Abstract

It is believed that some numerical technique must be employed for determining the system parameters of a visual binary or a star with a planet, because the relevant equations are not only highly nonlinear, but also transcendental owing to Kepler’s equation. Such common sense, however, is not true; we have discovered an analytic inversion formula, in which the original orbital parameters are expressed as elementary functions of the observable quantities, such as the location of four observed points and the time interval between these points. The key thing is that we use the time interval, but not the time of each observation, in order to avoid treating Kepler’s equation. The present formula can be applied even in cases where the observations cover a short arc of the orbit during less than one period. Thus, the formula will be useful in future astrometric missions, such as SIM, GAIA, and JASMINE.

Introduction

The astrometric observation of binaries gives much invaluable information on the mass of the component stars and the orbital parameters, such as the ellipticity (Danby 1988; Roy 1988). In the near future, space missions such as SIM,1 GAIA,2 and JASMINE3 will find a number of binaries nearly within 10 kpc. A problem in the astrometry arises from a fact that we make a measurement of the angular position of the celestial object, which is projected on a plane perpendicular to the line of sight. For simplicity, we call this plane the observed plane. In the case of a binary system, the plane of the Keplerian orbit is inclined with respect to the line of sight. As a result, the observed ellipse can be considerably different from the original Keplerian orbit. To determine the original orbital parameter, we thus need to make a kind of data fitting (Olevic, Cvetkovic 2004). It rather consumes time and computer resources, especially for a huge amount of data that will be available by the near-future missions. Obviously, it would be preferable to use an inversion formula, which would enable us to directly determine from the observed quantities the original orbital parameters and the inclination. However, the conditional equations that connect the observable quantities with the orbital parameters are not only highly nonlinear, but also transcendental because of Kepler’s equation. It is thus believed that an explicit solution is impossible, so that some numerical technique must be employed to solve the equations for the orbital parameters (Eichhorn, Xu 1990; Catovic, Olevic 1992). Such common sense in this field, however, is not true, as shown below.

The purpose of this letter is to derive the inversion formula. The key point is that we use the time interval between the observations, but not the time of each observation. First, we determine the observed ellipse from the positional data. Next, together with the time interval between observations, all of the ellipticity, the orbital period, the major and minor axes of the original Keplerian orbit, and the inclination are expressed in the measured quantities.

Inversion Formula

First, we determine the observed ellipse. Next, we discuss how to determine the original orbital parameters and the inclination angle. In this letter, we consider only the Keplerian motion of the binary by neglecting the motion of the observer and the galactic motion.

Observed Ellipse

We observe an ellipse on the plane perpendicular to the line of sight. The Cartesian coordinates on the plane are denoted by $$(\overline{x}, \overline{y})$$. A general form of the ellipse is  

(1)
$$\begin{equation} \alpha \overline{x}^2 + \beta \overline{y}^2 + 2\gamma \overline{x} \overline{y} + 2\delta \overline{x} + 2\varepsilon \overline{y} = 1, \end{equation}$$
which is specified by five parameters, since the center, the major/minor axes, and the orientation of the ellipse are arbitrary. We need to make at least five observations to determine the parameters. The location of each observed point is $$(\overline{x}_i, \overline{y}_i)$$ for $$i=1, \cdots, 5$$. Then, we find  
(2)
$$\begin{eqnarray} \left(\begin{array}{ccccc} \overline{x}_1^2 &\overline{y}_1^2 &2\overline{x}_1\overline{y}_1 &2\overline{x}_1 &2\overline{y}_1 \\ \overline{x}_2^2 &\overline{y}_2^2 &2\overline{x}_2\overline{y}_2 &2\overline{x}_2 &2\overline{y}_2 \\ \overline{x}_3^2 &\overline{y}_3^2 &2\overline{x}_3\overline{y}_3 &2\overline{x}_3 &2\overline{y}_3 \\ \overline{x}_4^2 &\overline{y}_4^2 &2\overline{x}_4\overline{y}_4 &2\overline{x}_4 &2\overline{y}_4 \\ \overline{x}_5^2 &\overline{y}_5^2 &2\overline{x}_5\overline{y}_5 &2\overline{x}_5 &2\overline{y}_5 \end{array} \right) \left(\begin{array}{c} \alpha \\ \beta \\ \gamma \\ \delta \\ \varepsilon \end{array} \right) = \left(\begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{array} \right). \end{eqnarray}$$

By using the inverse matrix, we can determine $$(\alpha, \beta, \gamma, \delta, \varepsilon)$$ in terms of $$(\overline{x}_i, \overline{y}_i)$$. Henceforth, we choose the Cartesian coordinates $$(x, y)$$ so that the observed ellipse can be reexpressed in the standard form as  

(3)
$$\begin{equation} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \end{equation}$$
where $$a\geq b$$. The ellipticity, $$e$$, is $$\sqrt{1-b^2/a^2}$$.

