Observed Redshift from Moving Source: Deriving the Result

[ergospherical]

If a galaxy is receding from us, then the 1+redshift observed on Earth is the product $(1+z_{pec})(1+z_{cosm})$ of the doppler redshift due to the peculiar motion of the galaxy and the cosmological redshift due to the FRW metric. It makes sense if we think about some intermediate observers (e.g. someone stationary w.r.t. hubble flow but at the same position instantaneously as the emitting galaxy, who measures the doppler part only). Could someone show me how to derive the result from the general definitions? i.e. the galaxy has some 4-velocity $u_{gal} = (u_{gal}^t, u_{gal}^r, 0,0)$, and an observer attached to the galaxy measures\begin{align*}
\omega_{em} = u_{gal} \cdot p = u_{gal}^t p^t - \frac{a^2}{1-Kr^2} u_{gal}^r p^r
\end{align*}where $p$ is the photon 4-momn. And the earth observer measures $\omega_{obs} = u_{earth}^t p^t$. The constraints are that both 4-velocities are normalised to $u \cdot u = 1$, and $p^t = E$ is conserved along the photon's path. That isn't enough constraits to derive the result, I think?

 

[vanhees71]

All we can observe in GR is expressed by local invariant quantities. Indeed the observed frequency of an electromagnetic wave is $\omega=u \cdot k$, where $u$ is the four-velocity of the observer and $k$ the wave-four-vector of the em. wave. For more about electrodynamics in GR, see

https://itp.uni-frankfurt.de/~hees/pf-faq/gr-edyn.pdf

 

[PeterDonis]

$p^t = E$ is conserved along the photon's path.

Are you sure? Remember cosmological redshift.

 

[vanhees71]

In a static spacetime, i.e., if there are coordinates, for which the $g_{\mu \nu}$ are independent of the time coordinate, then $p_t$ is conserved along the "photon's" path.

Take Schwarzschild spacetime,
$$L=\frac{1}{2} [(1-2m/r) \dot{t}^2 - (1-2m/r)^{-1} \dot{r}^2 - r^2 (\dot{\vartheta}^2+\sin^2 \vartheta \dot{\varphi}^2).$$
For an observer "at rest", observing light from a source "at rest", indeed all you need to know is that
$$p_t=\frac{\partial L}{\partial \dot{t}}=(1-2m/r) \dot{t}.$$
From this you get
The four-velocity of the observer at rest is
$$u_{\text{obs}}^{\mu}=(1-2m/r_{\text{obs}})^{-1/2}(1,0,0,0).$$
Then
$$\omega_{\text{obs}}=p_t u_{\text{obs}}^t=p_t (1-2m/r_{\text{obs}})^{-1/2}.$$
For the frequency at the source you get
$$\omega_{\text{source}}=p_t u_{\text{source}}^t=(1-2m/r_{\text{source}})^{-1/2},$$
i.e.,
$$\omega_{\text{obs}}=\sqrt{\frac{1-2m/r_{\text{source}}}{1-2m/r_{\text{obs}}}} \omega_{\text{source}},$$
which describes (for $r_{\text{source}}<r_{\text{obs}}$) the gravitational redshift.

 

[PeterDonis]

In a static spacetime

Which FRW spacetime is not.

 

[vanhees71]

FRW spacetime is conformal to a static spacetime. For details, how to exploit this to calculate the cosmological redshift, see the said manuscript,

https://itp.uni-frankfurt.de/~hees/pf-faq/gr-edyn.pdf