I've been going through a translation of Frank and Tamm's original theory on Cherenkov radiation published by Jelley in 1958, and the bottom line is that I'm stuck on one of the intuitive leaps that was performed in the derivation. Been looking around a number of different sources for nearly a week now, and just can't seem to break through this nearly last step.

The derivation is based upon an electron passing through a perfect dielectric medium of infinite dimensions, with no dispersion, with constant velocity upon a path of infinite length. The coordinates are 3D polar ($\rho$, $z$, $\phi$). All of the following mathematics is done in the frequency domain after applying a Fourier transform to Maxwell's equations.

My question is mostly mathematical so I think it's probably OK to skip the specifics of the derivation up to the following step where the magnetic field intensity $\mathbf{H_{\omega}}$ is being calculated from the magnetic vector potential $\mathbf{A_{\omega}}$. The problem occurs in the following section of the derivation:

$$ \mathbf{A_{\omega}} =[0]\hat{\rho} -\left[\frac{q_e}{c\sqrt{2\pi s \rho}}\exp\left(i\omega\left(t-\frac{z \cos(\theta)+\rho \sin(\theta)}{c/n}\right) + \frac{3 \pi}{4}i\right)\right]\hat{z} +[0]\hat{\phi} $$ Where $\theta$ is the half angle of the Cherenkov light cone from the characteristic coherence relation: $\cos(\theta)=\frac{1}{\beta n}$. Also, in this equation $n$ is the refractive index and $\beta$ is the ratio of the particle velocity to the speed of light in vacuum ($\beta = \frac{v_{particle}}{c}$). In this case $s$ is defined as $s^2 = \frac{\omega^2}{v_{particle}^2}(\beta^2 n^2 -1)$ for convenience. From Maxwell's Equations: $$ \mathbf{H_{\omega}}=\nabla \times \mathbf{A_{\omega}} $$

Jelley skips to the magnetic field intensity without showing intermediate steps. He gets:

$$ H_{\rho} = 0 \\ H_{z} = 0 \\ H_{\phi} = -\frac{a}{\sqrt{\rho}} \int \sqrt{s}\cdot \cos(\chi) d\omega $$

Where $ a = \frac{q_e}{c}\sqrt{\frac{2}{\pi}} $ and $ \chi = \omega[t-\frac{z \cos(\theta)+\rho \cos(\theta)}{c/n}]+\frac{\pi}{4}$. Unfortunately when I calculate the curl of $\mathbf{A_{\omega}}$, I get the following:

$$ H_{\phi}=-\frac{a}{\sqrt{\rho}}\int\frac{1}{\sqrt{s}}\left(\cos(\chi)+i \sin(\chi)\right)\left(\frac{1}{4\rho}-\frac{n\omega \sin(\theta)}{2c}\right)d\omega $$

I can't entirely rule out a math error, but aside from some of the algebraic simplification, I used wolfram alpha to corroborate my results. If it's of any use, the components of the electrical field were subsequently computed, once again applying Maxwell's equations to the newly calculated $\mathbf{H}$ field.

If anyone could offer some insight into the "then a miracle occurs" step, that would be fantastic. Also if anybody could suggest alternative presentations of Frank and Tamm's formula I'd be appreciative (I've skimmed the one on wikipedia, the one in Classical Electrodynamics by Jackson, the one in the confusingly titled Classical Electrodynamics by Puri).

In the regrettably likely event that I've failed to convey something clearly or some aspect needs clarifying, please let me know and I'll ameliorate the situation ASAP.

FYI, the derivation is published in it's entirety in Jelley's original paper at the following link: https://archive.org/details/cerenkovradiatio030980mbp The derivation extends from pages 15-19.

  • $\begingroup$ I glanced in Zangwill's Modern Electrodynamics, but he does most of his calculations in the time domain and focuses on the electric field, not the magnetic field. So that's not much help. $\endgroup$ – Michael Seifert Mar 19 '18 at 15:02

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