Two posts ago I argued that the exponent 2 in $1/r^2$ is doing far more work than it looks like it’s doing — it hides the dimensionality of space, and it’s the whole reason a spherical shell is hollow. Last post I turned the crank on the differential equation and got a conic section out. And at the end of both, the same nagging question was left sitting there: fine, $1/r^2$ gives closed ellipses — but is it the only one that does?
Joseph Bertrand answered that in 1873, in a note of about four pages. The answer is: almost. There are exactly two force laws in the whole infinite space of central forces whose bound orbits all close up. One of them is gravity. The other one is a spring.
0. What “closed” actually means
Let me be precise about the claim first, because the theorem is easy to state sloppily and then it becomes either trivial or false.
We’re looking at a central force: a force that always points along the line to some fixed centre, with a magnitude that depends only on the distance $r$. That’s it. Gravity qualifies. So does a spring anchored at the origin. So does $F = -k/r^{3.7}$, or $F=-k r^{0.5}$, or anything else you feel like writing down.
Last post we established that such a force conserves angular momentum, so the motion stays in a plane, and the trajectory is some curve $r(\theta)$.
Now, a bound orbit is one that stays trapped between an inner radius $r_{\min}$ and an outer radius $r_{\max}$ forever. The points where $r$ hits those extremes are called apsides (perihelion and aphelion, if you’re talking about a planet). A bound orbit is closed if after some finite number of trips the particle returns exactly to its starting position with its starting velocity — the curve bites its own tail and repeats forever.
Here’s the thing people underestimate: a generic bound orbit does not close. The default behaviour of a central force is that the particle oscillates in and out between $r_{\min}$ and $r_{\max}$ while going around, and the perihelion drifts a little every lap. The orbit is a rosette that never quite repeats, and given long enough it smears out into an annulus.
【Figure 1: three orbits under $F\propto -1/r^{2.3}$, $-1/r^{2}$, $-1/r^{1.7}$ over ~8 revolutions. The middle one is a clean ellipse; the outer two are rosettes precessing in opposite directions.】
Closing is the exception, not the rule. Bertrand’s theorem says the exception happens for exactly two laws:
$$ F = -\frac{k}{r^{2}} \qquad\text{and}\qquad F = -kr . $$
Inverse square, and Hooke’s law. Kepler and the spring. Nothing else — not approximately nothing else, exactly nothing else.
Two remarks before we start. First, the theorem demands that every bound orbit close, at every energy and every angular momentum. Individual closed orbits are cheap; we’ll see in §4 that plenty of force laws give you closed near-circular orbits. It’s the “all of them” that’s expensive. Second, both survivors produce ellipses, but not the same kind — one has the centre of force at a focus, the other has it at the geometric centre. That coincidence is not a coincidence, and I’ll come back to it in §8.
1. The one tool we need
Everything below runs on the equation we derived last post. Let me restate it in a slightly more general form.
Throughout, write $f(r)$ for the force per unit mass (so $f<0$ means attraction), let $h = r^{2}\dot\theta$ be the specific angular momentum, and substitute $u = 1/r$. The derivation is word-for-word the one from last time, just with $-GM/r^2$ replaced by a general $f(r)$:
$$ \boxed{\;\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta^{2}} + u \;=\; -\frac{f(1/u)}{h^{2}u^{2}} \;\equiv\; J(u).\;} $$
Sanity check: put $f = -GM/r^{2} = -GMu^{2}$ and the right-hand side becomes $GM/h^{2}$, a constant, which is exactly the equation we solved last post. Good.
The function $J(u)$ is positive for attractive forces, and it packages the entire force law. From here on, the whole theorem is a story about $J$.
I also want one piece of vocabulary. Define the apsidal angle $\Phi$ as the angle swept by the particle in going from one perihelion to the next aphelion — i.e. half a radial oscillation.
【Figure 2: a rosette orbit with $r_{\min}$ and $r_{\max}$ marked, and the apsidal angle $\Phi$ shown as the wedge between a perihelion and the following aphelion.】
The orbit closes iff $\Phi$ is a rational multiple of $\pi$. Reason: after $2N$ apsides the particle has swept $2N\Phi$ in angle and is back to the same radius with the same radial speed. It has genuinely returned to its starting point only if $2N\Phi$ is a whole number of full turns, i.e. $2N\Phi = 2\pi M$ for some integers, i.e. $\Phi/\pi = M/N \in \mathbb{Q}$.
