The slope a spin leaves in a spectrum

The cone a decay cannot leave found an exact result in the kinematics of a particle breaking into two while in flight. If the parent decays equally in every direction in its own frame, one product’s laboratory energy is spread evenly between two sharp edges, and the positions of the edges give the parent’s speed and mass. It flagged the assumption that makes the spectrum flat: that the parent has no spin pointing anywhere, so nothing in its rest frame prefers one direction to another.

Real parents are often not like that. A tau lepton made in the decay of a Z boson, a lambda hyperon made in a collision, a muon made when a pion decays — each can come out with its spin pointing preferentially one way. What that does to the rectangle is simple, exact and useful. The edges do not move. The top tilts.

A tilt between edges that do not move #

The three samples in the figure come from the same decay at the same speed and differ only in the direction of the taus’ spin. Every one of them fills the interval from 313 to 2667 MeV and nothing lies outside it, because the edges are fixed by the masses and the speed. The edges are kinematics and the tilt is dynamics, and the two separate cleanly. A spectrum’s end points still weigh the parent however it was polarised; its slope measures how the spin was pointing and nothing else.

The reason is the same one-line argument that made the spectrum flat. The laboratory energy is $\gamma(E^* + \beta p^* \cos\theta^*)$, a straight-line function of the cosine of the rest-frame angle, so the energy axis is the $\cos\theta^*$ axis stretched and shifted. Whatever distribution the decay has in $\cos\theta^*$ appears unchanged on the energy axis. A uniform distribution gave a rectangle. A distribution that rises linearly in $\cos\theta^*$ gives a rectangle with a sloping top, and for a fully polarised tau the slope takes the height from nothing at one edge to twice the average at the other.

The slopes read back from 60,000 decays, +0.993 and −1.006, are each about a standard error from ±1, and the unpolarised sample’s −0.010 is a standard error from zero. That the reading is statistical is not a detail. The rest of the subject is about how much a slope costs.

Which way the tilt goes follows from the boost. A pion emitted forwards in the tau’s frame is carried further forwards and gains energy; one emitted backwards loses it — the same boost that crowds a moving source’s light into a forward cone and shifts it to the blue. So a spin pointing along the flight, which favours forward emission, piles decays up at the high-energy edge, and a spin pointing back piles them at the low one. Nothing else distinguishes the three samples: the same edges, the same number of decays, the same decay.

A pattern that leans towards the spin #

To know what the slope measures, the rest-frame distribution has to be written down, and for a parent of spin one half decaying into two products it has a form fixed by angular momentum rather than by any model:

$$\frac{dN}{d\cos\theta^*} \propto 1 + \alpha P \cos\theta^*.$$

Here $\theta^*$ is measured from the spin axis, $P$ is the polarisation of the sample — the excess of spins pointing one way over the other, from −1 to +1 — and $\alpha$ is the decay’s analysing power, how strongly the product’s direction responds to the spin. The form is exact rather than a first term: a state of spin one half can carry an angular pattern only up to the first power of the cosine, for the same reason a spin-½ measurement goes as the cosine of half the angle and never as anything more elaborate.

The three patterns share a feature worth noticing: every one passes through the same point straight up. At right angles to the spin the spin makes no difference, and all of its effect is a transfer of rate from the backward hemisphere to the forward one, a fraction $(2 + \alpha P)/4$ ending up forward. With $\alpha P = 1$ that is three quarters, and the backward direction itself is empty.

The analysing power belongs to the decay, not to the sample. For a tau decaying to a pion and a neutrino it is exactly one, because the pion has no spin and the neutrino’s spin is locked to its direction, so the product’s direction carries the whole of the tau’s spin. For a lambda decaying to a proton and a pion it is about three quarters. For a muon decaying to an electron and two neutrinos, averaged over the electron’s energy, it is one third.

A decay used this way is an analyser in the sense that a magnet is in a chain of Stern–Gerlach devices, but a leaky one. The magnet sends each atom one way or the other according to its spin, so a single split of a beam reads its polarisation. A decay only tilts the odds for each product’s direction: a pion leaving at right angles to the spin is equally likely whichever way the spin pointed, and even a forward pion is only more likely, not certain, to have come from a forward spin. A polarisation is never read from one decay, only from an average over many, which is why the number of decays a reading costs, rather than the arithmetic of the reading, is what limits every measurement of this kind.

Why the lean could not be taken for granted #

Until 1957 it was assumed that no decay could lean at all.

