If photons have no mass, how they create electrons and positrons on colliding? [duplicate]

Question

How are electrons and positrons formed when photons collide, as photons are massless?

Answer 1

I am uncertain why mass is relevant here, but perhaps a nice way of looking at this sort of thing is to go the other way - as, other than thermodynamic physics laws, most physics laws are time-direction independent. Most people would not find it a stretch to imagine that when matter and antimatter collide, their masses would 'cancel out'. However, they don't, not really. The mass of the two particles is converted into energy via $E=mc^2$. As the mass of one particle is equal to the mass of the other, we can say $E=2m_ec^2$. This, however, would only take into account static collisions - rather an oxymoron - and so we add in an extra term to account for the speed, giving us:

$E=\frac{1}{2}m_ev_1^2+\frac{1}{2}m_ev_2^2+2m_ec^2$

This now fulfills conservation of energy, and so the next thing to be considered is conservation of momentum. If the two particles collided head-on, their net momentum would be 0. As photons have non-zero momentum, and move, there must be two photons emitted in opposite directions to conserve momentum.

Now, we simply take time, and reverse it!

If photons of high enough energy/momentum collide, then, fulfilling conservation of energy, they can convert their energy into mass. Having multiple photons helps to fulfill conservation of momentum, but interestingly, that's not the only way - recoil of a nearby mass is also a method of conserving momentum (although, in that instance, one could argue that a photon acts to transfer momentum to the mass, anyway).

At this point, we've pretty much fulfilled all physical laws, and shown that this logically could happen. We've also shown (via time-reversal) that, if annihilation is logical, then pair production is as well.

Answer 2

There is no law of conservation of mass.

The equation that is applicable for massive and massless particles is: $E^2 = p^2c^2 + m^2c^4$, where $p$ is the 3-momentum and $m$ is the invariant mass. $c$ is the usual speed of light.

Now the conservation law in a scattering process is the conservation of 4-momentum, which can be decomposed into the two familiar statements of: 1) conservation of energy and 2) conservation of 3-momentum.

So, as long as you satisfy the above conservation laws, you're allowed to have different masses before and after a collision process.

Answer 3

The conserved quantity is mass-energy.

1. A particle at rest has only (rest) mass.

2. A particle near the speed of light, such as a typical neutrino, has some rest mass (with $E_{rest}$ given by $m_{rest}c^2$) and kinetic energy $K \gg E_{rest}$.

3. A massless particle, such as a photon, has no rest mass, no $E_{rest}$, only its "kinetic" energy $pc = \frac{hc}{\lambda}$.

In a collision where some particles are destroyed and others created, the total mass-energy is conserved, but it can be distributed in various combinations of rest masses and kinetic energies.