Divergence of curvature scalars * metric

[JustinLevy]

How can one work out what terms like:
[tex](g^{cd}R^{ab}R_{ab})_{;d}[/tex]
are in terms of the divergence of the Ricci curvature or Ricci scalar?

One student noted that since:
[tex]G^{ab} = R^{ab} - \frac12 g^{ab}R[/tex]
[tex]{G^{ab}}_{;b} = 0[/tex]
that we could maybe use the fact that
[tex]G^{ab}G_{ab} = R^{ab}R_{ab} - \frac12 R^{ab}g_{ab}R - \frac12 R R_{ab}g^{ab} + \frac14 RRg^{ab}g_{ab} = R^{ab}R_{ab} [/tex]
to help? We weren't sure where to go next.

Can someone better with tensor manipulations show how we could work this out?

 

[nicksauce]

You have divergence (Ricci tensor) = 1/2 divergence (Ricci scalar). For your original expression just apply the product rule using that rule. Remember that the metric is divergence free.

 

[JustinLevy]

Basically, you are saying I should be able to use
[tex] {R^{ab}}_{;b} = \frac12 g^{ab}R_{,b}[/tex]
to help here.

I feel like an idiot, but I really don't see how to use that.

If I expand out, I get:
[tex](g^{cd}R^{ab}R_{ab})_{;d} = {g^{cd}}_{;d}R^{ab}R_{ab} + g^{cd}{R^{ab}}_{;d}R_{ab} + g^{cd}R^{ab}R_{ab;d}[/tex]
As you mention, the first of those is just zero, since the metric covariant derivatives are zero.

But how do I relate terms like
[tex]{R^{ab}}_{;b}[/tex]
to terms like
[tex]{R^{ab}}_{;d}[/tex]
?

 

[Mentz114]

Applying Liebnitz

[tex]
(R^{ab}R_{ab})_{;d}={R^{ab}}_{;d}R_{ab}+{R^{ab}}R_{ab}_{;d}
[/tex]

might help. I have to dash ...

[I just noticed this is what nicksauce said ]

 

[JustinLevy]

[tex](R^{ab}R_{ab})_{;d}={R^{ab}}_{;d}R_{ab}+{R^{ab}}R_{ab}_{;d}[/tex]

Yes, that is what I wrote above. I just don't know how to relate terms containing
[tex]g^{cd}{R^{ab}}_{;d}[/tex]
to the divergence
[tex]{R^{ab}}_{;b}[/tex]
which after that, I can of course relate to the divergence of the Ricci scalar.


What I'm actually trying to get at, is a friend asked what is the next highest order term which we can add to the EFE such that:
1) still shows T_uv is divergenceless
2) has the same vacuum equations (R_uv = 0)

I thought it would be a neat thing to work out for myself, but I'd need to know the divergence of more complicated terms. So I could put in something simple and hopefully a couple counter terms to make the divergence zero. But I quickly end up with terms I can't figure out how to simplify further so I can't tell if stuff cancels or not.

I'll admit I tried to sneak a peak at the answer, but searching papers I only found examples of people manipulating the Hilbert action. There's probably someone who already worked it out for the field equations directly, but if so I didn't find it. The reason I want to focus on the field equations is this sometimes allows simplier changes, for example imagine adding magnetic monopoles from the Maxwell E and B field equations point of view -- which is easy, vs trying to add it in an action formalism which no one has found a consistent local way to do it yet. So the simplest change in the action formalism, isn't necessarily the simplest change to the field equations, to meet the desired conditions.

 

[atyy]

Don't look if you want to have the fun of working it out for yourself (I believe these still give local energy conservation, but the vacuum equations are not the same):

http://www.gravityresearchfoundation.org/pdf/awarded/2008/Padmanabhan_2008.pdf
http://arxiv.org/abs/gr-qc/0305004
http://arxiv.org/abs/0805.1726

 

[JustinLevy]

Thanks atyy.
That first paper makes me a bit uncomfortable. For instance "Horizons are always observer dependent" and other claims; it all seems kind of sloppy. Especially declaring at the end that there have to be "atoms of spacetime".
As for the others, I've seen f(R) and Lovelock gravity before while searching, but those papers are from the point of view of modifying the action. Which I commented on my thoughts about above. It does give me an idea though...

if the action doesn't have explicit time or spatial dependence, I guess I can use it as a tool to "generate" more terms I know (from physical principles) have zero divergence. Right?

So going along that idea (and using some hints from wikipedia), does this look right?
(interspersed with blatant copying from wikipedia, to speed up the math)

Consider the action
[tex] S = \int (\frac{1}{2\kappa}R_{ab}R^{ab} + \mathcal{L}_m) \sqrt{-g} d^4x[/tex]

The variation of the determinant is:
[tex]\delta \sqrt{-g}= -\frac{1}{2} \sqrt{-g} g^{ab} \delta g_{ab}[/tex]

The variation of the curvature scalar is:
[tex]\begin{align*}
\delta (R_{ab}R^{ab}) &= R^{ab} \delta R_{ab} + R_{ab} \delta R^{ab} \\
&= 2 R^{ab} \delta R_{ab} \\
&= 2 R^{ab}(\nabla_c \delta \Gamma^c_{ab} - \nabla_b \delta \Gamma^c_{c a})
\end{align*}
[/tex]

Since [tex]\delta \Gamma^\lambda_{\mu\nu}\,[/tex] is actually the difference of two connections, it should transform as a tensor. Therefore, it can be written as
[tex]\delta \Gamma^\lambda_{\mu\nu}=\frac{1}{2}g^{\lambda d}\left(\nabla_\mu\delta g_{d\nu}+\nabla_\nu\delta g_{d\mu}-\nabla_d\delta g_{\mu\nu} \right)[/tex]

and substituting in the equation above one finds:
[tex]\begin{align*}
\delta R_{ab} &= [\nabla_c \delta \Gamma^c_{ab} - \nabla_b \delta \Gamma^c_{c a}] \\
&= [\nabla_c \frac{1}{2}g^{c d}\left(\nabla_a\delta g_{db}+\nabla_b\delta g_{da}-\nabla_d\delta g_{ab} \right)
-\nabla_b \frac{1}{2}g^{c d}\left(\nabla_c\delta g_{da}+\nabla_a\delta g_{dc}-\nabla_d\delta g_{ca} \right)] \\
&= \frac{1}{2}g^{c d}[\left(\nabla_c\nabla_a\delta g_{db}+\nabla_c\nabla_b\delta g_{da}-\nabla_c\nabla_d\delta g_{ab} \right)
-\left(\nabla_b\nabla_c\delta g_{da}+\nabla_b\nabla_a\delta g_{dc}-\nabla_b\nabla_d\delta g_{ca} \right)] \\
&= \frac{1}{2}g^{c d}[\nabla_c\nabla_a\delta g_{db}-\nabla_c\nabla_d\delta g_{ab} - \nabla_b\nabla_a\delta g_{dc} + \nabla_b\nabla_d\delta g_{ca}]
\end{align*}
[/tex]
(hmm... somehow the overall sign is off compared to the f(R) gravity wiki article. But magnitudes match.)

In the action, this term in integrated, and via integration by parts, we can move the derivatives from on the metric piece to on the R_ab piece.

