The nucleus that would make positrons from nothing

The X-ray line that counts the protons found that an atom’s innermost electrons behave like hydrogen’s electron with the nuclear charge turned up, and that their energies grow as the square of that charge, with relativity pulling them further as the charge rises. It ended on a question that relativity itself raises: whether there is a charge at which the counting stops.

For a single electron round a point nucleus of charge $Ze$, Dirac’s equation gives the energy of the innermost, 1s, state exactly:

$$E_{1s} = mc^2\sqrt{1 - (Z\alpha)^2},$$

where $\alpha \approx 1/137.036$ is the fine-structure constant. For hydrogen $Z\alpha$ is small, and the formula is the rest energy less 13.6 electronvolts. For heavier nuclei the square root falls faster and faster, and at $Z\alpha = 1$ — a nucleus of 137 protons — it reaches zero with a vertical slope. Beyond that the formula gives the square root of a negative number. Richard Feynman is supposed to have said that the periodic table must end there, and element 137 is sometimes called feynmanium on the strength of it.

The argument usually given for that is older than Dirac. In Bohr’s model the electron in the innermost orbit of a nucleus of charge $Z$ moves at $Z\alpha$ times the speed of light, so a nucleus of 137 protons would need an electron moving at the speed of light, and one of 138 an electron moving faster. Bohr’s model is wrong about almost everything, but its speeds are right in order of magnitude, and Dirac’s exact answer fails at the same place for a related reason. Both say that something must change before the 138th proton. Neither says what.

The level that does not stop #

The failure is real, but it belongs to the point nucleus. An electron in the 1s state of a heavy atom spends part of its time inside the nucleus, and for $Z$ near 137 that part is large. A point charge has a potential that goes to minus infinity at the centre; a real nucleus of radius about 7 femtometres for uranium spreads its charge through its volume, and the potential inside is finite. Isaak Pomeranchuk and Yakov Smorodinsky pointed out in 1945 that this changes the answer, and Walter Greiner’s group in Frankfurt, and Yakov Zeldovich and Vladimir Popov in Moscow, worked out the consequences in the late 1960s.

The solid curve is the same equation with the nucleus given a size — a uniformly charged sphere of radius $1.2A^{1/3}$ femtometres, with the mass number taken as two and a half times the charge — solved numerically by integrating the electron’s two radial wave functions outward and adjusting the energy until the solution falls off at large distances without crossing zero. For light and medium nuclei it lies on top of the point formula; the finite size shifts uranium’s innermost level by only a few hundred electronvolts out of a hundred and thirty thousand. Near 137 the two part. The point-nucleus curve plunges vertically and stops; the finite-nucleus curve keeps going smoothly, through zero at about 148, and on down until, at a charge of about 170, the electron is bound by twice its own rest energy and its total energy is $-mc^2$.

That number is not arbitrary. Below $-mc^2$ lie the negative-energy solutions of Dirac’s equation, the states that forced Dirac to propose in 1930 that they are all already occupied — a filled sea of electrons, whose holes would behave as particles of positive charge. The positron was found two years later, and the sea, reinterpreted in quantum field theory, became the vacuum. At $Z \approx 170$ the innermost bound level has fallen to the top of that sea.

What relativity was already doing #

The dive is the endpoint of a trend that is measurable long before it.

The non-relativistic answer, $(Z\alpha)^2mc^2/2$, grows as the square of the charge. Dirac’s grows faster, because an electron bound by a heavy nucleus moves at a large fraction of the speed of light — at about $Z\alpha$ times it, two-thirds for uranium — and its mass rises with its speed, pulling it closer to the nucleus and binding it harder. The metal that is yellow because it is heavy found the chemical consequence: gold’s inner electrons contract, its outer ones are screened differently, and the gap that sets its colour shifts into the blue. Here the effect is on the innermost electron directly, and it is large: seventeen kiloelectronvolts in uranium, a seventh of the binding.

The hydrogen-like uranium ion, U⁹¹⁺, with ninety-one of its electrons stripped away, has been made in storage rings and its 1s binding measured, by catching the photons emitted when a free electron drops into the empty level. The point-Dirac value, 132.3 kiloelectronvolts, is within half a kiloelectronvolt of the measurement. The remaining difference is the nucleus’s size, which raises the level by a few hundred electronvolts, and the radiative corrections of quantum electrodynamics, which raise it by a little more. Those are the same two effects that decide what happens near 170.

An electron wrapped round a nucleus #

As the level falls, the electron that occupies it shrinks towards the nucleus.

