Special Relativity pion decay problem

AI Thread Summary
The discussion revolves around calculating the momentum of a positron resulting from the decay of a pion at rest. The user correctly identifies that momentum conservation leads to the conclusion that the total final momentum must also be zero, indicating that the positron and neutrino have equal and opposite momenta. Initial calculations yielded a non-zero momentum, prompting confusion about whether this represented the system's momentum or just the positron's. The conversation emphasizes the importance of accounting for the energy of the massless neutrino and correctly applying conservation laws. Ultimately, the user is guided to refine their approach using energy-momentum relations to arrive at the correct solution.
Nate D
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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!
 
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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:
Thanks for the quick reply!

To answer your questions,

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 :smile:

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 :smile:
: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.
Then you will be able to answer your own questions more.

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 :smile:

I thought you said that Eν=pc?

I definitely did, I must have mistyped that earlier :frown:

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!
 

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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.
 
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