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I tried to find the reason by couldn't get the right answer. I wished to know that if both proton and electron have same acceleration, do they radiate same amount/amplitude of EM radiation. Also, since quarks have electric charge, do they radiate photons along with gluons (which I have not heard much about)?

kbg
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    The answer will be different when you say "subjected to the same electric or magnetic field", since in that case obviously the acceleration of the lighter particle will be (much) greater. That may be why one usually says "electrons emit more radiation" - because the comparison would be "in the same field". – Floris Sep 25 '17 at 21:36

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Proton and electron have exactly the same charge so they would radiate the same amount of EM radiation as per Larmor Formula. Relativistic generalization of it could be calculated through Liénard–Wiechert potential, but it is still going to be independent of the mass.

Quarks are charged particles with a charges equal to the fraction of the electron charge ($\frac{1}{3}$, $\frac{2}{3}$, $\cdots$), and thus are obliged to interact with Electromagnetic field too.

This answer might be useful too.

Darkseid
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  • Do the radiation depend upon the mass of the charged particles? I remember my Prof telling that proton emits less radiation than electron (Thus it is better to use protons for collision in LHC). As for as quarks(which have less mass and charge than electron), I guess its gluon radiation dominates over EM radiation (as we get jets which are then used to reconstruct quarks emitting them, I haven't yet come across case where we consider even photons for thier reconstruction. However my knowledge could be incomplete! – kbg Sep 25 '17 at 19:35
  • As in regular electrodynamics, the mass of the charged object doesn't affect electromagnetic field. Only charge can contribute to this. As for collider - I believe that they measure/track charged particles. See CMS or ATLAS wikipedia page for example. – Darkseid Sep 25 '17 at 19:41
  • Correction and update: they also use Hardon Calorimeters for tracking/measuring hadrons coming out of the collision. – Darkseid Sep 25 '17 at 19:45
  • Could be in regular electrodynamics. I agree with you that we use tracks to get the charged particles but those apply to the observable free particles (like electron, muons, etc) but the quarks aren't free particles and hence we don't see them directly.(they undergo hardronisation to give a stream of hadrons which then get detected) – kbg Sep 25 '17 at 19:46
  • Many hadrons are charged too. – Darkseid Sep 25 '17 at 19:47
  • Yup they are (like pions). – kbg Sep 25 '17 at 19:48
  • Let us continue this discussion in chat. – Darkseid Sep 25 '17 at 19:48
  • @Fedor My 5 Cent: Accelerating a subatomic particle by photons (for example with a laser) one need more power for protons to accelerate them in the same manner as electrons. The gained energy will be returned in the form of photons during deflection. So the radiation from protons has to be in relation to electrons like their masses? So I’m not sure your answer is right, isn’t it? – HolgerFiedler Sep 25 '17 at 20:18
  • The original question assumes the same acceleration of particles and the radiation that they would produce, not the superposition of EM fields from the particles and the source of their acceleration. As it stands for the radiation itself, it must be independent from the mass according to electrodynamics. – Darkseid Sep 25 '17 at 20:27
  • Oh, I do see what you mean now. I was wrong indeed. Sorry! – Darkseid Sep 25 '17 at 20:33
  • Flipped my opinion again after some further thoughts on this. – Darkseid Sep 25 '17 at 20:50
  • There would be a small difference because the electron and proton have different magnetic dipole moments, which woudl also be a changing field in the source, but I doubt it would matter much, considering how much smaller the magnetic fields are. I guess you might have to factor in the proton having an electric dipole moment from its substructure, too. – Zo the Relativist Sep 25 '17 at 21:31
  • Sorry, but the mass plays a role in radiation both in linear and circular accelerations. The smaller the mass the larger the power loss to radiation from the power provided. – anna v Sep 26 '17 at 07:12
  • at relativistic energies where both particles are fractions of the velocity of light the difference made by the acceleration is minimal, rather the lighter the particle the more energy is needed to increase its velocity due to the gamma factor, and that energy becomes part of radiation. – anna v Sep 26 '17 at 07:36
  • It depends upon the setup and the source of the acceleration. Original question didn't specify it, so synchrotron radiation formula doesn't quite apply here. – Darkseid Sep 26 '17 at 08:01
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Charged particles emit synchrotron radiation in the special case of moving perpendicularly to a magnetic field, and this radiation depends on mass.

The LEP electron synchrotron has a rated energy of 50 GeV and a radius of 4300 meters. This gives a relativistic gamma of about 98,000 compared to a gamma of 54 for a 50 GeV proton (see relativistic kinetic energy calculation). At that energy, the velocity is essentially c, so the synchrotron radiation power for a single electron can be calculated as follows:

syncr

Two-tenths of a microwatt may not sound like much loss, but per electron it is enormous! At this energy the proton velocity would also be essentially c, so the synchrotron radiation loss for the two particles scales like their gammas. So the loss rate for the electron is (97833/54)^4 or over 10^13 times the loss for a proton of the same energy in the same synchrotron.

So for circular accelerators the energy loss of an electron is much greater than for a proton, and that is why the future plans after LEP are talking of a linear collider, (also here page8).

enter image description here

By contrast the radiation in linear, one dimensional acceleration is smaller in its dependence on the mass of the particle :

linrad

Note the smaller power dependence. Thus ligher particles radiate more, but heavier one do also.

Quarks are not free , and radiation is expressed in Feynman diagrams and yes, there are electromagnetic interactions of quarks, but the strong ones dominate due to their strength. If you are discussing within the bound state of a proton or a neutron, that is taken into account with the lattice QCD models used.(example: Electromagnetic Splittings and Light Quark Masses in Lattice QCD A. Duncan, E. Eichten, and H. Thacker Phys. Rev. Lett. 76, 3894 (1996) )

I have toclarify that in this discussion I am talking about radiated energy, which is the important factor in constructing accelerators. And radiated energy depends on the mass of the particle. At the energies of accelerators the velocities are fractions of the velocity of light.

anna v
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  • Sir, what is the exact difference between the radiation cased by Liénard–Wiechert potential and the one which you stated for the case of linear accelerators. (As those emitted by Liénard–Wiechert potential seem to be mass independent). – kbg Sep 26 '17 at 20:14
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    You are asking about same acceleration, which involves velocity, and does not have a mass dependence. I am looking at energy input to reach the same energy of an accelerating charged particle, for example reach 10GeV for electron and proton. Due to the gamma factor it takes a lot more energy for an electron than for a proton, both in linear and much more in circular acceleration to reach the same momentum (and momentum is mass dependent). – anna v Sep 27 '17 at 03:51
  • I replied because of your statement in a comment " I remember my Prof telling that proton emits less radiation than electron (Thus it is better to use protons for collision in LHC)" – anna v Sep 27 '17 at 03:52
  • Yes, I was looking for this answer itself. Also, Sir, in your 1st statement of your answer "in the special case of moving perpendicularly to a magnetic field, and this radiation depends on mass". – kbg Sep 27 '17 at 07:05
  • According to the hyperphysics page you mentioned, it says that the charged particle only needs to be in circular orbit/curved to emit synchrotron radiation and hence if an electron gets defelected(curved) due to coloumb field of nucleus, it would emit synchrotron radiation (like slide 15 of https://indico.cern.ch/event/294651/contributions/671929/attachments/552041/760669/Delmastro_ESIPAP2014_3.pdf). Thus, requirement of magnetic field is not a necessity. Am I right? – kbg Sep 27 '17 at 07:05
  • yes, a dp/dt perpendicular to the direction of motion, any scattering at the particle level, with any field – anna v Sep 27 '17 at 08:03