1
$\begingroup$

I am starting to learn about Feynman diagrams and I wonder if this reaction could happen as a physical process on its own, as opposed to part of a more complicated diagram:
$e^- \to W^- + \nu_e$

enter image description here
An electron emits a $W^-$ boson and an electron neutrino.

Following the logic used by David Z here:
First-order EM Feynman diagram?
The question asked if a first order electromagnetic diagram could happen on its own. For example, an electron coming in and emitting a photon
$e^- \to \gamma + e^-$.
The answer given by David Z was that this was forbidden because in the rest frame of the initial state ($e^-$ on its own), the energy was $E=m_ec^2$, while the energy in the final state is at least $E=m_ec^2 + E_\gamma$. I am wondering why we need to look at the rest frame of the electron in the first place.
My logic tells me that if an electron highly energetic comes in, it should be able to emit a photon and end up with less kinetic energy. This is wrong but could someone tell me why?
Following David's logic, the reaction above could not happen on its own because the rest mass of the electron is obviously much smaller than the reduced mass of the $W^- + \nu_e$.
Could someone clarify my ideas?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ If you have "an electron highly energetic comes in" and it "emits" a photon, that photon is actually some photon that's reflected off the electron. I.e. you put an electron and a photon in, and you get an electron and a photon out (in different state). This even works when the electron is extremely fast (as in the X-FEL), and the photon comes from a static magnetic field (zero energy photon): The result is a slightly slower electron, and a very energetic X-ray photon due to our frame of reference. You absolutely need the magnets to get the X-rays, the electrons can't do it on their own. $\endgroup$ – cmaster - reinstate monica Dec 22 '18 at 0:23
  • $\begingroup$ @cmaster That sounds like it should be an answer, not a comment. $\endgroup$ – David Z♦ Dec 22 '18 at 2:23
  • $\begingroup$ @DavidZ It would have been if it were not for the fact that I'm not much of a particle physicist myself. I just don't feel confident that this argument holds up to any theoretic rigor. However, thinking about it like this seems to help me understand what's going on on a mental-model-level, so I shared it. $\endgroup$ – cmaster - reinstate monica Dec 22 '18 at 8:10
  • $\begingroup$ @cmaster None of that really matters, though. Comments are meant for suggesting improvements or requesting clarifications. Other things should generally be answers; they don't have to be rigorous (or even correct), they don't have to have references, you don't have to have credentials or experience or anything like that. Here's a rule of thumb I find useful: are you trying to get the asker to edit their post in response to your comment? If not, then it probably shouldn't be a comment. $\endgroup$ – David Z♦ Dec 22 '18 at 8:55
6
$\begingroup$

My logic tells me that if an electron highly energetic comes in, it should be able to emit a photon and end up with less kinetic energy. This is wrong but could someone tell me why?

Good question. Basically, the thing is that energy conservation and momentum conservation have to apply in every reference frame in order for the interaction to be valid. That's a consequence of relativity, that the laws of physics need to work just as well in every frame. So in this case, even though you're only considering things from a frame (presumably your own rest frame) in which the electron is moving quickly, whatever happens also needs to be "legal" from a reference frame in which the electron is at rest and you're the one moving. An observer in that reference frame would notice a huge violation of energy conservation if the electron turned into a W boson and a neutrino, despite the fact that you might not find that too weird.

I also expect that if you tried to actually do the calculation to figure out the final energies and momenta of the W boson and neutrino, even in the reference frame where the electron is moving quickly, you would find that the math doesn't work out - it'd be impossible to conserve both energy and momentum given the masses of the particles involved. But it's not really clear that you can't do that unless you actually go through the calculation. The benefit of considering the process from the electron's rest frame is that, in that rest frame, it is obvious that the math can't work, so you get to bypass actually doing it.

|cite|improve this answer
$\endgroup$
3
$\begingroup$

One need not go to the center of mass of the electron to see that the reaction is against energy and momentum conservation.

Write the four vector of the incoming electron, no mater how high its energy. It should be equal to the four vector of the sum of the two outgoing particles since energyand momentum are conserved quantities.

The invariant masses of the incoming four vector and the outgoing four vector are different: the outgoing invariant mass is at least the sum of the two masses, electron+ W, the incoming has invariant mass the mass of the electron.

This means that either energy or momentum is not conserved, because the two four vectors have to be identical in this diagram.

My logic tells me that if an electron highly energetic comes in, it should be able to emit a photon and end up with less kinetic energy.

You are describing a decay, and here the center of mass of the electron is clearer, since it is at rest and all the energy is in the invariant mass of the electron. As photons always have velocity c (cannot be at rest) the invariant mass of the outgoing system will be larger than the invariant mass of the electron, i.e. energy or momentum are not conserved as the incoming and outgoing four vectors are different.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.