# How is momentum conserved when a particle/antiparticle pair is created?

1. Sep 26, 2011

### andrewkirk

My understanding is that when a particle-antiparticle pair is created, they will have opposite spins and momenta. So the net momentum will be zero. But doesn't the momentum depend on the frame of reference? Say the particles both have mass m and velocities +v and -v. Then the net momentum is +mv-mv=0.

What about in a different reference frame that has a velocity boost of u in the positive direction? The velocities in that frame are u+v and u-v. So the net momentum is
m(u+v) + m(u-v) = 2mu. So it appears that, viewed in this new reference frame, momentum has been increased by 2mu by the creation of the pair, apparently violating the law of conservation of momentum.

No doubt I have made a silly mistake somewhere here. Can someone please explain what it is?

Thank you.

2. Sep 26, 2011

### xts

Particles are not created from nothing. You must provide some energy. So you may create e.g. pair of e+/e- in reaction like: gamma+nucleus->e+ + e- + nucleus.
Nucleus is needed to ensure momentum/energy conservation.
So now, in any reference frame you like, sum of momenta of all particles before the reaction is equal to sum of momenta of all particles after reaction.

3. Sep 26, 2011

### Ken G

As xts points out, the answer to your question requires that you consider the momentum present before the creation of the pair. I just want to stress how extremely astute your question is-- you have basically shown, by asking that question, that the presence of energy must always be associated with the presence of momentum, it's just that there is often a "center of momentum" frame in which all the momenta add to zero. If such a frame exists, it is usually the simplest frame to work in, but your question shows that if we transform to a different frame, we might have a net momentum showing up. Conservation of momentum requires that this net momentum must have also existed in the initial frame, regardless of the form the energy took in that frame. That is a very powerful constraint on the connection between energy and momentum! You might wish to explore its ramifications some more. For example, what does the center of momentum frame look like when the annihilating particles are two photons? How do the photon energies transform if you are in a frame moving in the direction that the photons collide? Analyzing the momentum of the outbound electron/positron, what is the momentum of a photon?

4. Sep 26, 2011

### stone

To add to previous posts, at relevant energies (high), it is the four-momentum which must be taken into account (E, px, py, pz). Further the momentum addition laws are also changed to their lorentz form.

5. Sep 26, 2011

### andrewkirk

Thank you very much for the replies. There's lots for me to think about there. I had forgotten about how the four-momentum can resolve problems that exist when you only work with three-momenta. Just on the question of whether particles can be created from nothing, I am way out of my depth here but I thought that Hawking Radiation was an instance of the creation of particle-antiparticle pairs from nothing, or at least without requiring prior interaction with a nucleus or other particles. What is it that causes the creation of particle-antiparticle pairs in Hawking Radiation?