mathfrak{g}_{ab}-\delta\sqrt{-g}R.$$

Now it all falls upon you to reveal the rest of the calculation.

AB

I have done this too but my answer is not quite what you have written, I get:

$$\delta \mathfrak{g}^{ab} R_{ab}=-\delta \mathfrak{g}_{ab}R^{ab}+\frac{1}{2}g^{ab}R\sqrt{-g}\delta g_{ab}-\delta\sqrt{-g}R.$$

This then reduces to

$$\delta \mathfrak{g}^{ab} R_{ab}=-\delta \mathfrak{g}_{ab}R^{ab}$$

because

$$\frac{1}{2}g^{ab}R\sqrt{-g}\delta g_{ab}-\delta\sqrt{-g}R = 0$$

(Once again my LaTex equation does not appear as I intend!)

I used a slightly different approach to (iii) in which I start with (i) and just expand

$$\delta\mathfrak{g}_{ab}$$ but it works out OK.

Reqards Terry

 

[Altabeh]

Hi Altabeh

Thank you for your reply and your patience with me!

Hi and my pleasure!

Ah! I see what you are getting at now. But I have looked back through D'Inverno and I cannot find anything that established the rather crucial result that the variation of a tensor wrt a density is always zero. Can you point me in the direction of a proof anywhere?

I think this is trivial! The variation of $$\sqrt{-g}$$ again contains a weight factor $$\sqrt{-g}$$ which when multiplied by a metric tensor would give a metric density, something that tensors and their variations don't have. That is, the lack of $$\sqrt{-g}$$ in tensors and consequently in their variations leads to the following general formula:

$$\frac{\delta T_{ab..}^{a'b'..}}{\delta \mathfrak{g}_{ab}}=0.$$

Imagine now $$T_{ab..}^{a'b'..}$$ is accompanied by a weight factor $$\sqrt{-g}$$. Thus we have a tensor density of order (p,q) where p and q represent respectively the number of lower and upper indices. We know that

$$\delta(\sqrt{-g}T_{ab..}^{a'b'..}) = \delta \sqrt{-g}T_{ab..}^{a'b'..}+\delta T_{ab..}^{a'b'..} \sqrt{-g},$$ *

and that

$$\delta \sqrt{-g}= \frac{1}{2}\sqrt{-g} g^{ab}\delta g_{ab}=\frac{1}{2} g^{ab}\delta \mathfrak{g}_{ab}-2\delta \sqrt{-g}\rightarrow$$

$$\delta \sqrt{-g}=\frac{1}{6}g^{cd}\delta \mathfrak{g}_{cd}.$$

So combining this last result with the equation * yields

$$\delta(\sqrt{-g}T_{ab..}^{a'b'..}) = \frac{1}{6}g^{cd}\delta \mathfrak{g}_{cd}T_{ab..}^{a'b'..}+\delta T_{ab..}^{a'b'..} \sqrt{-g},$$

and thus

$$\frac{\delta(\sqrt{-g}T_{ab..}^{a'b'..})}{\delta \mathfrak{g}_{ab}} =\frac{1}{6}g^{cd}T_{cd..}^{a'b'..}.$$

Which is obviously generally non-zero.

For the case of the Ricci tensor, it should be pointed out that the change of the contracted Christoffel symbols $$\Gamma^{a}_{ab}$$ with $$\frac{1}{2}\frac{\partial \ln(-g)}{\partial x^b}$$ may be tricky as it sounds like some squared weight factors would appear along with the metric tensors. But this is again like we are running in circles as in the end we will be left with the terms involving the product of the first derivatives of the metric tensor and itself, meaning that varying the Ricci tensor wrt the metric density is always zero.

I have done this too but my answer is not quite what you have written, I get:

$$\delta \mathfrak{g}^{ab} R_{ab}=-\delta \mathfrak{g}_{ab}R^{ab}+\frac{1}{2}g^{ab}R\sqrt{-g}\delta g_{ab}-\delta\sqrt{-g}R.$$

This then reduces to

$$\delta \mathfrak{g}^{ab} R_{ab}=-\delta \mathfrak{g}_{ab}R^{ab}$$

because

$$\frac{1}{2}g^{ab}R\sqrt{-g}\delta g_{ab}-\delta\sqrt{-g}R = 0$$

There you have written the second term on the right hand side of the first equation from top incorrectly. It would be great if I could see how you got this. But by my own calculation the most simplified equation we hit in the end is

$$\delta \mathfrak{g}^{ab} R_{ab}=-\delta \mathfrak{g}_{ab}R^{ab}+2\delta\sqrt{-g}R,$$

which leads to the same result as in (i) if you proceed to compute it wrt the variation $$\delta\mathfrak{g}_{ab}.$$

I used a slightly different approach to (iii) in which I start with (i) and just expand

$$\delta\mathfrak{g}_{ab}$$ but it works out OK.

Reqards Terry

If that works, so it's okay!

