The restricted three-body problem in cylindrical clouds

Abstract

The fragmentation of interstellar molecular clouds has been investigated with great effort by many authors. In this paper, a simple model is given to describe the dynamics of two fragments moving in a special cylindrical potential. Using a modified version of the restricted three-body problem and the corresponding Jacobian integral, some constraints are given for the motion of the fragments.

Appendix

Here, we investigate the linear stability of the collinear equilibrium point found above. First, Eqs. (24) and (25) are linearized (that is, we investigate their behavior at $x=x_0+\xi$ and $y=0+\,\eta$, $x_0$ and 0 being the coordinates of the equilibrium points):

$$\begin{aligned} \ddot{\xi } = -\frac{1}{2}b^2\xi + 2\dot{\eta }\sqrt{1-\frac{1}{2}b^2} + \frac{3}{2}x_0^2\xi + 2m_1 \frac{\xi }{|x_0-b|^3}, \end{aligned}$$

(33)

$$\begin{aligned} \ddot{\eta } = -\frac{1}{2}b^2\eta - 2\dot{\xi }\sqrt{1-\frac{1}{2}b^2} + \frac{1}{2}x_0^2\eta - m_1 \frac{\eta }{|x_0-b|^3}. \end{aligned}$$

(34)

Assuming the solution in the form $\xi = A e^{\lambda t}$ and $\eta = B e^{\lambda t}$ with $A \ne 0$ and $B \ne 0$, substitution yields

$$\begin{aligned} \begin{pmatrix} -\frac{1}{2}b^2 + \frac{3}{2}x_0^2 + 2m_1 \frac{1}{|x_0-b|^3} -\lambda ^2 &{} 2 \lambda \sqrt{1-\frac{1}{2}b^2} \\ -2 \lambda \sqrt{1-\frac{1}{2}b^2} &{} -\frac{1}{2}b^2+\frac{1}{2}x_0^2-m_1\frac{1}{|x_0-b|^3}-\lambda ^2 \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix}=0. \end{aligned}$$

(35)

This leads to the characteristic equation with $\Lambda = \lambda ^2$

$$\begin{aligned}&\Lambda ^2+\left( -b^2-2x_0^2-m_1\frac{1}{|x_0-b|^3}+4\right) \Lambda \nonumber \\&\quad +\left( -\frac{b^2}{2}+\frac{3x_0^2}{2}+\frac{2m_1}{|x_0-b|^3}\right) \left( -\frac{b^2}{2}+\frac{x_0^2}{2}-\frac{m_1}{|x_0-b|^3}\right) =0. \end{aligned}$$

(36)

An equilibrium point is stable if and only if $\Lambda _1, \Lambda _2 < 0$ and real.

Let

$$\begin{aligned} B=-\,b^2-2x_0^2-m_1\frac{1}{|x_0-b|^3}+4 \end{aligned}$$

(37)

and

$$\begin{aligned} C=\left( -\frac{b^2}{2}+\frac{3x_0^2}{2}+\frac{2m_1}{|x_0-b|^3}\right) \left( -\frac{b^2}{2}+\frac{x_0^2}{2}-\frac{m_1}{|x_0-b|^3}\right) , \end{aligned}$$

(38)

so that

$$\begin{aligned} \Lambda ^2+B\Lambda +C=0. \end{aligned}$$

(39)

1.1 $x_0>b$, the case of $L_1$

As $3x_0^2 > b^2$, the first bracket in C is positive. Using Eq. (30), the second bracket is

$$\begin{aligned}&-\frac{b^2}{2}+\frac{x_0^2}{2}-\frac{m_1}{|x_0-b|^3}\nonumber \\&\quad = -\frac{b^2}{2}+\frac{x_0^2}{2}-\frac{1}{2}\frac{1}{x_0-b}(x_0^3-b^2x_0)=-\frac{1}{2}b(x_0+b)<0. \end{aligned}$$

(40)

In conclusion, C is negative. As $\Lambda _{1,2}=(-B\pm \sqrt{B^2-4C})/2$ and $B^2-4C>|B|$, there is always a positive $\Lambda$ which results in instability.

1.2 $x_0<-\,b$, the case of $L_3$

The first bracket in C is positive again. However, in this case

$$\begin{aligned} -\frac{b^2}{2}+\frac{x_0^2}{2}-\frac{m_1}{|x_0-b|^3}=-\frac{1}{2}b(x_0+b)>0, \end{aligned}$$

(41)

which concludes that C itself is positive. Equation (39) determines a parabola, which has negative roots if both the position ($-B/2$) and the value ($-B^2/4+C$) of its minimum are negative. Then, the following equations are to be solved:

$$\begin{aligned} B=-b^2-2x_0^2-m_1\frac{1}{|x_0-b|^3}+4 = -b^2-2x_0^2-\frac{1}{2}x_0(x_0+b)+4>0, \end{aligned}$$

(42)

and

$$\begin{aligned} B^2-4C= & {} \left( -b^2-2x_0^2-\frac{1}{2}x_0(x_0+b)+4 \right) ^2\nonumber \\&-4\left( -\frac{1}{2}b(x_0+b) \right) \left( -\frac{b^2}{2}+\frac{3x_0^2}{2} + x_0(x_0+b) \right) >0. \end{aligned}$$

(43)

Assuming again that $b=0.1$ (see Sect. 4), the condition of stability is $-\,1.13< x_0 <-\,0.1$ or $m_1<1.08$, which seems physically reasonable.