Part 1 · full derivation
Deriving the Frank–Tamm Formula
Part 1 obtained the Cherenkov angle from a wavefront construction and then quoted the photon yield. This page derives the yield, and in doing so recovers the angle and the threshold as well — all three fall out of a single denominator.
Gaussian units throughout, a non-magnetic medium, and a permittivity ε(ω) = n²(ω) taken as real except where causality requires otherwise. The target is:
\[ \frac{d^2N}{dx\,d\lambda} = \frac{2\pi\alpha}{\lambda^{2}}\left(1 - \frac{1}{\beta^{2}n^{2}}\right) \]
Step 1 — The source, in Fourier space
A charge e moves with constant velocity v = βc along z. Its charge and current densities are
\[ \rho(\mathbf{x},t) = e\,\delta^{3}(\mathbf{x} - \mathbf{v}t), \qquad \mathbf{J} = \mathbf{v}\rho \]
Transforming with \(f(\mathbf{k},\omega) = \int d^3x\,dt; f\,e^{-i(\mathbf{k}\cdot\mathbf{x}-\omega t)}\), the time integral is just a phase and collapses to a delta function:
\[ \rho(\mathbf{k},\omega) = 2\pi e\,\delta(\omega - k_z v) \]
This is the whole physical input. A charge in uniform motion drives the medium only at frequencies tied to the wavenumber along the track by kz = ω/v. Everything that follows is a consequence.
Step 2 — The potentials, and one denominator
In the medium the wave equations for the potentials become algebraic:
\[ \left(k^{2} - \varepsilon\frac{\omega^{2}}{c^{2}}\right)\phi = \frac{4\pi\rho}{\varepsilon}, \qquad \left(k^{2} - \varepsilon\frac{\omega^{2}}{c^{2}}\right)A_z = \frac{4\pi J_z}{c} \]
Since Jz = vρ, the two are related by Az = εβ φ. Both carry the same denominator,
\[ D(\mathbf{k},\omega) = k^{2} - \varepsilon\frac{\omega^{2}}{c^{2}} \]
Everything interesting is in where D vanishes. A vanishing denominator means a field that propagates on its own — radiation.
Step 3 — The threshold and the angle, for free
The delta function has already fixed kz = ω/v. Writing k² = kz² + k⊥²:
\[ D = k_\perp^{2} + \frac{\omega^{2}}{v^{2}} - \varepsilon\frac{\omega^{2}}{c^{2}} = k_\perp^{2} - \frac{\omega^{2}}{c^{2}}\left(\varepsilon - \frac{1}{\beta^{2}}\right) \]
So D can vanish for real k⊥ only if
\[ \varepsilon - \frac{1}{\beta^{2}} > 0 \quad\Longleftrightarrow\quad n^{2}\beta^{2} > 1 \quad\Longleftrightarrow\quad \beta > \frac{1}{n} \]
Below that speed k⊥ is imaginary: the field clings to the track and decays exponentially away from it. Nothing escapes. The threshold is not an extra assumption — it is the condition for a wave rather than a bound disturbance.
And when it is satisfied, the radiated wave has k⊥² = (ω²/c²)(ε − 1/β²), so |k|² = εω²/c² as any wave in the medium must, and its direction follows immediately:
\[ \cos\theta_c = \frac{k_z}{|\mathbf{k}|} = \frac{\omega/v}{n\omega/c} = \frac{1}{n\beta} \]
The same line of algebra that produced the threshold produces the angle. Part 1's Huygens picture was right, but this shows it was not a special construction: it is what the pole of the propagator says.