After the integration by parts, we have:

[tex] \begin{align*}
\delta [\sqrt{-g} R_{ab} R^{ab}] &= \sqrt{-g}[ - \delta g_{ab}\frac12 g^{ab} R^{cd}R_{cd} +
[ \delta g_{db}\nabla_c\nabla_a g^{c d} R^{ab}-\delta g_{ab}\nabla_c\nabla_d g^{c d} R^{ab} - \delta g_{dc}\nabla_b\nabla_a g^{c d} R^{ab} + \delta g_{ca}\nabla_b\nabla_d g^{c d} R^{ab}
]] \\
&= \sqrt{-g}[ - \delta g_{ab}\frac12 g^{ab} R^{cd}R_{cd} +
[ \delta g_{ab}\nabla_c\nabla_d g^{c a} R^{db}-\delta g_{ab}\nabla_c\nabla_d g^{c d} R^{ab} - \delta g_{ab}\nabla_c\nabla_d g^{b a} R^{dc} + \delta g_{ba}\nabla_c\nabla_d g^{b d} R^{ac}
]] \\
&= \sqrt{-g}\delta g_{ab}[ - \frac12 g^{ab} R^{cd}R_{cd} +
\nabla_c\nabla_d g^{c a} R^{db} - \nabla_c\nabla_d g^{c d} R^{ab} - \nabla_c\nabla_d g^{b a} R^{dc} + \nabla_c\nabla_d g^{b d} R^{ac}
] \\
&= \sqrt{-g}\delta g_{ab}[ - \frac12 g^{ab} R^{cd}R_{cd} +
\nabla^a\nabla_d R^{db} - \nabla^d\nabla_d R^{ab} - g^{b a} \nabla_c\nabla_d R^{dc} + \nabla_c\nabla^b R^{ac}
] \\
&= \sqrt{-g}\delta g_{ab}[ - \frac12 g^{ab} R^{cd}R_{cd} +
\nabla^a\frac12 g^{db}R_{;d} - \nabla^d\nabla_d R^{ab} - g^{b a} \nabla_c\frac12 g^{dc}R_{;d} + \nabla^b \frac12 g^{ac} R_{;c}
] \\
&= \sqrt{-g}\delta g_{ab}[ - \frac12 g^{ab} R^{cd}R_{cd} +
\frac12\nabla^a\nabla^b R - \nabla^d\nabla_d R^{ab} - g^{b a} \frac12 \nabla^d\nabla_d R + \frac12 \nabla^a\nabla^b R
] \\
&= \sqrt{-g}\delta g_{ab}[ - \frac12 g^{ab} R^{cd}R_{cd} +
\nabla^a\nabla^b R - \nabla^d\nabla_d ( R^{ab} + \frac12 g^{ab} R)
]
\end{align*}
[/tex]

So, the field equations for this "theory" are:
[tex]- \frac12 g^{ab} R^{cd}R_{cd} +
\nabla^a\nabla^b R - \nabla^d\nabla_d ( R^{ab} + \frac12 g^{ab} R) = \kappa T^{ab}[/tex]

The whole point of this was to find the divergence of metric * scalar invariant. So this method seems to have generated the result that:

[tex]- (\frac12 g^{ab} R^{cd}R_{cd})_{;b} +
\nabla^a\nabla^b R_{;b} - \nabla^d\nabla_d ( {R^{ab}}_{;b} + \frac12 g^{ab} R_{;b}) = 0[/tex]
[tex]- (\frac12 g^{ab} R^{cd}R_{cd})_{;b} +
\nabla^a\nabla^b R_{;b} - \nabla^d\nabla_d ( \frac12 g^{ab} R_{;b} + \frac12 g^{ab} R_{;b}) = 0[/tex]
which simplifying is
[tex] \begin{align*} (g^{ab} R^{cd}R_{cd})_{;b}
&= 2 (\nabla^a\nabla^b R_{;b} - \nabla^d\nabla_d g^{ab}R_{;b}) \\
&= 2 (\nabla^a\nabla^b\nabla_b R - \nabla^d\nabla_d \nabla^a R) \\
&= 0
\end{align*}
[/tex]

Is this really correct?
That's a pretty neat way of generating divergence equations. But it seems pretty involved.

As an aside, does:
[tex]- \frac12 g^{ab} R^{cd}R_{cd} + \nabla^a\nabla^b R - \nabla^d\nabla_d ( R^{ab} + \frac12 g^{ab} R) = \kappa T^{ab}[/tex]
have compatible vacuum solutions with:
[tex]R^{ab} - \frac12 g^{ab} R = \kappa T^{ab}[/tex]
If so, I guess I answered the other goal of playing with these as well. But I'm not sure how to say difinitively either way. I mean
[tex]R^{ab} = 0[/tex]
clearly solves
[tex]- \frac12 g^{ab} R^{cd}R_{cd} + \nabla^a\nabla^b R - \nabla^d\nabla_d ( R^{ab} + \frac12 g^{ab} R) = 0[/tex]
[tex]- 2 R^{cd}R_{cd} + \nabla^a\nabla_a R - \nabla^d\nabla_d ( R + 2 R) = 0[/tex]
[tex]R^{cd}R_{cd} = -\nabla^a\nabla_a R[/tex]
but I don't know if boundary conditions, or something else I'm ignoring, would screw things up.

EDIT: Needed to fix quite a few mistakes and typos.

 

[haushofer]

Are you looking for Gauss-Bonnet terms in the action, perhaps?

 

[JustinLevy]

I was using that action merely as a way to calculate the divergence of some terms. Just an idea that popped in my head after thinking about the f(R) theories some more after atyy mentioned them. I'm not even sure if that is a valid method to calculate divergences ... but if so, it is an interesting method for generating divergence relations.

Can someone skim through my reasoning in the last post and tell me if that is even a valid method?

And even better, while it may be tedious, would someone mind checking my math there? It seems really strange to me that:
[tex](g^{cd}R^{ab}R_{ab})_{;d} = 0[/tex]
and therefore [tex]\nabla^c(R^{ab}R_{ab}) = 0[/tex]
The curvature scalar is covariantly constant!? How can that be true? And if it is, what does that mean? Should this have been obvious from a different line of attack (maybe something along the lines of the openning post, which I couldn't figure out what to do with)?

 

[Altabeh]

The variation of the curvature scalar is:
[tex]\begin{align*}
\delta (R_{ab}R^{ab}) &= R^{ab} \delta R_{ab} + R_{ab} \delta R^{ab} \\
&= 2 R^{ab} \delta R_{ab} \\
&= 2 R^{ab}(\nabla_c \delta \Gamma^c_{ab} - \nabla_b \delta \Gamma^c_{c a})
\end{align*}
[/tex]

If we calculate the variation of the scalar curvature [tex]R=R_{\mu\nu}g^{\mu\nu}[/tex], one gets

[tex]\begin{align*}
\delta (R_{\mu\nu}g^{\mu\nu}) &= R_{\mu\nu} \delta g^{\mu\nu}+g^{\mu\nu}\delta R_{\mu\nu} \\
&= R_{\mu\nu} \delta g^{\mu\nu}+\nabla_{\kappa}(g^{\mu\nu}\delta\Gamma_{\mu\nu}^{\kappa}-g^{\mu\kappa}\delta\Gamma_{\mu\rho}^{\rho}). \\
\end{align*}
[/tex]
So the further calculations all have to be incorrect!

AB

 

[JustinLevy]

So the further calculations all have to be incorrect!

