# Derivation of the Lorentz transform of brightness ($dP/d\Omega$)


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: $$\frac{dP}{d\Omega}=\left(\f{\omega}{\omega_0}\right)^4\frac{dP_0}{d\Omega_0}$$

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?

• Can you define all used quantities. Particulary what is meant by 'u'? Apr 28 '17 at 14:08
• @lalala when I get chance I will update the question with a list of all quantities. For now $u$ is the energy density. Apr 28 '17 at 14:39
• 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.
– Void
May 3 '17 at 9:51
• @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? 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).
• 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). May 4 '17 at 18:49
• 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. May 4 '17 at 19:40