1
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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 '19 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. $\endgroup$ – liu111111119 Oct 29 '19 at 17:18
  • $\begingroup$ See if my answer helps/ $\endgroup$ – Andrea Oct 29 '19 at 18:16
0
$\begingroup$

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 $\frac{d}{d\tau} = \frac{d x^\alpha}{d\tau} \frac{\partial}{\partial x^\alpha}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.