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A charged particle in a uniform magnetic field will travel in a circular trajectory. We know this from solving Newton's 2nd law with the Lorentz law, i.e. solving

$$\textbf{F} = m\textbf{a} = q(\textbf{v}\times \textbf{B}),$$

which, if $\textbf{B}$ is exactly perpendicular to $\textbf{v}$ gives us the circular path we know and love.

However, we also know a charged particle undergoing acceleration radiates energy according to the Larmor formula, so this situation cannot exist in reality; the charge radiates energy and spirals inwards. This is due to a 'radiation reaction' or 'self' force, arising from the interaction of the charge with its own field.

Now, according to Griffiths' Electroydynamics textbook, this self-force is the Abraham–Lorentz force

$$\textbf{F}_{rad} = \frac{\mu_0 q^2}{6\pi c}\dot{\textbf{a}}.$$

What I'm asking is

  1. Is the above reasoning correct and
  2. Is the route to obtaining the 'spiralling-in' trajectory equation of motion 'simply' incorporating this $\textbf{F}_{rad}$ force into Newton's 2nd law? I.e. solving the differential equation

$$\textbf{F} = m\textbf{a} = q(\textbf{v}\times \textbf{B}) + \frac{\mu_0 q^2}{6\pi c}\dot{\textbf{a}} ?$$

Assuming this is correct, and the uniform magnetic field points in the z-direction with magnitude $B$, we get the coupled system of differential equations

$$ \begin{cases} & q\dot{y}B + \frac{\mu_0 q^2}{6\pi c}\dddot{x} = m\ddot{x} \\ & -q\dot{x}B + \frac{\mu_0 q^2}{6\pi c}\dddot{y} = m\ddot{y} \\ & \frac{\mu_0 q^2}{6\pi c}\dddot{z} = m\ddot{z} & \end{cases} $$

so my third question would be

  1. How on Earth do I go about numerically solving these? Is there any literature on this?

thanks!

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  • $\begingroup$ Take it the other way. The Larmor precession happens due to the alignment of the electrons magnetic dipole moment and this deflection is an acceleration and let to the emission of a photon. The photons momentum disalignes the magnet dipole moment again and the cycle repeats until the electrons kinetic energy is exhausted. The Lorentz force is the expression for this process. See for example here $\endgroup$ – HolgerFiedler Nov 4 '17 at 19:04
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  1. Is the above reasoning correct

Yes, provided the Larmor formula applies. Larmor's formula was obtained based on some assumptions:

1) The Poynting theorem is applicable and can be interpreted as work-energy theorem, so the surface integral of the Poynting vector gives EM energy that is lost from a region of space. This is so when the material particles interacting with the EM field have finite charge and current density so the involved integrals make sense. It is NOT valid for point particles, because the mathematical derivation of the Poynting theorem for region containing such particles breaks down. If used anyway, as unfortunately many textbooks do, some of the familiar Poynting expressions have either infinite value or the expression does not make sense which is a confirmation that for point particles the theorem is of no use.

2) The EM field far enough from the accelerated charged body is given by the retarded solution of Maxwell's equations. This is well established for total EM field in macroscopic cases like field of an antenna or field of fast moving electron/proton bunches in a cyclotron. It is also believed to be valid for lone microscopic particles, but this was never properly tested and it is possible that microscopic field of a microscopic particle such as electron is not given by purely retarded solution.

In simple terms, using Larmor's formula or the Lorentz-Abraham formula for microscopic particles may be erroneous, may be not. The effect predicted by them on the motion of microscopic particles is so minuscule that it is difficult to directly experimentally measure and as far as I know, it was never successfuly done. When motion of large collections of particles is studied (charges in an antenna, electron bunches in an accelerator), the radiation resistance is measurable and is consistent with the Larmor and the Lorentz-Abraham formula.

  1. Is the route to obtaining the 'spiralling-in' trajectory equation of motion 'simply' incorporating this $\textbf{F}_{rad}$ force into Newton's 2nd law? I.e. solving the differential equation ...

No, the resulting equation of motion incorporating the radiation reaction term was studied decades ago and people realized that it has weird, unwanted solutions, such as particle accelerating before external field acts on it, or run-aways, where particle keeps accelerating indefinitely due to its own field, or small but systematic non-conservation of energy.

This is because, as it was actually already known right from the start, the Lorentz-Abraham term is not the entire story about the self-force of charged body on itself. The Lorentz-Abraham force is actually one term of an infinite series developed as an approximate solution to an idealized scenario, where the charged particle has spherical symmetry and does not deform too much. These other terms are often not mentioned because they are more complicated (higher derivatives of particle velocity) and hard to use but without them, the Lorentz-Abraham expression gives only an approximation to total self-force on the particle.

Unfortunately, this approximation, when taken as sole contribution to total force, leads to equations that have unphysical solutions.

  1. How on Earth do I go about numerically solving these? Is there any literature on this?

One way to make use of the Lorentz-Abraham term sensibly without getting too far from the physical solutions is the following:

  • calculate trajectory of the particle as if there was not any self-force
  • evaluate the Lorentz-Abraham force for the above calculated trajectory and use it to calculate correction to it

In all common cases involving charged particles, the radiation reaction as given by the Lorentz-Abraham formula is a very minor correction of the total force, so this works well. This may be another example from physics where theory of an effect works well if the effect is a relatively small perturbation, but should not be trusted in situations when the effect stops being small.

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A charged particle in a uniform magnetic field will travel in a circular trajectory. We ...

describe and calculate this with

... the Lorentz law, i.e. solving $$\textbf{F} = m\textbf{a} = q(\textbf{v}\times \textbf{B}),$$ which, if $\textbf{B}$ is exactly perpendicular to $\textbf{v}$ gives us the circular path we know and love.

In reality the Lorentz force does not draw the full picture. As you mentioned ...

... a charged particle undergoing acceleration radiates energy ... and spirals inwards. This is due to a 'radiation reaction' or 'self' force, arising from the interaction of the charge with its own field.

Take it the other way. The Larmor precession happens due to the alignment of the electrons magnetic dipole moment from the external magnetic field and this precession let to a deflection. This deflection is an acceleration and let to the emission of photons. The photons momentum disalignes the electrons magnet dipole moment again and the cycle repeats until the electrons kinetic energy is exhausted. Under this view the introduction of a self-force seems not to be necessary.

Since the external field $\textbf{B}$, the charge $q$ and the mass $m$ are fixed values, the velocity $\textbf{v}$ only seems to depend from the magnetic field over time $\textbf{v} = f(\textbf{B},t)$

Is the route to obtaining the 'spiralling-in' trajectory equation of motion 'simply' incorporating this $\textbf{F}_{rad}$ force into Newton's 2nd law? I.e. solving the differential equation $$\textbf{F} = m\textbf{a} = q(\textbf{v}\times \textbf{B}) + \frac{\mu_0 q^2}{6\pi c}\dot{\textbf{a}} ?$$

I like your idea to complete the Lorentz force equation in a way to get the right calculations for the spiral path. But why not to take in consideration the lost of kinetic energy due to the EM radiation? So perhaps $\frac{m\textbf{v}^2}{2} = f(\textbf{B},t)$ would be a good way to get the additional term to the Lorentz force equation?

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