# Flat Roberston-Walker metric particle geodescics

Consider a Robertson-Walker metric with flat spatial section (k= 0),

$$ds^2=dt^2−a^2(t)[dx^2+dy^2+dz^2]$$

Check that particles staying at $$x,y,z=$$ constant follow geodesic motion.

Thoughts as to first steps:

just plug $$t = \alpha$$ and $$x,y,z =$$ constant into the geodesic equation,

$$0=\frac{d}{d \tau} (g_{\alpha \beta} \frac{dx^\beta}{d \tau}) - \frac{1}{2} \partial_\alpha g_{\mu\nu} \frac{dx^\mu}{d \tau} \frac{dx^\nu}{d \tau}$$

to see is the RHS equals $$0$$?

Having trouble believeing this is correct approach, since the $$\alpha=t$$ equation tells me $$\frac{\partial^2 t}{\partial t^2}=0$$. I do not think this is generally true...

• Hi! Welcome to the site. That sounds like a good idea. Don't expect anyone to do the calculation for you however. This is clearly a homework problem. Good luck! – Andrea Oct 29 '19 at 17:00
• Old exam problem from test bank — studying for exam. Sorry not clear, plugging into the geodesic doesn't yield particularly useful results. I'll edit to mak that clear. – liu111111119 Oct 29 '19 at 17:18
• See if my answer helps/ – Andrea Oct 29 '19 at 18:16

Since the metric does not depend on the coordinates $$x,y,z$$, we can assume that the partcle is located at $$(x,y,z) = (0,0,0)$$.
The trajectory of the particle in question is then given in these coordinates by a curve $$x^\mu = x^\mu(\tau) = (\tau,0,0,0)$$. Note we cannot assume $$\tau$$ is the proper time along the geodesic: we must check that.
Let $$\dot x^\mu = \frac{d}{d\tau}x^\mu = (1,0,0,0)$$ and check that it is indeed true that $$g_{\mu\nu}\dot x^\mu \dot x^\nu = +1$$ Now that you have checked that you know that the trajectory is affinely parametised and that thus if it is a geodesic, it will satisfy the geodesic equation as in your post.
$$\frac{d}{d\tau}g_{\mu\nu}.$$ In these it is implicit that $$g_{\mu\nu}=g_{\mu\nu}(x^\alpha)$$ so you can use the chain rule $$\frac{d}{d\tau} = \frac{d x^\alpha}{d\tau} \frac{\partial}{\partial x^\alpha}$$.