R is not the only curvature scalar. It should be very clear from what I wrote that I was talking about the curvature scalar [tex]R_{ab}R^{ab}[/tex] not R. So I think you misunderstood what I trying to do.

Anyway, I did find some errors. There are a couple places where I was sloppy with order of operations, which is important since covariant derivatives don't commute. I also dropped a term in the variation of R_ab R^ab. I also tracked down where that sign error was coming from, which is a minor fix. So I need to rewrite those calculations.

I realize tracking though my math is probably tedious, but the general line of argument is fairly simple:
1) I take a Lagrangian that doesn't have explicit time or spatial dependence.
2) I therefore expect the stress energy should be divergence-less, and hence can get a new relation on the divergence of some curvature terms.

So even without trudging through the math, I hope some people here can comment.
Is this method valid? If not, I don't really want to take the time to add in those extra terms to fix the calculation.

Either way, the point of this thread was to learn how to calculate the divergence of terms like (as I said in the openning post)
[tex]g^{\mu\nu}R_{ab}R^{ab}[/tex]
if someone knows another way to generate divergence relations of curvature terms besides starting with the second Bianchi identity, please do let me know.

 

[Altabeh]

R is not the only curvature scalar. It should be very clear from what I wrote that I was talking about the curvature scalar [tex]R_{ab}R^{ab}[/tex] not R. So I think you misunderstood what I trying to do.

Yet again it is not true, to the best of my knowledge! The variations of [tex]R^{ab}[/tex] and its covariant form are not the same at all! You are putting both equal and this gives rise to an incorrect expression for the so-called "curvature scalar" you've defined here!

AB

 

[JustinLevy]

Yet again it is not true, to the best of my knowledge! The variations of [tex]R^{ab}[/tex] and its covariant form are not the same at all! You are putting both equal and this gives rise to an incorrect expression for the so-called "curvature scalar" you've defined here!

AB

I thank you for responding to my threads, but I wish you would take a bit more time reading what I wrote. As I said in the last post: "I also dropped a term in the variation of R_ab R^ab." Also, I am not sure why you feel the need to refer to it as a 'so-called "curvature scalar"' ... it IS a curvature scalar. There are many curvature scalars
http://en.wikipedia.org/wiki/Carminati-McLenaghan_invariants
http://en.wikipedia.org/wiki/Curvature_invariant_(general_relativity)

Regarding my comment about dropping a term, if we look at:
[tex] \delta (R^{ab}R_{ab}) = \delta (R_{cd}g^{ca}g^{db}R_{ab}) = 2 R_{cd}g^{ca}g^{db} \delta R_{ab} + 2 R_{cd}g^{ca} R_{ab} \delta g^{db} = 2 R^{ab}\delta R_{ab} + 2 g^{ca}R_{cd}R_{ab} \delta g^{db}[/tex]
you can see that I dropped the second term, which being already proportional to the metric variation adds a term directly to the field equations.

Anyway, I found an article which states the result of considering the curvature scalars:
[tex]R, R^{ab}R_{ab}, R^{abcd}R_{abcd}[/tex]
http://arxiv.org/abs/astro-ph/0410031

So let's skip the math and return to my question:
The general line of argument is fairly simple,
1) I take a Lagrangian that doesn't have explicit time or spatial dependence.
2) I therefore expect the stress energy should be divergence-less, and hence can get a new relation on the divergence of some curvature terms.

So even without trudging through the math, I hope some people here can comment.
Is this method valid?

 

[Altabeh]

I thank you for responding to my threads, but I wish you would take a bit more time reading what I wrote. As I said in the last post: "I also dropped a term in the variation of R_ab R^ab." Also, I am not sure why you feel the need to refer to it as a 'so-called "curvature scalar"' ... it IS a curvature scalar. There are many curvature scalars
http://en.wikipedia.org/wiki/Carminati-McLenaghan_invariants
http://en.wikipedia.org/wiki/Curvature_invariant_(general_relativity)

Yes, but look at the the notation! Later on in your calculations, you make use of [tex]R[/tex] as the curvature scalar within the field equations and I can't understand how that is going to be possible to introduce the fundamental laws of GR with this [tex]R.[/tex] Either you've written the post in a bad way or I'm getting it all wrong, this confusion in the text must be somehow cleared out so I can tell you if the method works or not!

Regarding my comment about dropping a term, if we look at:
[tex] \delta (R^{ab}R_{ab}) = \delta (R_{cd}g^{ca}g^{db}R_{ab}) = 2 R_{cd}g^{ca}g^{db} \delta R_{ab} + 2 R_{cd}g^{ca} R_{ab} \delta g^{db} = 2 R^{ab}\delta R_{ab} + 2 g^{ca}R_{cd}R_{ab} \delta g^{db}[/tex]
you can see that I dropped the second term, which being already proportional to the metric variation adds a term directly to the field equations.

Your problem is that you are just on a roll when doing all calculations and seems like you don't even trouble yourself to check what you've written clearly right after everything is done! I don't see in your post #7 anything impying the fact that you dropped the second term in the equations!

So even without trudging through the math, I hope some people here can comment.
Is this method valid?

Anyways, if the math is seamless, why not? We can admot a new field equation with a new curvature invariant as once Einstein published his own and we read it and started to believe in it! But there are so many other problems with the math: the following is completely erroneous and misleading:

[tex]- (\frac12 g^{ab} R^{cd}R_{cd})_{;b} + \nabla^a\nabla^b R_{;b} - \nabla^d\nabla_d ( {R^{ab}}_{;b} + \frac12 g^{ab} R_{;b}) = 0.[/tex]

This is the first time I see something like this.

AB

 

[JustinLevy]

Ugh. Altabeh, again, I appreciate that you take the time to write a response... but it is incredibly frustrating when you reply without apparently taking the time to read what I have even written or am asking. It is very frustrating. Every post of yours in this thread, and a couple in other threads have been like that.

Let me give examples in this thread:
From the opening post, and from the very start of my calculations in post #7, it should have been clear that I was trying to work out a divergence relation involving the divergence of a term metric * the scalar R_ab R^ab.

You write a post saying everything is wrong because you feel the scalar I want is R. And therefore dismiss everything. Clearly showing you didn't even read enough to get the basics of what I am even asking.

---
I then respond explaining again that the curvature scalar I'm focussing on is R_ab R^ab. I also point out that I've redone some of the calculations and found multiple errors. I list some of those errors. I then point out that one can ignore the math for now, as the more helpful information is whether this method for obtaining divergence relations is valid at all.

You again seemingly ignore most of that and focus on the old math instead ... and again ignoring what I wrote, you even complain about an error I already pointed out. And yet again complain about the scalar I'm focussing on, even casting doubt on it by referring to it as a 'so-called "curvature scalar"'.

-----
I respond again. Stressing yet again the curvature scalar I want to focus on is R_ab R^ab. Since you complained about errors, I stated more explicitly this time one of the errors. I again stress that focussing on the math is not as helpful as telling me whether the general method is viable. I summarize this method yet again. I even post a paper I found that gives the result I should find if I go back and fix my errors, so that we can stop focussing on the details of working out the field equations.

What do you do? You yet AGAIN complain about the scalar. You AGAIN refer to errors in my math all the way back in post 7 (which I've already pointed out myself I found errors in), and again ignoring the important parts of the discussion.
-----

Do you see the pattern here?
You seem very knowledgeable. And again, I thank you for taking the time to respond. But it is not useful, and in fact incredibly frustrating, when you seemingly ignore what I write. So please, read a bit more carefully otherwise we'll just be talking past each other every time.

