Numerical integration of a relativistic two-body problem via a multiple scales method

Abstract

We offer an analytical study on the dynamics of a two-body problem perturbed by small post-Newtonian relativistic term. We prove that, while the angular momentum is not conserved, the motion is planar. We also show that the energy is subject to small changes due to the relativistic effect. We also offer a periodic solution to this problem, obtained by a method based on the separation of time scales. We demonstrate that our solution is more general than the method developed in the book by Brumberg (Essential Relativistic Celestial Mechanics, Hilger, Bristol, 1991). The practical applicability of this model may be in studies of the long-term evolution of relativistic binaries (neutron stars or black holes).

Appendix

1.1 A.1 Mechanical energy-like invariant of motion

Here we derive an energy-like integral of motion, by taking the scalar product of Eq. (14) with $\lambda\dot{\underline{r}}$ instead of $\dot{\underline{r}}$, where $\lambda=\lambda(t)$ is a scalar function of time which can be established for the time of the invariant integral. Thus we obtain

$$ \lambda\ddot{\underline{r}}.\dot{\underline{r}}=- \biggl[\dfrac {1}{r^{2}}- \dfrac{ \varepsilon}{ r^{3}} \bigl(4+3v^{2}r \bigr) \biggr]\lambda\dot{r}; $$

(A.1)

the above equation can be rewritten in the form

$$ \dfrac{1}{2}\dfrac{\mathrm{d}}{\mathrm{d}t} \bigl(\lambda v^{2} \bigr)=- \biggl(\dfrac{1}{r^{2}}-\dfrac{ 4 \varepsilon}{ r^{3}} \biggr)\lambda\dot{r} + \biggl( \dfrac{1}{2}\dot{\lambda } +\dfrac{ 3 \varepsilon\lambda\dot{r}}{ r^{2}} \biggr)v^{2}. $$

(A.2)

In order to obtain the integration of Eq. (A.2) in closed form, we must eliminate the second term in the right hand side of this equation. This implies that

$$ \dfrac{1}{2}\dot{\lambda} +\dfrac{ 3 \varepsilon\lambda\dot{r}}{ r^{2}}=0 $$

(A.3)

after integration of Eq. (A.3), the function $\lambda$ is given by

$$ \lambda(t)=A \mathrm{e}^{-6\varepsilon/r}, $$

(A.4)

where $A$ is an arbitrary constant of integration not equal to zero. Since $\varepsilon$ is a very small quantity, we can expand the right hand side of Eq. (A.4) and restrict ourselves to the first order of $\varepsilon$. Therefore this equation can be rewritten in the form

$$ \lambda(t)=A \biggl(1-\dfrac{6\varepsilon}{r} \biggr) $$

(A.5)

substituting Eq. (A.5) into Eq. (A.2) and on integration with neglecting the terms of $O(\varepsilon^{2})$ or higher, we obtain

$$ \dfrac{1}{2}\lambda v^{2}-\dfrac{A}{r}+\dfrac{7\varepsilon A}{ r^{2}}= \bar {E.} $$

(A.6)

$\bar{E}$ is the integration constant and $\bar{v}^{2}=(1-\frac{6 \varepsilon}{r})v^{2}$. Thus Eq. (A.6) will take the new form

$$ \dfrac{1}{2}\bar{v}^{2}-\dfrac{1}{r}+\dfrac{7\varepsilon}{ r^{2}}=E_{l}. $$

(A.7)

It is clear that Eq. (A.7) represents an expression of the energy-like invariant integral with an extra term which characterizes the relativistic effect where $E_{l}=\bar{E}/A$.

1.2 A.2 Proof of Theorem 2

Since $\bar{v}^{2}/2$ and $1/r^{2}$ can be written in the form

$$\begin{aligned} &\dfrac{1}{2}\bar{v}^{2}=\dfrac{1}{2}v^{2}-\dfrac{3\varepsilon}{h^{4}} \bigl[1+3e \cos f+e^{2} \bigl(1+2 \cos^{2} f \bigr) \\ &\phantom{\dfrac{1}{2}\bar{v}^{2}=}{}+e^{3} \cos f \bigr] , \end{aligned}$$

(A.8)

$$\begin{aligned} &\dfrac{1}{r^{2}}-\dfrac{1}{h^{4}} \bigl[1+2e\cos f +e^{2} \cos^{2} f \bigr] , \end{aligned}$$

(A.9)

substituting Eqs. (A.8), (A.9) into Eq. (A.7), one obtains

$$\begin{aligned} &\dfrac{1}{2}{v}^{2}-\dfrac{1}{r}+\dfrac{4 \varepsilon}{ h^{4}} \biggl(1+\dfrac{5}{4}e \cos f -\dfrac{1}{ 4}e^{2} \bigl(3- \cos^{2} f \bigr) \\ &\quad{} -\dfrac{3}{4}e^{3} \cos f \biggr)=E_{l}. \end{aligned}$$

(A.10)

Neglecting terms of the coefficient $O(\varepsilon e)$, Eq. (A.10) is reduced to

$$ \dfrac{1}{2}{v}^{2}-\dfrac{1}{r}=E, $$

(A.11)

where $E=E_{l}-4\varepsilon/h^{4}$.

Hence the energy-like approximate integral of motion becomes as in the unperturbed motion.

To investigate the accuracy of the invariant-like motion, it must be compared with the exact invariant motion. Let $\mbox{err}(E)=E-E_{l}$ be the difference between both constants of motion; one obtains

$$ \mbox{err}(E)=\varepsilon \biggl[\dfrac{3}{8}v^{4} + \dfrac{9}{2r}v^{2}-\dfrac {13}{2r^{2}} \biggr]. $$

(A.12)

According to Eq. (A.12), we can conclude that this error will vanish when $v^{2}=\pm v^{2}_{r}$ decreases for $v^{2} \in(-v^{2}_{r}, v^{2}_{r})$ and increases for $v^{2} \in(-\infty, -v^{2}_{r})\cup(v^{2}_{r}, \infty)$, where $v^{2}_{r}= 2(2\sqrt{30}-9)/3r$. On the other hand, Eq. (A.12) can be written in the form

$$\begin{aligned} \mbox{err}(E)&=-\dfrac{13 \varepsilon}{8h^{4}} \biggl[1-\dfrac{16}{13}e \cos f- \dfrac{42}{13} e^{2} \biggl(1+\dfrac{16}{21} \cos^{2} f \biggr) \\ &\quad{} -\dfrac {48}{13} e^{3} \cos f -\dfrac{3}{13}e^{4} \biggr]. \end{aligned}$$

(A.13)

Consequently, the deviation of the energy-like approximate integral of motion from the exact integral is of the order of $O(\varepsilon)$.