1
$\begingroup$

I see in page 31 of Martin, Hanzel "Quarks and Leptons", that the fine structure constant is the probability for photon emission by a charged particle. Also I read from Lubos Motl's answer on this similar question that there are more calculations to do to actually find out the probability.

So my question is: who is correct? And how do I calculate the probability for photon emission by a charged particle? Should we have to specify how long or how far do we measure, or do we look for the probability as time go to infinity and all space? If anyone can have the extra time to give me an example, that'd be awesome!

Thank you!

enter image description here

$\endgroup$
3
$\begingroup$

To start with, the answer by Motl covers the subject.

The statements are order of magnitude statements depending on the Feynman diagram expansion of the particular interaction under consideration. The first order diagrams when calculated are proportional to the coupling constants, the internal variations of the variables within the integral making a contribution do not change the order of magnitude.

And how do I calculate the probability for photon emission by a charged particle?

As Motl states, a single stable charged particle cannot emit anything real, because in its center of mass the momentum is zero, and there is not internal energy to be radiated. You need an interaction of the charge particle with a field,

brems

In these two lowest order diagrams the incoming electron interacts with an off mass shell, virtual, photon with a nucleus radiating a real gamma. The field of the nucleus could be any electric field generated from electrons at the surface of matterials, but quantum mechanically that is the only way a charged particle can radiate, transforming kinetic energy to a radiated photon. To calculate the probability you need to write down the integrals represented by these feynman diagrams and integrate them, with input the energy momentum vector, the calculation giving the probability distribution for this energy.

For exact calculations higher order diagrams have to be added , but as the α is very small, and higher orders will have it in higher powers, the contributions will be very small.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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