Gauss - Bonnet Gravity -> Curvature variations

[Breo]

So I am working on the next quadratic Lagrangian:

$$ L = \alpha R_{\mu\nu}R^{\mu\nu} + \beta R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma} + \gamma R² $$

I have already derived $$ \delta (R_{\mu\nu}R^{\mu\nu}) = [- \frac{1}{2}g_{\mu\nu}R_{\alpha\beta}R^{\alpha\beta} + 2R^{\alpha\beta}R_{\mu\alpha\nu\beta}]\delta g^{\mu\nu} $$ right?

How can I compute the other variations in order to compute the equations of motion?

 

[Breo]

Do not ask for results, just a tip to start deriving. Not sure if the most appropiate here would be using the Leibniz rule.

 

[ChrisVer]

you can use the Leibniz rule...

 

[Breo]

But lots of Christoffel symbols appear and I do not know to simplificate the equations. I will try again.

What about R², should I set it as $$ \delta R R + R\delta R$$

or

$$ \delta (g^{\mu\nu}R_{\mu\nu}) R + g^{\mu\nu}R_{\mu\nu} \delta R $$ ?

Another way?

 

[ChrisVer]

http://en.wikipedia.org/wiki/Einstein–Hilbert_action#Variation_of_the_Riemann_tensor.2C_the_Ricci_tensor.2C_and_the_Ricci_scalar

I hope this can help.

As for the second question... you have written the same thing twice...but yes, the second way, by replacing the other ricci scalar can help... All in all, what you will mainly need is the variation of the Riemann tensor, and all the rest can come by contractions...

eg
$\delta R^2 = R ~\delta R + \delta R~ R= g^{\mu \nu} R_{\mu \nu} \delta (g^{\alpha \beta}R_{\alpha \beta} ) + \delta (g^{\alpha \beta}R_{\alpha \beta} ) g^{\mu \nu} R_{\mu \nu} = 2 g^{\mu \nu} R_{\mu \nu} \delta (g^{\alpha \beta}R_{\alpha \beta} )=$
$=2 g^{\mu \nu} R_{\mu \nu} (g^{\alpha \beta} \delta R_{\alpha \beta} + R_{\alpha \beta} \delta g^{\alpha \beta})$

 

[Breo]

I do not see clear what it says (bolds):

In the second line we used the previously obtained result for the variation of the Ricci curvature and the metric compatibility of the covariant derivative,

.

The last term,

, multiplied by

becomes a total derivative, since

and thus by Stokes' theorem only yields a boundary term when integrated. Hence when the variation of the metric

vanishes at infinity, this term does not contribute to the variation of the action. And we thus obtain,

 

[ChrisVer]

I don't understand what is not clear? Maybe if I call it Gauss theorem?
$\int_V d^4 x \partial_\mu A^\mu = \oint_{\partial V} d \sigma_{\mu}A^{\mu}$

So here if you set a boundary condition that $\delta g^{\mu \nu}=0$ at the limits of your integral, that integral can vanish...so the last term in $\delta R$ won't contribute...

 

[Breo]

Oh right! So we are anticipating the action integral, right?.

 

[ChrisVer]

Yes... If you don't anticipate the action integral, but keep carrying that term, you'll be carrying a lagrangian that comes with a total derivative...
$L'= L + \partial_\mu A^\mu$
you can work with both $L',L$ since they are equivalent...so it's better (convenient) to work with $L$.

 

[Breo]

I guess $$ \delta (R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}) $$ should vanish due to the indices location (the variation of all up or all down indices should be zero? I notice this because of the Christoffels symbols obtained...). Otherwise I tried to do:

$$ \delta (g_{\mu\alpha}R^{\alpha}_{\nu\rho\sigma})R^{\mu\nu\rho\sigma} + R_{\mu\nu\rho\sigma}\delta(g^{\nu\beta}g^{\rho\gamma}g^{\sigma\delta}R^{\mu}_{\beta\gamma\delta} = (\delta g_{\mu\alpha}R^{\alpha}_{\nu\rho\sigma})R^{\mu\nu\rho\sigma} + R_{\mu\nu\rho\sigma}(\delta g^{\mu\alpha}R_{\alpha}^{\nu\rho\sigma}) $$

This seems wrong... :-/

Assuming it must vanish then, after a few calcs I get the next equations of motion (avoiding the constants of the original lagrangian):

$$ (\frac{3}{2}g_{\mu\nu}R + 2R_{\mu\nu}) = 0 $$

This seems better if we choose in the Lagrangian of the statement: $$\alpha = \frac{1}{2} \ \space ; \space \ \gamma= -\frac{1}{8} $$ or multiples. So we obtain:

$$ (R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R) = 0 $$

Is this correct?