Your problem is that you are just on a roll when doing all calculations and seems like you don't even trouble yourself to check what you've written clearly right after everything is done!

I felt the need to respond to this though. Yes, I've made mistakes. But please don't take this as "you don't even trouble yourself to check what you've written". I put a lot of time into writing up those calculations. And I have checked them and I've caught many mistakes myself, and I can see how to fix them. Since I found a paper that gave the final answer (albeit without showing the work), and I found where my mistakes were, I didn't feel the need to retype up all the calculations.


So, back to the discussion:
I've checked by hand for f(R) gravity that the field equations are divergenceless in general (and not just in solutions to the equations), as I expected. But it would be nice to know that this method works in general. To put in explicitly...

If I do the following steps:
1) put a function of any scalars I want in replace of f(R) in the f(R) lagrangian
2) solve for the field equations
3) look at the vacuum case (T_uv = 0)
4) take the divergence of the vacuum field equation, thus obtaining a relationship between the divergence of many curvature parts

Is the relation I obtain from this method true in general or only true for solutions of those field equations. I've already made heuristic arguments for why I expect them to be true in general, and therefore this is a valid method to generate relationships between the divergence of curvature terms ... but it would be nice to know for sure (and explicitly why) this is the case.

 

[Altabeh]

Let me give examples in this thread:
From the opening post, and from the very start of my calculations in post #7, it should have been clear that I was trying to work out a divergence relation involving the divergence of a term metric * the scalar R_ab R^ab.

You write a post saying everything is wrong because you feel the scalar I want is R. And therefore dismiss everything. Clearly showing you didn't even read enough to get the basics of what I am even asking.

You seem to have forgotten to take a quick peep at your post #7 in this thread: All you said there was that you introduced an scalar but called it inattentively "scalar curvature" and assigned to it the same symbol we use for the "scalar curvature" [tex]g^{\mu\nu}R_{\mu\nu}[/tex] in GR so when I asked for the reason behind defining R in this way, you suggested me to take a look at this http://en.wikipedia.org/wiki/Curvature_invariant_(general_relativity)" where the curvature invariants are discussed briefly without any reference to the fact that all of them use distinctive symbols other than [tex]R.[/tex] So am I given this right to get confused when seeing your posts and misleading notations!?

You again seemingly ignore most of that and focus on the old math instead ... and again ignoring what I wrote, you even complain about an error I already pointed out. And yet again complain about the scalar I'm focussing on, even casting doubt on it by referring to it as a 'so-called "curvature scalar"'.

The reason is that you looked like you got us led to the EFE starting from this [tex]R[/tex] ending up with the scalar curvature included in the equations. I don't know what makes it all silly to you to have questions like this formed in mind after reading your convoluted post!

What do you do? You yet AGAIN complain about the scalar. You AGAIN refer to errors in my math all the way back in post 7 (which I've already pointed out myself I found errors in), and again ignoring the important parts of the discussion.

Right after you were born, what was your language to be understandable to your parents!? Nothing and it is the reason the baby language is just an inspiration of the inward feelings under the circumstances the baby sees at that moment! Speaking of which, for physics to be understood well, at this level of prestige, the standard language is mathematics but your calculations are all leaky and all we can do about the procedure is to lean on our inward feelings about it unless you provide us with a great indefectible mathematical framework because you're not talking about something really elementary but fully scientific and professional in the real sense of the words! I think the answer I gave in an early post gets everything straight:

Anyways, if the math is seamless, why not? We can admit a new field equation with a new curvature invariant as once Einstein published his own and we read it and started to believe in it! But there are so many other problems with the math:

Do you see the pattern here?
You seem very knowledgeable. And again, I thank you for taking the time to respond. But it is not useful, and in fact incredibly frustrating, when you seemingly ignore what I write. So please, read a bit more carefully otherwise we'll just be talking past each other every time.

As the title of this thread says all we are going to talk about is mathematical so if I got the time to spend on checking the math, I'd be glad to announce the results soon! Remember that all you're doing here is strongly backed up by mathematics and if this part isn't seamless, I can't even predict what'll happen to the theory in the end!

If I do the following steps:
...
3) look at the vacuum case (T_uv = 0)
...

Your theory has to be divergence-less and this can be obtained by taking the covariant derivative of the left-hand side of the field equations without even looking at the other side. This step compels the theory 1) go astray from the main road, i.e. the generality of equations for any material distribution given, and 2) nullifies the effect of "divergence" because it's already zero! It is mandatory for this step to be modified to give the correct equations! Also

4) take the divergence of the vacuum field equation, thus obtaining a relationship between the divergence of many curvature parts

will only be valid then in a vacuum spacetime.

Is the relation I obtain from this method true in general or only true for solutions of those field equations. I've already made heuristic arguments for why I expect them to be true in general, and therefore this is a valid method to generate relationships between the divergence of curvature terms ... but it would be nice to know for sure (and explicitly why) this is the case.

If your theory makes the following general demands met, then it works fine:

1- The modification I'm taking about is a necessity for your theory to be "general";
2- The relation you obtained, i.e.

[tex] \begin{align*} (g^{ab} R^{cd}R_{cd})_{;b} &= 2 (\nabla^a\nabla^b R_{;b} - \nabla^d\nabla_d g^{ab}R_{;b}) \\&= 2 (\nabla^a\nabla^b\nabla_b R - \nabla^d\nabla_d \nabla^a R) \\&= 0\end{align*}[/tex]

has many wrong things with it! If this was flawless mathematically, your theory at least in a limited sense would rock!

Are these helpful to you or yet you think I'm stuck on nagging at your [tex]R[/tex]?

AB

 

[JustinLevy]

You seem to have forgotten to take a quick peep at your post #7 in this thread: All you said there was that you introduced an scalar but called it inattentively "scalar curvature" and assigned to it the same symbol we use for the "scalar curvature" [tex]g^{\mu\nu}R_{\mu\nu}[/tex] in GR so when I asked for the reason behind defining R in this way, you suggested me to take a look at this http://en.wikipedia.org/wiki/Curvature_invariant_(general_relativity)" where the curvature invariants are discussed briefly without any reference to the fact that all of them use distinctive symbols other than [tex]R.[/tex] So am I given this right to get confused when seeing your posts and misleading notations!?

I am using standard notation.
[tex]R_{ab}[/tex] is used to refer to the Ricci curvature. The curvature scalar I am talking about is, once again,
[tex]R^{ab}R_{ab}[/tex]

which is of course not
[tex]R=g^{ab}R_{ab}[/tex]
the Ricci scalar.

It is clear from the very first line in this thread what curvature scalar I was referring to. It baffles me that you are complaining that I said in post #7 "variation of the curvature scalar" and then I wrote the curvature scalar I was referring to once again ... how can you complain that is unclear even now? Fine, you misunderstood. But I've tried to correct your misunderstanding several times now.

The fact that you are now starting to use "R" to refer to the curvature scalar I was referring to, is only making me worry you still don't understand. Especially since I am using standard tensor notation here. There should be no ambiguity.

There are many curvature scalars. If you need to give this one a name, wikipedia refers to it as "The principal quadratic invariant of the Ricci tensor" in that link I gave you. R^{ab}R_{ab} is unambiguous and shorter.

