5
$\begingroup$

The Wikipedia page on the Larmor formula says that

the Larmor formula makes the unavoidable assumption that the charged particle is orbiting in a circle.

This quoted sentence isn't true, correct? I just reread the derivation of the Larmor formula in a couple sources and never saw any assumption of circular motion. The only assumptions that I could see are (a) no quantum effects, and (b) nonrelativistic speeds $v \ll c$ (with the Liénard formula removing the latter assumption). I think that whoever wrote that sentence may have been thinking of the Abraham-Lorentz force instead, whose most common derivation assumes periodic (not circular) motion.

$\endgroup$

2 Answers 2

4
$\begingroup$

The Larmor formula derivation does not assume circular motion, it can hold for any accelerated motion.

The major assumptions in its derivation are: that the field around the particle by which we judge the radiated energy is just due to that single particle (not valid in , e.g., antenna, where many fields of many close accelerated charges combine to produce net field); EM field of the particle is the retarded solution to Maxwell's equations, so the EM wave goes away from the particle, not coming into the particle (the usual assumption in macroscopic EM theory, but not strictly necessary in microscopic theory); and that EM energy flow is given correctly by the Poynting expressions (this is not valid for point particles).

$\endgroup$
4
  • $\begingroup$ Could you clarify your very last point? Why isn’t the Poynting expression valid for point particles? $\endgroup$
    – tparker
    Commented Jun 17 at 13:07
  • $\begingroup$ The Poynting theorem involves the expression $\mathbf j \cdot \mathbf E$. Energy interpretation of the Poynting formulae for energy and energy flow comes from interpreting this expression as work of electric field on current. This works for finite $\mathbf j$, but the expression is mathematically undefined for point charged particle, because $\mathbf E$ is singular at the particle, $\mathbf E\cdot \mathbf j$ is undefined, and the actual work of external forces is finite. $\endgroup$ Commented Jun 17 at 13:21
  • $\begingroup$ Ah, gotcha. This is the usual subtlety that for the Lorentz force law for classical point particles, you should really only be considering the external fields produced by all the charge other than the charge in question that is experiencing the force - but as you say, the expression ${\bf E} \cdot {\bf J}$ also incorporates the ${\bf E}$ field of the charge itself, as discussed at physics.stackexchange.com/questions/808231/…. Thanks! $\endgroup$
    – tparker
    Commented Jun 17 at 15:29
  • 1
    $\begingroup$ @tparker Indeed, we must be cognizant of whether we're talking about the usual not too-singular sources, so we can work with terms like $\mathbf E\cdot \mathbf j$ or $\rho\mathbf E + \mathbf j\times \mathbf B$ with net fields, or about line/point sources, where these terms do not make sense mathematically. In the latter case, we have to use some other fields, not too singular at the sources. The force on electrons in EM field, and even macroscopic magnetic force $BIL$ on a line current-carrying wire, reveal external fields due to other bodies, so the usual formulae really refer to those. $\endgroup$ Commented Jun 29 at 21:41
4
$\begingroup$

The Larmor formula was derived in 1897 and would therefore apply to what was known about physics prior to quantum mechanics (and relativity$^1$). The prevailing idea at the time was that electrons move in circular orbits. In fact, Larmor introduced "the solar system model of the atom" in 1897.

The formula quantifies the generation of electromagnetic radiation by an accelerated charge, whether moving in a circle or not. But because an electron in atom was thought to move in a circular orbit, it is therefore undergoing centripetal acceleration and therefore should be radiating electromagnetic waves (why the electron never loses energy and spirals to the nucleus was a problem, but that is another story and spurred ideas that would eventually lead to quantum theory).

the Larmor formula makes the unavoidable assumption that the charged particle is orbiting in a circle.

These words and the word "unavoidable" would be in reference to the prevailing view at the time that electrons in atoms follow circular orbits.

$^1$ The Larmor formula $$P=\mu_0 \frac{q^2a^2}{6\pi c}$$ contains the speed of light, $c$ which was a result of Maxwell's work in 1865. Even though the speed of light was established, this was still prior to special relativity. See also Liénard–Wiechert potential for the relativistic version of the equation.

$\endgroup$
4
  • $\begingroup$ Well maybe Larmor himself originally assumed circular motion in deriving his formula. But he turned out to have gotten lucky and derived a formula that is exactly correct for arbitrary nonrelativistic trajectories, correct? Note that the article does not say that "Larmor made the unavoidable assumption that...", but "The Larmor formula makes the unavoidable assumption that...". $\endgroup$
    – tparker
    Commented Jun 17 at 6:06
  • $\begingroup$ I realize that the quoted sentence is a bit ambiguous as written, because formulas can't "make" assumptions - they can have been originally derived under certain assumptions, or they can hold under certain assumptions, which are not equivalent statements. I interpreted the sentence to mean the latter. $\endgroup$
    – tparker
    Commented Jun 17 at 6:06
  • $\begingroup$ Yes, it is accurate for non-relativistic charges. I don't know that luck had anything to do with it, as how electrons "move" as an assumption is secondary to what the experimental data shows, which was that electrons have fixed energies/ang. momentum etc. in atoms. I think that "Larmor made the unavoidable assumption that..." and "The Larmor formula makes the unavoidable assumption that..." are assumed to be equivalent by whoever wrote the sentence...... $\endgroup$
    – joseph h
    Commented Jun 17 at 6:20
  • $\begingroup$ ...and when Larmor was deriving the equation, that was his assumption and that of other physicists. Cheers. $\endgroup$
    – joseph h
    Commented Jun 17 at 6:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.