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Sandbox Physics
Kozai–Lidov cycles from torque to planetary orbits

AC09 · L1

Angular momentum connects orbital shape and tilt

A distant circular companion preserves one component of the inner orbit’s angular momentum in the doubly averaged model. That constraint explains why a growing eccentricity accompanies a changing inclination. Energy fixes the size of the ellipse; angular momentum controls its shape and plane.

Start with the planet’s sideways motion

Use a planet of negligible mass around a primary star. Its position and velocity are measured relative to that star, in a frame whose axes do not rotate. The specific angular momentum is their cross product: “specific” means per unit planet mass. It points perpendicular to the instantaneous orbit plane.

At periapsis the radial velocity vanishes, so the speed is entirely sideways. The angular-momentum size is the periapsis distance times that speed. Substitute this speed into the two-body orbital energy, then use the ellipse’s periapsis distance. Cancelling the common factor gives the angular-momentum identity below.

A circle has the largest angular momentum at a given semimajor axis. An ellipse with eccentricity eight-tenths has only six-tenths of that circular value. Its planet spends a long time far out, then sweeps rapidly through periapsis. The cross product keeps track of that sideways motion throughout the orbit.

h=r×v,E=v22−GMr=−GM2a\boldsymbol h=\boldsymbol r\times\boldsymbol v,\qquad E=\frac{v^2}{2}-\frac{GM}{r}=-\frac{GM}{2a}
Position, velocity and energy here are relative to the primary; the energy identity is exact for an isolated Kepler orbit.
q=a(1−e),vq=hq,h22q2−GMq=−GM2a⟹h2=GMa(1−e2)q=a(1-e),\quad v_q=\frac{h}{q},\quad \frac{h^2}{2q^2}-\frac{GM}{q}=-\frac{GM}{2a}\quad\Longrightarrow\quad h^2=GMa(1-e^2)
The periapsis substitution uses the ellipse’s own velocity and orbital energy.

Reading: Tremaine & Yavetz, Why do Earth satellites stay up?, arXiv:1309.5244v1

Keep the component along the outer orbit axis

Average over the fast motion of the planet and its circular outer companion. The outer orbit then behaves like an axisymmetric ring. Rotating the entire inner ellipse around that ring’s axis changes no averaged interaction energy. The associated angular-momentum component is conserved.

Normalize the angular momentum by that of a circular orbit with the same semimajor axis. Its length becomes the square root of one minus eccentricity squared. The mutual inclination is the angle between its direction and the fixed outer axis. Their projection is therefore a product of that length and the inclination cosine.

For the default initial eccentricity of 0.05 and mutual inclination of 65 degrees, the conserved projection is approximately 0.422090. If eccentricity later reaches 0.8, angular-momentum length is 0.6 and the inclination must fall to about 45.3 degrees. The outer star has twisted the orbit while reducing its angular-momentum size.

The same calculation sets a geometric ceiling. Angular-momentum length cannot be smaller than the magnitude of one of its components. That ceiling alone would allow eccentricity about 0.907. The gravitational Hamiltonian imposes an additional restriction; the actual default averaged maximum is about 0.838909. The next chapter derives the stronger result.

j=hGMa,j=1−e2,j∥=j⋅n^2=1−e2cos⁡i\boldsymbol j=\frac{\boldsymbol h}{\sqrt{GMa}},\quad j=\sqrt{1-e^2},\quad j_\parallel=\boldsymbol j\cdot\hat{\boldsymbol n}_2=\sqrt{1-e^2}\cos i
The outer unit normal defines the projection. Inclination here is the angle between the two orbital planes.
j∥=1−0.052cos⁡65∘=0.422090,cos⁡i=0.4220900.600j_\parallel=\sqrt{1-0.05^2}\cos65^\circ=0.422090,\qquad \cos i=\frac{0.422090}{0.600}
The numerical example assumes a prograde orbit and the conserved projection of the doubly averaged circular-outer problem.
e≤1−j∥2e\leq\sqrt{1-j_\parallel^2}
This necessary geometric bound is weaker than conservation of the full averaged Hamiltonian.

Reading: Tremaine & Yavetz, Why do Earth satellites stay up?, arXiv:1309.5244v1 · Liu, Muñoz & Lai, Short-range forces in triple systems, arXiv:1409.6717v2

Stretching is different from dissipative migration

In this conservative averaged model the semimajor axis stays fixed. When eccentricity grows from 0.05 to 0.8, periapsis shrinks from 0.95 to 0.2 times the semimajor axis, while apoapsis expands from 1.05 to 1.8 times it. A close stellar passage is possible without the whole orbit becoming smaller.

To turn such passages into a short-period planet requires another process that removes orbital energy, usually tides in the migration scenario. A conservative relativistic periapsis precession rotates the ellipse but dissipates no energy. Neither a high eccentricity nor a precessing ellipse alone establishes that migration occurred.

The direct three-body problem also contains small changes in the osculating semimajor axis. That is an instantaneous Kepler description fitted to the current position and velocity, not the constant action of the averaged approximation. Keep these two meanings separate when reading the data plots.

q=a(1−e),Q=a(1+e)q=a(1-e),\qquad Q=a(1+e)
At fixed semimajor axis, the nearest and farthest distances move in opposite directions.

Reading: Liu, Muñoz & Lai, Short-range forces in triple systems, arXiv:1409.6717v2 · Gupta et al., A hot-Jupiter progenitor on a super-eccentric retrograde orbit, Nature 632 (2024)

Estimate how long the torque takes

The companion’s differential acceleration scales with the inner orbit size divided by the cube of the outer separation. Its orbit-averaged potential per unit planet mass therefore has the scale below. Divide the circular angular momentum by that potential to obtain a characteristic secular time.

