# How is the energy distributed in a proton-antiprotion annihilation?

I know the products of the annihilation, but I don't know how much energy each particle has or gets. For example I know that 1876 MeV is released for each annihilation. Now, this energy is distributed among a number of pions, so how much energy does each pion get ?

Also, this energy is distributed between kinetic and rest mass energy, how is it distributed ? and where does the kinetic energy go when the pions decay ?

And what about the electron we get from the muon decay ? how much energy does it get and is this energy easy to obtain or it will just fly away similar to a neutrino ? I mean how far will the electron go in a medium until its energy is absorbed ?

EDIT : I asked this question because I am confused where the kinetic energy of the particles go. Firstly, the neutral pion gets 390 MeV total energy, of which 135 MeV is the rest mass energy and 255 MeV is its kinetic energy. As some people said before, the decay of the neutral pion result in 2 gamma rays each with 67 MeV, which is only the rest mass energy over two, so where did the kinetic energy go ? shouldn't the pion have resulted in 2 gamma rays with 195 MeV each considering energy conservation ?

Secondly, the charged pions. as well as the neutral pion, they get 395 MeV of energy too. When they decay, the product is a muon and a neutrino which I dont know how the energy is distributed here. Then the muon decays into an electron or a positron and 2 more neutrinos which I also don't know how much energy they get.

Thirdly, the kinetic energy of the charged pions. How do the charged pions lose their kinetic energy ? and what will happen if the charged pion decayed before losing its kinetic energy ?

All these three questions haven't been discussed in the related question, and thus this is not a duplicate.

• – user4552 Aug 13 '13 at 21:20
• Nope, the previous question didn't discuss the kinetic energy part, neither the energy of the muons nor the electrons. – Abanob Ebrahim Aug 13 '13 at 22:54
• Abanob, if the suggested duplicate doesn't address what you want to know, you should edit your question to say so and also explain what exactly you're looking for that the other question doesn't include. – David Z Aug 14 '13 at 0:20
• The 67 MeV energy for the neutral pions is only if they stop and then decay at rest, which is very possibly not what happens. – user4552 Aug 14 '13 at 4:37
• I desagree with the question @BenCrowell linked to being a dublicate since that one just asks how much energy can be extracted for everyday usage whereas this question asks in detail about how the collision energy is redistributed among the new particles created. They are not the same at all. – Dilaton Aug 14 '13 at 13:36

: I asked this question because I am confused where the kinetic energy of the particles go. Firstly, the neutral pion gets 390 MeV total energy, of which 135 MeV is the rest mass energy and 255 MeV is its kinetic energy.

We are talking of average numbers. On average there are five pions, on average they each have 390 MeV, on average ...

As some people said before, the decay of the neutral pion result in 2 gamma rays each with 67 MeV, which is only the rest mass energy over two, so where did the kinetic energy go ? shouldn't the pion have resulted in 2 gamma rays with 195 MeV each considering energy conservation ?

The 67 MeV is the energy at the center of mass system of the pi0. As the pi0 is in the lab, you are right the two gamma rays will on the average have 195 MeV each.

Secondly, the charged pions. as well as the neutral pion, they get 395 MeV of energy too. When they decay, the product is a muon and 2 neutrinos which I dont know how the energy is distributed here.

Pions decay into two bodies, they go to a muon and a muon anti/neutrino ( to conserve lepton number). A two body decay and at the center of mass of the pion the neutrino gets most of the kinetic energy. What happens in the lab will depend on the angular distribution of the decay versus the direction of motion of the pion.

Then the muon decays into an electron or a positron and 2 more neutrinos which I also don't know how much energy they get.

These are three body decays and their distribution can be found in the literature of some decades back .

Thirdly, the kinetic energy of the charged pions. How do the charged pions lose their kinetic energy ?

Charged particles interact with the field of the molecules of the matter they go through and lose a bit of kinetic energy by ionizing them, that is why we can see their tracks in bubble chamber pictures. In dense matter they lose energy fast.

Have a look at this lecture.

and what will happen if the charged pion decayed before losing its kinetic energy ?

Here is a pion stopping and going into a muon etc. In flight most of the momentum gets carried on by the muon. It will decay in flight into a muon and a neutrino which will share all the energy ( mass and kinetic) of the pion.

I think you are trying to get in a nutshell information that needs a course in elementary particle physics. It cannot be done. Knowledge cannot be transmitted by hand waving words. To understand such matters problems must be solved and time spent in reading relevant material. Students who sit in courses are not stupid, they are interested in learning this stuff that you are trying to absorb by osmosis. It cannot be done. If you need details you have to spend the time necessary to understand details.

• thank you very much. I just have one more question, and I promise no any question of this kind will be asked again, just this one. I tried to understand as much as I could from the lecture you gave me, but there is one thing I could not get a number for which I might need your help with. The number I need is the percentage of the the energy a pion with kinetic energy of 250 MeV will lose when it goes through 5 cm of lead. Can you please help me with this ? – Abanob Ebrahim Aug 14 '13 at 13:16
• Here is the relevant curve for lead also: page five, pdg.lbl.gov/2011/reviews/… . it is about 1 MeV to be multiplied by density of lead 11.34 gm/cm^3, times the cm of the path so about 50 Mev loss on 5 cm path in lead. As the energy diminishes it loses more, if you look at the curve. – anna v Aug 14 '13 at 16:23