For a Kepler ellipse, $\Phi = \pi$ exactly: perihelion and aphelion sit on opposite ends of the major axis. That’s the target we’re trying to reproduce — or fail to reproduce.
2. Circular orbits, and whether they survive being poked
Start with the simplest possible orbit. A circle has $u = u_{0}$ constant, so $\mathrm{d}^{2}u/\mathrm{d}\theta^{2} = 0$, and the equation reduces to
$$ u_{0} = J(u_{0}). $$
For any sensible attractive force there’s a circular orbit at every radius — you just choose $h$ to match. Nothing interesting yet.
The interesting question is what happens when you nudge it. Write
$$ u(\theta) = u_{0} + \delta(\theta), \qquad |\delta| \ll u_{0}, $$
and Taylor-expand $J$ about $u_{0}$, keeping only the linear term:
$$ J(u_{0}+\delta) \approx J(u_{0}) + J'(u_{0})\,\delta . $$
Substituting into the orbit equation and using $u_{0} = J(u_{0})$ to kill the constants:
$$ \delta” + \delta = J'(u_{0})\,\delta \quad\Longrightarrow\quad \boxed{\;\delta” + \beta^{2}\delta = 0,\qquad \beta^{2} \equiv 1 – J'(u_{0}).\;} $$
This is the same simple-harmonic equation from the very first post in this series, except the independent variable is $\theta$ rather than $t$. If $\beta^{2}>0$ the solution is
$$ \delta(\theta) = A\cos(\beta\theta + \varphi), $$
so $r$ wobbles sinusoidally as you go around, and the radial wobble completes $\beta$ cycles per revolution. If $\beta^{2}<0$ you get growing exponentials instead — the circular orbit is unstable, the particle either spirals into the centre or escapes, and there are no nearby bound orbits at all. So $\beta^{2}>0$ is a hard requirement.
Since $\delta$ goes from max to min in $\Delta\theta = \pi/\beta$, the apsidal angle of a near-circular orbit is
$$ \Phi = \frac{\pi}{\beta}. $$
3. Getting $\beta$ in terms of the force
Now let’s turn $\beta^{2} = 1 – J'(u_{0})$ into something about $f(r)$. There’s a slick way to do this using logarithmic derivatives, and I like it because it makes the answer feel inevitable instead of computed.
The trick is this: because the circular-orbit condition says $J(u_{0}) = u_{0}$, we can freely multiply by $u_0/J(u_0) = 1$:
$$ J'(u_{0}) \;=\; \frac{u_{0}}{J(u_{0})}J'(u_{0}) \;=\; \left[\frac{\mathrm{d}\ln J}{\mathrm{d}\ln u}\right]_{u_{0}} . $$
That bracket is just “the local power-law exponent of $J$” — if $J \propto u^{p}$ near $u_0$, then $\mathrm{d}\ln J/\mathrm{d}\ln u = p$. So all we have to do is read off the exponent of $J$.
From the definition, $J(u) = -f(1/u)/(h^{2}u^{2})$, so
$$ \ln J = \ln\bigl(-f(r)\bigr) – 2\ln u – \ln h^{2}, \qquad r = 1/u . $$
And $\ln r = -\ln u$, so $\mathrm{d}\ln u = -\,\mathrm{d}\ln r$. Differentiating:
$$ \frac{\mathrm{d}\ln J}{\mathrm{d}\ln u} \;=\; -\frac{\mathrm{d}\ln(-f)}{\mathrm{d}\ln r} – 2 \;=\; -\frac{r f'(r)}{f(r)} – 2 . $$
Therefore
$$ \boxed{\;\beta^{2} \;=\; 3 + \frac{r f'(r)}{f(r)}\Bigg|_{r_{0}} \;} $$
Let’s test it on a pure power law $f = -k/r^{n}$. Then $rf’/f = -n$, and
$$ \beta^{2} = 3-n, \qquad \Phi = \frac{\pi}{\sqrt{3-n}} . $$
| $n$ | force | $\beta = \sqrt{3-n}$ | $\Phi = \pi/\beta$ | closed? |
|---|---|---|---|---|
| $-1$ | $-kr$ (Hooke) | $2$ | $\pi/2 = 90^\circ$ | yes |
| $0$ | constant | $\sqrt{3}$ | $103.9^\circ$ | no |
| $1$ | $-k/r$ | $\sqrt{2}$ | $127.3^\circ$ | no |
| $2$ | $-k/r^{2}$ (Kepler) | $1$ | $\pi = 180^\circ$ | yes |
| $2.5$ | $-k/r^{2.5}$ | $\sqrt{0.5}$ | $254.6^\circ$ | no |
| $\ge 3$ | — | imaginary | — | circular orbit unstable |
Two things fall out of this table immediately, and they’re both worth pausing on.
