Bremstrahlung radiation - a stupid question

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why is it that the feynman diagram illustrating Bremstrahlung radiation has three vertices; basically an electron decays into a)a photon which recoils the nucleus and b)a virtual electron WHICH THEN decays into another photon and an electron of lower energy.

why can't the original electron just decay into another REAL electron (plus the photon), like in Compton scattering?
 

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Phlogistonian
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It can do that, but that's called Rutherford scattering, not bremsstrahlung.

By definition, bremsstrahlung involves 2 processes: an electron decelerates due to an external field, and it emits real radiation.
 
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nrqed
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why is it that the feynman diagram illustrating Bremstrahlung radiation has three vertices; basically an electron decays into a)a photon which recoils the nucleus and b)a virtual electron WHICH THEN decays into another photon and an electron of lower energy.

why can't the original electron just decay into another REAL electron (plus the photon), like in Compton scattering?

Phlogistonian gave the answer. But let me add that it's confusing when you use the term "decay" here. Decay is used when a particle transforms into other particles. Here the electron does not "decay" since it's still there after. The way people would normally describe what happens is that the electron "emits" or "radiates" a photon. So the process you are describing is that an electron emits a photon absorbs by the nucleus and then a second, real photon. If no real photon is emitted than this is is just Rutherford scattering, not Bremsthrahlung, as Phlogistonian said.
 
  • #4
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thanks Phlogistonian and nrqed...

(and yes ofcourse I was misusing the word 'decay' in this context)
 

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