# Homework Help: Stress-energy tensor proof (schutz ch7 q 8)

1. Mar 2, 2009

### Mmmm

1. The problem statement, all variables and given/known data

Using the (previously proved) equation :
if $g_{\alpha \beta}$ is independent of $$x^\mu[/itex] then [tex] \frac{1}{\sqrt{-g}}(\sqrt{-g}{T^\nu}_\mu)_{,\nu} = 0$$

Prove that
$$\int_{x_{0}=const} {T^\nu}_\mu (\sqrt{-g})n_\nu d^3 x$$

is independent of $x^0$ if $n_\nu$ is the unit normal to the hypersurface.

3. The attempt at a solution

Need to prove that the derivative of the integral wrt $x^0$ is zero.

as $n_\nu$ is the unit normal to the hypersurface, $n_\nu$= (1,0,0,0).

so

$$\frac{\partial}{\partial t}\int_{x_{0}=const} {T^\nu}_\mu (\sqrt{-g})n_\nu d^3 x$$

$$=\frac{\partial}{\partial t}\int {T^0}_\mu (\sqrt{-g}) d^3 x$$
$$=\int ({T^0}_\mu (\sqrt{-g}))_{,0} d^3 x$$

now
$$\frac{1}{\sqrt{-g}}(\sqrt{-g}{T^\nu}_\mu)_{,\nu} = 0$$
$$\Rightarrow(\sqrt{-g}{T^0}_\mu)_{,0} = -(\sqrt{-g}{T^i}_\mu)_{,i}$$

so
$$\int ({T^0}_\mu (\sqrt{-g}))_{,0} d^3 x$$
$$= -\int ({T^i}_\mu (\sqrt{-g}))_{,i} d^3 x$$

This is zero, but I don't know why. I have come across the same thing before but never really resolved it https://www.physicsforums.com/showthread.php?t=268733"

I've come a long way since then , but I'm still not getting this one!

Last edited by a moderator: Apr 24, 2017
2. Mar 4, 2009

### xboy

I checked the question on Schutz and it says that the stress energy tensor is non zero 'only in some bounded region of each spacelike hypersurface $$x^0 =$$ const. '. I don't understand what that's supposed to mean. Does it mean we're working with closed surfaces and Gauss' law applies. What do you make of it?

3. Mar 4, 2009

### turin

Combine this with the divergence theorem, and the problem is solved.

4. Mar 4, 2009

### xboy

Turin, If I combine this with divergence theorem shouldn't the integral just vanish, as a consequence of the conservation of energy-momentum?

5. Mar 4, 2009

### turin

Of course, and that's what this problem shows. However, you still need to do the last step. Mathematically, the divergence theorem has nothing to do with energy and momentum, but with the relation of what's "enclosed" to what's "flowing in and out". If you can find a clever surface ...

6. Mar 4, 2009

### Mmmm

$$\int_{x_{0}=const} {T^\nu}_\mu (\sqrt{-g})n_\nu d^3 x = \int_{x_{0}=const}\frac{1}{\sqrt{-g}}(\sqrt{-g}{T^\nu}_\mu)_{,\nu}\sqrt{-g}d^4x = 0$$
if$g_{\alpha \beta}$ is independent of $$x^\mu$$
....

but how does this show that the original integral is independent of $x^0$ ?

is what I did in my first post not the thing to do then?

7. Mar 5, 2009

### xboy

If it vanishes, that means it's a constant and independent of $$x^0$$

In your first post you did not use the fact that we are dealing here with a closed hypersurface, which is what Schutz seems to be implying.

8. Mar 5, 2009

### Mmmm

but if the integral is always zero it is independent of everything!
that can't be right can it?
surely I must show that only the differential wrt t vanishes???
and if this is the answer what is the point in having x^0 constant? the integral is zero no matter what!
I think I'm missing something here!

9. Mar 5, 2009

### xboy

I don't like this solution either. For one, it does not really use the formula mentioned. The only way out is to say that the surface isn't closed. But if that's not what Schutz meant, what did he mean by that sentence?

10. Mar 5, 2009

### turin

In in OP you just need to add one (or two) more steps. You need to show that the integral in the last step of the OP vanishes. In order to do that, combine what xboy said in post #2 with the divergence theorem. Instead of integrating over all space, integrate of a very large volume of space, and let the volume approach infinity. (This is how you should always think of an integral over all space.)

11. Mar 6, 2009

### Mmmm

So
$$-\int ({T^i}_\mu (\sqrt{-g}))_{,i} d^3 x = -\int {T^i}_\mu (\sqrt{-g})n_i d^2 x$$
Using Divergence thm. Is this what you mean?
now I'm stuck again...

