# Divergence of curvature scalars * metric

Hopefully some other people will respond as well.
And since this thread has become long, here is a summary of some of the preceeding.

Using the method described above to try to solve for the divergence of g^{cd}R_{ab}R^{ab}

Consider the action
$$S = \int (\frac{1}{2\kappa}R_{ab}R^{ab} + \mathcal{L}_m) \sqrt{-g} d^4x$$
I made mistakes when trying to work out the field equations, but this paper http://arxiv.org/PS_cache/astro-ph/pdf/0410/0410031v2.pdf
does it and gets:
$$-\frac{1}{2}g^{ab}R_{cd}R^{cd} + 2R^{ca}R_c{}^{b} -2 \nabla^c\nabla^d R_c{}^{(a} \delta_d{}^{b)} + \nabla^c \nabla_c R^{ab} + g^{ab} \nabla^c \nabla^d R_{cd} = \kappa T^{ab}$$

Applying the method, we then get:
$$\nabla_b\left(g^{ab}R_{cd}R^{cd}\right) = 2\nabla_b\left(2R^{ca}R_c{}^{b} -2 \nabla^c\nabla^d R_c{}^{(a} \delta_d{}^{b)} + \nabla^c \nabla_c R^{ab} + g^{ab} \nabla^c \nabla^d R_{cd} \right)$$

The problem is, unlike in the f(R) case I worked out explicitly above, I cannot check by hand whether the method gave something which is true in general or only true in solutions for this theory's field equations. There is, in my opinion, good heuristic reason to expect that the method gives divergence relations which are true in general (along with evidence in the form of other cases which were checked by hand being true in general). Yet here I cannot check by hand to verify, because that would require me to work out, by hand, what the divergence of $$g^{ab}R_{cd}R^{cd}$$ is ... which I don't know how to do, as that was the opening question.

If someone knows how to show mathematically that my method is valid then that would fully answer the question and we'd be done! On the other hand, if someone knows how to show mathematically that my method doesn't hold generally, that would at least put discussion on that to rest and would be immensely helpful as well (especially to see the mathematical details of such a proof).

Basically, if anyone has guidance on how to proceed from here, please do share.

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Sticking my neck out for Altabeh to chop

$$(g^{ab}P)_{;b} = (g^{ab})_{;b}P + g^{ab}(P)_{;b}$$

I think the first term on the right is zero which leaves $g^{ab}(P)_{;b}$.

$$(P)_{;b} = (g^{cp}g^{qd}R_{pq}R_{cd})_{;b} = g^{cp}g^{qd}\left[R_{pq}(R_{cd})_{;b}+R_{cd}(R_{pq})_{;b}\right] + R_{pq}R_{cd}\left[ g^{cp}(g^{qd})_{;b} +g^{qd}(g^{cp})_{;b}\right]$$

and given that

$$R_{cd}={R^\rho}_{c\rho d} = \partial_\rho\Gamma^\rho_{dc} - \partial_d\Gamma^\rho_{\rho c} + \Gamma^\rho_{\rho\lambda}\Gamma^\lambda_{dc} - \Gamma^\rho_{d\lambda}\Gamma^\lambda_{\rho c}$$

I give up.

I think the first term on the right is zero
Yep, even more generally
$$g^{ab}{}_{;c}=0$$
which can be seen if one writes out the covariant derivative explicitly in terms of partial derivatives and christoffel symbols (and then expand those in terms of partial derivatives of the metric).

