Bel-Robinson Tensor in empty spacetime

[TerryW]

Homework Statement

This is Exercise 15.2 in MTW - See attachment

Homework Equations

See attachment

The Attempt at a Solution

[/B]
My attempt at a solution is also in the attachment.

Are my initial assumptions OK? If not can someone nudge me in the right direction.

If my initial assumptions are OK, then I've gone wrong somewhere in my workings. Can anyone give me a pointer as to where I have gone wrong.TerryW

 

[Drakkith]

Hi Terry!

Normally we like members to type in their equations and the work they've done into the post itself, but due to the sheer amount of work involved here I think we can let that slide this time. However, could you type up what the last paragraph in your attachment says? The other mentors and I are having a difficult time reading it (which is one of the reasons we normally like people to type up everything into a post instead of attaching a handwritten attachment).

Thanks!

 

[TerryW]

Hi Drakkith,

My last paragraph sums up the sticky position I have arrived at. I've rewritten it to include some of the expressions referred to in the manuscript.

So - the process has delivered one part of the answer required, i.e. R_ανδσR_β^ν_γ^σ, but the second part (-½g_αβg_κδR_ωλγσR^ωλκσ) doesn't look very promising.

I have reworked this expression into:
-½g_αβg_κδR_ωλγσR^λκσ = -½g_αβg_γδR_ωλγσR^ωλκσ + ½g_αβg_γδR_ωλτσR^ωλτσ_(τ≠γ)

-½g_αβg_κδR_ωλτσR^ωλκσ_(τ≠κ).​

It would be nice if I could use the empty spacetime condition to make ½g_αβg_γδR_ωλτσR^ωλκσ_(τ≠γ) and ½g_αβg_κδR_ωλτσR^ωλκσ_(τ≠κ) disappear but I haven't so far been able to do this. Even if I do, I am still left with -½g_αβg_γδR_ωλκσR^ωλκσ instead of -(½*¼)g_αβg_γδR_ωλκσR^ωλκσ.

I hope this helps and look forward to your next post.Regards

TerryW

 

[TSny]

Terry,

I looked at your work and it looks good to me all the way through to your expressions (A) and (B). In the attachment below, I noted a couple of minor typos.

I believe your results for (A) and (B) are correct. They appear to be corroborated here: http://arxiv.org/pdf/1006.3168.pdf (see equation 1).

I don't follow how you got your expressions for (C) and (D).

So, it seems that you are left with showing $g_{\kappa \delta} R_{\omega \lambda \gamma \sigma} R^{\omega \lambda \kappa \sigma} = \frac{1}{4} g_{\gamma \delta} R_{\omega \lambda \tau \sigma} R^{\omega \lambda \tau \sigma}$ for the vacuum.

This identity is mentioned below equation (1) in the link above. I have spent quite a bit of time trying to prove it but have not been successful. The only vacuum condition that I can think of that would be useful is the condition that the Ricci tensor be zero. But, I haven't been able to use that to get anywhere.

If you crack this nut, please post it.

[EDIT: After looking further into the paper in the link given above, it appears that the paper does provide a proof of the identity $g_{\kappa \delta} R_{\omega \lambda \gamma \sigma} R^{\omega \lambda \kappa \sigma} = \frac{1}{4} g_{\gamma \delta} R_{\omega \lambda \tau \sigma} R^{\omega \lambda \tau \sigma}$ for the vacuum. The proof is somewhat sprawled out in the paper, but a key part of the proof is in the discussion surrounding equations (16) - (18).]

 

[TerryW]

Hi TSny,

Many thanks for spotting the typos, but most of all thanks for confirming that my process up to (A) and (B) is correct.

As for (C) and (D), I was trying to find a way forward and arrived at that equation by the brute force method of writing out the values for the indices for _κγ^κ in the expression (B) and comparing them with the indices _γκ^κ in the expression I need to arrive at. What I observed was that the two sets of indices had a common set of four components (000, 111,222,333) and that the remaining 12 components of expression (B) could be written as g_αβg_δR_ωλτσR^ωλτσ(τ≠γ).
So I add these 12 values to the (negative) expression (B), but then I only have four of the required components of my target expression so I have to subtract a further twelve components provided by g_αβg_κδR_ωλτσR^ωλκσ(τ≠κ).
Not very elegant I admit, and it didn't lead anywhere, so let's dump it!

