Conserved quantities for geodesics

[Brewer]

Homework Statement

In comoving coordinates, a one dimensional expanding flat universe has a metric $$ds^2 = -c^2dt^2 + at(t)^2dr^2$$. Derive an expression for a conserved quantity for geodesics in terms of $$a, \tau$$ and $$r$$, where $$\tau$$ is the time measured in the rest frame of the freely falling particle.

Homework Equations

The Attempt at a Solution

I have the answer to the question in front of me, I just don't follow one of the steps, so I just wondered if anyone could explain it to me.

After writing a Lagrangian as $$L = c^2(\frac{dt}{d\tau})^2 - a(t)^2(\frac{dr}{d\tau})^2$$

it can be seen that since r does not appear explicitly that it has something to do with it a constant.

The next line in the answer goes onto say
$$\frac{dL}{d\frac{dr}{d\tau}}$$ is a constant, but I don't know why this is.

At a guess I would say this is because the $$\frac{dL}{dr}$$ term in the Euler Lagrange equations is zero (because of the lack of dependence on r), and as such you get $$\frac{d}{d\tau}\frac{dL}{d\frac{dr}{d\tau}} = 0$$, and so by integrating with respect to $$\tau$$ you'll get a constant on the right hand side right? Or am I scratching at the wrong tree?

Any help would be appreciated.

 

[Jhero]

Hello,

You are on the right track! The reason why $$\frac{dL}{d\frac{dr}{d\tau}}$$ is a constant is because of the conservation of energy along geodesics in general relativity. This means that the Lagrangian L is a constant along the path of a freely falling particle. Since r is not explicitly present in L, $$\frac{dL}{d\frac{dr}{d\tau}}$$ is constant. In other words, the quantity $$\frac{dL}{d\frac{dr}{d\tau}}$$ remains the same no matter where the particle is along its path. This is a consequence of the symmetry of the metric in comoving coordinates.

Hope that helps clarify things for you!