A particle traveling at a speed V decays to form two photons (no mass)

[amr55533]

Conservation of relativistic momentum and energy, pion decays

Homework Statement

A small particle (pion), traveling at a velocity V, decays into two rays, γ1 and γ2. Find the Momentum and Energy of γ1 and γ2 if: a) γ1 is in line with V, and b) if γ1 is perpendicular to V.

I drew out the problem and listed all of the givens:

http://imageshack.us/a/img832/648/physprob.png

Homework Equations

Eπ=Eγ1+Eγ2

Pπ=Pγ1+Pγ2

E=P*c (for rays, no mass)

E=mc^2/sqrt(1-V^2/c^2) (for particle)

E=K+E0

P=mV/sqrt(1-V^2/c^2) (for particle)

E0=mc^2

The Attempt at a Solution

Below is my attempt at a solution so far:

http://img593.imageshack.us/img593/5444/solutionattempt.png

So far, I found:


m∏=2.406E-28 kg

V=2.827E8 m/s

Ebefore=405 MeV

Pbefore=382.1 MeV/c


I know that the total momentum after the particle decays has to equal the momentum before and same with the energy. I am just unsure how to finish the problem at this point.

Can I assume that Pγ1=Pγ2 after the disintegration?

I am also confused on how to handle part b) where γ1 is perpendicular to V.

Thanks in advance!

 

[Simon Bridge]

Can I assume that Pγ1=Pγ2 after the disintegration?

If the two photons had equal momentum, would they add up to the same as the momentum of the pion (remember that momentum is a vector)?

Note - you left off $E^2=E_0^2 + (pc)^2$

 

[Simon Bridge]

Could I assume that pγ1=pγ2=190.9 MeV/c

You can answer this for yourself - if the two photon momenta were the same, then what is the total momentum in fig (a) "after"? Compare with the total momentum in fig (a) "before". Is this sensible?

If you don't answer questions, nobody can help you.

 

[amr55533]

Since momentum is a vector, the total momentum would be zero in the "after" state if they had equal magnitudes, but opposite directions. However, the direction of γ2 is actually arbitrary. The only way that I can seem figure this out is if both photons are moving in the same direction with magnitudes equal to one half of the ∏ particle. This would make it so that each photon had a momentum of 190.9 MeV/c.

Using that value for the momentum and the equation that a massless particle has an associated energy of P*c, it would also come out that the energy of each photon was 190.9 MeV

Therefore, there would have to be a disintegration energy of 405MeV-381.8MeV=23.2MeV associated with the decay.

Am I moving in the right direction with this?

Thanks for the help!

 

[vela]

Since momentum is a vector, the total momentum would be zero in the "after" state if they had equal magnitudes, but opposite directions. However, the direction of γ2 is actually arbitrary. The only way that I can seem figure this out is if both photons are moving in the same direction with magnitudes equal to one half of the ∏ particle. This would make it so that each photon had a momentum of 190.9 MeV/c.

Using that value for the momentum and the equation that a massless particle has an associated energy of P*c, it would also come out that the energy of each photon was 190.9 MeV

Good so far.

Therefore, there would have to be a disintegration energy of 405MeV-381.8MeV=23.2MeV associated with the decay.

What is "disintegration energy"? Energy has to be conserved; it has to go somewhere. You can't have extra left over. What you've discovered is energy isn't conserved: before doesn't equal after. Therefore, your assumption that the photons both move in the same direction, each carrying half the total momentum is wrong.

Am I moving in the right direction with this?

Thanks for the help!

Let $E_\pi$ and $\vec{p}_\pi$ denote the energy and momentum of the pion. First, orient the coordinate system so that the pion moves in the +x direction. That means the y and z components of its momentum will be 0, so you have $\vec{p}_\pi = (p, 0, 0)$, where we take $p$ to be a positive quantity.

Let $E_1$ and $\vec{p}_1$ denote the energy and momentum of photon 1. In part a, you're given that photon 1 moves in the same direction, so the y and z components of its momentum will also be 0. You can write $\vec{p}_1 = (p_1, 0, 0)$, where $p_1$ is positive. You also know that $E_1 = p_1 c$.

Let $E_2$ and $\vec{p}_2$ denote the energy and momentum of photon 2. As you noted, photon 2 could go off in any direction. All you can say right now is that $E_2 = p_2 c$, where $p_2 = \|\vec{p}_2\|$.

So far you've used conservation of energy to say that $E_\pi = E_1 + E_2$. So far, so good.

Now apply conservation of momentum. What can you deduce about y and z components of $\vec{p}_2$? Come up with an equation that relates $p_\pi$, $p_1$, and $p_2$.

 

[amr55533]

I figured it out. Once I applied the fact that momentum is a vector and and energy is a scalar, I was able to set up 2 sets of equations for each part of the problem and solve for the momentum values and then the energies.

Thanks for all of the help!

 

[Simon Bridge]

Well done!