two identical point charges can't collide

Question

I've convinced myself intuitively that if you place two massless classical particles with the same charge in $\mathbb{R}^n$, with arbitrary initial velocities and (distinct) positions, they will never collide. However, I'm have a heck of a time trying to prove it, and would appreciate some help.

Formally, consider $q_1, q_2: \mathbb{R} \rightarrow \mathbb{R}^n$ satisfying $$\ddot{q_i} = \frac{1}{\|q_i - q_j\|^3} (q_i - q_j)$$With $q_1(0) \neq q_2(0)$. The claim is that $q_1(t) \neq q_2(t)$ for all $t > 0$.

So my questions are (i) is this true? (ii) what happens if we replace the exponent 3 in the denominator with say $\alpha > 0$ ?

N.B. The question's already a bit long, but I'd be happy to post my thoughts so far.

Edit All the answers were very helpful, thanks so much everyone!

Answer 1

First let me address the second question. Notice that the electric field is no longer inverse-square dependent in dimensions other than $3$. The fundamental equation here is $\nabla\cdot \mathbf E=\rho$ (the divergence of the field is the charge density.) In three dimensions the field caused by a point particle will be indeed radial field with magnitude $E(r)=q/4\pi r^2$. In other dimensions the field caused by a point-particle at the origin is still radial but one has $$\mathrm{vol}(S^{n-1})E(r)=\int_{S^{n-1}}\mathbf{E}\cdot d\mathbf{s} =\int_{n-ball} \,\nabla\cdot \mathbf{E}\,d^{n}x =\int_{n-ball}\rho d^{n}x= q,$$after using Gauss theorem. Then in if you want to consider the field in $\mathbb{R}^n$, its norm is $$ E(r)=\frac{\Gamma(n/2)}{2 \pi^{n/2}}\frac{q}{r^{n-1}}. $$

Notice that you still get energy conservation. Assuming $1<n\neq 2$, the potential energy goes as $r^{-(n-2)}$, whereas for $n=2$ the potential goes as $\mathrm{ln}(r)$. Since the charges are initially at different positions, $U_i$, the initial potential energy is finite. Since $T_i+U_i=T_f+E_f$ and the kinetic energy is always positive, you need an infinite initial kinetik energy to make them collide, which is impossible. Then the charges never collide. This answers the first question too, for arbitrary dimension.

Answer 2

By describing all motion with respect to the center-of-mass frame, we can restrict our attention to $ \mathbb{R}^{2} $ only. In what follows, $ \mathbf{q}_{1},\mathbf{q}_{2}: \mathbb{R} \to \mathbb{R}^{n} $ denote the displacement functions of two particles with respect to the center-of-mass frame, where the center-of-mass is fixed at the origin of $ \mathbb{R}^{n} $.

For central-force motion involving only two particles, the trajectories $ \mathbf{q}_{1} $ and $ \mathbf{q}_{2} $ are seen to lie strictly within a $ 2 $-dimensional subspace $ \Pi $ of $ \mathbb{R}^{n} $. If the affine vectors $ {\dot{\mathbf{q}}_{1}}(0) $ and $ {\dot{\mathbf{q}}_{2}}(0) $ are oriented such that they do not simultaneously point toward/away from the origin, then $ \Pi $ is uniquely determined.

What I have done above is to choose an isometry $ T \in \mathbf{O}(n,\mathbb{R}) $ in order to obtain $$ T[\Pi] \subseteq \mathbb{R}^{2} \times \underbrace{\{ 0 \} \times \cdots \times \{ 0 \}}_{\text{$ n - 2 $ times}}. $$ This allows us to shift our focus to $ \mathbb{R}^{2} $. Clearly, the chosen isometrically-linear coordinate transformation does not affect the physics that is being described by the equations of motion specified by the OP above.

With this in mind, note that for $ \alpha = 3 $, what we have is basically the well-studied Coulomb Collision Problem. Depending on the orientation of the affine vectors $ {\dot{\mathbf{q}}_{1}}(0) $ and $ {\dot{\mathbf{q}}_{2}}(0) $, the trajectories lie in

• non-intersecting hyperbolas or

• non-intersecting segments of a single straight line.

I find it rather interesting that the derivation of the Rutherford Scattering Formula in atomic physics relies upon this fact.

For $ \alpha \in \mathbb{R}_{> 0} \setminus \{ 3 \} $ in general, we no longer have a nice description of the trajectories involved. However, one can easily use an energy-conservation argument to prove that trajectories cannot collide, and this is precisely what Jorge has described in his solution.

Answer 3

This is as good a place as any to mention how you can derive conservation of energy from scratch, starting with nothing but a differential equation for $\ddot q_i$... as long as your force term is conservative, that is, it is the negative gradient of a scalar "potential energy" function. (I'll deal only with a one-particle system, but you can handle multiple particles simply by packing in all the position variables $q_1, q_2, \ldots, q_m$ into a single vector in $\mathbb R^{mn}$.)

Consider $\ddot q=f(q)$ where $f$ is conservative, i.e. $f(q) = -\frac{\mathrm d}{\mathrm dq} U(q)$ for some scalar-valued potential $U$. Introduce a momentum variable $p=\dot q$ so that $\dot p = f(q).$ Observe that $p = \frac{\mathrm d}{\mathrm dp} T(p)$ where $T(p) = \frac12\lVert p\rVert^2$. Define the energy function $H(q,p) = U(q) + T(p)$, and observe that $$\dot H(q,p) = \frac{\partial H}{\partial q}\cdot\dot q + \frac{\partial H}{\partial p}\cdot\dot p = -\dot p\cdot\dot q + \dot q\cdot\dot p = 0,$$ so $H$ is constant over time for any solution.

For your problem, your $U(q_1,q_2)$ will depend only on $\lVert q_1-q_2\rVert$, and you'll want to check whether $H$ is infinite in a colliding configuration. I think you need $\alpha>1$ for that to happen.