Thanks particularly for the reference to the paper. As you have pointed out in your edit, it does appear to provide a proof of the Lanczos identity, and given that Lanczos died in 1974, aged 81, it is likely that this identity was around when MTW was being written (1972), though I don't think MT&W would have been expecting the reader to know it!

All considered, a fairly tough problem, so I'm not too upset about not finishing it!Regards
TerryW

 

[TSny]

I think it is uncharacteristic of MTW to expect the reader to bring this much to the table to solve one of their exercises. For their tougher problems, they usually provide hints or guidance (by breaking up the problem into multiple parts). Makes me wonder if there is a better way.

 

[TerryW]

I agree. I am working my way through MTW and have completed almost all of the exercises (with the occasional bit of help from some great guys (and gals?) on PF, including your good self). This is the first exercise where the key is a fairly obscure formula, though the paper you referenced does say "there is a well known equation in empty space"!

I will probably have another look at this when I come to do my write up. If I find anything, I'll post it.RegardsTerryW

 

[TSny]

I will probably have another look at this when I come to do my write up. If I find anything, I'll post it.

Thanks.

 

[TerryW]

Hi TSny,

I've been doing my write up and have had a look at the outstanding bit of this problem again, i.e. to prove:

R_γσλμR_δ^σλμ = ¼g_γδR_ρσλμR^ρσλμ

As this expression is only valid in empty spacetime, I decided to see if G_σμ = R_σμ - ½g_σμR = 0 could help.

So

R_σμ = R^λ_σλμ = g^γλR_γσλμ = ½g_σμR

Therefore R_γσλμ = ½g_σμg_γλR

R^σμ = g^σαg^βμR_αβ = g^σαg^βμR^λ_αλβ = g^σαg^βμg^δλR_δαλβ = g^δτR_δ^σ_τ^μ = g^δτg_λτR_δ^σλμ = δ^δ_λR_δ^σλμ = ½g^σμR

Therefore R_δ^σλμ = ½g^σμδ^λ_δR

Now multiply the two expressions together to get:

R_γσλμR_δ^σλμ = ¼g_σμg_γλg^σμδ^λ_δRR

It is then simple to show that RR = R_ρσλμR^ρσλμ

Hope you like it!Regards and thanks for all th helpTerryW

 

[TSny]

So

R_σμ = R^λ_σλμ = g^γλR_γσλμ = ½g_σμR

Therefore R_γσλμ = ½g_σμg_γλR

Hi Terry,
You'll need to walk me through how you got the above.

The equation g^γλR_γσλμ = ½g_σμR may be written as

g^αβR_ασβμ = ½g_σμR

Then, multiplying both sides by g_γλ gives

g_γλg^αβR_ασβμ = ½g_σμg_γλR

I don't see how to simplify the left side to R_γσλμ.

EDIT: I don't see how the equation R_γσλμ = ½g_σμg_γλR can be true in general for the vacuum. R_γσλμ need not be zero in the vacuum while R must be is zero in the vacuum.

 

[TerryW]

Hi TSny,

The equation gγλRγσλμ = ½gσμR may be written as

gαβRασβμ = ½gσμR

Well, I prefer to stick with my own formulation of

g^γλR_γσλμ = ½g_σμR

So that I can multiply both sides by g_γλ to get

g_γλg^γλR_γσλμ = ½g_γλg_σμR

And as g_γλg^γλ = 1

then R_γσλμ = ½g_γλg_σμR

EDIT: I don't see how the equation Rγσλμ = ½gσμgγλR can be true in general for the vacuum. Rγσλμ need not be zero in the vacuum while R must be is zero in the vacuum.

If my maths is correct, then maybe this is not true. I've attached an example from MTW in which R is non-zero in empty spacetime.Regards

Terry

 

[TSny]

g^γλR_γσλμ = ½g_σμR

As you know, the indices γ and λ are being summed over on the left hand side. γ and λ will take on all values.

So that I can multiply both sides by g_γλ

This is where the trouble starts.

to get

g_γλg^γλR_γσλμ = ½g_γλg_σμR

Note the confusion on the left where the red γ and λ are being summed over but the blue γ and λ are not. The blue indices have some fixed values.

And as g_γλg^γλ = 1

This does not follow since the blue indices have fixed values. They are not summed with the red indices.

(In the case where you are summing indices, then I believe g_γλg^γλ = 4.)

I've attached an example from MTW in which R is non-zero in empty spacetime.

I don't believe this is an example of working with the Einstein equations in a vacuum.

