Can the Tensor Integral in GR be Bounded by a Region Outside of d^{3}x?

[Mr-R]

Dear all,

I am self studying GR and stuck on problem (23) on page 108/109. I am trying to do all of them.

First I will start with (a) so you guys can breath while laughing at my attempts at (b) and (c)

(a) Attempt

The tensor in the equation is bounded in the d^{3}x region. Outside the region T^{\alpha\beta}=0

\partial_t ∫T^{t\alpha}d^{3}x=0. I will replace the derivative of the tensor with a spatial Div

=-∫T^{\beta\alpha}_{,\beta}d^{3}x. Then using Gauss' law d^3x -> d^2x;

=-∫T^{\beta\alpha}d^{2}x. It is not bounded by a region now but a surface which extends to infinity (??). So, unbounded? Therefore T^{\beta \alpha}=0

I will continue to (b) and (c) after you guys set me up with (a).

Thanks in advance!

 

[Mr-R]

Since no one is replying, I will attempt part (b).

Problem: show

\frac{\partial^{2}}{\partial t^{2}}∫ T^{tt} x^{i} x^{j} d^{3}x = 2∫ T^{ij} d^{3}x

Attempt:

= \frac{\partial^{2}}{\partial t^{2}}∫ T^{tt} x^{i} x^{j} d^{3}x

operate using one time derivative and using identity:

= \frac{\partial}{\partial t}∫ T^{tt}_{,t} x^{i} x^{j} d^{3}x= -\frac{\partial}{\partial t}∫ T^{tk}_{,k} x^{i} x^{j} d^{3}x

rewriting:

= -\frac{\partial}{\partial t}∫ (T^{tk} x^{i} x^{j})_{,k} d^{3}x + \frac{\partial}{\partial t}∫ T^{tk} x^{i}_{,k} x^{j} d^{3}x + \frac{\partial}{\partial t}∫ T^{tk} x^{i} x^{j}_{,k} d^{3}x<br />

using the identity, the first term is zero and the other wo are symmetric so I will add them:

2\frac{\partial}{\partial t}∫ T^{tk} x^{i}_{,k} x^{j} d^{3}x=2\frac{\partial}{\partial t}∫ T^{tk} \delta^{i}_{k} x^{j} d^{3}x = 2∫ T^{ti}_{,t} x^{j} d^{3}x

using identity again and rewriting:

-2∫ T^{ik}_{,i} x^{j} d^{3}x = -2∫ (T^{ik} x^{j})_{,k} d^{3}x +2∫ T^{ik} x^{j}_{k} d^{3}x

First them is zero as above and I will have = 2∫ T^{ij} d^{3}x

Edit: uh about me calling it identity. I am just swapping with spatial divergence.

 

[Orodruin]

There is a reason that there is a homework template that starts with a heading under which you should fill in the problem statement. Many (most) people will not have the book you are referring to or not have it at hand or have it in the bookshelf and cannot be bothered to go to it to look it up. If you really want to get help on a problem, you should at least make the effort to write a proper problem statement or you are essentially asking us to make the effort of both finding out what the problem is and then ti help you.

 

[Mr-R]

Dear all,

I am self studying GR and stuck on problem (23) on page 108/109. I am trying to do all of them.

First I will start with (a).

Problem: Use the identity Tμν,ν = 0 to prove the following results for a bounded system (i.e. a system for which Tμν = 0 outside a bounded region of space):

(a)

\partial_t ∫T^{t\alpha}d^{3}x=0

Attempt:

\partial_t ∫T^{t\alpha}d^{3}x=0. I will replace the derivative of the tensor with a spatial Div

=-∫T^{\beta\alpha}_{,\beta}d^{3}x. Then using Gauss' law d^3x -> d^2x;

=-∫T^{\beta\alpha}d^{2}x. It is not bounded by a region now but a surface which extends to infinity (??). So, unbounded? Therefore T^{\beta \alpha}=0 which means that the former equation equals 0.

Is this correct?


