Dodelson - Problem 2.4

Chapter 2 - The Smooth, Expanding Universe

Exercise 4
Find how the energy of a massive, nonrelativistic particle changes as the universe expands. Recall that in the massless case we used the fact that $g_{\mu\nu}P^{\mu}P^{\nu} = 0$. In this case, it is equal not to zero, but to $-m^2$.

Solution
We consider the zeroth component of the geodesic equation to obtain $$\frac{d^2x^{\mu}}{d\lambda ^2} = - \Gamma_{\alpha\beta}^{\mu}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda}$$
Recall that we define $\lambda$ via the energy-momentum four vector: $$P^{\alpha} = \frac{dx^{\alpha}}{d\lambda}$$ Thus we obtain $$\frac{d}{d\lambda} = \frac{dx^{0}}{d\lambda}\frac{d}{dx^0} = E \frac{d}{dt}$$ This enables us to rewrite the zeroth component of the geodesic equation: $$E \frac{dE}{dt} = - \Gamma_{ij}^{0}P^{i}P^{j}$$ $$= -\delta_{ij}a^2P^{i}P^{j}$$ Another important fact that we are employing is that for a massive particle we have $$g_{\mu\nu}P^{\mu}P^{\nu} = -E^2 + \delta_{ij}a^2P^{i}P^{j} = -m^2$$ and so we can, again, rewrite the geodesic equation to read $$E \frac{dE}{dt} = - \frac{\dot{a}}{a}(E^2 - m^2)$$ $$\frac{E}{E^2 - m^2}\frac{dE}{dt} = -\frac{1}{a}\frac{da}{dt}$$ where we can cancel the dt terms and integrate making use of a substitution $u = m^2 + E^2$ to obtain $$\frac{1}{2}ln(m^2 + E^2) = - ln(a) + c$$ $$m^2 + E^2 \propto a^{-2}$$ and finally $$E \propto \frac{1}{a}$$ .