1
$\begingroup$

The term 'bremsstrahlung' is usually reserved for the emission of electromagnetic radiation caused by $(1)$ acceleration or deceleration of the charged particle by an electric field and $(2)$ in presence of matter/medium. See section $8.8$ of the book Quantum Field Theory by Mandl and Shaw, and section 11.5.3 of Perkins's Introduction to High Energy Physics.

On the contrary, in a cyclotron-like device, during most of its path, a charged particle is subjected to a centripetal acceleration $(1)$ caused by a magnetic field and takes place $(2)$ in a vacuum or near-vacuum situation. In this case, the total radiated power is also given by the classical Larmor formula. See this post and the answer by G. Smith.

Therefore, cyclotron emission does not qualify to be a bremsstrahlung process if I adhere to the strict definition above. However, if I'm not mistaken, the answer here by @JohnRennie, suggests to me that cyclotron radiation is also bremsstrahlung. Let me quote that for quick reference:

"Whenever you accelerate a charged particle it emits EM radiation known as Bremsstrahlung, and obviously charged particles moving in a circle are accelerating (towards the centre). This means that any circular collider emits a continual stream of Bremsstrahlung radiation."

Question Does John Rennie's answer (linked above) uses a broader definition of bremsstrahlung to include radiation by charged particles accelerated by magnetic fields in a vacuum and calculable by classical Larmor formula?

$\endgroup$
2
  • 1
    $\begingroup$ The understanding of bremsstrahlung that you call "strict definition" is not relativistically invariant. That alone should be enough to disqualify it. $\endgroup$ Sep 26 '20 at 11:56
  • $\begingroup$ @EmilioPisanty Fair enough. Thanks! $\endgroup$
    – SRS
    Sep 26 '20 at 12:34
2
$\begingroup$

I think your emphasis on electric only fields is wrong, both semantically, as bremsstrahlun comes from the german root "to brake" and has nothing specific to electric fields, and because Maxwell's equations are symmetric in electric and magnetic fields.

As for the objection on "acceleration". when one defines acceleration from the momentum vector, as a vector, dp/dt exists for a charged particle moving in a circle. It would radiate even if it were possible to turn it with a string over your head.There are studies .

So John's statement is OK , in my opinion.

$\endgroup$
2
  • $\begingroup$ If you look at the expression of the crosssection of Bremsstrahlung in Eq.(8.103) of Mandl and Shaw (linked in the question), it is proportional to the crosssection in Eq. (8.91) which is derived by assuming Eqs. (8.92a) and (8.92b), which in turn uses scattering from the Coulomb field of a nucleus. @annav $\endgroup$
    – SRS
    Sep 26 '20 at 10:26
  • $\begingroup$ I think this does not exclude other ways of getting acceleration, there is symmetry in electric and magnetic fields in QED, classical and quantum.. Bremms were first observed in interactions with electric fields , but this is not exclusive. If you turn a charged sphere over your head you would have to give the force turning it and supplying the energy of radiation. $\endgroup$
    – anna v
    Sep 26 '20 at 10:43

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.