Energy Momentum Tensor and Divergence Theorem

[latentcorpse]

In the notes attached to this thread:
https://www.physicsforums.com/showthread.php?t=457123

On page 110, how has he gone from equation (369) to eqn (370). He claims to have done it by "integration by parts using the divergence theorem to eliminate derivatives of $\delta g_{ab}$ if present".
(The divergence theorem is given in eqn (343) at the bottom of p105 if necessary)

To me it appears that all he has done is cross multiply so I appear to be missing something crucial here about the application of the divergence theorem!And secondly, where did equation (371) come from? The scalar field action that he talks about was discussed on p106.

 

[dextercioby]

Actually,

$$\delta S_{\mbox{matter}} = \int d^4 x \frac{\delta S_{\mbox{matter}}}{\delta g_{ab}} \delta g_{ab}$$ from which, using the convenient definition of the energy-momentum tensor, the equation (370) follows.

As for (371) it follows from (344) and (345) with the help of the 2 equations he mentions.

 

[latentcorpse]

Actually,

$$\delta S_{\mbox{matter}} = \int d^4 x \frac{\delta S_{\mbox{matter}}}{\delta g_{ab}} \delta g_{ab}$$ from which, using the convenient definition of the energy-momentum tensor, the equation (370) follows.

As for (371) it follows from (344) and (345) with the help of the 2 equations he mentions.

I still can't get (371) -

do I use $\delta S = \int_M d^4x \frac{\delta S}{\delta \sqrt{-g}} \delta \sqrt{-g} + \frac{\delta S}{\delta g^{ab}} \delta g^{ab}$?
I doubt it - I think one of them should be a $\frac{\delta S}{\delta \Phi}$ term but how else am I going to make use of the $\delta \sqrt{-g}$ and $\delta g^{ab}$ that he is talking about?

Also, in the equaiton you wrote down:
$$\delta S_{\mbox{matter}} = \int d^4 x \frac{\delta S_{\mbox{matter}}}{\delta g_{ab}} \delta g_{ab}$$
why is there no $\sqrt{-g}$ factor? surely there should be in order to keep our definitions chart independent, no?

Thanks!

 

[dextercioby]

Well, as I defined the functional derivative, it's a tensor density of weight 1/2. That way, using the smart definition of the energy-momentum tensor, the energy-momentum object with 2 indices is indeed a tensor. The volume element is a tensor density of weight -1/2, so that the product b/w the volume element and the functional derivative is tensor derivative of weight 0, i.e. a tensor.

 

[dextercioby]

As for the derivation of (371), follow in my footsteps

$$\delta S = \int d^4 x \left[\left(\delta \sqrt{\left|g\right|}\right)\mathcal{L}_{m}+\sqrt{\left|g\right|}\frac{\delta\mathcal{L}_{m}}{\delta g^{cd}}{}\delta g^{cd}\right] = ...$$

 

[latentcorpse]

As for the derivation of (371), follow in my footsteps

$$\delta S = \int d^4 x \left[\left(\delta \sqrt{\left|g\right|}\right)\mathcal{L}_{m}+\sqrt{\left|g\right|}\frac{\delta\mathcal{L}_{m}}{\delta g^{cd}}{}\delta g^{cd}\right] = ...$$

Ok. I think I can see how it is going to work but I don't know how to evaluate
$\frac{\delta L}{\delta g^{cd}}$

 

[dextercioby]

Well, the delta's can be taken as partial derivatives, so that the ratio you wrote is quite easy to compute.

 

[latentcorpse]

Well, the delta's can be taken as partial derivatives, so that the ratio you wrote is quite easy to compute.

Ok. I get a minus sign that I can't get rid of though?

$\sqrt{-g} \frac{\partial L}{\partial g^{ab}} \delta g^{ab}=\sqrt{-g} ( - \frac{1}{2} \nabla_a \Phi \nabla_b \Phi ) \delta g^{ab}=- \frac{1}{2} \sqrt{-g} \nabla^a \Phi \nabla^b \Phi \delta g_{ab}$

 

[dextercioby]

The last equality you wrote is wrong, because

$$\nabla_a \Phi \nabla_b \Phi \delta g^{ab} = - \nabla_a \Phi \nabla_b \Phi g^{ac} g^{bd} \delta g_{cd}$$.

That's how you get rid of the minus you mention.

 

[latentcorpse]

The last equality you wrote is wrong, because

$$\nabla_a \Phi \nabla_b \Phi \delta g^{ab} = - \nabla_a \Phi \nabla_b \Phi g^{ac} g^{bd} \delta g_{cd}$$.

