Where does mass come from in pair production?

Question

In pair production, two gamma rays with > .511Mev can come together to create a positron and an electron.

So two electromagnetic waves E and B fields, with No mass and No gravity and traveling at the speed of light, can create two objects with Mass, with Gravity and with opposite E fields, and with NO B field.

Is there any theories on how E and B fields go about creating mass? It seems the resultant objects both have opposite E fields, but the B field has vanished...where did it go?

In slow motion, as this happens, what exactly goes on to create mass from EM fields?

Can we use EM fields to manipulate this process further? i.e. change the rate of pair production? Or to manipulate the G field of the resultant objects?

Answer 1

"Rest mass" is probably a bit more abstract in modern science than what you seem to be thinking. It is simply this: if you can put yourself into an inertial frame such that a body is at rest relative to you, then that body's rest mass is defined as that body's total energy measured in this particular inertial frame - multiplied by $c^2$, if you want to express the energy in SI units, otherwise, we simply think of rest mass as a special kind of a body's total energy.

So, its that part of the body's energy that is not owing to the motion of the body's center of mass.

A photon is always observed to be travelling at $c$. There is no inertial frame wherein it is at rest, so it has, by definition, no rest mass.

Two photons, with opposite momentums, coming towards one another before the pair production experiment have a fixed center of mass - the system's total momentum is nought. By definition the pair has rest mass $E$ (or $E/c^2$ in SI units), where $E$ is the total energy, even before the collision, even though, one at a time, they both have a rest mass of nought. Beware of rest mass - it is not additive and conserved in the way that older physics expositions imply.

After the collision, we have two bodies with weird dispersion relationships $\omega = c\,\sqrt{k^2+ \frac{m^2\,c^2}{\hbar^2}}$ which means that they can be observed travelling at low speed. A wave packet with such a dispersion relationship can stay put for a bit, so we can define a rest mass for both the objects, whereas for photons $\omega = c\,k$ so that the group velocity is precisly $c$ at all times. The combined rest mass of positron and electron is still likely to be a little below the total system energy, though, as the particles generally recoil with nonzero velocities.

The body's rest mass is also the inertial mass measured by imparting a known impulse and dividing it by the change in velocity that results. But this measurement must be done with a small impulse with the body at rest in the measurement frame before the impulse to be sound. This is the kind of consideration that can come out of imagining light energy confined to a mirrored box, as in John Duffield's answer.

Answer 2

What is mass of elementary particles? It is the "length" of the four vector (p_x,p_y,-_z,E), for complex systems it is called rest mass. As the length of three dimensional vectors is not additive ( think adding two opposite momenta), rest masses are not additive, vector algebra has to be used.

The invariant mass of your two gammas must be larger than the added mass of the electron and positron because of energy conservation . The four vectors of the electron and the positron added vectorially by energy and momentum conservation, if the energy of the photons is just at threshold, twice the masses, will gave a four vector (0,0,0,2*.511) (convention c=1). Over the threshold the invariant mass of the system will be larger.

Photons interact with the gravitational field according to general relativity.

It is not the E and B that are creating the mass. Once enough energy is there particle antiparticle pairs can be created with some probability, as long as quantum numbers are conserved. The probability depends on the coupling constants and the higher order diagrams contributing to the creation (i.e. feynman diagram calculations)

Answer 3

Where does mass come from in pair production?

From the energy-momentum of the gamma photons. Think of photon momentum as resistance to change-in-motion for a wave propagating linearly at c. You've maybe heard of electron spin and the Einstein-de Haas effect which "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". And of the wave nature of matter, and of atomic orbitals wherein electrons "exist as standing waves". So think of electron mass as resistance to change-in-motion for a wave going round and round at c. Don't forget that when you trap a massless photon in a gedanken mirror-box, you increase the mass of that system. The box is harder to move when it's got the massless photon inside it. Then when you open the box, it's a radiating body that loses mass, just as like Einstein said in his E=mc² paper. The electron is like a photon in a box, but without the box. See Light is Heavy by van der Mark and 't Hooft (not the Nobel prizewinner).

In pair production, two gamma rays with > .511Mev can come together to create a positron and an electron. So two electromagnetic waves E and B fields, with No mass and No gravity and traveling at the speed of light, can create two objects with Mass, with Gravity and with opposite E fields, and with NO B field.

That's wrong on several counts I'm afraid. I'll try to explain: 1) The photon is an electromagnetic wave. The field concerned is the electromagnetic field. I know there's plenty of depictions of an E-field wave and an orthogonal B-field wave, but there aren't really two orthogonal waves. E represents is the spatial derivative of the electromagnetic four-potential, B the time-derivative. It's like you're in a canoe going over an ocean wave. The slope of your canoe represents E and the rate of change of slope represents B, and you're going over one wave, not two. 2) The photon has a non-zero "inertial mass" and a non-zero "active gravitational mass". A 511keV photon causes the same amount of gravity as an electron or a positron. 3) The electron and the positron each have an electromagnetic field. Not an E field and a B field.

Is there any theories on how E and B fields go about creating mass?

Special relativity. The mass of a body is a measure of its energy content. Again see Einstein's E=mc² paper.

the B field has vanished...where did it go?

It hasn't vanished. The electron has an electromagnetic field.

In slow motion, as this happens, what exactly goes on to create mass from EM fields?

Some people will tell you that one of the photons fluctuates into a fermion–antifermion pair, and the other photon couples to one of them. That's a cargo-cult tautology I'm afraid. Pair production does not occur because pair production occurred, spontaneously, like worms from mud. IMHO a better description is that each photon interacts with and displaces the other into a curved path, such that each encounters itself whereafter it continually displaces its own path into a closed path.

Can we use EM fields to manipulate this process further? i.e. change the rate of pair production?

Yes. See the Breit-Wheeler process and this recent paper.

Or to manipulate the G field of the resultant objects?

No. The electron's electromagnetic field is not something that is distinct from its gravitational field. It doesn't have two fields. As to why, that's one for another day.