 

[ChrisVer]

can you elaborate what you mean with the vanishing of the $\delta(R_{abcd}R^{abcd})$?

$\delta R_{abcd}= \delta (g_{ka} R^{k}_{~bcd} )=R^{k}_{~bcd} \delta g_{ka} + g_{ka} \delta R^{k}_{~bcd} = R^{k}_{~bcd} \delta g_{ak} + g_{ak} \nabla_{[c}\delta \Gamma_{d]~ b}^k$

Similarily:

$\delta R^{abcd}= \delta (g ^{am} g^{bn} g^{cw} g^{dr} R_{mnwr})$

I am sorry, I just can't see how can you notice that they should vanish because of the indicies...

 

[Breo]

My fault, I reached that conclusion before the trial. So what I calculated was correct?

δ(gμαRανρσ)Rμνρσ+Rμνρσδ(gνβgργgσδRμβγδ=(δgμαRανρσ)Rμνρσ+Rμνρσ(δgμαRνρσα)

I am stucked then... Will keep trying. But after the result I reached assuming it vanishes, I think it really should vanish... I will try to forget that idea...

Thank you very much ChrisVer for all your support!

EDIT: Oh wait! Maybe it vanish because a factor of $$ g^{\alpha\beta} $$ will remain in the action so it will vanish? I mean I am calculating: $$ \frac{\delta S}{\delta g^{\mu\nu}} $$ so I can not remove the metric variations obtained for the Riemann tensor calc, right?

 

[ChrisVer]

I also hope this can help...
http://www.scholarpedia.org/article/F(R)_theories_of_gravitation

 

[Breo]

Not really, I am stucked in the computation of the riemann tensor square variation. :(

 

[Breo]

I think still can not solve it. I get this:

$$ \delta (R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}) = (\delta g_{\mu\alpha} R^{\alpha}_{\nu\rho\sigma})R^{\mu\nu\rho\sigma} + R_{\mu\nu\rho\sigma}(\delta g^{\mu\alpha} R_{\alpha}^{\nu\rho\sigma})$$

Thus, the equations of motion:

$$ 0 = \alpha RR_{\mu\nu} + \gamma ( R^{\rho\sigma} R_{\mu\nu\rho\sigma} - \frac{1}{4}g_{\mu\nu}R_{\rho\sigma}R^{\rho\sigma}) + \beta (R_{\nu\alpha\rho\sigma}R_{\mu}^{\alpha\rho\sigma}) $$

:-/

 

[Breo]

I just read in the Zee's book "Quantum Field Theory in a Nutshell":

"Also, use the Gauss-Bonnet theorem to get rid of the $$ R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}$$ term" pag. 459

I am trying to find that fact but the most I found was a bit complex explanation in the Hawking - Ellis book. Can anyone give me a reasoning for beginners?

Thank you in advance.

 

[ChrisVer]

http://relativity.livingreviews.org/open?pubNo=lrr-2010-3&page=articlesu33.html

maybe tis can be of help

 

[Breo]

It says the equations of motion gives second derivatives, but I did not obtain any second one...