That's how you get rid of the minus you mention.

Wait a second...how did you get a minus in that equation? and how did you lower the indices on the $\delta g^{ab}$?And in the remark at the top of p116, he says an n dimensional amnifold is maximally symmetric if it has n(n+1)/2 linearly independent Killing vector fields and he calims that this 5 dimensional de Sitter spacetime is maximally symmetric even though it only has 10 Killing vector fields (but form the formula it should have 15)?

 

[dextercioby]

Well, putting down the indices on the delta g picks up a minus sign, because of the following relation

$$\delta \left(g^{ab}g_{bc}\right) = \delta \delta^{a}_{c} = 0$$

As for de Sitter part, the manifold dimension of the de Sitter spacetime is 4, not 5. So 4(4+1) / 2 = 4x 5 /2 =10 independent Killing vectors.

 

[latentcorpse]

Well, putting down the indices on the delta g picks up a minus sign, because of the following relation

$$\delta \left(g^{ab}g_{bc}\right) = \delta \delta^{a}_{c} = 0$$

As for de Sitter part, the manifold dimension of the de Sitter spacetime is 4, not 5. So 4(4+1) / 2 = 4x 5 /2 =10 independent Killing vectors.

Excellent. Thanks.

What about these next bits though:

a)In the example on p112, he says that "This satisfies (382) with K=1/a^2". How do I actually check this? Surely there's a faster way than computing the Riemann tensor and then comparing it with $g_{a[c}g_{d]b}$, no?b)And at the top of p188, why does $L^2 \cosh^2{\frac{t}{L}}=\frac{1}{\sin^2{\eta}}$? I've been struggling with this calculation all day!

c)In the last paragraph of p118, he claims that we can check that any timelike or null geodesic will reach $t=\pm \infty$ in the limit of infinite affine parameter. How would we actually do this though?

Thanks again.

 

[dextercioby]

For the first part, I don't see other option to check which value of K satisfies the equation. It suffices to pick one of the independent component of the Riemann tensor, though.

As for the issue on the de Sitter metric, the L^2 shouldn't be there, at all. So the author of the notes made an error.

I don't know the answer to issue c).

 

[latentcorpse]

For the first part, I don't see other option to check which value of K satisfies the equation. It suffices to pick one of the independent component of the Riemann tensor, though.

What do you mean? Just pick a particular component? How do I know which ones are independent? And why wouldn't it work if i picked a dependent one?

As for the issue on the de Sitter metric, the L^2 shouldn't be there, at all. So the author of the notes made an error.

Surely this affects the rest of the results on that page and, in fact, the remainder of the chapter?


Thanks again for your reply.

 

[dextercioby]

You're right. I should have said <non-zero> components. Judging by the symmetries of the Riemann curvature tensor, you can easily find a non-zero component, dependent or independent, it's irrelevant.

Well, it doesn't make any conceptual or interpretational difference if the L^2 changes the following formulas, because it's just a constant. The value for $\Omega$ should be the same, though.

 

[latentcorpse]

a)

You're right. I should have said <non-zero> components. Judging by the symmetries of the Riemann curvature tensor, you can easily find a non-zero component, dependent or independent, it's irrelevant.

Could you give me an example of how to pick one that we know will be non zero?

b)

Well, it doesn't make any conceptual or interpretational difference if the L^2 changes the following formulas, because it's just a constant. The value for $\Omega$ should be the same, though.

Ok. So I get that if everything is just scaled by a constant it won't make an interpretational difference but what did you mean by "the value for $\Omega$ should be the same? You mean (404) shouldn't change and then (405) should be
$\hat{g}=L^2 \left( - d \eta^2 + d \Omega_3^2 \right)$
and because this is just a scaling the subsequent results will not change.

c)I don't really understand the definition of spatially homogeneous on p121. Could you elaborate please. How is it different from the definition of homogeneous on p 115.

d) How would we go about the exercise on p123 where we are asked to show the worldlines of the comoving observers are geodesics?

e)Finally, in that paragraph after eqn (405), instead of saying that the de Sitter spacetime corresponds to the $\eta \in (0,\pi)$ portion of the Einstein static universe, should it say $(0,L \pi)$ or $(0, \frac{\pi}{L})$?

Cheers.

 

[latentcorpse]

The last equality you wrote is wrong, because

$$\nabla_a \Phi \nabla_b \Phi \delta g^{ab} = - \nabla_a \Phi \nabla_b \Phi g^{ac} g^{bd} \delta g_{cd}$$.

That's how you get rid of the minus you mention.

There is a $V(\Phi)$ term in (371) that it we haven't accounted for though...