$n_i = (1,0,0)$ if we must integrate over y and z, so

$$-\int ({T^i}_\mu (\sqrt{-g}))n_i d^2 x = -\int {T^1}_\mu (\sqrt{-g})dydz$$

.....??

or could I say that
$$-\int ({T^i}_\mu (\sqrt{-g}))_{,i} d^3 x$$
$$=-\left(\int ({T^x}_\mu (\sqrt{-g}))_{,x} dxdydz + \int ({T^y}_\mu (\sqrt{-g}))_{,y} dxdydz + \int ({T^z}_\mu (\sqrt{-g}))_{,z} dxdydz\right)$$

$$=-\left(\int ({T^x}_\mu (\sqrt{-g}))(x_2)-({T^x}_\mu (\sqrt{-g}))(x_1)dydz + \int ({T^y}_\mu (\sqrt{-g}))(y_2)- ({T^y}_\mu (\sqrt{-g}))(y_1)dxdz +$$
$$\int ({T^z}_\mu (\sqrt{-g}))(z_2)-({T^z}_\mu (\sqrt{-g}))(z_1) dxdy\right)$$

But here I havent used the divergence - or maybe I have but without knowing it...
Letting the volume -> infinity means x_1 x_2 etc are very far apart... why does this make the integral -> 0 ?

I'm afraid that I really need someone to hold my hand through the next steps as I'm feeling a bit shaky here!

12. Mar 7, 2009

### turin

Understand that I am also taking into consideration what xboy said about the problem statement in Schutz.

When you apply the divergence theorem, you get a closed surface. What is that surface? What do you know about the value of the stress tensor there? If you have trouble figuring it out, try writing the limits of integration explicitly.

13. Mar 9, 2009

### Mmmm

$$-\int ({T^i}_\mu (\sqrt{-g}))_{,i} d^3 x = -\int {T^i}_\mu (\sqrt{-g})n_i d^2 x=-\int {T^1}_\mu (\sqrt{-g})dydz$$
So the surface is $x^1=const$
I need to think about the value of ${T^1}_\mu$ on this surface...

${T^1}_\mu$ is the momentum per unit volume per second flowing across the surface - right?

My problem here is the translation from maths into physics... I can (usually) deal with the algebra but the interpretation I find very difficult.

Is the value of ${T^1}_\mu$ the same all along any particular given surface ($x^1=const$) ?.. so ${T^1}_\mu(y_2) = {T^1}_\mu(y_1)$ ? But that still won't make the integral disappear because of the $(\sqrt{-g})$ which can also be a function of y and/or z.

I still don't get it :(

14. Mar 9, 2009

### xboy

Turin, can you please tell me how you interpret ' bounded region of each spacelike hypersurface $$x^0 =$$const.

Does it mean that $$x^0 =$$ const. are closed hypersurfaces ? Or is it something else?

15. Mar 12, 2009

### turin

I interpret that as a simply-connected 3-D FINITE volume at a given time. As an anology, consider the xyz space. If someone says ' bounded region of each constant-z hypersurface z=const. ', then I would consider some finite area of each plane that is perpendicular to the z-axis. I would not interpret it as a closed surface, but I would assume the existence of a closed curve bounding the region (a reduction of 2 dimensions).

I don't know what that means, but I suspect not. (I'm disregarding a closed cosmology or an extremely warped spacetime.)

Last edited: Mar 12, 2009
16. Mar 12, 2009

### Mmmm

Ah..I thought it looked strange...

$$-\int {T^i}_\mu (\sqrt{-g})n_i d^2 x=-\int^{z_2}_{z_1}\int^{y_2}_{y_1} {T^1}_\mu (\sqrt{-g})dydz-\int^{z_2}_{z_1}\int^{x_2}_{x_1}{T^2}_\mu (\sqrt{-g})dxdz-\int^{y_2}_{y_1}\int^{x_2}_{x_1}{T^3}_\mu (\sqrt{-g})dxdy$$

But I'm afraid that I still don't see why it vanishes...

17. Mar 12, 2009

### turin

OK, next step. What are the values of those limits?

x1=?,x2=?,y1=?y2=?z1=?,z2=?

Hint: They are determined by the limits of the volume integral (that you did not show). So, what are the limits of integration of the volume integral (effectively same question)?

x1=?,x2=?,y1=?y2=?z1=?,z2=?

18. Mar 13, 2009

### Mmmm

But I thought it was an arbitrarily bounded space so any values for the limits would be ok..
Or do you mean choose the limits so that they are simple eg x1=y1=z1=0, x2=y2=z2=1?