I hope we have finally settled any notation issues.

I think the answer I gave in an early post gets everything straight:

Anyways, if the math is seamless, why not? We can admit a new field equation with a new curvature invariant as once Einstein published his own and we read it and started to believe in it! But there are so many other problems with the math:

That is completely missing the point of what I'm trying to get here. I'm generating field equations from a Lagrangian, and using the method I've described several times now, in hopes to generate general relations between curvature terms. I'm not using the Lagrangian to suggest new physics.

And in that post, as in even your latest post, you again continue to complain about the specific math in post 7 which is frustrating, considering I already pointed out myself that there were errors and therefore we should focus on whether the method is viable. Heck I even posted a paper containing the correct field equations that I would obtain if I went back and fixed the errors.

As the title of this thread says all we are going to talk about is mathematical so if I got the time to spend on checking the math, I'd be glad to announce the results soon! Remember that all you're doing here is strongly backed up by mathematics and if this part isn't seamless, I can't even predict what'll happen to the theory in the end!

If you insist on seeing the correct field equations, let me link the paper again:
http://arxiv.org/PS_cache/astro-ph/pdf/0410/0410031v2.pdf
equation 15 are the field equations

Going through all the math to get the field equations is unimportant if the method I'm proposing to use these to get general divergence relations is wrong. I have done no math to prove the method I suggest is valid, only given heuristic arguments. I don't know how to prove one way or the other (besides getting the equations and checking by hand that they have the property I expected). If you know how to do either of these mathematically:
1) How to calculate the covariant gradient of a non-tensor object (like a Christoffel symbol), in which case I could calculate any gradient of a curvature scalar built from the Reimann curvature directly.
or
2) How to prove that the method I'm proposing to generate divergence relations, which hold in general (and not just in some specific theory), will work.
Then that would be of great help, and I'd love to hear those math details.
Complaining about mistakes in post #7, that I've already pointed out myself and furthermore linked to the correct equations later, is not helpful.

[...in response to step 3...]
Your theory has to be divergence-less and this can be obtained by taking the covariant derivative of the left-hand side of the field equations without even looking at the other side. This step compels the theory 1) go astray from the main road, i.e. the generality of equations for any material distribution given, and 2) nullifies the effect of "divergence" because it's already zero! It is mandatory for this step to be modified to give the correct equations! Also

[...in response to step 4...]
will only be valid then in a vacuum spacetime.

I'm sorry. I've reread this multiple times. I cannot figure out what you are trying to say here.

Are you saying the field equations obtained this way "have to be divergence-less" due to the mathematical way in which they were derived? If so, can you please show me how you know that? That is what I've been asking about this method.

But then you go on to say that looking at the vacuum case "compels the theory 1) go astray from the main road". And that "It is mandatory for this step to be modified to give the correct equations!" Modified to what? Astray from what? I think you are still misunderstanding what this thread is about. I am trying to obtain relations between divergences of curvature terms. And I hit about this some-what bizarre method to generate them, and I'm pursuing it because I don't know how to find them another way (and no one has suggested an alternate way).

The final comment you make there is that what I obtain will NOT be relations which hold in general. So you seem to strongly disagree with my method, but this seems to contradict your first statement that the equations must be divergence-less.

Did you only mean they "have to be divergence-less" in the sense that for the field equations to make physical sense the solutions to the equations must be divergence-less?

 

[Mentz114]

Are you saying the field equations obtained this way "have to be divergence-less" due to the mathematical way in which they were derived? If so, can you please show me how you know that? That is what I've been asking about this method.

I believe this is the case but I can't get corroboration. I'm pretty sure the Einstein tensor obeys [itex]{G^{\mu\nu}}_{;\nu}=0[/itex] identically. Any canonical EMT obeys this, so I think G must also.

[Edit]Got it. It's confirmed here

http://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry

It follows from the Bianchi identity of course !

 

[JustinLevy]

I believe this is the case but I can't get corroboration.

Yeah, it really seems like it should, and every case I've been able to work out by hand has worked so far, but I don't know how to prove it.

I'm pretty sure the Einstein tensor obeys [itex]{G^{\mu\nu}}_{;\nu}=0[/itex] identically. Any canonical EMT obeys this, so I think G must also.

[Edit]Got it. It's confirmed here

http://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry

It follows from the Bianchi identity of course !

Yeah, that's probably the easiest case. Unfortunately that series of contractions on the Bianchi identity is the only useful one. So to get anymore equations relating the divergence of curvature terms, I need to resort to something more drastic.

The [itex]{G^{\mu\nu}}_{;\nu}=0[/itex] is the relation we get when using the gravitational action:
[tex]S= - {1 \over 2\kappa}\int R \sqrt{-g} d^4x[/tex]

I'm hoping that replacing R with any function of curvature scalars will yield "field equations" that are in general divergence-free. In vacuum the solutions must be, so I'm worried that all this is showing is that solutions are divergence free ... and thus I can't use this to get a new set of divergence relations for curvature terms.

However, since the action would have no explicit time or spatial dependence regardless of the function of curvature scalars chosen, I'd expect that the field equations from such a modified lagrangian would be divergence free. As I mentioned previously, I've check it a couple cases and it has been true so far. Maybe it is a simple step to show that this is always the case, so any guidance along that direction would be much appreciated.

 

[Altabeh]

I am using standard notation.
[tex]R_{ab}[/tex] is used to refer to the Ricci curvature. The curvature scalar I am talking about is, once again,
[tex]R^{ab}R_{ab}[/tex]

which is of course not
[tex]R=g^{ab}R_{ab}[/tex]
the Ricci scalar.

It is clear from the very first line in this thread what curvature scalar I was referring to. It baffles me that you are complaining that I said in post #7 "variation of the curvature scalar" and then I wrote the curvature scalar I was referring to once again ... how can you complain that is unclear even now? Fine, you misunderstood. But I've tried to correct your misunderstanding several times now.

Let's face it: You as someone who does not deal with physics on a regular basis (at least I can smell it from the extraordinarily flawed mathematics of yours), just defined something in your own sense and kept talking about it in some post! Okay and there is no problem with this, but what do you expect from the reader when you bring up the Einstein's field equations (EFE) with a completely different [tex]R[/tex] in the same post just a few lines right below an streak of equations involved with your own [tex]R[/tex]!? What does it mean and what should we do there to find a comparison between them while there is such huge ambigiuty and a nerve-wracking violation of energy\momentum conservation?

So your notation is NOT satandard and to keep notations in agreement with the standards established in that paper you just cited in your latest post, you would have used [tex]P[/tex] instead of [tex]R[/tex] which, as in the same paper, refers basically to the standard scalar curvature [tex]g^{ab}R_{ab}.[/tex]

The fact that you are now starting to use "R" to refer to the curvature scalar I was referring to, is only making me worry you still don't understand. Especially since I am using standard tensor notation here. There should be no ambiguity.

No quite contrary! You seem to first have gone way far from the topic, and secondly I don't assume you're trying to insult me by making a duffer out of me! People need to tease out things that sound unclear to them and if this bothers you, I'll be out of here because I don't want to be gone off on this way anymore! I'm here to help so just stay cool and keep to the subject!

That is completely missing the point of what I'm trying to get here. I'm generating field equations from a Lagrangian, and using the method I've described several times now, in hopes to generate general relations between curvature terms. I'm not using the Lagrangian to suggest new physics.

