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Consider a stationary solution with stress-energy $T_{ab}$ in the context of linearized gravity. Choose a global inertial coordinate system for the flat metric $\eta_{ab}$ so that the "time direction" $(\frac{\partial }{\partial t})^{a}$ of this coordinate system agrees with the time-like killing vector field $\xi^{a}$ to zeroth order.

(a) Show that the conservation equation, $\partial^{a}T_{ab} = 0$, implies $\int _{\Sigma}T_{i\nu} d^{3}x = 0$ where $i = 1,2,3$, $\nu = 0,1,2,3$, and $\Sigma$ is a $t = \text{constant}$ hypersurface (therefore it has unit future-pointing normal $n^{\mu} = \delta ^{\mu}_{t}$).

(there is also a part b but it is trivial given the result of part a so I don't think there is any need to list it here)

I am very lost as to where to start for this question. Usually for these kinds of problems, you would take the local conservation equation $\partial^{a}T_{ab} = 0$ and use the divergence theorem in some way but that doesn't seem to be of any use here given the form of $\int _{\Sigma}T_{i\nu} d^{3}x = 0$ (it isn't the surface integral of a vector field over the boundary of something nor is it the volume integral of the divergence of a vector field over something - it's just the integral over $\Sigma$ of a scalar field $T_{i\nu}$ for each fixed $i,\nu$). The only thing I've been able to write down that might be of use is that since the linearized field equations are $\partial^{\alpha}\partial_{\alpha}\gamma_{\mu\nu} = -16\pi T_{\mu\nu}$, we have that $\partial^{t}\partial^{\alpha}\partial_{\alpha}\gamma_{\mu\nu} = \partial^{\alpha}\partial_{\alpha}\partial^{t}\gamma_{\mu\nu} = 0 = \partial^{t}T_{\mu\nu}$ where I have used the fact that in this global inertial coordinate system with stationary killing field $\xi^{a} = (\frac{\partial }{\partial t})^{a}$, the perturbation cannot have any time dependence. This then reduces the conservation equation to $\partial^{\mu}T_{\mu\nu} = \partial^{i}T_{i\nu} = 0$ where again $i=1,2,3$. I really haven't been able to make much progress from here though. I would really appreciate any and all help, thanks.

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The solution can indeed be acquired by making use of the divergence theorem, which relates the integral over an $n$-dimensional surface to that over an $(n-1)$-dimensional one. Furthermore, one can also apply it to tensors. For your problem, this means that we can write the equality

$$\int_V\partial^\mu T_{\mu\nu}\,d^4x=\int_\Sigma n^\mu T_{\mu\nu}d^3x,$$

where we take $V$ to be the four-dimensional volume. Since the vector $n^\mu$ is a unit normal vector to the surface of constant $t$, one can view its contraction with the tensor as a projection on the remaining coordinates, essentially removing the zero-component. We can therefore rewrite the equality as

$$\int_V\partial^\mu T_{\mu\nu}\,d^4x=\int_\Sigma T_{i\nu}\, d^3x. $$

Since the integrand on the left hand side vanishes, the right hand side vanishes as well.

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