Application of the logarithmic Hamiltonian algorithm to the circular restricted three-body problem with some post-Newtonian terms

Abstract

An implementation of a fourth-order symplectic algorithm to the logarithmic Hamiltonian of the Newtonian circular restricted three-body problem in an inertial frame is detailed. The logarithmic Hamiltonian algorithm produces highly accurate results, comparable to the non-logarithmic one. Its numerical performance is independent of an orbital eccentricity. However, it is not when some post-Newtonian terms are included in this problem. Although the numerical accuracy becomes somewhat poorer as the orbital eccentricity gets larger, it is still much higher than that of the non-logarithmic Hamiltonian algorithm. As a result, the present code can drastically eliminate the overestimation of Lyapunov exponents and the spurious rapid growth of fast Lyapunov indicators for high-eccentricity orbits in the Newtonian or post-Newtonian circular restricted three-body problem.

Appendices

Appendix A: Construction of logarithmic Hamiltonian algorithms in the rotating frame

In terms of extended phase space variables $(x,y,q_{0}; p_{x}, p_{y}, p_{0})$, the Hamiltonian (16) is rewritten as

$$ \mathcal{H}=K-U, $$

(A.1)

where $K=(p_{x}^{2}+p_{y}^{2})/2+p_{0}+yp_{x}-xp_{y}$ is the kinetic energy. We have the logarithmic Hamiltonian

$$ \digamma=\log K-\log U. $$

(A.2)

The potential part $(-\log U)$ is only dependent on coordinates $x$ and $y$, so it is easily analytically solved. As to the kinetic part $\log K$, it corresponds to canonical equations

$$\begin{aligned} \begin{aligned}[c] &\frac{dq_{0}}{ds} = \frac{1}{K}, \\ &\frac{dx}{ds} = \frac{1}{K}(p_{x}+y), \\ &\frac{dy}{ds} = \frac{1}{K}(p_{y}-x); \\ &\frac{dp_{0}}{ds} = 0, \\ &\frac{dp_{x}}{ds} = \frac{p_{y}}{K}, \\ &\frac{dp_{y}}{ds} = -\frac{p_{x}}{K}. \end{aligned} \end{aligned}$$

(A.3)

If $K$ is viewed as a constant at the $(n-1)$th step of new time interval $[s_{n-1},s_{n}]$ and is labeled as $K_{n-1}=K(x_{n-1},y_{n-1},p_{x,n-1},p_{y,n-1},p_{0})$,⁷ then Eq. (A.3) turns out to be the sum of the linear motion and rotation and is easily solvable. Therefore, the leapfrog Algorithm (11) is also valid to solve the logarithmic Hamiltonian $\digamma$. We call it $S^{*}_{2}$, which has the discrete difference formulae:

$$\begin{aligned} &q^{*}_{0,n} = q_{0,n-1}+\frac{dt}{2}, \\ & \begin{aligned}[t] x^{*}_{n} &= x_{n-1}\cos\biggl(\frac{dt}{2} \biggr)+y_{n-1}\sin\biggl(\frac{dt}{2}\biggr)\\ &\quad {}+\frac{dt}{2} \biggl(p_{x,n-1}\cos\biggl(\frac{dt}{2}\biggr)+p_{y,n-1}\sin \biggl(\frac{dt}{2}\biggr)\biggr), \end{aligned} \\ & \begin{aligned}[b] y^{*}_{n} &= y_{n-1}\cos\biggl(\frac{dt}{2} \biggr)-x_{n-1}\sin\biggl(\frac{dt}{2}\biggr)\\ &\quad {}+\frac{dt}{2} \biggl(p_{y,n-1}\cos\biggl(\frac{dt}{2}\biggr)-p_{x,n-1}\sin \biggl(\frac{dt}{2}\biggr)\biggr), \end{aligned} \\ &p^{*}_{x,n} = p_{x,n-1}\cos\biggl(\frac{dt}{2} \biggr)+p_{y,n-1}\sin\biggl(\frac{dt}{2}\biggr), \\ &p^{*}_{y,n} = p_{y,n-1}\cos\biggl(\frac{dt}{2} \biggr)-p_{x,n-1}\sin\biggl(\frac{dt}{2}\biggr); \\ &p^{\sharp}_{x,n} = p^{*}_{x,n}+ \frac{h}{U(x_{n-1},y_{n-1})}\frac{\partial U(x^{*}_{n},y^{*}_{n})}{\partial x^{*}_{n}}, \\ &p^{\sharp}_{y,n} = p^{*}_{y,n}+ \frac{h}{U(x_{n-1},y_{n-1})}\frac{\partial U(x^{*}_{n},y^{*}_{n})}{\partial y^{*}_{n}}; \\ &q_{0,n} = q^{*}_{0,n}+\frac{dt}{2}, \\ & \begin{aligned}[t] x_{n} &= x^{*}_{n}\cos\biggl(\frac{dt}{2} \biggr)+y^{*}_{n}\sin\biggl(\frac{dt}{2}\biggr)\\ &\quad {}+ \frac{dt}{2}\biggl(p^{\sharp}_{x,n}\cos\biggl( \frac{dt}{2}\biggr)+p^{\sharp}_{y,n}\sin\biggl( \frac{dt}{2}\biggr)\biggr), \end{aligned} \\ &\begin{aligned}[t] y_{n} &= y^{*}_{n}\cos\biggl(\frac{dt}{2} \biggr)-x^{*}_{n}\sin\biggl(\frac{dt}{2}\biggr)\\ &\quad {}+ \frac{dt}{2}\biggl(p^{\sharp}_{y,n}\cos\biggl( \frac{dt}{2}\biggr)-p^{\sharp}_{x,n}\sin\biggl( \frac{dt}{2}\biggr)\biggr), \end{aligned} \\ &p_{x,n} = p^{\sharp}_{x,n}\cos\biggl( \frac{dt}{2}\biggr)+p^{\sharp}_{y,n}\sin\biggl( \frac{dt}{2}\biggr), \\ &p_{y,n} = p^{\sharp}_{y,n}\cos\biggl( \frac{dt}{2}\biggr)-p^{\sharp}_{x,n}\sin\biggl( \frac{dt}{2}\biggr). \end{aligned}$$

