The wall that angular momentum builds

Nothing in the two-body problem is harder to see than the thing that stops a body arriving. Gravity pulls inward at every distance, without exception and without limit, and yet the overwhelming majority of orbits never get anywhere near the centre. The usual explanation — it keeps missing — is true and unsatisfying, because it describes the outcome rather than the constraint.

There is a way of drawing the problem in which the constraint is the whole picture.

One equation, and one fewer dimension #

An orbit is a two-dimensional problem: the body has a distance and an angle, and both change. What makes it tractable is that the angle can be removed.

The removal is Kepler’s second law, written as a conservation law rather than as a statement about areas. The quantity $L = r^2\dot\theta$ does not change, so $\dot\theta = L/r^2$ can be substituted wherever it appears. Put that into the energy per unit mass,

$$E = \tfrac12\left(\dot r^2 + r^2\dot\theta^2\right) - \frac{GM}{r},$$

and the angular term becomes a function of $r$ alone:

$$E = \tfrac12 \dot r^2 + \underbrace{\frac{L^2}{2r^2} - \frac{GM}{r}}_{U_{\rm eff}(r)}.$$

What is left is a particle moving in one dimension, on a line, in the potential $U_{\rm eff}$. Every question about how close the body gets, how far out it reaches, and whether it comes back at all is now a question about a curve.

The second term is the price of the sideways motion. It is often called centrifugal, and the word carries an old argument about whether such a force exists; the drawing settles nothing about that and does not need to. What is on the page is arithmetic: the rotational kinetic energy $\tfrac12 r^2\dot\theta^2$, with $\dot\theta$ eliminated, is $L^2/2r^2$, and it is a decreasing function of radius. Squeezing an orbit inward at fixed angular momentum requires exactly that much energy.

Why nothing with angular momentum reaches the centre #

Compare the two terms as $r \to 0$. The attraction grows as $1/r$; the angular term grows as $1/r^2$. The second wins, always and by an ever-increasing margin, so the effective potential turns upward and goes to $+\infty$ at the origin no matter how small $L$ is, provided only that it is not exactly zero.

That is the wall. A body arrives with some total energy $E$, and it can only be where $U_{\rm eff}(r) \le E$, because the difference is $\tfrac12\dot r^2$ and that cannot be negative. The radius where the curve rises to meet the energy is a turning point: the radial velocity goes to zero there and reverses. The body has not stopped — it is moving at its fastest, all of it sideways — but it has stopped approaching.

The pair of turning points is the whole of the ellipse’s radial content. Solving $E = U_{\rm eff}(r)$ is solving a quadratic — $2Er^2 + 2GMr - L^2 = 0$ — and its two roots are $r_-$ and $r_+$, the near and far ends of the orbit. From those two numbers the eccentricity falls out as

$$e = \frac{r_+ - r_-}{r_+ + r_-},$$

and the semi-major axis is their average. An ellipse has been reconstructed without a single trigonometric function, from a picture of a curve and a horizontal line.

The circular orbit is a minimum, not a special case #

Textbooks usually introduce the circle as the ellipse with $e = 0$, which makes it a degenerate member of a family. The effective potential says something stronger and more useful: it is the bottom of a well.

Setting the derivative to zero,

$$\frac{dU_{\rm eff}}{dr} = \frac{GM}{r^2} - \frac{L^2}{r^3} = 0 \quad\Longrightarrow\quad r_{\rm circ} = \frac{L^2}{GM},$$

which is exactly the balance of the inward attraction against the angular term. The depth there is $U_{\rm min} = -G^2M^2/2L^2$, and since the total energy of an orbit is $-GM/2a$, a circular orbit’s energy is the least any orbit of that angular momentum can have. Of all the orbits with a given angular momentum, the circle is the one with the least energy. Anything else is that same angular momentum carrying an energy surplus, and the surplus is what does the radial oscillating.

That reframing is worth having because it explains a fact that is otherwise a coincidence: circular orbits are stable. A well has a bottom, and a body displaced from the bottom of a well returns to it and overshoots — which is precisely an eccentric orbit, seen radially. The period of that radial oscillation is the period of the orbit itself, and that the two agree exactly is the reason an ellipse closes rather than precessing. Nothing requires them to agree. They agree for the inverse-square law and for the harmonic-oscillator law and for nothing else, a result of Bertrand’s from 1873, and a millimetre’s worth of departure from the inverse square would show up as an orbit that never quite came back to where it started — which is the subject of what a residual of forty-three arcseconds a century turned out to mean.

