Calculating Particle Redshift at Time t_0 & t_1

[ergospherical]

A question about the FLRW solution has confused a few of us. At time $t_0$ a particle has radial speed $v_0^r$ relative to the fundamental observers, and at a later time $t_1$ it has radial speed $v_1^r$. The task is to show that\begin{align*}
\frac{a_0}{a_1} =\frac{v_1^r \gamma_1}{v_0^r \gamma_0}
\end{align*}where $a_{j}$ is the scale factor at $t_{j}$ and $\gamma_j = (1-(v_j^r)^2)^{-1/2}$. The metric is diagonal with $g_{00} = 1$ and $g_{11}(t_j) = -a_j^2$. The fundamental observers have 4-velocities $u^{\mu}=\delta^{\mu}_t$ so\begin{align*}
\gamma_j = g(u, v_j) = u^t v_j^t-a^2 u^r v_j^r = v_j^t
\end{align*}Normalisation of $v_j$ gives two constraints:\begin{align*}
1 = g(v_j, v_j) = v_j^t v_j^t- a_j^2 v_j^r v_j^r = \gamma_j^2 - a_j^2 (v_j^r)^2
\end{align*}which lead to\begin{align*}
\left( \frac{\gamma_0}{\gamma_1}\right)^2 = \frac{1 + a_0 (v_0^r)^2}{1 + a_1 (v_1^r)^2}
\end{align*}?

 

[Ibix]

I haven't followed through this in detail, but it looks to me like you are using $v_j^r$ for the $r$ components of both the particle's four velocity and its three velocity. There may also be some algebraic games to be played with $\gamma^2$s.

 

[Orodruin]

As @Ibix said, the $v_i$ are the relative velocities, not the components of the 4-velocity. You also need to take into account exactly how the 4-velocity evolves with time. Hint: Spacetime symmetry.

 

[ergospherical]

As before:\begin{align*}
\gamma = g(u, v) = g_{\mu \nu} u^{\mu} v^{\nu} = v^t
\end{align*}Normalise 4-momentum:\begin{align*}
g(p, p) = g^{\mu \nu} p_{\mu} p_{\nu} = (p_t)^2 - a^{-2} (p_r)^2 = m^2
\end{align*}then re-arrange to get \begin{align*}
(p_r)^2 = a^2 m^2 ((v_t)^2 -1) = a^2 m^2 (\gamma^2 - 1)\end{align*}where last equality follows because $v_t = g_{tt} v^t = v^t = \gamma$. From the geodesic equation,
\begin{align*}
\frac{dp_r}{d\lambda} = \frac{1}{2} \left( p^t p^t \partial_{r} g_{tt} + p^r p^r \partial_{r} g_{rr} \right) = 0 \implies p_r = \mathrm{const}
\end{align*}so also $a^2(\gamma^2-1) = \mathrm{const}$. But $\gamma^2-1 = (v^r)^2 \gamma^2$ via algebraic manipulation, so \begin{align*}
a_0^2 (v_0^r)^2 \gamma_0^2 &= a_1^2 (v_1^r)^2 \gamma_1^2 \\
a_0 v_0^r \gamma_0 &= a_1 v_1^r \gamma_1
\end{align*}

 

[Orodruin]

Easier solution:

The FLRW metric has a Killing tensor $K_{\mu\nu} = a^2(g_{\mu\nu} - u_{\mu} u_{\nu})$ so we have that
$$
K_{\mu\nu} v^\mu v^\nu = a^2 (v^2 - (u\cdot v)^2)
= a^2( 1 - \gamma^2) = a^2 v^2 \gamma^2
$$
is a constant of motion if $v$ is the tangent of a geodesic.

 

[Orodruin]

To expand a bit on that. Showing that $K_{\mu\nu}$ is indeed a Killing tensor field is not difficult, but many would struggle with it and try to insert the explicit form of the metric. In reality, all you really need to introduce is that the metric takes the form
$$
ds^2 = dt^2 - a^2 h_{\mu\nu} dx^\mu dx^\nu = (u_\mu u_\nu - a^2 h_{\mu\nu}) dx^\mu dx^\nu,
$$
where any temporal components of $h_{\mu\nu}$ equal to zero and $h_{\mu\nu}$ does not depend on $t$. (You do not even need the isotropy and homogeneity properties.) We note that $g_{\mu\nu} - u_\mu u_\nu = - a^2 h_{\mu\nu}$ for future reference and that the Christoffel symbols from the time component of the geodesic equations are given by
$$
\Gamma^0_{\mu\nu} = a a' h_{\mu\nu}.
$$
It then holds that
$$
\nabla_{(\sigma} K_{\mu\nu)} = -2a^3a'u_{(\sigma} h_{\mu\nu)} + 2a^2 u^{}_{(\sigma} \Gamma^0_{\mu\nu)}
= 0.
$$
So $K_{\mu\nu}$ is indeed a Killing tensor field.
It does not need to get rougher than that.

 

[ergospherical]

Thank you!