(A.4)

As a point to emphasize, $U(x_{n-1},y_{n-1})$ is used instead of $U(x^{*}_{n},y^{*}_{n})$ because of the theoretical result $K_{n-1}\equiv U(x_{n-1},y_{n-1})$. Additionally, the difference between the present numerical approach (A.4) and Eq. (12) lies in that these variables $(x,y,p_{x},p_{y},p_{0})$ in $K$ of Eq. (A.4) are taken as constants at time $s_{n-1}$ but are not in $T$ of Eq. (12). Replacing $S_{2}$ in Eq. (13) with $S^{*}_{2}$, we have a fourth-order integrator $S^{*}$.

If $K$ cannot be regarded as a constant at the $(n-1)$th step, then Eq. (A.3) has no analytical solution. In this case, the second-order implicit midpoint method is suggested to numerically solve Eq. (A.3). When the leapfrog method (11) is still used, $A$ is an implicit operator acting on $\log K$ and $B$ is an explicit operator acting on $(-\log U)$. Here the discrete difference formulae read

$$\begin{aligned} &q^{*}_{0,n} = q_{0,n-1}+\frac{h}{2} \frac{1}{K^{*}}, \\ &x^{*}_{n} = x_{n-1}+\frac{h}{2} \frac{1}{K^{*}}\biggl(\frac{p^{*}_{x,n}+p_{x,n-1}}{2}+\frac {y^{*}_{n}+y_{n-1}}{2}\biggr), \\ &y^{*}_{n} = y_{n-1}+\frac{h}{2} \frac{1}{K^{*}}\biggl(\frac{p^{*}_{y,n}+p_{y,n-1}}{2}-\frac {x^{*}_{n}+x_{n-1}}{2}\biggr), \\ &p^{*}_{x,n} = p_{x,n-1}+\frac{h}{2} \frac{1}{K^{*}}\frac{p^{*}_{y,n}+p_{y,n-1}}{2}, \\ &p^{*}_{y,n} = p_{y,n-1}-\frac{h}{2} \frac{1}{K^{*}}\frac{p^{*}_{x,n}+p_{x,n-1}}{2}; \\ & \begin{aligned}[c] &p^{\sharp}_{x,n} = p^{*}_{x,n}+ \frac{h}{U(x^{*}_{n},y^{*}_{n})}\frac{\partial U(x^{*}_{n},y^{*}_{n})}{\partial x^{*}_{n}}, \\ &p^{\sharp}_{y,n} = p^{*}_{y,n}+ \frac{h}{U(x^{*}_{n},y^{*}_{n})}\frac{\partial U(x^{*}_{n},y^{*}_{n})}{\partial y^{*}_{n}}; \end{aligned} \\ &q_{0,n} = q^{*}_{0,n}+\frac{h}{2} \frac{1}{K^{\star}}, \\ &x_{n} = x^{*}_{n}+\frac{h}{2} \frac{1}{K^{\star}}\biggl(\frac{p_{x,n}+p^{\sharp}_{x,n}}{2}+\frac {y_{n}+y^{*}_{n}}{2}\biggr), \\ &y_{n} = y^{*}_{n}+\frac{h}{2} \frac{1}{K^{\star}}\biggl(\frac{p_{y,n}+p^{\sharp}_{y,n}}{2}-\frac {x_{n}+x^{*}_{n}}{2}\biggr), \\ &p_{x,n} = p^{\sharp}_{x,n}+\frac{h}{2} \frac{1}{K^{\star}}\frac{p_{y,n}+p^{\sharp}_{y,n}}{2}, \\ &p_{y,n} = p^{\sharp}_{y,n}-\frac{h}{2} \frac{1}{K^{\star}}\frac{p_{x,n}+p^{\sharp}_{x,n}}{2}. \end{aligned}$$

(A.5)

