The one extra force that only turns the orbit

The orbit that does not come back to itself found that almost every central force fails to close its orbits, and that the failure can be read off as a slow turning of the long axis. It treated that turning as a defect in the force, a sign that a law is not the inverse square. Book I of the Principia contains a stranger way of looking at the same thing, and it comes well before any of Newton’s work on the Moon. Take an orbit. Decide that it should turn. Ask what force would make it turn, with nothing else about it changed. The answer is exact, it holds for any orbit under any central force, and it is simple enough to fit in a line.

Propositions 43 to 45 are known as the theorem of revolving orbits, and they are seldom taught. The result is that if a body moves on some orbit, then an identical body can be made to trace the same curve while the whole curve rotates, keeping pace with the first body’s radius at every moment, by adding only one extra central force. That extra force goes as the inverse cube of distance. No other power will do it. And there is a surprise in the arithmetic, since the extra term can be either attractive or repulsive and it can multiply the angle by any number at all.

Two bodies keeping the same appointment #

Write the angular momentum per unit mass as $L$. Any central force gives a radial equation in which $L$ appears only as a centrifugal term,

$$\ddot{r} = \frac{L^2}{r^3} + f(r).$$

The orbit that cannot be made smaller showed what that term does: it behaves as a repulsion, pushing a body back out from the centre, and it goes as the inverse cube of the radius. All of the angular part of the motion gets into the radial motion through that single term.

Now change the angular momentum, to $kL$. The body’s sideways motion is $k$ times as fast at every radius, so its angle grows $k$ times as fast. Its radial equation picks up $k^2 L^2/r^3$ where it had $L^2/r^3$, and the radial motion would ordinarily change. But the difference between the two is itself an inverse cube, so a force of that form can cancel it exactly. Add the force

$$F(r) = \frac{L^2\,(1 - k^2)}{r^3},$$

and the radial equation becomes the original one, with the same solution for the same start. The radial motion has no way of knowing anything happened. Meanwhile the angle has been multiplied by $k$ everywhere. That is the theorem, and its proof is that short. For $k$ above one, the added force points inward, and for $k$ below one it points outward.

What is striking about the picture is that nothing in it looks like an ellipse any more, and yet it is one. Take the curve, and at each moment rotate the body back through a quarter of the angle it has moved. The red ellipse is what comes out. One way of saying it is that the rosette is the ellipse viewed from a platform that turns at a rate proportional to the body’s own angular speed, and which therefore turns fast near the centre and slowly far out. The whole motion is equally a Kepler orbit with a clock on it. Newton’s phrase for the orbit’s motion is an orbit revolving, and the pictures make it clear what he meant.

Three forces and one history #

The derivation is an algebraic trick, and the way to be sure of an algebraic trick is to do the sum without it. So here the forced orbit is not computed from the theorem at all. It is integrated in Cartesian coordinates, with the force as a vector, from an initial position and a sideways speed multiplied by $k$. The integrator is told nothing about radial motion. Then the radius is measured off and compared.

The comparison is exact to the integrator’s accuracy. The body on the $k = 2$ orbit has an angular momentum twice the ellipse’s. It feels a strong extra attraction, three times the centrifugal term in strength, which is what it takes to hold it to the same radial schedule while it goes round twice for every radial period. The body on the $k = 1/2$ orbit feels a repulsion. All three reach pericentre at the same moment and apocentre at the same moment. Anyone with a rangefinder and no way of measuring angle could not distinguish them.

That equality has a use that goes further than this example. The arrow that says which way the orbit points showed that an ellipse under the inverse square has an extra conserved vector, and that it is this vector that holds the ellipse still. Under the added force that vector is no longer conserved; it precesses steadily, at a rate fixed by $k$. So the theorem is, among other things, an exact solution for the motion of the arrow, where most such calculations are approximate.

How strong the term has to be #

The extra force’s strength is set by the orbit’s own angular momentum, which is a peculiar feature. The law of force that makes one given orbit revolve by a given factor does not do the same to another orbit with a different $L$. For a body with a different $L$ the same inverse-cube force gives a different $k$, since the actual condition is

$$k^2 = 1 + \frac{\lambda}{L^2}$$

for an attraction $\lambda/r^3$. So a single force law gives a family of orbits that revolve at different rates, faster the less angular momentum they have.

