= Solution
For neighbouring <comoving observers> separated by a small physical distance $d\ell$, the recession velocity is $H(t)d\ell$, where $H=\dot a/a$ is the <Hubble parameter>. During the photon's travel time $dt=d\ell/c$, the local <Doppler effect> gives $d\lambda/\lambda=Hdt$. Integrating along the ray gives the <cosmological wavelength-redshift relation>:
$$
\log\frac{\lambda_0}{\lambda_e}=\int_{t_e}^{t_0}\frac{\dot a}a\,dt,
\qquad\boxed{1+z=\frac{a(t_0)}{a(t_e)}.}
$$
The same ratio stretches the arrival interval of successive wave crests and of successive emitted pulses: $dt_0=(1+z)dt_e$. Each <photon> loses a factor $1+z$ in energy because $E=hc/\lambda$. Thus energy arriving through the sphere per unit observer time has one factor from <cosmological redshift> and one from time dilation:
$$
\boxed{\frac{dE_0}{dt_0}=\frac L{(1+z)^2}.}
$$
This is the total power crossing the sphere; division by its area would give the observed <flux>.
Back to article page
