Energy-stress tensor integration proof (from schutz ch.4)

[Mmmm]

Homework Statement

Use the identity $$T^{\mu \nu}_{ ,\nu} = 0$$ to prove the following results for a bounded system (ie a system for which $$T^{\mu \nu} = 0$$
outside a bounded region of space),

$$\frac{\partial}{\partial t}\int T^{0\alpha}d^{3}x = 0$$

Homework Equations

The Attempt at a Solution

The integral obviously gives 4 equations (one for each $$\alpha$$) which must all be = 0.
I tried just working on the first and passing the partial derivative into the integral and writing $$\frac{\partial}{\partial t} T^{0 0} = -\left(\frac{\partial}{\partial x} T^{0 1} +\frac{\partial}{\partial y} T^{0 2} + \frac{\partial}{\partial z} T^{0 3} \right)$$

This gives
$$- \int T^{0 i}_{,i}d^{3}x$$

which doeesnt really seem to be getting me anywhere - also the same reasoning wouldn't work for the other equations because they must all be partially differentiated wrt t and this is only relevant for $$T^{0 0}$$ in $$T^{\mu \nu}_{ ,\nu} = 0$$

Another direction I thought of was to use Gauss' law but then there is no outward normal one-form and so maybe not...

There are another two parts to this question but I thought that if I had an idea of how to do the first part I could figure the others out by myself.

Thanks for any replies :)

 

[gabbagabbahey]

Since the system is bounded, why not use $T^{0 \alpha}=T^{\alpha 0}$? Then $$\frac{\partial}{\partial t}T^{0 \alpha}=\frac{\partial}{\partial t}T^{\alpha 0}=T^{\alpha 0}_{,0}$$...but what is this last quantity if $T^{\mu \nu}_{ ,\nu} = 0$? ;0)

 

[Mmmm]

Wow, thanks, that is a nifty little trick.

So
$$T^{\mu \nu}_{ ,\nu} = 0$$
$$\Rightarrow T^{\alpha 0}_{,0}= - \left( \frac{\partial}{\partial x} T^{\alpha 1} + \frac{\partial}{\partial y} T^{\alpha 2} + \frac{\partial}{\partial z} T^{\alpha 3} \right)$$

or
$$T^{\alpha 0}_{,0}= - T^{\alpha i}_{,i}$$ where $i = 1,2,3$ and $\alpha = 0,1,2,3.$

and so
$$\frac{\partial}{\partial t} \int T^{0 \alpha} d^{3}x = \int T^{\alpha i}_{,i} d^{3}x$$

which can be written as
$$\int \int T^{\alpha 1} (x_{1}) - T^{\alpha 1} (x_{2}) dydz + \int \int T^{\alpha 2} (y_{1}) - T^{\alpha 2} (y_{2}) dxdz + \int \int T^{\alpha 3} (z_{1}) - T^{\alpha 3} (z_{2}) dxdy$$

and I'm afraid that I have come to that dead end again - but this time all four equations are included which is nice... Is the last step right? why is this 0?

 

[gabbagabbahey]

Well, what do the components $$T^{\alpha i}$$ represent physically?

 

[Mmmm]

Well, what do the components $$T^{\alpha i}$$ represent physically?

The $\alpha$ momentum per second crossing a unit area of surface of constant $x^{i}$

So does the " constant $x^{i}$ " mean that $x_{1}=x_{2}$, $y_{1}=y_{2}$ and $z_{1}=z_{2}$ and therefore the integrands in the above integrals are all zero and so the integrals are also zero?

I feel a bit like I'm clutching at straws here...
Is that right or am I just making stuff up? :D

 

[Mmmm]

I have managed to complete the proof of the second question (Tensor Virial Theorem) but it requires that

$$\int \left[T^{0k}x^{i}x^{j} \right] ^{x^{k}_2}_{x^{k}_1} d^{2}x = 0$$

Which makes me rethink my previous post

I said that $x_{1}=x_{2}$ etc...

But the second proof made me realize that this doesn't really make sense... why should $x_{1}=x_{2}$? these values are unrelated to $T^{\alpha 1}$
What would make more sense (and satisfy both proofs) is that $T^{\alpha 1} (x_{1}) = T^{\alpha 1}(x_{2}) = 0$
Then both
$$\left[T^{0k}x^{i}x^{j} \right] ^{x^{k}_2}_{x^{k}_1} = 0$$ (satisfying the second proof)
and
$$T^{\alpha 1} (x_{1}) - T^{\alpha 1} (x_{2}) = 0$$ (satisfying the first)

But I'm not sure why $T^{\alpha 1} (x_{1}) = 0$.
My logic says this is incorrect - if we have a constant x surface, then $T^{\alpha 1}$ doesn't depend on x and so $T^{\alpha 1} (x_{1}) = T^{\alpha 1}(x_{2}) = T^{\alpha 1}$ but not necessarily 0. So I have a problem with my second proof.

Am i thinking in the right direction anywhere here?