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Cherenkov radiation is electromagnetic radiation from a charged particle exceeding the medium's phase velocity: $c/n(\omega )<v<c$. Here $n(\omega )$ is the positive refractive index at angular frequency $\omega$. The threshold uses phase velocity, not group velocity.^[ref]

The beauty of Cherenkov radiation is that uniform motion can generate a sharply directed electromagnetic wave: the medium's response along successive portions of the track adds with matching wave phase on a cone. The moving electric charge drives the medium's electric polarization; the resulting radiation depends on the collective refractive index. This differs from luminescence, where excitation of material is followed by light emission. Both mechanisms can contribute to the same irradiated sample. A visible blue glow alone therefore identifies neither Cherenkov radiation nor a particular radioactive source.

The Cherenkov cone angle turns a measured direction into a constraint on particle speed when the refractive index is known. The Frank-Tamm formula instead predicts the number of emitted photons in a specified wavelength interval. These are complementary observables: the finite-band Cherenkov yield model below shows why both depend on optical dispersion and on which wavelengths are collected. This makes Cherenkov radiation useful for particle detectors, with photon production, transport and detection treated separately.

For constant particle speed and electric charge $q=ze$, in a transparent, linear, nonmagnetic medium with isotropy, the Frank-Tamm formula gives the produced photon density:Set it to zero below threshold. Here $e$ is the positive elementary charge, $x$ is track length, $\lambda =2\pi c/\omega$ is vacuum wavelength, and $\alpha$ is the fine-structure constant. With SI lengths in metres, the density has units ${\mathrm{m}}^{-2}$; multiplying by $dxd\lambda$ gives a dimensionless expected count. The explicit SI convention isThese conventions agree.^[ref]

To check the convention, start from $E=\hslash \omega =hc/\lambda$ and $d^{2}N/(dxdE)=\alpha z^2{\sin }^2\theta /(\hslash c)$. Changing variables uses the positive density Jacobian $|dE/d\lambda |=hc/{\lambda }^2$; since $h/\hslash =2\pi$, the coefficient becomes $2\pi \alpha z^2$. No factor of the material refractive index belongs in this Jacobian: $\lambda$ is not the in-medium wavelength. Multiplication by $\hslash \omega$ gives the corresponding energy loss density in this SI nonmagnetic convention:

For an original constant-index example, stipulate $n=1.50$, $\beta =0.90$, $z=1$ and $\alpha =1/137.036$. Integration from $400$ to $700\mathrm{nm}$ givesThus a stipulated $1\mathrm{cm}$ track produces about $222$ photons in this band in the ideal model. This is an emitted count before absorption, collection losses and detector response, not a brightness measurement or a measured water yield.