Bertrand's Theorem

Contents

The orbit equation and its circular solutions
First stage: closure near a circle forces a power law
Second stage: the third-order solvability condition
The two laws, and that they really do close

This appendix proves Theorem 31.41 of Central Forces and Statics: among all central forces there are exactly two for which every bounded orbit closes, namely the inverse-square attraction \(f(r)=-k/r^{2}\) and the linear attraction \(f(r)=-\kappa r\) of the isotropic oscillator. Here \(k\) carries the unit \(\mathrm{N}\,\mathrm{m}^{2}\) and \(\kappa\) the unit \(\mathrm{N}/\mathrm{m}\), so that \(f\) is a force in newtons in both cases.

Nothing is quoted here. The whole argument is carried out from the Binet equation Equation (31.30) of Central Forces and Statics, Taylor's theorem with remainder (Theorem 11.38) and the elementary fact that a continuous rational-valued function on an interval is constant. There is no imported theorem and therefore no What is quoted here remark.

What the section does contain, and what the reader should watch for, is the division of labour between its two stages. The chapter's roadmap after Theorem 31.41 states it, and it is worth stating again because the point is easy to get wrong: the first-order stage does not isolate the two force laws. It narrows the candidates from all central forces to a one-parameter family of power laws, and no more. The third-order stage is where the two laws are picked out of that family, and it is the real content of the theorem.

The orbit equation and its circular solutions

Throughout, the motion is that of a particle of mass \(m\) (in \(\mathrm{kg}\); for a two-body problem read the reduced mass) in a central field \(f(r)\), with angular momentum \(L=mr^{2}\dot\phi\neq0\) of Equation (31.20), in \(\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}\). Write \(u:=1/r\), in \(/\mathrm{m}\).

Definition A55.1 (The orbit function).

For a central force \(f\) and a fixed angular momentum \(L\neq0\), the orbit function is

\begin{equation}\tag{A55.1} J(u) := -\frac{m}{L^{2}u^{2}}\,f\!\left(\frac{1}{u}\right)\ec \end{equation}

defined for those \(u>0\) at which \(f\) is defined, and carrying the unit \(/\mathrm{m}\). Combining \(ma_{r}=f(r)\) from Equation (23.4) with the Binet equation Equation (31.30) puts the shape of every orbit of that angular momentum in the form

\begin{equation}\tag{A55.2} \frac{\dd^{2}u}{\dd\phi^{2}}+u=J(u)\ep \end{equation}

The force is attractive where \(f<0\), that is where \(J>0\). Rests on Equation (31.30), Proposition 31.23 and Equation (23.4).

The time has been eliminated: Equation (A55.2) is an autonomous second-order equation for the shape alone, and closure of an orbit is the statement that its solution \(u(\phi)\) is periodic in \(\phi\) with a period commensurable with \(2\pi\).

Definition A55.2 (Circular orbit and its index).

A circular orbit of the field at angular momentum \(L\) is a constant solution \(u\equiv u_{0}>0\) of Equation (A55.2), that is a root of

\begin{equation}\tag{A55.3} u_{0}=J(u_{0})\ep \end{equation}

Its index is the dimensionless number

\begin{equation}\tag{A55.4} \beta^{2} := 1-J'(u_{0})\ep \end{equation}

Rests on Definition A55.1 and Equation (A55.2).

Lemma A55.3 (The index in terms of the force).

Let \(r_{0}=1/u_{0}\) be the radius of a circular orbit. Then

\begin{equation}\tag{A55.5} \beta^{2}=3+\frac{r_{0}f'(r_{0})}{f(r_{0})}\ec \end{equation}

which depends on the force law alone and not on \(L\) or \(m\) except through the radius at which it is evaluated. Rests on Definition A55.2 and Equation (A55.1).

Proof.

