Monday, July 26, 2021

The fine structure constant is determined by a semiclassical model?

In our July 14, 2021 blog post, we noticed that the bremsstrahlung approximation that is used in astrophysics can be explained by a semiclassical model where the electron comes very close to the nucleus, so that most of the kinetic energy of the electron is converted into vibration of its electromagnetic field.

Let the nucleus be a single proton.

Recall our "rubber plate" model of the static electric field of the electron. The nucleus suddenly pulls on the electron, which makes the "rubber plate" to vibrate. We have used the rubber plate model to explain qualitatively the birth of electromagnetic waves.

The model is not completely classical because the vibration frequency in the rubber plate would be much higher than the the frequency of the emitted bremsstrahlung photon. This is the "length scale problem" which we have discussed in many posts during this year.

In the astrophysical bremsstrahlung approximation, the classical Larmor formula is used to calculate the radiation, but the closest approaches to the nucleus are brutally cut off at the distance b, where b is the de Broglie wavelength of the electron.

How close must the electron come to the nucleus, so that most of the kinetic energy goes into the vibration of the rubber plate?

The mass of the electron resides in the energy of its static electric field at a distance

       > r₀ / 2,

from the pointlike electron, where

       r₀ = 2.8 * 10^-15 m

is the electron classical radius.

Let us assume that the electron is mildly relativistic, its total energy 1 MeV.

If the electron orbits about 90 degrees around the nucleus, traveling a distance 1.4 * 10^-15 m, then the complete field of the electron "lags behind" in the movement, and it might be that most of the kinetic energy is converted into vibrational energy. The radius of that orbit is 0.9 * 10^-15 m.

The cross section for emitting a large photon (containing most of the kinetic energy) is then something like

       A = π * (0.9 * 10^-15 m)^2
           = 2.5 * 10^-30 m^2
           = 25 millibarn.

We calculated earlier from the astrophysical approximation that the cross section for emitting a large photon is roughly

       A' = 10^-7 * π * (2.4 * 10^-12)^2
           = 2 * 10^-30 m^2
           = 20 millibarn.

Is it a coincidence that the semiclassical model roughly reproduces the right cross section?

The quantum mechanical way to calculate the cross section is the Bethe-Heitler formula. The cross section in the formula linearly depends on the fine structure constant, which is approximately 1 / 137.

If our semiclassical model is the "right" way to explain bremsstrahlung, then our semiclassical model fixes the value of the fine structure constant.

The fine structure constant depends on the Planck constant. If our model is "right", it fixes the value of the Planck constant.

The semiclassical model is about the "geometry" and mass-energy of the electric field. The Planck constant does not appear in the semiclassical model.

We need to study in more detail bremsstrahlung. Can we show that the semiclassical model qualitatively works for different energies and photon deflection angles, and different numbers of protons in the nucleus?

What would it mean if the Planck constant were determined by the semiclassical model? We need to investigate that.

The mass-energy of the electric field depends on vacuum permittivity ε₀. The Planck constant in our semiclassical model is determined by ε₀?

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