Time Interval

Before considering a general case, we study a special case to illustrate a problem due to the inclination; we assume that an ellipse is inclined around its major axis by the angle $$i$$. The lengths of the major and minor axes are denoted by $$a$$ and $$b'$$, respectively. The lengths of the major and minor axes of the observed ellipse become $$a$$ and $$b'\cos i$$, respectively. By only using the shape of the ellipse, we cannot determine $$b'$$ and $$i$$ separately. This degeneracy can be broken if we use not only the location of each point, but also its time. More rigorously speaking, we use the time interval between points as shown below. We adopt four observational data at the time of $$t_4 \gt t_3 \gt t_2 \gt t_1 t_4 \gt t_3 \gt t_2 \gt t_1$$. The star is observed at the location of $$\boldsymbol{P}_i = (x_i, y_i) = (a \cos u_i, b \sin u_i)$$ at each time $$t_i$$ for $$i=1, \cdots, 4$$, where $$a$$ and $$b$$ have been determined by fixing the observed ellipse in the preceding subsection, and $$u_i$$ denotes the eccentric anomaly. We assume anti-clockwise motion, such that $$u_i \gt u_j$$ for $$i \gt j$$. All we have to do in the case of the clockwise motion is to change the signature of equation (5) for the area in the following. We define the time interval as $$t_{ij}=t_i-t_j$$. The number of these additional quantities $$(t_{21}, t_{32}, t_{43})$$ is three, which equals the number of unknown parameters of the period, the orientation, and the angle of inclination.

The original Keplerian orbit is assumed to be parameterized by the length of the major and minor axes, $$a_{\mathrm{K}}$$ and $$b_{\mathrm{K}}$$, and the period $$T$$. The ellipticity of the orbit is denoted by $$e_{\mathrm{K}}$$. The focus of the original ellipse is projected onto the observed plane at $$\boldsymbol{P}_{\mathrm{e}} = (x_{\mathrm{e}}, y_{\mathrm{e}})$$. Figure 1 shows the geometrical configuration.

Fig. 1

Relation between the Keplerian ellipse and the observed one. The line of sight is along the $$z$$-axis.

Fig. 1

Relation between the Keplerian ellipse and the observed one. The line of sight is along the $$z$$-axis.

One of the important things about the Keplerian orbit is that the areal velocity is constant, where the area is swept by the line interval between the focus and the object in the Keplerian motion (for instance, Goldstein 1980). Henceforth, the object in the Keplerian motion is called star for simplicity. We should note that the projected focus is not necessarily the focus of the observed ellipse. Even after the projection, however, the law of constant-areal velocity still holds, where the area is swept by the line interval between the projected focus and the star. The area swept during the time interval, $$(t_{ij})$$, is denoted by $$S_{ij}$$. The total area of the observed ellipse, $$S$$, is $$\pi ab$$. The law of the constant areal velocity on the observed plane becomes  

(4)
$$\begin{equation} \frac{S}{T}=\frac{S_{ij}}{t_{ij}}. \end{equation}$$

Determining the Ellipticity and the Orbital Period

By elementary computations, the area $$S_{ij}$$ is obtained as  

(5)
$$\begin{eqnarray} S_{ij} = \frac{1}{2} ab &&\left[u_i-u_j-\frac{x_{\mathrm{e}}}{a}(\sin u_i-\sin u_j) \right. \nonumber\\ &&\left. {} + \frac{y_{\mathrm{e}}}{b}(\cos u_i-\cos u_j) \right]. \end{eqnarray}$$

It is noteworthy that $$S_{ij}$$ is linear in $$x_{\mathrm{e}}$$ and $$y_{\mathrm{e}}$$. By using equation (4), we obtain  

(6)
$$\begin{equation} \frac{S_{21}}{t_{21}} = \frac{S_{32}}{t_{32}},\end{equation}$$
 
(7)
$$\begin{equation} \frac{S_{32}}{t_{32}} = \frac{S_{43}}{t_{43}}. \end{equation}$$

