What is the Maximum Energy of a Photon in Positron-Electron Annihilation?

[tnho]

Homework Statement

A positron having a kinetic energy equal to its rest mass energy mec^2 collides with a stationary electron. The positron and the electron annihilate in the process and two photons are created. What is the maximum possible energy of a photon produced in the annihilation process??

Homework Equations

4-momentum vectors(??)
CM frame can help ??

The Attempt at a Solution

 

[Dick]

4-momentum vectors, absolutely. Though it's hard to be sure until you write some down and try them out. CM may help (at least to think about) but I suggest you try first in the rest frame of the electron since it seems that is the frame in which they want an answer.

 

[tim_lou]

I'll definitely get a boost to the CM frame so that the two final photons travel at opposite direction.

then look at momentum 4 vectors of each objects, take the 4 vector dot product. and you'll get something involving the usual dot product of the two momentum vector (of the photons). simplify things (put everything in terms of energy of the two photons, E1, E2) (the dot products can be simplified since the photons travel at opposite direction). then work some inequality out (complete the squares, work out AM-GM or Cauchy if you know those).

formulas to use:
recall:
$$E=pc=hf$$ for photons

and the CM frame moves at:
$$\vec{V}_{\text{CM}}=\frac{\sum_i{\vec{p}_ic^2}}{\sum_i E_i}$$

 

[tnho]

in fact, i don't know under what physical conditions, the energy of photon is at its maximum.

If the energy of photon is maximum in one frame (says CM frame), it is also maximum in another frame (says Rest frame), isn't it??

Thx

 

[Dick]

in fact, i don't know under what physical conditions, the energy of photon is at its maximum.

If the energy of photon is maximum in one frame (says CM frame), it is also maximum in another frame (says Rest frame), isn't it??

Thx

Not true at all. This why I suggested you stay in the rest frame. There is such a thing as a doppler shift.