Part 2
Why a Cubic Kilometre
IceCube is the size it is because of an arithmetic you can do yourself. Three numbers decide it: how likely a neutrino is to interact, how many of them arrive, and how long you are willing to wait. This part puts them together and arrives at a detector volume — and then at the number of events per year, which is the number that tells you whether the instrument was worth building.
Derivation: from a cross-section to a volume
Step 1 — Interaction probability
For a thin target, the chance of interacting while crossing a length L of material containing n targets per unit volume is:
\[ P = 1 - e^{-n\sigma L} \simeq n\,\sigma\,L \quad (n\sigma L \ll 1) \]
Step 2 — How many nucleons are in ice
Ice has density 0.92 g/cm³, and one gram of ordinary matter contains very nearly Avogadro's number of nucleons:
\[ n \approx 0.92\ \text{g/cm}^3 \times 6.02\times10^{23}\ \text{nucleons/g} \approx 5.5\times10^{23}\ \text{cm}^{-3} \]
Step 3 — The odds across a kilometre
With L = 1 km = 10⁵ cm, and the cross-section rising with energy:
| Energy | σ (cm²) | Chance of interacting in 1 km of ice |
|---|---|---|
| 1 TeV | 10⁻³⁵ | 1 in 1,800,000 |
| 100 TeV | 2 × 10⁻³⁴ | 1 in 90,000 |
| 1 PeV | 10⁻³³ | 1 in 18,000 |
Even at a PeV — a thousand times the energy of an LHC proton — a neutrino crossing a kilometre of ice almost certainly does not notice it. That is the number the detector size has to compensate for.
Step 4 — How many arrive
The astrophysical neutrino flux is measured to follow roughly an E⁻² spectrum, conventionally quoted as the energy-squared-weighted flux:
\[ E^2\frac{d\Phi}{dE} \approx 10^{-8}\ \text{GeV}\,\text{cm}^{-2}\,\text{s}^{-1}\,\text{sr}^{-1} \]
Integrating an E⁻² spectrum above some threshold is easy, and that is part of why the convention exists:
\[ \Phi(>E_0) = \int_{E_0}^{\infty} \frac{10^{-8}}{E^2}\,dE = \frac{10^{-8}}{E_0} \]
With E0 = 100 TeV = 10⁵ GeV, that is 10⁻¹³ neutrinos per cm² per second per steradian. Across a square kilometre (10¹⁰ cm²) and the whole sky (4π sr), about 10⁻²·⁵ neutrinos pass through every second — and almost all of them keep going.
Step 5 — Multiply, and read off the year
\[ R = \Phi(>E_0);\Omega;A;P \]
Putting in 10⁻¹³ cm⁻²s⁻¹sr⁻¹, 4π sr, 10¹⁰ cm² and P = 1.1 × 10⁻⁵:
\[ R \approx 1.4\times10^{-7}\ \text{s}^{-1} \approx 4\ \text{events per year} \]
That is the whole argument. A cubic kilometre of ice, watched for a year, yields a handful of astrophysical neutrinos above 100 TeV. Build something ten times smaller and you get one event every couple of years — not an observatory, an anecdote. Build the kilometre and within a few years you have a population you can do statistics on.
Why ice, specifically
Nothing above says ice. The target could be water, and other experiments use the Mediterranean and Lake Baikal. What Antarctic ice supplies is everything around the arithmetic: it is already there in cubic-kilometre quantities, it is optically clear at the depths that matter, it is dark, it does not move, and it contains almost no radioactivity or living things to glow.
It has one serious drawback: the sensors are frozen in permanently. Nothing can ever be repaired, upgraded or recovered. Every design decision had to be right the first time, for twenty years.
What this calculation leaves out
It treats the detector as a perfect square kilometre that registers every interaction, and the Earth as transparent. Neither is true: the real quantity is an effective area that depends on energy, direction and event type, and above a few PeV the Earth becomes opaque to neutrinos, so the highest-energy events can only arrive from near the horizon. The estimate lands within a factor of a few of what IceCube reports, which for a calculation this crude is the point rather than a coincidence.