- #1

jcap

- 170

- 12

##ds^2=-dt^2+a(t)^2(dx^2+dy^2+dz^2)##

As I understand it, since the metric does not depend on the the spatial co-ordinates, there exist Killing vectors in the ##x##,##y##,##z## directions.

Does this imply that the 3-momentum of particles traveling on geodesics is conserved (when one refers to the co-moving co-ordinates ##x##,##y##,##z##) ?

I have read elsewhere that the 3-momentum of free particles goes like ##1/a(t)## instead.

Is the difference due to proper co-ordinates being used instead of the co-moving co-ordinates ##x##,##y##,##z##?