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Exchange particles in two reactions

  • Thread starter astenroo
  • Start date
  • #1
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Homework Statement



Hi all!

I was wondering in two Feynman diagrams (1 and 2)
1) in an electron positron annihilation two photons are created. Now, is the virtual electron the exchanged particle? I'm having problems understanding the diagrams.

2) A proton decays to a neutron as a mu-neutrino collide and a muon (u-) is emitted. As far as I know the exchange particle should be a Boson, but still I'm clueless as it comes to which. I'm ruling out the Z boson, and guessing it should be the W-.

Homework Equations



1. e+ + e- -> y (photons) exchange particle?


2. v(u) + n -> p + u-
Here a down quark decays to an up quark and the exchange particle should be the W- boson. Am I on the right track here at all?

The Attempt at a Solution



Some help would be much appreciated.

-Alex
 

Answers and Replies

  • #2
tiny-tim
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Hi Alex! :smile:
1) in an electron positron annihilation two photons are created. Now, is the virtual electron the exchanged particle? I'm having problems understanding the diagrams.
the lowest-order feynman diagram (there are infinitely many higher order ones, of course) is the "H" diagram, with an electron and positron at the bottom, a line across the middle, and two photons at the top

yes, that line is a virtual electron (or positron, same thing) … that gives each vertex an electron in, an electron out (or positron in), and a photon :wink:
2) A proton decays to a neutron as a mu-neutrino collide and a muon (u-) is emitted. As far as I know the exchange particle should be a Boson, but still I'm clueless as it comes to which. I'm ruling out the Z boson, and guessing it should be the W-.
let's see …

yes, µ doesn't feel the strong interaction, so it has to be a Z or a W

and there's a change in charge, so it can't be a Z or W0 :smile:
 
  • #3
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Then to "conserve" charge it should be the W- boson :)
 
  • #4
tiny-tim
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yup! :biggrin:
 
  • #5
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yup! :biggrin:
Thank you for the help :biggrin:
 

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