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Decaying particle mass?

  • Thread starter Angel_Kate
  • Start date
  • #1
Hi there,

I'm struggling with a question about particle mass. I know that the particle is neutral and decays to two photons of energy 500Gev and 600Gev with an angle of 60o between them. I now have to work out the mass of the initial particle. The initial particle is in flight when it decays so I can't assume zero momentum. I know it must have something to do with momentum but I can't figure out where to go from there!

Thanks for any help!

Angel_Kate
 

Answers and Replies

  • #2
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5
There was no transverse momemtum before the decay, so there is no net transverse momentum afterward. So calculate the trajectory of the neutral particle between the two photons before decay. Now calculate the longitudinal momentum of the neutral particle. Now calculate the total energy of the neutral particle. Now calculate the rest mass of the neutral particle.
 
  • #3
Hi Bob,

Thanks for the reply. How do I know there is no transverse momentum before the decay? I thought the only way you could neglect momentum was to have the resultant photons at 180o to each other?

Thanks!

Angel_Kate
 
  • #4
4,662
5
Hi Bob,

Thanks for the reply. How do I know there is no transverse momentum before the decay? I thought the only way you could neglect momentum was to have the resultant photons at 180o to each other?
The two decay photons define a plane. The neutral particle is in that plane. Draw the neutral particle moving along x in the x-y plane from -x to zero, with the decay at zero. Now draw the two decay photons with the vertex at x = 0. Does the neutral particle have any momentum in the y direction? (your answer better be zero). Solve the transverse momenta of the two photons to set total transverse momenta to zero. (It has to be.) Do everything in MeV units, including momenta (pc units).
 
  • #5
Hi Bob,

I'm still being a bit dense! I understand that there can't be any transverse momentum after the decay (I've drawn a nice picture!) but I'm struggling to do the calculation. I've done it so that I'm resolving each photon along the y-axis to calculate its momenta. But then how do I just set these values to zero?

Apologies for my confusion!!

Angel_Kate
 
  • #6
diazona
Homework Helper
2,175
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By "resolving" I suppose you mean "projecting" - taking the y-component of each photon's momentum? If so, then the sum of the y-components needs to be zero. One photon flies up and to the right, the other flies down and to the right, so one y-component is up and the other is down. In order to sum to zero, they should have the same magnitude.
 
  • #7
181
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One photon flies up and to the right, the other flies down and to the right, so one y-component is up and the other is down. In order to sum to zero, they should have the same magnitude.
But we don't know (neither do we need to know) the angles the photons
make with the trajectory of the original particle.
Furthermore, it's not simple to solve
500sin[tex]\vartheta[/tex] = 600sin(60 - [tex]\vartheta[/tex]) for [tex]\vartheta[/tex]

We just want the magnitude of the vector sum of the momenta of the photons.

Let the momenta be 500i and 600cos(60)i + 600sin(60)j
Just add them and find the square of the magnitude to give p2c2

E2 = p2c2 + (m0c2)2

BTW If you needed [tex]\vartheta[/tex] (as above), you could find it easily from this as well

David
 
Last edited:
  • #8
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Apologies for my confusion!!

Angel_Kate
I understand your confusion.
I hope my post explains how inappropriate it is to
take the original direction to be the x axis, and that you
can understand the simple approach I gave.

David
 
  • #9
4,662
5
But we don't know (neither do we need to know) the angles the photons
make with the trajectory of the original particle.
Furthermore, it's not simple to solve
500sin[tex]\vartheta[/tex] = 600sin(60 - [tex]\vartheta[/tex]) for [tex]\vartheta[/tex]
500 sin(theta) = 600 sin(60 - theta) = 600 sin(60) cos(theta) - 600 sin(theta) cos(60)

[500 + 600 cos(60)] sin(theta) = 600 sin(60) cos(theta)

tan(theta) = [600 sin(60)]/[500 +600 cos(60)] = 0.64952
theta = 33.004 degrees

pc = 500 cos(33) + 600 cos(27) = 953.9 MeV (momentum in pc units)
E=500 + 600 = 1100 MeV (total energy)
(m0c2)2 = E2 - (pc)2 = 11002 - 953.92
m0c2 = 547.8 MeV rest mass
Eta particle (meson) has a rest mass of 547.3 MeV and a 2-gamma decay mode. Must be it.
 
  • #10
181
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OK Bob,
I'll take your word for it:rofl:
 
  • #11
181
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Let the momenta be 500i and 600cos(60)i + 600sin(60)j
Just add them and find the square of the magnitude to give p2c2
I'm sorry to have been so flippant. What I meant was:

(pc/GeV)2 = (500 + 600/2)2 + 6002*3/4 = 910000

David
 
Last edited:

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