Proving the tensor virial theorem In Schutz's Relativity Chapter 4, problem 23b) states:
Use the identity $T^{\mu\nu} _{~~~~~,\nu} = 0 $ to prove the following results for a bounded system (i.e. a system for which $T^{\mu\nu} = 0 $ outside a bounded region of space):
$$\frac{\partial^2}{\partial t^2}\int T^{00}x^i x^j d^3 x=2\int T^{ij}d^3x ~(\text{tensor virial theorem})$$
I have derived the following result:
\begin{align*}
\left(T^{\alpha\beta}x^ix^j\right)_{,\alpha ,\beta}&=\left(\left(T^{\alpha\beta}x^ix^j\right)_{,\alpha }\right)_{,\beta}\\
&=\left(T^{\alpha\beta}_{~~~~~,\alpha}~x^ix^j + T^{\alpha\beta}(x^i_{~,\alpha}x^j+x^j_{~,\alpha}x^i)\right)_{,\beta}\\
&=\left(T^{\alpha\beta}x^i_{~,\alpha}x^j+T^{\alpha\beta}x^j_{~,\alpha}x^i\right)_{,\beta} ~~~\text{(using }T^{\mu\nu} _{~~~~~,\nu} = 0 \text{)}\\
&=\left(T^{i\beta}x^j+T^{j\beta}x^i\right)_{,\beta}\\
&=T^{i\beta}x^j_{~,\beta}+T^{j\beta}x^i_{~,\beta} ~~~\text{(using }T^{\mu\nu} _{~~~~~,\nu} = 0 \text{ again)}\\
&=T^{ij}+T^{ji}=2T^{ij} ~~~\text{(by symmetry of } T\text{)}
\end{align*}
Expanding this expression spits out the integrand of the LHS of the theorem in the first term:
\begin{align*}
\left(T^{\alpha\beta}x^ix^j\right)_{,\alpha ,\beta}&=\frac{\partial^2}{\partial t^2}{T^{00}}x^i x^j + ~...
\end{align*}
I'm stuck here because I can't find a way to cancel out the other terms in the expansion. I'm not sure if it needs some version of the divergence theorem. I could also be going in a completely wrong direction.
 A: Start with $$\frac12\partial_0^2\int_\Sigma T^{00}x^i x^j \ d^3 x = \frac12\partial_0\left(\int_\Sigma \partial_0T^{00}x^i x^j \ d^3 x\right)$$
Since we're integrating over a spatial hypersurface, we can add total spatial derivatives without altering anything, since by the divergence theorem, $\int_\Sigma \partial_iA^{ijk...} \ d^3x = 0$.
$$
\frac12\partial_0\left(\int_\Sigma \partial_0T^{00}x^i x^j + \partial_k(T^{0k}x^ix^j)\ d^3 x\right)
$$
You might be a little concerned about this step, since the spatial derivative is also acting on two $x^i$'s, so the boundary conditions seem insufficient. Normally you'd have to do something fiddly involving making the E-M tensor fall off faster than $\frac{1}{r^2}$ (schematically), but here we can get away with it since the linearised quadrupole formula is usually applied to situations where the source of energy-momentum is localised (so there are no drop-off issues). Thus taking $\Sigma$ as a ball of radius $r$, we can just increase $r$ until all sources are within the hypersurface: $\int_\Sigma \partial_k(T^{0k}x^ix^j)\ d^3 x = \int_{\partial\Sigma} T^{0k}x^ix^j\ d\mathbf{S} = 0$, since $T^{\mu\nu}$ is cleanly zero on the boundary (alternatively, note that Schutz just flatly states in the question that $T^{\mu\nu}$ is zero outside a bounded region. Just extend $\Sigma$ to include this bounded region within it. If this step also sounds strange, recall that we perform a similar trick while working with localised charge distributions in electrostatics problems).
Next, turn the time derivative into a spatial one using conservation of the energy-momentum tensor (the linearised version, in this case):
$$
\frac12\partial_0\left(\int_\Sigma  -\partial_kT^{0k}x^i x^j + \partial_k(T^{0k}x^ix^j)\ d^3 x\right)
$$
You can verify for yourself that the integrand is equal to $(T^{0i}x^j + T^{0j}x^i)$, so we have
$$
\frac12\partial_0\int_\Sigma  T^{0i}x^j + T^{0j}x^i \ d^3 x
$$
Again, add a spatial derivative (using the same reasoning as before):
$$
\int_\Sigma  \partial_0 T^{0(i} x^{j)} + \partial_k(T^{ik} x^j) \ d^3 x
$$
Whereupon you effect a series of trivial relations to simplify the integrand:
$$ \partial_k(T^{ik}x^j) + \partial_0 T^{0i}x^j  = \partial_k(T^{ik}x^j) - (\partial_k T^{ik})x^j = T^{ik}\partial_k x^j = T^{ik}\delta^j_k = T^{ij}$$
