Annihilation, perhaps a defination problem

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

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

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

[tex]where \qquad K= \frac{2\pi}{\lambda} [/tex]

 

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

[tex]E^2 = p^2c^2 + m^2c^4[/tex]

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

[tex]\Rightarrow p = \frac{E}{c}[/tex]

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 travelling directly towards each other with the same energy). So to conserve momentum two photons are created, travelling in opposite directions with the same energy.

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(*) 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.

 

Sounds good to me. No problem