Derives Lemma A55.3. Differentiate Equation (A55.1), writing \(f\) and \(f'\) for the values at \(r=1/u\) and using \(\dd(1/u)/\dd u=-u^{-2}\):

\begin{equation}\tag{A55.6} J'(u)=-\frac{m}{L^{2}} \left[-2u^{-3}f-u^{-2}f'\,u^{-2}\right] =\frac{m}{L^{2}}\left[2u^{-3}f+u^{-4}f'\right]\ep \end{equation}

At a circular orbit Equation (A55.3) reads \(u_{0}=-\left(m/L^{2}\right)u_{0}^{-2}f(r_{0})\), so

\begin{equation}\tag{A55.7} \frac{m}{L^{2}}=-\frac{u_{0}^{3}}{f(r_{0})}\ec \end{equation}

which incidentally requires \(f(r_{0})<0\): a circular orbit exists only where the force attracts. Substituting Equation (A55.7) into Equation (A55.6) at \(u=u_{0}\),

\begin{equation}\tag{A55.8} J'(u_{0}) =-\frac{u_{0}^{3}}{f}\left[2u_{0}^{-3}f+u_{0}^{-4}f'\right] =-2-\frac{f'(r_{0})}{u_{0}f(r_{0})} =-2-\frac{r_{0}f'(r_{0})}{f(r_{0})}\ec \end{equation}

and Equation (A55.4) gives \(\beta^{2}=1-J'(u_{0})=3+r_{0}f'(r_{0})/f(r_{0})\), which is Equation (A55.5).

∎

First stage: closure near a circle forces a power law

Proposition A55.4 (The first-order narrowing).

Suppose every bounded orbit of the field \(f\) closes, and suppose \(f\) admits circular orbits at every radius of some interval \(0<r_{-}<r<r_{+}\) — which is the case as \(L\) ranges over an interval, by Equation (A55.7). Then \(\beta\) is one fixed rational number on that interval and

\begin{equation}\tag{A55.9} f(r)=-C\,r^{\beta^{2}-3}\ec\qquad C>0\ec \end{equation}

where \(C>0\) carries whatever unit makes the right-hand side a force in \(\mathrm{N}\), namely \(\mathrm{N}\) times a length raised to the power \(3-\beta^{2}\). Equivalently, in the orbit variable,

\begin{equation}\tag{A55.10} J(u)=A\,u^{1-\beta^{2}}\ec\qquad A=\frac{mC}{L^{2}}>0\ep \end{equation}

This is as far as the first-order calculation reaches: it leaves a one-parameter family, indexed by \(\beta\), and does not select any member of it. Rests on Lemma A55.3 and Equation (31.39).

Proof.

Derives Proposition A55.4. Perturb the circular orbit: put \(u=u_{0}+\delta\) in Equation (A55.2) and expand \(J\) about \(u_{0}\) by Theorem 11.38. Using Equation (A55.3) to cancel the constant terms and Equation (A55.4) to name the coefficient of \(\delta\),

\begin{equation}\tag{A55.11} \frac{\dd^{2}\delta}{\dd\phi^{2}}+\beta^{2}\delta =\tfrac{1}{2}J''(u_{0})\,\delta^{2} +\tfrac{1}{6}J'''(u_{0})\,\delta^{3}+O\!\left(\delta^{4}\right)\ep \end{equation}

To first order in \(\delta\) the right-hand side is absent and the solution is \(\delta=a\cos\left(\beta\phi\right)\) with the origin of \(\phi\) at an apsis. If \(\beta^{2}\le0\) the solution grows without bound and there are no bounded orbits in the neighbourhood at all, so \(\beta^{2}>0\) and \(\delta\) oscillates. Successive turning points of \(r\) are successive extrema of \(\delta\), separated by \(\Delta\phi=\pi/\beta\); that is, the apsidal angle Equation (31.39) of these orbits satisfies

\begin{equation}\tag{A55.12} \lim_{a\to0}\Phi=\frac{\pi}{\beta}\ep \end{equation}

An orbit closes exactly when the total angle swept between one apsis and the next repetition of the whole configuration is a whole multiple of \(2\pi\), that is when \(\Phi/\pi\) is rational. So \(\beta\) is rational for every circular radius in the interval. By Equation (A55.5), \(\beta\) is a continuous function of \(r_{0}\) wherever \(f\) is continuously differentiable and non-vanishing; a continuous function on an interval taking only rational values has connected image contained in \(\Q\), hence is constant. So \(\beta\) is one fixed rational number.

