Electron/Positron Annihilation

[Von Neumann]

I recently read a textbook that stated without explanation "When a positron collides head on with an electron, the energy of each photon is the sum of one particle's rest and kinetic energy." So E = m_o*c^2 + K. However my question is, why isn't the energy of each photon twice the mass and kinetic energy of each particle since the ratio is 2:1? I know it's the incorrect answer, but that's just what I intuitively think. Any help?

 

[Dick]

I recently read a textbook that stated without explanation "When a positron collides head on with an electron, the energy of each photon is the sum of one particle's rest and kinetic energy." So E = m_o*c^2 + K. However my question is, why isn't the energy of each photon twice the mass and kinetic energy of each particle since the ratio is 2:1? I know it's the incorrect answer, but that's just what I intuitively think. Any help?

The total energy of the two photons is the same as the sum of the mass and kinetic energy of the colliding electron positron pair. Two photons, two particles. What's this 2:1 ratio of which you speak?

 

[Von Neumann]

The way it's worded, to me it sounds as though one photon is created as the result of the pair colliding. Therefore the total kinetic energy of one photon being twice the energy of one colliding particle.

 

[Dick]

The way it's worded, to me it sounds as though one photon is created as the result of the pair colliding. Therefore the total kinetic energy of one photon being twice the energy of one colliding particle.

One photon could not be created. There is zero momentum total if they collide head on. It couldn't conserve momentum and energy. You need two. I think "each photon" doesn't mean each single photon created by the collision. It means each of the two photons created by the collision.

 

[Von Neumann]

Oh right! Thanks!