The mass that is missing

Two protons and two neutrons, weighed separately, come to 4.0319 atomic mass units. A helium-4 nucleus, weighed whole, comes to 4.0015. The difference is 0.0304 units — three-quarters of one per cent — and it is not a measurement error: it is larger than the uncertainty by a factor of about ten thousand, and it has been known since Aston’s mass spectrograph of 1919.

The missing mass is the binding energy. Taking the nucleus apart would require 28.3 MeV of work, and a bound system weighs less than its parts by exactly that energy divided by $c^2$.

Everything about nuclear energy is in the shape of that curve, and the shape comes from a competition between two effects.

Why the curve has a peak #

The nuclear force is short-ranged: a nucleon attracts only its immediate neighbours. So the binding energy is roughly proportional to the number of nucleons — each contributes the same amount — which gives a constant binding energy per nucleon and a flat curve. Two corrections bend it.

Nucleons on the surface have fewer neighbours. A nucleus of mass number $A$ has a radius proportional to $A^{1/3}$ and a surface proportional to $A^{2/3}$, so the fraction of nucleons on the surface goes as $A^{-1/3}$. Small nuclei are nearly all surface and are therefore under-bound, which is why the curve rises steeply at the light end. This is the same term that makes a small droplet cost more energy per unit volume than a large one, and the nuclear model that uses it is called the liquid drop model for exactly that reason.

Every proton repels every other. The Coulomb energy goes as $Z^2/A^{1/3}$, and since $Z$ grows roughly with $A$, the repulsion per nucleon grows as $A^{2/3}$. It is negligible for light nuclei and dominant for heavy ones, which is why the curve falls at the heavy end and why there is no stable nucleus beyond bismuth.

One term pulling the curve down at small $A$ and another pulling it down at large $A$ must produce a maximum in between, and it lands at $A \approx 58$. The peak’s existence is a consequence of two effects with opposite size-dependence, and its position is where they cross.

The five terms, and what each one is doing #

Written out, the semi-empirical mass formula is five terms and one line:

$$B(A,Z) = a_v A - a_s A^{2/3} - a_c\frac{Z(Z-1)}{A^{1/3}} - a_a\frac{(N-Z)^2}{A} + \delta,$$

with $a_v \approx 15.8$, $a_s \approx 17.8$, $a_c \approx 0.71$ and $a_a \approx 23.7$ MeV, and $\delta$ a pairing term of about $\pm 11/\sqrt{A}$.

The first two are the drop: volume and surface. The third is electrostatics, and it is the only term in the formula that comes from a force anybody had written down before nuclei were discovered. The fourth is the asymmetry term, which penalises any excess of neutrons over protons or the reverse; it has no classical analogue at all and is a consequence of the exclusion principle, since filling the neutron levels far above the proton levels costs kinetic energy for the same reason stacking fermions does anywhere else. The fifth is pairing: nuclei with even numbers of both are more tightly bound than odd-odd ones by about 2 MeV, which is why of the 288 naturally occurring nuclides, 166 are even-even and only four are odd-odd.

The competition between the third and fourth terms is what fixes the shape of the valley of stability. Coulomb repulsion prefers fewer protons; the asymmetry term prefers equal numbers. Minimising their sum at fixed $A$ gives a proton fraction that starts at one half for light nuclei and falls steadily — uranium is 39 per cent protons — which is why heavy elements are neutron-rich and why fission fragments, inheriting that neutron excess but landing at a mass where less of it is wanted, are radioactive and emit neutrons. Delayed neutrons from exactly that mismatch are what make a reactor controllable on a timescale of seconds rather than microseconds.

Both directions downhill #

A peak in the middle means that light nuclei joined and heavy nuclei split both move up the curve, and moving up releases the difference.

Fusing four hydrogen nuclei into helium climbs from zero to 7.07 MeV per nucleon, releasing 28.3 MeV for four particles — about 0.7 per cent of the rest mass. Splitting uranium-235 into fragments near mass 100 climbs from 7.59 to about 8.5, releasing roughly 0.9 MeV per nucleon, or about 200 MeV per fission — about 0.09 per cent of the rest mass.