For tin, $Z = 50$, the innermost electron is found most often about one and a half picometres out, hundreds of nuclear radii from the surface; the nucleus is a point to it, which is why the point formula is so good. By $Z = 160$ the peak has moved to a hundred and thirty femtometres, and at 169 to under a hundred — a cloud only a few nuclear diameters across, with a large share of its probability inside the nucleus itself. That is why the shape of the nuclear charge matters near the critical point and not before. A distribution that holds the charge further out, in a larger or more diffuse nucleus, gives the electron a shallower well at the centre, and it then takes more charge to pull the level down to the sea.

Why an atom is the size it is found the size of hydrogen to be a balance between the electron’s confinement energy and its electrostatic attraction. The balance is the same here, but the confinement energy is relativistic: squeezing an electron into a region smaller than its Compton wavelength, 386 femtometres, costs energy that grows only linearly with the inverse size rather than as its square. The Coulomb attraction also grows linearly, so once the charge is large enough the attraction wins at every size — that is the classical collapse the orbit special relativity cannot close found at $Z\alpha = 1$ — and only the nucleus’s finite size stops the fall.

The sea gives up a positron #

Suppose a nucleus of charge above 170 existed, stripped of its electrons. Its innermost level would lie below $-mc^2$, inside the energy range of the filled sea. The level is then no longer a separate bound state but a resonance — a state with the same energy as some of the sea’s states and coupled to them. An electron from the sea can fall into the empty bound level, giving up no energy, and the hole it leaves behind in the sea is a positron, which is repelled by the nucleus and flies off with a kinetic energy roughly equal to how far the level has dived. The process happens spontaneously, with no energy supplied from outside, because the final state — two electrons bound in the 1s level, two positrons gone — has lower energy than the bare nucleus. Once both 1s places are filled, the emission stops: the vacuum round the nucleus has acquired a charge of $-2e$, and it is stable.

The energy bookkeeping is worth doing explicitly, because it shows why $-mc^2$ is the threshold and not some other number. Making an electron and a positron out of nothing costs at least $2mc^2$, the rest energy of the pair, as mass is a form of energy insisted. If the electron is then placed in a bound level of total energy $E_{1s}$ — rest energy included — and the positron flies away with at least its rest energy $mc^2$, the final energy is $E_{1s} + mc^2$ or more. The vacuum’s energy is zero. The pair can appear for free exactly when $E_{1s} + mc^2 \le 0$, that is when the bound level has reached $-mc^2$. Above that the pair costs something, which a static nucleus cannot supply; below it the pair pays for itself, and the excess becomes the positron’s kinetic energy. In this accounting the dive is not mysterious at all: it is the point at which the binding energy of one electron reaches the rest energy of two.

The 1s level is only the first. The next level down, $2p_{1/2}$ — which in Dirac’s theory has exactly the same energy as $2s$ for a point nucleus, the coincidence the line that is really two found broken by quantum electrodynamics — dives too, at a charge of about 185, and higher levels follow at larger charges still. A nucleus well past the threshold would bind a growing number of electrons drawn from the sea, each adding one unit of negative charge to the vacuum and one positron to the world.

This is the most direct prediction there is of the reality of Dirac’s sea: a region of space with no particles in it decaying, in a strong enough field, into particles. It is a cousin of the effect the field nobody can transform away described for a uniform electric field strong enough to pull pairs out of the vacuum — Julian Schwinger’s critical field of about $10^{18}$ volts per metre. Near a supercritical nucleus the field at the surface is thousands of times stronger than that, but over a distance far shorter than the Compton wavelength a uniform field would need to do its work, and the mechanism is sharper: the binding of a discrete level, rather than a field acting over a distance, is what supplies the energy for the pair.

Two nuclei, for an instant #

No nucleus with 170 protons has ever been made, and the heaviest known, oganesson, has 118. The way round, proposed by Greiner’s group, is to bring two heavy nuclei close together. During a slow collision the electrons move much faster than the nuclei, so they see the two as a single quasi-molecule with the combined charge.

The figure keeps only the spherically averaged part of the two nuclei’s field, a charged shell of radius half their separation, which captures most of what decides the lowest level. Two uranium nuclei, 184 protons between them, pull the level below $-mc^2$ when they are within about 35 femtometres of each other — a distance they reach in a head-on collision at energies just below the barrier at which the nuclei touch. Lead on uranium has to come within about 19 femtometres, and lead on lead never gets there at all.

The trouble is time. In such a collision the nuclei spend about $10^{-21}$ seconds inside the critical distance. The spontaneous decay of the vacuum, if the level is empty, takes about a hundred times longer, so it has barely begun before the nuclei separate and the level rises back out of the sea. Meanwhile the rapidly changing field produces positrons on its own, dynamically, with a smooth spread of energies, and those swamp any signal from the vacuum. The clean signature would appear only if the two nuclei stuck together for a while before separating, which would add a narrow line at a well-defined energy to the smooth spectrum.

