0
$\begingroup$

My textbooks lists the exchange particle for a neutron-neutrino interaction as being the W- boson. Is this the only option, or can it also be a W+ boson? Nothing jumps out at me that would suggest it being impossible. In the scenario that it can only be a W- boson though, why?

And as a followup question, could anyone explain what exactly the role of the W boson is in this interaction? In beta decay, for example, it's fairly intuitive, as it's just the one particle (a neutron) that transforms into other particles, and so it's easy to visualise the W- boson as simply being "emitted" by the neutron. But with a neutron-neutrino interaction, it's hard to conceptualise what exactly is happening. Does the neutron "give" a W boson to the neutrino, or what?

$\endgroup$
2
  • $\begingroup$ What do you mean by n-n interaction? I sounds like the mentioned interaction is just the beta- decay? $\endgroup$
    – john
    Nov 15, 2015 at 17:01
  • $\begingroup$ @john I'm talking about a neutron and a neutrino turning into a proton and an electron. $\endgroup$
    – John Doe
    Nov 15, 2015 at 17:23

2 Answers 2

1
$\begingroup$

Neutrino can interact only by exchange of electroweak boson. So in each reaction with neutrino $W^\pm$ or $Z$ bosons must be involved.

Also, Standard Model neutrino is assumed to be massless, so there is defined handedness: neutrino is left-handed and antineutrino is right-handed. Consequence of it is that left-handed neutrino will interact only with negative weak current $W^-$ and right-handed antineutrino with positive current $W^+$. Or, in other words, neutrino will emit $W^+$ and antineutrino will emit $W^-$.

So in $\nu_l + B \to l^- + B'$ scattering neutrino will always emit $W^+$ leaving negatively charged lepton.

$\endgroup$
2
  • $\begingroup$ This is fine for time-like exchanges, but with space-like exchanges there is no clear understanding of "emit" or "receive". Draw the diagram with the exchange line slanted one way and you get one sign, draw it slanted the other way and you get the other. $\endgroup$ Nov 15, 2015 at 18:44
  • $\begingroup$ Sure, what I wanted to point out that is, if we assume that neutrino emits something, it must be $W^+$ not $W^-$, due to handedness. Op did not specified what he means by asking whether both bosons can participate. They can if you assume both directions, but only one can participate if direction is specified. $\endgroup$
    – Rafal
    Nov 16, 2015 at 7:40
0
$\begingroup$

In the reaction that you ask about the exchange boson is space-like (meaning that for that particle $E^2 - (pc)^2$ takes on a negative value.

In cases like that there is no unique way to decide if you have a $W^-$ going from the nucleon to the lepton or a $W^+$ going from the lepton to the nucleon, and the drawing is usually annotated only with a $W$.

Short answer: yes, both types are allowed, but you can't actually tell what type you had.

$\endgroup$
2
  • $\begingroup$ Okay, thank you. And just to confirm, is it also the case for the electron capture interaction (p+e⁻ -> n+v) that it can be either type? $\endgroup$
    – John Doe
    Nov 15, 2015 at 17:46
  • $\begingroup$ Yes, electron capture is the time-reversal of beta decay, so it has all the same features. $\endgroup$ Nov 15, 2015 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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