The Classical Atom
Bruce Harvey
4 Quarry Road Weoley Castle, Birmingham B29 5PB, UK
Telephone 44 121 476 1610
Copyright April 2003
Originally published on 12/09/03, this paper was found to contain an error or two in calculation. This revised file gives the correction. BH. 25/03/04
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If the quantisation of magnetic flux is added to the cannon of Classical Electromagnetic theory and Lorentz's theory of electromagnetic mass admitted, then Classical Physis predicts the quantised energy levels within the atom.
We can show that the quantized orbits of the one electron atom result from the laws of classical physics if we are allowed to claim the quantization of magnetic flux as a classical result. We do so on the basis that the experimental discovery of flux quantization by Deaver and Fairbank in 1961 (Phys.Rev.Letters 7,43 (1961) uses only classical methods.
We take
Lorentz's theory of the electron: that its motion generates a magnetic field which contains its Kinetic Energy.
By the Viriial Theorem the total energy (potential + kinetic) is one half of the time average of the potential energy. For a circular orbit: For a circular orbit:
Potential Energy = -2 Kinetic Energy
Total Energy = - Kinetic Energy
The principle of equipartitioning of energy from thermodynamics and assume that half the kinetic energy will be present in the magnetic field generated by the current loop of the orbiting electron while the other half will remain in that part of the electron's magnetic field which continues to move with it. We will call this the Magnetic Energy.
This gives us 2 equations:
We assume a circular orbit. Equations are in their SI form.
The variables are:
radius of orbit =
Frequency in orbits per second =
velocity =
Angular momentum = L
Quantum fluxoid =
Current =
Electron charge =
Atomic number =
The Magnetic energy in flux threaded by current is
The potential energy is that due to the Coulomb force between the electron and nucleus
From the geometry of the circular orbit
Substituting these expressions into the equations:
Solving for and n.
The total energy may be expressed in terms of the magnetic energy giving
Angular momentum
Error The term should be
On substitution of London's theoretical result we obtain the results first derived by Bohr, albeit disguised in their SI form.
By putting the discovery that magnetic flux is quantized on the same par as the discovery that electric charge is quantized and applying the laws of Classical physics, we predict the quantized energy levels of the one electron atom.
The Form of the magnetic flux
We might be open to the criticism that the above is just a neat bit of algebra. The reader may be familiar with the classical treatment of the current loop and know the formulae for the field strength at the centre and at distances large compared to the radius. Putting numbers into the problem we find that T and m which would give a flux of through a circle the size of the orbit, compared with the quantum fluxoid .
Textbooks do not give a formula for B off centre because the problem does not admit to an algebraic solution and we must use numerical integration.
The outer integral is summed until to find the value of . This is 95% of the radius if the sphere surrounding the part of the electron's magnetic field which moves with it.
It is evident that virtually all of the flux is concentrated in a narrow region close to the line of the current loop making it possible to find a fairly accurate approximate algebraic solution. For a point in the plane of a current loop radius r distance y from its centre:
Results from the numerical integration indicate that
If r "roe" is the tunnel radius, then the flux content is given approximately by taking the integral from r to infinity.
If the magnetic field contains n quantum fluxoids, we may solve for the tunnel radius r
On substitution of the results for the orbital radius and the current
Comparing the result with the radius of the Lorentz electron we see that the tunnel is times the size of the electron and independent of the number of quantum fluxoids. Our approximate solution is only 0.5% from the numerical result.
The quantum fluxoids forming the tunnel are stable on the time scale of the electron's orbit, by which we mean that the remain within the locus of the orbit. However the periodic passing of the electron through a section of tunnel does not constitute a steady current. If we increase the current in a current loop, the magnetic field moves outward with new flux emerging from the conductor We should expect the same effect to be present within the atom so that the tunnel diameter is not constant, but increases with the approach of the electron and then contacts. An average tunnel diameter of 95% of the diameter of the electron and its surrounding magnetic field could easily allow the passage of the electron.
Centrifugal force
The question arises as to the nature of the centrifugal force. In a planetary system, the circular orbit orbit results from the balance of centrifugal force against gravitational force. We have based our argument herein on the assumption that inertial mass is electromagnetic and that the electron is surrounded by a magnetic field which contains its kinetic energy. In the case of an orbiting electron, we assumed that the magnetic field surrounding the electron is divided in two, half continuing to move with the electron and the other half forming a stable magnetic field due to the orbital motion. It has been shown elsewhere that centrifugal force results from the rotation of the magnetic field. In this model, only half of the kinetic energy is now contained in that part of the magnetic field which moves with the electron. We would therefore expect the centrifugal force to be halved.
This leave us with the problem of explaining why the electron orbits as if the full centrifugal force were present. The relationship of the orbiting electron to the stable part of the magnetic field must generate the missing half of the force required to balance the Coulomb force.
