Accelerated Classical Atom

This is an animation of a pair of particles: a nucleus with a charge $+1$ and an electron with a charge $-1$. Initially, the nucleus is fixed, and the electron is in a circular orbit around it. Then the nucleus starts accelerating at a constant rate (that is, it's momentum increases as $p_n=(Ft,0,0)$).

The position of the nucleus solves $$ \dot r_n=\frac{p_n}{\sqrt{m_n^2+p_n^2}}=\frac{(Ft,0,0)}{\sqrt{m_n^2+F^2t^2}} $$ that is $$ r_n(t)=\frac1F\sqrt{m_n^2+F^2t^2}-\frac{m_n}F . $$

The electron is subjected to the electromagnetic force due to the nucleus. Now, the nucleus is accelerating, but we will approximate the field by one moving at a constant speed $\dot r_n(t)$. (For a computation of the full dynamics, with the field generated by the accelerated nucleus, see http://ian.jauslin.org/animations/bell_atom/fulleq.php.) When making this approximation, the system reduces to two dimensions, and we will choose the initial condition such that the motion is confined to the x,y plane. The electric and magnetic fields at position $r_n(t)+\rho$ are $$ E_x(\rho)=\frac{x'}{\sqrt{(x')^2+y^2}^3} ,\quad E_y(\rho)=\frac{y}{\sqrt{(x')^2+y^2}^3}\frac1{\sqrt{1-\dot r_n^2(t)}} $$ $$ B_z(\rho)=(\dot r_n(t))_x E_y(\rho) $$ with $$ (x,y)=\rho ,\quad x'=\frac x{\sqrt{1-\dot r_n(t)^2}} . $$ The position of the electron is $r_e(t)$, and the relative position is $\rho_e(t):=r_e(t)-r_n(t)$, which satisfies $$ \dot p_e=-E(\rho_e)-(\dot \rho_e+\dot r_n(t))\times B(\rho_e) ,\quad \dot \rho_e=\frac{p_e}{\sqrt{m_e^2+p_e^2}}-\dot r_n . $$

Stop!