Introduction
In a previous post:
I showed that elliptical motion in classical mechanics can be expressed by
\[ \small \begin{align*} &x(t) = r(t)\cos\omega(t) = \frac{a(1-\epsilon^2)}{1+\epsilon\cos \omega(t)}\cos\omega(t) \\ &y(t) = r(t)\sin\omega(t) = \frac{a(1-\epsilon^2)}{1+\epsilon\cos \omega(t)}\sin\omega(t) \\ &\frac{d\omega}{dt} = \frac{\sqrt{L^2}}{mr^2} \\ &a= -\frac{1}{2}\frac{GmM}{E} \\ &\epsilon = \sqrt{1+\frac{2EL^2}{G^2m^3M^2}}. \end{align*} \]
By differentiating with respect to time, the velocity and acceleration can be found as follows:
\[ \small \begin{align*} &\frac{dx}{dt}=-r^2(t)\frac{d\omega}{dt}\frac{\sin\omega(t)}{a(1-\epsilon^2)}= -\sqrt{\frac{GM}{a(1-\epsilon^2)}} \sin \omega(t) \\ &\frac{dy}{dt}=r^2(t)\frac{d\omega}{dt}\frac{\epsilon+\cos\omega(t)}{a(1-\epsilon^2)} = \sqrt{\frac{GM}{a(1-\epsilon^2)}} (\epsilon+\cos \omega(t)) \\ &\frac{d^2x}{dt^2} =-\sqrt{\frac{GM}{a(1-\epsilon^2)}} \cos \omega(t)\frac{d\omega}{dt}=-\frac{GM}{r^2}\cos\omega(t) \\ &\frac{d^2y}{dt^2} =-\sqrt{\frac{GM}{a(1-\epsilon^2)}} \sin \omega(t)\frac{d\omega}{dt}=-\frac{GM}{r^2}\sin\omega(t). \end{align*} \]
I would like to eventually extend this to the framework of the theory of relativity, but these calculations have the problem that the coordinates of the point mass with the larger mass are fixed at the origin, and the two point masses are not treated symmetrically. Since the theory of relativity requires treating motion relatively, I feel that this equation is inconvenient as it is. The purpose of this article is to relativize the above equations of motion and show how they can be formulated when the equations are rearranged to treat two masses as moving independently.
Two-body Problem
The problem in this article is to derive an equation of motion that expresses the same motion even if the coordinates of two masses are \(\small x_1,y_1,x_2,y_2\) (the \(\small z\) axis is set to 0) and the masses are \(\small m_1,m_2\), and these parameters are swapped. This type of problem is generally called the Two-body Problem. The famous Three-body Problem is an extension of this problem to three or more bodies. There may be several forms of solution to a two-body problem, but a common solution is to have the center of mass of the two masses at the origin. Specifically, this is a method for determining the coordinates of two mass points so as to satisfy
\[ \small \begin{align*} &m_1x_1(t)+m_2x_2(t) = 0 \\ &m_1y_1(t)+m_2y_2(t) = 0, \end{align*} \]
and is called a barycentric coordinate system. The solution to the equation of motion itself is the same as the elliptical motion in classical mechanics, and it is necessary to determine the equation of motion for the two masses so that it is consistent with this.
If we define the elliptical motion in classical mechanics as \(\small x(t),y(t)\), since it can be defined as:
\[ \small \begin{align*} &x(t) = x_1(t)-x_2(t) \\ &y(t) = y_1(t)-y_2(t), \end{align*} \]
to find an equation that satisfies the conditions of the barycentric coordinate system, simply define it as:
\[ \small \begin{align*} &x_1(t) = \frac{m_2}{m_1+m_2}x(t), \quad x_2(t) = -\frac{m_1}{m_1+m_2}x(t) \\ &y_1(t) = \frac{m_2}{m_1+m_2}y(t), \quad y_2(t) = -\frac{m_1}{m_1+m_2}y(t). \end{align*} \]
Velocity and acceleration are the differentiated values, but since the coefficients in \(\small x(t),y(t)\) are constants, they can be easily calculated.
