Is the decay of a muon always a 1 to n process?

[spaghetti3451]

Decay processes are quite common in particle physics.

Is the decay process always a $1 \rightarrow n$ process?

In other words, can we call the reaction $$\mu^{-} + \mu^{+} \rightarrow \phi,$$

where $\phi$ is some scalar particle, the decay of the muon?

 

[mathman]

As a matter of definition, decay processes start with one particle.

 

[spaghetti3451]

Can we have the incoming particle also in the set of outgoing particles?

 

[mfb]

That would violate energy/momentum conservation.

 

[stoomart]

Can we have the incoming particle also in the set of outgoing particles?

I believe this would require an intermediate stage, plus an external energy source. Consider neutron → proton → neutron transformation through beta decay:

β⁻ decay: when a free neutron decays into a proton

n → p + e⁻ + -ν_e​

β⁺ decay: when a proton inside a nucleus decays into a neutron

p → n + e⁺ + ν_e​

Note:

β⁺ decay cannot occur in an isolated proton because it requires energy due to the mass of the neutron being greater than the mass of the proton. β⁺ decay can only happen inside nuclei when the daughter nucleus has a greater binding energy (and therefore a lower total energy) than the mother nucleus.

 

[ChrisVer]

where ϕ\phi is some scalar particle, the decay of the muon?

you call it annihilation of muon-antimuon...
obviously you don't have 1 muon to call it decay of the muon.
Can you have the incoming particle also in the outgoing particles? In vacuum as already mentioned no... but in other cases, yes, like Brehmstralung $e \rightarrow e \gamma$.

 

[stoomart]

like Brehmstralung $e \rightarrow e \gamma$.

Since the charged particle is only losing kinetic energy and its invariant mass remains unchanged, is this really considered decay?

 

[ChrisVer]

Since the charged particle is only losing kinetic energy and its invariant mass remains unchanged, is this really considered decay?

I didn't call it a decay- I gave that as an example to that you can have the same incoming and outgoing particle.

 

[vanhees71]

Of course bremsstrahlung is not a decay process since you always need the electron to scatter with something since a free electron won't radiate. Only accelerated charges radiate. So the correct bremsstrahlung process is a scattering process like $\mathrm{e}^-+X \rightarrow \mathrm{e}^- + X +\gamma$, where $X$ is some particle or atomic nucleus scattering with the electron.