Variational methods on Lagrangians (D'Inverno Chapter 11) - continued

[TerryW]

This is a continuation of an original thread first posted by me on May 11th 2010. Altabeh has been very kindly trying to guide me towards a resolution.

I started the thread 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 Altabeh said 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:


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

for 11.7 (ii)

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.

 

[Altabeh]

Hello TerryW,

Some apologies are in order for the hiatus I took due to personal problems. But attached is what you've been originally seeking for as it helps you eliminate all the confusion about those problems from D'inverno's book!

AB

 

[TerryW]

Hi Altabeh,

Thanks for the time and effort you have put into this. I hope that others are able to appreciate this too.

I've carried on with my reading. My latest query on PF is about the Schwartzschild solution. Do you have any thoughts on this?

 

[TerryW]

Hi Altabeh,

I've been working through your paper and I think I have spotted an error. The expansion of (1.4) produces only three terms, the ordinary differential and two $$\Gamma$$ terms. I don't know where the fourth term $$\mathfrak{g}^{\mu\upsilon}\Gamma^{\alpha}_{\alpha\lambda}$$ comes from.RegardsTerryW

 

[Altabeh]

Hi Altabeh,

I've been working through your paper and I think I have spotted an error. The expansion of (1.4) produces only three terms, the ordinary differential and two $$\Gamma$$ terms. I don't know where the fourth term $$\mathfrak{g}^{\mu\upsilon}\Gamma^{\alpha}_{\alpha\lambda}$$ comes from.


Regards


TerryW

Hi

That is the term appearing in the covariant derivative of any weighted tensor! In the other words, if $$\mathfrak{P}_{\mu}=\sqrt{-g}P_{\mu}$$ be a vector-density (of course $$\mathfrak{P}$$ has weight w=1) and $$a^{\mu}$$ be an arbitrary contravariant vector, then

$$\mathfrak{P}_{\mu}a^{\mu}=\sqrt{-g}P_{\mu}a^{\mu}=\ scalar density=D.$$

Using

$$D_{;\beta}=D_{,\beta}-\Gamma^{\alpha}_{\alpha\beta}$$

you can show that

$$\mathfrak{P}_{\mu; \beta}=\mathfrak{P}_{\mu,\beta}-\Gamma^{\alpha}_{\mu\beta}\mathfrak{P}_{\alpha}-\mathfrak{P}_{\mu}\Gamma^{\alpha}_{\alpha\beta}.$$

AB

 

[TerryW]

Oh dear yes, I forgot about the tensor weight! Sorry

 

[TerryW]

Hi Altabeh,

A couple of small points first: In (1.4) shouldn't it be the covariant derivative wrt $$\lambda$$ rather than $$\nu$$? Then, having worked my way through to (1.7) I find that the last part of the expression should be -2 $$\mathfrak{g}^{\mu\nu}$$ which changes the signs in the expression for $$\bar{\mathfrank{L}}$$ in (1.8).

You will have to help me out though from (1.10) through to (1.11). I have had a try but I cannot find an elegant proof that the term on the rhs of (1.9) is equivalent to (1.11).

I look forward to hearing from you. And I do appreciate your patience with me.


Regards


Terry

 

[Altabeh]

Hi Altabeh,

A couple of small points first: In (1.4) shouldn't it be the covariant derivative wrt $$\lambda$$ rather than $$\nu$$?

No. If so, then it is not divergence anymore and the only thing you can get from $$\mathfrak{g}^{\mu\nu}_{,\lambda}=0$$ is that all components of metric density are constants!

Then, having worked my way through to (1.7) I find that the last part of the expression should be -2 $$\mathfrak{g}^{\mu\nu}$$ which changes the signs in the expression for $$\bar{\mathfrank{L}}$$ in (1.8).

You have to be more careful! I have said somewhere between Eq. (1.5) and (1.6) that we set $$\nu=\lambda.$$ This is to get the equation (1.4) involved in our equations (1.3) and (1.5)!

You will have to help me out though from (1.10) through to (1.11). I have had a try but I cannot find an elegant proof that the term on the rhs of (1.9) is equivalent to (1.11).

Well, just multiply Palatini equation by $$\mathfrak{g}^{\mu\nu}$$ and take a look at the observation we made in Eq. (1.4). At the same time bear in mind what contribution this result will make after taking the variation of Eg. (1.1). You also have to note that if you got a term of the form $$\mathfrak{D}^\mu_{,\mu}$$ in the action, what happens then!?

