Ordinarily we consider the path of a charged particle under the influence of a magnetic field to be curved. However, in order for the trajectory of the particle to change, it must emit a photon. Therefore, in principle, if we were able to view the path of the particle at sufficiently high resolution, would its path actually be polygonal, with a photon emitted at each vertex?

  • $\begingroup$ Note that the momentum of the emitted photons is not opposite the change of momentum of the bending particle in either the quantum or the quasi-classical view, so the presumed piecewise continuous trajectory fails to conserve momentum one way or the other. $\endgroup$ Commented Mar 10, 2013 at 21:06
  • $\begingroup$ Mark Allen, did you reach a progress in your question? Here I asked a related question about Description of the spiral path of an electron under the influence of a magnetic field. $\endgroup$ Commented Jun 2, 2019 at 5:48
  • $\begingroup$ @HolgerFiedler I'm pretty sure the original question was based on an elementary blunder of 2-body physics, and my reasoning ignored that, in the most general case, a curved trajectory for one body in an n-body system implies nothing in particular about the total momentum of the system. Regardless, as phrased, the question earned responses that led to me thinking more concretely about the distinction between 'the trajectory of a particle in some system at some time t' and 'the expected value of some measurement of a system at some time t'. (I think? Thereafter I abandoned physics) $\endgroup$
    – Mark Allen
    Commented Apr 1 at 0:46

3 Answers 3


You're asking what would happen if we could view things with an unlimited high resolution. You view the emissions of the synchotron photons as discrete events and you ask is the path linear between these emissions. The problem is - quantum particles do not have trajectories so it's not meaningful to ask about the actual path followed by the particle. All you can do is make measurements and ask about the sequence of measurement results. There is then a limit on the resolution of what you will find, such as the ~micron limit in the pictures in Anna's answer, where the measurements are the ionization events.


When a charged particle bends in a magnetic field ( an electric field also ) synchrotron radiations is released at the bend. In principle the bend can be approximated with a circle appropriate to the radius of curvature of the bend and the radiation released calculated, as in the link.

In a circular path there will be synchrotron radiation emitted incrementally on the curve.How small an increment? it will depend on the mass of the particle, and the energy of the photon. The higher the energy the greater the loss, the smaller the mass the greater the loss, that is why we have LHC and not LEPIII at the moment. The path in vacuum microscopically, will be erratic due also,to the Heisenberg Uncertainty Principle at that microscopic scale. Not a regular polygon then because probabilities determine the specific energy at each vertex.

When one sees a perfect circle starting in a bubble chamber for example

bubble chamber photo

we see also the effect of energy loss on the circular path from ionization scatterings, which also depends on the mass of the particle. We see on the right entering a high energy charged particle which decays in two short circular segment and ends an electron that spirals in the magnetic field as it loses energy from collisions mainly, since synchrotron radiation is very soft at the energies in the photo.

The paths are good circular fits even in the micron measurements of such photos for the massive particles and taking care of ionization loss fits well the electrons.


The motion of the electron in the constant uniform magnetic field is very much like the motion of an electron in the hydrogen atom. The difference is in the source of the centripetal acceleration.

EDITION: The comments that kindly have been written, which I welcome, show how difficult it is to formulate a question, and how easy it is to misinterpret it (it is still possible that I have not understood the question.) In view of these comments I thought I should add this edition:

It must be emphasised that the similarity between an electron in the hydrogen atom and in a uniform magnetic field is only in relation to the fact that, even in the case of an electron in a uniform magnetic field, the electronic energy levels can be discrete, i.e. the electron energy has quantum levels, the Landau levels. For example, an electron in a box of size $L$ and the uniform magnetic field pointing in the $z$-direction, the electron energy is given by the equation


where $k_z=\frac{2\pi n_z}{L}$, $n$, $n_z$ are quantum numbers (positive integers), and $\omega_c$ is the classical angular speed of the electron in the uniform magnetic field. This phenomenon is also known as orbit quantisation. The Landau levels are present regardless whether the electron is in a box or not, as one can see by letting $L\rightarrow\infty$. These Landau energy levels, n, are very important in condensed matter physics for the study of the geometry of the Fermi surface of metals.

The presence of the Landau levels show that the electron, being in the uniform magnetic field, cannot radiate energy unless it is forced to change from one Landau quantum level to another.

For the electron in hydrogen atom the centripetal acceleration is provided by the Coulomb field,

${\bf a}=-\frac{e^2}{4\pi\epsilon_0 m_er^3}{\bf r}$

For the electron in the uniform magnetic field the centripetal acceleration is provided by the Lorentz force as

${\bf a}=-\frac{e{\bf B}\times {\bf v}}{m_e}$.

As the speed remains constant, the kinetic energy of the electron in its circular path in the magnetic field remains fixed. This implies that the electron does not emit photons. One can understand this from the energy equation over one period (cycle)

$E=\oint_C e{\bf B}\times {\bf v}.d{\bf r}= \int_T e{\bf B}\times {\bf v}.{\bf v}dt=0$.

The situation in an accelerator is different in that, in this case the electrons are accelerated to increase their speed, hence the electrons emit Bremsstlahlung radiation. If the field is fixed as is suggested in this question, the electron cannot radiate as its energy remains fixed.

If the electrons did emit photons while in fixed circular orbit in the constant uniform magnetic field, we could use this as an endless source of energy!! In direct violation of the principle of conservation of energy.

  • $\begingroup$ In many accelerators the bending region is quite separate from the acceleration region. None the less the particles radiate around the bends. The difference between electrons in atoms and in, say CEBAF, has noting to do with the "purpose" of the acceleration. $\endgroup$ Commented Mar 10, 2013 at 4:19
  • 2
    $\begingroup$ I think you should rethink this answer, as the example you are giving of the hydrogen atom is wrong. That the hydrogen atom and atoms in general existed without the electrons spiraling into the nucleus was one of the basic experimental reasons that quantum mechanics was invented. In addition your argument about endless source of energy is also wrong, as the energy would be provided by the magnetic field. $\endgroup$
    – anna v
    Commented Mar 10, 2013 at 4:27
  • $\begingroup$ @annav Many thanks for your comment. Please see edited answer. I will appreciate further comments. $\endgroup$
    – JKL
    Commented Mar 11, 2013 at 5:33
  • $\begingroup$ @JKL The equations you are using are for an instantaneous classical motion, i.e. balancing the mechanical forces but not allowing for the radiation due to the classical solutions of maxwell's equations as far as the energy radiated goes. The calculations are not simple, see paragraph 20-4 in "classical electricity and magnetism" by Panofski and Philips, radiation from circular orbits. Your answer is still not correct. $\endgroup$
    – anna v
    Commented Mar 11, 2013 at 5:55
  • $\begingroup$ I believe that the confusion comes because you are treating the magnetic field as a quantum mechanical box but the dimensions with respect to h-bar have hbar=0 effectively. This means that the solutions are continuous in this macroscopic problem. As Maxwell's equations hold in both the QFT regime and the classical regime the formulae in the reference I gave above are what holds macroscopically. There is nothing magical in the energy emitted, it is taken from the magnetic field. If it is permanent magnets it would reduce their magnetization. $\endgroup$
    – anna v
    Commented Mar 11, 2013 at 6:46

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