(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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

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