Particle decay

Particle decay is the random process by which an unstable particle spontaneously splits into one or more particles. It is a process that is first and foremost governed by conservation laws. While a particle may decay in a large number of other particles, the outcome of a decay is heavily restricted by the fact that many intrinsic properties of the particle must be conserved. For instance, electric charge is conserved, so if the decaying particle has charge +e+e, the total charge across all decay products must also be +e+e.

Particles decay in order to reduce their rest energy, which makes them more stable. This is allowed because in relativity, total energy is split between rest energy and kinetic energy. As such, a decay process involves converting some of the parent particle's rest mass into kinetic energy of the products, leading to a smaller total mass. Indeed, the condition for decay to be spontaneous is

mparent>∑i=1Nmi-th productm_\text{parent}>\sum_{i=1}^{N} m_{i\text{-th product}}

The energy difference is converted to kinetic energy.

Particle decay is often explained through Feynman diagrams, which are standardized schematic representations of particle scattering. Being entire diagrams, they can carry much more information than a simple mathematical representation, which is sorely needed given the complexity of many-particle interactions.

It is similar to, but distinct from, nuclear decay, which occurs in nuclei and involves different mechanisms.

Common situations

The exact nature of a particle decay is entirely dependent on the particle itself and its properties. However, the kinematics of the process are by-and-large the same for all of them. Most decays happen in one of three situations, which are presented below.

Particle at rest

This is the simplest case. An unstable particle aa is at rest. At some point, it spontaneously decays into two particles, bb and cc:

a→b+ca\to b+c

The emission sends particles away from each other, in opposite directions. This is called back-to-back decay.

center

We'll use natural units. Conservation of energy and momentum read

{0=pb+pcma=Eb+Ec\begin{cases} 0=\mathbf{p}_{b}+\mathbf{p}_{c} \\ m_{a}=E_{b}+E_{c} \end{cases}

The momenta must go in opposite directions and have the same magnitude:

pb=−pc⇒∣pb∣=∣pc∣\mathbf{p}_{b}=-\mathbf{p}_{c}\quad\Rightarrow \quad \lvert \mathbf{p}_{b} \rvert =\lvert \mathbf{p}_{c} \rvert

In other words, momentum is evenly split between the two. Call pp the common magnitude. Then, using the relativistic energy formula E=p2+m2E=\sqrt{ p^{2}+m^{2} },

ma=p2+mb2+p2+mc2m_{a}=\sqrt{p^{2}+m_{b}^{2}}+\sqrt{p^{2}+m_{c}^{2}}

or

ma−p2+mb2=p2+mc2m_{a} - \sqrt{p^{2}+m_{b}^{2}} = \sqrt{p^{2}+m_{c}^{2}}

We want to solve for pp. Take the square of both sides

ma2+p2+mb2−2map2+mb2=p2+mc2m_{a}^{2} + p^{2} + m_{b}^{2} - 2m_{a}\sqrt{p^{2}+m_{b}^{2}} = p^{2} + m_{c}^{2}

Cancel out p2p^{2} and rearrange

ma2+mb2−mc2=2map2+mb2m_{a}^{2} + m_{b}^{2} - m_{c}^{2} = 2m_{a}\sqrt{p^{2}+m_{b}^{2}}

Square again to get the final root out of the way

(ma2+mb2−mc2)2=4ma2(p2+mb2)(m_{a}^{2} + m_{b}^{2} - m_{c}^{2})^{2} = 4m_{a}^{2}(p^{2} + m_{b}^{2})

Now solve for pp

p=12mama4+(mb2−mc2)2−2ma2(mb2+mc2)\boxed{p = \frac{1}{2m_{a}} \sqrt{ m_{a}^{4} + (m_{b}^{2} - m_{c}^{2})^{2} - 2m_{a}^{2}(m_{b}^{2} + m_{c}^{2}) }}

The interesting result here is that the final momentum of both products is exclusively determined by masses. But since masses are inherent properties of particles, so must be the decay momentum at rest. As such, as long as we have the masses of the particles involved, we can directly predict the momenta of a back-to-back decay at rest.

If you want, you can now also determine the energies. Since pp is a function of just mass, then so is all relativistic energy:

{Eb=p2+mb2=ma2+mb2−mc22ma=s+mb2−mc22sEc=p2+mc2=ma2−mb2+mc22ma=s−mb2+mc22s\begin{cases} E_{b} = \sqrt{p^{2} + m_{b}^{2}} = \dfrac{ m_{a}^{2} + m_{b}^{2} - m_{c}^{2} }{ 2m_{a} } = \dfrac{ s + m_{b}^{2} - m_{c}^{2} }{ 2\sqrt{s} } \\ E_{c} = \sqrt{p^{2} + m_{c}^{2}} = \dfrac{ m_{a}^{2} - m_{b}^{2} + m_{c}^{2} }{ 2m_{a} } = \dfrac{ s - m_{b}^{2} + m_{c}^{2} }{ 2\sqrt{s} } \end{cases}

where s\sqrt{ s } is the center-of-mass energy. In this case it's equal to s=ma\sqrt{ s }=m_{a}, since the mass energy of (resting) parent particle is the only energy available to the decay. Once s\sqrt{s} is fixed, the energies depend only on the masses.

Particle in flight

This scenario is similar to the above, in that it's a particle decaying in two bodies, but now the particle is moving. Hence, the center-of-mass energy also includes the kinetic energy of the parent. Since the particle is moving from our perspective, the choice of frame of reference becomes important.

In the laboratory frame, at rest with the scientists and detectors, the particle moves at some initial speed. Since momentum is conserved, the decay products will inherit it. This makes them also fly forward, but at an angle, with respect to the original line of motion.

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Conservation says

{pa=pb+pcEa=Eb+Ec\begin{cases} \mathbf{p}_{a}=\mathbf{p}_{b}+\mathbf{p}_{c} \\ E_{a}=E_{b}+E_{c} \end{cases}

Call θ1\theta_{1} and θ2\theta_{2} the two angles at which the products are emitted and θ=θ1+θ2\theta=\theta_{1}+\theta_{2} their sum. The interesting part of this decay is that if you go into the center-of-momentum frame, where ptot=0\mathbf{p}_\text{tot}=0, this goes back to being back-to-back decay. In fact, since ptot=pa′=0\mathbf{p}_\text{tot}=\mathbf{p}'_{a}=0, you get

{0=pb′+pc′ma=Eb′+Ec′=s\begin{cases} 0=\mathbf{p}'_{b}+\mathbf{p}'_{c} \\ m_{a}=E'_{b}+E'_{c}=\sqrt{ s } \end{cases}

which is back-to-back decay. The primes denote that the we are in a different frame, so the values change.