Mmmm
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Homework Statement
Using the (previously proved) equation :
if g_{\alpha \beta} is independent of x^\mu[/itex] then<br /> <br /> \frac{1}{\sqrt{-g}}(\sqrt{-g}{T^\nu}_\mu)_{,\nu} = 0<br /> <br /> Prove that <br /> \int_{x_{0}=const} {T^\nu}_\mu (\sqrt{-g})n_\nu d^3 x<br /> <br /> is independent of x^0 if n_\nu is the unit normal to the hypersurface.<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> Need to prove that the derivative of the integral wrt x^0 is zero.<br /> <br /> as n_\nu is the unit normal to the hypersurface, n_\nu= (1,0,0,0).<br /> <br /> so<br /> <br /> \frac{\partial}{\partial t}\int_{x_{0}=const} {T^\nu}_\mu (\sqrt{-g})n_\nu d^3 x<br /> <br /> =\frac{\partial}{\partial t}\int {T^0}_\mu (\sqrt{-g}) d^3 x<br /> =\int ({T^0}_\mu (\sqrt{-g}))_{,0} d^3 x<br /> <br /> now <br /> \frac{1}{\sqrt{-g}}(\sqrt{-g}{T^\nu}_\mu)_{,\nu} = 0<br /> \Rightarrow(\sqrt{-g}{T^0}_\mu)_{,0} = -(\sqrt{-g}{T^i}_\mu)_{,i}<br /> <br /> so<br /> \int ({T^0}_\mu (\sqrt{-g}))_{,0} d^3 x<br /> = -\int ({T^i}_\mu (\sqrt{-g}))_{,i} d^3 x<br /> <br /> This is zero, but I don't know why. I have come across the same thing before but never really resolved it <a href="https://www.physicsforums.com/showthread.php?t=268733"" class="link link--internal">https://www.physicsforums.com/showthread.php?t=268733"</a><br /> <br /> I've come a long way since then , but I'm still not getting this one!
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