So fusion is nearly eight times more productive per kilogram, which is a fact about the steepness of the curve at the two ends and nothing else.

That figure is the answer to why the mass change went unnoticed in chemistry for so long. Lavoisier’s balance would have had to detect one part in three billion; the best balances of his era managed one part in ten thousand. Conservation of mass was not a wrong law but a very good approximation, and the correction is the same size as the correction to the mass of a compressed spring or a hot cup of tea.

Why the release is not automatic #

A slope is not a rate. Both fusion and fission are downhill and neither happens spontaneously in ordinary matter, because both have a barrier in front of them.

Nothing about an energy difference says how fast a transition happens, or whether it happens at all. A barrier between a state and a lower one is the general shape of the obstacle, and the whole of the next section is about what gets a nucleus over or through it — because the binding-energy curve says only that the destination is lower, and that is a statement about thermodynamics rather than about rates.

For fusion, the barrier is the Coulomb repulsion between two positively charged nuclei, which must be overcome before the short-ranged attraction can act — a barrier that decides the rate and not the direction. For two deuterons the barrier is a few hundred keV, and thermal energies at the centre of the Sun are about 1.3 keV — a factor of a couple of hundred short.

A thermal distribution has a tail, and the particles that react are drawn from it rather than from the average. In the Sun’s core the mean thermal energy is about 1.4 keV and the Coulomb barrier is some hundreds of times higher — so nothing reacts at the average energy, and the rate is set entirely by how many particles are far out in the tail.

Nuclei do not have to clear the barrier at all: transmission through it falls exponentially with width, and that exponential is what makes stellar fusion possible at temperatures far below the classical requirement. Gamow’s calculation combines the two exponentials — one falling with energy and one rising — and their product peaks at an energy neither would pick alone.

The rate carries an exponential of the barrier over $kT$, so a modest change in temperature changes it by orders of magnitude. Stellar fusion is therefore a thermostat: too hot and the rate runs away until expansion cools it, too cold and it nearly stops. The steepness is what makes a star stable rather than what makes it fragile.

For fission, the barrier is the surface tension of the nuclear drop, which resists being deformed into two pieces. A uranium-235 nucleus that absorbs a neutron gains about 6.5 MeV of binding energy on the spot — enough to clear its own 6.2 MeV barrier — and splits within $10^{-14}$ seconds. Uranium-238 absorbing the same neutron gains only 4.8 MeV, which is not enough, and requires a fast neutron carrying the difference. That one comparison of two numbers is the whole of why one isotope is a reactor fuel and the other is not.

The measurement, and what it settles #

The numbers behind all of this are obtained by firing a beam at a target and measuring how often something happens, expressed as an effective area. Nuclear cross-sections at stellar energies are minute and the measurements are made at higher energies and extrapolated down — which is the dominant uncertainty in stellar models, and the reason underground accelerators were built to push the measurements lower.

The binding-energy curve is not a theoretical construction with data laid over it. The measured points come from mass spectrometry, which weighs nuclei against one another to parts in $10^9$, and the curve is a five-parameter fit whose parameters are adjusted to those measurements. Its value is that five numbers reproduce two thousand nuclear masses to within about one per cent, which is a strong statement about how little of the nucleus’s structure matters for its energy.

A nucleus with a lower binding energy than its neighbours has somewhere to go, and it goes there at a rate the barrier sets. That is what happens to the nuclei the curve says are not at the peak: they decay, on timescales from nanoseconds to longer than the age of the universe, and the curve says nothing about which. Position on it decides the direction; the barrier decides the wait.

Aston, and the number he almost had #

The measurements came first and the interpretation took a decade.

Francis Aston built a mass spectrograph in 1919 that could separate isotopes and weigh them against one another to about one part in a thousand, later improved to one in ten thousand. What he found was that nuclear masses are very nearly whole multiples of the hydrogen mass, but not exactly — and he defined a “packing fraction” to record the departure, plotting it against mass number in 1927. The curve he published is the binding-energy curve upside down.

