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The flat FRW metric can be written in conformal co-ordinates: $$ds^2=a^2(\eta)(d\eta^2-dx^2-dy^2-dz^2)$$ where $\eta$ is conformal time. Let us assume that $a(\eta_0)=1$ when $\eta_0$ is the present conformal time.

Now the energy of a massive particle $E$ is given by: $$E=P^\mu V_\mu=mg_{\mu \nu}U^\mu V^\nu$$ where the 4-momentum of the particle is $P^\mu=mU^\mu$ and $U^\mu$,$V^\nu$ are the 4-velocities of the particle and observer respectively.

Let us assume that both the particle and the observer are co-moving at the same conformal time $\eta$. Therefore the spatial components of their 4-velocities are zero. As the 4-velocities must also be normalised we have: $$g_{00}U^0 U^0=g_{00}V^0V^0=1$$ Therefore the 4-velocities of the co-moving particle and observer are given by: $$U^\mu=V^\mu=(\frac{1}{a(\eta)},0,0,0)$$ Thus the energy $E$ of a co-moving particle at time $\eta$, as measured by a co-moving observer at time $\eta$, is given by: $$E = m\ g_{00}\ U^0\ V^0=m\ a^2(\eta) \frac{1}{a(\eta)} \frac{1}{a(\eta)}=m$$

Thus, using this definition of energy, the energy of individual co-moving massive particles is constant. Therefore, for example, we can say that the mass density of cosmological "dust", used in the Friedmann equations, simply goes like $\rho_m \propto 1/a^3$. This is the conventional viewpoint.

But we can define an energy $E_0$ which is the energy of a comoving particle at time $\eta$ with respect to a comoving observer at the present time $\eta_0$ when $a(\eta_0)=1$:

$$E_0 = m\ g_{00}\ U^0\ V^0=m\ a^2(\eta) \frac{1}{a(\eta)} \frac{1}{1}=m\ a(\eta)$$

My question is this: Perhaps $E_0$ is the correct energy for a comoving particle with respect to the co-ordinates in which the metric is expressed?

It seems to me that the co-ordinate system, used in the FRW metric above, is the system of co-ordinates that corresponds not to an arbitrary co-moving observer at time $\eta$ but to ourselves who are co-moving observers at the present time $\eta_0$.

(In standard co-ordinates in which $g_{00}=1$ both $E$ and $E_0$ are the same. So it seems that one could then argue that the difference between $E$ and $E_0$ does not matter. But I think that in arguing this way one is implicitly assuming that the standard FRW co-ordinate system has a timelike Killing vector which it doesn't have. By contrast when one makes the argument in conformal co-ordinates one is implicitly assuming a conformal timelike Killing vector which is in fact a correct assumption.)

Thus we should say that the rest mass/energy of a comoving massive particle at time $\eta$ is given by $E_0=m\ a(\eta)$.

This would have consequences for Einstein's field equations $G_{\mu \nu}=8\pi G\ T_{\mu \nu}$ when applied to cosmology.

Newton's gravitational constant $G$ in natural units, where $\hbar=c=1$, is given by: $$G = \frac{1}{M_{Pl}^2}$$ where $M_{Pl}$ is the Planck mass.

If rest masses are increasing with the scale factor $a$ with respect to the co-ordinate system then the Planck mass should also increase with the factor $a$. This would imply that Newton's constant actually varies with the scale factor:

$$G \propto \frac{1}{a^2}$$

Thus the Friedmann equations, in either conformal or standard co-ordinates, would have to be modified.

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    $\begingroup$ I am not expert enough to give a full answer - but from my research I know that rest-mass is not well-defined in deSitter space, at least not from a Lagrangean. In order to find the mass of a test particle in dS, one has to solve the equation of motion and compare to test geodesics. So, the rest-mass in dS space cannot just be read off the first component of a four-vector. This is one of the reasons why most theorists try to have a Minkowski vacuum of their theory, even though our universe is slightly dS. $\endgroup$ – Neuneck Nov 16 '15 at 14:33
  • $\begingroup$ The FLRW metric has no timelike Killing vector so the energy you've calculated isn't a conserved quantity. You could I suppose interpret this as the mass changing, but that seems a determinedly eccentric way to view the situation. $\endgroup$ – John Rennie Nov 16 '15 at 14:57
  • $\begingroup$ However the FLRW metric does have a conformal timelike Killing vector so that the energy of a co-moving particle, with respect to the metric co-ordinates, increases with the scale factor. This leads to interesting consequences like a Universe with non-zero vacuum energy expanding linearly with cosmological time rather than exponentially. $\endgroup$ – John Eastmond Nov 16 '15 at 17:40

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