How does a photon mediate the electromagnetic force between a proton and an electron?

[Punoqllads]

For an example of what confuses me, consider this Feynman diagram: http://www.britannica.com/nobel/cap/ofeynmd002a4.html. Here, a proton emits a photon, which has some momentum and therefore pushes the proton in the opposite direction. The photon, in turn, reaches the electron where it is absorbed. Again, the electron’s path is adjusted to keep the momentum of the system the same.

But.

Protons and electrons are attracted to each other. The diagram has them moving away from one another. How do protons and electrons absorb and emit photons so that they pull each other together?

[Pleonast]

In simple terms: there are virtual photons with “negative” momentum.

In complex terms: I’m not sure.

[Punoqllads]

Negative momentum? p = E/c. They have negative energy?

[MikeS]

@Punoqllads

Negative momentum? p = E/c. They have negative energy?

Ah, but that’s the beauty of virtual photons: They can have any relation between energy and momentum they damn well please.

More precisely, when we try to calculate the interactions between two particles in Quantum Field Theory, we have to sum the “internal” lines over all momenta and energy independently, without limiting ourselves to those that would be allowed for a “free” particle. (The fancy physics terms for this are that virutal particles are “off mass shell” and real ones are “on mass shell”.) In fact, if you go through the math, you can show that the internal photon has to be off mass shell – it’s impossible for a massive particle to emit a massless particle (one satisfying p[sup]2[/sup]c[sup]2[/sup] = E[sup]2[/sup]) and still conserve energy and momentum.

[Chronos]

The diagram has them moving away from one another.

Actually, it doesn’t. Which way you have the legs of a Feynman diagram pointing does not imply anything about which way the particles are actually going. You could just as well draw that same diagram shaped like <-> instead of >-< , and it’d be just as correct (or even like <-< , or |-< ,or whatever). It’s just usually easier to draw everything all splayed out, so that’s what physicists do (physicists are notoriously lazy).

[Mathochist]

@Chronos

Actually, it doesn’t. Which way you have the legs of a Feynman diagram pointing does not imply anything about which way the particles are actually going. You could just as well draw that same diagram shaped like <-> instead of >-< , and it’d be just as correct (or even like <-< , or |-< ,or whatever). It’s just usually easier to draw everything all splayed out, so that’s what physicists do (physicists are notoriously lazy).

Indeed, though not as lazy as mathematicians…

Anyhow, yes, the diagram is just a graph, not a picture. The topology is all that matters. In fact, there are whole fields (no pun intended) of modern particle physics that are based on “the topology is all that matters”.

[Chronos]

The topology is all that matters.

Not quite… It does make a difference which external legs are pointing up and which are pointing down. A single diagram can represent, for instance, an interaction between two electrons, an interaction between an electron and a positron, or a (kinematically forbidden) decay of an electron into two electrons and a positron, depending on how you arrange the legs.

But other than that, yes, the topology is all that matters.

[Mathochist]

@Chronos

Not quite… It does make a difference which external legs are pointing up and which are pointing down. A single diagram can represent, for instance, an interaction between two electrons, an interaction between an electron and a positron, or a (kinematically forbidden) decay of an electron into two electrons and a positron, depending on how you arrange the legs.

But other than that, yes, the topology is all that matters.

Bah, that’s an orientation of the legs. Draw the arrows, but point the line in a different direction and it’s the same thing.

Okay, okay. Oriented topology.

[Ring]

As a point of interest; on the mass shell means that Einstein’s equation that relates mass, energy and momentum applies.

m[sup]2[/sup] = E[sup]2[/sup] - p[sup]2[/sup]

(c = 1)

Off the mass shell means that it doesn’t.

[Ring]

I guess my above post would have made slightly more sense if I would have quoted
MikeS.

(The fancy physics terms for this are that virutal particles are “off mass shell” and real ones are “on mass shell”.) In fact, if you go through the math, you can show that the internal photon has to be off mass shell – it’s impossible for a massive particle to emit a massless particle (one satisfying p2c2 = E2) and still conserve energy and momentum.