Energy-momentum conservation equation

1. Feb 28, 2007

cristo

Staff Emeritus
We have the metric $$ds^2=-e^{2\Phi}dt^2+e^{2\Lambda}dr^2+r^2d\Omega^2$$, and the energy momentum tensor takes the form $$t^{ab}=(\rho+p)u^au^b+pg^{ab}$$ where the 4-velocity is $$u=e^{-\Phi}\partial_t$$, and $\Phi$ and $\Lambda$ are functions of r only.

I'm asked to show that the ebergy-momentum conservation equation $$\nabla_aT^{ab}=0$$ is equivalent to $$(\rho+p)\frac{d\Phi}{dr}=-\frac{dp}{dr}$$.

Now, I'm after some advice on how to calculate this. I know all the connection coefficients (they were given in the question). I tried setting up the four equations, one for each value of b, and then attempting to solve these, but each of these equations has 12 terms in, so I figured there must be an easier way to do this!

Does anyone know whether there is another way, or if there are any simplifications I can make; or do I need to plough through the 4 equations?

Thanks

2. Feb 28, 2007

cristo

Staff Emeritus
Ok, I've got the answer. Just decided to compute the r equation, and it gave the result.