System of two particles: Prove motion along connecting line

[Cepterus]

Homework Statement

Given an isolated system of 2 particles in space, we can express the motion of both particles as follows:
$$m_1\ddot{\vec{x_1}}=-\frac\partial{\partial \vec{x_{1}}} V(\vec{x_1},\vec{x_2})\\
m_2\ddot{\vec{x_2}}=-\frac\partial{\partial \vec{x_2}} V(\vec{x_1},\vec{x_2}),$$ where $V$ shall be a potential and $\frac\partial{\partial \vec{x}}V$ shall denote its gradient.

Assuming both particles are at rest at first, prove that they will move on the line which connects both starting points.

Homework Equations

The Attempt at a Solution

In the end, we have to get a result of $\vec x_i = k_i(\vec x_{2,0} - \vec x_{1,0})$, where $k_i$ is a scalar. Usually I would find $\vec x$ by setting up a differential equation and solving it, but since the equations of motion given here are so general, I don't know how to do this.

 

[andrewkirk]

How to prove this depends on what properties we are allowed to infer from the statement

where V shall be a potential

If we are allowed to infer that the potential of the field created by particle 1 is a function solely of the distance from particle 1 (which is the case for point particles under classical gravitational and electrostatic forces), and we further assume that both particles are approximately point particles, then the result will be easy to prove.

If not, it may not even be true.

 

[kuruman]

I would try coordinate transformations involving the center of mass the motion of which you should be able to separate out. For example, let
$X=\frac{m_1x_1+m_2x_2}{m_1+m_2}$ and $\xi=x_2-x_1$.
With copious use of the chain rule, you should be able to recast the two equations in terms of the new coordinates. Since the system is isolated (no external forces) you expect one equation to be $\ddot{X}=0$. The other equation should involve $\ddot{\xi}$ which is along the line joining the particles.

Disclaimer: I have not worked this out, but my intuition says it's worth trying.