Energy and Momentum in Particle Decay

[*FaerieLight*]

If a particle decays via A →B + C, and A had some initial non-zero momentum, is it possible for either B or C to be stationary? I can't seem to find a restriction on this from energy conservation or momentum conservation.

From energy conservation, the stationary particle B still contributes energy from its rest mass, so C does not need to have the same mass as A. From momentum conservation, C can carry off all the momentum of A, leaving B free to be stationary.

I've never heard of this kind of thing occurring in Nature, and that makes me wonder if it is actually possible for a decay product to be stationary.

Thanks a lot.

 

[jtbell]

If a particle decays via A →B + C, and A had some initial non-zero momentum, is it possible for either B or C to be stationary?

It's possible in principle, but very unlikely in practice. First, the initial momentum of A has to be "just right." Second, the non-stationary decay product has to be emitted in exactly the same direction that A was moving.

 

[Bob S]

A good example is the two-body pion π⁺ decay into a muon μ⁺ and muon neutrino ν_μ. The angular distribution is isotropic. In the pion rest frame, the kinetic energy of the muon is about 4.12 MeV (the range is a few hundred microns in nuclear emulsion). If the pion had a kinetic energy of about 5.45 MeV and the muon decayed backwards, it would be nearly stationary. The muon neutrino would carry away all the pion momentum.

 

[tom.stoer]

The kinematics is rather simple. What you are asking for is

A \to B + C

and e.g.

\vec{p}_B = 0

Now go to the rest frame of particle A, i.e.

p^\mu_A = (m_A, 0) \to p^\mu_B + p^\mu_C = (E_B+E_C, \vec{p}_b+\vec{p}_C)

From energy and momentum conservation you can deduce that the momentum of B and C are antiparallel and add up to zero. From the p's and the masses you can calculate the E's.

What you know need to do is the following; take the momentum of B and construct a Lorentz boost to B's restframe (i.e. such the the new p' of B vanishes). Now apply this Lorentz transformation to A and C.