AB

 

[Altabeh]

If there is nothing ambiguous here with these questions, and of course if you allow me, I'd like to put your intelligence to test here and ask you the following neat question to only see if you are really good at dealing with this tensor-density thing:

Calculate the variation of the Lagrangian $$\mathfrak{L}_G=\sqrt{-g}g^{ab}R_{ab}$$ wrt $$g^{ab}.$$ How can we prove the variation of the Ricci tensor wrt the metric tensor vanishes if we want to have (iii) hold!?

AB

 

[TerryW]

Hi Altabeh,

This is quick reply to your post just to give you my workings on the equation where we have a different result.

I really do appreciate the time and effort you are putting in on my behalf. Your latest post has given me a lot to think about but I won't now be able to get back to you for a few weeks because I am going away on holiday. I will however take your posts and D'Inverno with me on holiday and get back in touch with you when I return.

Regards for now,

Terry

Hi and my pleasure!

There you have written the second term on the right hand side of the first equation from top incorrectly. It would be great if I could see how you got this. AB

Starting from the result you led me to (and expanding G^ab)

1. $$\delta \mathfrak{g}^{ab} R_{ab}= -\delta g_{ab}\sqrt{-g}(R^{ab}-\frac{1}{2}g^{ab}R)$$

Then expanding out $$\delta \mathfrak{g}_{ab} R^{ab}$$

2. $$\delta \mathfrak{g}_{ab} R^{ab}= \delta g_{ab}\sqrt{-g}R^{ab}+g_{ab}R^{ab} \delta\sqrt{-g}$$

(I'm still getting odd results when I look at the saved post. The LaTeX for the second equation is

\delta \mathfrak{g}_{ab} R^{ab}= \delta g_{ab}\sqrt{-g}R^{ab}+g_{ab}R^{ab} \delta\sqrt{-g}

But the result of the rendering seems to introduce an extra $$\sqrt{-g}$$ before the $$\delta\sqrt{-g}$$at the end.)

I am now looking at the post a day later and the second equation now reads correctly!

We can add the two equations together and reorganise the terms to get what I got!

 

[Altabeh]

Hi Altabeh,

This is quick reply to your post just to give you my workings on the equation where we have a different result.

I really do appreciate the time and effort you are putting in on my behalf. Your latest post has given me a lot to think about but I won't now be able to get back to you for a few weeks because I am going away on holiday. I will however take your posts and D'Inverno with me on holiday and get back in touch with you when I return.

Regards for now,

Terry

Have a good holiday.

We can add the two equations together and reorganise the terms to get what I got!

I still don't get how you manage to get your result from this. Hope to make it clear for after you are back from holiday.

AB

 

[TerryW]

Hi Altabeh,

I had another look at my post this morning and realized that when I add the two equations together, I had made an error and made one quantity -ve when it should have been +ve. So my equations added together now produce the same result as yours!RegardsTerry

 

[TerryW]

Hi Altabeh,

I'm back from my holiday and keen to get back on with our discussion.

If there is nothing ambiguous here with these questions, and of course if you allow me, I'd like to put your intelligence to test here and ask you the following neat question to only see if you are really good at dealing with this tensor-density thing:

Calculate the variation of the Lagrangian $$\mathfrak{L}_G=\sqrt{-g}g^{ab}R_{ab}$$ wrt $$g^{ab}.$$ How can we prove the variation of the Ricci tensor wrt the metric tensor vanishes if we want to have (iii) hold!?

AB

Let me see if I can rise to the challenge!

$$\mathfrak{L}_G=\sqrt{-g}g^{ab}R_{ab}$$

So $$\delta \mathfrak{L}_G = \delta \sqrt{-g}g^{ab}R_{ab} + \sqrt{-g}\delta g^{ab}R_{ab} + \sqrt{-g}g^{ab}\delta R_{ab}$$

$$\delta \mathfrak{L}_G = \frac{1}{2} \sqrt{-g}g^{cd}\delta g_{cd}g^{ab}R_{ab} + \sqrt{-g}\delta g^{ab}R_{ab} + \sqrt{-g}g^{ab}\delta R_{ab}$$

$$\delta \mathfrak{L}_G = \frac{1}{2} \sqrt{-g}g^{ab}\delta g_{ab}g^{cd}R_{cd} + \sqrt{-g}\delta g^{ab}R_{ab} + \sqrt{-g}g^{ab}\delta R_{ab}$$

$$\delta \mathfrak{L}_G = -\frac{1}{2} \sqrt{-g}g_{ab}\delta g^{ab}g^{cd}R_{cd} + \sqrt{-g}\delta g^{ab}R_{ab} + \sqrt{-g}g^{ab}\delta R_{ab}$$

$$\delta \mathfrak{L}_G = -\frac{1}{2} \sqrt{-g}g_{ab}\delta g^{ab}R + \sqrt{-g}\delta g^{ab}R_{ab} + \sqrt{-g}g^{ab}\delta R_{ab}$$

$$\delta \mathfrak{L}_G = \sqrt{-g}\delta g^{ab}(-\frac{1}{2}g_{ab}R + R_{ab}) + \sqrt{-g}g^{ab}\delta R_{ab}$$

$$\delta \mathfrak{L}_G = \sqrt{-g}\delta g^{ab}G_{ab} + \sqrt{-g}g^{ab}\delta R_{ab}$$

$$\frac {\delta \mathfrak{L}_G}{\delta g^{ab}} = \sqrt{-g}G_{ab} + \sqrt{-g}g^{ab}\frac{\delta R_{ab}}{\delta g^{ab}}$$

So if (iii) is to hold, then $$\sqrt{-g}g^{ab}\frac{\delta R_{ab}}{\delta g^{ab}} = 0$$

ie $$\frac{\delta R_{ab}}{\delta g^{ab}} = 0$$

But does this really prove that the change of the Ricci tensor wrt the metric tensor is zero?