Thanks in advance

 

[Mr-R]

There is a reason that there is a homework template that starts with a heading under which you should fill in the problem statement. Many (most) people will not have the book you are referring to or not have it at hand or have it in the bookshelf and cannot be bothered to go to it to look it up. If you really want to get help on a problem, you should at least make the effort to write a proper problem statement or you are essentially asking us to make the effort of both finding out what the problem is and then ti help you.

Yup sorry. I assumed too much. (a) should be good now. Could not edit my first post so I had to make a new post. Too late I think.

(b) is clear.

 

[Matterwave]

If you can justify going from your first line to your second line, you're already done since the problem statement says that $T^{\beta\alpha}_{~~~~~~,\beta}=0$.

Maybe I'm missing something, but it seems you just replaced the index $t$ with the index $\beta$ and used a different notation? [strike]Isn't this then a trivial problem?[/strike]

EDIT: I misinterpreted your meaning, I thought you had a sum on t because I saw a lower and an upper index t. But I looked up the problem in Schutz and there is certainly no sum on that index. In this case, I'm afraid I've forgotten how to actually go through this derivation. So I can't help further. Sorry for my comment.

 

[Mr-R]

@Matterwave

No worries mate. Its 3 am here so I will look at your post again tomorrow and reply properly. Thanks anyways for your comment Matterwave.

 

[Matterwave]

I've had time to look at the problem, and here are my thoughts. Your solution looks basically correct except for a few places.

1) Going from your first line to your second line, you should have an spatial divergence only. In other words, instead of $T^{\beta\alpha}_{~~~~~,\beta}$ you should have $T^{i\alpha}_{~~~~~,i}$ in the second line.

2) You come to the conclusion at the very end that $T=0$ (and without seeing any qualifiers, I assume you meant everywhere); however, this is far too strict of a conclusion to make. We certainly should not be able to prove that since $T=0$ outside of a certain bounded region, then it is 0 everywhere!

3) The boundary of a bounded region is certainly not unbounded and "going to infinity". For example, the boundary of a spherical volume, is a (2-D) sphere, which is certainly bounded! What we DO know however is that $T$ vanishes outside of a certain region of space; and so, what happens when I make the Gaussian surface to be outside of this region? :)

 

[Orodruin]

It will be correct once you take Matterwave's suggestions into account. However, let me also mention that there is a different way of doing the problem that involves a four-dimensional integral rather than a two-dimensional one.

 

[Mr-R]

1) Going from your first line to your second line, you should have an spatial divergence only. In other words, instead of $T^{\beta\alpha}_{~~~~~,\beta}$ you should have $T^{i\alpha}_{~~~~~,i}$ in the second line.

Yup, silly me. Still getting used to the *greek letters {0,1,2,3}*...

2) You come to the conclusion at the very end that $T=0$ (and without seeing any qualifiers, I assume you meant everywhere); however, this is far too strict of a conclusion to make. We certainly should not be able to prove that since $T=0$ outside of a certain bounded region, then it is 0 everywhere!

3) The boundary of a bounded region is certainly not unbounded and "going to infinity". For example, the boundary of a spherical volume, is a (2-D) sphere, which is certainly bounded! What we DO know however is that $T$ vanishes outside of a certain region of space; and so, what happens when I make the Gaussian surface to be outside of this region? :)

Oh I forgot that d^2x is a closed surface. I though of it as a sheet extending to infinity. To answer you question; well...T vanishes? But I am not sure on how to do that mathematically. I mean to make a surface outside that region.

It will be correct once you take Matterwave's suggestions into account. However, let me also mention that there is a different way of doing the problem that involves a four-dimensional integral rather than a two-dimensional one.

Thanks. I will search about that when I am done with this exercise. Problem is that I am going to be busy for a few days due to preparation for my flight and such. I will still check PF though.

 

[Matterwave]

Oh I forgot that d^2x is a closed surface. I though of it as a sheet extending to infinity. To answer you question; well...T vanishes? But I am not sure on how to do that mathematically. I mean to make a surface outside that region.

Yes, so T vanishes outside this region. If we draw the Gaussian surface outside this region, then the spatial divergence of T there also vanishes, after all the derivative of 0 is still 0! (We expect T to actually be 0 in some neighborhood of this Gaussian surface as well).