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Mmmm

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## Homework Statement

Using the (previously proved) equation :

if [itex] g_{\alpha \beta} [/itex] is independent of [tex]x^\mu[/itex] then

[tex] \frac{1}{\sqrt{-g}}(\sqrt{-g}{T^\nu}_\mu)_{,\nu} = 0[/tex]

Prove that

[tex]\int_{x_{0}=const} {T^\nu}_\mu (\sqrt{-g})n_\nu d^3 x[/tex]

is independent of [itex]x^0[/itex] if [itex]n_\nu[/itex] is the unit normal to the hypersurface.

## The Attempt at a Solution

Need to prove that the derivative of the integral wrt [itex]x^0[/itex] is zero.

as [itex]n_\nu[/itex] is the unit normal to the hypersurface, [itex]n_\nu[/itex]= (1,0,0,0).

so

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

[tex]=\frac{\partial}{\partial t}\int {T^0}_\mu (\sqrt{-g}) d^3 x[/tex]

[tex]=\int ({T^0}_\mu (\sqrt{-g}))_{,0} d^3 x[/tex]

now

[tex] \frac{1}{\sqrt{-g}}(\sqrt{-g}{T^\nu}_\mu)_{,\nu} = 0[/tex]

[tex]\Rightarrow(\sqrt{-g}{T^0}_\mu)_{,0} = -(\sqrt{-g}{T^i}_\mu)_{,i}[/tex]

so

[tex]\int ({T^0}_\mu (\sqrt{-g}))_{,0} d^3 x[/tex]

[tex]= -\int ({T^i}_\mu (\sqrt{-g}))_{,i} d^3 x[/tex]

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!

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