The orbit special relativity cannot close

An orbit under an exact inverse-square attraction comes back to the point it started from, and it does so because of a conserved vector pointing at the perihelion that no other force law has. Any change to the force turns that vector, and the orbit becomes a rosette. Mercury’s perihelion turns by 43 arcseconds a century more than the other planets explain, and the usual summary is that relativity is responsible. Which relativity is a question worth asking precisely, because the answer is not the one the summary suggests. Keep Newton’s attraction exactly as it is and change only one thing — let the orbiting body’s inertia grow with its energy, as special relativity insists it must — and the orbit already fails to close.

Heavier where it turns hardest #

The mechanism takes one sentence once the right quantity is watched. A body on an eccentric orbit is fastest at its closest approach, so that is where its inertia is largest.

A push across a body’s path turns it by an amount inversely proportional to its inertia. That is the content of the relativistic second law for a force at right angles to the motion: the rate at which the direction changes is the force divided by $\gamma m v$, not by $mv$. So at closest approach, where the attraction is strongest and the turn is sharpest, the relativistic body is turned less than a Newtonian one would be at the same speed. It swings wide, and the swing carries the whole orbit round. By the time the body comes back to its closest approach it has swept more than a full turn — 384.3 degrees in the figure — and the perihelion has advanced by the excess.

Nothing here requires curved spacetime, a field with energy of its own, or any modification of the inverse square. The attraction is Newton’s, the geometry is flat, and the only relativistic ingredient is that a moving body’s inertia is its energy. That is enough to break the one symmetry that made the Kepler orbit close.

It is worth being precise about which symmetry, because the answer says what kind of effect this is. Of all the central forces, only the inverse square and the linear spring close every bound orbit, and the inverse square does it because its orbits have a conserved vector over and above energy and angular momentum. Conserved quantities come from symmetries — each one is handed over by an invariance of the equations — and the one behind this vector is not a symmetry of space at all. It is a hidden symmetry of the Kepler problem’s equations in a four-dimensional space of positions and momenta, and it holds only because the kinetic energy is exactly $p^2/2m$ and the potential exactly $-k/r$. Replace the kinetic energy by $\sqrt{p^2c^2 + m^2c^4}$ and the hidden symmetry is gone, whatever the potential is. The rotation of space survives — angular momentum is still conserved, the orbit still lies in a plane — but the orientation of the ellipse within the plane is no longer fixed by anything, and it drifts.

An exact rosette #

The special-relativistic Kepler problem has an exact solution, and it is worth having, because it turns the qualitative story into a number and shows where the number stops making sense.

The body’s energy is $E = \sqrt{p^2c^2 + m^2c^4} - k/r$, with $k = GMm$ for gravity or $k = Ze^2/4\pi\varepsilon_0$ for an electron round a nucleus. Squaring and separating the radial motion gives

$$p_r^2 c^2 = \left(E + \frac{k}{r}\right)^2 - m^2c^4 - \frac{L^2c^2 - k^2}{r^2},$$

and the last term is the telling one. In Newtonian mechanics the centrifugal barrier is $L^2/r^2$; here it is $(L^2 - k^2/c^2)/r^2$. The attraction has eaten into the angular momentum’s barrier, because the attraction’s own energy $k/r$ is squared along with everything else and one piece of the square has the same $1/r^2$ shape as the barrier. The orbit that results is

$$\frac{1}{r} = A\left[1 + e \cos(\Gamma \varphi)\right], \qquad \Gamma = \sqrt{1 - \frac{k^2}{L^2c^2}},$$

an ellipse whose angle has been rescaled by $\Gamma$. The body returns to its closest approach after $\varphi = 2\pi/\Gamma$ rather than $2\pi$, so the perihelion advances by $2\pi(1/\Gamma - 1)$ every revolution, which for a weak attraction is $\pi k^2/L^2c^2$. For gravity, where $L^2 = GMm^2 a(1 - e^2)$ for an orbit of semi-major axis $a$, that is

$$\Delta\varphi = \frac{\pi GM}{c^2 a(1-e^2)} \quad \text{per revolution}.$$

General relativity’s answer, the one the Laplace–Runge–Lenz argument reconciles with Mercury, is $6\pi GM/c^2a(1-e^2)$. Special relativity alone supplies exactly one sixth of it.

Two properties of the exact rosette are not obvious from the approximate one. The advance depends on the angular momentum and on nothing else — not on the eccentricity, not on the energy — so a nearly circular orbit and a very eccentric one with the same $L$ precess at the same rate. And $\Gamma$ reaches zero at $k = Lc$, which is not a small correction becoming large but the end of the solution.

One curve, from a planet to the inside of a heavy atom #

The ratio $k/Lc$ is the only number the rosette depends on, and it runs over an enormous range across systems that obey the same equation.

