The right angle a fast collision closes

A snooker player cutting the object ball into a pocket knows where the cue ball will go without thinking about it: along the line at right angles to the object ball’s path. When a moving ball strikes an identical ball at rest and both roll away, they part at ninety degrees, whatever the angle of the cut. It is one of the few exact results in everyday mechanics, and it has a two-line proof. That proof uses the Newtonian formula for kinetic energy, and at the speeds of particles in a bubble chamber or an accelerator the formula is wrong. The right angle closes, by an amount set entirely by the incoming particle’s energy, and early photographs of fast particles scattering off the protons and electrons of a detector showed the narrowed angles long before anyone needed them.

Why slow collisions make right angles #

The billiard-table result follows from the two conservation laws that make collisions easier than forces. Momentum is conserved, so the incoming momentum $\mathbf{p}$ equals the sum of the outgoing ones, $\mathbf{p}_1 + \mathbf{p}_2$: the three vectors form a triangle. Energy is conserved too, if the collision is elastic, and for equal masses the Newtonian kinetic energy is $p^2/2m$, so

$$p^2 = p_1^2 + p_2^2.$$

That is Pythagoras’s theorem. A triangle whose sides satisfy it has a right angle, and the right angle is between $\mathbf{p}_1$ and $\mathbf{p}_2$. The same statement can be put geometrically: whatever the outgoing momenta are, their tips — drawn from the start of the incoming momentum — lie on the circle whose diameter is the incoming momentum, and any angle inscribed in a semicircle is a right angle. The collision can choose where on the circle to put the tip, which is the choice of cut; it cannot choose the angle between the two tracks.

What changes when the particle is fast #

Momentum conservation is exact at every speed; only the momenta are now $\gamma m v$ instead of $mv$. What changes is the energy. The kinetic energy of a fast particle is $(\gamma - 1)mc^2$, and in terms of momentum it is $\sqrt{p^2c^2 + m^2c^4} - mc^2$, which is not $p^2/2m$. The sum of squares no longer balances, the triangle of momenta is no longer right-angled, and the angle between the outgoing tracks can be anything the new energy condition allows.

The cleanest way to find out what it allows is to change frames, in the manner that four-momentum makes painless. In the frame in which the total momentum is zero, the two particles approach each other with equal and opposite momenta, and after an elastic collision they recede with the same speeds, back to back, in some direction. Every direction is allowed, so the possible outgoing momenta form a circle round the origin. Now boost back to the laboratory, in which the target was at rest. In Newtonian mechanics the boost just shifts the circle along the beam, and the shifted circle is the Thales circle of the slow result. In relativity the boost does more than shift: it compresses directions towards the beam and multiplies the momentum components along the beam by $\gamma$ while leaving the transverse ones alone. The circle becomes an ellipse.

The ellipse has the same two endpoints as the circle — the incoming momentum’s start and end, where one particle takes everything and the other nothing — and lies inside it everywhere between. An angle inscribed in something narrower than a semicircle is acute. So at any energy above the Newtonian limit, and for every possible split of the momentum between the two, the tracks open at less than a right angle.

Where the ellipse comes from, in numbers #

The ellipse is not a sketch; its shape follows from two numbers that describe the collision in the centre-of-momentum frame, and working them out shows exactly where the incoming particle’s energy goes.

In the laboratory the incoming particle has energy $\gamma mc^2$ and the target $mc^2$, so the pair has total energy $(\gamma + 1)mc^2$ and total momentum $\gamma m v$. The combination that every frame agrees on — the invariant mass of the pair — is $mc^2\sqrt{2(\gamma + 1)}$, and the centre of momentum moves at the speed that makes the total momentum vanish, with a Lorentz factor of $\sqrt{(\gamma + 1)/2}$. For $\gamma = 3$ that is $\sqrt2$. In the centre-of-momentum frame each particle has energy $mc^2\sqrt{(\gamma + 1)/2}$ and the same Lorentz factor, and after the collision each recedes with the same speed in some direction.