In the 1980s experiments at the GSI laboratory in Darmstadt, colliding uranium with uranium and with curium, reported exactly such narrow positron lines, near 300 kiloelectronvolts. They caused a sensation, and a decade of effort, because their energies did not depend on the combined charge in the way supercritical decay required, and some suggested new particles. Later experiments with better detectors, in Darmstadt and at Argonne in the 1990s, did not see them. The lines are now regarded as unconfirmed, and the spontaneous decay of the vacuum has not been observed in heavy-ion collisions.

A charge spread thin, and a charge spread wide #

The dependence on how the charge is distributed can be drawn directly.

As the nucleus shrinks, the well inside it deepens, and the critical charge falls towards the point-nucleus value; as it grows, the critical charge rises, and a sphere sixty femtometres across needs over two hundred protons. That is the same statement the colliding nuclei make: two uranium nuclei thirty femtometres apart are a charge of 184 spread over a large region, and they are just supercritical; pressed closer, they become more so.

The idea has been tested in a laboratory that has nothing to do with nuclei. In graphene, electrons near the Fermi level move as if they were massless relativistic particles with a speed three hundred times lower than light’s, and the effective fine-structure constant is about two rather than 1/137. A cluster of a few charged calcium atoms on a sheet of graphene is then supercritical, and in 2013 scanning-tunnelling measurements found the predicted collapse resonances appearing as the cluster’s charge was raised past the threshold. The arithmetic of the dive, with $\alpha$ replaced by graphene’s coupling, works at a charge of a few electrons.

What the table already shows #

The elements that do exist stop well short of the dive, but relativity is already reshaping their chemistry. The innermost levels of the heaviest atoms contract and deepen, the outer $s$ and $p_{1/2}$ electrons are pulled in with them, and the $d$ and $f$ electrons, screened more effectively, expand. The order in which shells fill — the order the order the shells fill explained by screening alone — starts to be rearranged. Copernicium, element 112, is predicted to be a volatile metal rather than a typical member of the zinc group, and oganesson, element 118, which sits below radon in the noble gases, is predicted not to behave like a noble gas at all: its outer electrons are smeared into a nearly uniform cloud by relativity, and calculations make it a solid semiconductor at room temperature. Only a handful of atoms of each have ever been made, each living for milliseconds or less, so these predictions have been tested only in single-atom chemistry experiments that confirm the trend without settling the details. The same equation that dives at 170 is already, at 112 and 118, deciding what kind of element an element is.

One electron, a uniform sphere and no vacuum polarisation #

The figures solve Dirac’s equation for one electron in the electrostatic field of a uniformly charged sphere, with the mass number fixed at two and a half times the charge. Real nuclei have a diffuse surface rather than a sharp edge, and their sizes at these charges are an extrapolation; the critical charge moves by a few units with the model of the nucleus. Quantum electrodynamics adds corrections of its own. Vacuum polarisation strengthens the field close to the nucleus and lowers the level; the electron’s self-energy raises it; together they shift the critical charge by about one unit. The other electrons of a neutral atom screen the nucleus and raise the level slightly too, though the innermost electron sits so deep that screening matters little. The collision figure keeps only the spherically averaged field of two nuclei and treats them as frozen at each separation; the true two-centre problem, and the nuclei’s motion, change the critical distance by a few femtometres. The domain is a single electron in a static field, nuclear charges up to about 230, and separations of two nuclei from fifteen to a hundred and twenty femtometres.

Still open: whether the vacuum’s decay can be seen #

The theory of the supercritical vacuum is not seriously in doubt, but it has never been tested where it was meant to be. The obstacle is the collision time, and the only known way past it is a collision in which the two nuclei stick for longer than they would in a simple Coulomb fly-by — a delay that some nuclear reactions near the barrier may provide and that has not been demonstrated at the length needed. New heavy-ion accelerators, including the FAIR facility under construction in Darmstadt, will reach collision energies and beam intensities beyond those of the 1980s, and recent calculations have suggested that the shape of the dynamically produced positron spectrum, rather than a narrow line, may carry a measurable signature of the dive even in short collisions. Whether any experiment will see the vacuum around a heavy nucleus give up a positron of its own accord, and whether the threshold is where the calculations put it, is not yet known.

The arithmetic leading there is short. For a point nucleus Dirac’s 1s energy $mc^2\sqrt{1-(Z\alpha)^2}$ fails at $Z = 137$; for a nucleus of real size the level continues, crosses zero near 148 and reaches $-mc^2$ near $Z = 170$, the top of the filled sea, after which an empty level would be filled from the sea and a positron emitted from nothing — a threshold two uranium nuclei, at 184 protons, cross for about $10^{-21}$ s when they come within 35 fm. The vacuum is stable only because nuclei are not large enough to unsettle it.