Finding the total energy of a Pi meson

AI Thread Summary
The discussion focuses on calculating the total energy of a moving Pi meson after a K meson decays into two Pi mesons, with one remaining at rest. The initial energy is derived from the rest mass and kinetic energy of the K meson, while the final energy involves the rest mass of both Pi mesons and the kinetic energy of the moving one. Participants suggest using conservation of energy and momentum principles, along with relevant equations, to solve for the energy of the moving Pi meson. There is a caution against using gamma directly due to complexity, and instead, a systematic approach involving squaring energy equations is recommended. The conversation emphasizes the importance of careful algebraic manipulation to arrive at the solution.
AsadaShino92
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Homework Statement



A K meson (an elementary particle with approximately 500 Mev rest mass) traveling through the laboratory breaks up into two Pi mesons (elementary particles with 140 Mev rest energies). One of the Pi mesons is left at rest. What is the total energy of the remaining Pi meson?

Homework Equations



E=mc^2
KE=(γ-1)mc^2

The Attempt at a Solution



My first step was trying to use energy conservation E(initial)=E(final)
The K meson is traveling so the total E at the start is E(rest mass of K)+KE(K meson)
E(initial)=500 Mev+(γ-1)mc^2

In the end, the K meson breaks into 2 Pi mesons, each with 140 Mev rest mass. One stops and the other keeps on moving.
So the total E at the end is E(rest mass Pi)+E(rest mass 2nd Pi)+(γ-1)mc^2
E(final)=140 Mev+140Mev+(γ-1)mc^2

E(initial)=E(final)
500 Mev+(γ-1)mc^2=140 Mev+140Mev+(γ-1)mc^2

But now I don't know what to do next. I don't have a value for gamma.
 
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AsadaShino92 said:

Homework Statement



A K meson (an elementary particle with approximately 500 Mev rest mass) traveling through the laboratory breaks up into two Pi mesons (elementary particles with 140 Mev rest energies). One of the Pi mesons is left at rest. What is the total energy of the remaining Pi meson?

Homework Equations



E=mc^2
KE=(γ-1)mc^2

The Attempt at a Solution



My first step was trying to use energy conservation E(initial)=E(final)
The K meson is traveling so the total E at the start is E(rest mass of K)+KE(K meson)
E(initial)=500 Mev+(γ-1)mc^2

In the end, the K meson breaks into 2 Pi mesons, each with 140 Mev rest mass. One stops and the other keeps on moving.
So the total E at the end is E(rest mass Pi)+E(rest mass 2nd Pi)+(γ-1)mc^2
E(final)=140 Mev+140Mev+(γ-1)mc^2

E(initial)=E(final)
500 Mev+(γ-1)mc^2=140 Mev+140Mev+(γ-1)mc^2

But now I don't know what to do next. I don't have a value for gamma.

What about momentum?
 
PeroK said:
What about momentum?

Then I would use p=γmv but I don't think I have the value for those. I just know that momentum would also be conserved.
 
AsadaShino92 said:
Then I would use p=γmv but I don't think I have the value for those. I just know that momentum would also be conserved.

In general it's messy to use ##\gamma## in this sort of problem. Can you think of a way to tackle the problem without it?
 
PeroK said:
In general it's messy to use ##\gamma## in this sort of problem. Can you think of a way to tackle the problem without it?
P=E/C ?
 
AsadaShino92 said:
P=E/C ?

That's for a massless particle. Do you know:

##E^2 = p^2c^2 + m^2c^4##?
 
PeroK said:
That's for a massless particle. Do you know:

##E^2 = p^2c^2 + m^2c^4##?
So am I using this equation to solve for momentum?
 
AsadaShino92 said:
So am I using this equation to solve for momentum?

The general approach to these particle problems is:

1) Conservation of energy
2) Conservation of momentum.
3) For each particle: ##E^2 = p^2c^2 + m^2c^4##

Hint: stick with ##M## for the initial K meson mass and ##m## for the pi meson mass and leave the numbers out until the end.

The notation is up to you, but I would use ##E_0, p_0## for the Energy & Momentum of the initial K meson; ##E_1, p_1 (p_1= 0)## for the pi meson that ends up at rest and ##E_2, p_2## for the second pi meson.

You're trying to solve for ##E_2## of course.
 
PeroK said:
The general approach to these particle problems is:

1) Conservation of energy
2) Conservation of momentum.
3) For each particle: ##E^2 = p^2c^2 + m^2c^4##

Hint: stick with ##M## for the initial K meson mass and ##m## for the pi meson mass and leave the numbers out until the end.

The notation is up to you, but I would use ##E_0, p_0## for the Energy & Momentum of the initial K meson; ##E_1, p_1 (p_1= 0)## for the pi meson that ends up at rest and ##E_2, p_2## for the second pi meson.

You're trying to solve for ##E_2## of course.
Using the notation, I have the conservation of energy written as E0=E1+E2
Then using the energy equation for each particle

For the K meson
E02=P02C2+M2C4

For the Pi stopped
E12=m2C4 since P1=0

For Pi moving
E22=P22C2+m2C4

Momentum conservation
P0=P1+P2
Since P1=0
P0=P2

Do I just plug all this into E2=E0-E1?
 
  • #10
AsadaShino92 said:
Do I just plug all this into E2=E0-E1?

Why not? You may want to square that equation first.
 
  • #11
PeroK said:
Why not? You may want to square that equation first.
You mean instead I should use E22=E02-E12?
 
  • #12
AsadaShino92 said:
You mean instead I should use E22=E02-E12?

##(E_0 - E_1)^2 \ne E_0^2 - E_1^2##
 
  • #13
PeroK said:
##(E_0 - E_1)^2 \ne E_0^2 - E_1^2##
My mistake. Then squaring both sides should be E22=(E0-E1)2
 
  • #14
AsadaShino92 said:
My mistake. Then squaring both sides should be E22=(E0-E1)2
##E_2^2 = (E_0 - E_1)^2 = \dots##?
 
  • #15
PeroK said:
##E_2^2 = (E_0 - E_1)^2 = \dots##?
(E0-E1)2=E02-2E0E1+E12
 
  • #16
AsadaShino92 said:
(E0-E1)2=E02-2E0E1+E12

Okay, but you may need to do more than one line of algebra at a time. I'm going off-line now, but maybe someone else can provide further help if you need it.
 

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