Particle decay

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A Feynman diagram of the common decays of the tau lepton. Here, it decays into a W boson and an anti-neutrino. The W boson, also being an unstable particle, will decay, for example, into an electron and anti-electron neutrino, or any of the other pairs of particles listed.

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 a W boson. The W boson 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.

Note that this article uses natural units, where

$c=\hbar=1. \,$

1 Table of particle lifetimes
2 Probability of survival
3 Decay rate
- 3.1 3-body decay
4 Four-momentum
- 4.1 Conservation of four-momentum
  - 4.1.1 In two-body decays
5 Two-body decays
- 5.1 From two different frames
- 5.2 Decay rate
6 See also
7 References

[edit] Table of particle lifetimes

All data are from the Particle Data Group.

Type	Name	Symbol	Mass (MeV/c²)	Mean lifetime
Lepton	Electron / Positron	$e^- \, / \, e^+$	0.511	$> 4.6 \times 10^{20} \ \mathrm{years} \,$
	Muon / Antimuon	$\mu^- \, / \, \mu^+$	105.6	$2.2\times 10^{-6} \ \mathrm{seconds} \,$
	Tau lepton / Antitau	$\tau^- \, / \, \tau^+$	1777	$291 \times 10^{-15} \ \mathrm{seconds} \,$
Meson	Neutral Pion	$\pi^0\,$	135	$8.4 \times 10^{-17} \ \mathrm{seconds} \,$
Meson	Charged Pion	$\pi^+ \, / \, \pi^-$	139.6	$2.6 \times 10^{-8} \ \mathrm{seconds} \,$
Baryon	Proton / Antiproton	$p^+ \, / \, p^-$	938.2	$> 10^{29} \ \mathrm{years} \,$
Baryon	Neutron / Antineutron	$n \, / \, \bar{n}$	939.6	$885.7 \ \mathrm{seconds} \,$
Boson	W boson	$W^+ \, / \, W^-$	80,400	$10^{-25} \ \mathrm{seconds} \,$
Boson	Z boson	$Z^0 \,$	91,000	$10^{-25} \ \mathrm{seconds} \,$

[edit] Probability of survival

The mean lifetime of a particle is labeled $τ$ , 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

γ = E / m

is the Lorentz factor (Energy divided by mass) of the particle.

[edit] Decay rate

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.

[edit] 3-body decay

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

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

[edit] Four-momentum

Main article: Four-momentum

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

This is defined as the difference between the square of its energy and the square of its three-momentum:

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

The square of the four momentum of two particles is

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

[edit] Conservation of four-momentum

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

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

[edit] In two-body decays

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

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

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

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

[edit] Two-body decays


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.