I have some homework that I am really struggling with (this class is above my level), and I would hope I could get some clarification.

I am given the geodesic equation $$\frac{\mathrm{d}^{2}x^{\mu}}{\mathrm{d}\lambda^{2}} + \Gamma ^{\mu}{}_{\alpha\beta}\frac{\mathrm{d}x^{\alpha}}{\mathrm{d}\lambda}\frac{\mathrm{d}x^{\beta}}{\mathrm{d}\lambda} = 0,$$ the FRW metric in the form $$\mathrm{d}s^{2} = \mathrm{d}t^{2} - a^{2}(t) \left[\frac{\mathrm{d}r^{2}}{1-\kappa r^{2}} + r^{2} \left( \mathrm{d}\theta^{2} + \sin^{2}\theta \mathrm{d}\phi^{2}) \right)\right],$$ and we are looking at a photon that has a 4-momentum of the form $$ \rho^{\mu} = \frac{\mathrm{d} x^{\mu}}{\mathrm{d}\lambda},$$ with the photon's 3-momentum being $ \rho^{i} = 0 $, and $ E = \rho^{0} $.

We are assuming that the photon travels in the radial direction, and we are working in null geodesics.

The point of the exercise is proving that $ E^{2} - \frac{a^{2}}{1-\kappa r^{2}}(\rho ^{1})^{2} = 0.$

Taking $E = \rho^{0} = \frac{\mathrm{d}x^{0}}{\mathrm{d}\lambda}$, then the first term of the geodesic equation is equal to $E^{2}$. With a similar approach, the terms to right of the Christoffel symbol can be $(\rho^{1})^{2}$ if $ \alpha = \beta = 1 $ (and because in the 4 momentum, as it is traveling radially, the other terms are 0), we are left with $$E^{2} + \Gamma^{0}{}_{1,1} (\rho^{1})^2) = 0$$

This is fairly close to the answer I am looking for, but the Christoffel symbol for $\Gamma^{0}{}_{1,1}$ is $\frac{a\dot{a}}{1-\kappa r^{2}}$.

I know that working backwards is not the best idea, but I am quite lost. I think that I am either quite lost and there is something very obvious that I am not seeing, or that my approach is fundamentally wrong. I hope you can help me, and thank you.


1 Answer 1


Radial geodesics have $\frac{\mathrm{d}\theta}{\mathrm{d}\lambda} = \frac{\mathrm{d}\phi}{\mathrm{d}\lambda} = 0$. Furthermore, for null geodesics $\mathrm{d}s^2=0$ so $$ 0 = \left( \frac{\mathrm{d}t}{\mathrm{d}\lambda} \right)^2 - \frac{a(t)^2}{1-\kappa r^2} \left( \frac{\mathrm{d}r}{\mathrm{d}\lambda}\right)^2 = E^2 - \frac{a^2}{1-\kappa r^2} (\rho^1)^2. $$


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