Relativistic energy and momentum question

Click For Summary

Homework Help Overview

The discussion revolves around a decay process involving a lambda particle that decays into a proton and a pion, with the proton at rest. Participants are exploring the conservation of energy and momentum in the context of relativistic physics, specifically focusing on calculating the energies of the pion and the original lambda particle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using conservation laws and Lorentz transformations to analyze the problem. There are attempts to derive relationships between the energies and momenta of the particles involved. Questions arise regarding the use of four-vectors and the implications of the proton being at rest.

Discussion Status

The discussion is active, with participants providing insights and corrections to each other's reasoning. Some have identified mistakes in their calculations and are working through the implications of their findings. There is a collaborative effort to clarify the relationships between the energies and momenta of the particles.

Contextual Notes

Participants are working under the constraints of given masses for the lambda particle, proton, and pion, and are trying to derive energies without explicit velocity values. The discussion reflects uncertainty regarding the application of relativistic equations and the proper handling of terms in their calculations.

doublemint
Messages
138
Reaction score
0
1. A lambda particle decays into a proton and pion and it is observed that the proton is left at rest.
a. what is the energy of the pion?
b. what is the energy of the original lambda? m of lambda = 1116MeV/c^2, m of proton = 938 MeV/c^2, and m of pion = 140 MeV/c^2



Homework Equations


Conservation of energy and momentum. E[tex]\lambda[/tex] = E [tex]\rho[/tex] + E[tex]\pi[/tex], P [tex]\lambda[/tex] = P[tex]\rho[/tex] + P[tex]\pi[/tex]
Lorentz Transformation



The Attempt at a Solution


I tried breaking down conservation of energy and momentum so solve for what i need but i can't seem to get it right. I've tried using lorentz transformation with the energy and momentum but no lucky either.
Any help is appreciated!
Thanks
 
Physics news on Phys.org
The energy and momentum of the proton are straightforward to write down.

Use [itex]m^2 = E^2-p^2[/itex].

There's no need for a Lorentz transformation because you only have one frame in this problem.

You can save yourself a bit of algebra work if you know how to work with four-vectors.
 
So you mean m[tex]^{2}[/tex]c[tex]^{2}[/tex]=E[tex]^{2}[/tex] - p[tex]^{2}[/tex]c[tex]^{2}[/tex] ?
so,
E[tex]\lambda[/tex] = E[tex]\rho[/tex] + E[tex]\pi[/tex]
and since the proton is not moving, p = 0
the energy of the pion is,
E[tex]\pi[/tex] = E[tex]\lambda[/tex] - mc[tex]^{2}[/tex]

i don't know anything about doing work in 4 vectors...

for part b, what do i need to do to get the energy of the lambda particle? i have no values for velocity.
 
OK, so you have, so far, these relations:

[tex]E_\lambda = m_pc^2+E_\pi[/tex]
[tex]p_\lambda = p_\pi[/tex]

Multiply the second by c, square both equations, and then subtract the second from the first. Hopefully, you'll see where to go from there.
 
I see what you are doing! Thanks!
If i did it right, it simplifies to m[tex]_{\lambda}[/tex]c[tex]^{2}[/tex] = m[tex]_{\pi}[/tex]c[tex]^{2}[/tex] + m[tex]_{p}[/tex]c[tex]^{2}[/tex] = E of the original lambda
where E = 1078 MeV
but i think i did something wrong though
 
Last edited:
What happened to the cross term on the RHS? That's the one that'll allow you to solve for the energy of the pion. Also, the other terms should be of the form [itex](mc^2)^2[/itex], not just [itex]mc^2[/itex].
 
I found my mistake! I knew there was a problem with my answer.
So, from the energy of the pion which i got as 184.4MeV, I use that to solve for lambda which i got to be 1122MeV!
I hope that is right!

Thanks a bunch vela!
 

Similar threads

Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
1
Views
2K
Replies
13
Views
3K
  • · Replies 8 ·
Replies
8
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K