They are reexpressed as  

(8)
$$\begin{equation} A_3-\frac{x_{\mathrm{e}}}{a} A_1 + \frac{y_{\mathrm{e}}}{b} A_2 = 0,\end{equation}$$
 
(9)
$$\begin{equation} B_3-\frac{x_{\mathrm{e}}}{a} B_1 + \frac{y_{\mathrm{e}}}{b} B_2 = 0,\end{equation}$$
where  
(10)
$$\begin{equation} A_1 = t_{21}\sin u_3 + t_{32}\sin u_1-t_{31}\sin u_2,\end{equation}$$
 
(11)
$$\begin{equation} A_2 = t_{21}\cos u_3+t_{32}\cos u_1-t_{31}\cos u_2,\end{equation}$$
 
(12)
$$\begin{equation} A_3 = t_{21}u_3+t_{32}u_1-t_{31}u_2,\end{equation}$$
 
(13)
$$\begin{equation} B_1 = t_{32}\sin u_4+t_{43}\sin u_2-t_{42}\sin u_3,\end{equation}$$
 
(14)
$$\begin{equation} B_2 = t_{32}\cos u_4+t_{43}\cos u_2-t_{42}\cos u_3,\end{equation}$$
 
(15)
$$\begin{equation} B_3 = t_{32}u_4+t_{43}u_2-t_{42}u_3. \end{equation}$$

These equations are solved as  

(16)
$$\begin{equation} x_{\mathrm{e}} =-a \frac{A_2 B_3-A_3 B_2}{A_1 B_2-A_2 B_1},\end{equation}$$
 
(17)
$$\begin{equation} y_{\mathrm{e}} = b \frac{A_3 B_1-A_1 B_3}{A_1 B_2-A_2 B_1},\end{equation}$$
which determine the location of the projected focus.

The original major axis is projected onto the observed ellipse at $$\boldsymbol{P}_{\mathrm{L}} \equiv (x_{\mathrm{L}}, y_{\mathrm{L}}) = (a \cos u_{\mathrm{L}}, b \sin u_{\mathrm{L}})$$. The ratio of the major axis to the distance between the origin and the focus remains the same, even after the projection. Hence, we find  

(18)
$$\begin{equation} \boldsymbol{P}_{\mathrm{L}} = \frac{1}{e_{\mathrm{K}}} \boldsymbol{P}_{\mathrm{e}}. \end{equation}$$

The positional vector $$\boldsymbol{P}_{\mathrm{L}}$$ is located on the observed ellipse given by equation (3). We thus obtain the ellipticity as  

(19)
$$\begin{equation} e_{\mathrm{K}} = \sqrt{\frac{x_{\mathrm{e}}^2}{a^2}+\frac{y_{\mathrm{e}}^2}{b^2}}. \end{equation}$$

The parameter $$u_{\mathrm{L}} \in [0, 2\pi)$$ is thus given by  

(20)
$$\begin{equation} \cos u_{\mathrm{L}} = \frac{x_{\mathrm{e}}}{ae_{\mathrm{K}}},\end{equation}$$
 
(21)
$$\begin{equation} \sin u_{\mathrm{L}} = \frac{y_{\mathrm{e}}}{be_{\mathrm{K}}}. \end{equation}$$

The location of the projected focus is given by equations (16) and (17), so that we can determine the area $$S_{ij}$$ from equation (5). By using $$S_{21}$$, for instance, in equation (4), we obtain the orbital period as  

(22)
$$\begin{equation} T=\frac{S t_{21}}{S_{21}}. \end{equation}$$

Determining the Major Axis and the Inclination Angle

First, from the above result, we determine the location of the intersection of the observed ellipse and the projected minor axis, denoted by $$\boldsymbol{P}_{\mathrm{S}} \equiv (x_{\mathrm{S}}, y_{\mathrm{S}}) = (a \cos u_{\mathrm{S}}, b \sin u_{\mathrm{S}})$$. It is not necessary to make an observation of the intersection. The major and minor axes divide the area of the ellipse in quarters. Even after projecting the ellipse, the projected major and minor axes, which are not those of the observed ellipse, divide the observed ellipse into quarter areas. This fact implies that $$u_{\mathrm{S}} = u_{\mathrm{L}} + \pi/2$$, which does not mean that $$\boldsymbol{P}_{\mathrm{L}}$$ is perpendicular to $$\boldsymbol{P}_{\mathrm{S}}$$.

