I'm reading Arnol'd for self study. I'm struggling with this question: "Show that any system of two particles will remain on the same line that connected them at the initial moment, if they started at rest." My argument is the following:

Set up coordinates such that the $x$ axis points along the line, and the $y$ and $z$ axis are arbitrarily set up to form an orthonormal coordinate system. Supposing that the particles' positions and velocities lie entirely along the $x$ axis, we may apply any one of the following coordinate transformations 1. $(t, x, y, z) \to (t, x, -y, z)$ and 2. $(t, x, y, z) \to (t, x, y, -z)$. These transformations don't change relative positions and velocities since $y = z = 0$ for both particles, and similarly they don't change relative velocities. The transformations obviously preserve time intervals, and the distances between simultaneous events so they are Galilean. Since forces can only depend on relative positions and velocities, and forces are invariant under Galilean transformations the component of the forces on each particle along the y, and z axes is zero since this is the unique number $x$ such that $-x = x$. So the particles' forces point along the $x$-axis at the initial moment. Later when the particles have moved along the $x$-axis, gaining some velocity along this axis the forces still point along the $x$-axis. "Therefore," the particles stay on the $x$ axis.

The issue with this argument is the last "Therefore,". For example, let $y(t)$ be the $y$-component of the relative distance between the particles at time $t$. Then, we know that $y(0) = 0$, $y'(0) = 0$ and the above argument shows that $\forall t \in \mathbb{R}, y(t) = 0 \land y'(t) = 0 \implies y''(t) =0$, but obviously one can't conclude that $y$ is constantly 0 since third derivatives may introduce changes in $y$. For example the function $y = t^3$, satisfies both conditions. This seems to suggest that the theorem is false, since if we name the particles 1 and 2, we may define $\mathbb{f}_1 = R(\mathbb{x}_2 -\mathbb{x}_1)$ and $\mathbb{f}_2= R^{-1}(\mathbb{x}_1 - \mathbb{x}_2)$ where $R$ is some rotation matrix that doesn't keep any axis fixed. These forces are invariant under time translations, rotations and spacial translations therefore, they are invariant under all Galilean transformations. However, an appropriate choice for R allows the particles to gain speed along any axis.


Your reasoning is essentially correct, apart from the last paragraph.

Rotational invariance, in Arnold's book, is expressed by the requirement that if $\{\mathbf r _i (t)\}$ is a solution of Newton's EOM, and $B$ is a rotation, then $\{B\,\mathbf r _i (t)\}$ is also a solution.

Consider your problem: you have two particles with an initial condition which we may, without loss of generality, take to be: $$\mathbf r_1 (0)=(0,0,0),\qquad \mathbf r _2 (0) =(d,0,0), \qquad \dot {\mathbf r_1} (0)=\dot {\mathbf r_2} (0)=\mathbf 0 .$$

Let $\boldsymbol \varphi (t)=(\mathbf r _1 (t),\mathbf r _2 (t))$ denote the actual trajectory of the two particles. Then $B\boldsymbol \varphi (t)\equiv (B\mathbf r _1 (t),B\mathbf r _2 (t))$ is another possible trajectory.

Now let $B$ denote a rotation about the $x$ axis. This is another solution with the same initial conditions, $B\boldsymbol \varphi (0)=\boldsymbol \varphi (0)$ and $B \dot {\boldsymbol \varphi} (0)=\dot {\boldsymbol \varphi} (0)$. Therefore, by the principle of determinism (that is, by requiring the uniqueness of the solution), we obtain $B \boldsymbol \varphi (t) = \boldsymbol \varphi (t)$.

This tells you that the solution is invariant under rotations about the $x$-axis. In particular, its "components" $\mathbf r _i (t)$ are invariant. It's easy to see that this implies that the particles lie on the $x$-axis at all times $t$.

Regarding your couple of forces, note that these don't satisfy galileian invariance, since $\mathbf f _2 \neq -\mathbf f _1$ (can you prove this?).

I suspect that there's a typo and you really meant $\mathbf f _2 =-\mathbf f_1$. In this case, note that rotational invariance is not satisfied, since for these force vectors the relation $$\mathbf f (B\mathbf x )=B\mathbf f (\mathbf x)$$ is not satisfied for a general rotation $B$.

Side note: If you were wondering if the theorem is true or false, which is perfectly legitimate, I suggest to pause a moment and think about what does galilean invariance mean. As referred to space, it means that there is no preferred point (homogeneity) and no preferred direction (isotropy).

If the pair of particle is found at some time $t>0$ with a separation $\mathbf r(t)$ skew to $\mathbf r (0)$, then it must have somehow selected a particular direction in space different from the only given one, $\mathbf r _0$, therefore breaking the axial simmetry of the initial situation.

  • $\begingroup$ Could you elaborate on why Newton's equation can be solved by defining y(t) and z(t) to be zero? How do you know that this results in a solution to Newton's equation. $\endgroup$ – 11Kilobytes Jul 18 '16 at 7:15
  • $\begingroup$ Just plug $\boldsymbol \varphi (t) = (x(t),0,0)$, with $x(t)$ a function to be determined, into Newton's equation. If you can find an $x(t)$ such that $\boldsymbol \varphi (t)$ is a solution, you're done. In this case it is easy to show that such a function exists, since the problem is reduced to a one-dimensonal one: $$\ddot x = f(x).$$ $\endgroup$ – pppqqq Jul 18 '16 at 8:27
  • $\begingroup$ It seems that I didn't translate the conclusion of my argument into a sufficiently general proposition. I currently think that my argument shows that, "At every point in time where the particles don't have position or velocity components in the y or z directions, their acceleration doesn't have a component in the y or z directions." But I don't know how to conclude, "therefore the particles don't ever have position components in the y or z directions," even given the assumption that they start at rest. @pppqqq could you please elaborate on this part of your answer? $\endgroup$ – 11Kilobytes Jul 19 '16 at 7:39
  • $\begingroup$ Hi @11Kilobytes, I've tried to give a different argument. I've noticed that the previous one (v2) would break if velocity dependent forces were considered, so I rewrote it from scratch. Let me know if there's anything not clear. $\endgroup$ – pppqqq Jul 19 '16 at 11:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.