If we return to Faraday's understanding of induction, the force on a conductor in a magnetic field due to the current passing through it may be understood in terms of the work which would be done if the conductor a small distance in the direction of the force. Lines of magnetic flux cut by the conductor will decrease their current linkage by the current in the conductor. Since their energy content is where is the current linkage, the loss of energy will be available to work resulting in the force . However, as the lines of flux are cut by the conductor, a reverse emf is generated which must be countered by an increased emf in the circuit to maintain the current which does an equal amount of work. [ In most practical examples to be found in text books, the reverse emf is small compared with emf of the circuit which is required to overcome resistance and is therefore neglected. ]
Considering the orbiting electron as a current loop generating the stable magnetic field of energy content , we can apply the same principle.
Which is the missing half of the centrifugal force :
Similar arguments can be applied to the case of elliptical orbits.
Elliptical orbits
We can rewrite the equation for the energy in terms of the the radius :
This equation is equivalent to the planetary orbit equation:
From the theory of orbits, we know that planets have elliptical orbits and that their energy is determined by , the semimajor axis of the ellipse. We had assumed a circular orbit, now we can generalise to an elliptical orbit simply by substituting for .
However, our results for the angular momentum and magnetic moment both depend on the formula for the area of a circle. We need to change this for the area of an ellipse . Writing the equations become:
A circuit as two descriptors of its magnetic properties, Inductance and Magnetic Moment. Both result from the geometrical properties of the circuit. The inductance is largely determined by the tightness with which the flux can be wrapped around the circuit while the magnetic moment is largely determined by the area enclosed. The magnetic field far from the circuit is determined by its magnetic moment while inductance is largely determined by the close up view of the flux. When we consider the atom, we are dealing with a very few quantum fluxoids.
We make the bold assumption that the geometry of the orbit determines how the quantum fluxoids fit into it. For an orbit enclosing two fluxoids, there are two ways: each wrapping half the orbit or one tightly wrapping the orbit and the other passing through its centre. For three fluxoids: each might wrap a third of the orbit, one might wrap the orbit with the two passing through the centre, or two might each wrap half the orbit with the third passing through the centre. If only fluxoids which pass through the centre influence the far field, we find that the magnetic moment is quantized
With the implication these correspond to elliptical orbits with semiminor axis where
This results in a quantized angular momentum.
Thus, the electron is confined to move in Somerfield orbits.
Quantization of angular momentum vector
The interaction of the orbital magnetic field with an external magnetic field depends on flux linkage. There can only be an interaction if one or more fluxoids link with the external field. This quantizes the interaction. Since at least one fluxoid must wrap the orbital path, the ground state orbit with only one fluxoid cannot interact. Hence the result of Quantum mechanics, that there is a quantum state with zero angular momentum.
There are two possible ways of fitting two electrons into the helium atom. They may share the same flux tunnel or they may each have there own slightly elliptical orbit with the two orbits linked.
Suppose two electrons share the same tunnel. The equation derived from PE of electron = now becomes
Where the second term is included to account for the potential energy due to the repulsion of the two electrons.
Single electron result for comparison
Insert out of order.
These solutions relate to the section "Protons and Neutrons" below. They are perfomred here before numerical values are defined below which would effect their evaluation.
Error The term should be
End of correction
The energy of the system of two electrons orbiting a helium nucleus is half the potential energy
Using the subscripts 1 to denote the single electron solution and 2 to denote the two electron solution, comparing answers gives:
For helium the radius is increased by a factor and the frequency reduced by . Note the effect on magnetic moment where N is the number of electrons is unaffected by the alteration in radius and frequency and is simply doubled as N changes from 1 to 2.
Let us now calculate the energy
The energy of the helium atom with two electrons in the same tunnel is
The term being introduced to convert the units from Joules to electron volts
This result of 83.33 eV is only is 4.2 different from the accepted value of -79.0 electron volts, a favourable outcome at this early stage in the development of a new theory.
Solutions involving linked orbits offer more hope of a correct prediction of the ionisation energy, but are mathematically far more complicated introducing two new parameters, the eccentricity of the orbits and the angle between their planes. These parameters control the average distance between the electrons and the mutual inductance of the two current loops formed by the orbits.
Protons and Neutrons
When we apply the helium model to the motion of quarks within the nucleons, we get some very interesting results. Both nucleons have three body solutions with the two similar quarks orbiting the odd quark such that they remain on opposite sides of a circular orbit. We can take the theory developed above and substitute the quark masses and charges.
(The algebraic solutions given below are performed in the collapsed region of the worksheet above. Because numerical values have been defined above, if the solve blocks are placed here, they would ignore the equations and supply the previously defined result)
Error The term should be
FIND r, n
Taking the mass of the quark as 1/3 nucleon mass
Radius of proton
Magnetic moment
FIND r, n
Radius of neutron
Magnetic moment
Error The term should be
This compares with the accepted value
FIND r, n
Taking the mass of the quark as 1/3 nucleon mass
Radius of proton
Magnetic moment
FIND r, n
Radius of neutron
Magnetic moment
End of correction
This compares with the accepted value
We can describe these results as interesting; not conclusive.

The reader should bear in mind the vast amount of work which went into developing the quantum mechanical model of the atom. In comparison, these are the first tentative steps on the way to developing a full model. It is therefore remarkable that we are achieving so much with so little effort. With but a small fraction of the effort put into the development of the QM model, this Classical Theory has the potential to explain the inner workings of the atom in a much more satisfactory manor.