The adopted units set the gravitational constant, primary mass, companion mass and initial inner semimajor axis to one. The inner mean motion is then one, so its period is about 6.28 time units. With an outer separation of twelve, the companion-primary relative period is about 184.687 and the secular time is 1728. The saved default eccentricity cycle takes about five secular times.

The companion’s relative Kepler clock uses the sum of both stellar masses. The inner test particle’s clock uses the primary mass. Omitting that distinction changes the fast phase and the numerical comparison. Increasing outer separation slows the torque much more rapidly than it lengthens the outer period.

Averaging is most comfortable when both orbital periods are short compared with the change in orbital angular momentum. High eccentricity can make the angular momentum change sharply during one outer revolution, even when the semimajor axes look well separated. Saved trajectories will test how both hierarchy and angular-momentum changes affect this approximation.

Φ0=Gm2a12a23,tK=GMa1Φ0=1n1Mm2(a2a1)3\Phi_0=\frac{Gm_2a_1^2}{a_2^3},\qquad t_K=\frac{\sqrt{GMa_1}}{\Phi_0}=\frac{1}{n_1}\frac{M}{m_2}\left(\frac{a_2}{a_1}\right)^3
The outer orbit is circular. This definition fixes the time normalization used by every plotted averaged equation.
n1=GMa13,n2=G(M+m2)a23,τ=ttKn_1=\sqrt{\frac{GM}{a_1^3}},\quad n_2=\sqrt{\frac{G(M+m_2)}{a_2^3}},\quad \tau=\frac{t}{t_K}
The full cycle period is measured separately in units of this characteristic time.

Reading: Tremaine & Yavetz, Why do Earth satellites stay up?, arXiv:1309.5244v1 · REBOUND 5.2.2 official Kozai example and IAS15 integrator

Saved numerical example and method

Synthetic dynamics in units with gravitational constant, primary mass, companion mass and initial inner semimajor axis all equal to one. Raw-vector RK4 uses a maximum secular step of 0.001; the direct reference uses frozen REBOUND 5.2.2 IAS15 states at 0.005 secular-time cadence.

Initial eccentricity and mutual inclinatione0=0.0500,i0=65.0∘e_0=0.0500,\quad i_0=65.0^\circ
Semimajor-axis ratioa2/a1=12.0a_2/a_1=12.0
Inner period, outer period, secular timeP1=6.28319,P2=184.687,tK=1728P_1=6.28319,\quad P_2=184.687,\quad t_K=1728
Representative replay timet/tK=1.800t/t_K=1.800
Eccentricity and mutual inclinatione=0.503185,i=60.762649∘e=0.503185,\quad i=60.762649^\circ
Periapsis and axial angular momentumq/a1=0.496815,j∥=0.422090q/a_1=0.496815,\quad j_\parallel=0.422090

The representative frame is taken at 1.8 secular times. The finite initial eccentricity and the near-circular analytic limit are kept distinct. The downloadable analysis reproduces the maxima and supplies actual direct-state records.

Try the orbit exchange

  1. Open the low-tilt preset, then the exchange preset. Compare the amplitude of the eccentricity curve and the changing inner plane.
  2. Pause near the eccentricity maximum. Use the readouts to check that the axial angular-momentum component remains steady while the total length decreases.

Exercises with explanations

  1. At eccentricity 0.6, how do periapsis, apoapsis and angular-momentum length compare with a circular orbit of the same semimajor axis?

    Read the explanation

    The nearest distance is 0.4 times the semimajor axis, the farthest is 1.6 times it, and angular momentum is 0.8 of the circular value. All three follow from the ellipse and energy identities derived above.

  2. Double the companion mass while keeping the primary mass and both semimajor axes fixed. Does every relevant time simply halve?

    Read the explanation

    The adopted secular time halves because the torque scale doubles. The outer period changes according to the square root of the total stellar mass: for an initially equal-mass pair its ratio becomes the square root of two-thirds. The inner Kepler period is unchanged.

  3. Could a prograde orbit with the default conserved projection reach an inclination of 90 degrees during this averaged evolution?

    Read the explanation

    No. Its positive axial projection stays about 0.422. A right angle would make the projection zero. A flip requires terms or fast orbital changes that can alter the axial projection.

Sources and calculation scope

  1. Tremaine & Yavetz, Why do Earth satellites stay up?, arXiv:1309.5244v1

    Fixed 20 September 2013 version, DOI 10.1119/1.4874853. Potential 18, vector equations 23, constraints 24 and section IV give the quadrupole model and near-circular critical-angle limit.

  2. Liu, Muñoz & Lai, Short-range forces in triple systems, arXiv:1409.6717v2

    Fixed 7 January 2015 version, DOI 10.1093/mnras/stu2396. Sections 2–4, quadrupole potential 7, octupole 8, GR 32–33/42 and Figures 3–4/7. The browser retains the circular-outer, test-particle quadrupole; the guide reads the wider model.

  3. REBOUND 5.2.2 official Kozai example and IAS15 integrator

    The official example motivates an independent direct reference. Frozen trajectories use REBOUND 5.2.2, source commit c34134716cd484ed9cdb9f7d54ef8d859bdf547d, IAS15 accuracy parameter one hundred-billionth, with separately stated initial conditions and stopping rules. REBOUND software is GPL-3.0-only; no publisher code is copied into the Lab.

  4. Gupta et al., A hot-Jupiter progenitor on a super-eccentric retrograde orbit, Nature 632 (2024)

    Version of record 17 July 2024, DOI 10.1038/s41586-024-07688-3, CC BY 4.0. Observed transits/RVs are Figures 1–2; Extended Data 5–6 give model-dependent initial conditions and an illustrative tidal-migration history. Full publisher-deposited XML, PMC11291287, was read for these distinctions.

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