# Energy-momentum tensor identity - linearized gravity

1. Apr 25, 2013

### WannabeNewton

1. The problem statement, all variables and given/known data
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)

3. The attempt at a solution
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). 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.

2. Apr 26, 2013

### andrien

that is actually very easy and it has nothing to do with linearized gravity.The integral form is most conveniently written as
∂ΩTμvnvdσ=0
where ∂Ω is the boundary of Ω,which is a hypersurface with constant t,nv is outward normal vector of ∂Ω.you just use gauss integral theorem to obtain the result for an orbitrary Ω.You might have seen something similar while deriving schwinger dyson eqn.

3. Apr 26, 2013

### WannabeNewton

Hi andrien! Thanks for replying. I assume you did $\int _{\partial\Sigma}T_{\mu\nu}n^{\nu}d^{2}x = \int _{\Sigma}\partial^{\mu}T_{\mu\nu}d^{3}x = 0$ by using the divergence theorem and the fact that $\partial^{\mu}T_{\mu\nu} = 0$? That was what I originally did but the book says to show $\int _{\Sigma}T_{i\nu}d^3x = 0$ i.e. that the integral over $\Sigma$ of the scalar field $T_{i\nu}$, for each fixed $i,\nu$, vanishes; applying the divergence theorem on the other hand just gives $\int _{\partial\Sigma}T_{i\nu}n^{i}d^2x = 0$ which isn't exactly what Wald wants. Hopefully I didn't misinterpret what you wrote down.

4. Apr 26, 2013

### andrien

dσ is a three dimensional surface,not the two dimensional one.∂Ω is same as your Ʃ, a hypersurface with t=constant.Also in your phrase d3x=nvdσ.

5. Apr 26, 2013

### WannabeNewton

But what would $\Sigma$ be the single boundary of? If I take a hypercube then even with all the spatial sides off at infinity (so that on account of asymptotic flatness the integral over these sides vanishes) I'm still left with the two temporal sides which are $t,t'=const.$ so the integral over the boundary will in general have terms on both $\Sigma_{t}$ and $\Sigma_{t'}$ so I won't end up with just a single hyperplane as my boundary.

Also, as far as I can tell Wald uses $d\sigma$ interchangeably with $d^{3}x$ for $\mathbb{R}^{4}$ so I'm not seeing why $\int T_{\mu\nu}n^{\nu}d\sigma$ is the same thing as $\int T_{\mu\nu}d^{3}x$ where in the second expression, $\mu,\nu$ are fixed. If it was simply $\int T_{\mu\nu}n^{\nu}d\sigma = 0$ by applying the divergence theorem then it would be a sum over the second index so I will end up with a sum of different components of $T_{\mu\nu}$ in the integral adding up to zero whereas the problem wants me to show that the individual components $T_{\mu\nu}$ integrate to zero over $\Sigma$ for all indices except $\mu = \nu = 0$. Thanks again andrien.

Last edited: Apr 26, 2013