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Particle decay is the spontaneous process of one elementary particle transforming into other elementary particles. During this process, an elementary particle becomes a different particle with less mass and an intermediate particle such as W boson in muon decay. The intermediate particle then transforms into other particles. If the particles created are not stable, the decay process can continue.

The process of particle decay is distinct from radioactive decay, in which an unstable atomic nucleus is transformed into a smaller nucleus accompanied by the emission of particles or radiation.

$c=\hbar=1. \,$

All data are from the Particle Data Group.

Type Name Symbol Mass (MeV/c2) Mean lifetime
Lepton Electron / Positron $e^- \, / \, e^+$ 0.511 $> 4.6 \times 10^{26} \ \mathrm{years} \,$
Muon / Antimuon $\mu^- \, / \, \mu^+$ 105.6 $2.2\times 10^{-6} \ \mathrm{seconds} \,$
Tau lepton / Antitau $\tau^- \, / \, \tau^+$ 1777 $2.9 \times 10^{-13} \ \mathrm{seconds} \,$
Meson Neutral Pion $\pi^0\,$ 135 $8.4 \times 10^{-17} \ \mathrm{seconds} \,$
Charged Pion $\pi^+ \, / \, \pi^-$ 139.6 $2.6 \times 10^{-8} \ \mathrm{seconds} \,$
Baryon Proton / Antiproton $p^+ \, / \, p^-$ 938.2 $> 10^{29} \ \mathrm{years} \,$
Neutron / Antineutron $n \, / \, \bar{n}$ 939.6 $885.7 \ \mathrm{seconds} \,$
Boson W boson $W^+ \, / \, W^-$ 80,400 $10^{-25} \ \mathrm{seconds} \,$
Z boson $Z^0 \,$ 91,000 $10^{-25} \ \mathrm{seconds} \,$

## Probability of survivalEdit

The mean lifetime of a particle is labeled $\tau$, and thus the probability that a particle survives for a time greater than t before decaying is given by the relation

$P(t) = e^{-t/(\gamma \tau)} \,$
where
$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$ is the Lorentz factor of the particle.

## Decay rateEdit

For a particle of a mass M, the decay rate is given by the general formula

$d \Gamma_n = \frac{(2\pi)^4}{2M}\left|\mathcal{M} \right|^2 d \Phi_n (P; p_1, p_2,\dots, p_n) \,$
where
n is the number of particles created by the decay of the original,
$\mathcal{M}\,$ is the invariant matrix element that connects the initial state to the final state,
$d\Phi_n \,$ is an element of the phase space, and
$p_i \,$ is the four-momentum of particle i.

The phase space can be determined from

$d \Phi_n (P; p_1, p_2,\dots, p_n) = \delta^4 (P - \sum_{i=1}^n p_i) \left( \prod_{i=1}^n \frac{d^3 \vec{p}_i}{(2\pi)^3 2 E_i} \right) \,$
where
$\delta^4 \,$ is a four-dimensional Dirac delta function.

### Complex massEdit

Further information: Resonance#Resonances in quantum mechanics

The mass of an unstable particle is formally a complex number, with the real part being its mass in the usual sense, and the imaginary part being its decay rate in natural units. When the imaginary part is large compared to the real part, the particle is usually thought of as a resonance more than a particle. This is because in quantum field theory a particle of mass M (a real number) is often exchanged between two other particles when there is not enough energy to create it, if the time to travel between these other particles is short enough, of order 1/M, according to the uncertainty principle. For a particle of mass $M+i\Gamma$, the particle can travel for time 1/M, but decays after time of order of $1/\Gamma$. If $\Gamma > M$ then the particle usually decays before it completes its travel.

### 3-body decayEdit

As an example, the phase space element of one particle decaying into three is

$d\Phi_3 = \frac{1}{(2\pi)^9} \delta^4(P - p_1 - p_2 - p_3) \frac{d^3 \vec{p}_1}{2 E_1} \frac{d^3 \vec{p}_2}{2 E_2} \frac{d^3 \vec{p}_3}{2 E_3} \,$

## Four-momentumEdit

Main article: Four-momentum

The four-momentum of one particle is also known as its invariant mass.

The square of the four-momentum of one particle is defined as the difference between the square of its energy and the square of its three-momentum (note that all units from here on are chosen such at the speed of light is equal to 1):

$p^2 = E^2 - (\vec{p})^2 = m^2 \quad \quad \quad \quad (1) \,$

The square of the four momentum of two particles is

$p^2 = \left(p_1 + p_2 \right)^2 = p_1^2 + p_2^2 + 2 p_1 p_2 = m_1^2 + m_2^2 + 2(E_1 E_2 - \vec{p}_1 \cdot \vec{p}_2)\,$

### Conservation of four-momentumEdit

Four-momentum must be conserved in all decays and all particle interactions, so

$p_\mathrm{initial} = p_\mathrm{final} .\,$

#### In two-body decaysEdit

If a parent particle of mass M decays into two particles (labeled 1 and 2), then the condition of four-momentum conservation becomes

$p_M = p_1 + p_2 .\,$

Re-arrange this to

$p_M - p_1 = p_2 \,$

and then square both sides

$p_M^2 + p_1^2 - 2p_M p_1 = p_2^2 .\,$

Now use the very definition of the square of four-momentum, eq (1), to see

$M^2 + m_1^2 - 2 \left(E_M E_1 - \vec{p}_M \cdot \vec{p}_1 \right) = m_2^2 . \quad \quad \quad \quad (2) \,$

If we enter the rest frame of the parent particle, then

• $\vec{p}_M =0 \,$, and
• $E_M = M \,$

Plug these into eq (2):

$M^2 + m_1 ^2 - 2 M E_1 = m_2^2. \,$

Now we have arrived at the formula for the energy of particle 1 as seen in the rest frame of the parent particle,

$E_1 = \frac{M^2 + m_1^2 - m_2^2}{2 M} \,$

Similarly, the energy of particle 2 as seen in the rest frame of the parent particle is

$E_2 = \frac{M^2 + m_2^2 - m_1^2}{2 M}. \,$

## Two-body decaysEdit

 File:2-body Particle Decay-CoM.svg File:2-body Particle Decay-Lab.svg In the Center of Momentum Frame the decay of a particle into two equal mass particles results in them being emitted with an angle of 180 degrees between them. ...while in the Lab Frame the parent particle is probably moving at a speed close to the speed of light so the two emitted particles would come out at angles different than that of in the center of momentum frame.

### From two different framesEdit

The angle of an emitted particle in the lab frame is related to the angle it's emitted in the center of momentum frame by the equation

$\tan{\theta'} = \frac{\sin{\theta}}{\gamma \left(\beta / \beta' + \cos{\theta} \right)}$

### Decay rateEdit

Say a parent particle of mass M decays into two particles, labeled 1 and 2. In the rest frame of the parent particle,

$|\vec{p}_1| = |\vec{p_2}| = \frac{[(M^2 - (m_1 + m_2)^2)(M^2 - (m_1 - m_2)^2)]^{1/2}}{2M}. \,$

Also, in spherical coordinates,

$d^3 \vec{p} = |p|^2\, dp d\Omega = p^2\, d \phi\, d\left( \cos \theta \right). \,$

Use this with knowledge of the phase-space element for a two-body decay, to see that the decay rate in the frame of the parent particle is

$d\Gamma = \frac{1}{32 \pi^2} \left| \mathcal{M} \right|^2 \frac{|\vec{p}_1|}{M^2}\, d\phi_1\, d\left( \cos \theta_1 \right). \,$