I start with the spatially flat FRW metric in conformal co-ordinates:(adsbygoogle = window.adsbygoogle || []).push({});

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

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 foranymassive particle moving along a geodesic andnotjust constant for a particle that is at rest in the co-moving frame.

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?

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# An argument that massive particles don't redshift?

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