 

[ChrisVer]

$$R_{abcd} \delta (g^{ia}g^{jb}g^{kc}g^{wd}R_{ijkw})=R_{abcd} ( g^{jb}g^{kc}g^{wd}R_{ijkw} \delta g^{ia} + g^{ia}g^{kc}g^{wd}R_{ijkw} \delta g^{jb} +$$
$$ +g^{jb}g^{ia}g^{wd}R_{ijkw} \delta g^{kc}+ g^{jb}g^{kc}g^{ia}R_{ijkw} \delta g^{wd} + g^{ia} g^{jb}g^{kc}g^{wd} \delta R_{ijkw})$$

The first term is:

$$R_{abcd} g^{jb}g^{kc}g^{wd}R_{ijkw} \delta g^{ia}$$

The second term is giving:

$$R_{abcd}g^{ia}g^{kc}g^{wd}R_{ijkw} \delta g^{jb}= R_{bacd} R_{jikw} g^{ia}g^{kc}g^{wd} \delta g^{jb}=R_{abcd} g^{jb}g^{kc}g^{wd}R_{ijkw} \delta g^{ia}$$

where at first equation I changed the indicies position using the antisymmetry of the Riemann tensor and then at the second equation I just renamed null indices.

The third term is again playing with indices:
$$R_{abcd}g^{jb}g^{ia}g^{wd}R_{ijkw} \delta g^{kc} =R_{cdab}g^{wd}g^{kc}g^{jb}R_{kwij} \delta g^{ia}= R_{abcd}R_{ijkw} g^{wd}g^{kc}g^{jb} \delta g^{ia}$$

The fourth:
$$ g^{jb}g^{kc}g^{ia} R_{abcd}R_{ijkw} \delta g^{wd}= g^{kc}g^{jb}g^{wd} R_{dcba}R_{wkji} \delta g^{ia}= R_{abcd}R_{ijkw} g^{kc}g^{jb}g^{wd} \delta g^{ia} $$

Using also that $$R_{abcd}g^{ai}g^{bj}g^{ck}g^{dw}= R^{ijkw}$$

These terms in the above first equation will give:

$$R_{abcd} \delta R^{abcd} = R_{abcd}R_{ijkw} (g^{jb}g^{kc}g^{wd}+g^{jb}g^{kc}g^{wd}+g^{wd}g^{kc}g^{jb}+g^{kc}g^{jb}g^{wd} ) \delta g^{ia} +R^{ijkw} \delta R_{ijkw} $$

or

$$R_{abcd} \delta R^{abcd}= 4 R_{abcd}R_{ijkw} g^{jb}g^{kc}g^{wd} \delta g^{ia} + R^{ijkw} R^{s}_{~jkw} \delta g_{is} + R^{ijkw}g_{is} \nabla_{[k} \delta~\Gamma^{s}_{w]j}$$

Now you can play again with indices and use that $\delta g_{is}= - g_{ib} g_{sp} \delta g^{bp}$

$$R_{abcd} \delta R^{abcd}= 4 R_{abcd}R_{ijkw} g^{jb}g^{kc}g^{wd} \delta g^{ia} - R^{rjkw} R_{ijkw} g_{ra} \delta g^{ia} + R^{ijkw}g_{is} \nabla_{[k} \delta~\Gamma^{s}_{w]j}$$

$$=(4 R_{abcd}R_{ijkw} g^{jb}g^{kc}g^{wd} - R^{rjkw}R_{ijkw} g_{ra} ) \delta g^{ia}+ R^{ijkw}g_{is} \nabla_{[k} \delta~\Gamma^{s}_{w]j}$$

To this you just add then the $R^{abcd}\delta R_{abcd}$ I gavee in a previous post and work out the result...

Did you get this result by your calculations? or something similar...I did it really fast I may have done mistakes with the renaming of indicies and so on...

 

[ChrisVer]

To my last not about adding the previous result, you don't have to do it... because ere it appears the same quantity so you can just add a 2 in the appropriate terms.
However do the calculations also by yourself to check whether I did any mistake or not...

 

[Breo]

Yes it is right. So the final contribution to the equations of motion would be:

$$ (4R_{abcd}R_{ijkw}g^{jb}g^{kc}g^{wd} -2R^{rjwk}R_{ijkw}g_{ra})g_{\mu}^i g_{\nu}^a = 2R_{\nu} ^{jkw}R_{\mu jkw} $$.

Right?

 

[ChrisVer]

What is $\delta \Gamma$ like?
Also I don't know how you got those mixed up-low indices metric tensors ...

 

[Breo]

While multiplied by $\sqrt{-g}$ it becomes a total derivative so it vanishes for the equations of motion.