Where the barrier is low enough to cross #

The wall’s height depends on $L$, so the question of whether a body can reach a given small radius is a question about angular momentum rather than about energy. A comet with a large angular momentum cannot reach the Sun however fast it is moving; a comet with almost none reaches it at almost any speed.

This has a consequence that sounds like a mistake and is not. Reaching the Sun is harder than leaving the solar system.

The arithmetic is short. A spacecraft in the Earth’s orbit is already travelling at 29.78 kilometres a second sideways, and that number is its angular momentum. To leave the solar system it must be raised to $\sqrt2$ times as fast, which costs 12.33 kilometres a second — the escape condition, applied to the Sun rather than the Earth. To reach the Sun the sideways speed has to be removed, and almost all of it: an orbit whose far end is at 1 AU and whose near end grazes the solar surface at 0.00465 AU has a speed at 1 AU of only 2.85 kilometres a second, so the burn needed is 26.9. More than twice as much.

That is why the Parker Solar Probe, whose entire purpose is to get close to the Sun, was launched on a Delta IV Heavy and then spent seven years making seven flybys of Venus. Each flyby is a gravity assist run backwards: instead of stealing speed it hands angular momentum to the planet, a little at a time, because the direct burn is beyond any rocket that could be built. The mission profile is a picture of this wall.

What was actually measured #

The effective potential itself is never observed. It is a rearrangement of Newton’s law, and what is observed is a set of positions and times, from which a distance and an angle are computed, from which the energy and angular momentum are inferred.

The place where the inference is at its most direct is the Sun-grazing comets, of which the SOHO spacecraft has found more than five thousand since 1995. Their orbits are reconstructed from a few days of positions before they disappear, and the family called the Kreutz group has perihelion distances clustered around 0.005 to 0.008 AU — inside the corona. These are objects whose angular momentum is small enough to put the turning point below the solar surface, and the observable consequence is that they do not come back. The count is the measurement: the rate at which comets are seen to arrive and not depart is the rate at which orbits with $r_- < R_\odot$ occur in the population, which is an angular-momentum distribution read off a survival statistic.

The second measurement is more familiar and less obviously about this at all. Every planet’s semi-major axis is a statement about its energy and every planet’s eccentricity is a statement about how far its energy sits above the floor of its own well. The eight planets are nearly circular — the largest eccentricity among them is Mercury’s 0.206 — and in the language of this picture that is the claim that the solar system’s planets sit close to the bottoms of their wells. Something took the surplus out, and the current best answer is that the gas the planets formed in did the damping.

Where the picture stops #

Three limits, and the third is the interesting one.

It says nothing about the angle. A body’s radial history can be read off exactly, and the picture is silent about where in its orbit it is at any given moment. That question is Kepler’s equation, and it is transcendental. The effective potential’s clarity is bought by discarding precisely the part of the problem that has no closed-form answer.

It assumes two bodies. With three, angular momentum about the centre of mass is still conserved for the system but not for any one body, so there is no $L$ to eliminate and no such curve. The nearest equivalent is the potential of the rotating frame, and it is a genuinely different object. It is Newtonian, and the correction removes the wall. In general relativity the same reduction can be carried out, and the effective potential acquires a third term:

$$U_{\rm eff} = -\frac{GM}{r} + \frac{L^2}{2r^2} - \frac{GML^2}{c^2r^3}.$$

The new term is negative and goes as $1/r^3$, so at small enough radius it beats the barrier. The wall is not infinitely high after all; it has a summit, and beyond the summit the curve plunges. A body that gets over the top does not turn around — it arrives.

Everything peculiar about orbits near a black hole follows from that one change of shape. The barrier has a maximum, so there is a smallest radius at which a stable circular orbit exists, at $6GM/c^2$ for a non-rotating hole, and inside it the well has no bottom. Matter spiralling inward through an accretion disc gets that far and then falls, which is why the inner edge of the disc is where it is and why the efficiency of accretion — some 6 per cent of the rest mass for a non-rotating hole, up to 42 for a maximally rotating one — is a fixed number rather than a property of the fuel. It is also why the growth of a black hole is capped by the light it emits rather than by the supply.