With \(\beta\) constant, Equation (A55.5) is the separable first-order equation \(rf'/f=\beta^{2}-3\) on the interval, whose solution is \(\log\abs{f}=\left(\beta^{2}-3\right)\log r+\text{const}\), that is Equation (A55.9), with \(C>0\) because \(f<0\) at a circular orbit. Substituting into Equation (A55.1),

\begin{equation}\tag{A55.13} J(u)=\frac{m}{L^{2}}u^{-2}\,C\,u^{-\left(\beta^{2}-3\right)} =\frac{mC}{L^{2}}\,u^{1-\beta^{2}}\ec \end{equation}

which is Equation (A55.10). Every \(\beta>0\) produces a power law satisfying the hypotheses used so far, so nothing has yet been excluded.

∎

Remark A55.5 (Why the first stage cannot finish the proof).

The condition just used is closure of the orbits infinitesimally near a circle, which is strictly weaker than closure of all bounded orbits. Equation (A55.12) is a limit as the amplitude \(a\) tends to zero, and it fixes the apsidal angle only in that limit; at finite amplitude the apsidal angle of a power law generally depends on \(a\), and an orbit whose apsidal angle drifts with amplitude cannot close at every amplitude even though it closes in the limit. The second stage is precisely the computation of that drift.

Lemma A55.6 (Derivatives of the power law).

For \(J\) of the form Equation (A55.10) with circular orbit at \(u_{0}\),

\begin{align} J'(u_{0})&=1-\beta^{2}\ec\tag{A55.14}\\ u_{0}J''(u_{0})&=-\beta^{2}\left(1-\beta^{2}\right)\ec \tag{A55.15}\\ u_{0}^{2}J'''(u_{0})&=\beta^{2}\left(1-\beta^{2}\right) \left(1+\beta^{2}\right)\ep\tag{A55.16} \end{align}

Rests on Equations (A55.3) and (A55.10).

Proof.

Derives Lemma A55.6. Equation (A55.3) applied to Equation (A55.10) reads \(u_{0}=Au_{0}^{1-\beta^{2}}\), that is

\begin{equation}\tag{A55.17} A\,u_{0}^{-\beta^{2}}=1\ep \end{equation}

Differentiating Equation (A55.10) three times,

\begin{align*} J'(u)&=A\left(1-\beta^{2}\right)u^{-\beta^{2}}\ec\\ J''(u)&=A\left(1-\beta^{2}\right)\left(-\beta^{2}\right) u^{-\beta^{2}-1}\ec\\ J'''(u)&=A\left(1-\beta^{2}\right)\left(-\beta^{2}\right) \left(-\beta^{2}-1\right)u^{-\beta^{2}-2}\ec \end{align*}

and evaluating at \(u_{0}\) with Equation (A55.17) gives Equations (A55.14), (A55.15) and (A55.16) in turn. Equation (A55.14) is a consistency check on Equation (A55.4): it returns \(\beta^{2}=1-J'(u_{0})\), as it must.

∎

Second stage: the third-order solvability condition

The apsidal angle at finite amplitude is computed by the standard device for a nonlinear oscillator: allow the frequency itself to depend on the amplitude, and fix that dependence by demanding that the solution stay periodic. The alternative bookkeeping — keeping the frequency at \(\beta\) and requiring that no term proportional to \(\phi\sin\left(\beta\phi\right)\) appear — gives the same condition, because such a term is exactly the first-order Taylor expansion of the frequency shift.

Theorem A55.7 (Bertrand's condition).

Let \(J\) be the power law Equation (A55.10) with \(\beta^{2}>0\), and let the radial oscillation about the circular orbit \(u_{0}\) have amplitude \(a\). Then its angular frequency is

\begin{equation}\tag{A55.18} \Omega^{2}=\beta^{2}+a^{2}\omega_{2}+O\!\left(a^{4}\right)\ec \qquad \omega_{2}=-\frac{5\left(J''\right)^{2}}{24\beta^{2}} -\frac{J'''}{8}\ec \end{equation}

with \(J''\), \(J'''\) evaluated at \(u_{0}\), and the apsidal angle is \(\Phi=\pi/\Omega+O(a^{4})\). If every bounded orbit closes then \(\omega_{2}=0\), and

\begin{equation}\tag{A55.19} 5\left(J''\right)^{2}+3\beta^{2}J'''=0 \quad\Longleftrightarrow\quad 2\beta^{4}\left(1-\beta^{2}\right)\left(4-\beta^{2}\right)=0\ec \end{equation}

so that \(\beta^{2}=1\) or \(\beta^{2}=4\). Rests on Lemma A55.6, Proposition A55.4 and Equation (A55.11).

