I had a question about a particle (say a proton) with relativistic energies interacts with a magnetic field (in the z direction). As it is accelerates the particle emits synchrotron radiation. Naturally I assume that this emission of photons reduces the particle's energy. My question is as the energy decreases is there a change of orbit (orbit decay) associated with the emission of synchrotron radiation?
$\begingroup$
$\endgroup$
3
-
2$\begingroup$ possible duplicate of Why doesn't the LHC accelerate electrons? $\endgroup$– John RennieSep 25, 2015 at 10:36
-
$\begingroup$ Hi Pierre. The question I've linked isn't an obvious duplicate, but it does answer your question. $\endgroup$– John RennieSep 25, 2015 at 10:37
-
$\begingroup$ Thank you @JohnRennie, the link is very helpful. I can't comment on DarioP's answer. I'm interested in how to derive the energy loss for any particle on an arbitrary curved path. Any advice? $\endgroup$– QuantumGhostMachineFeb 7, 2016 at 22:41
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
Yes, there is an orbit change which is however rather small for one turn (one revolution) in a electron synchrotron (Protons almost don't radiate at usual energies around 1-10GeV and don't do it at even higher energies , however they will radiate in the Future Circular Collider) . Actually, all e-synchrotrons are equipped with a set of radio-frequency cavities which restore the lost energy. So due this compensation the orbit change is very small and only reflected in a small variation of the beam energy of the order of 1 promille of the total beam energy. Orbits that leave dipoles actually vary in function of the beam energy (dipoles generate "dispersion", i.e. an orbit that varies with energy) , but as the energy deviation is very small it actually only leads to a small increase of the horizontal beam size which is subsequently corrected by other beam optic elements. The exact effect of the emission of synchrotron photons on a particle orbit is a bit more involved because the considered particle usually exerts transverse "betatron" oscillations which get actually damped by a combined effect of synchrotron radiation emission and energy compensation in the radio-frequency cavities.
If the radio-frequency cavities, however, are switched off, the orbit will "slowly" spiral inside, "slowly" means that if you start with a electron beam of 3GeV during 1 turn each electron looses for instance 1MeV around 3000turns are needed for the beam to loose its full energy (this is an approximation), 3000turns correspond about 2.7 milliseconds. However, the spiral motion typically is interrupted as soon as the beam particle hits the vacuum chamber wall of the vacuum system.
answered Apr 4, 2018 at 16:41
-
$\begingroup$ Wonderful answer, what do you mean by FCC and promille. Also do you have a particular text where I can find the derivation of synchrotron and betatron oscillations? $\endgroup$ Apr 5, 2018 at 22:55
-
$\begingroup$ A rather good introduction is written by K.Wille: "The physics of accelerators: An Introduction". The original edition is in German. $\endgroup$ Apr 6, 2018 at 8:46
-
$\begingroup$ I forgot: the FCC is the "Future Circular Collider", the successor project of the LHC at CERN. $\endgroup$ Apr 6, 2018 at 9:42
-