# LFRW Universe - equivalence principle and Hubble flow

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?

• 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. $k \ne 0$) ? – Cham Jul 8 '18 at 13:26