Generally speaking, choosing an inertial frame as the reference frame, the equations of motion for the n-body problem are listed as: \ddot{\vec{r}}_k=\sum_{i=1,i \neq k}^{n} Gm_i\frac{\vec{r}_i-\vec{r}_k}{|\vec{r}_i-\vec{r}_k|^3}
Historically, many different forms of transformations have appeared, giving the equations of motion for the three-body problem various forms, such as the Lagrange form, Jacobi form, Hamilton form, etc. These transformation forms each have their own characteristics and can all simplify the three-body problem to a certain extent. While BoJone was studying and playing with the equations of motion for the equal-mass three-body problem, he also discovered a very interesting transformation, which is posted here to share with everyone.
Let \vec{R}_1=\vec{r}_1-\vec{r}_2, \vec{R}_2=\vec{r}_2-\vec{r}_3, \vec{R}_3=\vec{r}_3-\vec{r}_1, then the equations of motion for the three-body problem become:
\ddot{\vec{r}}_1=-Gm_2\frac{\vec{R}_1}{R_1^3}+Gm_3\frac{\vec{R}_3}{R_1^3} \tag{35} \ddot{\vec{r}}_2=-Gm_3\frac{\vec{R}_2}{R_2^3}+Gm_1\frac{\vec{R}_1}{R_1^3} \tag{36} \ddot{\vec{r}}_3=-Gm_1\frac{\vec{R}_3}{R_3^3}+Gm_2\frac{\vec{R}_2}{R_2^3} \tag{37}
where R_1, R_2, R_3 are the magnitudes of the corresponding vectors. Subtracting (35)-(36), (36)-(37), and (37)-(35) yields:
\begin{aligned} \ddot{\vec{R}}_1 &= -G(m_1+m_2)\frac{\vec{R}_1}{R_1^3}+Gm_3\left(\frac{\vec{R}_3}{R_1^3}+\frac{\vec{R}_2}{R_2^3}\right) \tag{38} \\ \ddot{\vec{R}}_2 &= -G(m_2+m_3)\frac{\vec{R}_2}{R_2^3}+Gm_1\left(\frac{\vec{R}_1}{R_1^3}+\frac{\vec{R}_3}{R_3^3}\right) \tag{39} \\ \ddot{\vec{R}}_3 &= -G(m_1+m_3)\frac{\vec{R}_3}{R_3^3}+Gm_2\left(\frac{\vec{R}_1}{R_1^3}+\frac{\vec{R}_2}{R_2^3}\right) \tag{40} \end{aligned}
Obviously, this is the “perturbation form” of the three-body equations of motion. Of course, this does not yet show anything “interesting,” but it becomes different when switched to the case of equal masses. Selecting appropriate units such that G, m_1, m_2, m_3 = 1, and subtracting (38)-(39), (39)-(40), and (40)-(38) gives:
\ddot{\vec{R}}_1-\ddot{\vec{R}}_2=-\frac{3\vec{R}_1}{R_1^3}+\frac{3\vec{R}_2}{R_2^3} \tag{41} \ddot{\vec{R}}_2-\ddot{\vec{R}}_3=-\frac{3\vec{R}_2}{R_2^3}+\frac{3\vec{R}_3}{R_3^3} \tag{42} \ddot{\vec{R}}_3-\ddot{\vec{R}}_1=-\frac{3\vec{R}_3}{R_3^3}+\frac{3\vec{R}_1}{R_1^3} \tag{43}
(41) (43) possess a high degree of symmetry! However, whether transforming them into this form simplifies or complicates the problem, BoJone does not know. From the general perspective of solving differential equations, having two second-derivative terms on the left side of each equation suggests it is a complication. I only know that this is a quite interesting form because it removes the terms where two vectors are subtracted in the denominators and numerators, representing them with a single new vector instead. And in the case of equal masses, it is so symmetric that it at least gives us a sense of “aesthetic beauty.”
After this transformation, the original ten integrals of the three-body problem can be transformed accordingly. For example, the conservation of momentum becomes \vec{R}_1+\vec{R}_2+\vec{R}_3=\vec{0}, etc. The angular momentum integral and the conservation of energy can also be transformed; interested readers might want to try it themselves.
Just recording this!
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