How to calculate the four-momentum of a photon in FRW Metric

[JohnH123]

Homework Statement

Find the 4-momentum of a photon moving in the x-direction. That is, find dt/dλ
and dx/dλ as functions of a. Note that dt/dλ at a = 1 is the present-day frequency f0.

Relevant Equations

The spatially flat Robertson-Walker Metric: ds^2 = -dt^2 + a^2(t)[dr^2+r^2(dtheta^2 + sin^2(theta)dphi^2)]

I have calculated the Christoffel symbols for the above given metric, but I don't understand how to calculate a photon's four-momentum using this information. I believe it has something to do with the null geodesic equation but I can't understand how to put that information into the problem. Thank you.

 

[Abhishek11235]

This problem can be done using geodesic equation of motion. But there is a simpler way to do using Lagrangian mechanics. The Lagrangian of the given metric is:
$L= g_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda}= -\dot t^2+ a^2(t)(\dot r^2+ r^2 \dot \theta^2+ r^2 \sin^2\theta \dot \phi^2)$

Now, the momentum is :
$p_i=\frac{\partial L}{\partial \dot x_i}$
Now, simplify it for example using the fact that metric is isotropic to evaluate $r_{th}$ component of momentum. For $t$ component, you will get same result as geodesic equation. Solve it (again using the fact that metric is isotropic to remove $\theta$ and $\phi$ components from metric).

 

[JohnH123]

This problem can be done using geodesic equation of motion. But there is a simpler way to do using Lagrangian mechanics. The Lagrangian of the given metric is:
$L= g_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda}= -\dot t^2+ a^2(t)(\dot r^2+ r^2 \dot \theta^2+ r^2 \sin^2\theta \dot \phi^2)$

Now, the momentum is :
$p_i=\frac{\partial L}{\partial \dot x_i}$
Now, simplify it for example using the fact that metric is isotropic to evaluate $r_{th}$ component of momentum. For $t$ component, you will get same result as geodesic equation. Solve it (again using the fact that metric is isotropic to remove $\theta$ and $\phi$ components from metric).

Just so I’m clear, $\frac{\partial L}{\partial \dot x_i}$ means the partial of the lagrangian with respect to each coordinate right? So the four momentum has components in coordinates {t, r, $\theta$, $\phi$}, so those components are found by finding $\frac{\partial L}{\partial \dot t}$, $\frac{\partial L}{\partial \dot r}$ and so on, correct?

 

[JohnH123]

This problem can be done using geodesic equation of motion. But there is a simpler way to do using Lagrangian mechanics. The Lagrangian of the given metric is:
$L= g_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda}= -\dot t^2+ a^2(t)(\dot r^2+ r^2 \dot \theta^2+ r^2 \sin^2\theta \dot \phi^2)$

Now, the momentum is :
$p_i=\frac{\partial L}{\partial \dot x_i}$
Now, simplify it for example using the fact that metric is isotropic to evaluate $r_{th}$ component of momentum. For $t$ component, you will get same result as geodesic equation. Solve it (again using the fact that metric is isotropic to remove $\theta$ and $\phi$ components from metric).

Also, could you please expand on how I could calculate the four-momentum using the geodesic equation? I believe that’s the method my professor would like me to use, as stated in the problem. Thank you.