We showed last week that we can explain the anomalous magnetic moment of the electron with the rubber plate model. The far field of the electron does not have time to react to the very fast spinning of the electron charge in the zitterbewegung loop. That reduces the effective mass of the electron.
We believe that the Lamb shift is caused by the same phenomenon, but it is harder to calculate an approximation, since the acceleration depends on the distance of the electron from the proton.
Furthermore, the zitterbewegung movement complicates things. How should we handle the assumed light speed orbit of the electron?
Let us assume that the electron on the 2s-orbital is on a very narrow elliptical path whose focus is the proton.
The speed of the electron close to the proton is the speed of light, if we assume zitterbewegung to be real motion.
The reduced mass might follow a similar formula to the calculation of the anomalous magnetic moment:
m_eff = (1 - r_e / R) m_e,
where r_e is the classical radius of the electron, R is the distance where light can reach during one "cycle" of the accelerated movement, and m_eff is the reduced mass.
The formula is nonsensical when R < r_e. Let us ignore that for now.
Reduced mass reduces the momentum
p = m v,
but how much? If the mass is 1% less, then the momentum at a certain distance r from the proton is in the non-relativistic case 0.5% less because the electron has less time to collect momentum.
A 0.5% reduction in momentum in non-relativistic movement requires a 1% reduction in the effective potential V, if the mass were not reduced.
Then
V_eff = (1 - r_e / R) V.
The effect raises the energy of both 2s and 2p, but since 2p has zero probability density at the nucleus, the effect is larger for 2s.
--->
_________________
/
| ● proton
\__________________ e- at distance r
<---
What is the "cycle" in our case? The electron does a back-and-forth movement close to the proton in its infinitesimally narrow elliptical orbit.
If the electron starts at a radius r, the cycle might be 2 r. But is that right? It really is just half a cycle because the electron does not return to the same state.
Should we multiply the 2 r by 2?
Let us not multiply.
The Darwin term
The Darwin term is the result of smearing the potential by the radius
λ_e / (2 π) = 4 * 10^-13 m,
where λ_e is the Compton wavelength of the electron.
We know that
λ_e = 137 r_e.
Then
r_e / (λ_e / 2 π) = 4.6%.
Let us check the numerical values of the Darwin term and the Lamb shift.
Darwin = 90.57 μeV
Lamb = 1.057 GHz * h / e
= 4.371 μeV
The Lamb shift is 4.8% on top of the Darwin term.
It is probably not a coincidence that the percentages are that close.
A very crude calculation
Let us calculate what is the average correction to the 2s potential relative to the 2p potential, when we take into account the reduced mass of the electron.
We start by calculating the effect for r < 5 * 10^-12 m.
Zitterbewegung does not have too much an effect on the electron at this distance, because the zitterbewegung radius is only 4 * 10^-13 m.
The speed of the electron (ignoring zitterbewegung) is roughly 0.03 c at a distance
r = 5 * 10^-12 m
from the nucleus.
The potential there is roughly -300 eV.
The effective mass reduction: when the electron moves the distance 2 r near the nucleus, its field gets a message at the speed of light, of the moving electron 67 r away:
R = 67 r.
We can reduce the electron mass by the mass-energy of the far field > 67 r away. The reduction is
1 / 67 * 300 eV = 4.5 eV,
which is 10^-5 of the electron mass.
The effective potential is reduced by the factor 10^-5 from the value -300 eV.
In the link there are formulae for the radial wave functions for 2s and 2p by Jim Branson, 2013.
The 2s electron wave function squared at r = 0.1 is 0.4 if r is given in units of the Bohr radius. That is the probability density.
The volume of the sphere r < 5 * 10^-12 m is
4 * 10^-3 cubic units
and the integrated probability
1.6 * 10^-3.
Thus, the average correction of the potential for the 2s orbital is
ΔE = 1.6 * 10^-3 * 10^-5 * 300 eV
= 4.8 μeV.
The speed of the electron is ~ sqrt(1 / r). The effect of the potential is ~ 1 / r^2. The integrated probability is ~ r^3. The contribution grows as sqrt(r) when r is very small.
The contribution from
r = 5 * 10^-11 m
is
ΔE = 0.8 μeV,
because the probability density of 2s there is only 0.045 and the probability density of 2p is already around 0.02
The contribution from
r = 7 * 10^-11 m ... 20 * 10^-11 m
is negative because 2p has there larger probability density than 2s. The probability density of 2p minus 2s is 0.02.
The contribution is
ΔE = -1.3 μeV.
The contribution from
r > 20 * 10^-11 m
is only
ΔE = 0.1 μeV,
because there the integrated probability grows slowly, and 2s and 2p have roughly equal probability densities.
Zitterbewegung affects the contribution of very short distances. The contribution of
r = 5 * 10^-13 m
would be
ΔE = 1.8 μeV
if zitterbewegung has no effect. If we put a "cutoff" at the zitterbewegung radius 4 * 10^-13 m, then the contribution is
ΔE = 0.9 μeV.
The contribution from 5 * 10^-14 m might be 0.6 μeV. It is not clear what happens at distances shorter than the classical radius 3 * 10^-15 m.
The sum of the contributions, if we put the "cutoff" at the zitterbewegung radius, is
ΔE = 5.3 μeV,
which is close to the Lamb shift 4.4 μeV.
In his back-of-the-envelope calculation, Hans Bethe in 1947 did put a cutoff at the the zitterbewegung radius h-bar / (m_e c) (see the linked document by Philip O. Koch (1956), page 18).
We made a lot of assumptions in the calculation. The calculated figure could vary 4-fold with different assumptions.
Considerations about classical versus quantum phenomena
The mass reduction is a classical phenomenon. Generally, quantum mechanics is aware of classical physics - quantum mechanics does not ignore classical phenomena. Therefore, the mass reduction should be visible in the hydrogen spectrum. Since the order of magnitude of this classical phenomenon agrees with the Lamb shift, it is probable that it is the same effect.
If we make the particles very heavy and increase their charges in the right proportion, then the "super-electron" passes the "super-proton" along the same orbit as the ordinary electron. The superparticles will behave according to classical physics, if we believe in the correspondence principle. Mass reduction will happen. How does QED handle such superparticles? There is no upper limit for the mass and energy in Feynman diagrams. They should calculate the right approximation for the classical case.
Conclusions
We believe that the Lamb shift can be explained with the rubber plate model of the electron electric field. The field is elastic and the effective mass of the electron is reduced close to the nucleus, because of the large acceleration and speed.
Therefore, the Lamb shift is a classical effect which affects the Schrödinger equation solutions of the hydrogen atom.
Since the effect is classical, there is no need for regularization or renormalization of the electron mass. Hans Bethe in 1947 regularized the self-energy of the electron by calculating the difference between a free electron and the bound electron.
The rubber plate model has turned out to be fruitful: it qualitatively describes the reaction of the electron to its own field, and can explain how a radio transmitter works. It can explain the electron anomalous magnetic moment. It may explain the vertex correction in QED.
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