For a vacuum you always have G_σμ = R_σμ - ½g_σμR = 0.

As a little exercise, start with R_σμ - ½g_σμR = 0 and raise the σ index. Then contract σ with μ to show that R = 0.
Then conclude from R_σμ - ½g_σμR = 0 that R_σμ = 0 for all values of σ and μ

 

[TerryW]

Hi TSny,

Yes, my critical faculties were overridden by the tantalising prospect of a solution to the problem!

I've had another look and this time got as far as

g^γδR_γσλμR_δ^σλμ = ¼RR

but no further because of the original problem with summation indices.
I agree with your argument that leads to R = 0 for empty spacetime, so why have we been bothering with this latter part of the exercise which is essentially creating a complex expression for something which has the value of 0!?Regards
TerryW

 

[TSny]

I've had another look and this time got as far as

g^γδR_γσλμR_δ^σλμ = ¼RR

but no further because of the original problem with summation indices.

I don't see how you get the above result.

I agree with your argument that leads to R = 0 for empty spacetime, so why have we been bothering with this latter part of the exercise which is essentially creating a complex expression for something which has the value of 0!?

We need to show R_γσλμR_δ^σλμ = ¼g_γδR_ρσλμR^ρσλμ. Here, the two sides need not be zero in a vacuum.

In post #9 it looked like you were going to construct a proof of this based on G_σμ = R_σμ - ½g_σμR = 0, which you rearranged as R_σμ = ½g_σμR. Here, both sides are each equal to zero in a vacuum. I wasn't sure where you were going with it. This equation doesn't seem to contain any more information than the equation R_σμ = 0.

As I mentioned back in post #4, I tried using R_σμ = 0 to derive R_γσλμR_δ^σλμ = ¼g_γδR_ρσλμR^ρσλμ but had no luck. I don't think I have my notes any more.

 

[TerryW]

Hi TSny,

My attempt to get a resolution to this problem by using R_αβ = ½g_αβ was based as much as anything on the fact that I could (or at least I thought I could) generate expressions for R_γσλμ and R_δ^σλμ, which I could multiply together and (inter alia) produce the factor of ¼ which was needed.

It is now pretty clear that this just doesn't work.

I've just revisited my earlier workings and realized that A is only one third of the answer! I need to wrangle B into a sum of two more expressions -

+ R_αργσR_β^ρ_δ^σ and - 1/8g_αβg_γδR_ρσλμR^ρσλμ

So it's back to the drawing board. I'll post again if I make any progress

BTW, I've realized that my earlier supposition that R_ρσλμR^ρσλμ = RR isn't true!

So thanks for all your help so farTerryW

 

[TSny]

I'll post again if I make any progress

BTW, I've realized that my earlier supposition that R_ρσλμR^ρσλμ = RR isn't true!

OK. Good luck with it.

 

[TerryW]

Hi TSny,

Is this a proof?

Suppose
R_γσμλR_δ ^σλμ = ¼g_δγR_ρσλμR^ρσλμ

Then

g^γδR_γσλμR_δ^σλμ = ¼g^γδg_γδR_ρσλμR^ρσλμ

So R_γσλμR^γσμ = R_ρσλμR^ρσλμTerryW

 

[TSny]

You've shown that if R_γσμλR_δ^σλμ = ¼g_δγR_ρσλμR^ρσλμ is assumed to be true, then it follows that R_γσλμR^γσλμ = R_ρσλμR^ρσλμ is true. (You inadvertently left out the superscript λ on the left side.)

The conclusion R_γσλμR^γσλμ = R_ρσλμR^ρσλμ is clearly a true statement. But this argument doesn't prove that your initial assumption R_γσμλR_δ^σλμ = ¼g_δγR_ρσλμR^ρσλμ is true.

If you could show the converse, then you would have a proof of what you want. Starting with the true statement R_γσλμR^γσλμ = R_ρσλμR^ρσλμ you can easily work back to g^γδR_γσλμR_δ^σλμ = ¼g^γδg_γδR_ρσλμR^ρσλμ. But I don't see how to go from this to the conclusion
R_γσμλR_δ^σλμ = ¼g_δγR_ρσλμR^ρσλμ .

 

[TerryW]

It would be nice to be able to divide both sides by g^γδ but we can't do that. Also, this line of attack hasn't involved the empty spacetime consideration so something is missing along the way.