The mixed indices come from: $\frac{\delta g^{ia}}{\delta g{\mu\nu}}$ as I am calculating: $\frac{\delta S}{\delta g{\mu\nu}}$ which define the equations of motion.

 

[ChrisVer]

Also , since you work with variations it's better to work with the action, rather than the lagrangian. The action will be for this $R_{abcd}R^{abcd}$ only term [Kretschmann scalar]:
$S = \int d^4 x \sqrt{-g} R_{abcd}R^{abcd}$

So when you vary it with respect to the metric, you will get an additional term from the variation of the $\sqrt{-g}$ which will also exist in the equations of motion... I would try to bring it to a form:

$\delta S = \int d^4 x [A]_{ij} \delta g^{ij}$

And thus the EoM will be $\frac{\delta S}{\delta g^{\mu \nu}}=A_{ij} \delta^{i}_\mu \delta^j_\nu=A_{\mu \nu}=0$

with $A$ given by the terms coming from the variation of the Kretschmann scalar and the square root of the determinant of the metric...

In your case you have addiitional terms coming from the Ricci scalar^2 and the "Ricci curvature^2"...

 

[Breo]

That means that the terms coming from $R_{abcd}R^{abcd}$ will vanish in the EoM?

 

[ChrisVer]

I don't see why it should vanish in the EoM... sorry...
and without $\delta \Gamma$ in terms of $\delta g$ I cannot proceed with the calculations. From what I've read online though, in the sources that I gave you, there is no reason for it to vanish but it will give contributions to the EoM that will contain higher order derivatives of the metric tensor...

 

[Breo]

Then I do not understand what was your point in your previous post.

But when you multiply it by $\sqrt{-g}$ becomes a total derivative, right? that was what we talk yesterday at first because the first link you gave me stated that. I am lost with this computation :/

 

[ChrisVer]

I think you misunderstood the link... The total derivative talk in the wiki-link I sent was about the Ricci Scalar part...
The Ricci Scalar variation gives:
$\delta R= R_{\mu \nu} \delta g^{\mu \nu} + \nabla_\sigma (g^{\mu \nu} \delta~\Gamma^{\sigma}_{\mu \nu} -g^{\mu \sigma} \delta~\Gamma^{\rho}_{\rho \nu})$

Then in the Action this will be:

$S=a \int d^4 x \sqrt{-g}R$
$\delta S=a \int d^4 x \sqrt{-g} \delta R +a \int d^4 x R \delta \sqrt{-g}$

The second term of $\delta R$ in the first integral will give the surface integral and thus can vanish by configuring $\delta g^{\mu \nu}=0$ at infinity... In general instead of trying to see total derivatives in the simple sense of Gauss law, that means $\int d^4 x \partial_\mu J^\mu$ you can generalize this to the general 4-volume element in gravitational coupling existence (metric existence) $g$ as: $dV'= \sqrt{-g} dV$ and instead of normal partial derivatives have covariant derivatives. Thus the generalized surface integrals appear as $\int d^4x \sqrt{-g} \nabla_{\sigma} J^\sigma$...I (personally) can't see any total covariant derivative in the expression you got for the Kretschmann scalar.
and thus the equation of motion will contain only the first term from $\delta R$ which will be the $R_{\mu \nu}$ but that's not the full equation of motion. You will also get from the 2nd integral a term that is proportional to the Ricci scalar:

$R \delta \sqrt{-g}= - R \frac{1}{2} \sqrt{-g} g_{\mu \nu} \delta g^{\mu \nu}$

and thus the full equation of motion will be:

$a \sqrt{-g} R_{\mu \nu} -a R \frac{1}{2} \sqrt{-g} g_{\mu \nu} =0$
or
$R_{\mu \nu}- \frac{1}{2} g_{\mu \nu} R =0$
(no matter Einstein equations)

EDIT
The same argument for the total derivative does not apply for the $R^2$ case, so watch out...
Because the covariant derivative of $\delta R$ will then appear in the action variation as:
$\delta S = \int d^4 x \sqrt{-g} R \nabla_{\sigma} A^{\sigma}$
and if $\nabla_{\sigma} R \ne 0$ you can bring the $R$ in the covariant derivative and drop this term as a surface integral...

 

[ChrisVer]

^editted.

