Two bodies initially at rest with arbitrary interaction remain in the line that connects them

Question

I am working through VI Arnold "Mathematical Methods of Classical Mechanics". One of the first problems, after defining Galilean structures and Newton's equation of motion, it is to prove 'For a mechanical system of two points with initial velocity zero, the points will remain in the line that connected them in the begining'

I tried to prove as follows: denote $\vec{x} = (\vec{x}_1, \vec{x}_2)$ the position of the system, and $\ddot{\vec{x}} = \vec{F}(\vec{x},\dot{\vec{x}})$ the equations of motion. By galilean invariance I can prove that

$$ \ddot{\vec{x}_1} = \vec{F}_1(\vec{x}_1-\vec{x}_2,\dot{\vec{x}_1}-\dot{\vec{x}_2})\\ \ddot{\vec{x}_2} = -\vec{F}_1(\vec{x}_1-\vec{x}_2,\dot{\vec{x}_1}-\dot{\vec{x}_2}), $$

and, thus, proving that the center of mass $\vec{x}_1+\vec{x}_2$ stays static if $\dot{\vec{x}_i}(t_0) = \vec{0}$.

In order to prove that they remain in the line that connected them I thought I could try to prove that for any instant $t$, $\vec{x}_1(t)-\vec{x}_2(t)\propto\vec{x}_1(t_0)-\vec{x}_2(t_0)$. Rearrange the previous equations, calling $\vec{R}=\vec{x}_1+\vec{x}_2$; $\vec{r}=\vec{x}_1-\vec{x}_2$:

$$ \ddot{\vec{R}} = \vec{0}\\ \ddot{\vec{r}} = 2\vec{F}_1(\vec{r},\dot{\vec{r}}), $$

with initial conditions $\vec{R}(t_0)=\vec{R}_0$, $\vec{r}(t_0)=\vec{r}_0$, $\dot{\vec{R}}(t_0)=0$ ,$ \dot{\vec{r}}(t_0)=0$.

How can I prove that $\vec{r}(t)\propto\vec{r}(t_0)$? If it is not possile in this way, how can I prove that the points remain in the same line without invoking angular momentum conservation?

Edit: To include the insights of @Alexander : We can perform a galilean transformation such that the center of mass $R_0 = \vec{0}$, in that way the dynamics simplifies to $\ddot{\vec{r}} = \vec{F}(\vec{r},\dot{\vec{r}}) $

Because the equation must be invariant under the galilean transformations group, it follows that $\vec{F}(M\vec{r},M\dot{\vec{r}}) = M\vec{F}(\vec{r},\dot{\vec{r}})$ where $M$ is an orthogonal matrix (the rest of galilean transformations have been alreagy gauged away by previous choice $\vec{R}=\vec{0}$). This implies (I think) that the force is radial, therefore $\vec{F}(\vec{r},\dot{\vec{r}}) = f(|\vec{r}|,|\dot{\vec{r}}|)\vec{r}$ where $f$ is a scalar function. Thus $\ddot{\vec{r}}\propto\vec{r}$. Given that the initial velocity is zero it is enough to conclude that $\dot{\vec{r}}\propto\vec{r}$, and the bodies remain in the same line.

Is this logic flawed?

Answer 1

Two bodies initially at rest with arbitrary interaction remain in the line that connects them ?

The EOM's are:

$$m_1\,\ddot{x}_1=F_{12}\tag 1$$ $$m_2\,\ddot{x}_2=-F_{12}\tag 2$$

The center of mass equation is:

$$R=\frac{m_1\,x_1+m_2\,x_2}{m_1+m_2}$$

Thus

$$\dot{R}=\frac{m_1\,\dot{x}_1+m_2\,\dot{x}_2}{m_1+m_2}\tag 3$$ $$\ddot{R}=\frac{m_1\,\ddot{x}_1+m_2\,\ddot{x}_2}{m_1+m_2}\tag 4$$

equation (1) plus equation (2) and with equation (4) you obtain

$$m_1\,\ddot{x}_1+m_2\,\ddot{x}_2=\left(m_1+m_2\right)\,\ddot{R}=0$$

thus $\ddot{R}=0$ and $\quad \dot{R}=v=0$

with equation (3) you get:

$$m_1\,\dot{x}_1=-m_2\,\dot{x}_2$$

thus

momentum of mass one is equal to the negative momentum of mass two , this means that the parts remain in the line that connects them.

Edit

$$m_1\,\dot{x}_1=-m_2\,\dot{x}_2$$ Thus: $$x_2=-\frac{m_1}{m_2}\,x_1+c$$