Up to this point, we have not specified the angle and orientation of the inclination. Let the Keplerian ellipse inclined with the angle $$i$$ in such a way that the angle at the origin between the periastron and the ascending node is $$\omega$$, being the angular distance of the periastron. We can assume that the inclination angle is in $$[0, \pi/2)$$, though it takes a value within $$[0, \pi)$$ in the standard context of celestial mechanics. This is because in the present paper we do not specify whether the motion of the star in the Keplerian orbit is prograde or retrograde. Hence, rigorously speaking, the ascending node, which we have mentioned above for convenience, may be the descending node. In short, our inclination angle is between the line of sight and a unit normal vector to the orbital plane, where there exist two unit normal vectors, and we choose one such as $$\cos i \geq 0$$. Only in this paragraph, we adopt another Cartesian coordinates $$(x', y')$$ so that the ascending node is located on the $$x'$$-axis. The periastron of the original ellipse is projected at $$\boldsymbol{P}_{\mathrm{L}} \equiv (x_{\mathrm{L}}', y_{\mathrm{L}}') = (a_{\mathrm{K}}\cos\omega, a_{\mathrm{K}}\sin\omega\cos i)$$. Similarly an intersection of the ellipse and the minor axis is projected at $$\boldsymbol{P}_{\mathrm{S}} \equiv (x_{\mathrm{S}}', y_{\mathrm{S}}') = (-b_{\mathrm{K}} \sin\omega, b_{\mathrm{K}} \cos\omega \cos i)$$. The component of the vector depends on the adopted coordinates. Therefore, it is useful to consider the invariants, such as $$\vert\boldsymbol{P}_{\mathrm{L}}\vert$$, $$\vert\boldsymbol{P}_{\mathrm{S}}\vert$$ and $$\vert\boldsymbol{P}_{\mathrm{L}}\times\boldsymbol{P}_{\mathrm{S}}\vert$$. We find  

(23)
$$\begin{equation} |\boldsymbol{P}_{\mathrm{L}}| = a_{\mathrm{K}}\sqrt{\cos^2\omega+\sin^2\omega\cos^2 i},\end{equation}$$
 
(24)
$$\begin{equation} |\boldsymbol{P}_{\mathrm{S}}| = b_{\mathrm{K}}\sqrt{\sin^2\omega+\cos^2\omega\cos^2 i},\end{equation}$$
 
(25)
$$\begin{equation} |\boldsymbol{P}_{\mathrm{L}} \times \boldsymbol{P}_{\mathrm{S}}| = a_{\mathrm{K}}b_{\mathrm{K}}\cos i. \end{equation}$$

From equations (23) and (24), we obtain  

(26)
$$\begin{equation} C^2+D^2 = a_{\mathrm{K}}^2 (1+\cos^2 i),\end{equation}$$
 
(27)
$$\begin{equation} C^2-D^2 = a_{\mathrm{K}}^2 \cos 2\omega \sin^2 i. \end{equation}$$

Here, we use $$b_{\mathrm{K}} = a_{\mathrm{K}} \sqrt{1-e_{\mathrm{K}}^2}$$ and define  

(28)
$$\begin{equation} C = |\boldsymbol{P}_{\mathrm{L}}|,\end{equation}$$
 
(29)
$$\begin{equation} D = |\boldsymbol{Q}_{\mathrm{S}}|,\end{equation}$$
by introducing  
(30)
$$\begin{equation} \boldsymbol{Q}_{\mathrm{S}} = \frac{\boldsymbol{P}_{\mathrm{S}}}{\sqrt{1-e_{\mathrm{K}}^2}},\end{equation}$$
where $$\boldsymbol{Q}_{\mathrm{S}}$$ denotes a positional vector on a circle that is made by stretching the observed ellipse $$a_{\mathrm{K}}/b_{\mathrm{K}}$$ times along its minor axis. Equation (25) is rewritten as  
(31)
$$\begin{equation} |\boldsymbol{P}_{\mathrm{L}}\times\boldsymbol{Q}_{\mathrm{S}}| = a_{\mathrm{K}}^2 \cos i.\end{equation}$$

By using another expression of $$\boldsymbol{P}_{\mathrm{L}} = (a \cos u_{\mathrm{L}}, b \sin u_{\mathrm{L}})$$ and $$\boldsymbol{P}_{\mathrm{S}} = (-a \sin u_{\mathrm{L}}, b \cos u_{\mathrm{L}})$$, $$C$$, $$D$$ and $$\vert\boldsymbol{P}_{\mathrm{L}} \times \boldsymbol{Q}_{\mathrm{S}}\vert$$ are rewritten as  

