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 Jan14 comment two identical point charges can't collide I agree with Haskell's claim. To clarify, I've been talking only about two inertial frames: $S$, w.r.t which the particles have position $\mathbf q_i$, and $S'$, defined by having the origin at $Q:=(\sum_i m_i \mathbf q_i)/(\sum_i m_i)$. If you write down Newton's second law for $Q$, it's equation of motion is $\ddot{Q}=F_{12}+F_{21}\equiv 0$ (by third law), whence $Q(t)=Vt$ for some constant vector $V$. If one of them is inertial, both are. Jan14 revised two identical point charges can't collide added $n=2$ case. Jan14 comment two identical point charges can't collide @uncookedfalcon Yes, but if you don't like that frame, you can replace $\mathbf{r}$ everywhere by $\mathbf q_1-\mathbf q_2$. The same analysis caries over to this "more general" frame. Regarding your question, both frames are inertial (because setting the force to zero, you get a particle that follows a stright line with constant velocity as the solution to the differential equation). Does that answer your question about frames? Jan13 comment two identical point charges can't collide @uncookedfalcon if you wish to analyze the problem in arbitrary dimension, the field is no longer inverse-square dependent. Jan13 revised two identical point charges can't collide added 196 characters in body Jan13 answered two identical point charges can't collide Jan13 comment two identical point charges can't collide If your law is to have physical meaning, it must be $n=3$ and in Newton+Coulomb law it's not the norm squared, but to the third power. Jan12 comment Metric tensor of complex numbers & Hamiltonian Mechanics I've found it. It's "Introduction to Mechanics and Symmetry" by Marsden and Ratiu, Exercise 2.1-1. Jan11 comment Metric tensor of complex numbers & Hamiltonian Mechanics the $2$ factor comes from the definition of the partial with respect to $\bar{z}$. // These equations are $N$ complex equations, so $2N$ real eqs. I think I saw that once on one of Marsden's books. Let me search... Jan11 revised Metric tensor of complex numbers & Hamiltonian Mechanics added info Jan11 comment Metric tensor of complex numbers & Hamiltonian Mechanics Ah, I see. But curvature has double derivatives of the metric tensor. In this case $\eta$ is the metric tensor on the nose :) en.wikipedia.org/wiki/Riemann_curvature_tensor Jan11 comment Metric tensor of complex numbers & Hamiltonian Mechanics @AimForClarity I'm not sure I understand why you write "curvature" for $\eta$. I do not understand that part of the question. Jan11 revised Metric tensor of complex numbers & Hamiltonian Mechanics added 21 characters in body Jan11 answered Metric tensor of complex numbers & Hamiltonian Mechanics Jan11 answered Definition for Covariant Derivative Jan5 revised Earth-Sun distance equation corrected spelling Jan5 suggested approved edit on Earth-Sun distance equation Jan5 revised Earth-Sun distance equation spelling correction Jan5 suggested approved edit on Earth-Sun distance equation Jan5 comment Earth-Sun distance equation