Part 1
Seeing a Neutrino: Cherenkov Light
You cannot detect a neutrino directly. What you detect is the charged particle it creates — usually a muon — and you detect that by the light it leaves behind. The light exists for a reason worth deriving, because the same derivation fixes the angle of the cone, the brightness of the track, and the minimum speed below which there is nothing to see.
The idea in one sentence
Light in a medium travels at c/n, which is slower than c. A particle can therefore move faster than the light it emits, and the wavefronts it leaves behind pile up into a cone — the optical equivalent of a sonic boom.
Derivation: the Cherenkov angle
Step 1 — Two distances
In a time t, the particle travels βct. Light emitted at the start of that interval has travelled ct/n in every direction.
Step 2 — The tangent
A coherent wavefront exists where the spherical wavelets share a common tangent. That tangent meets the track at an angle θc whose cosine is the ratio of the two distances, since the wavefront radius is the adjacent side of a right triangle with the particle's path as hypotenuse:
\[ \cos\theta_c = \frac{ct/n}{\beta c t} = \frac{1}{n\beta} \]
Step 3 — A threshold falls out for free
A cosine cannot exceed 1, so the construction only works when nβ > 1. There is a minimum speed:
\[ \beta_{\min} = \frac{1}{n} \]
Deep Antarctic ice has n ≈ 1.31 for the blue light the sensors are most sensitive to, so βmin ≈ 0.76. A slower particle emits nothing at all, which is a useful filter: IceCube is blind to anything that is not relativistic.
Step 4 — The angle in ice
For a particle with β → 1, which every particle of interest here is:
\[ \theta_c = \arccos\!\left(\frac{1}{1.31}\right) \approx 40.2^\circ \]
This is the single most useful number in the detector. Every relativistic track in the ice makes the same cone, so the arrival times of light across the array encode the geometry of the track and nothing else.
How bright is the cone?
The Frank–Tamm result gives the number of photons emitted per unit track length per unit wavelength by a particle of charge e. It is stated here and derived in full on its own page, where the threshold and the angle above turn out to follow from the same denominator:
\[ \frac{d^2N}{dx\,d\lambda} = \frac{2\pi\alpha}{\lambda^2}\left(1 - \frac{1}{\beta^2 n^2}\right) = \frac{2\pi\alpha}{\lambda^2}\sin^2\theta_c \]
Two things are worth reading off before computing anything. The bracket is exactly sin²θc, so the cone closes and dims together — at threshold both vanish. And the 1/λ² means the spectrum rises steeply toward the blue, which is why Cherenkov light looks blue and why the sensors are tuned there.
Integrating over the band the sensors actually see, 300 to 600 nm:
\[ \frac{dN}{dx} = 2\pi\alpha\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)\sin^2\theta_c \approx 3.2\times10^{4}\ \text{photons per metre} \]
The derivation starts from a charge in uniform motion, finds that one denominator controls everything, and recovers the threshold, the angle and this yield from it. It also shows that the 1/λ² is not a property of the emission at all — it is the Jacobian of changing variable from frequency to wavelength.
Thirty thousand photons per metre sounds like a great deal, and over the hundreds of metres a high-energy muon travels it is. But they spread through ice, scatter, and are collected by sensors occupying a vanishing fraction of the volume — which is why Part 3's reconstruction has to work so hard with so few detected photons.
Check your understanding
- Water has n ≈ 1.33. Is the Cherenkov angle larger or smaller than in ice, and why?
Answer: larger. cos θc = 1/(nβ) falls as n rises, so θc ≈ 41.2°. A denser medium also lowers the threshold speed. - Why does the yield formula contain no particle mass?
Answer: Cherenkov emission depends only on charge and speed. An electron and a muon at the same β radiate identically — which is precisely why the light alone cannot tell you which particle made it. - The sensors are most sensitive near 400 nm. Given 1/λ², is that a coincidence?
Answer: no. The spectrum is brightest in the blue and the ice is most transparent there, so the photomultipliers were chosen to match.
What this leaves out
The refractive index varies with wavelength, so strictly θc does too and the cone is slightly smeared. Deep ice also scatters light over tens of metres, which blurs arrival times far more than the dispersion does — the dominant limit on angular resolution in Part 3, and the reason the ice's optical properties had to be mapped before any of this worked.