I'm reading the Cosmology lecture notes Daniel Baumann and there they describe the path of a freely-falling particle along a geodesic, which is denoted by the curve $X^{\mu}(\tau)$, $\tau$ being proper time and $\mu$ denoting a spacetime coordinate. The geodesic equation is given by $$ \frac{d^2 X^{\mu}}{d \tau^2} = - \Gamma^{\mu}_{\alpha \beta} \frac{dX^{\alpha}}{d\tau} \frac{dX^{\beta}}{d\tau}, $$ with $\Gamma^{\mu}_{\alpha \beta}$ the Cristoffel symbols. Defining $U^{\mu} := dX^{\mu}/d\tau$ and $P^{\mu} := m U^{\mu}$, they say that due to the homogeneity of the universe, one has $d P^{\mu}/dX^{i} = 0$ (the $i$ representing a spatial coordinate). This seems to be equivalent to saying that $\frac{d^2 X^{\mu}}{dX^{i} d \tau} = 0$, however I don't see how this follows from homogeneity. I would say $X^{\mu}(\tau)$ is only a function of $\tau$, hence the derivative would vanish. However, it seems that this reasoning is too simple and there is more at play here. What am I missing in my understanding?


1 Answer 1


First, $\frac{d}{dX^{i}}= 0$ can be considered as "generator" of displacement in $X^i$ direction so the statement $d P^{\mu}/dX^{i} = 0$ implies the velocity/momentum is independent of displacement along the $X^i$ axis, which is just the definition of homogeneity. I used quote when saying generator since Hamiltonian formulation of GR is bit involved. But the essence is what I just said.

Second, the fact that $P^{i}$ is only a function of $t$ or $x^0$ comes not just from homogeneity but also from isotropy of space. It's another way to say Hubble's law while excluding some parts.


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