Proof.

Derives Theorem A55.7. Introduce the stretched angle \(\theta=\Omega\phi\), so that \(\dd/\dd\phi=\Omega\,\dd/\dd\theta\) and \(\delta\) is sought as a \(2\pi\)-periodic function of \(\theta\). Writing a dot for \(\dd/\dd\theta\), Equation (A55.11) becomes

\begin{equation}\tag{A55.20} \Omega^{2}\ddot\delta+\beta^{2}\delta =\tfrac{1}{2}J''\delta^{2}+\tfrac{1}{6}J'''\delta^{3} +O\!\left(\delta^{4}\right)\ep \end{equation}

Expand in the amplitude,

\begin{equation}\tag{A55.21} \delta=a\delta_{1}+a^{2}\delta_{2}+a^{3}\delta_{3}+\cdots\ec \qquad \Omega^{2}=\beta^{2}+a^{2}\omega_{2}+O\!\left(a^{4}\right)\ec \end{equation}

the frequency carrying only even powers of \(a\) because Equation (A55.20) is unchanged by \(a\mapsto-a\) together with \(\theta\mapsto\theta+\pi\).

Order \(a\). \(\beta^{2}\left(\ddot\delta_{1}+\delta_{1}\right)=0\), so \(\delta_{1}=\cos\theta\), the amplitude being carried by \(a\) and the phase fixed by putting an apsis at \(\theta=0\).

Order \(a^{2}\). Using \(\cos^{2}\theta=\tfrac{1}{2}\left(1+\cos2\theta\right)\),

\begin{equation}\tag{A55.22} \beta^{2}\left(\ddot\delta_{2}+\delta_{2}\right) =\tfrac{1}{2}J''\cos^{2}\theta =\frac{J''}{4}\left(1+\cos2\theta\right)\ec \end{equation}

which has no resonant forcing — neither \(1\) nor \(\cos2\theta\) is a solution of the homogeneous equation — and is solved by

\begin{equation}\tag{A55.23} \delta_{2}=\frac{J''}{4\beta^{2}} \left(1-\tfrac{1}{3}\cos2\theta\right)\ec \end{equation}

as is checked by applying \(\ddot{\ }+1\): the constant reproduces itself, and \(-\tfrac{1}{3}\cos2\theta\) returns \(-\tfrac{1}{3}\left(-4+1\right)\cos2\theta=\cos2\theta\). The homogeneous solution that could be added here is absorbed into the definition of \(a\).

Order \(a^{3}\). Collecting the terms of that order in Equation (A55.20), and using \(\ddot\delta_{1}=-\cos\theta\),

\begin{equation}\tag{A55.24} \beta^{2}\left(\ddot\delta_{3}+\delta_{3}\right) =\omega_{2}\cos\theta +J''\delta_{1}\delta_{2} +\tfrac{1}{6}J'''\cos^{3}\theta\ep \end{equation}

Reduce the two nonlinear terms with \(\cos\theta\cos2\theta =\tfrac{1}{2}\left(\cos3\theta+\cos\theta\right)\) and \(\cos^{3}\theta =\tfrac{3}{4}\cos\theta+\tfrac{1}{4}\cos3\theta\):

\begin{align} J''\delta_{1}\delta_{2} &=\frac{\left(J''\right)^{2}}{4\beta^{2}} \left[\cos\theta-\tfrac{1}{3}\cos\theta\cos2\theta\right] \nn\\ &=\frac{\left(J''\right)^{2}}{4\beta^{2}} \left[\tfrac{5}{6}\cos\theta-\tfrac{1}{6}\cos3\theta\right]\ec \tag{A55.25}\\ \tfrac{1}{6}J'''\cos^{3}\theta &=\frac{J'''}{8}\cos\theta+\frac{J'''}{24}\cos3\theta\ep \tag{A55.26} \end{align}

The operator \(\ddot{\ }+1\) annihilates \(\cos\theta\), so Equation (A55.24) has a \(2\pi\)-periodic solution only if the total coefficient of \(\cos\theta\) on its right-hand side vanishes. By Equations (A55.25) and (A55.26) that coefficient is

\begin{equation}\tag{A55.27} \omega_{2}+\frac{5\left(J''\right)^{2}}{24\beta^{2}} +\frac{J'''}{8}=0\ec \end{equation}

which is the second member of Equation (A55.18). (The \(\cos3\theta\) terms are non-resonant and merely fix \(\delta_{3}\); they play no part in what follows.) The apsidal angle is half the period of \(\delta\) in \(\phi\), that is \(\pi/\Omega\).

