Annihilation, perhaps a defination problem

[Hevonen]

[SOLVED] Annihilation, perhaps a defination problem

Problem:
An electron and an anti-electron undergo mutual annihilation.
Which of the following does not have to be conserved?
A. Electric charge
B. Kinetic energy
C. Lepton number
D. Momentum

Attempt to question:
The answer is B but I why not also d. I would consider that the mass of photon is zero. So in both B and D, kinetic energy and momentum are perhaps zero. Problem is: it is impossible to divide by zero. Therefore, this problem seems to a definatition problem. It must be that momentum conserves but why? What is the assumption in the theory?

Thank you for your answers!

 

[neu]

Photons are massles, yes, but they still carry momentum (equal to \hbar k)

E^2 = m^2c^4 + P^2c^2 \qquad so \quad E_{\gamma} = Pc \quad

where \quad E = h \nu = \frac{h c}{\lambda} \qquad so \quad P = \hbar K

where \qquad K= \frac{2\pi}{\lambda}

 

[Hootenanny]

Whilst the photon is indeed massless^* it most certainly has momentum. Consider the expression for relativistic energy,

E^2 = p^2c^2 + m^2c^4

As you correctly say, the photon is massless and hence the final term disappears,

\Rightarrow p = \frac{E}{c}

Hence, the magnitude of the photon's momentum is it's energy divided by the speed of light in a vacuum. Notice that since a photon has no 'rest energy', all it's energy must be kinetic energy.

Returning to your problem, before the collision the total momentum of the system is zero (assuming that both particles are traveling directly towards each other with the same energy). So to conserve momentum two photons are created, traveling in opposite directions with the same energy.

\hline
(*) For those proponents of so-called relativistic mass, let us not get into another discussion regarding the pros and cons of the subject as the concept of relativistic mass is not nesscary to explain the OP's query.

 

[Hevonen]

Whilst the photon is indeed massless^* it most certainly has momentum. Consider the expression for relativistic energy,

E^2 = p^2c^2 + m^2c^4

As you correctly say, the photon is massless and hence the final term disappears,

\Rightarrow p = \frac{E}{c}

Hence, the magnitude of the photon's momentum is it's energy divided by the speed of light in a vacuum. Notice that since a photon has no 'rest energy', all it's energy must be kinetic energy.

Returning to your problem, before the collision the total momentum of the system is zero (assuming that both particles are traveling directly towards each other with the same energy). So to conserve momentum two photons are created, traveling in opposite directions with the same energy.

\hline
(*) For those proponents of so-called relativistic mass, let us not get into another discussion regarding the pros and cons of the subject as the concept of relativistic mass is not nesscary to explain the OP's query.

Thank you!
Your both answers solve this problem: photon is massless and it has momentum due to relativity.

 

[Hootenanny]

Thank you!
Your both answers solve this problem: photon is massless and it has momentum due to relativity.

Sounds good to me. No problem