# Feynman Diagrams and free electrons

Richard Feynman in his Physics lectures and diagrams show a free electron emitting a photo and/ or absorbing a photon. He (They?) always show the electron being deflected. My question is if the photon has no mass how can or why does the electron react by being deflected in motion? If it is a way to show the change in the energy state of the electron which would change its velocity or mass wouldn't it be an instantaneous step change rather than a change it the angle of motion?

## Answers and Replies

chroot
Staff Emeritus
Gold Member
Photons carry momentum, but have no mass. Remember that the definition of momentum is somewhat different relativistically than in the Newtonian limit.

- Warren

Thanks, chroot. I'm not sure that I really understand the difference.
I understand that total momentum is conserved so if the photon carries momentum from or to the electron it changes the momentum of the electron thus its motion, direction and/or velocity?
I,m not yet fully grounded relativistcally yet. I feel like I have a split personally, one Newtonian and the other Einstienian and now I'm trying to learn about QED. I may shatter completely in a Quantum Relativistic way and enter a parallel universe.
Royce

damgo
Yup -- the final momentum of the electron + photon momentum = initial momentum of electron.

One thing to remember is that those Feynman diagrams lines don't indicate a particular direction. They just tell you whether you're dealing with a particle or anti-particle, and (very roughly) the time-ordering of events. You can draw the lines straight, bent, whatever way you want. Some complicated QCD/QED diagrams are intentionally drawn oddly -- there's one called the "penguin diagram" because the original paper deliberately drew it to look like a penguin.

Thanks again. I didn't know for sure if the lines really indicated direction in space or time or both. In his book "QED" Feynman shows an electron "deflection" everytime a photon was emitted or absorbed.
Since he was so consistant in doing so I wondered if it was really significate and if so why if a photon has no mass. Thanks for clearing it up for me. I which I could say that now I understand but from what Feynman says nobody really does. We just have to except what is.

drag
Greetings !

Royce, I think it's worth noting that
Feynmann diagrams talk about "virtual
photons" - the virtual particles
"transferring" electromagnetic interactions.
They do not really exist.

You see, in QM there is no such thing as
a "continuing" wave or interaction. All
phenomena are quantified - seperated into
pieces - separate wave-particles. So the
virtual photons in these diagrams are actually
quantified interactions of the particles.

Live long and prosper.

Staff Emeritus
Gold Member
Dearly Missed
I think they show the lines diverging just to emphasize that the particles have interacted and they now have (at least) new momenta.

You know the diagrams (the real ones, that is, not the ones in popularizations) are coded methods for determining the integrands of the integrals used in calculating such things as momenta, cross section, and such. Each kind of line and vertex corresponds to a particular expression, and they are all multiplied together under the integral sign. For this purpode it's only the connectivity of the diagram (bearing in mind there are different sorts of lines) that matters, not the slant of the in and out lines.

Hi,

as Feynman describes in his very good divulgatory book "QED", his diagrams have the horizontal axis indicating space, and the vertical that is time. Choosing the proper scales, a 45° line indicates a particle travelling at the speed of light in timespace. A line connecting two points indicates a contribution to the probability ampliture that the particle in the starting point in time-space can be found in the arrival point.
A "vertex" point, connecting two electron lines with a photon line (the "coupling" diagram) can be drawn in many possible different ways, each one indicating the probability amplitude to that initial-to-final state scattering. So, the electron lines are not shown either vertically, or horizontally, simply because vertical means being "steady" in one place, while horizontal means distributed over space.
Interestingly, a hydrogen atom is represented by Feynman as a straight vertical line (the proton) linked by many randomly drawn photons to an electron oscillating around another parallel vertical line. One more nice note - an electron line with the initial point higher in the diagram than the final one is not going "back in time" but it actually represents a positron going forward.
Hope it clarifies something...
bye

umm... where could I find the written versions of his lecture?

chroot
Staff Emeritus