Boosting back multiplies the component of each momentum along the beam by the centre’s Lorentz factor, and adds the same forward push to both; the transverse components are untouched. So a circle of radius $p^*$ becomes an ellipse whose semi-axis along the beam is $\sqrt{(\gamma + 1)/2}$ times its transverse semi-axis — $\sqrt2$ times for $\gamma = 3$ — and the two outgoing momenta, still summing to the incoming one, are chords of that ellipse from its two ends. The tangent product follows in two lines of trigonometry: each laboratory angle is $\tan\theta = \sin\chi/(\gamma_{\text{cm}}(1 \pm \cos\chi))$ for a centre-of-momentum angle $\chi$, and multiplying the two gives $\sin^2\chi/(\gamma_{\text{cm}}^2 \sin^2\chi) = 1/\gamma_{\text{cm}}^2 = 2/(\gamma + 1)$, with the scattering angle cancelling completely.

That cancellation is why a single law covers every collision at a given energy. The dependence on how hard the particles hit — the angle $\chi$ — drops out of the product and survives only in how the product is split between the two angles. The motion of the centre of energy carries all the information about the incident speed, and the right angle of the billiard table is the special case in which that centre’s Lorentz factor is one.

One law for every split #

The ellipse can be turned into an equation connecting the two angles, and the equation is remarkably simple. Measuring each particle’s angle from the incoming direction,

$$\tan\theta_1 \tan\theta_2 = \frac{2}{\gamma + 1},$$

where $\gamma$ is the incoming particle’s Lorentz factor. When $\gamma = 1$ the product is 1, which means the two angles add to 90°. As $\gamma$ grows the product falls, and both angles must shrink towards the beam.

The curves in the figure are drawn from the boosted kinematics — every pair of angles comes from choosing a direction in the centre-of-momentum frame and transforming the two momenta — and they lie on the closed-form curves to a part in a billion. The shape carries the practical point. A glancing collision, in which the target barely moves, has the incoming particle deflected by a small angle and the target leaving at nearly a right angle to the beam in both regimes; a hard collision in which the two share the momentum evenly is where the difference is largest. And any measured pair of angles, from a single photograph of two tracks, gives $\gamma$ directly and so the incident energy — without any measurement of speed, curvature or range.

The universal curve #

For the symmetric case, where the two particles leave at equal angles, the opening angle has a closed form of its own.

The curve depends only on $\gamma$, which is to say on the ratio of kinetic energy to rest energy, and so it is the same for every particle. A proton needs a gigaelectronvolt to be where an electron is at a megaelectronvolt, because its rest energy is 1,836 times larger. That universality is what made the effect useful before it was understood as anything other than a check of relativity. In cloud-chamber photographs of fast electrons knocking electrons out of atoms, the angles between the two tracks were measured in the 1930s and matched the relativistic formula rather than the right angle; in bubble-chamber photographs of protons scattering off the chamber’s hydrogen, the opening angle of an elastic scatter identified the event and measured the beam energy in the same glance.

The first careful test was made in 1932 by F. C. Champion, who photographed fast electrons from a radioactive source crossing a Wilson cloud chamber and picked out the rare frames in which one struck an electron of the chamber’s gas head on enough to send it off on a visible track of its own. With speeds of a few tenths to nine-tenths of the speed of light, the Newtonian prediction was a right angle for every such fork and the relativistic one a smaller angle that depended on the incident energy, which the curvature of the tracks in a magnetic field measured independently. The forks followed the relativistic curve. It was one of the earliest direct confirmations that the energy of a moving particle is $\gamma mc^2$ and not $\tfrac12 mv^2$ plus a constant, made with no clocks and no light, only the angle between two lines of droplets — and it depended on mass being a form of energy in exactly the sense that the incoming electron’s kinetic energy adds to its inertia.

At the highest energies the curve falls towards zero: two protons scattering elastically at the energies of the largest colliders — seen from the frame of one of them — would leave within a fraction of a degree of each other and of the beam. Colliders avoid that frame for the reason a fixed target wastes most of the energy: almost everything in the laboratory is carried forward by the motion of the centre of momentum, and only what is left in that frame is available to the collision. The narrowing of the right angle and the waste of a fixed-target experiment are the same fact seen from two sides — the incoming particle’s energy is mostly inertia, and inertia goes forward.

How the energy divides #

The angles tell half the story. The other half is how the kinetic energy is shared between the two outgoing particles, and it too departs from the billiard table.

In the Newtonian case the share of energy a particle takes when it leaves at angle $\theta$ is $\cos^2\theta$: a ball deflected by 45° keeps half the energy, and a ball that goes straight on keeps all of it. At $\gamma = 3$ a particle deflected by 45° keeps well under half. The reason is the same compression towards the beam: a given laboratory angle corresponds to a larger angle in the centre-of-momentum frame, where the scattering was harder, and a harder scatter hands more of the energy to the target. Equal sharing of the energy now happens at a smaller laboratory angle, which is the symmetric opening of the previous figure.

