How far a neutrino gets

The mean free path of a molecule of air at room temperature is about seventy nanometres. The mean free path of a neutrino in solid lead is about a light-year and a half. These two numbers come from the same formula, with the same structure and the same derivation, and the entire difference between them is one factor.

$$\lambda = \frac{1}{n\sigma}.$$

Two numbers go into that formula and only one of them is remarkable. The number density of targets varies over a modest range in ordinary matter — from about $10^{25}$ per cubic metre in air to about $10^{31}$ nucleons per cubic metre in a heavy solid, six decades in all. The cross-section varies over nearly thirty. So in any comparison of this kind it is the cross-section that decides, and the density is a detail; which is a useful thing to know before starting, because the instinct is to reach for density first.

The formula, and why it does not care what is flying #

The gas rung of this ladder derived the free path by drawing a straight line through a field of targets and asking how far it goes before it hits one. The derivation used nothing about molecules: a particle sweeping out a tube of cross-sectional area $\sigma$ encounters, on average, one target when the tube’s volume has grown to hold one — that is, when $n\sigma\ell = 1$.

That is why the law travels. It is a statement about a straight line through a random field, and the physics enters through exactly two numbers: how many targets there are per cubic metre, and how big each one is for the process in question.

A cross-section is not a size #

The second of those numbers is the one that carries all the surprise, and the word for it is misleading.

For hard spheres, the cross-section really is a geometrical area. For anything else it is a probability wearing the units of area: define it as the rate of interactions divided by the incident flux, and it comes out in square metres whether or not anything of that size exists. A cross-section can be far larger than the target — a slow neutron at a resonance sees a nucleus thousands of times bigger than its geometry — and far smaller.

For a neutrino at a few MeV interacting with a nucleon, the cross-section is around $10^{-47}$ m². The nucleon’s geometrical area is about $10^{-30}$ m². The ratio is seventeen orders of magnitude, and it is a statement about how feebly the weak interaction couples rather than about anything’s size.

Where the smallness comes from is worth one paragraph, because it is not arbitrary. The weak interaction is carried by particles about ninety times as heavy as a proton, and a heavy carrier makes an interaction short-ranged and feeble at low energies: the amplitude picks up a factor of $1/M^2$ from the carrier’s mass, and the cross-section, being an amplitude squared, picks up $1/M^4$. That fourth power of a large mass is the whole of the number. The interaction is not weak because it is somehow reluctant; it is weak because its carrier is heavy, and at energies large enough to make the carrier’s mass irrelevant it stops being weak at all.

The scaling is worth carrying, because the number is not a constant. At these energies $\sigma$ grows roughly as the square of the neutrino’s energy, so a neutrino a thousand times more energetic is a million times easier to stop. Everything about detection follows from that: high-energy neutrinos are comparatively easy, and the low-energy ones that are produced in vastly greater numbers are the hard case.

A light-year of lead #

Put the numbers in. Lead has $3.3\times10^{28}$ atoms per cubic metre and 207 nucleons in each, so $n = 6.8\times10^{30}$ nucleons per cubic metre. With $\sigma = 10^{-47}$ m²,

$$\lambda = \frac{1}{6.8\times10^{30} \times 10^{-47}} = 1.5\times10^{16}\ \text{m},$$

which is 1.55 light-years. The figure computes it rather than repeating it, which matters because the number is usually quoted with no density attached and is meaningless without one.

The number deserves one more comparison, because a light-year is not a quantity anybody has intuition for. A metre of lead is about $6\times10^{-17}$ of a free path. A thickness equal to the distance from the Earth to the Sun is about $10^{-5}$ of one. To stop half of a beam of low-energy neutrinos requires roughly a light-year of solid lead, and to stop all but a thousandth requires about fifteen.

There is a corollary that is easy to state and worth stating. Anything that interacts this feebly is also produced this feebly, in the sense that the same coupling governs both — so a process that emits neutrinos loses energy to them slowly, unless it involves an enormous number of reactions. Where a great many reactions are involved, the feebleness reverses its role entirely: neutrinos are then the fastest way for a dense region to lose energy, because everything else is trapped and they are not.

What survives a slab #

The free path is a mean, and what a detector cares about is the fraction of a beam that gets through a given thickness. That is the same exponential that governs every process where each step is independent of the last.

For thin targets the exponential is a straight line: the probability of an interaction is $L/\lambda = n\sigma L$, which is simply the fraction of the slab’s area that the targets block. A metre of lead is $10^{-16}$ free paths, so it stops about one neutrino in $10^{16}$.

What survives a slab falls exponentially with its thickness, and the reason is worth separating from the reason a sample decays exponentially with time. Here each unit of distance carries the same chance of an interaction, so the surviving fraction falls by a constant factor per unit length. There the same statement is made about waiting rather than travelling. The two produce identical curves from different variables, and running them together is how “half-life” ends up misapplied to shielding.