19. Mar 17, 2009

### turin

I understand your confusion, probably more than you know. This is one of those things that physicists do (including those physicists that write textbooks, unfortunately): If an integral doesn't have limits, you have to guess what they are from context. In this case the integral is over all space, which you have to guess from the clue that they give you (x^0=constant). So, what to a math person would be an INdefinite integral is to a physicist a definite integral. You are right to be confused, but unfortunately you have to get used to this practice in order to understand a lot of the literature. I would not call the limits arbitrary, but if you're unfamiliar with this practice, then they certainly are ambiguous. I'm sorry to tell you that the confusion gets much MUCH worse when (if?) you start dealing with distributions and renormalization (properly).

Yes.
No. The values of the limits are more or less fixed (to "very large" values), but you can sometimes make things simple if you pick a nice coordinate system. In this case, it doesn't really matter, because you don't need to evaluate the integral; you only need to recognize that the integrand has a certain constant value when its coordinates are in a certain range.

Last edited: Mar 17, 2009
20. Mar 18, 2009

### Mmmm

Thanks..you made me feel a little less stupid! :D

So if the integral is over all space I should choose x2=a and x1=-a where a is a very large number, and similarly for y and z. is that right?

also should I have the limits in my integrand here? like this:
$$-\int {T^i}_\mu (\sqrt{-g})n_i d^2 x=$$
$$-\int^{z_2}_{z_1}\int^{y_2}_{y_1} {T^1}_\mu (\sqrt{-g})(x_2) - {T^1}_\mu (\sqrt{-g})(x_1)dydz-$$
$$\int^{z_2}_{z_1}\int^{x_2}_{x_1}{T^2}_\mu (\sqrt{-g})(y_2)-{T^2}_\mu (\sqrt{-g})(y_1)dxdz-$$
$$\int^{y_2}_{y_1}\int^{x_2}_{x_1}{T^3}_\mu (\sqrt{-g})(z_2)-{T^3}_\mu (\sqrt{-g})(z_1)dxdy$$

Last edited: Mar 18, 2009
21. Mar 19, 2009

### Mmmm

Is it the case that as $x\rightarrow \infty, {T^\alpha} _\beta (x) \rightarrow0$?

A number of things I've read recently imply this is the case but I'm not sure if this is true, and if it is I have no idea why..

22. Mar 22, 2009

### turin

Alright. It looks like you've got your limits sorted out pretty well, now.

Yes. Well, actually, it's better than that. A localized distribtution (i.e. bounded support), f(x,y,z), means that there exists some finite r such that f(x,y,z) vanishes exactly whenever x^2+y^2+z^2>r^2. I don't intend that to be some rigorous or profound statement. I am just sharing with you the way that I think about it, so maybe that can be one less confusion for you in the future. Notice that this implies that, if the magnitude of ANY ONE of the coordinates is greater than r, then f(x,y,z)=0. Don't get too hung up on what the value of r actually is. All you need is for it to be finite.

That's another one of those things that seems arbitrary when you first encounter it. In fact, by now you're probably so used to assuming the existence of things that don't really exist for the sake of solving problems in previous courses (e.g. plane waves) that it may seem strange and frustrating when you impose these conditions that actually make much more physical sense (e.g. bounded support).

I restate yet again that I don't know that this assumption is valid for your particular problem. It is just what I would assume, and this was further supported by xboy's input.

23. Mar 26, 2009

### Mmmm

Which means that I have the answer...
Thanks Turin.. You've really helped me. and thanks for explaining things in the way that you have..It's been really useful.
So as long as T is bounded this applies right? Is that the assumption you are talking about? The question is as Xboy stated, I just foolishly missed that part out in the OP thinking that it was unimportant.

24. Mar 26, 2009

### turin

Here you need to be careful about the language you use. For example, if a problem states that "T is bounded", I would interpret that to mean that |T|<Tmax for some arbitrary finite value, Tmax. However, I would not interpret this to mean that T vanishes outside of some finite sphere, which is the interpretation of, "T has bounded support". The first case regards the possible values of T itself (sort of like the range of a function), whereas the second case regards the values of the argument of T for which T is nontrivial (sort of like the domain of a function).

At any rate, you definitely learned something from this problem, which is the main point. Homework problems are used to maximize Δ(understanding) rather than demonstrate maximum understanding - the latter is what examination is for.

25. Mar 27, 2009

### Mmmm

Ah... yes I see...
The question actually says:
Suppose that $T^{\alpha \beta} \neq 0$ only in some bounded region of each spacelike hypersurface.....

So I think that fits our bill nicely eh?