# 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.

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: $$\partial_k(T^{ik}x^j) + \partial_0 T^{0i}x^j = \partial_k(T^{ik}x^j) - (\partial_k T^{ik})x^j = T^{ik}\partial_k x^j = T^{ik}\delta^j_k = T^{ij}$$

• There might be a couple of minor errors since I'm doing it off the top of my head, but the principle is correct: take only time derivatives rather than 4-derivatives, use linearised energy conservation and integration by parts. Jan 11, 2021 at 17:31
• Could you elaborate on why $\int \partial_kT^{0k}x^ix^jd^3x =0$? Are you using divergence theorem here? Jan 11, 2021 at 22:04
• @HexiangChang Yes, $\int_{\Sigma} \partial_k(T^{0k}x^ix^j) \ d^3 x$ is zero due to the divergence theorem. Jan 12, 2021 at 3:02
• I'm sorry, could you expand on the portion about the divergence theorem? I can't see how you can apply the boundary conditions with the factors of $x^ix^j$ in the integrand. Is it possible to show the steps you took fully? Jan 12, 2021 at 14:23
• @HexiangChang yes, that is correct. Oh, and the linearised quadrupole formula is just the formula for calculating the linearised trace-reversed metric in terms of the quadrupole tensor $\int_\Sigma T^{00}x^i x^j \ d^3 x$ for gravitational waves, the problem that you're asking about is usually used as an intermediate step in simplifying calculations (the name, of course, is not relevant to the proof above) Jan 13, 2021 at 13:51