Let us consider how to define the parameters \(\small a,\epsilon\) that determine these motions. Since the semimajor axis \(\small a\) is a length, it can be determined by multiplying it by the weight of the barycentric coordinate system, just like coordinates. That is,
\[ \small a_1 =\frac{m_2}{m_1+m_2}a, \quad a_2 =\frac{m_1}{m_1+m_2}a, \quad a =a_1+a_2 \]
holds. To be more specific, it can be written as:
\[ \small \begin{align*} &a_1 = -\frac{m_2}{m_1+m_2}\frac{1}{2}\frac{Gm_1m_2}{E} \\ &a_2 = -\frac{m_1}{m_1+m_2}\frac{1}{2}\frac{Gm_1m_2}{E} \\ &E = -\frac{1}{2}\frac{Gm_1m_2}{a_1+a_2}. \end{align*} \]
Next, note that the eccentricity \(\small \epsilon\), period \(\small T\), and true anomaly \(\small \omega(t)\) must be common to the motion of the two mass points. The orbital angular momentum when the coordinates of one mass point are fixed is:
\[ \small L^2 = Gm^2Ma(1-\epsilon^2), \]
but it is:
\[ \small \begin{align*} &L_1^2 = Gm_1^2m_2a_1(1-\epsilon^2) \\
&L_2^2 = Gm_1m_2^2a_2(1-\epsilon^2) \end{align*} \]
and \(\small L^2 = L_1^2+L_2^2\) for mass points 1 and 2, respectively. By calculating it,
\[ \small \begin{align*} L^2 &= Gm_1m_2(1-\epsilon^2)(m_1a_1+m_2a_2) \\ &= -\frac{1}{2E}\frac{G^2m_1^3m_2^3}{(m_1+m_2)}(1-\epsilon^2) = \frac{Gm_1^2m_2^2}{m_1+m_2}a(1-\epsilon^2) \end{align*} \]
can be obtained. We can get
\[ \small \epsilon =\sqrt{1+\frac{2EL^2(m_1+m_2)}{G^2m_1^3m_2^3}}=\sqrt{1-\frac{L^2}{Gm_1^2m_2^2}\frac{m_1+m_2}{a_1+a_2}} \]
by working backwards from \(\small \epsilon\).
Finally,
\[ \small \begin{align*} &r_1^2\frac{d\omega}{dt} = r^2\frac{d\omega}{dt}\left(\frac{m_2}{m_1+m_2}\right)^2 = \left(\frac{m_2}{m_1+m_2}\right)^2\frac{L}{m} = \frac{L_1}{m_1} \\ &r_2^2\frac{d\omega}{dt} = r^2\frac{d\omega}{dt}\left(\frac{m_1}{m_1+m_2}\right)^2 = \left(\frac{m_1}{m_1+m_2}\right)^2\frac{L}{m} = \frac{L_2}{m_2} \end{align*} \]
holds true according to the law of constant areal velocity. \(\small m\) is the mass of a small mass point in classical mechanical elliptical motion, but since this is not a symmetric equation, we will try to guess what the equation for \(\small m\) is in the barycentric coordinate system. If we consider the total angular momentum of two mass points to be
\[ \small m_1L_1+m_2L_2 =\left(\frac{m_1m_2}{m_1+m_2}\right)^2\frac{L}{m}=mL, \]
we can obtain
\[ \small \frac{1}{m}= \frac{m_1+m_2}{m_1m_2} = \frac{1}{m_1}+\frac{1}{m_2}. \]
By substituting it,
\[ \small \begin{align*} &r_1^2\frac{d\omega}{dt} = \frac{1}{m_1+m_2}\frac{m_2}{m_1}L \\ &r_2^2\frac{d\omega}{dt} = \frac{1}{m_1+m_2}\frac{m_1}{m_2}L \end{align*} \]
