Experiments to distinguish 3-body and 2-body decays?

In summary, there are experiments that can distinguish between 2 and 3-body decays by measuring the energy spectrum of the visible decay products. In a 2-body decay, all muons are produced at the same energy, while in a 3-body decay, there is a continuous energy spectrum due to the ambiguity in momenta and energies. This difference was a key factor in demonstrating the existence of neutrinos. Additionally, in a 2-body decay, there are two unknowns and two conservation laws, resulting in a single solution, while in a 3-body decay, there are more unknowns and fewer conservation laws, leaving room for more than one solution.
  • #1
unscientific
1,734
13
I was wondering are there any experiments to distinguish between 2 and 3-body decays? For example, consider decay of the muon and the pion:
7RsFo.jpg


The pion only emits 1 muon neutrino ("missing energy") and 1 muon. The muon however, emits 1 muon neutrino, 1 electron neutrino and 1 electron.

How is it established experimentally that the pion only emits 1 neutrino whereas the muon emits 2 neutrinos?
 
  • Like
Likes Buzz Bloom
Physics news on Phys.org
  • #2
Well, there are experiments to see the neutrinos, but you can measure the energy spectrum of the visible decay product: a two.body decay produces all muons at the same energy (in the pion rest frame), in a three-body decay you get a continuous energy spectrum.
 
  • Like
Likes Buzz Bloom
  • #3
In the rest frame of the pion the muon will carry an energy of [itex]E_\mu=\dfrac{m_\pi}{2}[/itex].
In the same result for the muon the electron will have to carry less energy...
 
  • #4
mfb said:
there are experiments to see the neutrinos,
^ also that ^_^ however I don't know whether there are experiments actually going on with muon decays...
 
  • #5
mfb said:
Well, there are experiments to see the neutrinos, but you can measure the energy spectrum of the visible decay product: a two.body decay produces all muons at the same energy (in the pion rest frame), in a three-body decay you get a continuous energy spectrum.

Why is it that in a 2 body decay all muons are produced at the same energy? Since ##E_{cm} = E_\nu + E_\mu##, shouldn't there be a spectrum as well?
 
  • #6
In the rest frame of the pion you initially have a four-momentum:

[itex]P_{in} = \begin{pmatrix} m_\pi \\ \vec{0} \end{pmatrix}~~,~~P_{fin}= \begin{pmatrix} E_\mu + E_\nu \\ \vec{p}_\mu + \vec{p}_\nu \end{pmatrix}~~,~~ E_\nu = \sqrt{|p_\nu|^2 + m_\nu^2} \approx |p_\nu|[/itex]

conservation of momentum tells you that:
[itex]\vec{p}_\nu = -\vec{p}_\mu \Rightarrow E_\nu = |\vec{p}_\mu|[/itex]

So you have:
[itex]P_{fin}= \begin{pmatrix} E_\mu + |\vec{p}_\mu| \\ \vec{0} \end{pmatrix}[/itex]

Take the square equality (I am not sure whether my equations are correct BUT by the fact that I already wrote that [itex]\vec{p}_\nu = - \vec{p}_\mu[/itex] you should already get your answer- in particular they should be fixed numbers for the given decay since you can only have 1 variable):
[itex] E_\mu^2 + |p_\mu|^2 + 2 E_\mu |p_\mu| = m_\pi^2[/itex]
[itex] 2 |p_\mu|^2 + 2E_\mu |p_\mu| = m_\pi^2 - m_\mu^2[/itex]
[itex] \sqrt{m_\mu^2 + |p_\mu|^2}= \dfrac{ m_\pi^2 - m_\mu^2}{2|p_\mu|}- |p_\mu|[/itex]
[itex] m_\mu^2 + |p_\mu|^2 = \Big( \dfrac{ m_\pi^2 - m_\mu^2}{2|p_\mu|}\Big)^2+ |p_\mu|^2- (m_\pi^2 - m_\mu^2)[/itex]
[itex] m_\mu^2 +(m_\pi^2 - m_\mu^2)= \dfrac{ (m_\pi^2 - m_\mu^2)^2}{4|p_\mu|^2}[/itex]

[itex] |p_\mu|= \frac{1}{2} \sqrt{\dfrac{ (m_\pi^2 - m_\mu^2)^2}{m_\mu^2 +(m_\pi^2 - m_\mu^2)}}[/itex]
which is a fixed number and so is the muon energy...
Forget my m_pi/2 that would be the case for equally massive produced particles.
 
Last edited:
  • Like
Likes Buzz Bloom and unscientific
  • #7
In a 3 body decay [itex]1 \rightarrow 2+3+4[/itex] the condition for momentum conservation in the rest frame of 1 would be:
[itex]\vec{p}_2= - \vec{p}_3 - \vec{p}_4[/itex]
Which doesn't give a fixed number anymore , you can change 3 and 4 freely to keep the equality. In particular if I say that [itex]\vec{p}_3,\vec{p}_4[/itex] are such to give the equality right, then [itex]\vec{p}_3^\prime= \vec{p}_3 +\vec{a}[/itex] and [itex]\vec{p}_4^\prime= \vec{p}_4 - \vec{a}[/itex] can still keep the equality. This gives an ambiguity in the momenta and so the energies which results in a continuous spectrum.

That is a general fact difference between 2 and 3 body decays, and it was the reason why neutrinos were first demonstrated.
 
Last edited:
  • Like
Likes Buzz Bloom and unscientific
  • #8
ChrisVer said:
That is a general fact difference between 2 and 3 body decays, and it was the reason why neutrinos were first demonstrated.
Good point.

Another way to see the difference:
2-body decay: let the x-direction be along the motion of one particle, the other will then move in negative x direction. You get two unknowns (the momenta of the two particle), and two conservation laws - energy and momentum. The result is a single solution.
3-body decay: suddenly you get 1+2+2 unknowns (you have to take the second dimension into account), and only 3 conservation laws (energy, and momentum in two dimensions). That leaves two degrees of freedom for the decay mechanism.
 
  • Like
Likes Buzz Bloom

1. What is the difference between 3-body and 2-body decays?

The main difference between 3-body and 2-body decays is the number of particles involved in the decay process. In a 3-body decay, three particles are produced as a result of the decay, while in a 2-body decay, only two particles are produced.

2. How can we experimentally distinguish between 3-body and 2-body decays?

One way to distinguish between 3-body and 2-body decays is by looking at the energy and momentum distributions of the decay products. In a 3-body decay, the energy and momentum will be shared between three particles, resulting in a more complex distribution compared to a 2-body decay.

3. What factors influence the rate of 3-body and 2-body decays?

The rate of 3-body and 2-body decays is influenced by a variety of factors, including the masses of the particles involved, the strength of the decay interaction, and the available energy for the decay process. Additionally, the spin and angular momentum of the particles can also affect the decay rate.

4. How do we measure the branching ratio for 3-body and 2-body decays?

The branching ratio for 3-body and 2-body decays can be measured by comparing the number of events with three particles in the final state to the number of events with two particles in the final state. This ratio is then compared to the theoretical predictions to determine the branching ratio.

5. What are some examples of 3-body and 2-body decays?

Some examples of 3-body decays include the decay of a neutral pion into two photons and an electron-positron pair, and the decay of a neutral kaon into three pions. Examples of 2-body decays include the decay of a neutral B meson into a J/psi meson and a photon, and the decay of a neutron into a proton and an electron-antineutrino.

Similar threads

  • High Energy, Nuclear, Particle Physics
Replies
2
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
17
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
13
Views
3K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
11
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
1K
Replies
13
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
2
Views
919
Back
Top