Regularized phase-space volume for the three-body problem

Abstract

The micro-canonical phase-space volume for the three-body problem is an elementary quantity of intrinsic interest, and within the flux-based statistical theory, it sets the scale of the disintegration time. While the bare phase-volume diverges, we show that a regularized version can be defined by subtracting a reference phase-volume, which is associated with hierarchical configurations. The reference quantity, also known as a counter-term, can be chosen from a 1-parameter class. The regularized phase-volume of a given (negative) total energy, ${{\bar{\sigma }}}(E)$, is evaluated. First, it is reduced to a function of the masses only, which is sensitive to the choice of a regularization scheme only through an additive constant. Then, analytic integration is used to reduce the integration to a sphere, known as shape sphere. Finally, the remaining integral is evaluated numerically and presented by a contour plot in parameter space. Regularized phase-volumes are presented for both the planar three-body system and the full 3d system. In the test mass limit, the regularized phase-volume is found to become negative, thereby signaling the breakdown of the non-hierarchical statistical theory. This work opens the road to the evaluation of ${{\bar{\sigma }}}(E,L)$, where L is the total angular momentum, and in turn to comparison with simulation determined disintegration times.

Appendices

A Compensator evaluation

In this appendix, we detail the evaluation of the compensator (35). In fact, we provide two derivations.

The first way involves a reflection symmetry

$$\begin{aligned} \Delta {{\hat{\sigma }}}= & {} \frac{8}{\pi ^2} \int _{1/2}^\infty du_B \int _{1/2}^\infty du_F \int _{-1/2}^{u_F-1} d\epsilon \, \frac{\sqrt{u_B+\epsilon } \,\sqrt{u_F-1-\epsilon }}{u_B^4 \, u_F^4} \nonumber \\= & {} \frac{1}{2}\cdot \frac{8}{\pi ^2} \int _{1/2}^\infty du_B \int _{1/2}^\infty du_F \int _{-u_B}^{u_F-1} d\epsilon \, \frac{\sqrt{u_B+\epsilon } \,\sqrt{u_F-1-\epsilon }}{u_B^4 \, u_F^4} \nonumber \\= & {} \frac{1}{2}\cdot \frac{8}{\pi ^2} \cdot \frac{5 \pi }{9} = \frac{20}{9\pi } ~. \end{aligned}$$

(67)

The second way involves a direct integration with a convenient ordering of the three integrations

$$\begin{aligned} \Delta {{\hat{\sigma }}}= & {} \frac{8}{\pi ^2} \int _{1/2}^\infty \frac{du_B}{u_B^4} \int _{-1/2}^\infty d\epsilon \, \sqrt{u_B+\epsilon } \int _{1+\epsilon }^\infty du_F \, \frac{\sqrt{u_F-1-\epsilon }}{u_F^4} \nonumber \\= & {} \frac{8}{\pi ^2} \int _{1/2}^\infty \frac{du_B}{u_B^4} \int _{-1/2}^\infty d\epsilon \sqrt{u_B+\epsilon } \, \frac{\pi }{16} (1+\epsilon )^{-5/2} \nonumber \\= & {} \frac{8}{\pi ^2} \int _{1/2}^\infty \frac{du_B}{u_B^4} \, \frac{\pi }{24} \frac{(2 u_B-1)^{3/2}-1}{u_B-1} \nonumber \\= & {} \frac{32}{3 \pi } \int _0^\infty du \, \frac{u (u^2+u+1)}{(u^2+1)^4(u+1)} = \frac{20}{9 \pi } \end{aligned}$$

(68)

where in passing to the fourth line, we have changed variables according to $u^2=2 u_B-1$.

B Planar reduction

In this appendix, we derive (36). We start with the following rewriting of the LHS:

$$\begin{aligned} \begin{aligned} \left( \prod _{c=1}^3 d^3 r_c \right) \, \delta ^{(3)}(\vec R_\textrm{CM})&= \left( \prod _{c=1}^3 d^2 r_c \right) \, dz_1 \, dz_2 \, dz_3 \, \delta ^{(2)}(\vec R_\textrm{CM}) \, \delta \left( \frac{1}{M}(m_1 z_1 + m_2 z_2 + m_3 z_3)\right) \\&=\left( \prod _{c=1}^3 d^2 r_c \right) \, dz_1 \, dz_2 \, \delta ^{(2)}(\vec R_\textrm{CM}) \left( \frac{M}{m_3}\right) ~. \end{aligned} \end{aligned}$$

(69)