The hardest place to go is the middle #

The wall has a practical consequence that surprises everybody the first time they meet it: the most expensive destination in the solar system is the Sun.

A spacecraft leaving the Earth inherits the Earth’s motion — thirty kilometres a second around the Sun — and that motion is almost entirely transverse. It is angular momentum, and the barrier it builds has its inner turning point near the Earth’s own orbit. Pointing the spacecraft inward does not help; adding speed towards the Sun raises the energy without removing the angular momentum, and the body arrives at a perihelion barely closer than where it started.

To reach the Sun the transverse motion has to be cancelled, which means removing something close to the whole thirty kilometres a second. That is more than twice the velocity change needed to leave the solar system entirely, and far beyond any launch vehicle.

So the missions that go close do not do it by burning. They do it by handing their angular momentum to a planet: a spacecraft passing Venus can leave with less angular momentum than it arrived with, and repeating the encounter walks the perihelion down. The solar probe now flying inside ten solar radii took seven Venus flybys over seven years to get there, and each one lowered its perihelion by a step that no engine of that size could have delivered.

The same statement in reverse explains why a small body does not simply fall into the Sun either. Anything in orbit has angular momentum, and losing it requires a torque; a comet arrives at a small perihelion because it was given a nearly radial orbit far away, not because it spiralled in. The barrier is why the solar system is a set of orbits rather than a queue of infalling material, and it is the reason a disc has to have a way of transporting angular momentum outward before anything can accrete at all.

The wall a photon feels #

Light has no mass and it has an impact parameter, and in general relativity the same reduction applies to it.

A photon’s effective potential has only two terms — the barrier and the relativistic correction — and no Newtonian well at all. The result is a single hump, peaking at $3GM/c^2$, which is one and a half times the horizon radius.

That peak is the photon sphere: the radius at which light can orbit, unstably, going round and round until the smallest disturbance sends it in or out. It has no analogue in Newtonian gravity, and it is entirely a feature of the shape of this curve.

Whether a photon aimed at a black hole escapes or not is decided the same way as for a massive body: by whether it clears the hump. The condition works out to an impact parameter of $\sqrt{27}\,GM/c^2$, and the capture cross-section is that squared times $\pi$ — some six and three-quarter times the area of the horizon itself.

A black hole is therefore a larger target for light than it is for anything travelling slowly, and the disc of sky it removes is the shadow that has now been imaged. Its radius is fixed by the mass alone, which is what makes those images a mass measurement, and the whole of it follows from adding one term to the curve this essay is about.

Both statements are the same one seen from opposite ends. Angular momentum is what keeps a body out, and the only way in — for a spacecraft, for a comet, or for the gas in a disc — is to give it to something else.

That is also why the barrier has a place in this collection at all. It is not a subtlety of the reduction; it is the reason the material in a galaxy did not all fall to the middle, and the reason a mission to the Sun is harder than a mission to the stars.

It is also the reason a disc exists at all rather than a sphere: material can lose energy by radiating and cannot easily lose angular momentum, so it settles to the lowest-energy state at fixed angular momentum, which is a circle.

The same curve, one field over #

The construction generalises the moment the central mass stops being a point. Replace $GM$ with $GM(r)$, the mass enclosed within the radius in question — which the shell theorem says is the right substitution for any spherical distribution — and the effective potential describes a star orbiting inside a galaxy just as well as a planet orbiting outside a sun.

What changes is that the well is no longer of a shape that makes the radial period equal the orbital period. In a galaxy they differ, the orbit does not close, and a star traces a rosette rather than an ellipse. The ratio of the two periods is a measurable property of the mass distribution, and it is one of the ways a rotation curve is turned into a mass profile. The same picture, with the same two terms, in a system fifteen orders of magnitude larger.

Where the ladder goes next #

The obvious next rung is the one this essay has been leaning on and not stating: the shape of the well decides whether an orbit closes, and that is a question about the force law rather than about the orbit. The rung after it is the perturbed case, where the well itself moves slowly because a third body is pulling on it — the elements stop being constant and start being functions of time, and the effective potential becomes a curve that drifts under the body sliding along it.