Neutrino detection

Why must a detector built to catch particles that pass straight through the Earth be buried a mile underground?

Neutrino detectors sit at the bottom of mines, under mountains, beneath a kilometre of Antarctic ice. The stated reason is shielding — but shielding from what? The particle being hunted is the one thing a mile of rock demonstrably does not stop, which is the whole reason it is interesting. Burying the apparatus cannot help the signal get in.

So the depth is not for the neutrinos. It is against everything else, and the question is why that trade is worth the expense of digging. It only makes sense once you look at how faint the signal is and how loud the surface is; the ratio between them turns out to be the entire design problem.

Start with how many neutrinos there are, because that follows from something anyone can measure. The Sun runs on fusing four protons into a helium nucleus, releasing about 26.7 MeV and emitting two neutrinos — both standard recalled figures. Sunlight at the top of the atmosphere delivers roughly 1361 watts per square metre, or 0.1361 W/cm².

Divide one by the other. 26.7 MeV is 26.7 × 10⁶ × 1.602 × 10⁻¹⁹ = 4.28 × 10⁻¹² joules, so the number of helium nuclei made per second per square centimetre of sky at our distance is 0.1361 / 4.28 × 10⁻¹² ≈ 3.2 × 10¹⁰. Two neutrinos apiece gives about 6.4 × 10¹⁰ solar neutrinos crossing every square centimetre of you, every second — a figure derived from the sunlight, not looked up.

Now ask how often one interacts. The weak cross-section at these energies is of order 10⁻⁴⁴ cm² per target — a recalled order of magnitude that rises steeply with neutrino energy, so treat it as a scale. Water holds about 3.3 × 10²² molecules per cm³ with ten electrons each, so roughly 3.3 × 10²³ electron targets per cm³. The mean free path is one over density times cross-section: 1 / (3.3 × 10²³ × 10⁻⁴⁴) ≈ 3 × 10²⁰ cm, or 3 × 10¹⁸ metres — some three hundred light-years of water.

That number settles the design. You cannot make a neutrino stop; you can only put so many targets in front of it that the multiplication works out. A kiloton of water holds of order 10³² molecules, and a huge flux times a minuscule probability lands, for the detectable high-energy minority of solar neutrinos, at a handful of events per day. Not per second. Per day.

Now look at what else arrives. Cosmic rays striking the upper atmosphere produce muons, and at sea level the flux is famously about one per square centimetre per minute — another recalled standard figure, near enough 170 per square metre per second. A detector presenting 100 m² to the sky therefore takes some 1.5 × 10⁹ muon crossings a day. Against a signal of a few.

So the problem is a ratio of roughly one in a billion, and the question becomes what depth buys. Two things, and the second justifies the mine.

The first is obvious. Rock absorbs muons and does nothing to neutrinos, so every metre of overburden improves the ratio for free. A kilometre of rock at about 2700 kg/m³ is 2.7 × 10⁶ kg per square metre — 2700 metres of water equivalent — and the muon flux falls very roughly by a decade per thousand of those metres, faster near the surface.

The second is statistical. Background counts fluctuate: a mean of B events comes with a random scatter of about √B, and a signal of S events is only convincing if it exceeds that scatter, so what matters is S/√B. Cut B by 10⁶ and the significance improves by √10⁶ = 10³ — with no change to the detector, the target mass, or the signal at all. That square root is the whole argument for digging.

And it has a test attached. The claim is that the residual events underground are cosmic-ray muons, not something intrinsic to the apparatus. If so, their rate must depend on the slant depth — the rock along the direction each one arrived from — so a detector under an irregular mountain should record more from the thin directions and fewer from the thick, tracing the profile of the rock above it. That anisotropy is observed. If the residual rate were isotropic, or independent of overburden, the background account would be wrong and something else — detector noise, radioactivity in the materials — would be doing it.

Three refinements tie the reasoning together.

First, this is the same logic a chemist uses in reporting a limit of detection, but with the levers reversed. In a laboratory you can usually reduce the blank and enlarge the signal by concentrating the sample. Here you can do neither, so every available lever acts on the background — which is why the field's engineering effort looks so lopsidedly defensive.

Second, depth is not a single number. What counts is metres of water equivalent — mass per unit area along the path — so a kilometre of dense rock shields far better than a kilometre of ice, and direction matters as much as nominal depth.

Third, depth alone never suffices. Beneath the muons lies a second tier: neutrons knocked out of the rock by the muons that do get through, and gammas from natural radioactivity in the surrounding material. These are fought with veto layers flagging anything that crosses the outer shell, with ultrapure water and copper, and with signatures a background cannot fake — a flash pointing back at the Sun, a burst coincident with a supernova.

• Why detectors hunting very high-energy astrophysical neutrinos use the whole Earth as a filter and look downward, at particles that came up through the planet.
• How the deficit in the measured solar neutrino rate, long blamed on the detectors, turned out to be a property of the particle itself.

Every term the collection defines is gathered in the glossary.