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So I know that when you collide particles with high enough kinetic energy, (kinetic energy = at least the rest mass of the particles you are making), you get particles.

How come potential energy cannot make particles? Say you have an electron held at a potential such that the amount of potential energy it has equals at least the rest mass of a particle you want to make. How come potential energy does not convert into rest mass energy directly? Would this require whatever is creating the potential to disappear because the potential energy turns into a particle? This shouldn't violate any conservation of energy laws does it?

Another question that kind of relates to my previous question is, say we have a 512 keV photon. What prevents this photon from turning into an electron that then has kinetic energy of 1 kev?

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3 Answers

up vote 6 down vote accepted

A strong enough static electric field can create real particle-antiparticle pairs out of the vacuum, so yes, potential energy can contribute to creating particles. This is called the Schwinger effect and was predicted by Schwinger in the 1950s. The original paper is here http://prola.aps.org/abstract/PR/v82/i5/p664_1.

Schwinger's formula gives $$\frac{N}{VT}=\frac{e^2E_0^2}{4\pi^3}\sum_{n=1}^{\infty}\frac{1}{n^2}\mbox{exp}(-\frac{nm^2\pi}{eE_0})$$ where $N$ e+e- pairs are produced in a volume $V$ and time $T$ by a static uniform electric field $E_0$. This means that the threshold to see any pair production is about $m^2/e$ or approximately $1.3$ x $10^8$ V/m.

As for your last question, a photon cannot turn into an electron because of charge conservation (and several other conserved numbers). It cannot turn into an e+e- pair in vacuum either because energy and momentum cannot be conserved simultaneously - this is a consequence of the masslessness of the photon. A photon can however decay into e+e- in the presence of some other matter that it scatters off, which allows it to conserve both energy and momentum.

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Analogous to pair creation by electric field in vacuum, there is an atom ionization with a strong electric field. The threshold of ion-electron creation is much smaller here. –  Vladimir Kalitvianski Mar 16 '11 at 21:46
+1 for "On Gauge Invariance and Vacuum Polarization", it's a beautiful paper. –  Simon Mar 17 '11 at 0:27
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Not that the other answers are wrong or anything, but here's something a different way to think about it:

When a particle is created, the process that creates the particle is (1) local, meaning that the particle is created at the same location that the energy used to create it disappears from, and (2) subject to conservation laws for various quantites (generically called "charges") like charge, total energy, etc. But potential energy is often a function of position and one of these conserved quantities. So, since particle-creating processes don't change either the location or the amount of conserved charge, the end products (including the newly created particle) will have the same amount of potential energy as the initial "reactants." Therefore, there isn't any extra potential energy you can take to use for particle creation.

Of course, it's certainly possible to come up with exceptions to this, as included in some of the other answers. It depends to some extent on what you consider a particle. (I think this argument works best in a pure QFT viewpoint where you consider the force carriers as proper particles, but then again in that case there is no such thing as potential energy anyway.)

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In principle you can create particles with potential, the prototypical case is color confinement. But I believe you'll need field densities that put the necessary energy into a very small space, which is why it works with the strong force, but we don't see it done with others.

A (on-shell) photon can't simply turn into an (on-shell) electron because such a process would not be able to conserve both energy and momentum.

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What do you mean by "on-shell"? Also can you elaborate a little more about why a photon would not be able to conserve both energy and momentum? –  QEntanglement Mar 16 '11 at 20:22
@QEntanglement: "On-shell" means a real (i.e. non-virtual) particle that can exist outside of a quantum fluctuation. This is equivalent to saying it has the "right" mass, where mass is properly understood to be the square of the energy--momentum Lorentz 4-vector, which should point you at why $\gamma \to e$ must violate one or the other. –  dmckee Mar 16 '11 at 20:29
Plus, there's charge conservation to worry about –  David Z Mar 16 '11 at 20:36
@David: ...and lepton number and spin. Details, details. –  dmckee Mar 16 '11 at 20:45
Thanks! I forgot about charge conservation and lepton number and spin. Okay so taking these details into account. I would like to rephrase my question. What determines the probability that a photon with sufficient energy will not turn into an electron positron pair? –  QEntanglement Mar 16 '11 at 20:50
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