The $n\ge3$ row is the same wall we hit last post. Remember I said at the end of the ellipse post that if you try $1/r^{3}$ the differential equation turns into a mess? Here’s the reason, stated properly: at $n=3$ the restoring effect vanishes entirely and beyond it the circular orbit is unstable. There are no bound non-circular orbits to ask about. The universe of candidate force laws is capped at $n<3$ before we’ve done any real work.
Kepler and Hooke already look distinguished. They are the only two rows where $\beta$ is a whole number. $\beta=1$ means the radius oscillates once per revolution — one perihelion, one aphelion, and the ellipse’s long axis stays put. $\beta=2$ means twice per revolution — two perihelia and two aphelia, which is exactly what an ellipse centred on the origin looks like.
But “already look distinguished” is not a proof. $\Phi$ doesn’t have to be $\pi$ or $\pi/2$; it just has to be a rational multiple of $\pi$. Take $\beta = 3/2$, i.e. $n = 3 – 9/4 = 0.75$. Then $\Phi = 2\pi/3$, so $\Phi/\pi = 2/3 = M/N$, and the near-circular orbits close after 2 laps and 3 radial wobbles. They’re perfectly good closed orbits. So $F \propto -r^{-3/4}$ is still in the running, and so is an infinite family of others.
4. The rationality trap: the force has to be a power law
Here’s my favourite step in the argument, because it’s the one place where a piece of pure mathematics does violence to physics for free.
So far $\beta$ was computed at one particular radius $r_{0}$. But the theorem demands that every bound orbit close, which includes near-circular orbits at every radius. So for each $r_0$ we need
$$ \beta(r_{0}) = \frac{p(r_0)}{q(r_0)} \in \mathbb{Q}. $$
Now: $\beta(r_{0}) = \sqrt{3 + r_{0}f'(r_{0})/f(r_{0})}$ is a continuous function of $r_{0}$ (any force law we’d take seriously is smooth). And a continuous function on an interval that only ever takes rational values must be constant.
Why? Because a continuous function on an interval takes every value in between — that’s the intermediate value theorem. If $\beta$ ever changed at all, it would have to pass through the irrationals in between, which it isn’t allowed to do. The rationals are riddled with holes; a continuous curve can’t hop across them. So $\beta$ is stuck.
I find this genuinely delightful. We asked a question about orbits and got told about the topology of the number line, and the number line answered back with a constraint on the force law. Setting $\beta^{2} = \text{const}$:
$$ \frac{r f'(r)}{f(r)} = \beta^{2}-3 \equiv -n \quad(\text{a constant}) \;\Longrightarrow\; \frac{\mathrm{d}\ln(-f)}{\mathrm{d}\ln r} = -n \;\Longrightarrow\; f(r) = -\frac{k}{r^{n}} . $$
Result so far. The force must be a pure power law $-k/r^{n}$ with $n<3$, and its apsidal angle for near-circular orbits is $\pi/\sqrt{3-n}$, with $\sqrt{3-n}$ rational.
That’s already a dramatic narrowing. But we’re not done: we’ve only looked at orbits that are barely distinguishable from circles.
5. Two extreme orbits finish the job
The remaining leverage is this. $\Phi$ is a function of the orbit’s energy $E$ as well as its radius, and it varies continuously with $E$. By exactly the same rationality argument, $\Phi$ cannot change with energy either. It’s pinned to a single value for the entire force law.
So here’s the plan: compute $\Phi$ for an orbit as far from circular as we can possibly get, and demand that it equal $\pi/\sqrt{3-n}$. Two extremes are available, depending on whether the potential falls off or grows.
