Why Does the Stress-Energy Tensor Conservation Lead to a Surface Integral?

[mason]

Homework Statement

Hello I'm trying to self study A First Course in General Relativity (2E) by Schutz and I've come across a problem that I need some advice on.

Here it is:
Use the identity T^μ$\nu$_,$\nu$=0 to prove the following results for a bounded system (ie. a system for which T^μ$\nu$=0 outside of a bounded region
a)
$\frac{\partial}{\partial t}$$\int$T^0$\alpha$d³x=0

Homework Equations

T is a symmetric tensor so T^μ$\nu$=T^{$\nu μ$}

The Attempt at a Solution

The Integral is over spatial variables so I brought the integral inside making
$\frac{\partial}{\partial t}$$\int$T^0$\alpha$d³x
=$\int$$\frac{\partial}{\partial t}$T^0$\alpha$d³x
=$\int$T^0$\alpha$_,0d³x
and then I would say I use the identity given to say T^0$\alpha$_,0=0

In the solution manual though, Schutz says the identity gives us that
T^0$\alpha$_,0=-T^j0_,j for a reason that completely eludes me and then used gauss' law to convert it to a surface integral, then said that since the region of integration is unbounded the integral can be taken anywhere (ie outside of the bounded region where T=0).

Does anybody know why I can't just say that T^0$\alpha$_,0=0 from the identity T^μ$\nu$_,$\nu$=0 ?

 

[TSny]

Hello, Mason. Welcome to PF!

Does anybody know why I can't just say that T^0$\alpha$_,0=0 from the identity T^μ$\nu$_,$\nu$=0 ?

Keep in mind the Einstein summation convention: a repeated index denotes summation over that index. So, the left hand side of T^μ$\nu$_,$\nu$=0 is actually the sum of 4 terms.

In the solution manual though, Schutz says the identity gives us that
T^0$\alpha$_,0=-T^j0_,j

Note that this equation can't be correct. There is a lone $\alpha$ index on the left, but no $\alpha$ index on the right. Did you copy this equation correctly?