can be obtain. Since
\[ \small \begin{align*} &S_1 = \pi a_1 b_1 = \pi a_1^2 \sqrt{1-\epsilon^2} \\ &S_2 = \pi a_2 b_2 = \pi a_2^2 \sqrt{1-\epsilon^2} \\ &\frac{dS_1}{dt} = \frac{1}{2}r_1^2\dot{\omega} = \frac{1}{2}\frac{1}{m_1+m_2}\frac{m_2}{m_1}L \\ &\frac{dS_2}{dt} = \frac{1}{2}r_2^2\dot{\omega} = \frac{1}{2}\frac{1}{m_1+m_2}\frac{m_1}{m_2}L, \end{align*} \]
if we calculate the period using Kepler’s third law, the period is:
\[ \small \begin{align*} T&=\frac{S_1}{dS_1/dt}=\frac{S_2}{dS_2/dt} \\ &=\frac{m_1(m_1+m_2)}{m_2}\frac{2\pi a_1^2\sqrt{1-\epsilon^2}}{L}= \frac{m_2(m_1+m_2)}{m_1}\frac{2\pi a_2^2\sqrt{1-\epsilon^2}}{L} \\ &=\frac{m_1m_2}{m_1+m_2}\frac{2\pi a^2\sqrt{1-\epsilon^2}}{L}. \end{align*} \]
Since
\[ \small L^2 = \frac{Gm_1^2m_2^2}{m_1+m_2}a(1-\epsilon^2), \]
we can get
\[ \small T= \frac{2\pi}{\sqrt{G(m_1+m_2)}}(a_1+a_2)^{3/2}. \]
With the above steps, we have been able to rewrite the equations into ones that are symmetric with respect to the two mass points. You can confirm that the formula does not change even if the subscripts 1 and 2 are swapped.
Summery of Results
Finally, let us summarize the coordinate equations of motion. The results are
\[ \small \begin{align*} &x_1(t)=\frac{m_2}{m_1+m_2}x(t) = r_1(t)\cos\omega(t) = \frac{a_1(1-\epsilon^2)}{1+\epsilon\cos \omega(t)}\cos\omega(t) \\ &x_2(t)=-\frac{m_1}{m_1+m_2}x(t) = -r_2(t)\cos\omega(t) = -\frac{a_2(1-\epsilon^2)}{1+\epsilon\cos \omega(t)}\cos\omega(t) \\ &y_1(t) = \frac{m_2}{m_1+m_2}y(t) = r_1(t)\sin\omega(t) = \frac{a_1(1-\epsilon^2)}{1+\epsilon\cos \omega(t)}\sin\omega(t) \\ &y_2(t)=-\frac{m_1}{m_1+m_2}y(t) = -r_2(t)\cos\omega(t) = -\frac{a_2(1-\epsilon^2)}{1+\epsilon\cos \omega(t)}\sin\omega(t). \end{align*} \]
\(\small a_1,a_2,\epsilon\) can each be calculated in
\[ \small \begin{align*} &a=a_1+a_2 = -\frac{1}{2}\frac{Gm_1m_2}{E}, \quad a_1 =\frac{m_2}{m_1+m_2}a, \quad a_2 =\frac{m_1}{m_1+m_2}a \\ &\epsilon =\sqrt{1+\frac{2EL^2(m_1+m_2)}{G^2m_1^3m_2^3}}=\sqrt{1-\frac{L^2}{Gm_1^2m_2^2}\frac{m_1+m_2}{a_1+a_2}} \end{align*} \]
when \(\small E,L^2\) are exogenously given. The equation for angular velocity is:
\[ \small r^2\frac{d\omega}{dt} = \sqrt{L^2}\left(\frac{1}{m_1}+\frac{1}{m_2}\right)=\sqrt{G(m_1+m_2)(a_1+a_2)(1-\epsilon^2)}. \]
By calculating the velocity, we get
