# About potential energy and matter-antimatter creation

1. ### zoteman94

10
Hi, I'm not sure if this is the correct section to post this. As far as I know, it is widely accepted that antimatter has "normal" gravity, and that for creating a matter-antimatter pair, you need to input the rest mass of the particle in energy (Einstein's equation).
But as potential gravitational energy is, as far as I know dependent on other masses, if I create a pair on an accelerator, wouldn't the input of energy include the potential energy that the particle will have because of all the mass in the universe?
That would mean a lot more energy is needed to conserve the energy of the Universe.
What I'm missing here? Thanks.

2. ### BruceW

3,600
I'm not totally sure what you mean. But I think you are asking about where does the gravitational potential energy come from when we create a matter/antimatter pair. The answer is that before we made the matter/antimatter pair, we must have had two photons collide together. And although a single photon doesn't have mass, two photons going in different directions do have mass. So no mass is created. It is a little bit more complicated, because there are two versions of mass: 'relativistic mass' and 'invariant mass'. But this is the rough idea of what is going on.

3. ### Simon Bridge

15,299
Probably you missed GR.
Gravity comes from energy-density, not restricted to just mass-energy.

You do need more than just the rest-mass energy of the two particles to make the pair, since, as you point out, you still have to conserve everything else as well. So if you think that picture incomplete, you are right. i.e. Pair production must take place near another particle, just to take care of the momentum. Additional energy is needed to separate them too. You can search for pair production experiments to see what actually goes into them.

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4. ### BruceW

3,600
I'm not sure what you mean. I thought pair production can happen with just two photons, isolated from anything else. the 3-momentum is zero before and after in the centre-of-momentum frame.

edit: haha, it seems that I'm not sure what anybody means. (in my 2nd post as well, I repeat the phrase).

### Staff: Mentor

There are two interestingly different cases: the single-photon cases in which high-energy gamma radiation hits a target and produces particle/anti-particle pairs; and the two-photon case that you're thinking of. It's a lot easier to produce high-energy gamma rays than it is to collide photons, so we tend to pay more attention to the first case.

(But google for "photon-photon pair production" - you'll get some interesting stuff).

6. ### BruceW

3,600
I see, the single-photon case (with adjacent matter) is more common here on earth. that does makes sense.

7. ### zoteman94

10
Ok, so the system of anti parallel photons does have mass, most now makes sense to me (I hope I'm right). By the way, which equations establish this geometry?

### Staff: Mentor

The photons don't have to be moving anti-parallel, that is, in exactly opposite directions. They just have to be moving in different directions. It's merely easier to calculate the result if they're moving anti-parallel.

9. ### zoteman94

10
Thanks for the answers. I would still like to know which equations predict these results though.

### Staff: Mentor

The mass of a single particle is
$$mc^2 = \sqrt{E^2 - (pc)^2}$$

Similarly, the mass of a system of particles is given by
$$mc^2 = \sqrt{E_{total}^2 - (|\vec p_{total}|c)^2}$$
where ##E_{total}## is the total energy, ##\vec p_{total}## is the total vector momentum, and ##|\vec p_{total}|## is its magnitude. That is, add up the momenta of the particles as vectors (this takes into account their different directions), then find the magnitude of the resulting total vector.

For a single photon, m = 0 so E = pc. (here p is the magnitude of the photon's vector momentum)

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