# Energy of comoving particle in FRW spacetime goes like scale factor?

The line element for the flat FRW metric using conformal time $\eta$, Cartesian spatial coordinates $(x,y,z)$ and scale factor $a(\eta)$, is given by $$ds^2=-d\tau^2=a^2(\eta)(-d\eta^2+dx^2+dy^2+dz^2).$$ A particle of unit mass moving on a geodesic, parametrized by $\sigma$ from $\sigma=0$ to $\sigma=1$, maximizes the proper time $\tau$ given by $$\tau=\int_0^1 L\ d\sigma$$ where the Lagrangian $L$ is given by $$L=\frac{d\tau}{d\sigma}=a(\eta)\Big[\Big(\frac{d\eta}{d\sigma}\Big)^2-\Big(\frac{dx}{d\sigma}\Big)^2-\Big(\frac{dy}{d\sigma}\Big)^2-\Big(\frac{dz}{d\sigma}\Big)^2\Big]^{1/2}.$$ Consider the time-component Euler-Lagrange equation $$\frac{\delta\tau}{\delta\eta}=\frac{d}{d\sigma}\Big(\frac{\partial L}{\partial(d\eta/d\sigma)}\Big)-\frac{\partial L}{\partial\eta}=0.$$ Now we have $$\frac{\partial L}{\partial(d\eta/d\sigma)}=-\frac{g_{0\mu}}{L}\frac{d x^\mu}{d\sigma}=-g_{\alpha\mu}\eta^\alpha u^\mu=-\eta_\mu u^\mu$$ where $d\tau=L\ d\sigma$ and $\eta^\alpha=(1,0,0,0)$.

Now we also have $$\frac{\partial L}{\partial \eta}=\frac{\dot a}{a}L=\frac{\dot a}{a}\frac{d\tau}{d\sigma}.$$ Let us assume that the particle is at rest in the comoving frame so that we have $d\tau=a(\eta)\ d\eta$.

Therefore we have $$\frac{\partial L}{\partial \eta}=\frac{d\eta}{d\tau}\frac{da}{d\eta}\frac{d\tau}{d\sigma}=\frac{da}{d\sigma}.$$ The time-component Euler-Lagrange equation becomes $$\frac{d}{d\sigma}(-\eta_\mu u^\mu)-\frac{da}{d\sigma}=0.$$ Integrating and using $a(t_0)=1$ we obtain $$-\eta_\mu u^\mu=a(\sigma).$$ The left-hand side can be identified with the energy, $\epsilon$, of the unit mass particle. Therefore we have $$\epsilon = a(\sigma).$$ Is this right? Is the energy of a unit mass comoving particle, in an expanding FRW spacetime, proportional to the scale factor?

• $\eta^\alpha$ is not of norm $-1$ so you cannot interpret $\eta_\mu u^\mu$ (or rather $\eta^\mu p_\mu$, with $p_\mu = m u_\mu$ the momentum) as the particle's energy – Christoph Mar 8 '17 at 19:18
• If $\eta^\alpha$ had norm $-1$ then I would be calculating the energy of the particle as measured by a local observer with that 4-velocity. In the above post I think I'm calculating the energy of the particle with respect to the global timelike vector field $\eta^\alpha$. – John Eastmond Mar 8 '17 at 20:35
• Why yould you want to calculate such an 'energy'? The result you'll get will have more to do with your choice of $\eta^\mu$ that the particle's kinematics – Christoph Mar 8 '17 at 21:34
• As far as I know $\eta^\alpha$ is unique in FRW space time as it is the only timelike conformal Killing vector field. The isometry of this field defines a global notion of energy proportional to scale factor a. I want this as I believe that the Einstein field equations in natural units in FRW space time should read $G_{\mu\nu}=1/(m_p^2 a^2) T_{\mu\nu}$. – John Eastmond Mar 8 '17 at 22:45
• Eta is not the comoving time. They are related, and you have an equation for the relation, so that'll introduce another factor. Anyway, eta is light like, it's clear from the metric. and it doesn't look like it's Killing, tthe metric depends on eta, but maybe. Why not look this up in a standard GR book? – Bob Bee Mar 9 '17 at 6:05