\[ \small \begin{align*} &\frac{dx_1(t)}{dt} =-r_1^2(t)\frac{d\omega}{dt}\frac{\sin\omega(t)}{a_1(1-\epsilon^2)}= -\frac{m_2}{m_1+m_2}\sqrt{\frac{G(m_1+m_2)}{a(1-\epsilon^2)}} \sin \omega(t) \\ &\frac{dx_2(t)}{dt} =r_2^2(t)\frac{d\omega}{dt}\frac{\sin\omega(t)}{a_2(1-\epsilon^2)}= \frac{m_1}{m_1+m_2}\sqrt{\frac{G(m_1+m_2)}{a(1-\epsilon^2)}} \sin \omega(t) \\ &\frac{dy_1(t)}{dt} =r_1^2(t)\frac{d\omega}{dt}\frac{\epsilon+\cos\omega(t)}{a_1(1-\epsilon^2)}= \frac{m_2}{m_1+m_2}\sqrt{\frac{G(m_1+m_2)}{a(1-\epsilon^2)}} (\epsilon+ \cos \omega(t)) \\ &\frac{dy_2(t)}{dt} =-r_2^2(t)\frac{d\omega}{dt}\frac{\epsilon+\cos\omega(t)}{a_2(1-\epsilon^2)}=-\frac{m_1}{m_1+m_2}\sqrt{\frac{G(m_1+m_2)}{a(1-\epsilon^2)}} (\epsilon+ \cos \omega(t)). \end{align*} \]
By calculating the difference,
\[ \small \begin{align*} &\frac{dx(t)}{dt} = \frac{dx_1(t)}{dt}-\frac{dx_2(t)}{dt} =-\sqrt{\frac{G(m_1+m_2)}{a(1-\epsilon^2)}} \sin \omega(t) \\ &\frac{dy(t)}{dt} = \frac{dy_1(t)}{dt}-\frac{dy_2(t)}{dt} =\sqrt{\frac{G(m_1+m_2)}{a(1-\epsilon^2)}} (\epsilon+ \cos \omega(t)) \end{align*} \]
can be obtained.
Differentiating once more, we calculate the acceleration to be:
\[ \small \begin{align*} &\frac{d^2x_1(t)}{dt^2} = -\frac{m_2}{m_1+m_2}\sqrt{\frac{G(m_1+m_2)}{a(1-\epsilon^2)}} \cos \omega(t) \frac{d\omega}{dt} =-\frac{m_2}{m_1+m_2}\frac{G(m_1+m_2)}{r^2}\cos \omega(t) \\ &\frac{d^2x_2(t)}{dt^2} = \frac{m_1}{m_1+m_2}\sqrt{\frac{G(m_1+m_2)}{a(1-\epsilon^2)}} \cos \omega(t) \frac{d\omega}{dt} =\frac{m_1}{m_1+m_2}\frac{G(m_1+m_2)}{r^2}\cos \omega(t) \\ &\frac{d^2y_1(t)}{dt^2} = -\frac{m_2}{m_1+m_2}\sqrt{\frac{G(m_1+m_2)}{a(1-\epsilon^2)}} \sin \omega(t)\frac{d\omega}{dt} = -\frac{m_2}{m_1+m_2}\frac{G(m_1+m_2)}{r^2}\sin \omega(t) \\ &\frac{d^2y_2(t)}{dt^2}=\frac{m_1}{m_1+m_2}\sqrt{\frac{G(m_1+m_2)}{a(1-\epsilon^2)}} \sin \omega(t) \frac{d\omega}{dt} = \frac{m_1}{m_1+m_2}\frac{G(m_1+m_2)}{r^2}\sin \omega(t). \end{align*} \]
As with velocity, if we calculate the difference, it comes to
\[ \small \begin{align*} &\frac{d^2x(t)}{dt^2} = \frac{d^2x_1(t)}{dt^2}-\frac{d^2x_2(t)}{dt^2} =-\frac{G(m_1+m_2)}{r^2}\cos \omega(t) = -\frac{G(m_1+m_2)}{r^2}\frac{x}{r} \\ &\frac{d^2y(t)}{dt^2} = \frac{d^2y_1(t)}{dt^2}-\frac{d^2y_2(t)}{dt^2} =-\frac{G(m_1+m_2)}{r^2}\sin \omega(t) = -\frac{G(m_1+m_2)}{r^2}\frac{y}{r}. \end{align*} \]
You might think that it is a strange equation to calculate, but if you calculate it from
\[ \small \begin{align*} &m_1\frac{d^2x_1(t)}{dt^2} = -\frac{Gm_1m_2}{r^3}(x_1(t)-x_2(t)) \\ &m_2\frac{d^2x_2(t)}{dt^2} = -\frac{Gm_1m_2}{r^3}(x_2(t)-x_1(t)) \\ &m_1\frac{d^2y_1(t)}{dt^2} = -\frac{Gm_1m_2}{r^3}(y_1(t)-y_2(t)) \\ &m_2\frac{d^2y_2(t)}{dt^2} = -\frac{Gm_1m_2}{r^3}(y_2(t)-y_1(t)), \end{align*} \]
you will see that it is indeed a valid equation.