$K^{*}$ and $K^{\star}$ are

$$\begin{aligned} & \begin{aligned}[t] K^{*} &= K\biggl(\frac{x^{*}_{n}+x_{n-1}}{2},\frac{y^{*}_{n}+y_{n-1}}{2}, \frac{p^{*}_{x,n}+p_{x,n-1}}{2},\\ &\quad {}\frac {p^{*}_{y,n}+p_{y,n-1}}{2},p_{0}\biggr), \end{aligned} \\ & \begin{aligned}[t] K^{\star} &= K\biggl(\frac{x_{n}+x^{*}_{n}}{2},\frac{y_{n}+y^{*}_{n}}{2}, \frac{p_{x,n}+p^{\sharp}_{x,n}}{2},\\ &\quad {}\frac{p_{y,n}+p^{\sharp }_{y,n}}{2},p_{0}\biggr). \end{aligned} \end{aligned}$$

In fact this scheme, marked as $S^{+}_{2}$, belongs to a second-order mixed symplectic integrator of explicit and implicit operators (Zhong et al. 2010). If the operators $A$ and $B$ in Eq. (11) are exchanged, more CPU times would be saved. In most cases, this kind of mixed symplectic integrator has less cost than the implicit midpoint method applied to the whole system. Of course, it is inferior to the explicit method (12) in the computational efficiency. However, it has a wider application; for example, it is suitable for solving PN Hamiltonian formulations (Mei et al. 2013a, 2013b). Using $S^{+}_{2}$ instead of $S_{2}$ in Eq. (13), we obtain another fourth-order symplectic integrator $S^{+}$.

Appendix B: PN Lagrangian of relativistic restricted three-body problem

All the scaled variables $(t,x,y,\dot{x},\dot{y})$ in Eq. (14) or Eq. (17) are still used. Based on the first order post-Newtonian (1PN) gravitational equations of Einstein, Infeld, and Hoffmann in the inertial frame, the Lagrangian of the third body of the relativistic restricted three-body problem in the rotating frame is expressed as

$$\begin{aligned} \tilde{L} =& \frac{1-\mu}{r_{1}}+\frac{\mu}{r_{2}}+\frac{1}{2}\bigl( \dot{x}^{2}+\dot{y}^{2}\bigr)+\biggl(1+\frac{\omega_{1}}{c^{2}} \biggr) (x\dot{y}-\dot{x}y) \\ &{}+\frac{1}{2}\biggl(1+2\frac{\omega_{1}}{c^{2}}+ \frac{\omega^{2}_{1}}{c^{4}}\biggr) \bigl(x^{2}+y^{2}\bigr)+ \frac{1}{c^{2}}\hat{L}. \end{aligned}$$

(B.1)

In the above equation, $\tilde{L}$ is given after the scale transformation $\tilde{L}\rightarrow\tilde{L}/a$, the velocity of light $c$ is only used to mark the order of PN contributions,⁸ and $c=1$ is taken in practical numerical computations. The 1PN angular speed is $\omega_{1}=[(1-\mu)\mu -3]/(2a)$. The PN part $\hat{L}$ is of the form

$$\begin{aligned} \hat{L} =& \frac{1}{8a}\bigl[\dot{x}^{2}+\dot{y}^{2}+2( \dot{y}x- \dot{x}y) +x^{2}+y^{2}\bigr]^{2} \\ &{}- \frac{(1-\mu)\mu}{a}\biggl(\frac{1}{r_{1}}+\frac{1}{r_{2}}\biggr)- \frac{1}{a} \biggl[\frac{(1-\mu)^{2}}{2r^{2}_{1}} \\ &{}+\frac{\mu ^{2}}{2r^{2}_{2}}+\frac{(1-\mu)\mu}{r_{1}r_{2}} \biggr] +\frac{3}{2a}\biggl(\frac{1-\mu}{r_{1}}+\frac{\mu}{r_{2}}\biggr)\bigl[ \dot{x}^{2}+\dot{y}^{2} \\ & {}+2(\dot{y}x- \dot{x}y) +x^{2}+y^{2}\bigr]+\frac{3}{2a}\biggl[ \frac{(1-\mu)\mu^{2}}{r_{1}} \\ & {}+\frac{\mu(1-\mu)^{2}}{r_{2}}\biggr] - \frac{7}{2a}(\dot{y}+x) (1-\mu)\mu\biggl(\frac{1}{r_{1}}- \frac{1}{r_{2}}\biggr) \\ & {}- \frac{1}{2a}y(\dot{x}-y) (1-\mu)\mu\biggl[ \frac{x-\mu}{r^{3}_{1}} -\frac{x+1-\mu}{r^{3}_{2}}\biggr] \\ & {}- \frac{1}{2a}y^{2}(\dot{y}+x) (1-\mu)\mu\biggl( \frac{1}{r^{3}_{1}}-\frac{1}{r^{3}_{2}}\biggr). \end{aligned}$$

(B.2)

The Lagrangian $\tilde{L}$ can be found in the article of Huang and Wu (2014a), and is obtained from the paper of Chandrasekhar and Contopoulos (1967) or the paper of Maindl and Dvorak (1994). The difference lies in that the former uses the related scale transformations and the latter takes directly $a=1$. Because of this, the relation between the separation $a$ and the PN contributions can be seen much clearly for the former than for the latter.