In the book 'Relativity made Relatively Easy' by A.Steane their is a derivation of the Lorentz transform of the 'brightness': $$ \newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left|#1\right>} \frac{dP}{d\Omega}=\left(\f{\omega}{\omega_0}\right)^4\frac{dP_0}{d\Omega_0}$$ An overview of the derivation is given below (my wording):
Let $S_0$ be the rest frame of the isotropically emitting object and $S$ out moving frame:
- The intensity of a plane wave is given by: $$I=uc=\f{E}{A\lambda}c$$ since the area $A$ of a wavefront is invariant under a Lorentz transform: $$I=\l\f{\omega}{\omega_0}\r^2I_0$$
- We consider $dN$ rays (or plane wave components) that travel in a solid angle $d\Omega_0$ in our frame those same rays then travel in the solid angle $d\Omega$ where: $$d\Omega=\l\f{\omega_0}{\omega}\r^2d\Omega_0$$
- Using the above We therefore must have: $$ \frac{dP}{d\Omega}=\left(\f{\omega}{\omega_0}\right)^4\frac{dP_0}{d\Omega_0}$$
There three things that I do not understand about this derivation:
- The need for an isotropic emitter.
- Why we take $u=E/(A\lambda)$ and not $u=E/(AL)$ where $L$ is some fixed length (that transforms like the length of a fixed body).
- Why the power $dP$ of the $dN$ 'plane wave components' must transform in the same way as $I$ - surly some area effects come into play (and even interference).
Does anyone have a simple explanation for these points?