For Mercury the advance is $8.4 \times 10^{-8}$ radians a revolution. The planet goes round 415 times a century, so special relativity alone gives 7.2 arcseconds a century — a sixth of the 43 that are observed after the other planets are subtracted, which is exactly what the formulas above say it must be.

For hydrogen the ratio $k/Lc$ in the lowest orbit is the fine-structure constant, $\alpha \approx 1/137$, and the advance is $\pi\alpha^2$, a part in six thousand of a turn every revolution. This is the rosette Arnold Sommerfeld computed in 1916, three years after Bohr had put the electron on a circle whose radius survives in the modern atom as the most probable distance and nothing else. Quantising it — requiring the rosette’s radial and angular motions each to hold a whole number of quanta — split every level of the old Bohr atom by an amount of order $\alpha^2$, and gave a formula for hydrogen’s fine structure that agrees with measurement. It also agrees, exactly, with the formula Dirac derived twelve years later from a relativistic wave equation with the electron’s spin built in — although Sommerfeld’s electron had no spin at all. The agreement is not a sign that the rosette was right. Two omissions in the older calculation, the missing spin and the crude quantisation rule, cancel term by term, and the cancellation has since been traced to a hidden symmetry the two calculations share. It remains one of the strangest coincidences in the history of the subject, and it is the reason the fine structure constant has its name.

For the innermost electron of a heavy atom the ratio is $Z\alpha$, and at $Z = 80$ it is 0.58. The rosette advances 83 degrees a turn; relativity is not a correction here but the dominant fact about the orbit. That is where the heavy elements start behaving in ways their lighter relatives do not, with inner orbits pulled in and everything outside them rearranged.

Past $k = Lc$ there is no orbit #

At $k/Lc = 1$ the effective centrifugal barrier, $(L^2 - k^2/c^2)/r^2$, vanishes, and beyond it the term changes sign and joins the attraction.

The Newtonian barrier is invincible: at small enough radius the $L^2/r^2$ term beats any $1/r$ attraction, so an orbit with any angular momentum at all has a closest approach. The relativistic barrier is not, because the attraction’s square has the same $1/r^2$ shape and can outweigh it. When it does, nothing stops the fall. A classical body with $k > Lc$ is not in an orbit with a large precession; it is not in an orbit.

In an atom the lowest possible $L$ is fixed by quantum mechanics, and the classical condition $k > Lc$ becomes $Z\alpha > 1$ for the innermost state — a nuclear charge near 137. The full quantum treatment moves the threshold and softens it, and a nucleus of finite size moves it again, to a charge of about 170, where the innermost level would plunge into the negative-energy continuum and the vacuum near the nucleus would be expected to produce electron–positron pairs spontaneously. The rosette’s collapse at $k = Lc$ is the classical shadow of that limit, and it appears from nothing more than an inertia that grows with energy and an inverse-square attraction.

Three relativistic theories, three answers #

The claim that special relativity “supplies one sixth” needs its qualification stated plainly, because it is where the argument actually lands. The rosette above assumed one particular coupling: the attraction is exactly Newton’s $k/r$, and only the body’s inertia changes with speed. That is the natural model for an electron round a nucleus. It is not the only relativistically consistent model of gravity, and the obvious alternatives give different answers.

Gunnar Nordström’s theory of 1913 is the instructive case. It is a perfectly Lorentz-invariant field theory of gravity with a single scalar field, it reproduces Newton in the weak static limit, and it was taken seriously by Einstein himself for a year. In it, a moving body’s coupling to the field is its energy measured in its own rest frame, which falls as $1/\gamma$ relative to its total energy. At closest approach, where the body is fastest, the pull is weakened more than the inertia is increased. The body is turned more sharply than a Newtonian one, not less, and the rosette runs backwards.

Nordström’s theory failed on exactly this count, among others. It predicts that Mercury’s perihelion should regress by 7 arcseconds a century, where the measured residual is an advance of 43. It also predicts no bending of light at all, since its geometry differs from flat spacetime only by an overall scale factor, and a light ray’s path does not care about the scale of the geometry it crosses. The 1919 eclipse and the perihelion together ruled it out.

Which of the three is the honest special-relativistic theory of gravity is the wrong question, and the reason is instructive. The Sommerfeld coupling, natural for an electron round a nucleus, is unnatural for gravity: its attraction stays fixed while the body’s inertia grows, so a fast body is pulled with a force that does not grow with its energy, and it accelerates more slowly in the field than a slow one. That violates the universality of free fall — the principle, tested to parts in $10^{15}$, that everything falls alike whatever it is made of and however much of its mass is kinetic energy. For an electric charge there is no such principle, since charge and inertia are different quantities. For gravity there is, and it says the gravitational charge is the energy. Nordström’s theory respects it — it is a theory in which every body follows the geodesics of one geometry — and fails instead because its geometry is the wrong shape, a flat spacetime rescaled point by point, which bends no light and turns the perihelion backwards. General relativity respects it too, and is the only one of the three that also bends light by the measured amount.