He drew the physical conclusion in his Nobel lecture: the departure meant that assembling helium from hydrogen would release energy, and he estimated the amount correctly. Eddington had already used the same idea in 1920 to propose that stars are powered by hydrogen fusion, against the objection that stellar temperatures were far too low to overcome the Coulomb barrier — an objection that was arithmetically correct and was answered only when Gamow’s tunnelling calculation arrived in 1928.

What makes the sequence worth recording is its order. The energetics were measured before any mechanism was known; the mechanism was refused on grounds that were valid within the physics of the day; and the resolution came from a quantum effect that had nothing to do with either the measurement or the objection. A measured energy difference constrains what is possible without saying anything about what is available, and the gap between those two questions was, in this case, eight years wide.

Where ordinary mass actually comes from #

The helium nucleus weighs less than its parts by three-quarters of a per cent. Ask the same question one level down — how much a proton weighs compared with what it is made of — and the answer runs the other way and is much larger.

A proton contains three valence quarks. Their masses, as they appear in the theory, come to about nine million electronvolts between them. A proton weighs 938. So roughly ninety-nine per cent of a proton’s mass is not the mass of anything inside it; it is the energy of the gluon field binding the quarks together and the kinetic energy of the quarks confined in a region a femtometre across, converted at the going rate.

The sign is opposite to the nuclear case and the reason is instructive. A nucleus weighs less than its parts because pulling it apart costs energy and the parts, once separated, are free. A proton weighs more than its parts because separating quarks costs energy without limit — the force between them does not fall off with distance — so there is no state of free quarks to compare with. What is being measured is a confined system whose energy is dominated by the confinement, and confinement adds.

The size of that claim is worth stating plainly. Almost all of the mass of every ordinary object — a person, a planet, a star — is this. The Higgs mechanism, which is what gives the quarks and electrons the masses they have, accounts for about one per cent of the mass of an atom and for all of the electron’s. Turning the Higgs coupling off would leave a proton weighing very nearly what it does now.

That is not a speculative statement. Lattice calculations compute the nucleon’s mass from the theory, with the quark masses as inputs and nothing else fitted, and they reproduce the measured value to a few per cent — a calculation that took thirty years and a great deal of computing and which is the direct confirmation that the mass is where this section says it is.

Two smaller cases put the nuclear number in perspective from the other direction. The hydrogen molecule is bound by 4.5 electronvolts, so it weighs about two parts in a thousand million less than two hydrogen atoms. And the Earth–Moon system, bound gravitationally, weighs about four hundred million tonnes less than the Earth and the Moon weighed separately — seven parts in a hundred million million of the total, and a mass defect in exactly the same sense as helium’s.

None of those is a different phenomenon. There is one relation, it applies to every bound system, and the only thing that varies is how large a fraction of the total the binding is.

The neutron that decays alone and not in company #

The binding curve has a consequence for stability that is worth spelling out, because it inverts the usual way of thinking about which particles are stable.

A free neutron is unstable. It weighs 939.565 MeV, and a proton plus an electron weigh 938.783 between them, so there is 0.78 MeV to spare and the neutron decays into them with a half-life of about ten minutes.

A neutron inside a helium nucleus, or a carbon nucleus, or an oxygen nucleus, does not decay at all, and has not in the age of the universe. Nothing about the neutron has changed. What has changed is what the decay would produce: turning one of carbon-12’s neutrons into a proton gives nitrogen-12, which sits some seventeen million electronvolts higher in total energy than carbon-12 does. The 0.78 MeV available is nowhere near enough, so the decay is forbidden by energy conservation and the neutron is stable indefinitely.

The reverse case is equally sharp and less expected. A free proton is stable — it is the lightest baryon and has nothing to decay into. A proton bound in a proton-rich nucleus is frequently not: it converts into a neutron, either by emitting a positron or by capturing one of the atom’s own electrons, because the daughter nucleus is more tightly bound and the difference pays for the conversion.

