Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.