 

[haushofer]

No, it doesn't. The action principle with the Hilbert action states that the variation of the Hilbert action wrt the metric is zero. The Hilbert action is
$$S[g_{\mu\nu}] = \int \sqrt{|g|}g^{\mu\nu}R_{\mu\nu}d^{4}x$$
Varying the Lagrangian L (including the square root of the metric in the measure) gives
$$\delta L = \sqrt{|g|}(\frac{1}{2}Rg^{\alpha\beta}-R^{\alpha\beta})\delta g_{\alpha\beta} + \sqrt{|g|}g^{\mu\nu}(\nabla_{\alpha}\delta\Gamma^{\alpha}_{\mu\nu} - \nabla_{\nu}\delta\Gamma^{\alpha}_{\mu\alpha})$$
One can show that for a large class of Lagrangians the variation reads
$$\delta L = E^{\alpha\beta}\delta g_{\alpha\beta} + \nabla_{\alpha}\Theta^{\alpha}$$
The first term on the RHS, the E, is the equations of motion, and the second term is a boundary term. With suitable boundary conditions this term becomes a boundary term in the action via Stokes, and the boundary conditions put this term to zero. However, if you don't impose these boundary terms, you need the Hawking-Gibbons boundary term in your action, which is constructed from the extrinsic curvature. For your Hilbert action this Theta becomes, written as a three-form,
$$\Theta_{\mu\nu\rho}(g, \delta g) =\frac{1}{16\pi} \epsilon_{\alpha\mu\nu\rho}g^{\alpha\lambda}g^{\sigma\theta}(\nabla_{\sigma}\delta g_{\lambda\theta} - \nabla_{\lambda}\delta g_{\sigma\theta})$$

 

[Altabeh]

But does this really prove that the change of the Ricci tensor wrt the metric tensor is zero?

Hi

In fact you had to be a little bit careful: D'inverno shows that the equation (11.33) reduces to (11.34) [and thus $$R_{ab}\delta{\mathfrak{g}^{ab}}=-\sqrt{-g}{G}^{ab}\delta{g_{ab}}$$] through imposing a boundary condition and Stokes theorem to eliminate the term including the variation of Ricci tensor in the action integral. This tells us that applying (iii) in the equation

$$\frac {\delta \mathfrak{L}_G}{\delta g^{ab}} = \sqrt{-g}G_{ab} + \sqrt{-g}g^{ab}\frac{\delta R_{ab}}{\delta g^{ab}}$$

calls for the vanishing of $$\frac{\delta R_{ab}}{\delta g^{ab}}$$ as a boundary condition in the action formula not by itself, as described by haushofer.

AB

 

[TerryW]

Hi Altabeh and haushofer,

I've had a bit more time to look through all the posts so far and I'd like to refresh the thread with a summary of where it has taken me so far:

I started when I realized that in producing an answer to (i) of Problem 11.7 in D'inverno, I had ignored the term

$$\mathfrak{g}^{ab} \delta R_{ab}$$

This produced an extra term in the result for $$\delta \mathfrak{L}_G$$ but Altabeh has pointed out that

$$\frac{\delta R_{ab}}{\delta \mathfrak{g}^{ab}} \mathfrak{g}^{ab} = 0$$

I kind of understand what you are saying Altabeh, about why this is the case but I'd be happier if I could see a rigorous mathematical proof.

My next problem was with the answer to part (ii) and on looking through all the posts, I still have a problem with this.

We worked through and have got as far as proving:

$$\delta \mathfrak{g}^{ab} R_{ab} = - \delta \mathfrak{g}_{ab} R^{ab} +2 \delta \sqrt{-g} R$$

Now in Altabeh's post dated May 17, he shows that:

$$\delta \sqrt{-g} = \frac{1}{6} g^{ab}\delta \mathfrak{g}_{ab}$$

which leads me to:


$$\frac{\mathfrak L_G}{\delta \mathfrak{g}_{ab}} = - R^{ab} + \frac{1}{3} g^{ab} R$$


I understand the process which D'Inverno uses to reduce equation (11.33) to equation (11.34) but I still do not really understand why, for example,

$$\frac{\delta R_{ab}}{\delta g^{ab}}$$ is not part of the answer to 11.7 (iii) as it isn't part of an integral expression.


Sorry if I don't seem to be getting this. I hope you will persist with me until we have a resolution.


Regards



Terry