To set up either one, we need the orbit integral. Conservation of energy per unit mass:
$$ \tfrac12\dot r^{2} + \frac{h^{2}}{2r^{2}} + U(r) = E, $$
and dividing by $\dot\theta = h/r^{2}$ to trade $t$ for $\theta$:
$$ \left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^{2} = \frac{r^{4}}{h^{2}}\left[2\bigl(E-U(r)\bigr) – \frac{h^{2}}{r^{2}}\right]. $$
Substituting $u = 1/r$ (so $\mathrm{d}u = -\mathrm{d}r/r^{2}$) tidies this into
$$ \boxed{\;\Phi = \int_{u_{\min}}^{u_{\max}} \frac{h\,\mathrm{d}u}{\sqrt{\,2\bigl(E-U(1/u)\bigr) – h^{2}u^{2}\,}}\;} $$
with the limits being the turning points, where the square root vanishes.
5.1 Case A: $1<n<3$, the orbit that barely escapes
Here $U(r) = -\alpha/r^{s}$ with $s = n-1 \in (0,2)$ and $\alpha = k/s$. The potential goes to zero at infinity, so orbits are bound only for $E<0$. Take the limit $E \to 0^{-}$: the aphelion runs off to infinity and the orbit becomes marginally unbound. Set $E=0$ exactly and compute the swept angle from perihelion out to infinity.
$$ \Phi = \int_{0}^{u_{\max}} \frac{h\,\mathrm{d}u}{\sqrt{2\alpha u^{s} – h^{2}u^{2}}}, \qquad u_{\max} = \left(\frac{2\alpha}{h^{2}}\right)^{1/(2-s)} . $$
Now the nice part. Rescale to $x = u/u_{\max}$. Since $2\alpha u_{\max}^{s} = h^{2}u_{\max}^{2}$ by definition of $u_{\max}$, the whole prefactor cancels and every constant in the problem — $h$, $\alpha$, everything — vanishes at once:
$$ \Phi = \int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{x^{s}-x^{2}}} . $$
All that’s left is the exponent. Which is precisely the phenomenon this whole series is about, so let’s finish it. Pull out $x^{s/2}$ and substitute $y = x^{(2-s)/2}$, so that $\mathrm{d}x = \frac{2}{2-s}\,y^{s/(2-s)}\mathrm{d}y$ and $x^{s/2} = y^{s/(2-s)}$:
$$ \Phi = \int_{0}^{1}\frac{\mathrm{d}x}{x^{s/2}\sqrt{1-x^{2-s}}} = \frac{2}{2-s}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{1-y^{2}}} = \frac{2}{2-s}\cdot\frac{\pi}{2}, $$
$$ \Phi_{E\to0} = \frac{\pi}{2-s} = \frac{\pi}{3-n} . $$
One substitution and an $\arcsin$. Compare with the near-circular value and set them equal:
$$ \frac{\pi}{\sqrt{3-n}} = \frac{\pi}{3-n} \quad\Longrightarrow\quad \sqrt{3-n} = 3-n \quad\Longrightarrow\quad 3-n = 1 \quad\Longrightarrow\quad \boxed{n=2}. $$
The only number equal to its own square root (other than zero, which is excluded) is 1. That’s the entire content of the inverse-square law’s uniqueness, in this half of the argument.
5.2 Case B: $n\le1$, the orbit that’s going way too fast
Here the potential grows with $r$ — for $n<1$, $U = \alpha r^{1-n}$ up to constants; at $n=1$ exactly, $U = k\ln r$. Either way the particle can never escape, so every orbit is bound and $E$ can be taken arbitrarily large. That’s the other extreme.
The argument is almost physical rather than computational. When $E$ is enormous, the potential energy is negligible compared to $E$ over the entire inner part of the trajectory, so near perihelion the particle just travels in a straight line past the centre. And for a straight line, the angle swept between the point of closest approach and “very far away” is exactly $90^\circ$.
