In our electrodynamics course we have learned that accelerating a charged particle will lead to a loss of energy in the particle due to Bremsstrahlung. The exact amount of power radiated away is given by the Larmor formula.

However, in a particle physics course I have learned that the following Feynman diagram is not possible without the presence of a nucleus. Feynman diagram of Bremsstrahlung of an electron

Hence my question: How does Bremsstrahlung, which is apparently a problem at large circular particle accelerators, even occur?

For example in the LHC the beam pipe vacuum seems to be around 10^-8 to 10^-9 Pa (LHC Vacuum ). According to this table, a vacuum that empty corresponds to a mean free path of around 1000km. While this might be enough collisions for the accelerated protons to fire a photon every once in a while, it is definitely not what I imagined the Bremsstrahlung in the LHC to be - namely a continuous loss of energy during the acceleration. Does that mean, that the Bremsstrahlung-issue could be completely resolved by creating an even better vacuum?

  • $\begingroup$ 1000km mean free path would mean 300 interactions per second... but that's mean free path for slow atoms/molecules. The cross sections with the high energy beam particles should be orders of magnitude smaller, leading to 100-1000h beam loss timescales. $\endgroup$
    – CuriousOne
    Apr 12, 2016 at 7:44
  • $\begingroup$ Did I understand you right, that the cross section does get smaller for high energy particles? Intuitively I thought it's the other way round, but I can't tell you why and never thought about it. $\endgroup$
    – user114116
    Apr 13, 2016 at 17:57
  • $\begingroup$ If I remember correctly, the total pp cross section drops somewhat and the differential cross section, which describes how much energy is lost in the collisions, drops like a stone, i.e. high momentum loss is rare. I don't think one can make a storage ring work, at all, without that effect. The beam loss would be way too high. $\endgroup$
    – CuriousOne
    Apr 13, 2016 at 18:36

1 Answer 1


The quality of the vacuum at LHC is pretty good but particles are constantly accelerated/bent by strong electromagnetic fields along the circumference of the accelerator. Thus the situation is not the one felt by a free particle. So conceptually, in your diagram, you can simply replace the photon coming from the nucleus by a photon from the electromagnetic field.

  • $\begingroup$ The beam energy loss due to synchrotron radiation can be compensated for by the accelerator, the particle loss due to gas bremsstrahlung can not be. For the LHC the total synchrotron radiation power is on the order of 10kW or so (a mere 7keV per turn). For LEP, because it was a lepton storage ring, it was in the MW, I believe. Synchrotron radiation is actually a good thing, because it dampens beam oscillations. If it isn't strong enough, then we have to create it artificially to dampen the machine. $\endgroup$
    – CuriousOne
    Apr 12, 2016 at 8:13
  • $\begingroup$ actually it is the magnetic field of the bending magnets and the quadrupoles , not electromagnetic. $\endgroup$
    – anna v
    Apr 12, 2016 at 10:41
  • $\begingroup$ for the acceleration, one needs electric field... $\endgroup$
    – Paganini
    Apr 12, 2016 at 12:06
  • $\begingroup$ No, a magnet works just fine to accelerate a charged particle. It just won't go any faster. $\endgroup$
    – Jon Custer
    Apr 12, 2016 at 12:33
  • $\begingroup$ The particle is working against its own field. The radiation in the Larmor formula is as a result of the retarded potential during the acceleration of a particle. $\endgroup$
    – Peter R
    Apr 12, 2016 at 13:37

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