The picture shows that these forces are not small adjustments, for the multipliers drawn. An orbit made to revolve by a quarter of a turn per radial cycle has had its force altered by a third of its value at the far point and four-fifths at the near point. The theorem does not depend on the term being small. Its exactness at any strength is what separates it from an approximate treatment, and it is why it can be used as a check on other methods: a numerical code for orbits can be asked to reproduce a revolving ellipse at any $k$, and a code that drifts can be caught by a test with a known exact answer.

The share of the force grows inward as $1/r$, and this is the sense in which the term is close to invisible when a body is far out. Well outside its pericentre a body with a small extra term is orbiting a nearly Keplerian force, and the turn in its orbit gets added almost entirely in the fast, close passage. This is true of almost every perturbation that turns an orbit. It holds whether the cause is an inner planet, an oblate star or general relativity.

Curves no one would guess are ellipses #

Since $k$ can be any number, an ellipse produces an unlimited family of curves. If $k$ is a ratio of whole numbers $p/q$, the orbit closes after $q$ radial periods and $p$ turns. If it is irrational, the orbit never closes and fills a ring. Every one of these has the same radial history as the ellipse it came from.

The cases with small $k$ are the least expected. With $k = 1/3$ the angle advances slowly, and the three radial cycles needed for one turn draw a smooth three-lobed curve around the centre, of a kind that looks as though it belongs to some other force entirely. With $k = 1/2$ the two radial cycles make a single loop, a closed oval with two far points opposite each other, and the centre at its middle, which looks very much like an ellipse centred on the attractor. That similarity deserves a look, because an orbit like that is what a linear spring gives, with an actual ellipse centred on its centre. The revolved $k = 1/2$ orbit is not one, and the curve it draws differs in shape. The orbit that does not come back to itself reported Bertrand’s theorem, by which only the spring and the inverse square close every bound orbit. That theorem is not broken by any of these closed curves. They close only for one chosen value of $L$. A body with a little more angular momentum would have an irrational $k$ and a curve that never closes.

The family also has an edge. For a force of fixed strength $\lambda$, a body’s actual angular momentum sets how much centrifugal term is left after the attraction has taken its share. If the attraction’s $\lambda$ exceeds the square of that angular momentum, the sum is negative, the centrifugal barrier is gone, and the body spirals into the centre on one of the curves called Cotes’s spirals. Those are the orbits under a pure inverse cube, and as the orbit that cannot be made smaller showed by the effective potential, this power is the one at which a force and the centrifugal term scale alike, so that no stable circle exists under it alone.

What a precession can and cannot say #

The most famous precession in physics is Mercury’s. Its perihelion advances by about 43 arcseconds a century beyond what the other planets can produce, and in 1915 general relativity accounted for it. The correction to Newton’s force, for a slow body in the Schwarzschild solution, is an extra attraction of a different form, $3GML^2/(c^2 r^4)$ per unit mass. It goes as the inverse fourth power, and its strength depends on the body’s own $L$. It is not the inverse cube, and so it does not give a revolving ellipse in Newton’s exact sense.

Yet the theorem gives a clean way of asking how much this matters. The result for the turn per orbit from the relativistic term is the textbook $6\pi GM/(c^2 a(1 - e^2))$. An inverse-cube attraction $\lambda/r^3$ gives, for small $\lambda$, a turn $\pi\lambda/L^2$, and for an ellipse $L^2 = GMa(1 - e^2)$. Setting the two equal gives $\lambda = 6(GM/c)^2$, a strength that depends only on the central mass. With that one choice the dependence on $a$ and on $e$ is identical.

In other words, a universe with Newton’s gravity and an extra inverse-cube force of exactly that strength would precess all its planets at the observed rate, for any orbit size and eccentricity, at this order. Mercury’s anomaly, by itself, does not distinguish them. What distinguishes them is everything else general relativity predicts with no further adjustment: the bend Newton got half right, the delay that is not a bend, and gravitational redshift. A force law invented to fit a precession has no reason to deliver those, and the inverse-cube law delivers none of them. There are differences within a single orbit too, since a body under the inverse-fourth-power term does not follow the same $r(t)$ as the Keplerian one, by a small amount, whereas under the inverse-cube term it follows it exactly. A long and careful enough tracking of a close orbit can in principle separate the two; for a binary pulsar the separation is a matter of many terms that a real analysis has to fit at once.

This is the point of the exercise, and it is also what the theorem can teach. The orbit special relativity cannot close showed that even a mild change to the dynamics turns an orbit, and gets the rate wrong by a factor of six. A precession measures one number. Any force with the right average effect over an orbit will fit that number. That is why fits of alternative gravity to one secular rate are unimpressive, and why Newton’s own attempt to read the Moon’s apsidal motion off a modified force law gave a result he must have found unsatisfying: the rate that the theorem and the simplest force law suggest, for the Sun’s perturbation of the Moon, is half what the Moon actually does, a discrepancy that took until Clairaut to resolve.

