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:

  1. The need for an isotropic emitter.
  2. 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).
  3. 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?

  • $\begingroup$ Can you define all used quantities. Particulary what is meant by 'u'? $\endgroup$
    – lalala
    Apr 28 '17 at 14:08
  • $\begingroup$ @lalala when I get chance I will update the question with a list of all quantities. For now $u$ is the energy density. $\endgroup$ Apr 28 '17 at 14:39
  • $\begingroup$ As is, the derivation must be simply wrong because the result does not seem to include effects such as the relativistic Doppler effect. I have to say that for a "Relatively easy" book the derivation you report here is very non-transparent to me. $\endgroup$
    – Void
    May 3 '17 at 9:51
  • $\begingroup$ @Void This is very true, since I don't think anyone is going to answer this question as it stands could you (or someone else) provide a more transparent derivation based on the same lines? $\endgroup$ May 4 '17 at 14:44

I can make a connection to some astrophysical literature where this topic often arises. There, the "brightness" is usually referred to as the specific intensity, and is denoted by $I_{\nu}$. This has units of energy per frequency interval per area per time per solid angle (steradian).

The specific intensity is not a Lorentz invariant, but the quantity $$I_\nu / \nu^3$$ is. This is a well-known result that goes back at least as far as the classic paper by L. H. Thomas (section 3). For a more recent discussion, see for example section 90 of Foundations of Radiation Hydrodynamics by Mihalas & Mihalas. Or, if you perform a search for "invariant intensity," you'll find other references such as these online notes (see section 1.6).

A set of thought experiments can be used to derive this result, and the paper by Thomas goes through them (and the Mihalas textbook restates them). Relativistic effects such as length contraction, time dilation, and the Doppler effect including relativistic aberration must be accounted for. Ultimately it is a statement of photon number conservation, and the linked online notes provide a derivation from a consideration of the photon phase-space density.

  • $\begingroup$ Thanks for this answer. Here is what I can add: the relation between the $I_\nu$ in your answer and $I$ in the question is: $$I_\nu=\frac{dI}{d\Omega d\nu}$$ therefore we get $\frac{dI}{d\Omega}=I_\nu d\nu$ which accounts for the fourth factor of $\nu$ (I think). The only bit I am then confused with is why does $P$ (the power of the wave) transform in the same way as $I$ (the intensity). $\endgroup$ May 4 '17 at 18:49
  • $\begingroup$ I see. If I make one slight amendment to what you wrote, it is that the relation of interest is actually $I= I_\nu d\nu$. The $1/d\Omega$ is already built into the definition of both $I$ and $I_\nu$. I suppose the texbook definition is then indeed correct, I'm just so used to thinking in terms of $I_\nu$ rather than $I$. I'll have to think more about why the argument given in the book is correct in order to answer your specific question. $\endgroup$ May 4 '17 at 19:40

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