It's all very frustrating!TerryW

 

[Fred Wright]

In the literature the dual in the Bel Robinson tensor is defined as:
$*R_{\alpha \rho \gamma \sigma} := \frac {1} {2} \epsilon_{\alpha \rho} {} ^{\lambda \mu} R_{\lambda \mu \gamma \sigma}$
$*R_\beta {} ^ \rho {} _\delta {} ^\sigma := \frac {1} {2} \epsilon_\beta {} ^{\rho \tau} {} _\phi R_\tau {} ^ \phi {} _\delta {} ^ \sigma$
Thus,
$*R_{\alpha \rho \gamma \sigma}*R_\beta {} ^ \rho {} _\delta {} ^\sigma = \frac {1} {4} \epsilon_{\alpha \rho} {} ^{\lambda \mu} \epsilon_\beta {} ^{\rho \tau} {} _\phi R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma$
$= - \frac {1} {4} \epsilon_{\alpha } {} ^{\lambda \mu}\epsilon_\beta {} ^{ \tau} {} _\phi R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma$ (contracting on the $\rho$ index)
$= - \frac {1} {4} \begin{vmatrix} {\delta_{\alpha \beta}} & {\delta_\alpha ^ \tau} & {\delta_{\alpha \phi}} \\ {\delta_\beta ^ \lambda} & {\delta ^ {\lambda \tau}} & {\delta ^ {\lambda} _\phi} \\ {\delta_\beta ^ \mu} & {\delta ^{\mu \tau}} & {\delta_ \phi ^ \mu} \end{vmatrix}R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma$
$= - \frac {1} {4} \delta_{\alpha \beta}(\delta^{\lambda \tau}\delta_\phi ^ \mu - \delta^{\mu \tau}\delta_\phi ^\lambda)R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma$
$+ \frac {1} {4} \delta_{\alpha}^\tau(\delta_\beta^\lambda \delta_\phi ^\mu - \delta_\phi ^\lambda \delta_\beta ^\mu)R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma$
$+ \frac {1} {4} \delta_{\alpha \phi}(\delta_\beta ^ \mu \delta^{\lambda \tau}-\delta_\beta^\lambda \delta^{\mu \tau})R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma$
$= \frac {1} {2}(-g_{\alpha \beta}g_{\gamma \delta}R_{ \gamma \sigma \lambda \mu }R^{ \gamma \sigma \lambda \mu }+ R_{\beta \phi \gamma \sigma} R_\alpha {} ^ \phi {} _\delta {} ^ \sigma + R_{\lambda \beta \gamma \sigma}R^{\lambda \alpha}{} _\delta {} ^ \sigma)$
In the following I have used the anti-symmetry property of the Riemann tensor for adjacent indices, the symmetry property for index pairs and $\delta_{\alpha \beta}=g_{\alpha \beta}$ (free space).
Examining the first term in the sum I note,
$\delta_\rho ^\gamma \delta _\gamma ^ \rho R_{ \gamma \sigma \lambda \mu }R^{ \gamma \sigma \lambda \mu } = 4R_{ \gamma \sigma \lambda \mu }R^{ \gamma \sigma \lambda \mu } = R_{ \rho \sigma \lambda \mu }R^{ \rho \sigma \lambda \mu }$
and the first term in the sum becomes,
$- \frac {1}{8}g_{\alpha \beta}g_{\gamma \delta}R_{ \rho \sigma \lambda \mu }R^{ \rho \sigma \lambda \mu }$
Examining the second term,
$\frac {1}{2}(\delta^{\phi \rho} \delta_\sigma ^ \sigma \delta_{\phi \rho} \delta_\sigma ^ \sigma)R_{\beta \phi \gamma \sigma} R_\alpha {} ^ \phi {} _\delta {} ^ \sigma = \frac {1}{2}(1)R_{\beta \phi \gamma \sigma} R_\alpha {} ^ \phi {} _\delta {} ^ \sigma = \frac {1}{2}R_{\alpha \rho \delta \sigma}R_\beta{}^\rho{} _\gamma {} ^\sigma$
Examining the third term,
$\frac {1}{2}(\delta_\alpha ^ \alpha \delta^{\lambda \rho} \delta_\rho ^ \rho \delta_{\lambda \rho} \delta_\rho ^ \rho) R_{\lambda \beta \gamma \sigma}R^{\lambda \alpha}{} _\delta {} ^ \sigma = \frac {1}{2}(1) R_{\lambda \beta \gamma \sigma}R^{\lambda \alpha}{} _\delta {} ^ \sigma = \frac {1}{2}R_{\alpha \rho \delta \sigma}R_\beta{}^\rho{} _\gamma {} ^\sigma$
and adding the terms we find,
$*R_{\alpha \rho \gamma \sigma}*R_\beta {} ^ \rho {} _\delta {} ^\sigma = R_{\alpha \rho \delta \sigma}R_\beta{}^\rho{} _\gamma {} ^\sigma - \frac {1}{8}g_{\alpha \beta}g_{\gamma \delta}R_{ \rho \sigma \lambda \mu }R^{ \rho \sigma \lambda \mu }$