AB

 

[TerryW]

Hi Altabeh,

I've acquired a decent Latex Editor so I can produce PDFs and avoid the uncertainties of the PF editor, which seems to have some caching problems still.

I've produced a response to your latest comments (attached). I am now happy with the derivations of (i) and (iii) - thank you for your assistance with this. But I still have a problem with (ii). I'll have another look at this sometime and if I do resolve it, I'll let you know. Please don't labour anymore on my behalf on this small issue (which hasn't stopped my progress) but if you do find a proof, please do let me know.

Many thanks again for your help so far.


Regards


Terry

 

[TerryW]

Hi Altabeh,

I've been doing a quick look back at some of the bits of D'Inverno I didn't feel entirely comfortable with and I took the opportunity to look at our old friend. In a very short time, I spotted what I feel is a resolution to the outstanding problem of proving (ii).

My solution is attached.


Regards


Terry

 

[dextercioby]

But isn't the Lagrangian $\mathcal{L} = \sqrt{|g|} R$, so that its variation comprises 3 terms ?

 

[TerryW]

Hi Dextercioby,

Hope the attached file clarifies this for you.



Regards


TerryW

 

[dextercioby]

Ah, alright, now it makes sense. I haven't seen the frak(g) notation in a new book before.

 

[Nikos]

I use the notation
${{\bar{g}}_{ab}}={{(-g)}^{1/2}}{{g}_{ab}}$
and ${{\bar{g}}^{ab}}={{(-g)}^{-1/2}}{{g}^{ab}}$

So, ${{\bar{g}}^{ab}}{{\bar{g}}_{bc}}=\delta _{c}^{a}\Rightarrow \delta {{\bar{g}}^{ab}}{{\bar{g}}_{bc}}+{{\bar{g}}^{ab}} \delta{{\bar{g}}_{bc}}=0$

$\delta {{\bar{g}}^{ab}}{{\bar{g}}_{bc}}=-{{\bar{g}}^{ab}}\delta {{\bar{g}}_{bc}}\Rightarrow \delta {{\bar{g}}^{ab}}{{\bar{g}}_{bc}}{{\bar{g}}^{cd}}=-{{\bar{g}}^{ab}}\delta {{\bar{g}}_{bc}}{{\bar{g}}^{cd}},\delta {{\bar{g}}^{ab}}\delta _{b}^{d}=-{{\bar{g}}^{ab}}\delta {{\bar{g}}_{bc}}{{\bar{g}}^{cd}}\Rightarrow \delta {{\bar{g}}^{ad}}=-{{\bar{g}}^{ab}}{{\bar{g}}^{cd}}\delta {{\bar{g}}_{bc}}\Rightarrow$

$\delta {{\bar{g}}^{ab}}=-{{\bar{g}}^{ac}}{{\bar{g}}^{bd}}\delta {{\bar{g}}_{cd}}$ or better $\delta {{\bar{g}}^{ab}}=-\frac{1}{2}({{\bar{g}}^{ac}}{{\bar{g}}^{bd}}+{{ \bar{g}}^{ad}}{{\bar{g}}^{bc}})\delta {{\bar{g}}_{cd}}$

$\delta {{\bar{g}}^{ab}}=-\frac{1}{2}({{\bar{g}}^{ac}}{{\bar{g}}^{bd}}+{ {\bar{g}}^{ad}}{{\bar{g}}^{bc}})\delta {{\bar{g}}_{cd}}\Rightarrow \delta {{\bar{g}}^{ab}}=-\frac{1}{2}{{(-g)}^{-1}}({{g}^{ac}}{{g}^{bd}}+{{g}^{ad}}{{g}^{bc}}) \delta {{\bar{g}}_{cd}}$

$\int{\delta {{{\bar{g}}}^{ab}}{{R}_{ab}}d\Omega }=-\int{\frac{1}{2}{{(-g)}^{-1}}({{g}^{ac}}{{g}^{bd}}+{{g}^{ad}}{{g}^{bc}}){{R}_{ab}}\delta {{{\bar{g}}}_{cd}}d\Omega }$

$\int{\delta {{{\bar{g}}}^{ab}}{{R}_{ab}}d\Omega }=\int{\frac{{{R}^{cd}}}{g}\delta {{{\bar{g}}}_{cd}}d\Omega }$

(The integral $\int{{{{\bar{g}}}^{ab}}\delta {{R}_{ab}}d\Omega }$ vanishes)

And finally $\frac{\delta {{L}_{G}}}{\delta{{{\bar{g}}}^{cd}}}=\frac{{{R}^{cd}}}{g}$ instead of $-{{R}^{cd}}$. There is something really wrong here but I cannot see it. For example $\frac{{{R}^{cd}}}{g}$ is not a symmetric tensor density of weight +1

Please Help!