Next, we express $dz_1$ and $dz_2$ in terms of rotations in the $x-y$ plane

$$\begin{aligned} \begin{aligned} dz_1&= d\vec \phi \times \vec r_1 \\ dz_2&= d\vec \phi \times \vec r_2 ~, \end{aligned} \end{aligned}$$

(70)

where $d\vec \phi$ is an infinitesimal vector in the $x-y$ plane. Since

$$\begin{aligned} \begin{bmatrix} dz_1 \\ dz_2 \end{bmatrix} = \begin{bmatrix} y_1 &{} -x_1 \\ y_2 &{} -x_2 \end{bmatrix} \begin{bmatrix} \hbox {d}\phi _x \\ \hbox {d}\phi _y \end{bmatrix} ~, \end{aligned}$$

(71)

we have

$$\begin{aligned} dz_1 dz_2 = (x_1y_2-y_1x_2)\hbox {d}\phi _x \hbox {d}\phi _y ~. \end{aligned}$$

(72)

Area of the triangle formed by the three bodies is given by the determinant

$$\begin{aligned} A= \frac{1}{2}\begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1 \end{vmatrix} ~, \end{aligned}$$

(73)

which can be reduced using the following steps²

$$\begin{aligned} \begin{aligned} A&= \frac{1}{2}\frac{M}{m_3}\begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}&\frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M}&1 \end{vmatrix} \\ {}&= \frac{1}{2}\frac{M}{m_3}\begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ 0&0&1 \end{vmatrix}\\ {}&= \frac{1}{2}\frac{M}{m_3} (x_1 y_2-y_1 x_2) \end{aligned} \end{aligned}$$

(74)

Using (72) and (74) we get

$$\begin{aligned} dz_1 dz_2 = 2A ~\frac{m_3}{M} ~\hbox {d}\phi _x \hbox {d}\phi _y \end{aligned}$$

(75)

Substituting (75) in (69), we get

$$\begin{aligned} \begin{aligned} \left( \prod _{c=1}^3 d^3 r_c \right) \, \delta ^{(3)}(\vec R_\textrm{CM})&= 2A \left( \prod _{c=1}^3 d^2 r_c \right) \, \delta ^{(2)}(\vec R_\textrm{CM})~ \hbox {d}\phi _x \hbox {d}\phi _y\\ {}&= 2A \left( \prod _{c=1}^3 d^2 r_c \right) \, \delta ^{(2)}(\vec R_\textrm{CM})~ \hbox {d}\Omega \end{aligned} \end{aligned}$$

(76)

where we rewrite $\hbox {d}\phi _x\hbox {d}\phi _y$ as $\hbox {d}\Omega$, which is the differential solid angle on the unit sphere. We have thus derived (36).

C Expressions for potential and triangle area

In this appendix, we derive the expressions for A, V in terms of $r,\theta ,\phi ,\psi$ coordinates. For this, we need the expressions for the relative distances between the three bodies, i.e., $r_{12},r_{13},r_{23}$. First, we express them in terms of w variables defined in (38)

$$\begin{aligned} r_{12} = ~\left| \frac{w-\eta {\bar{w}}}{1-\eta } \right| \quad ,\quad r_{13} = ~\left| \frac{w-{\bar{\eta }} {\bar{w}}}{1-{\bar{\eta }}} \right| \quad ,\quad r_{23} = ~\left| \frac{w- {\bar{w}}}{\eta -{\bar{\eta }}}\right| \end{aligned}$$

(77)

where we use the following notation: $\eta =e^{j\frac{2\pi }{3}},~{\bar{\eta }}=e^{-j\frac{2\pi }{3}},~\bar{w}=w/.\{j\rightarrow -j\}$ and the absolute values are calculated with respect to i. Using (42), we rewrite in terms of $r,\theta ,\phi ,\psi$ coordinates

$$\begin{aligned} \begin{aligned} r_{12}&= r \sqrt{1-\sin \theta \cos \left( \phi +\frac{2\pi }{3}\right) } \quad , \quad r_{13} = r \sqrt{1-\sin \theta \cos \left( \phi -\frac{2\pi }{3}\right) }\\ r_{23}&= r \sqrt{1-\sin \theta \cos \phi } \end{aligned} \end{aligned}$$

(78)

From (78), it is straightforward to derive expression for V in (46).

To derive the expression for A in (45), we use the formula

$$\begin{aligned} A = \frac{1}{4} \sqrt{(r_{12}^2+r_{13}^2+r_{23}^2)^2-2(r_{12}^4+r_{13}^4+r_{23}^4)} \end{aligned}$$

(79)

and substitute (78) to derive (45).