Now impose closure. For each amplitude \(a\) the orbit closes only if \(\Phi/\pi=1/\Omega(a)\) is rational; \(\Omega\) is a continuous function of \(a\) by Equation (A55.18), and a continuous rational-valued function on an interval is constant — the same step used in Proposition A55.4, now applied in the amplitude rather than in the radius. Hence \(\Omega\) is independent of \(a\) and \(\omega_{2}=0\). Multiplying Equation (A55.27) by \(-24\beta^{2}\) gives the first member of Equation (A55.19).

It remains to evaluate it. Substituting Equations (A55.15) and (A55.16),

\begin{align} 5\left(J''\right)^{2}+3\beta^{2}J''' &=\frac{5\beta^{4}\left(1-\beta^{2}\right)^{2}}{u_{0}^{2}} +\frac{3\beta^{4}\left(1-\beta^{2}\right) \left(1+\beta^{2}\right)}{u_{0}^{2}}\nn\\ &=\frac{\beta^{4}\left(1-\beta^{2}\right)}{u_{0}^{2}} \left[5\left(1-\beta^{2}\right)+3\left(1+\beta^{2}\right)\right] \nn\\ &=\frac{2\beta^{4}\left(1-\beta^{2}\right) \left(4-\beta^{2}\right)}{u_{0}^{2}}\ec \tag{A55.28} \end{align}

since \(5-5\beta^{2}+3+3\beta^{2}=8-2\beta^{2}\). With \(\beta^{2}>0\) the factor \(\beta^{4}\) cannot vanish, so \(\beta^{2}=1\) or \(\beta^{2}=4\).

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Remark A55.8 (The two roots are not accidents of the algebra).

Each root switches off the condition for a different reason, and it is worth seeing both. At \(\beta^{2}=1\) the power law is \(J(u)=A\), a constant: every derivative of \(J\) vanishes, the right-hand side of Equation (A55.11) is empty to all orders, and the orbit equation is exactly linear — which is Equation (31.47), and is why the Kepler problem has an apsidal angle independent of amplitude rather than merely to third order. At \(\beta^{2}=4\) the derivatives do not vanish; by Equations (A55.15) and (A55.16), \(u_{0}J''=12\) and \(u_{0}^{2}J'''=-60\), and the two terms of Equation (A55.19) cancel against each other: \(u_{0}^{2}\left[5\left(J''\right)^{2}+3\beta^{2}J'''\right] =5\cdot12^{2}+3\cdot4\cdot\left(-60\right)=720-720=0\). The oscillator therefore closes by a cancellation between the second-order feedback of Equation (A55.23) and the direct cubic term, not because its orbit equation is linear.

The two laws, and that they really do close

The argument so far is one of necessity: no force law other than these two can have every bounded orbit closed. Sufficiency has to be checked separately, and must be checked for all bounded orbits and not only for those near a circle, since that is the property claimed.

Proposition A55.9 (Both exceptional laws close every bounded orbit).

For \(f(r)=-k/r^{2}\) every bounded orbit is an ellipse with the force centre at a focus, with apsidal angle \(\Phi=\pi\); for \(f(r)=-\kappa r\) every orbit is an ellipse with the force centre at its centre, with apsidal angle \(\Phi=\pi/2\). In both cases \(\Phi\) is independent of the energy and of the angular momentum. Rests on Equations (31.47) and (A55.5).

Proof.

Derives Proposition A55.9. The inverse square. With \(f=-k/r^{2}\), \(rf'/f=r\left(2k/r^{3}\right)/\left(-k/r^{2}\right)=-2\), so Equation (A55.5) gives \(\beta^{2}=1\), consistent with Theorem A55.7. The orbit is Equation (31.47), \(1/r=\left(mk/L^{2}\right) \left(1+e\cos\phi\right)\), obtained there without any expansion in amplitude. It is \(2\pi\)-periodic in \(\phi\) for every \(e\), and bounded precisely for \(e<1\); \(r\) is least at \(\phi=0\) and greatest at \(\phi=\pi\), so \(\Phi=\pi\) exactly, at every eccentricity. The force centre sits at \(r=0\), which is a focus of the conic.