 

[Breo]

Oh! So I forgot the index contraction so it is not the same the $\delta \Gamma$ term for each case xD Then I have all the calculations, the thing is I will obtain in the EoM those term I were vanishing wrongly, right?

 

[ChrisVer]

again you talk about $\delta \Gamma$ while I don't know it...
In general it's good to bring $\delta \Gamma$ in terms of $\delta g$ in the last expression I gave in the previous page (with the factors of 2 you added) and take the $\delta g$ out as a common factor...
Also get a factor from $\sqrt{-g}$ although that's the easiest part in your Lagrangian, since it will come in complete analogy:
$\delta_g S_1 = \int d^{4} x L \delta \sqrt{-g} = -\frac{1}{2}\int d^{4} x \sqrt{-g}L g_{\mu \nu} \delta g^{\mu \nu}$

with $L$ your Lagrangian..and so you don't have to care about it aver again...in your last overall Eom just add that term: $-\frac{1}{2}\sqrt{-g}L g_{\mu \nu} \delta g^{\mu \nu}$...

 

[Breo]

Ok I will (what ugly EoM :s).

So my last question about this: I do not understand the post #24 :P It would be possible that you are mixing the terms of the variation of square root of -g times L with the variation of L times square root of -g? Because the integral you write is for the latter, right? but you introduced it as the former.

 

[ChrisVer]

It's supposed to be ugly since it has so many higher order terms... and you have them all in your lagrangian (I wonder where did you find it).
Not that they don't exist (eg the Kretschmann scalar as I read has applications in quantum gravity), but the Lagrangian looks awful.

As for the post

Also , since you work with variations it's better to work with the action, rather than the lagrangian. The action will be for this $R_{abcd}R^{abcd}$ only term [Kretschmann scalar]:
$S = \int d^4 x \sqrt{-g} R_{abcd}R^{abcd}$

So when you vary it with respect to the metric, you will get an additional term from the variation of the $\sqrt{-g}$ which will also exist in the equations of motion... I would try to bring it to a form:

$\delta S = \int d^4 x [A]_{ij} \delta g^{ij}$

And thus the EoM will be $\frac{\delta S}{\delta g^{\mu \nu}}=A_{ij} \delta^{i}_\mu \delta^j_\nu=A_{\mu \nu}=0$

with $A$ given by the terms coming from the variation of the Kretschmann scalar and the square root of the determinant of the metric...

In your case you have addiitional terms coming from the Ricci scalar^2 and the "Ricci curvature^2"...

I just sketched the idea of working this out... you will have:
$S= \int d^4 x \sqrt{-g} L$
take variation wrt the metric field:
$\delta _g S = \int d^4 x \sqrt{-g} \delta_g L + \int d^4 x L \delta_g \sqrt{-g}$
you will somehow try (I don't know whether that is possible or not) to bring $\delta_g L= A_{ij} \delta g^{ij}~~(1)$
and similar for $\delta_g \sqrt{-g} = B_{ij} \delta g^{ij}~~(2)$ so that:

$\delta _g S = \int d^4 x A_{ij} \sqrt{-g} \delta g^{ij}+ \int d^4 x L B_{ij} \delta g^{ij}$
or
$\delta _g S = \int d^4 x (A \sqrt{-g}+BL)_{ij} \delta g^{ij}= \int d^4 x (C)_{ij} \delta g^{ij}$
The EoM will be then $C_{ij}=0$ where you determine C by A,B which you determine by the variations $(1),(2)$.

In case you have the covarant derivatives of the variations of the metric, you can bring them out using:
$\nabla_\alpha ( A B)=A \nabla_\alpha B + (\nabla_\alpha A) B$

 

[Breo]

Ah! it is clear! I got lost because I expected to see the $\sqrt{-g}$ term explicitly out of $A_{ij}$ (in your last post $B_{ij}$).

I found this in my gravitation course notes. I am studying a masters right now.

I can not express how much I appreciate your efforts and help, ChrisVer :)

Thank you very very much!

 

[ChrisVer]

well I have a script right now where I derive the EoM for
$R_{abcd}R^{abcd}$ -term...
If you want I can send it to you...however there may be mistakes...