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Lets see the reaction:

$e^- \mu^- \to e^- \pi^- \nu_\mu \;\;\;\;\;\;\;\;\;\;\; {(1)}$

I suppose, that this reaction occurs as follows

$e^- \mu^- \to e^- \mu^- \pi^+ \pi^- \to e^- \pi^- \nu_\mu$

Is it possible at energy less than 2*140 MeV?

The same is for the analogous proton-muon reaction

$p^+ \mu^+ \to p^+ \pi^+ \bar{\nu}_\mu \;\;\;\;\;\;\;\;\;\;\;{(2)}$

Once more reaction:

$p^+ p^+ \to p^+ p^+ \pi^- \pi^+ \to p^+ n \pi^+ \;\;\;\;\;\;\;\;\;\;\;{(3)}$

What are the experimental data?

P.S. This question is important enough. The main solar reaction is

$p^+ p^+ \to d^+ e^+ \nu_e $

If this reaction occurs as follows

$p^+ p^+ \to p^+ p^+ e^- e^+ \to p^+ n \nu_e e^+ \to d^+ e^+ \nu_e$

then it would require the energy of more than 2*0.511 MeV to take place. Cross section of this reaction will be much less, so the main solar reaction should be

$p^+ p^+ e^- \to d^+ \nu_e $

Update 20.02.11

Why I think so? I suppose that new particles are created in pairs particle-antiparticle. So the reaction

$e^- e^- \to e^- e^- \pi^- \pi^+ $

requires the energy of more than 2*140 MeV to take place

As well as the reaction

$e^- e^- \to e^- e^- \pi^- e^+ \nu_e $

from symmetry considerations, since there is a decay

$\pi^+ \to e^+ \nu_e$

And the result is

$e^- e^- \to e^- e^- \pi^- e^+ \nu_e \to e^- \pi^- \nu_e $

contrary to

$e^- e^- \to e^- \nu_e W^- \to e^- \pi^- \nu_e$

with the minimum reaction energy of 140 MeV

The same is valid for the reactions (1) (2) (3)

So what is the experimental data on minimum energy of these reactions?

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Kinetic energy or total energy, do you mean? – Noldorin Nov 14 '10 at 19:23
(In relativity, this makes a big difference.) – Noldorin Nov 14 '10 at 19:30
@Noldorin, kinetic energy – voix Nov 14 '10 at 20:00
I TeXified your reactions for you, hope you don't mind – David Z Nov 15 '10 at 5:48
$p^+$ is proton or positron (anti-electron)? If it is positron the symbol should be $e^+$... – kennytm Nov 15 '10 at 7:21

There is no need for any intermediate state involving multiple pions. The leading Feynman diagram has a photon exchanged between the electron and muon, and the muon radiating a (very off-shell) W that decays into quarks that hadronize to a pion. The only constraint is that the initial energy is larger than the sum of the masses of the final state particles. (The pion in the final state, of course, is unstable and will soon decay back to a muon and antineutrino.)

(If you don't want to think in terms of the W and quarks, you can just consider that there's an effective pion–muon–neutrino vertex which is what allows the pion to decay, and the muon is using this interaction to turn into a pion and neutrino.)

More generally, for any process you can consider, the minimum energy needed is just the sum of the masses of the final-state particles.

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I guess voix doesn't know about off-shell momenta (otherwise he'd solve the problem himself), so I'll just rephrase that: virtual particles need not obey the conservation of momentum (and in particular energy). – Marek Nov 14 '10 at 19:58
@voix: if you need a more precise than this one, you should first explain why you introduced the intermediate states, and why you think it is "2*140 MeV". – Cedric H. Nov 14 '10 at 19:59
so you think that 35 MeV (140-105) is enough. And what is the experimental data? – voix Nov 14 '10 at 20:15
@Cedric, I suppose, that this reaction may involve real, not virtual particles. 140 Mev- mass of pion – voix Nov 14 '10 at 20:31
+1, although I also think a complete answer to this question should include some experimental data. If I can find a reference for it I will make a comment here. – David Z Nov 15 '10 at 5:50

The "main solar reaction" you are talking about, $p^+ p^+ \to d^+ e^+ \nu_e $ is the first step of the pp-chain allowing the fusion of 4 protons to form Helium. This "pp-chain" has different "branches" (possible reactions) but all of them start with the process you mentioned.

This process is exoenergetic and releases 0.420 MeV even if deuton has a very low binding energy the reaction is possible because of the small difference in mass between the proton and the neutron.

It is a process involving the weak interaction: one of the proton "decay" ($\beta^+$ process) into a neutron, a positron and a neutrino.

Now about your second equation with an intermediate state where an electron-positron pair appear: you are making a mistake because you take the difference between the two sides, but you artificially introduced this electron-positron pair: this has to be taken into account:

You wrote

$p^+ + p^+ \rightarrow p + p + e^- + e^+$

but that's wrong, it should be

$p + p + \textrm{“some kinetic energy allowing the production of the pair = 0.511 MeV”} \rightarrow p + p + e^- + e^+$

and now, it should be right.

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you are right, and the output of reaction in my case is 2*0.511+0.42 MeV – voix Nov 16 '10 at 5:58

To complement Matt's answer, we just compare the total energy in both sides:

$M_e$ = 0.5 MeV, $M_\mu = 106 MeV$, $M_\pi = 140 MeV, the mass of the neutrino can be neglected so we obtain that kinematically this process is possible only if we give 34 MeV in the form of kinetic energy.

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