So neither particle has a stability that belongs to it. What is stable or unstable is the whole system, and the question asked is always the same one this essay is about: is there a state of lower total mass that conservation laws allow it to reach?

That reframing is worth carrying, because it removes an apparent puzzle. Nothing about a nucleon is altered by being inside a nucleus; what is altered is the arithmetic of the comparison. And it explains the neutron-rich edge of the chart of nuclides, where adding one more neutron eventually produces a nucleus in which that neutron can decay — which is exactly where the line of stability ends.

Where the model stops #

The liquid drop is a caricature. It gets the smooth trend and misses everything structural. Nuclei with 2, 8, 20, 28, 50, 82 or 126 protons or neutrons are bound about 1 to 2 per cent more tightly than the formula predicts — the magic numbers, which are shell closures and require a quantum treatment of the individual nucleons rather than a drop.

The formula fails at both ends. For very light nuclei there are not enough nucleons for a surface term to mean anything: the formula gives deuterium a binding energy several times its measured 1.1 MeV per nucleon, which is why that point sits so far below the curve in the first figure. For very heavy nuclei the assumption of a spherical drop fails, since deformation is exactly what fission is.

Nothing here is a rate. The curve says which way is downhill and how far. Whether anything moves is a separate calculation involving barriers, cross-sections and available energy, and the two questions are related only in that the first has to be answered before the second is worth asking.

Weighing an energy, as an instrument #

The equivalence is more useful as a measuring technique than as a source of power, and the reason is that a mass can be compared far more precisely than an energy can be counted.

A modern Penning trap compares the cyclotron frequencies of two ions in the same magnetic field, and frequencies are the most precisely measurable quantities there are. Mass ratios come out to parts in $10^{11}$, which converts to binding energies good to a few hundred electronvolts on a nucleus bound by hundreds of MeV. That is precise enough to weigh a chemical bond in principle, and precise enough in practice to test the equivalence directly: comparing the mass difference between a nucleus before and after neutron capture against the energy of the gamma ray emitted, measured independently by crystal diffraction, agrees to about four parts in ten million. It is one of the sharpest tests the relation has.

The same technique settles arguments that no amount of theory could. Whether a particular nucleus can decay by a given route is a question about a mass difference of a few keV out of tens of GeV, and the answer decides whether an isotope is a candidate for a neutrinoless double beta decay experiment or is simply forbidden. The whole of that experimental programme rests on a weighing.

What the pictures cannot show #

The binding curve is drawn as a smooth function of mass number, and nuclear mass is not a continuous variable. Each integer $A$ has several isobars with different proton numbers and different binding energies, and the curve shows only the best of them — so the real data is a scatter of points at each abscissa, of which one is plotted.

The energy landscape figure is a caricature with one coordinate, and a fissioning nucleus moves through a space of deformation parameters with several. The barrier’s height depends on which path is taken through that space, and the one-dimensional picture cannot show that some paths are cheaper.

And nothing here shows the strong interaction. Every statement about binding is thermodynamic: energies in, energies out, and a fit. The force responsible does not appear in the formula at all — its short range appears as a volume term, its saturation as the constancy of that term, and the actual interaction between two nucleons is nowhere.

Where the ladder goes next #

This ladder began with mass and energy being one quantity, continued with the invariant that survives a change of frame, and has now used the equivalence as an instrument: a weighing that reveals an energy. The rungs beyond it are the shell model, which explains the departures the drop cannot; the valley of stability and what decides which way a nucleus decays off it; and the astrophysical question of where elements past the peak come from at all, since fusion cannot make them and something evidently did.

The habit worth carrying away is about what a defect is. Nothing has been lost from a helium nucleus — no matter was destroyed and none converted. What happened is that a bound system was assembled, and the energy that came out took its mass with it. A binding energy is not a subtraction from the parts; it is the weight of what left, and the same statement applies exactly, if unmeasurably, to a molecule, a box with light bouncing about inside it, a planet in orbit and a book on a shelf.