# An argument that massive particles don't redshift?

I start with the spatially flat FRW metric in conformal co-ordinates:

$$ds^2=a^2(\eta)(d\eta^2-dx^2-dy^2-dz^2)$$

This metric has the following non-zero Christoffel symbols:

\begin{eqnarray} \Gamma^0_{\alpha \beta} &=& \frac{\dot{a}}{a} \delta_{\alpha \beta} \\ \Gamma^i_{0j} &=& \Gamma^i_{j0} = \frac{\dot{a}}{a} \delta^i_j \end{eqnarray}

where $\alpha,\beta=\{0,1,2,3\}$ and $i,j=\{1,2,3\}$.

Let us assume the following:

1. A massive particle travels on a geodesic $x^\mu(\lambda)$ with tangent vector $U^\mu=dx^\mu/d\lambda$ where $U^\mu U_\mu=1$.
2. The 4-momentum of the particle is given by $P^\mu = m U^\mu$.
3. A timelike vector field (not assumed to be a Killing field) $\eta^\mu=\delta^\mu_0$, $\eta_\mu=\eta^\nu g_{\mu\nu}=\delta^\nu_0g_{\mu\nu}=g_{\mu0}$.

The energy of the particle, $E$, with respect to the co-ordinate system is given by:

$$E=\eta_\mu P^\mu=m\eta_\mu U^\mu$$

As the vector $\eta^\mu$ is not assumed to be a Killing vector then the energy $E$ is not necessarily constant.

Now the rate of change of the energy $E$ along the worldline $x^\mu(\lambda)$ is given by:

\begin{eqnarray} U^\nu \nabla_\nu(m\eta_\mu U^\mu)&=& mU^\nu U^\mu \nabla_\nu \eta_\mu + m\eta_\mu U^\nu \nabla_\nu U^\mu \\ &=& mU^\nu U^\mu \nabla_\nu \eta_\mu \end{eqnarray} where the last term on the right-hand side is zero as the particle moves on a geodesic.

We now evaluate the covariant derivative in the above equation using $\eta_\mu=g_{\mu0}$:

\begin{eqnarray} \nabla_\nu \eta_\mu &=& \partial_\nu \eta_\mu - \Gamma^\lambda_{\nu \mu} \eta_\lambda \\ &=& \partial_\nu g_{\mu0} - \Gamma^0_{\nu\mu} g_{00} \\ &=& \partial_\nu g_{\mu0} - \frac{\dot{a}}{a} \delta_{\nu\mu} g_{00} \\ &=& \mbox{diag}(a \dot{a},-a\dot{a},-a\dot{a},-a\dot{a}) \\ &=& \frac{\dot{a}}{a}g_{\nu\mu} \end{eqnarray}

Thus the rate of change of the energy of the particle along the geodesic is given by: \begin{eqnarray} U^\nu \nabla_\nu(m\eta_\mu U^\mu) &=& m\frac{\dot{a}}{a}U^\nu U^\mu g_{\nu\mu}\\ &=& m\frac{\dot{a}}{a} \\ &=& m \frac{(da/d\tau)(d\tau/d\eta)}{a} \\ &=& m \frac{da}{d\tau} \end{eqnarray}

If I integrate both sides of the above equation I find that the energy $E$, with respect to the metric co-ordinates, of any massive particle in geodesic motion is given by: \begin{eqnarray} E = m\eta_\mu U^\mu &=& \eta_\mu P^\mu \\ &=& \eta^\mu P_\mu \\ &=& \delta^\mu_0 P_\mu \\ &=& P_0 \\ &=& ma \end{eqnarray}

Now let us calculate the energy $E_{obs}$ that a co-moving observer with 4-velocity $U^\mu_{obs}=\frac{1}{a}\delta^\mu_0$ measures when he observes the massive particle.

\begin{eqnarray} E_{obs} &=& U^\mu_{obs} P_\mu \\ &=& \frac{1}{a}\delta^\mu_0 P_\mu \\ &=& \frac{1}{a} P_0 \\ &=& \frac{1}{a} m a \\ &=& m \end{eqnarray}

Notice that the energy $E_{obs}$, that a co-moving observer measures, is constant for any massive particle moving along a geodesic and not just constant for a particle that is at rest in the co-moving frame.

By the way, just to clarify, the co-moving observer is not co-moving with the particle but is co-moving with the expansion of the Universe. There can be a local relative velocity between the particle and the observer.

Therefore it seems that massive particles do not redshift as they travel on geodesics through the expanding Universe which is contrary to received wisdom.

Where have I gone wrong?

• What do you mean by a "massive" particle: just one with mass? And, by stating that a massive particle "does not redshift", do you mean that as a particle travels across an expanding Universe, it will not lose velocity? – Daniel Griscom Nov 21 '15 at 1:32
• Yes. It seems that as a particle with mass freely moves through expanding space its total energy, as measured by a co-moving observer, $E=\sqrt{m^2+p^2}$, does not change. Thus its momentum/velocity does not change either. – John Eastmond Nov 21 '15 at 9:43
• You are assuming that other local forces like electric or large gravitational masses are not influencing the particle. – Peter R Jan 24 '16 at 17:32

## 2 Answers

The problem is that you have put the integration constant to zero. In your equations you have $$\frac{dE_{coord}}{d\tau} = m \frac{d a}{d\tau}$$ which only justifies the integration to $$E_{coord} = ma + T_0$$ $T_0$ can be understood as the coordinate kinetic (nonrest) energy of the particle at the instant when $a=1$. When we then compute the comoving energy we get $$E_{comov} = m + \frac{T_0}{a}$$ I.e., as the scale factor grows, the comoving observers will observe an energy closer and closer to rest energy (rest mass) with the kinetic contribution going to zero. In terms of comoving observers, this can be well understood as the particle "slowing down" or "redshifting", even though for massive particles this is not quite kosher terminology.

Comoving observers move along with the Hubble flow, and perceive the universe as having no Hubble expansion, due to an increasing scale factor. So, when you invoke a comoving observer, it's not surprising that the observer sees no redshifting.

• I accept that comoving observers measure massless photons redshifting. The accepted viewpoint is that comoving observers also measure the momentum of massive particles redshifting. For instance in his GR lecture notes Sean Carroll states that a gas of particles cools down as the Universe expands. In the calculation above I'm questioning whether that latter effect actually does take place. Of course I'm almost certainly wrong. – John Eastmond Nov 21 '15 at 21:31
• Imo, we are comoving observers and we see the redshift. – Helder Velez Apr 5 '16 at 13:49
• @JohnEastmond Perhaps it would help to visualize a gas in an expanding box? As the volume increases the temperature of the gas goes down, but the energy of any individual particle, in any frame, is not dependent on the box's volume. – Jold May 10 '16 at 4:19