The linear law. With \(f=-\kappa r\), \(rf'/f=r\left(-\kappa\right)/\left(-\kappa r\right)=1\), so \(\beta^{2}=4\) and \(\beta=2\). Here the direct route is better than the orbit equation. The force is \(\vect{F}=-\kappa\vect{x}\), so each Cartesian component of Equation (23.4) is an independent harmonic oscillator of the same angular frequency \(\omega_{0}=\sqrt{\kappa/m}\), in \(\mathrm{rad}/\mathrm{s}\). In the plane of the motion,

\begin{equation}\tag{A55.29} x(t)=a_{1}\cos\left(\omega_{0}t\right)\ec\qquad y(t)=a_{2}\cos\left(\omega_{0}t-\gamma\right)\ec \end{equation}

which is a closed ellipse centred on the origin, traversed once per period \(2\pi/\omega_{0}\), for every \(a_{1}\), \(a_{2}\) and phase \(\gamma\). Every orbit is bounded and every orbit closes. Its radius \(r=\sqrt{x^{2}+y^{2}}\) is a function of \(\cos^{2}\) and \(\sin^{2}\) of \(\omega_{0}t\) and therefore has period \(\pi/\omega_{0}\), half the orbital period: the particle passes two pericentres and two apocentres per revolution, one pair at each end of each principal axis, and the apsidal angle is a quarter of a revolution,

\begin{equation}\tag{A55.30} \Phi=\frac{2\pi}{4}=\frac{\pi}{2}=\frac{\pi}{\beta}\ec \end{equation}

in agreement with Equation (A55.12) — and here the agreement holds at every amplitude, not merely in the limit, which is the content of \(\omega_{2}=0\).

The contrast between the two cases is not a detail of geometry. For the inverse square the centre of force is a focus of the ellipse and the two apsidal distances differ; for the oscillator it is the centre, the two axes are traversed symmetrically, and the orbit closes after half as much angle. That is the same fact as \(\beta=1\) against \(\beta=2\).

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Remark A55.10 (What ``closed'' is doing in the statement).

Three distinct properties are in play and only the strongest is Bertrand's. (i) Some bounded orbit closes: true for every attractive power law, since the circular orbits themselves close. (ii) Every orbit infinitesimally near a circle closes: true for every power law with \(\beta\) rational — a countable but infinite family, including \(f\propto r^{-5/4}\) with \(\beta=\tfrac{1}{2}\) and \(f\propto r^{6}\) with \(\beta=3\). (iii) Every bounded orbit closes: true only for \(\beta=1\) and \(\beta=2\). The whole distance between (ii) and (iii) is Equation (A55.27), which measures how the apsidal angle drifts as the orbit is made less circular. For any other power law, \(\Phi\) varies continuously with the amplitude, takes irrational multiples of \(\pi\) on a dense set of amplitudes, and those orbits fill the annulus \(r_{-}\le r\le r_{+}\) without ever repeating, exactly as Theorem 31.41 asserts.

Remark A55.11 (Attribution).

The theorem is Bertrand's, announced in the Comptes rendus of the Paris Academy in 1873. That memoir is not among the sources verified for this treatise and has no key in references.bib, so the attribution is made here in words, on the same footing as Darboux's memoir in Remark A12.15. Nothing above rests on it: the proof given is self-contained, and the third-order stage in particular is carried out from Equation (A55.11) alone.

Remark A55.12.

Bertrand's Theorem discharges the derivation owed at Theorem 31.41 of Central Forces and Statics. The chapter uses the theorem twice. It uses it to say what the closure of planetary orbits is evidence for — the inverse-square law and nothing weaker — and it uses it, through Phenomenon 31.42, to turn Mercury's residual perihelion advance into a measurement: by Equation (A55.5) an apsidal angle differing from \(\pi\) is a force law differing from \(r^{-2}\), and the size of the difference is the size of the departure. The reader returning there should carry back the division of labour set out at the head of this section, since the chapter's roadmap states it and this appendix is where it is earned: Proposition A55.4 narrows all central forces to one parameter, and Theorem A55.7 spends that parameter.