How something that never interacts was found anyway #

If the probability per particle is $10^{-16}$, the way to see an interaction is to have vastly more than $10^{16}$ particles. That is the whole strategy, and it is worth spelling out because it converts an impossible experiment into an engineering problem.

The flux from a nuclear reactor at a few metres is of order $10^{17}$ neutrinos per square metre per second. Reines and Cowan, in 1956, put a target of a few hundred litres of water beside a reactor and looked for a coincidence: a positron annihilation followed a few microseconds later by a gamma cascade from a neutron capture on dissolved cadmium. That double signature, separated by a known delay, is what made a handful of events per hour distinguishable from a background of everything else.

The design principle in that experiment is the one every detector since has used. A rate of a few per hour cannot be extracted by being sensitive; it has to be extracted by being specific. A signature that background cannot imitate — two events, in the right order, at the right separation, in the right place — turns a hopeless signal-to-noise ratio into a countable one.

The scale of the compensation is worth writing out. Reines and Cowan’s target held about $10^{28}$ protons; the flux was $10^{17}$ per square metre per second across a target of a few square metres; the probability per proton per neutrino is around $10^{-44}$. Multiplying, the expected rate is a few events an hour, which is what they saw. Every factor in that product is extreme in one direction or the other, and the experiment works because the two extremes are of comparable size.

That arithmetic also explains why the detectors have grown the way they have. Doubling the sensitivity means doubling the target, so the mass has climbed from Reines and Cowan’s few hundred kilograms to tens of thousands of tonnes, and the more recent designs stop building a target at all and instrument a volume of something that was there already — a lake, a block of ice, the sea. The instrument is the only one in physics whose design principle is to find a large volume of transparent matter and put photomultipliers around it.

The neutrino had been proposed by Pauli in 1930 for a completely different reason: beta decay’s electrons emerge with a continuous range of energies, and a two-body decay cannot do that. Something invisible was carrying away the balance. Pauli’s own comment was that he had done a terrible thing in postulating a particle that could not be detected, and he was very nearly right about the second half.

The opposite regime, where the free path is the short quantity #

It is worth turning the same formula the other way, because the interesting physics is at both ends and only one end is famous.

When $\lambda$ is very short compared with the size of the region, a particle does not travel through — it random-walks. The number of steps needed to cross a distance $L$ goes as $(L/\lambda)^2$ rather than $L/\lambda$, which is the diffusion result and is an enormous factor when the ratio is large. A photon in a dense medium takes not $L/\lambda$ steps but the square of it, and the escape time is correspondingly longer.

The dimensionless ratio $L/\lambda$ — the number of free paths across the system — decides which regime applies, and it is given its own name in every field that meets it: optical depth in radiative transfer, Knudsen number in gas dynamics, and the mean-free-path-to-size ratio in a detector. When it is small the medium is transparent and the exponential is nearly one; when it is large the medium is opaque and the transport is diffusive. A neutrino in lead sits at $10^{-16}$, which is about as far into the transparent regime as anything in physics gets.

That is the sense in which this ladder’s five rungs are one subject. The gas rung asks what happens when the ratio is around one and the continuum picture is beginning; this rung asks what happens when it is $10^{-16}$; and the arithmetic in between is unchanged.

Both regimes, in one collapsing star #

The two ends of this essay — transparent when $n\sigma L$ is tiny, diffusive when it is large — were seen in the same object within thirteen seconds of each other.

When a massive star’s core collapses, essentially all of the energy released leaves as neutrinos. Their free path in the material just outside the collapsing core is enormous, so they escape the moment they are made. The photons do not: the star’s envelope is thoroughly opaque, and the shock takes hours to reach the surface and make it bright.

So the neutrinos arrive first. In February 1987 three detectors on two continents recorded a couple of dozen events within a few seconds of one another, and the visible star brightened about three hours later. Two messengers, one event, a hundred and sixty thousand light-years of travel, and the order of arrival decided entirely by which of them had the shorter free path in the last few hundred million metres of the journey’s start.

The duration of the burst is the other regime, and the arithmetic closes neatly. Inside the newly formed neutron star the density reaches some $10^{17}$ kilograms per cubic metre, so the nucleon number density is about $10^{44}$ per cubic metre — thirteen orders above lead — and at the energies involved the free path falls to about ten metres. Against a core radius of ten kilometres that is a thousand free paths, so the neutrinos are trapped, and they leave by diffusion rather than by flight, taking $(L/\lambda)^2$ steps.

That predicts an escape time of a few seconds. The observed burst lasted about thirteen. A particle whose whole reputation is for passing through everything spends ten seconds diffusing out of an object ten kilometres across, because the one factor this essay is about has moved by thirteen decades.