【Figure 3: a high-energy orbit — nearly straight past the centre, sweeping $\pi/2$ from perihelion before the far-away potential finally turns it around.】
To confirm it from the integral: drop $U$ next to $E$, so $u_{\max} \to \sqrt{2E}/h$ and
$$ \Phi \to \int_{0}^{\sqrt{2E}/h}\frac{h\,\mathrm{d}u}{\sqrt{2E – h^{2}u^{2}}} = \left[\arcsin\frac{hu}{\sqrt{2E}}\right]_{0}^{\sqrt{2E}/h} = \frac{\pi}{2}. $$
Matching to the near-circular value:
$$ \frac{\pi}{\sqrt{3-n}} = \frac{\pi}{2} \quad\Longrightarrow\quad 3-n = 4 \quad\Longrightarrow\quad \boxed{n=-1}, $$
which is $f = -kr$: Hooke’s law. And note this also kills $n=1$ on the way past — the logarithmic potential is confining, so it’s subject to this same limit, and $\pi/\sqrt{2} \ne \pi/2$.
5.3 The verdict
Collecting the cases:
- $n \ge 3$: circular orbits unstable, no bound orbits to speak of. Dead.
- $1 < n < 3$: the $E\to0$ limit forces $n=2$. Kepler survives.
- $n \le 1$: the $E\to\infty$ limit forces $n=-1$. Hooke survives.
Every other central force in existence has at least one bound orbit that fails to close.
【Note to self: be honest in the post that Bertrand’s original argument, and Goldstein’s textbook version, go by expanding the orbit equation to third order in the perturbation and matching Fourier coefficients. That route is more self-contained but three pages of algebra. The two-limits argument above is Landau’s, and it’s cleaner, but it does lean on the continuity-in-$E$ step, which I’ve asserted rather than proved. Flag it as a shortcut, don’t pretend it’s the standard proof.】
6. Checking that the survivors actually do survive
We’ve shown at most two laws work. We should confirm they really do, otherwise the theorem might be about the empty set.
Kepler. Done last post. Solving $u”+u = GM/h^{2}$ gives $r = p/(1+e\cos\theta)$, which is periodic in $\theta$ with period exactly $2\pi$. Perihelion at $\theta=0$, aphelion at $\theta=\pi$: $\Phi = \pi$, matching $\pi/\sqrt{3-2}$. The ellipse has the centre of force at a focus.
Hooke. Much easier — don’t even use polar coordinates. $\vec F = -k\vec r$ separates completely in Cartesians:
$$ \ddot x = -\omega^{2}x, \qquad \ddot y = -\omega^{2}y, \qquad \omega = \sqrt{k/m}, $$
two independent SHMs with identical frequency:
$$ x = A\cos(\omega t + \varphi_{1}), \qquad y = B\cos(\omega t + \varphi_{2}). $$
Same $\omega$, so both coordinates return to their starting values after one period $T = 2\pi/\omega$, no matter what $A,B,\varphi_1,\varphi_2$ are. Every orbit closes after exactly one period. The curve is an ellipse with the centre of force at the geometric centre, and $r$ reaches its maximum twice per lap — $\Phi = \pi/2$, matching $\pi/\sqrt{3-(-1)}$. ✓
Notice how differently the two mechanisms feel. Hooke closes because the equations decouple into two clocks that happen to tick at the same rate. Kepler closes because of a much subtler conspiracy in a coupled nonlinear system. Yet they land on the same theorem.
7. What this looks like when you point a telescope at it
Here’s why Bertrand’s theorem isn’t just a curiosity. It converts a question about shape into a question about drift, and drift accumulates.
Suppose the gravitational exponent were not 2 but $2+\varepsilon$ for some tiny $\varepsilon$. Then
$$ \Phi = \frac{\pi}{\sqrt{1-\varepsilon}} \approx \pi\left(1+\frac{\varepsilon}{2}\right), $$
so each full radial cycle sweeps $2\Phi$ instead of $2\pi$, and the perihelion advances by
$$ \Delta\varpi \approx \pi\varepsilon \quad\text{per orbit}. $$
Now run it backwards on Mercury. Mercury’s perihelion advances about $43”$ per century beyond what the other planets account for. Its year is $0.2409$ of ours, so it completes about $415$ orbits per century, giving $0.104”$ per orbit, i.e. $5.0\times10^{-7}$ radians. Setting $\pi\varepsilon = 5.0\times10^{-7}$:
$$ \varepsilon \approx 1.6\times10^{-7}, \qquad n \approx 2.00000016 . $$
This is not a hypothetical exercise. Asaph Hall proposed exactly this in 1894 — keep Newton, just nudge the exponent to $2.00000016$, and Mercury is explained. It’s a beautifully economical idea, and it was wrong: the same nudge makes a prediction for the Moon’s perigee that doesn’t fit, and it’s ugly in a way that’s hard to articulate but everyone felt.