The same geometry with a massless partner #

The method that produced the ellipse — find the circle of possibilities in the frame where nothing is moving on average, then boost it — works for any two-body collision, and applying it where one of the two has no mass gives a result that helped establish that light carries momentum.

When a photon scatters off an electron at rest, the same two conservation laws apply with the photon’s energy equal to its momentum times $c$. The photon cannot lose energy to an electron without also changing direction, and the bookkeeping gives the shift in its wavelength as a function of the scattering angle alone: $\lambda' - \lambda = (h/mc)(1 - \cos\theta)$. That is the Compton shift, and it is this essay’s kinematics with one mass set to zero. The electron recoils forward, never beyond 90° from the photon’s original direction, and the angle between the scattered photon and the recoiling electron is fixed by the energies in the same way the angle between two equal masses is fixed by $\gamma$.

The general lesson is the one the relativistic second law teaches in a different setting: with energy no longer quadratic in momentum, the geometry of motion stops being Euclidean in the familiar way, and results that looked like facts about space — a right angle, a push along the direction of acceleration — turn out to have been facts about slow bodies.

What relativistic mechanics adds and what it keeps #

It is worth being precise about which parts of the billiard-table result survive, because it is easy to overstate what fails.

Momentum conservation survives unchanged. The outgoing momenta still add to the incoming one, and the three still form a closed triangle.

The collision is still planar. Two particles scattering elastically leave in a plane containing the incoming direction, because the transverse momenta must cancel. Nothing about the geometry leaves that plane.

What fails is the one statement that depended on the form of kinetic energy. The right angle was never a consequence of momentum conservation; it was Pythagoras applied to a sum of squares that holds only when energy goes as momentum squared. Speeds do not add and energies are not quadratic in momentum, and the right angle goes with them.

And unequal masses behave differently again. A heavy particle striking a light one at rest cannot be deflected by more than a maximum angle, set by the mass ratio, in Newtonian mechanics too; in relativity that maximum depends on the energy as well. The equal-mass case is the one where Newtonian mechanics has an exact, parameter-free prediction, and so it is the case where relativity’s departure is cleanest.

Where the picture stops #

The collision is elastic. At high enough energies a collision between protons usually is not: it makes new particles, and the elastic events are a minority whose share falls with energy. The kinematics here describe the elastic ones, and in a photograph they have to be picked out from the rest — which is what the opening angle does.

The target is free and at rest. A proton in a bubble chamber is the nucleus of a hydrogen atom and is effectively free; a proton in a heavier nucleus is bound and moving, and scattering off it smears the angles by its own motion.

Spin and the details of the force are left out. The kinematic curves say which angles are allowed and how they are linked; they say nothing about how likely each angle is. That depends on the force between the particles and on their spins, and it is what the rate of events at each angle, rather than their geometry, measures.

Momenta the straight tracks hide #

The tracks in the first figure are straight lines of equal length, drawn only to show directions. A real track’s length and curvature in a magnetic field record the particle’s momentum, and the two outgoing tracks of an asymmetric scatter carry very different momenta; the drawing of a symmetric case hides that entirely. Nor does a two-dimensional picture show the plane of the collision tilted out of the page, which in a real chamber means the measured angles have to be reconstructed in three dimensions before they can be compared with the curves.

Still open: how elastic collisions behave at the highest energies #

At the energies of the Large Hadron Collider, the elastic scattering of protons is measured by detectors placed hundreds of metres down the beam line, catching protons deflected by microradians, and the results have been surprising. The total rate of proton–proton collisions keeps rising with energy, the elastic fraction behaves differently from long-standing expectations, and the pattern of the elastic scattering at the smallest angles has been read by some groups as evidence of an exchange process that had been predicted decades ago and never isolated. The kinematics here are certain; what the protons do within them, at the highest energies, is still being worked out.

The next question in relativistic dynamics is what happens to a body that is pushed back and forth rather than scattered once — a mass on a spring driven to speeds where its inertia grows through every swing. The habit worth carrying from here is to ask which of a classical result’s ingredients it actually depends on. The right angle of a billiard-table cut rests entirely on kinetic energy being momentum squared over twice the mass — and the moment that stops being true, so does the geometry built on it.