# What is the expression of the rate of loss of energy of an accelerated charge in a magnetic field?

The equation of motion of a particle of charge $$q$$ and mass $$m$$ in a magnetic field $$m\frac{d{\vec v}}{dt}=q({\vec v}\times {\vec B}).$$ If the charged particle has an initial velocity perpendicular to a uniform magnetic field, it moves in a circle, called the cyclotron motion. Since cyclotron motion is accelerated, it radiates and therefore must lose energy. However, if we take the dot product with $${\vec v}$$, the EoM says, $$m{\vec v}\cdot\frac{d\vec v}{dt}=0\Rightarrow\frac{1}{2}mv^2={\rm constant}.$$ This fails to capture the dissipation of energy because the EoM is incomplete; the force of radiation reaction is not taken into consideration.

The complete EoM in the nonrelativistic limit, $$m\frac{d{\vec v}}{dt}=q({\vec v}\times {\vec B})+\frac{\mu_0 q^2}{6\pi c}\ddot{\vec v}$$ where the second term on the right is the Abraham-Lorentz formula for the radiation reaction force. Again considering the dot product of the EoM with $${\vec v}$$, $$\frac{d}{dt}\left(\frac{1}{2}mv^2\right)=\frac{\mu_0q^2}{6\pi c}{\vec v}\cdot\ddot{\vec v}.$$

• This equation cannot be correct (as @Duepietri pointed out) because it tells that if the acceleration is constant, i.e., $$\ddot{\vec v}=0$$, there is no energy loss. In reality, this is not the case. So how do we find the correct expression for $$\frac{d}{dt}\left(\frac{1}{2}mv^2\right)$$?

• Second, when the particle accelerates, it loses energy. Thus we expect the rate of change of kinetic energy to be negative. However, the right side of the expression I got is not manifestly so. What's going on?

• Why do you think that ${\vec v}\cdot\ddot{\vec v}$ is zero?
– user303670
Commented Jun 7, 2021 at 16:14
• @Duepietri I made no such claim. Commented Jun 7, 2021 at 16:16
• Oh, no indeed! I thought that you said that kinetic energy has to be conserved, but that's only the case for a perpendicular force as you stated in the first part. So you want to know how to work out the expression?
– user303670
Commented Jun 7, 2021 at 16:19
• Yes, precisely. Commented Jun 7, 2021 at 16:20
• The formula for the AL force holds locally, so for every moment, there is a force acting backward on the electron, causing it to go slower. How can you see this from ${\vec v}\cdot\ddot{\vec v}$? I see now that it's a nice question!
– user303670
Commented Jun 7, 2021 at 16:25

$$\frac{d}{dt}\left(\frac{1}{2}mv^2\right)=\frac{\mu_0q^2}{6\pi c}{\vec v}\cdot\ddot{\vec v}$$

This tells that when the acceleration is zero, the energy of the particle is conserved, as expected

But the formula talks about $${\vec v}\cdot\ddot{\vec v}$$, that is, the derivative of the acceleration wrt time. This means that if the acceleration is constant, energy will be conserved too. Which tells you that the formula can't be right (if the acceleration is zero the formula could be right).

The Abraham-Lorenz force talks about the derivative of acceleration. This derivative can be in the direction of motion. So the (quoted) assumption you made is not right. Energy can change.The derivative wrt to time is not zero.

• Oops! I will edit the question. Commented Jun 7, 2021 at 15:47
• @mithusengupta123 I couldn't comment (yet), that's why I wrote an answer. ;)
– user303670
Commented Jun 7, 2021 at 15:50

I think the formula for the radiative energy loss is correct, but to avoid contradictions you have to consider it in more detail. Consider a charged particle moving in a magnetic field perpendicular to the velocity. The trajectory will be circular ("around" the magnetic field direction). In circular motion, the acceleration is directed towards the center of the circle, and it is changing every time moment: even if its absolute value is constant, the acceleration direction obviously rotates together with the particle. Now if you consider two consecutive moments of time and look at the acceleration change between, you'll see that the vector $$\vec{a_2}-\vec{a_1}$$ is tangential to the circle and is directed "backwards", i.e. antiparallel to the velocity. As such, $$\vec{v}\cdot\vec{v}^{\prime\prime}<0$$, resulting in the negative change of energy, consistent with the common sense.