Step 4 — The field that does the work
The energy the particle loses is the work done against its own induced field, so we need the longitudinal electric field at the charge. From E = −∇φ − (1/c)∂A/∂t, in Fourier space,
\[ E_z = -ik_z\phi + i\frac{\omega}{c}A_z = i\phi\left(\frac{\omega}{c}\varepsilon\beta - \frac{\omega}{v}\right) = i\phi\,\frac{\omega}{c}\,\frac{\varepsilon\beta^{2}-1}{\beta} \]
The bracket is worth pausing on. Using ε = n²,
\[ \varepsilon\beta^{2} - 1 = n^{2}\beta^{2}\left(1 - \frac{1}{n^{2}\beta^{2}}\right) = n^{2}\beta^{2}\sin^{2}\theta_c \]
The factor that will appear in the final answer is already here, and it vanishes at threshold exactly as the cone closes. The angle and the brightness were never independent facts.
Step 5 — The one step this course quotes
What remains is to integrate Ez over all k to get the field at the particle, and that integral runs straight through the pole found in Step 3. Handling it properly means giving ε a small positive imaginary part — causality, the medium must absorb rather than amplify — and closing the contour, with the radiated energy appearing as the residue. In cylindrical coordinates the same calculation is usually done with the asymptotic form of Hankel functions at large radius.
That machinery sits above this course. The result is standard:
\[ -\frac{dE}{dx} = \frac{e^{2}}{c^{2}}\int_{\,n\beta>1}\omega\left(1 - \frac{1}{\beta^{2}n^{2}}\right)d\omega \]
Quoted from J. D. Jackson, Classical Electrodynamics, 3rd ed., §13.4, and originally I. Frank and I. Tamm, Dokl. Akad. Nauk SSSR 14, 107 (1937). Note the integral is restricted to frequencies where n(ω)β > 1, which is why a real spectrum is cut off rather than diverging.
Step 6 — From energy to photons
Each quantum carries ℏω, so the number radiated per unit length per unit frequency is the energy expression divided by ℏω:
\[ \frac{d^{2}N}{dx\,d\omega} = \frac{e^{2}}{\hbar c^{2}}\left(1 - \frac{1}{\beta^{2}n^{2}}\right) = \frac{\alpha}{c}\left(1 - \frac{1}{\beta^{2}n^{2}}\right) \]
using the fine-structure constant α = e²/ℏc. The ω has cancelled: the photon number per unit frequency is flat. All the spectral shape comes from changing variable.
With ω = 2πc/λ and |dω| = (2πc/λ²)|dλ|:
\[ \frac{d^{2}N}{dx\,d\lambda} = \frac{\alpha}{c}\left(1 - \frac{1}{\beta^{2}n^{2}}\right)\frac{2\pi c}{\lambda^{2}} = \frac{2\pi\alpha}{\lambda^{2}}\sin^{2}\theta_c \]
That is the Frank–Tamm formula. The famous 1/λ² — the reason Cherenkov light is blue — is not a property of the emission at all. It is the Jacobian of the change from frequency to wavelength.
Does it give the right number?
Integrating over the band the sensors see, with n = 1.31 and β = 1:
\[ \frac{dN}{dx} = 2\pi\alpha\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)\sin^{2}\theta_c \approx 3.19\times10^{4}\ \text{m}^{-1} \]
Doing the same integral for the energy instead gives about 990 eV per centimetre. Dividing the two is a check that costs nothing and catches a great deal:
\[ \frac{dE/dx}{dN/dx} \approx \frac{9.88\times10^{4}\ \text{eV/m}}{3.19\times10^{4}\ \text{m}^{-1}} \approx 3.10\ \text{eV} \]
The mean photon energy. It must lie between the band edges, 2.07 eV at 600 nm and 4.13 eV at 300 nm, and it must sit above the midpoint because the spectrum favours the blue. It does both. If a sign or a factor of 2π had gone astray in Step 6, this ratio would have landed outside the band.
What this leaves out
ε was treated as real, which is what lets a pole sit on the contour and forces the causal prescription in Step 5; a genuinely absorbing medium needs the full complex treatment. The charge was taken as moving at constant velocity for all time, so there is no bremsstrahlung and no formation-length correction for a track of finite length. And dispersion — n(ω) rather than a constant n — both smears the cone and supplies the high-frequency cutoff that keeps the integral finite.