Considering all the mathematics in your post #7 has serious problems in it, the relation you're giving, i.e.

[tex]- (\frac12 g^{ab} R^{cd}R_{cd})_{;b} + \nabla^a\nabla^b R_{;b} - \nabla^d\nabla_d ( {R^{ab}}_{;b} + \frac12 g^{ab} R_{;b}) = 0[/tex]

has only one way of being interpreted, say, logically and physically when compared to the field equations obtained in the paper "The Cosmology of Generalized Modified Gravity Models" as follows: The mathematics, as I examined this discreetly myself, does not prove the relation right and so you have to go with some other alternative which is the same method the authors of the article above take advantage of. The method is strongly based on a new parameter [tex]\mu[/tex] appearing in a sort of famous Lagrangian used earlier by many other authors (see, for instance, the equation (1) in http://arxiv.org/PS_cache/astro-ph/pdf/0306/0306438v2.pdf" This value since is present as a multiple in the correction terms, can be adjusted the way that

1) the amount of non-conserved quantities must get really small (this is shown to be of order 10^{-33}eV for a slow-expanding universe so the violation is really negligible);

2) the model keeps to be on a par with both observational data and the property of self-consistency of EFE (this means that invoking the conservation laws to hold in the new theory has nothing to do with the energy tensor being zero or not. They must be satisfied in any form).

Besides these, the EFE must be extractable from the new Lagrangian by setting the new parameter equal to some constant, basically, zero! But in your Lagrangian none of these are obeyed. Specially that your Lagrangian does not reduce to the Einstein-Hilbert Lagrangian.Also, we don't have a degree of freedom to adjust it according to the model under study. I feel like you changed your mind about using the primitive idea of getting some free values involved in the field equations.

Going through all the math to get the field equations is unimportant if the method I'm proposing to use these to get general divergence relations is wrong. I have done no math to prove the method I suggest is valid, only given heuristic arguments. I don't know how to prove one way or the other (besides getting the equations and checking by hand that they have the property I expected). If you know how to do either of these mathematically:
1) How to calculate the covariant gradient of a non-tensor object (like a Christoffel symbol), in which case I could calculate any gradient of a curvature scalar built from the Reimann curvature directly.
or
2) How to prove that the method I'm proposing to generate divergence relations, which hold in general (and not just in some specific theory), will work.
Then that would be of great help, and I'd love to hear those math details.

For what reason do we need to calculate all of them when we know that the theory is vacuous? The basic Lagrangian must be modified in such a way that the EFE be extracted from it when all the correction terms vanish. Yours is not even right in the sign of the second term on the left hand side of the EFE.

Complaining about mistakes in post #7, that I've already pointed out myself and furthermore linked to the correct equations later, is not helpful.

If you have no knowledge of math, why did you write all those void equations in post #7? Was that a practice essay or a play-around with equations? If you don't know that we spend time to read them, you better never think of writing nonsense again!

Are you saying the field equations obtained this way "have to be divergence-less" due to the mathematical way in which they were derived? If so, can you please show me how you know that? That is what I've been asking about this method.

They have to be, at the very very least, minimally divergence-less as they are in Carroll's papers. But in your Lagrangian this is not possible because first a mathematical derivation cannot be obtained and secondly there is nothing to be adjusted and the Lagrangian itself does not reduce to the fundamental Lagrangian of GR. This property is indeed essential to constructing a self-consistent field equation and if it is not cared about, you have to force the material distribution to consist of dark energy to compensate for the probable lost mass\momentum in interactions with ordinary matter. Looking for a relation between scalar curvature terms in the way you are into, is not feasible unless a necessary modification of the proposed Lagrangian is carried out!

Astray from what?

Going astray from keeping the theory at minimal level of the violation of the conservations laws.

Modified to what?

Talked about it above.

The final comment you make there is that what I obtain will NOT be relations which hold in general . So you seem to strongly disagree with my method, but this seems to contradict your first statement that the equations must be divergence-less.

What do you mean? Explain elaborately.

Did you only mean they "have to be divergence-less" in the sense that for the field equations to make physical sense the solutions to the equations must be divergence-less?

As I said, being divergence-free must be at least MINIMALLY satisfied. If not, the theory, the relations to be obtained and anything else you want to get from this method will be incorrect. How does your theory guarantee the occurring of such thing within it!?

AB

 

[Altabeh]

I believe this is the case but I can't get corroboration. I'm pretty sure the Einstein tensor obeys [itex]{G^{\mu\nu}}_{;\nu}=0[/itex] identically. Any canonical EMT obeys this, so I think G must also.

[Edit]Got it. It's confirmed here

http://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry

It follows from the Bianchi identity of course !

Thanks for sharing this with us. But to be a little bit more stringent, let's say that the modified theories or those cosmological solutions with a distance-varying cosmological constant (see the two-body models for the universe), where the cosmological constant varies as the square of the distance of the bodies, the conservation law of GR, [tex]{T^{\mu\nu}}_{;\nu}=0[/tex] or, equivalently, [tex]{G^{\mu\nu}}_{;\nu}=0[/tex] does not hold. So two theories come to mind as the solution:

[tex]\Lambda[/tex] is either strictly zero or so small that the term [tex]\Lambda g_{\mu\nu}[/tex] in the field equations has negligible consequences in the planetary motions. This slight non-conservation of energy and momentum does not have any remarkable effect on the whole of the theory so the modified gravity theories are to be given the right to be proposed.

AB

 

[JustinLevy]

Altabeh,
I don't know what your problem is. You either have atrocious reading comprehension, or you are just picking for a fight.

Let's face it: You as someone who does not deal with physics on a regular basis (at least I can smell it from the extraordinarily flawed mathematics of yours), just defined something in your own sense and kept talking about it in some post!

I made many math mistakes in post #7. I found them later and pointed them out myself in a later post. If it makes you feel better to insult me and continue bringing up past math mistakes I've made, so be it. This however does NOT make your argument correct. That is called argument by ad hominem and is prohibited on this forum.

What I WANT to know, and would be HELPFUL to this discussion is whether the METHOD I'm trying to apply is mathematically correct. If you could prove mathematically that the method is correct or flawed, that would be HELPFUL. Continuing to harp on mistakes not even pertinent to the discussion (because I already found a paper, and posted a free link to it, that worked out the equations I was trying to get in post #7) is not helpful.

what do you expect from the reader when you bring up the Einstein's field equations (EFE) with a completely different [tex]R[/tex] in the same post just a few lines right below an streak of equations involved with your own [tex]R[/tex]!?

How can you still not understand this?
I am not using non-standard notation. Please stop calling the curvature scalar I am referring to as R, for I never have.

Look at the openning post. Look even at post #7. It is clear what curvature scalar I am referring to. Seriously, let's look at post #7. In post 7 I refer explicitly to a couple theories of gravity which start with actions, which involve more complicated curvature scalars (not just R), before I write my first equation which was:

Consider the action
[tex] S = \int (\frac{1}{2\kappa}R_{ab}R^{ab} + \mathcal{L}_m) \sqrt{-g} d^4x[/tex]

There is no non-standard notation in that equation. If you are saying otherwise, then you don't understand the notation. Is this whole thing just a drawn out argument due to the fact that you don't understand this notation?

