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


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...

  • $\begingroup$ 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! $\endgroup$
    – Andrea
    Oct 29, 2019 at 17:00
  • $\begingroup$ 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. $\endgroup$ Oct 29, 2019 at 17:18
  • $\begingroup$ See if my answer helps/ $\endgroup$
    – Andrea
    Oct 29, 2019 at 18:16

1 Answer 1


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.

To check that it does, you will have to evaluate expressions such as:

$$\frac{d}{d\tau}g_{\mu\nu}.$$ In these it is implicit that $g_{\mu\nu}=g_{\mu\nu}(x^\alpha)$ so you can use the chain rule $\frac{d}{d\tau} = \frac{d x^\alpha}{d\tau} \frac{\partial}{\partial x^\alpha}$.


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