This also lets you simplify your equation, since it means the term
$$R_{pq}R_{cd}\left[ g^{cp}(g^{qd})_{;b} +g^{qd}(g^{cp})_{;b}\right]=0$$

The other term can be simplified too, since there is no significance in the label we give dummy indices
\begin{align*} (P)_{;b} &= (g^{cp}g^{qd}R_{pq}R_{cd})_{;b} \\ &= g^{cp}g^{qd}R_{pq}(R_{cd})_{;b}+g^{cp}g^{qd}R_{cd}(R_{pq})_{;b} \\ &= g^{cp}g^{qd}R_{pq}(R_{cd})_{;b}+g^{pc}g^{dq}R_{pq}(R_{cd})_{;b} \\ &= 2 g^{cp}g^{qd}R_{pq}(R_{cd})_{;b} \\ &= 2 R^{cd}(R_{cd})_{;b} \end{align*}

But, just like you, I have no idea where to go next in working it out by hand. If we go this route:
$$R_{cd}={R^\rho}_{c\rho d} = \partial_\rho\Gamma^\rho_{dc} - \partial_d\Gamma^\rho_{\rho c} + \Gamma^\rho_{\rho\lambda}\Gamma^\lambda_{dc} - \Gamma^\rho_{d\lambda}\Gamma^\lambda_{\rho c}$$
then when we try to calculate the covariant derivative of that, we'd have to ask what is the covariant derivative of a non-tensor object (Christoffel symbols are not themselves tensors) ... which I have even less of an idea of how to work out.

Thanks much for trying.
I really don't know how to progress further with that calculation myself. Anyone else want to take a crack at it?

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(Christoffel symbols are not themselves tensors) ... which I have even less of an idea of how to work out.
I give you a clue for how you can proceed to calculate the covariant derivative of Christoffel symbols. Have you ever heard of the "Palatini equation" in GR? I bet you have not! It is given by

$$\delta R_{ab}=-(\delta\Gamma^{c}_{ab})_{;c}+(\delta\Gamma^{c}_{ac})_{;b}$$

Use the detailed form of the Ricci tensor and calculate its variation to have this equation proven. What do you see, then?

Sticking my neck out for Altabeh to chop
Don't be scared; you survived!

AB

Altabeh said:
Have you ever heard of the "Palatini equation" in GR? I bet you have not!
This kind of one-upmanship ill-behoves you. If you're going to use this thread to show off how much you know, I for one can do without your help. Please, tone it down.

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I give you a clue for how you can proceed to calculate the covariant derivative of Christoffel symbols. Have you ever heard of the "Palatini equation" in GR? I bet you have not! It is given by

$$\delta R_{ab}=-(\delta\Gamma^{c}_{ab})_{;c}+(\delta\Gamma^{c}_{ac})_{;b}$$

Use the detailed form of the Ricci tensor and calculate its variation to have this equation proven. What do you see, then?
I've seen that before in the derivation of GR from the Einstein-Hilbert action, as it shows up in the metric variation based theory ( http://en.wikipedia.org/wiki/Einste...r.2C_the_Ricci_tensor.2C_and_the_Ricci_scalar ) and not just the connection variation / Palatini theory. (It looks like you might be using a different sign convention than Wikipedia, or you or wiki has a sign error.) But to answer your question, no I have not heard of the "Palatini equation". I was not aware that equation had a name.

Following that article they show:
$$\nabla_\lambda (\delta \Gamma^\rho_{\nu\mu} ) = \partial_\lambda (\delta \Gamma^\rho_{\nu\mu} ) + \Gamma^\rho_{\sigma\lambda} \delta\Gamma^\sigma_{\nu\mu} - \Gamma^\sigma_{\nu\lambda} \delta \Gamma^\rho_{\sigma\mu} - \Gamma^\sigma_{\mu\lambda} \delta \Gamma^\rho_{\nu\sigma}$$
But I'm not sure how this helps here, as I don't know how to relate $$\nabla_\lambda (\delta \Gamma^\rho_{\nu\mu} )$$ to the covariant derivative of a Christoffel symbol. I mean, the only way they were able to write that in the first place was because $$\delta \Gamma^a_{bc}$$ transforms as a tensor object, even if $$\Gamma^a_{bc}$$ does not.

If the math really is that simple to you, can you please just show the solution here?

