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By energy conservation, show the 4-velocity of dust satisfies this...

  1. Oct 31, 2017 #1
    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. jcsd
  3. Nov 5, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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