Which W Boson Mediates e^- + ν_μ → μ^- + ν_e?

  • Thread starter Thread starter QuantumCosmo
  • Start date Start date
QuantumCosmo
Messages
29
Reaction score
0
Hi,
lets say we have the scattering process e^- + \nu_{\mu} \rightarrow \mu^- + \nu_e via the weak interaction. This would then be mediated by a W-Boson, right? Now my question ist: Which one?

The Feynman-Diagramm should then look like the one on the right:

http://accessscience.com/loadBinary.aspx?aID=9704&filename=450700FG0010.gif"

(in the line on top, one has to exchange e \rightarrow \mu

I know that this diagramm actually represents two processes (which could be drawn as slightly "skewed" diagramms):
1) The electron emits a W^- and thereby "changes" into an \nu_e. This W^- is then absorbed by the \nu_{\mu} which therefore "changes into a \mu^-
2) The \nu_{\mu} emits a W^+ and changes into a \mu^-. The W^- is then absorbed by the e^- and changes into a \nu_e

So if I draw the Feynman-diagramm as in the link, is it ok to write W^+? Because wouldn't that mean that the diagramm only represents option 2)? Or am I wrong to assume that W^+ and W^- both have to do with the process and only one of the processes 1) and 2) is the "right" one?

I have a big exam in particle physics soon. If the professor asks me to draw the Feynman-diagramm for this process I simply wouldn't know what to write ( W^+, W^- or maybe forget about the charge (just write W) and hope the professor doesn't ask which of the W's it is?).

I know that my question is not very precise, so here is the short version:

What belongs in the diagramm?
a) W+
b) W-
c) simply W, because W+ and W- both have to do with the process
d) other

Thank you very much,
QuantumCosmo
 
Last edited by a moderator:
Physics news on Phys.org
The W^+ and W^- are related by a particle/anti-particle relationship, so for a virtual W, wether its charge positive or negative depends on how you implement conservation of momentum. You've perhaps heard the Feynman idea that an anti-particle is equivalent to a particle moving backwards in time? Have you studied charge conjugation symmetry?

What this means is that wether its a particle or anti-particle depends on which way you consider its momentum to be flowing. In the diagram you've drawn, the momentum of the internal W must be flowing towards the bottom node in the diagram for charge conservation to work.

If you want to draw a W^-, you also will reverse the direction of the four momentum. The two diagrams are completely equivalent, so ultimately it doesn't matter.
 
As a side note that may also be helpful, the same rules apply to internal fermion lines, but you just have to make sure that whatever you decide for a convention you keep it. I think the general rule is for internal fermions you draw the momentum lines in the same direction as the particle arrow on the propagator. This way all of your propagators are for the particle, not the anti-particle.
 
What belongs in the diagramm?

Label the line W±.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Back
Top