# Homework Help: By energy conservation, show the 4-velocity of dust satisfies this...

1. Oct 31, 2017

### whatisreality

1. The problem statement, all variables and given/known data
Using conservation of energy, show that the 4-velocity $u^{\mu}$ of dust satisfies:
$u^{\mu}\nabla_{\mu}u^{\nu} = f u^{\nu}$

Explicitly identify $f$, which is some function of $u^{\mu}$, $\rho$ and their derivatives. And show that this equation becomes

$\dot{\rho} + 3 \left(\frac{\dot{a}}{a}\right)\rho = 0$

For comoving observers in a homogeneous Friedmann-Robertson-Walker Universe.

2. Relevant equations

3. The attempt at a solution
I don't think I've done the first bit right, I'm not really sure how to treat the $\rho$. For dust, $T^{\mu\nu} = \rho u^{\mu}u^{\nu}$ and energy conservation means:
$\nabla_{\mu}T^{\mu\nu} = 0$
So
$\nabla_{\mu} (\rho u^{\mu}u^{\nu})$
Using the product rule,
$\rho u^{\mu} (\nabla_{\mu}u^{\nu}) + \rho u^{\nu} (\nabla_{\mu}u^{\mu}) + u^{\mu}u^{\nu}\nabla_{\mu}\rho = 0$
I think $\nabla_{\mu} \rho =0$, but if it is then I could just divide through by $\rho$ and I could rearrange to get the LHS
I want:
$u^{\mu}\nabla_{\mu}u^{\nu}$
But the RHS wouldn't be a function of $\rho$. So if $\nabla_{\mu} \rho$ isn't zero, then I get:
$u^{\mu}\nabla_{\mu}u^{\nu} = -u^{\nu}\nabla_{\mu}u^{\mu} - \frac{u^{\nu}u^{\mu}}{\rho} \nabla_{\mu}\rho$
$= \left(-\nabla_{\mu}u^{\mu}-\frac{u^{\mu}}{\rho}\nabla_{\mu}\rho \right)u^{\nu}$
Is that right? Is $\nabla_{\mu}\rho$ not zero?

2. Nov 5, 2017

### PF_Help_Bot

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