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If a particle travels on a geodesic with 4-momentum $P^\mu$ in a spacetime with a Killing vector $K_\mu$ then we have a constant of motion, $K$, given by: $$K=K_\mu P^\mu$$ Using the relationships: $$P^\mu=mU^\mu$$ and $$K_\mu=g_{\mu\nu}K^\nu$$ we obtain: $$K=mg_{\mu\nu}K^\nu U^\mu$$ Let us assume that $K^\nu$ is a timelike Killing vector so that we have: $$K^\nu=(1,0,0,0)$$ Then the constant $K$ is the total energy $e$ of the particle (including gravitational energy) given by: $$e=-mg_{00}\frac{dt}{d\tau}$$ The above argument is my generalisation from @StanLiou's answer to the question Potential Energy in General Relativity where he gives an expression for the total energy of a particle in geodesic motion in Schwarzschild spacetime. I hope I have got the algebra correct!

My question is: could this definition of total particle energy be carried over to cosmology where one has the FRW metric?

The FRW metric does not have a timelike Killing vector so that one cannot expect the total energy $e$ of a co-moving particle to be constant. But the FRW metric does have a timelike conformal Killing vector so that it seems reasonable that the particle energy should scale in some way with conformal time $\eta$.

We can write the flat FRW metric in conformal co-ordinates: $$ds^2=a(\eta)^2(-d\eta^2+dx^2+dy^2+dz^2)$$ Thus we have: $$g_{00}=-a(\eta)^2$$ $$\frac{d\eta}{d\tau}=\frac{1}{a(\eta)}$$ Therefore $$e=m\ a(\eta)$$ Thus it seems that the total energy of a co-moving particle, when expressed in conformal co-ordinates, scales with the Universal scale factor $a(\eta)$.

Is this reasoning valid?

  • what about the redshift ? –  Jun 13 '15 at 00:05
  • The redshift would be due to observers' clocks running faster with respect to conformal time as the Universe expands. See one of @MichaelSeifert's replies in http://physics.stackexchange.com/a/186483/22307. (I think he meant clocks run faster rather than slower). – John Eastmond Jun 13 '15 at 00:21
  • at least , redshift decreases the radiation energy as universe expands... I don't understand the last calculation. –  Jun 13 '15 at 00:31
  • Perhaps the radiation energy actually stays constant and it's the observer's energy which is increasing, with respect to conformal time. – John Eastmond Jun 13 '15 at 00:44
  • I don't agree ou disagree , this question is very interesting, I was about to post a similar one. Could you please detail the calculation ? –  Jun 13 '15 at 01:00
  • Sorry - which calculation are you referring to? – John Eastmond Jun 13 '15 at 01:02
  • how you found $e=m a(η)$ ... –  Jun 13 '15 at 01:04

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