I'll try to explain it if that would help. Maybe you are confused because the letter "R" is used in standard notation for several things. If there are four tensor indices on it in an equation, it is referring to the Reimannian curvature tensor. There is only one independent tensor that can be obtained from the Reimannian tensor with a contraction with the metric which can be non-zero, and this is called the Ricci curvature tensor. So if there are two tensor indices on a letter "R" in an equation, it is referring to the Ricci curvature tensor. Contract this again with the metric tensor and you get the Ricci curvature scalar. So if there are no tensor indices on a letter "R" in an equations it is referring to the Ricci curvature scalar.

Therefore note that it is the Ricci curvature tensor, NOT the Ricci curvature scalar in that equation.

If this is still confusing to you, feel free to ask. But please start listening. It is frustrating that you still are making the same incorrect claims.

What does it mean and what should we do there to find a comparison between them while there is such huge ambigiuty and a nerve-wracking violation of energy\momentum conservation?

So you ARE claiming that action leads to field equations (written correctly in that paper I linked previously) which are not divergenceless?

Now, can you prove that mathematically? If so, then you've given a counter-example to the method I was using.

But this would ALSO mean you are claiming that an action which has no explicit time or spatial dependence leads to equations of motion with violation of energy and momentum conservation. Do you at least see why I am skeptical?

The method is strongly based on a new parameter [tex]\mu[/tex] appearing in a sort of famous Lagrangian used earlier by many other authors (see, for instance, the equation (1) in http://arxiv.org/PS_cache/astro-ph/pdf/0306/0306438v2.pdf" ) which has units of mass and appears in the new paper within the function [tex]f.[/tex] As you can see, the authors rely on it to compensate for the violation of energy\momentum conservation by means of fine-tuning.

I can prove that those field equations are divergenceless regardless of to what value you set [itex]\mu[/tex]. So I highly doubt that.

I think you are now misunderstanding what someone else wrote as well.

So your notation is NOT satandard and to keep notations in agreement with the standards established in that paper you just cited in your latest post, you would have used [tex]P[/tex] instead of [tex]R[/tex] which, as in the same paper, refers basically to the standard scalar curvature [tex]g^{ab}R_{ab}.[/tex]

Using the letter "P" to stand for that is not all that common of notation. If I used that here, no one would know what I mean, so I wrote it out in clear standard notation which is exactly what they did when they introduced P
[tex]P=R_{ab}R^{ab}[/tex]In the rest of your post you again complain about my math mistakes posts ago that I already pointed out myself. And you still don't seem to even understand I am talking about. So let's start with your momentum conservation comment above to try to get this all back on track.

Are you claiming that actions of functions of scalar curvatures can lead to field equations which are not divergenceless?

Now, can you prove that mathematically?
If so, that would actually help answer questions here instead of furthering argument.

 

[JustinLevy]

Since it is becoming obscured for people just now reading the thread, let me repeat the method I'm talking about:

If I do the following steps:
1) put a function of any curvature scalars I want in replace of f(R) in the f(R) lagrangian
2) solve for the field equations
3) look at the vacuum case (T_uv = 0)
4) take the divergence of the vacuum field equation, thus obtaining a relationship between the divergence of many curvature parts

Is the relation I obtain from this method true in general or only true for solutions of those field equations. I've already made heuristic arguments for why I expect them to be true in general, and therefore this is a valid method to generate relationships between the divergence of curvature terms ... but it would be nice to know for sure (and explicitly why) this is the case.

I previously mentioned that I checked by hand for f(R) gravity and the field equations are divergenceless in general (and not just in solutions to the equations), as I expected.
This would give an explicit example of the method (besides [itex]{G^{ab}}_{;b}=0[/tex] which is the special case from EFE, and was already mentioned).

Also, Altabeh made a comment which would contradict that math.
So let me write it out here.

Alright, f(R) starts from the action (a special case of the more general http://arxiv.org/PS_cache/astro-ph/pdf/0410/0410031v2.pdf)
[tex]S = \int \sqrt{-g}({1 \over 2\kappa} f(R) +\mathcal{L}_M) \, \mathrm{d}^4x[/tex]
giving the field equations:
[tex]f'(R)R^{ab}-\frac{1}{2}f(R)g^{ab}+\left[g^{ab} \nabla^c \nabla_c-\nabla^a
\nabla^b \right]f'(R) = \kappa T_{\mu\nu}[/tex]
where the prime denotes partial derivative with respect to R
[tex]f'(R) = \frac{\partial}{\partial R} f(R)[/tex]Alright, first I'll start by repeating as noted previously that from the Bianchi identities we get:
[tex]{R^{ab}}_{;b} = \frac{1}{2}g^{ab}R_{;b}[/tex]
Now I'll evaluate:
[tex]\nabla_b\left(f'(R)R^{ab}-\frac{1}{2}f(R)g^{ab}+\left[g^{ab} \nabla^c \nabla_c-\nabla^a \nabla^b \right]f'(R)\right)[/tex]
by starting with the term
[tex]\begin{align*}
\nabla_b \nabla^a \nabla^b f'(R) &= \nabla^a \nabla_b \nabla^b f'(R) +R^b{{}_{db}}^a \nabla^d f'(R) \\
&= \nabla^a \nabla^c \nabla_c f'(R) + R^{bd}{{}_{b}}^a \nabla_d f'(R) \\
&= \nabla^a \nabla^c \nabla_c f'(R) + R^{da} f''(R) R_{;d}
\end{align*}[/tex]

which gives:
[tex]\begin{align*}
\nabla_b\left(f'(R)R^{ab}-\frac{1}{2}f(R)g^{ab}+\left[g^{ab} \nabla^c \nabla_c-\nabla^a \nabla^b \right]f'(R)\right) =& \\
\nabla_b\left(f'(R)R^{ab}-\frac{1}{2}f(R)g^{ab}\right) + g^{ab} \nabla_b \nabla^c \nabla_c f'(R) - \nabla^a \nabla^c \nabla_c f'(R) - R^{da} f''(R) R_{;d} =&\\
f''(R)R_{;b}R^{ab} + f'(R){R^{ab}}_{;b} -\frac{1}{2}g^{ab}f'(R)R_{;b} - R^{da} f''(R) R_{;d} =&\\
f'(R){R^{ab}}_{;b} -\frac{1}{2}g^{ab}f'(R)R_{;b}=& \ \ 0
\end{align*}[/tex]

So the method works for f(R) theory as I stated previously.

 

[Altabeh]

Heck that I don't know how on Earth someone who can't even calculate the covariant derivative of a simple tensor of rank 2, is giving us a ridiculously flawed mathematical "nonsense" and just rambles about its relation to a highly prestigious paper! I'm flabbergasted how you can even say confidently this other rambling of yours about what the "clear" standard notation is:

Using the letter "P" to stand for that is not all that common of notation. If I used that here, no one would know what I mean, so I wrote it out in clear standard notation which is exactly what they did when they introduced [tex]P=R_{ab}R^{ab}[/tex]

Everyone hitting this will be sure enough to say you're not even familiar with what a clear standard notation is. At least Carroll et al know better than you and I to introduce a new quantity with a different symbol assigned to it not a common symbol which every physicist adapts it to the scalar curvature in GR.

How come the newly introduced [tex]R[/tex] can be tracked in the Bianchi identity to have the following hold?