 

[TerryW]

Hi Nikos,

I think I've spotted an error in your work!


Regards


TerryW

 

[Nikos]

Hi Nikos,

I think I've spotted an error in your work!


Regards


TerryW

TerryW thank you so much for your answer!
As you've spotted: ${{\mathfrak{g}}_{ab}}$ is not defined by d'inverno as the inverse of ${{\mathfrak{g}}^{ab}}$=${{(-g)}^{1/2}}{{g}^{ab}}$,
but as ${{\mathfrak{g}}_{ab}}$=${{(-g)}^{1/2}}{{g}_{ab}}$.
Thank you again!

 

[Nikos]

Hello again! I have some problem with the variation of the legrangian that contains Kretschmann scalar.
$I=\int{{{(-g)}^{1/2}}{{R}^{abcd}}{{R}_{abcd}}d\Omega }$
I start as usually, but with a trick making use of curvature tensor symmetries
$L={{(-g)}^{1/2}}{{R}^{ab}}_{cd}{{R}_{ab}}^{cd}={{(-g)}^{1/2}}{{R}^{ab}}_{cd}{{R}^{cd}}_{ab}$
So we have:
$\delta \mathfrak{L}=\delta {{(-g)}^{1/2}}{{R}^{abcd}}{{R}_{abcd}}+{{(-g)}^{1/2}}\delta \left( {{R}^{ab}}_{cd}{{R}^{cd}}_{ab} \right)$
the first term gives a part of the d'inverno's solution, so let's take a closer look to the second one:
$\delta \left( {{R}^{ab}}_{cd}{{R}^{cd}}_{ab} \right)=\delta \left( {{g}^{mb}}{{g}^{dn}}{{R}^{a}}_{mcd}{{R}^{c}}_{nab} \right)={{g}^{dn}}{{R}^{a}}_{mcd}{{R}^{c}}_{nab} \delta{{g}^{mb}}+{{g}^{mb}}{{R}^{a}}_{mcd}{{R}^{c}}_{nab}\delta{{g}^{dn}}+{{g}^{mb}}{{g}^{dn}}\delta{{R}^{a}}_{mcd}{{R}^{c}}_{nab}$