A spin is unchanged if every direction in space is inverted through a point: like any angular momentum it is a product of a position and a momentum, and the inversion reverses both. A product’s direction is reversed by the inversion. So a correlation between the two — more products along the spin than against it — would look different in the inverted world, where the products would go more against the spin. If the laws of physics are the same in that world, no such correlation can exist and $\alpha$ must be zero for every decay. That symmetry is parity. It is a discrete symmetry, and where a continuous one hands over a conserved quantity — the argument of the conservation law a symmetry hands over — a discrete one hands over a rule forbidding patterns like this one.

The strong and electromagnetic interactions obey it. The weak interaction does not, and in 1957 two experiments showed it within weeks of each other. Chien-Shiung Wu’s group aligned the spins of cobalt-60 nuclei a few thousandths of a kelvin above absolute zero and found beta-decay electrons emitted preferentially against the spin. Garwin, Lederman and Weinrich stopped muons from pion decays and found the decay electrons preferentially along the muon’s spin — which established both that the muons were polarised and that their decay analysed the polarisation. Every nonzero $\alpha$ in the list above belongs to a decay by the weak interaction, and it is nonzero because that interaction treats a spinning particle and its mirror image differently.

The same fact makes polarised samples easy to produce. A pion decaying at rest into a muon and a neutrino must give the two opposite spins along their common line, and the neutrino’s spin always points against its motion, so every muon is produced with its spin pointing a definite way along its own. Beams of such muons are nearly fully polarised, and it is their decay electrons, emitted preferentially along the spin, that a storage ring reads to follow a muon’s spin precessing.

A spin measured without its particle’s direction #

The linear relation between energy and rest-frame angle has a consequence that made one of the precision measurements of the electroweak theory possible.

At the LEP collider, electrons and positrons annihilated at the mass of the Z boson, and some of the Zs decayed into tau pairs. The Z couples differently to particles spinning left-handed and right-handed along their motion, so the taus came out polarised along their flight, by an amount that measures how different the couplings are. But a tau lives for about three tenths of a trillionth of a second and decays into a neutrino that no detector sees, so its direction is not measured.

Its energy is. The collision happened at rest, each tau carries half the energy of the collision, and the pion it decays into is measured. So the natural variable is the fraction $x$ of the tau’s energy that the pion takes.

For a tau this fast the energy fraction runs from 0.0065 to 0.9996, nearly the whole of zero to one, and the spectrum is the straight line $1 + P(2x - 1)$ to within the pion’s small mass. The parent’s speed has dropped out. At a Lorentz factor of 25.66 or of a quarter of a million the curve is the same to better than a tenth of a per cent, so the measurement needs no correction for how fast the tau was going and no knowledge of its direction at all — only the energy of the collision, which the accelerator sets.

The measured polarisation at the Z is near −0.14, and 100,000 decays read it with a statistical error of about 0.0055: the sample in the figure returns −0.150, within two standard errors. That number is a direct reading of the weak mixing angle, the parameter that sets how the Z’s couplings divide between left- and right-handed particles, whose sine squared came out near 0.23. The tau’s own decay is the polarimeter, and the invariant that survives a boost is not needed — the energy fraction is not invariant, but in the limit that matters it has stopped depending on the boost.

The price of a weak analyser #

A slope is read from a sample, and the sample’s size sets the error.

The error on a polarisation has a closed form. The average of $\cos\theta^*$ over a sample is $\alpha P/3$, so the polarisation is three times that average divided by $\alpha$, and the variance of a cosine drawn from $1 + \alpha P\cos\theta^*$ gives a standard error of $\sqrt{(3 - \alpha^2P^2)/N}/\alpha$. The dots, each the scatter among 300 independent samples, fall on the lines across three decades of sample size.

The cost of a measurement grows as the inverse square of the analysing power. A decay with $\alpha = 1$ reaches ±0.01 with 27,500 decays; one with $\alpha = 1/3$ needs 268,000 for the same precision, nearly ten times as many. That is why the tau’s decay to a pion is the favoured channel at the Z despite being only about a tenth of tau decays: its decay to a rho meson is more common but has an analysing power near 0.45, and each of those decays is worth about a fifth as much.

It is also why the analysing power has to be known independently, and precisely. In 2019 the lambda’s was remeasured with entangled lambda–antilambda pairs and came out at 0.75 rather than the 0.64 that had been used for four decades. Every lambda polarisation derived from the old value was too large by about 17 per cent, since the slope measures the product $\alpha P$ and the division by $\alpha$ had been done with the wrong number.

