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## 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[itex]\nu[/itex]}

_{,[itex]\nu[/itex]}=0 to prove the following results for a bounded system (ie. a system for which T

^{μ[itex]\nu[/itex]}=0 outside of a bounded region

a)

[itex]\frac{\partial}{\partial t}[/itex][itex]\int[/itex]T

^{0[itex]\alpha[/itex]}d

^{3}x=0

## Homework Equations

T is a symmetric tensor so T

^{μ[itex]\nu[/itex]}=T

^{[itex]\nu μ[/itex]}

## The Attempt at a Solution

The Integral is over spatial variables so I brought the integral inside making

[itex]\frac{\partial}{\partial t}[/itex][itex]\int[/itex]T

^{0[itex]\alpha[/itex]}d

^{3}x

=[itex]\int[/itex][itex]\frac{\partial}{\partial t}[/itex]T

^{0[itex]\alpha[/itex]}d

^{3}x

=[itex]\int[/itex]T

^{0[itex]\alpha[/itex]}

_{,0}d

^{3}x

and then I would say I use the identity given to say T

^{0[itex]\alpha[/itex]}

_{,0}=0

In the solution manual though, Schutz says the identity gives us that

T

^{0[itex]\alpha[/itex]}

_{,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[itex]\alpha[/itex]}

_{,0}=0 from the identity T

^{μ[itex]\nu[/itex]}

_{,[itex]\nu[/itex]}=0 ?