# Special Relativity pion decay problem

• Nate D
In summary: Okay so I took a break and came back to it, here goes round two :smile:The following image has my new work attached to it, but as far as that specific question is concerned, based on conservation of energy, Ee = m∏*c^2 - pe*c. From that I think I was able to solve for the correct answer, hopefully I didn't make a glaring mistake in the initial stepsYour steps are correct but there is a mistake in the final answer. You forgot to take the square root. The energy of the positron has to be positive, but you get a negative value without the square root. So the correct answer is Ee = 136.85 MeV
Nate D

## Homework Statement

In the rare decay ∏+ → e+ + ve , what is the momentum of the positron (e+)? Assume the ∏+ decays from rest. (m+ = 139.6 MeV/c^2, mv ≈ 0, me = 0.511 MeV/c^2)

## Homework Equations

Conservation of Energy: E = Ee + Ev

Conservation of momentum: p = pe + pv
0 = pe + pv
pe = -pv

Invariant mass: E^2 = (pc)^2 + (mc^2)^2

## The Attempt at a Solution

My first step was to ensure momentum was conserved, stating that once the pion decayed, the positron and neutrino went off in opposite directions. This yields the equation pe = -pv.

Next I went about finding the rest energy of the pion, which following the equation, E = mc^2. Using this I found the rest energy of the pion to simply be 139.6 MeV.

After that I began to use the formula, E^2 = (pc)^2 + (mc^2)^2.

I thought that E in this case is the energy of the system, 139.6 MeV, m is the rest mass of the positron, .511MeV/c^2, and p is the momentum of the system that I am asked to solve for.

Solving that equation I found p to be 139.599 MeV/c.

This is where I am confused/unsure and could use some help if possible. Is that momentum
the momentum of the entire system or of the positron after the decay? Also how do I factor in the neutrino considering it is massless? I know the equation E = pc can be used to solve for the energy of a massless particle, however I am not sure if that needs to be used for this problem.

This is my first post here, and I am new to forums in general so I am sorry if I forgot to follow a certain protocol of these forums! Thank you for any help that you may be able to give me!

Welcome to PF;
Usually you want to specify a reference frame when you do relativity problems - don't leave it implied.

Anyway:
Is that momentum
the momentum of the entire system or of the positron after the decay?
What is the initial momentum of the system?
What, therefore, is the final momentum of the system?
How does that compare with the value you calculated for momentum?
Therefore, did you compute the momentum of the system?

However - in your energy calculation, how did you account for the total energy of the neutrino?
Surely conservation of energy goes something like this:

##E_i=E_f=E_{e^+}+E_\nu##

Last edited:

What is the initial momentum of the system?

Because the Pion is at rest the initial momentum is therefore zero.

What, therefore, is the final momentum of the system?

Due conservation of momentum, the final momentum is also zero.

How does that compare with the value you calculated for momentum?

Well evidently the value I calculated was not zero, so it could not be the final momentum of the system

Therefore, did you compute the momentum of the system?

No, it appears I did not.

However - in your energy calculation, how did you account for the total energy of the neutrino?

I was unsure of how to include the total energy of the neutrino in my energy calculation. I know that energy of a massless particle can be found by the equation E = pc, however I do not know the momentum of the neutrino, I only know that it is equal and opposite to that of the positron, correct?

If I continue on with the conservation of momentum I am able to get to this point,

E = Ee + Ev
139.6 MeV = Ee + pv/c

At which point I do not know the energy of the positron, or the momentum of the neutrino.

I'm sure I am missing a small detail or being very thick about this problem in some way, but I am just not sure what my next step should be.

Well evidently the value I calculated was not zero, so it could not be the final momentum of the system
:D It is very easy to lose sight of these things. Sometimes you have to sit back and take a break and then go back to first principles.

You are doing your math in the lab frame BTW.
Another common one is the center of mass frame - which can be fun when one of your particles is massless.

139.6 MeV = Ee + pv/c
I thought you said that ##E_\nu=pc##?
I do not know the momentum of the neutrino, I only know that it is equal and opposite to that of the positron, correct?
... for the total lab-frame momentum to be zero - that seems reasonable. So you can write ##p_\nu = p_{e^+}=p## ? How does the energy of the positron relate to it's momentum?

Okay so I took a break and came back to it, here goes round two

I thought you said that Eν=pc?

I definitely did, I must have mistyped that earlier

How does the energy of the positron relate to it's momentum?

The following image has my new work attached to it, but as far as that specific question is concerned, based on conservation of energy, Ee = m*c^2 - pe*c.

From that I think I was able to solve for the correct answer, hopefully I didn't make a glaring mistake in the initial steps!

#### Attachments

• IMG_1023.jpg
21.9 KB · Views: 645
Ee = m*c^2 - pe*c.

Um - This says to me that the total energy of the positron is equal to the difference between the rest-mass energy of the pion and the positron momentum term from the energy-momentum formula?

Surely: ##E_{e^+}^2=m_{e^+}^2c^4+p^2c^2##

----------------------------------

Aside:
... you know that ∏ is an upper-case pi right? The lower case pi is "π" - not to be confused with "n".
It is easier to write Greek letters like this: ##\Pi## and ##\pi##.

Notice how my equations are so much easier to read than yours - this is because I use LaTeX.
This is very good to learn. Just saying.

## 1. What is the Special Relativity pion decay problem?

The Special Relativity pion decay problem refers to the discrepancy between the predicted and observed lifetimes of subatomic particles called pions. According to Einstein's theory of Special Relativity, particles moving at high speeds should experience time dilation, meaning their perceived lifetimes should be longer than those at rest. However, this is not the case for pions, as their observed lifetimes are shorter than predicted.

## 2. How was the Special Relativity pion decay problem discovered?

The problem was first observed in the 1940s by physicists studying cosmic rays. They noticed that pions, which are produced in collisions between high-energy particles in the Earth's atmosphere, decayed faster than expected based on their known lifetimes at rest.

## 3. What is the proposed solution to the Special Relativity pion decay problem?

The proposed solution is the introduction of a new concept called the "pion cloud." According to this theory, pions are surrounded by a cloud of virtual particles that affect their decay rate. This cloud is not affected by time dilation and therefore explains the observed discrepancy in pion lifetimes.

## 4. How has the Special Relativity pion decay problem been tested?

The pion cloud theory has been tested through various experiments, including the CERN Muon g-2 experiment. This experiment measured the magnetic moment of muons, which are similar to pions, and found that they decayed at a rate consistent with the pion cloud theory.

## 5. What implications does the Special Relativity pion decay problem have for our understanding of physics?

The Special Relativity pion decay problem has significant implications for our understanding of the fundamental laws of physics. It challenges the assumptions of Special Relativity and has led to the development of new theories and concepts, such as the pion cloud theory, to explain the observed phenomenon. It also highlights the need for further research and exploration in this area to gain a deeper understanding of the nature of particles and their behavior at high speeds.

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