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...and not just the connection variation / Palatini theory. (It looks like you might be using a different sign convention than Wikipedia, or you or wiki has a sign error.) But to answer your question, no I have not heard of the "Palatini equation". I was not aware that equation had a name.
That is not an error. If the Ricci tensor is generated by contracting the second lower indice and the only upper indice of the Riemann tensor, then you lead to what the Wiki's article gives. Since the Riemann tensor is anti-symmetric in the second and third lower indices, so doing the same calculation with the third lower indice leads to a flip in the sign of the Palatini equation. Hope that this is clear!

If the math really is that simple to you, can you please just show the solution here?
The math is by no means simple to anybody to be followed by hand. First off you didn't get the whole point I made in my latest post. The Christoffel symbols are not tensors so they cannot be covariantly differentiated in the same way as a tensor of the same rank. But since the action has a variational nature, so all the terms appearing in the action are accompanied by a delta in them. This has an advantage, that is, the quantity $$\delta \Gamma^\rho_{\nu\mu}$$ is a tensor (remember that the difference of two connections is a tensor because in the transformation law the extra terms cancel out due to the symmetry of connections in their lower indices.) This is what you need in calculating the action of the proposed Lagrangian. But there is a very huge problem with the first term in the Lagrangian $$S.$$ That is it CANNOT be cast into the form of a scalar density by any means and this gives rise to the inefficacy of the Lagrangian to reduce to the Hilbert-Einstein action.

Are you still insisting this method can work?

This kind of one-upmanship ill-behoves you. If you're going to use this thread to show off how much you know, I for one can do without your help. Please, tone it down.
I don't care if you can do this with or without my help and I know better than everyone else in this place that a bunch of things from the tensor calculus cannot make a Plato out of me. So no reason to show off this way and please don't send such irrelevant messages in this thread. Neither we are playing games with the smilies nor are we threatening the other users by the use of knowledge!!

AB

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That is not an error. If the Ricci tensor is generated by contracting the second lower indice and the only upper indice of the Riemann tensor, then you lead to what the Wiki's article gives. Since the Riemann tensor is anti-symmetric in the second and third lower indices, so doing the same calculation with the third lower indice leads to a flip in the sign of the Palatini equation. Hope that this is clear!
Yes, I understand that due to the symmetry of the Riemann tensor
$$R^a{}_{bac} = - R^a{}_{bca}$$

So you are saying it is a choice of whether one defines: $$R_{bc} = R^a{}_{bac}$$ or $$R_{bc} = R^a{}_{bca}$$ ?

But due to the symmetry of the Riemann tensor, this choice in definition reduces to a sign convention. As I said, "you might be using a different sign convention than Wikipedia, or you or wiki has a sign error". So it looks like the answer is: you are using a different sign convention.

First off you didn't get the whole point I made in my latest post. The Christoffel symbols are not tensors so they cannot be covariantly differentiated in the same way as a tensor of the same rank. But since the action has a variational nature, so all the terms appearing in the action are accompanied by a delta in them. This has an advantage, that is, the quantity $$\delta \Gamma^\rho_{\nu\mu}$$ is a tensor (remember that the difference of two connections is a tensor because in the transformation law the extra terms cancel out due to the symmetry of connections in their lower indices.) This is what you need in calculating the action of the proposed Lagrangian.
I didn't get your intended point in that post, because I assumed you were trying to answer my question. It seems that instead we are still having communication problems. Initially it seemed directly related since you said "you can proceed to calculate the covariant derivative of Christoffel symbols". Which would have been interesting and directly useful. But now you are switching the intent to explaining how to do a different calculation.

I was NOT asking how to solve for the variation of R_{ab}, but instead for the covariant derivative of R_{ab} (ie. $R_{ab;c}$) in terms of the Reimann tensor (besides the trivial relation of course).

I don't need to work out the variation of the action because we already have the field equations, (since all the way back in post 13), from here http://arxiv.org/abs/astro-ph/0410031. From there it is trivial to apply the method I was suggesting to obtain a relation for the divergence of $$g^{ab}R_{cd}R^{cd}$$.

Thus the question is whether this relation is true in general or true only for solutions of those field equations. I'm starting to repeat myself again. Please refer to post #26.