A stranger reading of the same algebra #

There is another way of looking at the theorem, closer to modern physics. The term $L^2/r^3$ in the radial equation is the centrifugal term of the force a coordinate cannot see. Switching $L$ to $kL$ and adding the inverse cube is a statement that this term’s coefficient and an inverse-cube force’s coefficient cannot be told apart by the radial motion. So inverse-cube forces are unlike any other. Any other power changes the shape of the effective potential; this one only relabels it.

The same thing happens in quantum mechanics. The inverse-square potential, which gives an inverse-cube force, appears in the radial Schrödinger equation with the same form as the centrifugal term $\ell(\ell+1)/r^2$. Adding it shifts the effective value of $\ell$ to something that need not be a whole number and leaves the rest of the equation unchanged. The hydrogen spectrum, under such an extra term, keeps its Rydberg form with a shifted quantum number, a pattern that real atoms show in their outer electrons as the quantum defect. Newton’s revolving orbit and a modern calculation of an alkali metal’s spectrum use the same observation, two centuries apart.

There is also a limit where the theorem is telling a joke at the expense of a simple idea of force. Suppose someone records a body’s radial motion and nothing else, as astronomers often do when they measure a star’s line-of-sight speed. From that record alone it is not possible to recover the force law, because a whole family of force laws, one for each $k$, gives the same record. Only the direction of motion breaks the tie. Observers measuring a binary star by its radial velocity alone would have to supply something else to separate them, and in practice they assume a law and fit its parameters.

Where the construction stops #

The theorem needs a planar motion under a purely central force, a point mass and no dissipation. Add a non-central term, for instance the quadrupole of an oblate planet, and the orbit no longer lies in a plane or conserves its $L$, and the equality fails. Add drag, and $L$ changes, so $k$ is no longer constant for a given force. Nothing in the pictures includes any effect of the body on the centre; the two-body problem reduces to this form with a reduced mass, but only for two bodies.

The figures also leave out the question of what would make a force of this form. None of the fundamental forces in nature has an exact inverse-cube term. They appear as effective terms: a uniform spheroid’s quadrupole gives an inverse-fourth-power one in the plane; a dipole interacting with a charge gives an inverse-cube force at fixed orientation; and an attraction of this form appears in the radial motion of light near a black hole, whose unstable photon orbit is the same inverse-cube instability, discussed in the circle light cannot leave. A pure inverse-cube law of the required strength for planets has no physical basis, and is used here to make a point about what one number can show.

And the numerical demonstration is limited to the integrator. The agreement in the radial test is to one part in a million in a run of two orbital periods. That is enough to establish the theorem, since the algebra has already established it, and it does not constitute any kind of independent test of mechanics.

Still open: what the shape of an orbit can prove #

A precession gives one number, and the theorem shows how little that fixes. The natural question is how much more a whole orbit fixes. If an orbit is known completely as a curve, with its timing, Newton’s central force law can be recovered uniquely, and the theorem’s family of equivalents is broken by the angle. But real tracking is partial, noisy and full of other perturbations. In the strong-field case of stars around the black hole at the centre of the Milky Way, the orbit of S2 has been followed through two pericentres, and its Schwarzschild precession has been reported at the expected rate, with the shape and timing together contributing. How much of the alternative space a single such star excludes, and how many orbits it would take to rule out a hidden distribution of mass that could precess it the other way, is still a working question for the instruments that track those stars.

The theorem also suggests a question about method. A numerical integration can be checked against a revolving orbit at any strength, with an exact answer available at every step, and it is an unusually severe test because it demands both the radial motion and the angle. Whether long integrations of many-body systems keep that accuracy under their usual symplectic schemes is known in the cases that have been tried, and each new scheme needs the test again.

Add a force falling as the inverse cube to any central force, at the strength set by the orbit’s own angular momentum, and every radius keeps its moment while the angle is multiplied by a constant: the bare ellipse and its rosettes keep one $r(t)$ to one part in a million, and a force of strength $6(GM/c)^2$ gives the same precession as general relativity’s term across every eccentricity tested. A rate is an average, and an average can be met by many forces. The habit worth carrying is to ask, of any measurement that seems to establish a law, how many other laws would have produced the same reading.