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W+ or W- in neutrino collisions

by mrcotton
Tags: collisions, neutrino
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mrcotton
#1
Apr18-14, 08:32 AM
P: 118
All the Feynman diagrams I have seen so far for a neutron colliding with a neutrino have a w+ with an arrow from the neutrino to the neutron.
Would it not also be possible with a W- leaving the neutron taking away negative charge for it to become a positive proton or is there some quantum rule I am not aware of that forbids this?

Any help gratefully received.
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jtbell
#2
Apr18-14, 08:37 AM
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The W in a Feynman diagram like that one is "virtual" and can be interpreted either as a W+ going one way, or a W- going the other way. There's no difference as far as I know.
mrcotton
#3
Apr18-14, 09:02 AM
P: 118
Hi jtbell,
thanks for the rapid response,

What confuses me is that for example in say beta minus decay. The proton turns to a neutron by emitting a W- and this W- decays to become an electron and a anti-electon neutrino. The electron in a sense is formed from the "negativness" of the W-. So in this type of decay it must be a W-.
Also in electron capture it seems that the proton captures the electron by emitting a W+ to turn the electron into a neutrino, with this process happening to protons in a neucleus. Yet if a free electron and a proton collide via the weak interaction then a W- leaves the electron.

So do you mean there is no difference in the initial and final states or do you mean its the same W we just assigne a sign to it.

I hope this garbled rant makes sense
confused of planet Earth

mfb
#4
Apr18-14, 09:26 AM
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W+ or W- in neutrino collisions

What confuses me is that for example in say beta minus decay. The proton turns to a neutron by emitting a W- and this W- decays to become an electron and a anti-electon neutrino. The electron in a sense is formed from the "negativness" of the W-. So in this type of decay it must be a W-.
You can write the same process as W+, electron and antineutrino appearing out of nowhere, and then an interaction where the neutron "absorbs" the W+ and becomes a proton.
That is an unconventional way to draw the Feynman diagram, but it is the same physics and does not change the calculation at all.
So do you mean there is no difference in the initial and final states or do you mean its the same W we just assigne a sign to it.
It is the same process.


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