Appendix: Three-body Problem in Barycentric Coordinate System
The three-body problem involves finding the equations of motion for coordinates and velocity (18 in total) when nine acceleration equations:
\[ \small \begin{align*} &m_1\frac{d^2q_1(t)}{dt^2} = -\frac{Gm_1m_2}{r_{12}^3}(q_1(t)-q_2(t))-\frac{Gm_1m_3}{r_{13}^3}(q_1(t)-q_3(t)) \\ &m_2\frac{d^2q_2(t)}{dt^2} = -\frac{Gm_1m_2}{r_{12}^3}(q_2(t)-q_1(t))-\frac{Gm_2m_3}{r_{23}^3}(q_2(t)-q_3(t)) \\ &m_3\frac{d^2q_3(t)}{dt^2} = -\frac{Gm_1m_3}{r_{13}^3}(q_3(t)-q_1(t))-\frac{Gm_2m_3}{r_{23}^3}(q_3(t)-q_2(t)), \; q=x,y,z \\ &r_{ij} = \sqrt{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2} \end{align*} \]
that can be calculated from the energy equation in classical mechanics are given. Because this is a second-order differential equation, 18 integral constants are required to determine the solution. However, it is said that there are 10 integral constants given as natural conditions (4 for energy \(\small E\) and angular momentum \(\small l_x,l_y,l_z\), and 6 for how the center of the coordinate system is determined), and since there are 8 extra unknowns, it is a problem for which a general solution cannot be determined. It has been known as an unsolved mathematical problem for centuries. Of course, if we specify eight additional conditions on the coordinates and velocities (using some approximations, etc.), we can get a special solution, but that is probably too special to be considered a general-purpose solution.
