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I've recently been working on relative equilibria for some systems of particles. (ie. studying equilibrium solutions in a rotating frame. Saturn's rings for example.) This has evolved into some classical notions of atomic physics and questions of stability for Coulomb interactions there. Entertain for a few moments a classical(planetary) model for the atom. Take an isolated hydrogen atom for example. It seems to me that the $q\left(v\times B\right)$ term of the Lorentz force provides an intrinsic stabilization for the two particle system. Naively play the right hand rule game with the system and you'll see. If we assume that the electron is in a classical orbit and not radiating, then could we establish the stability of atoms with the Lorentz force? The non-radiating electron is a topic of its own and appears to be far from solved. For example, if you would like to make the statement that accelerating charges radiate, therefore...... Try to go one step further and show why this must be true in general and you imediately run into difficulties. Gyro-stabilization might be a relevant topic here.

Order of Magnitude investigation:

\begin{align} &\text{Proton Charge:}\qquad&q &\approx 1.602\times 10^{-19}\: C\\[2mm] &\text{Permeability of free space:}\qquad &\mu_0 &\approx 4\pi\times10^{-7}\frac{volt\cdot s}{amp\cdot m}\\[2mm] &\text{Permitivity of free space:}\qquad &\epsilon_0 &\approx 8.854\times10^{-12}\frac{farad}{m}\\[2mm] &\text{Electron Speed: }\qquad &v &= \alpha c\approx c/137 \approx 2.19\times10^6\frac{m}{s}\\[2mm] &\text{Electron Radius:(Bohr) }\qquad &r & = \frac{\hbar}{m_ec\alpha}\approx 5.29\times10^{-11}m\\[2mm] \end{align}

Maximum Fields(Im thinking of the fields generated by the electron in a linear approximation for an infinitesimal section of the circular orbit:

\begin{align} \text{Biot - Savart}\qquad{\bf B}_{\text{max}} &= \frac{\mu_0 q v}{4\pi r^2}&\approx& 12.537\:\text{T}\\[2mm] \text{Coloumb}\qquad{\bf E}_{\text{max}} &= \frac{q}{4\pi\epsilon_0 r^2}&\approx& 5.145\times 10^{11}\frac{N}{C} \end{align}

With Force ratio:

\begin{align} \frac{q(v\times{\bf B})_{max}}{q{\bf E}_{max}} =\frac{\frac{\mu_0 q^2 v^2}{4\pi r^2}}{\frac{q^2}{4\pi\epsilon_0 r^2}} = v^2\mu_0\epsilon_0 = \frac{v^2}{c^2}\approx\frac{(2.19\times10^6)^2}{(3\times10^8)^2}\approx 5.329\times10^{-5} \end{align}

This shows the magnetic force is 5 orders of magnitude smaller than the electrostatic force for this crude model of the ground state hydrogen. Not sure what this conclusively says about the internal dynamics but at least I think I can foresee some problems with this force stabilizing the system.

A colleague of mine who does work with NMR(nuclear magnetic resonance) informed me that the magnetic fields they work with in the lab are on the order of 3$T$ and that they do not observe the large fields I am envisoning in practice. The concept he quoted as the reasoning behind this was "orbital quenching". This appears to be a sort of averaging to zero of this field. However, it was also stated that the magnetic field that I am interested in can be measured in beam type experiments.

JEM
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  • So, the rotation around the nucleus creates a "ring current" with magnitude roughly $e v$ and radius equal to, let's say, the Bohr radius? What's $v$ of the electron, assuming things are dominated by the electrostatic field? If it's not that close to $c$ then the generated magnetic fields just won't be that big. – webb Apr 30 '14 at 16:20
  • That could be fine if only the ground state of the hydrogen atom did not have a zero angular momentum – gatsu Apr 30 '14 at 16:32
  • I've thought somewhat about the relative magnitudes of the electrostatic part to the dynamic part. What can we say about the speed of the electron in its orbit? A discussion about this is here: http://physics.stackexchange.com/questions/20187/how-fast-do-electrons-travel-in-an-atomic-orbital. Looks to me to be somewhere near $10^6\frac{m}{s}$. So now I compare the Biot-Savart law and Coulomb's law with this data? – JEM Apr 30 '14 at 16:42
  • @gatsu setting the total angular momentum equal to zero was prior to the introduction of the Maslov index. Born,Heisenberg,Schrodinger,etc didn't know about it. I'm learning about this topological idea currently and it seems to me that the total angular momentum of the ground state hydrogen might not be 0. – JEM Apr 30 '14 at 16:55
  • @JEM: thanks for pointing these things to me, I had no idea they even existed! – gatsu Apr 30 '14 at 19:58
  • @gatsu: your welcome, Im currently working out an order of magnitude estimate on these forces for the stated velocity. Ill post it when I get a chance. – JEM Apr 30 '14 at 21:48
  • The magnitude of the force difference turns out to be the fine structure constant squared. $\alpha^2$ for this approximation. I should have been able to see that one coming.. – JEM May 01 '14 at 01:54
  • I think the permeability of free space is defined to be $ 4\pi\times10^{-7}\frac{volt\cdot s}{amp\cdot m}$. So the value is exact. – AHB Mar 27 '18 at 12:33