$={{R}^{a}}_{mcd}{{R}^{cd}}_{ab}\delta {{g}^{mb}}+{{R}^{ab}}_{cd}{{R}^{c}}_{nab}\ delta {{g}^{dn}}+{{g}^{mb}}{{g}^{dn}}\delta {{R}^{a}}_{mcd}{{R}^{c}}_{nab}$
$={{R}_{amcd}}{{R}^{aucd}}{{g}_{bu}}\delta {{g}^{mb}}+{{R}^{ab}}{{_{c}}^{m}}{{R}^{c}}_{nab}{{g}_{md}}\delta {{g}^{dn}}+{{g}^{mb}}{{g}^{dn}}\delta \left( {{R}^{a}}_{mcd}{{R}^{c}}_{nab} \right)$
$={{R}_{cdam}}{{R}^{cdau}}{{g}_{bu}}\delta {{g}^{mb}}+{{R}^{abcm}}{{R}_{abcn}}{{g}_{md}} \delta{{g}^{dn}}+{{g}^{mb}}{{g}^{dn}}\delta \left( {{R}^{a}}_{mcd}{{R}^{c}}_{nab} \right)$
$={{R}^{abcu}}{{R}_{abcm}}{{g}_{nu}}\delta {{g}^{mn}}+{{R}^{abcu}}{{R}_{abcn}}{{g}_{um}} \delta{{g}^{mn}}+{{g}^{mb}}{{g}^{dn}}\delta \left( {{R}^{a}}_{mcd}{{R}^{c}}_{nab} \right)$
$={{R}^{abcd}}{{R}_{abcm}}{{g}_{dn}}\delta {{g}^{mn}}+{{R}^{abcd}}{{R}_{abcn}}{{g}_{dm}} \delta{{g}^{mn}}+{{g}^{mb}}{{g}^{dn}}\delta \left( {{R}^{a}}_{mcd}{{R}^{c}}_{nab} \right)$
The first two terms are the remaining part of d'inverno's solution.
Anyway the problem is that I cannot prove that the other term vanishes:
${{g}^{mb}}{{g}^{dn}}\delta \left( {{R}^{a}}_{mcd}{{R}^{c}}_{nab} \right)={{g}^{mb}}{{g}^{dn}}\delta {{R}^{a}}_{mcd}{{R}^{c}}_{nab}+{{g}^{mb}}{{g}{dn}}{{R}^{a}}_{mcd}\delta {{R}^{c}}_{nab}$
and using palatini's equation
${{g}^{mb}}{{g}^{dn}}\delta \left( {{R}^{a}}_{mcd}{{R}^{c}}_{nab} \right)={{g}^{mb}}{{g}^{dn}}({{\nabla }_{c}}\delta \Gamma _{md}^{a}-{{\nabla }_{d}}\delta \Gamma _{mc}^{a}){{R}^{c}}_{nab}+{{g}^{mb}}{{g}^{dn}}{{R}^{a}}_{mcd}({{\nabla }_{a}}\delta \Gamma _{nb}^{c}-{{\nabla }_{b}}\delta \Gamma _{na}^{c})$
$=({{\nabla }_{c}}\delta \Gamma _{md}^{a}-{{\nabla }_{d}}\delta \Gamma _{mc}^{a}){{R}^{cd}}{{_{a}}^{m}}+{{R}^{ab}}{{_{c}}^{n}}({{\nabla }_{a}}\delta \Gamma _{nb}^{c}-{{\nabla }_{b}}\delta \Gamma _{na}^{c})$
$=-\delta \Gamma _{md}^{a}{{\nabla }_{c}}{{R}^{cd}}{{_{a}}^{m}}+\delta \Gamma _{mc}^{a}{{\nabla }_{d}}{{R}^{cd}}{{_{a}}^{m}}-\delta \Gamma _{nb}^{c}{{\nabla }_{a}}{{R}^{ab}}{{_{c}}^{n}}+\delta \Gamma _{na}^{c}{{\nabla }_{b}}{{R}^{ab}}{{_{c}}^{n}})$
$=2\delta \Gamma _{mc}^{a}{{\nabla }_{d}}{{R}^{cd}}{{_{a}}^{m}}+2\delta \Gamma _{na}^{c}{{\nabla }_{b}}{{R}^{ab}}{{_{c}}^{n}}=4\delta \Gamma _{mc}^{a}{{\nabla }_{d}}{{R}^{cd}}{{_{a}}^{m}}$
I just cannot get rid of the last term!
In fact using palatini's approach, that is $\mathfrak{L}=\mathfrak{L}({{g}^{ab}},{{\Gamma }^{\alpha }}_{bc},{{\Gamma }^{\alpha }}_{bc;d})$ i just get the relation
${{\nabla }_{d}}{{R}^{cd}}{{_{a}}^{m}}=0$
I really need your help with this!

 

[TerryW]

Hi Nikos,

First the good news, I've already cracked this problem so I will be able to send you a solution.

The bad news is that it's about a page of detailed working out which will take quite a bit of time to produce in Latex, so it will take a few days to produce.

It might also be possible to spot something in your work which can resolve the problem so I'll look at that too.

Regards


Terry

 

[dextercioby]

I think it's easier if you scan your work sheet with your calculations and attach it here, than wasting some time by writing all that code.

 

[TerryW]

Hi Dextercioby,

That's a good idea, but you ain't seen my handwriting!.

I've had a look at the problem though and thought it might be an idea to see if I can help Nikos get to the solution himself by giving him the first line of my solution to see if he can take it on to a conclusion himself. That is what other contributors have done for me and it feels much better at the end.

Bear in mind though that I have been doing all this on my own with a 40 year gap between my original physics degree and starting work on D'Inverno so my work may all be rubbish!


Regards


TerryW

Terry

 

[TerryW]

Hi Nikos,

Here is my approach to the problem, see if you can take it to a conclusion.

 

[Nikos]

Hi TerryW!
Thank you for your response!
I'll try it and I'll post here soon my conclusions!
Thank you again!