General relativity got the same $43”$ without touching the exponent at all. In GR the orbit equation picks up one extra term:
$$ u” + u = \frac{GM}{h^{2}} + \frac{3GM}{c^{2}}u^{2}, $$
and that little $u^{2}$ — an effective $1/r^{4}$ correction to the force — breaks the closure and makes the ellipse rotate. Same observable, completely different bookkeeping.
The moral I’d take: because closure is so fragile, precession is an exquisitely sensitive probe of the force law. If lots of exponents produced closed orbits, a non-closing orbit would tell you nothing much. Because only $n=2$ does, every arcsecond of unexplained drift is a measurement of how the law departs from inverse-square. Mercury’s $43”$ per century — an angle roughly like the width of a human hair at 100 metres, accumulated over a century — was enough to help kill Newtonian gravity.
8. What’s hiding under the closure
One last layer, because this is the part that made me want to write the post.
Ask why closure is unusual in the first place. A bound orbit has two independent periodicities: the radial in-and-out, and the going-around. In general those two periods are incommensurable, and over time the trajectory wanders over a two-dimensional annulus, coming arbitrarily close to every point in it. A closed orbit is one that stays trapped on a one-dimensional curve instead.
Being confined to a smaller set than you’d expect is the signature of an extra constraint. There must be an additional conserved quantity, beyond energy and angular momentum, doing the trapping.
For Kepler, that quantity is the Laplace–Runge–Lenz vector:
$$ \vec A = \vec v\times\vec L – GMm\,\hat r . $$
You can verify $\mathrm{d}\vec A/\mathrm{d}t = 0$ directly, and it takes only a few lines provided the force is exactly inverse-square — put any other exponent in and the cancellation collapses. Geometrically $\vec A$ points from the focus along the major axis toward perihelion. So “the LRL vector is conserved” and “the perihelion doesn’t move” and “the orbit closes” are three phrasings of one fact. Hooke’s law has its own analogue, a conserved symmetric tensor rather than a vector.
And there’s a punchline I only learnt recently: the two exceptions are secretly the same exception. If you write the orbital plane as the complex plane and apply the map $z \mapsto z^{2}$, Hooke ellipses turn into Kepler ellipses. (This is the Bohlin–Kasner transformation.) Squaring the plane converts the centre into a focus and $-kr$ into $-k/r^{2}$. Bertrand’s two survivors are one object, viewed through two different charts.
So here is the full accounting of what’s packed into that superscript in $F = GMm/r^{2}$. From the shell theorem post: the dimensionality of space, the hollowness of shells, Gauss’s law, the masslessness of the photon. From the last post: conic sections. And from this one: closed orbits, a conserved vector nobody wrote into the law, a hidden symmetry, and the fact that Mercury’s perihelion drifting by an arcsecond a year is a signal at all rather than noise.
Feynman’s point again, and I keep finding it more true than I expect. $F = GMm/r^{2}$ looks like a sentence about two masses tugging on each other. It is not. It’s a compressed archive, and every time we’ve unzipped a bit of it something has fallen out that the sentence never mentioned.
Next up: I want to take the Laplace–Runge–Lenz vector seriously — prove it’s conserved, and then use it to re-derive the whole orbit equation in about five lines, with no differential equation at all. Same ellipse, third road there.
References / further reading
- J. Bertrand, Théorème relatif au mouvement d’un point attiré vers un centre fixe, C. R. Acad. Sci. 77 (1873), 849–853. The original — short, and worth looking at even if your French is bad.
- Goldstein, Classical Mechanics, §3.6. The standard third-order-expansion proof.
- Landau & Lifshitz, Mechanics, §14 and problems. Where the two-limits shortcut in §5 comes from.
- Arnold, Huygens and Barrow, Newton and Hooke. For the $z\mapsto z^2$ correspondence in §8.
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