What the angle folds away #

The energy is not the only variable carrying the rest-frame angle, but it is the only one that carries all of it at every speed.

The folding is the cone again. While a product moves faster in the parent’s frame than the parent moves in the laboratory, every rest-frame direction lands on its own laboratory angle, and the angle is as good a record of $\cos\theta^*$ as the energy is. Once the parent is faster, the laboratory angles are confined to a forward cone, and each angle inside it is reached from two rest-frame directions, one forward and one backward. A spin that sends more products forward and fewer backward sends them to the same laboratory angles, and the excess on one branch is cancelled by the deficit on the other.

How much cancels depends on how the two branches share the events. A tau’s pion moves at 0.988c in the tau’s frame, so the backward branch holds a sliver of its decays and the angle keeps essentially everything. A lambda’s proton moves at 0.107c, the two branches are nearly mirror images, and at high energy its angle keeps 5.7 per cent of the information — a measurement from that angle alone would need about eighteen times as many decays as one from the proton’s energy. In practice a lambda’s polarisation is measured by reconstructing both its products, adding their four-momenta, and boosting back into the lambda’s own frame, where the angle is unfolded again; the figure is the argument for why that effort is necessary.

Where a slope is not a spin #

Only the polarisation along the line of flight tilts the energy spectrum. The energy depends on the cosine of the angle to the flight direction. A spin at right angles to the flight produces an asymmetry around the flight direction, in the azimuthal angle, and the energy spectrum is blind to it. Measuring transverse polarisation needs angles, and the reconstruction back into the rest frame.

A spin greater than one half allows a second power of the cosine. A parent of spin one can be aligned without being polarised — its spin along the axis in either sense rather than at right angles to it — which adds a $\cos^2\theta^*$ term, and the spectrum curves rather than tilts. The linear form is a property of spin one half, not of decays.

A three-body decay has an analysing power that depends on energy. The muon’s one third is an average: its most energetic electrons are emitted along the spin with an analysing power of one, and its least energetic ones preferentially against it. Folding that into a spectrum whose shape is already set by three-body phase space — the continuous spectrum the energy that did not all arrive describes — means the slope must be extracted from a curve rather than read off a rectangle.

The detector can make a slope of its own. A detector whose efficiency changes with energy distorts a flat spectrum into a tilted one, and a polarisation of −0.14 changes the height of the spectrum by 28 per cent from one edge to the other. Separating a real polarisation from an efficiency slope is most of the systematic work of such a measurement, usually done by comparing channels whose analysing powers differ.

And the parent’s energy has to be what it is assumed to be. At a collider, a photon radiated by the incoming beams before they annihilate lowers the energy each tau carries, and a pion’s energy fraction computed from the nominal energy is then wrong. The correction is calculable, and it is one of the reasons the edge near $x = 1$ is not as sharp in data as in the figure.

What the figures leave out #

The spectrum figures show sampled decays for polarisations of +1, 0, −1 and −0.14. The first three are cartoons chosen to make the tilt visible; the physically interesting polarisations are a few per cent to a few tens of per cent, where the tilt is invisible to the eye in any single histogram and exists only as a fitted number with an error.

The information figure uses the lambda’s proton and pion separately, as though an experiment recorded one and discarded the other. None does; the figure isolates what the angle of one product can say, in order to show that the answer can be almost nothing.

Still open: the spin of the fastest-turning fluid #

Colliding heavy nuclei slightly off-centre makes a droplet of quark–gluon plasma with an enormous angular momentum, and the lambdas that condense out of it are polarised along that angular momentum by about a per cent. In 2017 the STAR experiment measured that global polarisation using the lambda’s own decay as the analyser — at an analysing power of three quarters and a polarisation of a per cent, a measurement needing more than a million lambdas — and inferred a vorticity of about $10^{22}$ per second, the fastest rotation of any fluid measured.

What is not settled is the pattern around the beam direction. Hydrodynamic calculations of the plasma’s local vorticity predicted a polarisation along the beam whose sign alternates around the collision in one sense, and the measured pattern alternates in the opposite sense. Proposals that shear in the flow, not only rotation, polarises the spins can reverse the sign, but whether they account for the data, and whether spin in such a fluid is in equilibrium with its flow at all, is still argued.

The habit worth carrying away is to find the variable in which a boost is a straight line. A linear map moves a distribution without deforming it, so whatever a spin did to the rest-frame pattern arrives intact on the energy axis at any speed — and a variable that folds, however natural, can lose nearly all of it.