2
$\begingroup$

If forces are mediated by the exchange of bosons, and bosons travel out from their "parent" fermion, wouldn't they be carrying momentum which pushes anything that absorbs them away from the "parent" fermion? Shouldn't two particles always diverge because they would recoil from shooting bosons at each other? Why is it that, for example, when two nucleons shoot mesons at each other, they come closer instead of repelling?

$\endgroup$
3
3
$\begingroup$

Photons are bosons and they certainly make like charges repel, and unlike ones attract. Spinless particles only mediate attractive forces: their charge conjugation couplings don't accommodate two opposite charges.

You should not take the pictorial metaphor of "particle exchange" too literally. If you are in a classical boat and shoot a ball at a nearby canoe, your boat will move away from that, and when the ball is caught in the canoe, it will move away from the origin of the ball too.

The quantum situation only comes that close to this geometric metaphor/fantasy when the force is repulsive. In an attractive interaction, however, the virtual particle "emitted" by one source particle (boat) may actually impart momentum to it making it "recoil" towards the target particle (canoe), and likewise suck momentum from the target particle when it hits it!

The rules of covariant perturbation theory dictate that momentum and energy are conserved at each vertex, and all matches up in the end; all such diagrams are in momentum space, not coordinate space, of course.

So, the boats-on-a-lake metaphor is more trouble than it's worth, if you took it too seriously, unless you can imagine throwing the ball away from the target, and the ball likewise hitting it from behind, in the line connecting you to it. People may talk about delocalization in quantum mechanics and such, but ultimately, you just follow the math in your calculation, and if such pictures confused you and hampered your calculation instead of helping it, you need not really stick to them.

At the end of the day, you have computed an amplitude with two incoming and two outgoing momenta. There is a semiclassical connection of a Born amplitude to a classical potential, detailed in several answers on this site, translating the charge conjugation features of the amplitude to the sign of the corresponding classical potential/force.

$\endgroup$
11
  • $\begingroup$ Very arguably linked, since you are asking about mesons without that freedom. $\endgroup$ – Cosmas Zachos Mar 14 at 22:36
  • $\begingroup$ Could you explain a little what charge conjugation coupling is? $\endgroup$ – eaeaa1232 Mar 15 at 2:08
  • $\begingroup$ Basically there is no i which can flip charges upon complex conjugation. Contrast photon to scalar coupling to fermions and antifermions. $\endgroup$ – Cosmas Zachos Mar 15 at 2:35
  • 1
    $\begingroup$ Yes, that's my very point! Your semi-classical "direction of travel" assignment is a bit too abstract for visualization: Feynman diagrams are not spacetime diagrams! They are lines in momentum space conducting momentum from one vertex to the next, and can go either way. $\endgroup$ – Cosmas Zachos Mar 15 at 13:01
  • 1
    $\begingroup$ Of course. What makes you suspect they couldn't? They, in fact do, as, often you integrate over all intermediate momenta... $\endgroup$ – Cosmas Zachos Mar 15 at 17:56
0
$\begingroup$

I think the reason goes something like this from what I was told many years ago: The sum of the probabilities of the Feynman paths for virtual bosons that push one particle further away from the other is less than the sum that would bring them together Also virtual bosons osons can travel around the back of a particle and then come back and strike it and so there are more attractive paths than you would think.

$\endgroup$

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.