 

[TSny]

Hi, Fred. Thanks for joining this discussion. (Warning, this thread can generate headaches.)

I believe I follow your work to here:

$= - \frac {1} {4} \delta_{\alpha \beta}(\delta^{\lambda \tau}\delta_\phi ^ \mu - \delta^{\mu \tau}\delta_\phi ^\lambda)R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma$
$+ \frac {1} {4} \delta_{\alpha}^\tau(\delta_\beta^\lambda \delta_\phi ^\mu - \delta_\phi ^\lambda \delta_\beta ^\mu)R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma$
$+ \frac {1} {4} \delta_{\alpha \phi}(\delta_\beta ^ \mu \delta^{\lambda \tau}-\delta_\beta^\lambda \delta^{\mu \tau})R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma$
$= \frac {1} {2}(-g_{\alpha \beta}g_{\gamma \delta}R_{ \gamma \sigma \lambda \mu }R^{ \gamma \sigma \lambda \mu }+ R_{\beta \phi \gamma \sigma} R_\alpha {} ^ \phi {} _\delta {} ^ \sigma + R_{\lambda \beta \gamma \sigma}R^{\lambda \alpha}{} _\delta {} ^ \sigma)$

In the last line above, the index $\gamma$ appears three times in the first term inside the parentheses. It seems to me that two of these $\gamma$'s should be replaced by a different dummy index. Thus, I think this term should be

$-g_{\alpha \beta}g_{\kappa \delta}R_{ \gamma \sigma \lambda \mu }R^{ \kappa \sigma \lambda \mu }$

where a pair of $\gamma$'s in your expression has been replaced by the dummy summation index $\kappa$. $\gamma$ is a fixed index in the Bel-Robinson tensor and should never be summed. With this change, I don't see how you can follow through with your later arguments.

In the last term of your last line above, I believe the superscript $\alpha$ should be a subscript, so the last term in the parentheses would be
$R_{\lambda \beta \gamma \sigma} R^{\lambda}{} _ {\alpha \delta} {} ^ \sigma$

In the following I have used the anti-symmetry property of the Riemann tensor for adjacent indices, the symmetry property for index pairs and $\delta_{\alpha \beta}=g_{\alpha \beta}$ (free space).

I don't see how free space enters here. I'm not sure I understand the notation $\delta_{\alpha \beta}$. The MTW text doesn't seem to use this notation (at least I can't find it in their discussion of the various $\delta$ symbols introduced on pages 87 and 88. Does $\delta_{\alpha \beta}$ mean the same as lowering an index on $\delta^{\alpha}_{\beta}$ ? If so, then $\delta_{\alpha \beta} \equiv g_{\alpha \rho} \delta^{\rho}_{\beta}$. This is easily seen to equal $g_{\alpha \beta}$. So, $\delta_{\alpha \beta} = g_{\alpha \beta}$, as you say. But I don't see any need to invoke the condition of free space here.

 

[TerryW]

Hi Fred,

I agree with TSny. Your matrix produces six equations which are more or less the same as the six equations in my original manuscript post. If you take one of the six elements e.g. -¼δ_αβδ^λτδ^μ_φR_λμγσR_τ^φ_δ^σ, you worked this forward to produce -¼g_αβg_γδR_λμγσR^ϒσλμ. But it doesn't work.

Taking out the -¼g_αβ, δ^λτδ^μ_φR_λμγσR_τ^φ_δ^σ leads to δ^μ_φR_λμγσR^λφ_δ^σ which leads to R_λμγσR^λμ_δ^σ.

The index γ in R_λμγσ means you can't use g_γδ as an index lowering device, you have to use g_κδ on R^λμκσ, which is where I came in originally!

As TSny said, it's all very headache inducing!RegardsTerryW