2 Answers2

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Could we establish the stability of atoms following this line of reasoning?

Magnetic forces do not seem to be enough if the fields are retarded (most natural assumption), for the system proton+electron will radiate some energy away and cannot be stable in the Newtonian sense. Background radiation or non-electromagnetic forces are needed to make the system stable. If the fields are half-retarded, half-advanced, stable orbits were reported to be possible, cf.

J. Frenkel, Zur Elektrodynamik punktfoermiger Elektronen, Zeits. f. Phys., 32, (1925), p. 518-534. | http://dx.doi.org/10.1007/BF01331692

and

L. Page, Advanced Potentials and their Application to Atomic Models, Phys. Rev. 24, 296 (1924) | http://journals.aps.org/pr/abstract/10.1103/PhysRev.24.296

Ján Lalinský
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"The non-radiating electron is a topic of its own and appears to be far from solved.". That's an interesting statement, since there is ample evidence, that there are no such things as "non-radiating electrons".

The only thing that "exists" in nature at that scale is a quantum field, one solution of which are electrons. Under certain circumstances (when energy levels stay below the rest energy of the next higher excitations i.e. muons), electrons and photons can be treated using a simplified theoretical framework called quantum electrodynamics (QED). The problem of atomic stability is completely resolved within that framework, even though other (numerical) problems still remain.

CuriousOne
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  • "The problem of atomic stability is completely resolved within that framework" That's also an interesting statement. Can you post a reference for it please? – Ján Lalinský Aug 19 '14 at 16:38
  • The atomic physics section of any physics department library will do... it should contain a couple of thousand mainstream textbooks on that matter, alone. – CuriousOne Aug 20 '14 at 03:15
  • Most of the books deals only with non-relativistic Schroedinger equation. You referred to QED, which is relativistic theory of field. The question of stability is much more difficult there. – Ján Lalinský Aug 20 '14 at 17:51
  • So what you are really trying to tell me is, that you haven't found a single textbook in the library that predicts that atoms should be unstable based on either theory? – CuriousOne Aug 21 '14 at 01:08
  • No, I am not. I am saying that your claim "The problem of atomic stability is completely resolved within that framework" is questionable, because relativistic theory of bound states is mathematically very difficult subject and I do not know of any work that proves atoms are stable in that theory. Perhaps there is such a proof - then I would like to see it. – Ján Lalinský Aug 21 '14 at 19:24
  • Why is relativistic theory of bound states difficult? You can calculate the proper matrix elements and predict decay times of states and with that line widths of transitions. The ground state is stable because it represents the energetic minimum of the system. – CuriousOne Aug 21 '14 at 20:57
  • 'Although Bethe and I put in these separate times in our 1950 preparation for the 1951 paper, because we had to, we never felt comfortable with it. A famous theorist, when young and not yet famous, confided in Hans Bethe "I don’t understand the Bethe-Salpeter equation". Hans’ reply was: "Don’t worry, neither do I really".' http://www.scholarpedia.org/article/Bethe-Salpeter_equation_%28origins%29 – Ján Lalinský Jul 31 '15 at 18:32
  • "Let us recall more precisely the difficulties encountered in treating systematically the bound states in QED by means of the BSE. ... To find its eigenvalues one has to specify a lowest-order equation - we will see the freedom present in this specification - and carry out on it a perturbative calculation. The main difficulty in pursuing this approach has been the absence of known analytic solutions to the equation even for an approximate Coulomb-like interaction kernel."

    p. 414, http://dx.doi.org/10.1016/0550-3213(78)90036-6

     – Ján Lalinský Jul 31 '15 at 18:36