I wrote that the six integral constants can be reduced by how the center of the coordinate system is determined, and this appendix explains how this can be expressed when the coordinate system is written in a barycentric coordinate system. Of course, we can reduce the number of parameters to six by assuming that one mass point does not move at the origin, but let us define the same thing using the barycentric coordinate system. Since the center of gravity of the three masses is the origin,
\[ \small \begin{align*} &m_1x_1(t)+m_2x_2(t)+m_3x_3(t) = 0 \\ &m_1y_1(t)+m_2y_2(t)+m_3y_3(t) = 0 \\ &m_1z_1(t)+m_2z_2(t)+m_3z_3(t) = 0 \\ \end{align*} \]
must be satisfied. Similarly, in the case of two mass points, let us consider determining the coordinates of each mass point from the difference in coordinates between mass point 1 and mass point 2, and the difference in coordinates between mass point 1 and mass point 3:
\[ \small \begin{align*} &x_{12}(t) = x_1(t)-x_2(t) \\ &x_{13}(t) = x_1(t)-x_3(t). \end{align*} \]
If we determine the coefficient correctly and set it as:
\[ \small \begin{align*} &x_1(t) = \frac{m_2x_{12}+m_3x_{13}}{m_1+m_2+m_3} \\ &x_2(t) = \frac{-m_1x_{12}+m_3x_{23}}{m_1+m_2+m_3} \\ &x_3(t) = \frac{-m_1x_{13}-m_2x_{23}}{m_1+m_2+m_3}, \end{align*} \]
we can confirm that the center of gravity is the origin. The difference in coordinates between mass 2 and mass 3, \(\small x_{23}\), is:
\[ \small x_{23}(t) = x_2(t)-x_3(t) = x_{13}(t)-x_{12}(t). \]
If we calculate and put
\[ \small \begin{align*} &x_1(t) = \frac{m_2x_{12}+m_3x_{13}}{m_1+m_2+m_3} \\ &x_2(t) = \frac{-(m_1+m_3)x_{12}+m_3x_{13}}{m_1+m_2+m_3} \\ &x_3(t) = \frac{-(m_1+m_2)x_{13}+m_2x_{12}}{m_1+m_2+m_3}, \end{align*} \]
we can define the coordinates of each mass point as a function of \(\small x_{12},x_{13}\). The three acceleration equations can be combined into two equations:
\[ \small \begin{align*} &\frac{d^2x_{12}(t)}{dt^2} = -\frac{G(m_1+m_2)}{r_{12}^3}x_{12}-\frac{Gm_3}{r_{13}^3}x_{13}-\frac{Gm_3}{r_{23}^3}(x_{13}-x_{12}) \\ &\frac{d^2x_{13}(t)}{dt^2} = -\frac{G(m_1+m_3)}{r_{13}^3}x_{13}-\frac{Gm_2}{r_{12}^3}x_{12}+\frac{Gm_2}{r_{23}^3}(x_{13}-x_{12}). \end{align*} \]
This will reduce the number of unknowns to six.
The three-body problem is not so much that we don’t know the solution, but rather that it has been proven to be unsolvable in the first place, so you might think it’s impossible to imagine how the three celestial bodies move. However, that does not mean that it is impossible to imagine what a special solution might be. Considering the relationship between the Sun, Earth, and Moon in the realistic motion of the planets, this would be a good solution to the three-body problem. However, this is because the mass of the Sun is so large compared to the Moon and Earth, and the distance between the Moon and Earth is so small compared to their distance from the Sun. A big interest is what kind of motion will occur when \(\small m_1\approx m_2\approx m_3\) and \(\small r_{12}\approx r_{13} \approx r_{23}\).
A major problem with classical mechanics is that it actually misunderstands the concepts of time and space. The real solution would be to formulate it as a problem in the theory of relativity. In the theory of relativity, space is assumed to be a three-dimensional sphere, so the problem may be that some force (or no force at all) acting on a three-dimensional sphere is mistaken for gravitation in Euclidean space. In other words, the difficulty with the three-body problem is that the problem itself is wrong, so we must first rethink what the correct problem setting is. In order to do this, we must understand the true nature of gravity, but this is currently a huge challenge in the field of physics.
What seems questionable is the assumption that the effects of gravity are independent for each coordinate axis; if this is incorrect, a degree of freedom would be lost that might make it possible to find a solution. One hypothesis would be that the problem with the three-body problem is that it is impossible to find a solution because degrees of freedom that should not actually exist have been introduced. Another hypothesis is that, like quantum mechanics, the motion of large objects such as planets is not actually definite, but can only be determined probabilistically. In cases such as \(\small m_1\approx m_2\approx m_3\), it is not possible to determine the orbit with certainty, and it is possible that we are only observing a randomly sampled orbit from an infinite number of possible orbits. In this case, the solution to the three-body problem must be formulated not as a deterministic equation of motion, but as a probability distribution problem, like the Schrödinger equation. In any case, my honest impression is that this is a problem that I cannot even imagine at this point in time…
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