3 body problem using only math

Question

This question was suggested to be placed in the math forum. 3 particles are at the corners of an equilateral triangle with side $a$. Assume that particle 1 is at $(0,0)$, particle 2 is at $(a,0)$ and particle 3 is at $(a/2, a\sqrt 3/2)$. They all start moving simultaneously with a velocity $v$ constant in modulus, but with the first particle heading towards the second one, the second towards the third, and the third towards the first particle.

The typical question is how soon will they meet? I can easily answer this question with symmetry and relative speed considerations (they meet at the centroid after a time $2a/3v$). My question is a bit more complicated.

• Can we solve this without invoking symmetry, but purely mathematically?
• Second, can we describe the velocity vector of particle 1 as a function of time $t$ assuming it started at the origin?
• Third, can we describe the trajectory of particle 1 as a curve mathematically?

Answer 1

It makes sense to arrange the three particles right from the start in a way that exhibits the symmetry of the problem. Therefore at time $t=0$ they are at $$R\>\omega^k\quad(0\leq k\leq2),\qquad R:={a\over\sqrt{ 3}},\quad \omega:=e^{2\pi i/3}\ .$$ The uniqueness part of the fundamental theorem about systems of ODEs implies that the three particles form an equilateral triangle centered at $0$ at all times. This means that for all $t$ and all $k$ (mod 3) we have $$z_k(t)=\omega^k\>z(t)$$ for a unique function $$t\mapsto z(t)=r(t)\>e^{i\phi(t)}\ .$$ (Note that the map $z\mapsto \omega\>z$ amounts to a rotation by $120^\circ$ around $0$.)

It remains to determine the function $t\mapsto z(t)$. The constituent equation $$\dot z_k={z_{k+1}-z_k \over| z_{k+1}-z_k|}\>v\qquad(0\leq k\leq2)$$ translates into $$(\dot r+ ir\dot\phi)e^{i\phi}={r e^{i\phi}(\omega-1)\over r|\omega-1|}\>v\ ,$$ so that we obtain $$\dot r+i r\dot\phi =\left(-{\sqrt{3}\over2}+{i\over2}\right)v\ .\tag{1}$$ It follows that $$r(t)=R-{\sqrt{3}\over2}v \>t\qquad(0\leq t\leq T)$$ with $T={2a\over3v}$. I leave it to you to determine the function $t\mapsto\phi(t)$ by looking at the imaginary part of $(1)$. The resulting three curves are logarithmic spirals.

Answer 2

Starting with the origin at one of the particles just makes the problem harder for no good reason. You pick up extra terms that have to be dealt with. Much easier to put it at the center of the triangle. Once you have that solution, it is simple to move it to one of points, if you really feel you must.

Let $p_1, p_2, p_3$ be the position vectors of the three particles. At time $t = 0$, we then have $$\|p_1\| = \|p_2\| = \|p_3\| = \frac a{\sqrt 3}$$ Also at $t = 0$, $$\|p_2 - p_1\| = \|p_3 - p_2\| = \|p_1 - p_3\| = a$$ From this, you can easily deduce that $$p_1\cdot p_2 = p_2\cdot p_3 = p_3\cdot p_1 = -\frac{a^2}6$$

Now, the initial velocity condition can be expressed as $$\dot p_1(0) = b(p_2(0) - p_1(0))\\\dot p_2(0) = b(p_3(0) - p_2(0))\\\dot p_3(0) = b(p_1(0) - p_3(0))$$ for some constant $b$. And the gravitational attraction can be expressed as $$\ddot p_1 = c\frac{p_2 - p_1}{\|p_2 - p_1\|^3} + c\frac{p_3 - p_1}{\|p_3 - p_1\|^3}$$ $$\ddot p_2 = c\frac{p_1 - p_2}{\|p_1 - p_2\|^3} + c\frac{p_3 - p_2}{\|p_3 - p_2\|^3}$$ $$\ddot p_3 = c\frac{p_1 - p_3}{\|p_1 - p_3\|^3} + c\frac{p_2 - p_3}{\|p_2 - p_3\|^3}$$

for some constant $c$. Define $p = p_1 + p_2 + p_3$. Because the origin is in the center of the triangle, $p(0) = 0$. Also $$\dot p(0)= \dot p_1(0) + \dot p_2(0) +\dot p_3(0) = 0$$ and $$\ddot p= \ddot p_1 + \ddot p_2 +\ddot p_3 = 0$$

Therefore $p = 0$ for all $t$.

You can continue in this vane to solve the entire system. By choosing expressions that are symmetric in their use of $p_1, p_2, p_3$, you can show that they behave simply, then use this information to deduce $p_1, p_2, p_3$ themselves. If I have time, I may take it farther.

Answer 3

Here is my take: Particles at A, B and C. O the center of the triangle. A move to B, B to C, and C to A, with a constant velocity $v$. $v_t$ the tangent component and $v_r$ the radial component.

$\frac{dR}{dt}=-v_r$ and $\frac{d\theta}{dt}=\frac{v_t}{R}$ from this $\frac{d\theta}{dR}$ may be determined and solved for $R=f(\theta)$

can you take it from here? If not will get the solution for you.