But there is a very huge problem with the first term in the Lagrangian $$S.$$ That is it CANNOT be cast into the form of a scalar density by any means and this gives rise to the inefficacy of the Lagrangian to reduce to the Hilbert-Einstein action.
I do not care that the Lagrangian doesn't reduce to the Hilbert action. I've explained my method many times. I don't understand why there is still confusion. I'm merely using the action to help me generate a relationship between the divergence of many curvature terms.

Are you still insisting this method can work?
Yes.
It makes intuitive sense to me that this method should work. And furthermore, I've already checked by hand, and proved an entire class of actions does lead to divergence relations which are true in general for any Lorentzian manifold.

I haven't figured out how to prove that this method will always work, but of course I still insist it could work. For it has worked in every case I could check by hand so far, and there is (in my opinion) reasonable heuristics for why it probably always works.

If someone can prove that the method will always work, that would be immensely helpful.

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I'm trying out a program called TTC (Tensor Tools for Calculus), which can help prove some divergence relations.
So far it can't verify/simplify the relation
Applying the method, we then get:
$$\nabla_b\left(g^{ab}R_{cd}R^{cd}\right) = 2\nabla_b\left(2R^{ca}R_c{}^{b} -2 \nabla^c\nabla^d R_c{}^{(a} \delta_d{}^{b)} + \nabla^c \nabla_c R^{ab} + g^{ab} \nabla^c \nabla^d R_{cd} \right)$$
But I have been able to get some relations involving the divergence of g^{ab}R_{cd}R^{cd} using the program. So I'll keep fiddling with it and see if I can prove it one way or another.

On another note, I found a paper proving there are only 3 independent rank 2, symmetric, divergenceless tensors of second order in the Reimann curvature in >= 5 dimensions. In d=4, this reduces to 2 due to additional constraints. This surprisingly means if you used Carrol's generalized gravity equations http://arxiv.org/abs/astro-ph/0410031 (using his notation here), with f(R,P,Q) = a R R + b P + c Q (where a,b,c are constants), the theory doesn't actually have three free parameters in spacetimes with 4 dimensions. They are related somehow, for instance the f=Q theory could be equated to some form of f = a R R + b Q theory. Strange.

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Thanks atyy! That is great!

They show the method would work for an even larger class of theories. Unfortunately this does not include the scalar R_{ab}R^{ab}.

However, I have been able to get "Tensor Tools for Calculus" to simplify the relation obtained by my method. And it verified that the method works in that case as well!

As that paper you found mentions, there is good heuristic reason to expect this method to always work. It falls short of proving it in complete generality though.

I guess that answers the openning question of the thread, as well as (to current best ability) the question about the validity of the method I used to obtain further divergence relations. And at the very least, we now know the relation I obtained for the divergence of g^{ab} R_{cd}R^{cd} is correct.

Thanks atyy! That is great!

They show the method would work for an even larger class of theories. Unfortunately this does not include the scalar R_{ab}R^{ab}.

And it verified that the method works in that case as well!
I'm so curious to see how the method is verified. Put the results in this thread or otherwise I sure know that if I can call this a "method", it does not work at all! The reason I already gave umpteen times before is that the approach does not include any appropriate function or parameter like, $$f(R)$$ in the $$f(R)$$ gravity theories, which has this advantage that cancels out with the extra non-vanishing terms of the divergence of the field equations on a purely geometrical (or mathematical) basis. (See for example the explanation below the equation 8 in http://arxiv.org/abs/gr-qc/0505128). This does not occur by accident or by the use of a cheap guess as in your theory.

