# Lorentz Boost of a photon?

I'm a bit confused about a remark given to me by one of my professors, I appreciate any help you may provide even though my question may be unclear.

When discussing a decay channel $\rm e^+e^- \to \eta_c \gamma \pi^+ \pi^-$, I was told that I need to make sure that the photon ($\gamma$) I have contributing to the decay channel is MONOCHROMATIC. I can do that by boosting the photon into the mother's particle frame. I don't understand why the photon should be monochromatic? why boosting it could do the trick?

Please let me know if I should provide you with any more details.

• Sorry, but to what extend can a single photon not be monochromatic? Does this make sense? One photon one color, is it not? Mar 27, 2017 at 13:03
• $e^{+}e^{-} \to \eta_c \gamma \pi^{+} \pi^{-}$ is not a decay channel. Is this the process you really mean? Mar 27, 2017 at 13:21
• @dukwon you're right. I called it a decay channel just because there's a certain particle (That I don't know). It decays to the final state I mentioned. Mar 27, 2017 at 13:58
• Is the photon from the decay of the particle, or is it produced associatively? e.g. $e^{+}e^{-} \to X\gamma \to \eta_c\pi^{+}\pi^{-}\gamma$? Then $X$ could be something like the $\eta_c(2s)$ and the photon would be monochromatic. Mar 27, 2017 at 14:06
• Ah, good, I was hoping for a 2-body decay at some point. You can use the $h_c$ rest frame: the momenta of the $\eta_c$ and the $\gamma$ will be the same (in opposite directions), but the energy will not because they have different masses. In general, keeping the photon monochromatic will give you a frame with some fixed velocity relative to the $h_c$. Mar 27, 2017 at 14:50

The most famous example of a problem like this is electromagnetic annihilation of para-positronium into photons: $$\rm e^+e^- \to \gamma\gamma$$ In that decay, there always exists a reference frame where the total momentum of the electron-positron pair is zero. In that reference frame, the photons must carry equal and opposite momentum, and so annihilation photons are "monochromatic" with an energy of 511 keV.

Contrast that with ortho-positronium, which must decay to three photons to conserve angular momentum. In that case, the extra degree of freedom of an angle between the photons means there are many sets of photon energies which can conserve energy and momentum. In general, two-body decays are monochromatic, while many-body decays have energies that fill some phase space.

You see the same division between alpha decays, which have two-body final states and produce alpha particles with definite energy; beta decays, with three-body final states and an energy spectrum for all particles; and electron-capture decays, with a two-body final state again and a definite neutrino energy.

I'm not familiar with the process you're discussing in your question, but if the photon is monochromatic that suggests it's emitted from some intermediate state which is a two-body decay. If that were the case, your treatment of the intermediate state would conserve momentum in its rest frame --- but that probably wouldn't be the rest frame of the initial electron-positron pair.

• Out of couriosity...is it 511 keV+ kinetic? Mar 27, 2017 at 13:40
• @rob thank you very much for your answer it helped. I did not mention this before, the intermidiate state is $$h_c \to \eta_c \gamma$$. Just to make sure that I understood well, If I need to select the monochromatic photons (photons coming from the 2 body decay). I should first boost all photons to hc rest frame and then keep the photons with energy equal to eta_c energy? Mar 27, 2017 at 14:11
• @mikuszefski You'll note that I wrote "postironium"; I'm a low-energy kind of guy. Of course you're correct that in an in-beam annihilation the photons must also carry the kinetic energy.
– rob
Mar 27, 2017 at 16:21
• @thephoenix I'd go the other way: produce the $h_c$ with whatever momentum it gets, compute the $h_c\to\eta_c\gamma$ decay in the rest frame of the $h_c$; then boost back into the lab frame to find the energy and momentum of the photon from your detector's perspective.
– rob
Mar 27, 2017 at 16:23
• Agreed...just mentioned it because - with $\eta$ etc involved - the OP is likely to be the high-energy kind of guy. Cheers. Mar 28, 2017 at 5:16