A Fluid Flow Symmetry: Showing $\pi_b G^b$ is Constant

[ergospherical]

I don't know where to start with this problem. If $\pi_a = (\mu + TS) u_a$ then show that \begin{align*}
u^a \nabla_{a} (\pi_b G^b) = 0
\end{align*}where the field $G^a$ is a symmetry generator. [$S$ is entropy/baryon, $T$ is temperature, $u_a$ is a one-form field corresponding to a fluid-comoving observer and $\mu$ is chemical potential].

 

[PeterDonis]

Moderator's note: Thread level changed to "A".

 

[robphy]

What have you tried? Killing's equation?

 

[ergospherical]

How would I use Killing's equation here? The only thing I wrote down so far was $L_{G} \pi = 0$, the statement of the invariance of $\pi$ under the symmetry transformation corresponding to the Killing vector $G$, but I don't know how to obtain the target equation.

 

[robphy]

Did you expand out the left hand side? I think Killing’s equation will kill off some terms involving \nabla_a G_b. What remains may vanish due to other properties of, say, u^a that are implicit in your problem statement.