Making the Earth opaque on purpose #

The cross-section grows with energy, and pushing it far enough turns the argument round: there is an energy at which the Earth stops being transparent to neutrinos.

The crossover is around a few tens of teraelectronvolts, where the interaction length falls to the planet’s diameter. Above that, a neutrino arriving from below has a substantial chance of being absorbed on the way through, and the higher its energy the worse its chances.

That makes the Earth’s own absorption a measurement. Neutrinos produced in the atmosphere all over the world arrive at a detector from every direction; those coming from directly below have crossed the whole planet, those from the horizon have crossed almost none of it, and the deficit as a function of arrival angle and energy reports how much matter lies along each path. Inverting it gives the density profile of the Earth.

It has been done, with a cubic kilometre of Antarctic ice instrumented as a detector, and the answer agrees with what seismology has said for a century: a dense core, a lighter mantle, and the total mass of the planet recovered to within the measurement’s accuracy.

Two entirely independent probes of the same interior — sound waves from earthquakes, and the absorption of particles that were famous for not being absorbed. The second is possible only because $\sigma$ is a function of energy, and the essay’s headline number is the value at one end of a very long curve.

What it costs, and where the model stops #

The cross-section is not a constant. Everything above uses a fixed $\sigma$, and the real one depends on energy, on the target, and on which of several processes is involved. At a few MeV it rises as $E^2$; at very high energies the rise flattens; at particular energies resonances appear. A free path quoted without an energy is incomplete.

The independence assumption can fail. The exponential requires each target to act on its own. When the wavelength associated with the incident particle is long compared with the spacing of the targets, they scatter coherently and the cross-section per nucleus grows as the square of the number of nucleons rather than in proportion to it. Coherent elastic neutrino–nucleus scattering, predicted in 1974 and measured in 2017, is exactly this effect, and it gives cross-sections a hundred times larger than the naive sum.

The targets are assumed to be at rest and randomly placed. Neither is exactly true. Motion of the targets modifies the answer by a factor of order one — the $\sqrt2$ the gas rung derives — and any ordering of the targets, as in a crystal, replaces the random-field argument with a diffraction problem in which the answer depends on direction. Neutrons in a single crystal exploit exactly that, and the free path along a crystal axis is not the free path across it.

A mean free path is a mean. The distribution of actual path lengths is exponential, so a substantial fraction of particles travel much less than $\lambda$ and a substantial fraction much more. Quoting the mean as though it were a range is the same error as quoting a half-life as a lifetime.

Neutrinos change identity in flight. They exist in three flavours that mix, so a beam that starts as one flavour arrives as a mixture, and a detector sensitive to only one type sees a deficit that has nothing to do with cross-sections. Two decades of a “missing neutrino” problem turned out to be this rather than a failure in any of the arithmetic above.

Nothing here is a claim about what is out there. This is the physics of a cross-section and a number density. What is inferred from neutrinos arriving from a distance — and a great deal is, precisely because they are not stopped — belongs to the collection that measures things in the sky.

Why the number is worth carrying about #

There is a habit this rung is meant to install, and it is worth stating plainly because it generalises past neutrinos.

Whenever a question of the form “does this get through” comes up, the useful move is not to reason about the mechanism but to compute $n\sigma L$ and see whether it is large or small. That single dimensionless product decides everything: below one, the medium is transparent and the interaction rate is simply proportional to the thickness; above one, the medium is opaque and the physics is transport rather than transmission. The mechanism enters only through $\sigma$, and it enters as one number.

The habit is cheap and unusually reliable. It answers why a sheet of paper stops alpha particles and a metre of concrete does not stop neutrons; why a gas at low pressure conducts electricity differently from one at high pressure; why an X-ray sees through flesh and not through bone. In each case the answer is a product of three quantities, two of which are geometry.

What makes the neutrino the memorable case is that its $\sigma$ is so far outside the range that the product stays microscopic even when $n$ and $L$ are made as large as anything permits. That is worth carrying as a calibration: $10^{-47}$ m² is what it takes to make a light-year of lead transparent, so any cross-section within twenty orders of magnitude of an atomic one is, for practical purposes, enormous.

The ladder from here #

Later rungs on this anchor: the energy dependence of the weak cross-section, and what its growth says about the interaction’s structure; coherent scattering, where the independence assumption breaks in the useful direction; the optical depth as the same quantity used backwards, counting free paths through a medium rather than metres; scattering as a random walk when the free path is short, which is the diffusion problem with a different name; and neutrino oscillation, in which the particle’s identity is not conserved along the path.

The neighbouring ladders are the gas rung, which is this arithmetic at nineteen orders of magnitude larger cross-section, and radioactive decay, whose exponential is the same independence assumption applied to time instead of distance.