In order to make the theory obey the conservation law, the proposed Lagrangian must be in a suitable form, meaning that after taking the divergence of its field equation all the terms appearing in the equation have to vanish somehow. You don't put the non-vanishing terms equal to zero by hand, as for the same paper this is obvious from the equation 8 that authors do not seek out some relation between non-zero terms such as $$(\nabla^{\mu}F) R_{\mu\nu}$$ and $$(\square\nabla_{\nu}-\nabla_{\nu}\square)F$$, to get a conserved matter tensor but rather this happens all by a mathematical reasoning where your theory seems to have a hole in it: We cannot write a relation or relations between the terms to get this result. You have to show us after taking the divergence that all terms vanish or cancel out each other so you're done then with a claim regarding the success of the theory or method or whatever.

AB

I'm so curious to see how the method is verified.
Then how about you reread the line you purposely deleted when quoting me in your post. "Tensor Tools for Calculus" is a freely available program. If you have access to Mathematica, give it a try yourself.

To make sure this isn't another reading comprehension issue, please note I said it verified the method in this case (ie. the divergence term I was working on). Nowhere did I claim I have been able to verify the method in generality.

Not trusting my math, and not trusting a computational package is one thing, but this comment is just offensive:
Put the results in this thread or otherwise I sure know that if I can call this a "method", it does not work at all!
You are sure the method doesn't work here (or at all)!?

You have repeatedly stated how easy it is to do the calculation I am struggling with (the divergence of $$g^{ab}R_{cd}R^{cd}$$). Yet instead of showing me how to do it, you deridingly give me "hints" which seem to lead no-where, or just insult me.

Look, I can't show the math here because I had to resort to using a computational package. I still don't know how to work it out by hand, except for using the method I suggested, which I can't prove in generality either. But if you are going to disagree with my result even after a computational package verified it... then please PLEASE, just show some math here. If it is so simple for you, and you are so sure the result is wrong, then please just work out the divergence of $$g^{ab}R_{cd}R^{cd}$$ as that will actually answer the openning question of this thread.

The reason I already gave umpteen times before is that the approach does not include any appropriate function or parameter like, $$f(R)$$ in the $$f(R)$$ gravity theories, which has this advantage that cancels out with the extra non-vanishing terms of the divergence of the field equations on a purely geometrical (or mathematical) basis. (See for example the explanation below the equation 8 in http://arxiv.org/abs/gr-qc/0505128). This does not occur by accident or by the use of a cheap guess as in your theory.
I stated that I worked out that my method works in general for f(R) theory, and you instead claimed that f(R) theory leads to violation of energy-momentum conservation unless careful choices of the function f(R) are made. I then wrote out the math explicitly showing that my method worked for f(R) theory in general. You called the math complete nonsense and dismissed my result without consideration. Now you've come all the way around and seem to agree with me on f(R) theory, but still after all this time you are not listening.

In trying to work out that divergence relation, I was using what amounts to f(R,P,Q) = P - R, in the paper mentioned previously http://arxiv.org/PS_cache/astro-ph/pdf/0410/0410031v2.pdf.

What you are claiming amounts to:
There are choices of f(R,P,Q) that lead to field equations which violate energy-momentum conservation, despite the action having no explicit time or spatial dependence.

I find that claim much MUCH more unlikely than the claim I made to arrive at the divergence relation:
Applying the method, we then get:
$$\nabla_b\left(g^{ab}R_{cd}R^{cd}\right) = 2\nabla_b\left(2R^{ca}R_c{}^{b} -2 \nabla^c\nabla^d R_c{}^{(a} \delta_d{}^{b)} + \nabla^c \nabla_c R^{ab} + g^{ab} \nabla^c \nabla^d R_{cd} \right)$$
If you believe the divergence relation I obtained is wrong, then finally stop leading us around and actually show the math which makes you so "sure" I am wrong. Showing the math will also answer the opening question of this thread, so will be directly useful as well.

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then please PLEASE, just show some math here. If it is so simple for you, and you are so sure the result is wrong, then please just work out the divergence of $$g^{ab}R_{cd}R^{cd}$$ as that will actually answer the openning question of this thread.
Oh God, I never ever again get involved with the threads whose OP is JustinLevy!!