So the famous 43 arcseconds are not special relativity with a correction added. Special relativity is a constraint every candidate satisfies, and the candidates it admits give answers ranging from minus one to plus six in the natural unit. What the measurement actually determines is how gravity couples to a moving body and how the field behaves at second order in its strength — the parameters that the post-Newtonian framework calls $\gamma$ and $\beta$, for which general relativity’s values combine to give six. The measured advance is one of the two or three numbers that pin them down.

The dependence on second-order structure is sharper than it sounds. Nordström’s field for a point mass is $\varphi = 1 - GM/rc^2$. Replace it with $e^{-GM/rc^2}$, which agrees with it to first order and therefore gives exactly the same Newtonian gravity, the same light bending and the same everything else a first-order experiment can see — and the backward precession doubles, from −1 unit to −2. A precession is a second-order effect, so the $(GM/rc^2)^2$ term in the field matters as much as the leading one. A theory of gravity is not fixed by its Newtonian limit, and the perihelion is one of the few measurements that looks past it.

Where the rosettes stop being the answer #

The attraction is taken to be instantaneous. Every coupling here uses a static field centred on a fixed source. A source that moves, or a field that propagates at a finite speed, adds velocity-dependent terms of the same order as the effects computed, and for two bodies of comparable mass those terms are not optional.

The body does not radiate. A charge on the rosettes of the fourth figure would radiate electromagnetically and spiral inward in a fraction of a nanosecond; a classical atom is not stable at all, and the Sommerfeld orbits are a calculation performed on an object that cannot exist classically, redeemed only by the quantum rules applied to it. A planet radiates gravitationally, far too weakly to matter.

Spin is left out. An electron’s magnetic moment couples to the field it sees in its own frame, and that coupling is as large as the rosette’s own effect in hydrogen — it is why the missing spin and the crude quantisation happened to cancel. The frame the electron sees is itself rotating, because two boosts in different directions leave a rotation behind, and that kinematic rotation halves the naive magnetic coupling. A spinning planet’s spin couples to the curvature, negligibly for any real planet.

The general-relativistic bar is the test-particle geodesic. It assumes the orbiting body has no gravity of its own and that the central mass does not rotate. Both are excellent for Mercury and the Sun and neither is excellent for two neutron stars. General relativity has its own version of the collapse at $k = Lc$, and it arrives sooner: below six gravitational radii no circular orbit is stable at any angular momentum, because the curvature adds an attractive $1/r^3$ term that eventually beats any barrier. The special-relativistic collapse needs the attraction to outweigh $Lc$; the general-relativistic one needs only that the body come close enough.

The clock carried round the rosette #

The drawings show one plane and one orbit. What they cannot show is the time along it. A body on the special-relativistic rosette carries a clock that runs slow in proportion to its speed, most at closest approach, and a body on the general-relativistic orbit carries one slowed by both its speed and its depth in the field. The rosettes are the same shape as they would be on a distant observer’s clock, but a traveller on each counts a different number of its own seconds per revolution, and that difference is a second observable that none of these pictures carries.

Nor can the pictures show that the exaggeration is an exaggeration. Every rosette is drawn at a strength where the orbit is relativistic in every sense; Mercury’s special-relativistic rosette, drawn at the same scale, would be an ellipse whose closest approach takes seventy-five million revolutions to go once round — and the full general-relativistic one, twelve million.

Still open: how well the second-order term is known #

The advance of Mercury’s perihelion is now measured by radio ranging to spacecraft in orbit around it, and the post-Newtonian parameters it constrains are known to parts in $10^5$: the space-curvature parameter from the deflection and delay of radio signals past the Sun, the second-order parameter from the perihelion and from the Moon’s orbit measured by laser ranging. Both agree with general relativity’s value of exactly one. Missions now in progress aim to improve the perihelion-based measurement by a further order of magnitude, and one of the obstacles is the Sun’s own flattening, whose contribution to the precession has a similar form and has to be measured separately. Whether the second-order coupling departs from one at a part in a million is not known, and theories that modify gravity at large distances generally predict that it should somewhere.

The natural question beyond this one is what happens when the orbiting body is not a test particle. Two bodies of comparable mass each move in the other’s field while radiating, and the orbit neither closes nor keeps its size. There is no rosette with a fixed step there, and computing what there is took most of a century. The habit worth carrying from here is to ask what a famous residual is a residual of. The 43 arcseconds are not relativity versus Newton; they are one theory’s coupling against the alternatives special relativity leaves open, and a sixth of them would have been there in a universe where space did not curve at all.