[tex]{R^{ab}}_{;b} = \frac{1}{2}g^{ab}R_{;b}.[/tex]

Weren't you the one who said [tex]R[/tex] has another definition in your clear standard notation!? What happened you made use of Ricci scalar [tex]R[/tex] in your calculations , considering you've not misunderstood mathematics of GR this time? That's completely nonsense.

Having received many private messages concerning this thread and how it is managed by the original poster in a very twisted way, I really feel the need for a quick stop in here to perhaps not let it go astray from the main topic.

AB

 

[JustinLevy]

Heck that I don't know how on Earth someone who can't even calculate the covariant derivative of a simple tensor of rank 2

I do not consider it simple to solve the opening problem in this thread. If you consider it simple, then please do show us how to work out (from the very openning lines of this thread)

How can one work out what terms like:
[tex](g^{cd}R^{ab}R_{ab})_{;d}[/tex]
are in terms of the divergence of the Ricci curvature or Ricci scalar?

If you STILL don't understand that notation, then consider it:
[tex](g^{cd}P)_{;d}[/tex]
if substituting one symbol defined as [tex]P=R_{ab}R^{ab}[/tex] somehow helps you understand.

If you feel it is simple, go ahead!
Please just solve it here. It would be incredibly helpful and answer the opening question directly.

[JustinLevy] is giving us a ridiculously flawed mathematical "nonsense" and just rambles about its relation to a highly prestigious paper!

Look at my last post. I wrote out the math showing explicitly that my method works for f(R) theory, in contrast to a claim you made. Either retract your claim above that f(R) theory can lead to violation of energy-momentum conservation, or show where that math is wrong.

You are avoiding direct questions by just insulting me. That is unreasonable. If you feel it is obvious, then please just show the math.

I'm flabbergasted how you can even say confidently this other rambling of yours about what the "clear" standard notation is:

Using the letter "P" to stand for that is not all that common of notation. If I used that here, no one would know what I mean, so I wrote it out in clear standard notation which is exactly what they did when they introduced [tex]P=R_{ab}R^{ab}[/tex]

Everyone hitting this will be sure enough to say you're not even familiar with what a clear standard notation is.

That definition equation is directly from their paper! It is clear what R_{ab}R^{ab} means in standard notation. I've even explained some of the standard notation several times to help clear up any miscommunication.

You seem to not even be listening. Please, let us work towards consensus on notation so we can move on.

At least Carroll et al know better than you and I to introduce a new quantity with a different symbol assigned to it not a common symbol which every physicist adapts it to the scalar curvature in GR.

Argh! Do you still not understand that R_{ab}R^{ab} is NOT referring in anyway to the Ricci curvature scalar? ... the meaning of that term is CLEAR from standard notation.

How come the newly introduced [tex]R[/tex] can be tracked in the Bianchi identity to have the following hold?

[tex]{R^{ab}}_{;b} = \frac{1}{2}g^{ab}R_{;b}.[/tex]

Weren't you the one who said [tex]R[/tex] has another definition in your clear standard notation!? What happened you made use of Ricci scalar [tex]R[/tex] in your calculations , considering you've not misunderstood mathematics of GR this time? That's completely nonsense.

The math I wrote in the previous post is NOT complete nonsense. The nonsense is inserted by your misunderstanding of what I'm saying due to your misunderstanding of standard notation. For the umpteenth time: I am NOT, I repeat NOT introducing a new definition of R. I am NOT defining R = R_{ab}R^{ab} anywhere. I have only ever used R (with no tensor indices) to refer to the Ricci curvature scalar, and R_{ab} (with two tensor indices) to refer to the Ricci curvature tensor, and R_{abcd} (with four tensor indices) to refer to the Riemann curvature tensor. THIS IS STANDARD NOTATION.

I'm starting to worry you are a troll. Please please start listening. To give the benefit of the doubt and believing that you are being sincere and STILL misunderstanding after all this time is really really requiring a stretch of my imagination.

Having received many private messages concerning this thread and how it is managed by the original poster in a very twisted way

Considering how bad your reading comprehension is here and in reading the paper you linked to above (claiming the author needed to fix "u" to prevent/reduce violation of local energy conservation), I don't trust you to have understood them either. Regardless, I don't like the attempt to argue by "hidden comments from authority".

Please, people writing to Altabeh in private, please just write here in public helping explain what the standard notation is. If you feel there is a clearer way to write that first equation in this entire thread (copied at the top of this post), then please do explain.

I want to get closer to solving this question, but it seems impossible to move forward until Altabeh is satisfied what the notation means. He seems to think R_{ab} is referring to the Ricci curvature scalar, and not the Ricci curvature tensor.

 

[JustinLevy]

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
[tex] S = \int (\frac{1}{2\kappa}R_{ab}R^{ab} + \mathcal{L}_m) \sqrt{-g} d^4x[/tex]

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:
[tex]-\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}[/tex]

Applying the method, we then get:
[tex]\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)[/tex]

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 [tex]g^{ab}R_{cd}R^{cd}[/tex] 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.

 

[Mentz114]

Sticking my neck out for Altabeh to chop

[tex]
(g^{ab}P)_{;b} = (g^{ab})_{;b}P + g^{ab}(P)_{;b}
[/tex]

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

[tex]
(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]
[/tex]

and given that

[tex]
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}
[/tex]

I give up.

 

[JustinLevy]

I think the first term on the right is zero

Yep, even more generally
[tex]g^{ab}{}_{;c}=0[/tex]
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 let's you simplify your equation, since it means the term
[tex]R_{pq}R_{cd}\left[ g^{cp}(g^{qd})_{;b} +g^{qd}(g^{cp})_{;b}\right]=0[/tex]

The other term can be simplified too, since there is no significance in the label we give dummy indices
[tex]
\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*}
[/tex]

But, just like you, I have no idea where to go next in working it out by hand. If we go this route:
[tex]
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}
[/tex]
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?

 

[Altabeh]

(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

[tex]\delta R_{ab}=-(\delta\Gamma^{c}_{ab})_{;c}+(\delta\Gamma^{c}_{ac})_{;b}[/tex]

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

 

[Mentz114]

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.

 

[JustinLevy]

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

[tex]\delta R_{ab}=-(\delta\Gamma^{c}_{ab})_{;c}+(\delta\Gamma^{c}_{ac})_{;b}[/tex]

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:
[tex]\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}[/tex]
But I'm not sure how this helps here, as I don't know how to relate [tex]\nabla_\lambda (\delta \Gamma^\rho_{\nu\mu} )[/tex] 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 [tex]\delta \Gamma^a_{bc}[/tex] transforms as a tensor object, even if [tex]\Gamma^a_{bc}[/tex] does not.

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

 

[Altabeh]

...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 [tex]\delta \Gamma^\rho_{\nu\mu} [/tex] 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 [tex]S.[/tex] 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

 

[JustinLevy]

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
[tex]R^a{}_{bac} = - R^a{}_{bca}[/tex]

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

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 [tex]\delta \Gamma^\rho_{\nu\mu} [/tex] 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.

(emphasis added)
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. [itex]R_{ab;c}[/itex]) 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 [tex]g^{ab}R_{cd}R^{cd}[/tex].

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 [tex]S.[/tex] 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.

I've already explained this multiple times. Please reread post #23.

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.

 

[JustinLevy]

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:
[tex]\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)[/tex]

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.

 

[atyy]

Maybe relevant http://arxiv.org/abs/gr-qc/0505128