Look, the divergence of $$g^{ab}R_{cd}R^{cd}$$ is going to nowhere because you're not doing any index-related action on the pair $$R_{cd}R^{cd}$$ and that $$g^{ab}$$ is just a spectator is the calculation of the divergence. Neither me nor any other physicist can do this by any means unless a contraction of one of the indices of the Ricci tensor with the only index of the divergence operator, $$\nabla_{a}$$, is produced which in this case is impossible! I want to make this point here that in calculating the divergence of the Ricci tensor, remember the Bianchi Identity, we are not getting Christoffel symbols involved directly as you can see http://en.wikipedia.org/wiki/Proofs_involving_Christoffel_symbols#Proof_1"! So if you were even an avreage physicist, you would be able to realize what choice of correction terms in Einstein-Hilbert action could be giving a divergence-free field equations and this is just a matter of realization to ignore $$g^{ab}R_{cd}R^{cd}$$ for not being influenced by the $$\nabla_{a}$$!

On the contrary, the situation is totally different for the Ricci scalar curvature: This has a relation with the Ricci tensor which together form the divergence-free Einstein tensor and this is taken advantage of in the paper http://arxiv.org/abs/gr-qc/0505128 to prove a same property for the proposed field equations. Besides, if they are claiming any conservation of energy-momentum in the paper http://arxiv.org/PS_cache/astro-ph/pdf/0410/0410031v2.pdf for the general field equations (15), it just falls upon the choice of $$f(R)$$ to determine whether the equation is conserved or not. As in this paper, I also believe and have been saying all this time in this thread that

"... it is that it is crucial to pick, not only the action, but also the fundamental fields which will be varied and assign to them a precise meaning."

I strongly suggest you to first learn the preliminaries of the f(R) theories (see the latest paper I cited above) and then try something new! All you're doing is wasting everyone's time and your own and if you don't take this seriously, I think I don't have anything more to say!

Period.

AB

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Neither me nor any other physicist can do this by any means
BY ANY MEANS!?
Again we seem to have some serious miscommunication issues. You seem not to even read what anyone writes. I've stated multiple times now that I have verified the divergence relation with a computational package.

You are still arguing against my result without showing any math. You again insult me without backing up your claims. This is unacceptable.

So if you were even an avreage physicist, you would be able to realize what choice of correction terms in Einstein-Hilbert action could be giving a divergence-free field equations and this is just a matter of realization to ignore $$g^{ab}R_{cd}R^{cd}$$ for not being influenced by the $$\nabla_{a}$$!
LISTEN! Will you please just actually listen.
I have verified that the field equations containing that term are indeed divergence-free. So don't be deriding me with phrases like "if you were even an average physicist", when I actually have proof behind my statements.

Please, you need to read more carefully. It seems the vast majority of our communication issues can be traced to you completely ignoring or misreading what has been written.

Besides, if they are claiming any conservation of energy-momentum in the paper http://arxiv.org/PS_cache/astro-ph/pdf/0410/0410031v2.pdf for the general field equations (15), it just falls upon the choice of $$f(R)$$ to determine whether the equation is conserved or not.
You are STILL claiming these actions, despite having no explicit time or spatial dependence, can lead to violation of local energy momentum conservation!?

This is just wrong.
We can debate about whether the assumption in my method is generally valid, but there can be NO debate about your claim. You are just WRONG. It is in direct violation of Noether's theorem.

As in this http://arxiv.org/PS_cache/arxiv/pdf/0810/0810.5594v1.pdf" [Broken], I also believe and have been saying all this time in this thread that

"... it is that it is crucial to pick, not only the action, but also the fundamental fields which will be varied and assign to them a precise meaning."
The thing being varied in ALL the actions we discussed in this thread, is the metric. These were ALL metric theories. And the metric has the same physical meaning in all these theories. Please read more carefully.

If you are going to continue to claim that a computational package is giving incorrect results, or continue claiming these actions yield field equations which violate local energy-momentum conservation. Provide the math which backs up your statements, or stopping claiming them. For it appears clear to me at this point in time that you are just plain wrong on multiple accounts.

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atyy