Free-fall atomic model: Bohr with magnetic corrections

1. Sep 14, 2013

jarekd

We still often meet Bohr classical atomic model, e.g. for nearly classical Rydberg atoms. However, this model ignores the fact that electron has very strong magnetic moment: is tiny magnet (it wasn't known when Bohr introduced his model).
To understand why there are needed corrections, let us change the reference frame for a moment: imagine that it is nucleus moving in electron's magnetic field - Lorentz force says that there appears perpendicular force, proportional to the speed.
So if electron would fall straight toward the nucleus, there would appear perpendicular force, making that it should finally miss the nucleus and return to the initial distance - here is article with links about this not well correction of Bohr model: http://en.wikipedia.org/wiki/Free-fall_atomic_model
Here is the basic Lagrangian for single electron: $\mathbf{L} = \frac{1}{2}m\mathbf{v}^2+\frac{Ze^2}{r}+\frac{Ze}{c}\left[ \mathbf{v}\cdot\left( \frac{\mu\times \mathbf{r}}{r^3}\right)\right]$
the last term is the correction for Bohr model - interaction between magnetic moment of electron and charge of the nucleus. It is classical analogue of the spin-orbit interaction and generally should make circular Bohr's orbits unstable.
One of consequences is that there is relatively large probability of electron backscattering - so it could jump between close proton and nucleus, screening their repulsion and so making such electron-assisted fusion less improbable ...

Similar, but relatively much weaker correction appears in gravitation - as so called frame-dragging, caused by internal rotation of stars and planets. To understand it, there is used gravitomagnetism ( http://en.wikipedia.org/wiki/Gravitoelectromagnetism ) as practical approximation of GRT - it says that like rotation of charge creates magnetism, rotation of mass creates gravitational analogue of magnetism.

What do you think about it?
Should e.g. Rydberg atoms be circular, or maybe free-falling?