Sundman’s Power Series Solution to the Three-Body Problem (Part II)

Summer research 2024: the three-body problem · Part 8 of 11

A weekly research log from my summer project at the University of Edinburgh, supervised by Prof. Jenni Smillie.

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Continuing from part 7, this post estimates how many terms of Sundman’s series are needed in a simple example, and shows why the series is useless in practice.

Lagrange’s equilateral solution

In 1772 Lagrange found a family of solutions in which the three bodies form an equilateral triangle at every instant. In the restricted problem, the small body then sits at one of the \(L_4\) or \(L_5\) Lagrange points, also called Trojan points.

The configuration is linearly stable only when one mass dominates. In the restricted problem, the two large masses must differ by a factor of more than about \(25\) (Routh’s criterion: a mass ratio above \(24.96\)). Otherwise any small perturbation grows and the triangle breaks up:

Estimating the number of terms

In the unperturbed solution each body moves on a circle, which we parametrise as \((x,y) = (\cos t, \sin t)\).

The mutual distances stay constant, say equal to \(r\), so Sundman’s factor

\[\Gamma = \left(1-e^{-r/l}\right)^3\]

is constant too, and \(t=\Gamma w\). Using the map from the previous post,

\[\begin{aligned} t&=\frac{2 \Omega \Gamma}{\pi} \log \frac{1+\tau}{1-\tau} \\ &= 2A\left(\tau + \frac{\tau^3}{3} + \frac{\tau^5}{5}+ O(\tau^7)\right), \\ \text{with}\quad A &= \frac{2\Omega\Gamma}{\pi} . \end{aligned}\]

Substituting into the Taylor series of \(\cos\) and \(\sin\) gives the coordinates as power series in \(\tau\):

\[\begin{aligned} \cos t &= 1 - 2A^2 \tau^2 - \left(\tfrac{4}{3} A^2 - \tfrac{2}{3} A^4\right) \tau^4 \\ &\quad - \left(\tfrac{46}{45}A^2 - \tfrac{8}{9} A^4 + \tfrac{4}{45} A^6\right) \tau^6+O(\tau^8), \\ \sin t &= 2A \tau + \left(\tfrac{2}{3}A - \tfrac{4}{3} A^3\right) \tau^3 \\ &\quad + \left(\tfrac{2}{5}A - \tfrac{4}{3}A^3 + \tfrac{4}{15} A^5\right) \tau^5 + O(\tau^7) . \end{aligned}\]

Now take masses \((m_0, m_1, m_2) = \left(\tfrac{1}{200}, \tfrac{1}{200}, \tfrac{99}{100}\right)\). For this configuration \(\Omega < 9 \times 10^{-8}\), which gives \(A < 4 \times 10^{-8}\).

How many terms do we need to follow the motion up to \(t = 1\)? Truncating the series for \(t\) after \(n\) terms, to get within \(0.1\) of \(t = 1\) we need

\[\begin{gathered} 2A \sum_{k=1}^{n} \frac{\tau^{2k-1}}{2k-1} > 0.9, \\ \text{i.e.}\quad S_\tau := \sum_{k=1}^{n} \frac{\tau^{2k-1}}{2k-1} > \frac{9}{20A} . \end{gathered}\]

With \(A< 4 \times 10^{-8}\) this requires \(S_\tau > 10^7\).

Compare with the partial sums of the odd harmonic series,

\[S_p=\sum_{k=1}^{p} \frac{1}{2k-1} \approx \tfrac12 \ln p + \text{const}.\]

Reaching \(S_p > S\) requires \(p>e^{2S-2.4}\), so \(S_p > 10^7\) requires \(p > 10^{8 \times 10^{6}}\). Since \(\lvert\tau\rvert < 1\), each term of \(S_\tau\) is smaller than the corresponding term of \(S_p\), so \(S_\tau\) needs at least as many terms: \(n > p > 10^{8 \times 10^{6}}\).

So even in this simple, collision-free example, about \(10^{8\,000\,000}\) terms are needed for one decimal place of accuracy. Sundman’s series converges, but far too slowly ever to be used.

Progress

The coding side of the project is now essentially complete, so I can concentrate on the theory, which will also form the core of my final report. Things have gone more smoothly than I expected.

Credits

This post is based on D. K. Yeomans’s exposition of Sundman’s work and on Belorizky’s paper (1930).

Changes on 25 October 2026

What was wrong.

• In the series for \(\cos t\), the coefficient \(\tfrac{46}{45}\) was written \(\tfrac{45}{46}\).
• The factor \(\Gamma\) was written with \(e^{1/l}\) instead of \(e^{-r/l}\).
• The sum \(\sum_{k=1}^{p} \frac{1}{2k-1}\) had \(2p-1\) in the denominator, and the error term after \(\tau^5\) is \(O(\tau^7)\), not \(O(\tau^6)\).
• A reference link no longer worked.

What changed. All are fixed, and the estimate of the number of terms is explained step by step.