# clarification of the Kozai mechanism

As Wikipedia says,

In celestial mechanics, the Kozai mechanism, or the Lidov–Kozai mechanism, is a perturbation of the orbit of a satellite by the gravity of another body orbiting farther out, causing libration (oscillation about a constant value) of the orbit's argument of pericenter. As the orbit librates, there is a periodic exchange between its inclination and its eccentricity.

My questions are:

Question A
Which is the least massive object? The tertiary object, which is the furthest object, or the satellite in the inner binary? It seems it is not necessary that the tertiary object be the least massive one, which violates what Wikipedia says.

Question B
How does the three-body system evolve?

There is a periodic exchange between its inclination and its eccentricity.

Whose inclination and whose eccentricity? Please specify them using m0, m1 or m2 in the figure below.

The orbit of the inner binary should become and more circular. Can it become circular, eccentric, circular, eccentric?

Question C
The inner binary will lose energy during the whole process, right? Which is the least massive object?

Quoting Wikipedia,

In the hierarchical, restricted three-body problem, it is assumed that the satellite has negligible mass compared with the other two bodies (the "primary" and the "perturber"), . . .

This is the case studied in Kozai (1962), specifically, the case of asteroids being perturbed by Jupiter. While not massless, the difference in mass is large enough that the asteroid's mass is negligible.

How does the three-body system evolve? . . . Whose inclination and whose eccentricity?

Wikipedia is again rather direct in this, stating that the conserved quantity $L_z$ depends on the satellite's orbit's eccentricity and inclination: $$L_z=\sqrt{1-e^2}\cos i$$ As the satellite's mass is negligible, it will not have any significant effect on its perturber.

Can it become circular, eccentric, circular, eccentric?

This is basically asking if the eccentricity (and therefore inclination) can be described by some periodic function. Again, this is given in the Wikipedia article. A slightly different but just as simple xpresison is given in Takeda & Rasio (2005): $$\text{Kozai Period}=P_{\text{perturbed}}\left(\frac{m_{\text{star}}+m_{\text{perturbed}}}{m_{\text{perturber}}}\right)\left(\frac{a_{\text{perturber}}}{a_{\text{perturbed}}}\right)^3(1-e_{\text{perturber}}^2)^{3/2}$$ In the approximation discussed above, in cases of extreme mass difference, $m_{\text{perturbed}}\to0$.

The inner binary will lose energy during the whole process, right?

The whole thing is periodic, so no energy is lost.

• Thank you, but you give no infomation at all. I see papers which talk about Kozai mechanism. It seems the mass of the tertiary object is between the satellite and the primary. The reason I ask this question is not I did not read Wiki carefully enough. Oct 23, 2015 at 1:41
• @questionhang I don't see how this gives "no information". I directly answer each point you made. Oct 23, 2015 at 1:41
• Sorry. Most of them are in Wiki. Wiki only gives a general case. Oct 23, 2015 at 1:57
• OK. Kozai mechanism has nothing to do with with the inner binary. The change is in the tertiary object? Oct 24, 2015 at 5:46
• Could you please specify which correspnds to the 'perturber' of the formula you give? m0 m1, or m2? Oct 24, 2015 at 5:54

The most simple Lidov-Kozai model is that of a massless object ($m_1$ in your diagram) rotating a massive object ($m_0$), which is itself in an orbit with another massive object ($m_2$).

This is a hierarchical 3-body system ($m_2$ is assumed to always be far enough from $m_0$ and $m_1$). It is easier to look at as two 2-body orbits:

Inner orbit - $m_0$ and $m_1$

Outer orbit - $m_2$ and $m_1$+$m_0$

Since $m_1$ is massless, the outer orbit is not affected by it and is a simple Kepler orbit of 2 bodies ($m_0$ and $m_2$) with fixed parameters. Because of that, the outer orbit defines the coordinates system and lie in XY plane, with its angular momentum $\vec{L}_{out}=L_{out}\hat{z}$. What we really have here is a motion of a test-particle ($m_1$ a.k.a the perturbed) around a massive object ($m_0$) in a binary system (with $m_2$). The inner orbit can be viewed as a Kepler orbit with a perturbation due to $m_2$ (a.k.a the perturber). Its parameters do change with time and are described by the Lidov-Kozai mechanism.

Using this model (which is probably what you're asking for):

Question A

The least massive object is the massless object $m_1$

Question B

What evolve are the parameters of the inner orbit ($m_0$ and $m_1$) - its eccentricity, inclination, angular momentum etc. How? in a periodic manner. The periodic change in the eccentricity literally means that the inner orbit becomes more circular, then more eccentric, then more circular again and again. The eccentricity-inclination change is easier to see because of the following constant of motion:

$$\sqrt{1-e^2}\cos i = const.$$

(It is not exactly $L_z$, but a scaled version of it $\frac{L_z}{\mu\sqrt{GMa_{in}}}$. $\mu,M$ being the reduced and total mass of the inner orbit respectively)

The fact of this being constant is not easy to see, but if given as a fact and the fact the $e$ is periodic, you can see that the inclination $i$ is periodic.

Question C

When deriving this model one is "averaging" out (over one full period) the true-anomaly (the exact position of the mass inside the orbit) of $m_1$ around $m_0$ -> meaning we refer to the inner orbit as an "elliptic ring", and the same goes for the outer orbit. We also assume no energy exchange between both orbits (rings), so both inner/outer semi-major axes are also fixed (from the relation $E=-\frac{GM}{2a}$, where M is the total mass of the orbit)

Generally speaking (without any averaging) - this is still a chaotic 3-body problem and everything can happen - the inner orbit might just completely be destroyed by m1 being thrown out of the system, for an instance.

• "...a massless object (m1 in your diagram)..." Since m1 and m0 orbit around a common center of mass outside of m0, m1 can not be massless. I think there's a bit of problem here, but the problem may only be with the diagram.
– uhoh
Jan 18, 2019 at 14:53
• true, in the diagram the center of m1 and m0 should have been inside m0 Jan 20, 2019 at 6:58
• in any event, it's a very nicely written answer about a challenging (or at least difficult to picture) problem.
– uhoh
Jan 20, 2019 at 7:05