Anyway, I'd not worry about the sign too much; you already know that the energy cannot increase, it can only decrease, so you can choose the sign appropriately.

Again considering the dot product of the EoM with $${\vec v}$$, $$\frac{d}{dt}\left(\frac{1}{2}mv^2\right)=\frac{\mu_0q^2}{6\pi c}{\vec v}\cdot\ddot{\vec v}.$$

• This equation cannot be correct (as @Duepietri pointed out) because it tells that if the acceleration is constant, i.e., $$\ddot{\vec v}=0$$, there is no energy loss. In reality, this is not the case. So how do we find the correct expression for $$\frac{d}{dt}\left(\frac{1}{2}mv^2\right)$$?

That equation indeed isn't even self-consistent during constant acceleration but for a different reason: you forgot to include the external force that is maintaining the constant acceleration. When you include it, the work-energy equation for the particle is fine: work of external force equals increase in its kinetic energy.

What you probably meant is that there is a problem with the idea of conservation of total energy (including EM energy), because we believe that due to acceleration, EM energy of radiation is increasing according to the Larmor formula, and also kinetic energy of the particle is increasing, and it looks like total energy is increasing beyond what work of the Lorentz-Abraham force (of unclear origin!) can provide. This observation is correct; the Lorentz-Abraham force does not explain how particle can increase its energy and also increase energy of EM radiation a la Larmor while obeying local conservation of energy.

Dirac and others tried to salvage the idea of Lorentz-Abraham force for point particles. It didn't work well - the resulting equation of motion has nonphysical solutions like runaways to infinity and preaccelerations (acceleration before external force acts).

The problem with this idea of radiation damping force for point particles is best seen in rest frame of the particle: since it accelerates, energy of EM radiation should increase a la Larmor, but work done on the particle is zero, because particle has zero velocity. So no damping force ever can explain how energy $$ka^2$$ per unit time gets added to energy of EM radiation - this energy increase rate is frame-independent, while work of any force is frame dependent and can be made zero by jumping to the rest frame.

The Lorentz-Abraham force for point particles is, from a theoretical standpoint that aims to describe each point of time consistently with local conservation of energy, a misguided idea and a failed project. Unfortunately it is hard to show this experimentally, say for electrons, because Larmor-predicted energy of radiation of a single accelerated electron is extremely small and its expected effect on particle trajectory is not measurable. Unfortunately we do not have a direct way to check whether something like Lorentz-Abraham force exists for elementary particles.

On the theoretical level, there are two obvious ways to solve this puzzle.

1. particles are extended charged bodies with parts, with internal degrees of freedom, Lorentz-Abraham force on the charged body is not accurate description of what goes on inside and thus mutual interaction of parts has to be described in more detail; this is very hard to do because it is basically a problem of relativistic continuous mechanics of compact structure that holds together due to internal forces of unclear properties; there seem to be too many degrees of freedom when formulating these models;

2. particles are points, but energy of EM radiation is not given by the Poynting formulae and thus Larmor's formula does not apply. This second way was investigated by Tetrode, Fokker and Frenkel in 1925 [1] and rediscovered by others later. This approach provides consistent formulation of theory of charged point particles obeying local conservation of energy. These particles radiate EM radiation when accelerated, but this radiation is associated with energy in a different way than one would expect based on the Poynting formulae. The good thing about this is when many such point particles move together in the same way, energy of EM radiation is close to what Poynting formula would predict, so the theory is consistent with observations of macroscopic synchroton radiation in particle accelerators and radiowave radiation by antennas.

[1] Frenkel J., Zur Elektrodynamik punktfoermiger Elektronen, Zeits. f. Phys., 32, (1925), p. 518-534. DOI: http://dx.doi.org/10.1007/BF01331692 English translation: http://www.neo-classical-physics.info/uploads/3/4/3/6/34363841/frenkel_-_ed_of_point-like_electrons.pdf