(32)
$$\begin{equation} C = \frac{1}{e_{\mathrm{K}}}\sqrt{x_{\mathrm{e}}^2 + y_{\mathrm{e}}^2},\end{equation}$$
 
(33)
$$\begin{equation} D = \frac{1}{abe_{\mathrm{K}}} \sqrt{\frac{a^4 y_{\mathrm{e}}^2 + b^4 x_{\mathrm{e}}^2}{1-e_{\mathrm{K}}^2}},\end{equation}$$
 
(34)
$$\begin{equation} |\boldsymbol{P}_{\mathrm{L}}\times\boldsymbol{Q}_{\mathrm{S}}| = \frac{ab}{\sqrt{1-e_{\mathrm{K}}^2}}.\end{equation}$$

From equations (26) and (31), we obtain  

(35)
$$\begin{equation} \cos^2 i-\xi \cos i+1=0,\end{equation}$$
where we define  
(36)
$$\begin{equation} \xi = \frac{C^2+D^2}{|\boldsymbol{P}_{\mathrm{L}}\times\boldsymbol{Q}_{\mathrm{S}}|}.\end{equation}$$

By using $$C^2 + D^2 \geq 2CD \geq 2 \vert\boldsymbol{P}_{\mathrm{L}} \times \boldsymbol{Q}_{\mathrm{S}}\vert$$, we can show that  

(37)
$$\begin{equation} \xi \geq 2.\end{equation}$$

Equation (35) is solved as  

(38)
$$\begin{equation} \cos i=\frac{1}{2} \bigl(\xi \pm \sqrt{\xi^2\hbox{--}4} \:\bigr).\end{equation}$$

We can show that $$\xi + \sqrt{\xi^2\hbox{--}4} \geq 2$$. Since $$\cos i$$ must be no more than unity, we find the only solution to be  

(39)
$$\begin{equation} \cos i=\frac{1}{2} \bigl(\xi-\sqrt{\xi^2\hbox{--}4} \:\bigr),\end{equation}$$
which satisfies $$\cos i \in [0, 1)$$. Hence, we can uniquely determine the inclination angle, $$i$$, for $$i\in [0, \pi/2)$$, which is discussed in the subsection 2.4. Here, by substituting equations (32), (33), and (34) into equation (36), $$\xi$$ is expressed in terms of $$a$$, $$b$$, $$e_{\mathrm{K}}$$, $$x_{\mathrm{e}}$$, and $$y_{\mathrm{e}}$$.

Next, equation (26) determines the length of the major axis as  

(40)
$$\begin{equation} a_{\mathrm{K}}=\sqrt{\frac{C^2+D^2}{1+\cos^2 i}}.\end{equation}$$

Finally, equation (27) determines the orientation of the inclination as  

(41)
$$\begin{equation} \cos 2\omega = \frac{C^2-D^2}{a_{\mathrm{K}}^2 \sin^2 i}.\end{equation}$$

Conclusion

We derive the inversion formula for astrometric observations of binaries. It is summarized as follows. First, we fix the observed ellipse by using the location of five points. Second, we choose four points and use their locations and the time intervals between these points. The projected focus is determined by equations (16) and (17). The ellipticity is given by equation (19). The orbital period is determined by equation (22). The inclination angle is given by equation (39). The length of the major axis is computed from equation (40). Finally, the orientation of the inclination is given by equation (41).

Moreover, our result proves that the mapping between a point in the Keplerian motion and the observed point (in time and space) is one-to-one if the number of the observations is more than four. For instance, a sixth observed point must satisfy our equations with the determined values of the parameters. In practice, the observation inevitably associates errors so that we must make a kind of fittings, for instance, by the least-squares method. Even in such a case, because our formula would give likely values of parameters, we could quickly save CPU time for fittings.

We would like to thank N. Gouda, T. Yano, and M. Yoshikawa for useful comments.

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1
Space Interferometry Mission, $$\langle$$http://sim.jpl.nasa.gov/$$\rangle$$.
2
Global Astrometric Interferometer for Astrophysics, $$\langle$$http://astro.estec.esa.nl/GAIA/$$\rangle$$.
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Japan Astrometry Satellite Mission for INfrared Exploration, $$\langle$$http://www.jasmine-galaxy.org/$$\rangle$$.