4
$\begingroup$

After a suitable change of coordinates, the metric for the (flat) FLRW universe becomes

$$ ds^2 = (1 + 2 \Phi(\vec{x},t))dt^2 - (1-2\Psi(\vec{x},t))d\vec{x}^2, $$ where $$ \Phi(\vec{x},t) = -\frac{1}{2}(\dot{H}(t)+ H^2(t))|\vec{x}|^2, \text{and} \\ \Psi(\vec{x},t) = \frac{1}{4}H^2(t)|\vec{x}|^2. $$ Notice the metric has the form $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$, which is in agreement with the equivalence principle.

Now my problem: for stationary bodies $(\dot{\vec{x}}(t)=0)$ away from $\vec{x}(t)=0$, there is a gravitational force. I need to show that this is equivalent to the Hubble flow, i.e. $\vec{v}(t) = H \vec{x}(t)$.

My attempt: I believe I need to start with the geodesic equation $$ \frac{d U^{\mu}}{d \tau} = -\Gamma^{\mu}_{\ \alpha \beta}U^{\alpha}U^{\beta}, $$ use the fact the $\dot{\vec{x}}(t)=0$ to simplify the equation, and compute the $\Gamma$'s from the metric. Then the $H$ would appear, since it's in the $\Phi(\vec{x},t)$ and $\Psi(\vec{x},t)$.

I believe I need to prove $$ \frac{d}{dt}(a(t)\vec{x},(t)) = \dot{a}(t)\vec{x}(t) = Ha(t)\vec{x}(t) = \vec{v}(t). $$

However, in the geodesic equation, I have the derivative of $\vec{x}$ with respect to the proper time, and the only thing I know is that the derivative of $\vec{x}$ with respect to the time coordinate $t$ vanishes. How can I use this information in the geodesic equation?

Improve this question
$\endgroup$
1
  • $\begingroup$ It's trivial : $\frac{dx}{d\tau} = 0$ if $\frac{dx}{dt} = 0$, since $d\tau = \sqrt{g_{00}} \, dt$. Could you show the coordinates transformation that gives your metric above from the standard FLRW metric? What about the non-flat metric (i.e. $k \ne 0$) ? $\endgroup$
    – Cham
    Jul 8, 2018 at 13:26

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.