The fine structure constant: an update

In our previous blog post we claimed that in a classical model an electron might lose most of its kinetic energy if it passes the proton at a distance

       < 0.9 * 10^-15 m.

The idea was that all of the inertia of the electron is located in its static electric field farther than

       r₀ / 2 = 1.4 * 10^-15 m

from the pointlike electron. There

       r₀ = 2.8 * 10^-15 m

is the classical radius of the electron.

A more detailed analysis reveals that it is enough that a half of the inertia of the electron "lags behind" in the movement. Imagine that a half of the mass of the electron is attached to it with an elastic rubber band. The rest of the mass is rigidly fixed to the pointlike electron.

              ● half of the electron mass

               |

               |  rubber band

               |

              ● e- electron

              v  ------->

Let the initial velocity vector of the system be v.

If the pointlike electron suddenly bounces back in the field of a proton, so that its velocity vector becomes -v, then all of the original kinetic energy of the system will go to stretching the rubber band. That is, the kinetic energy is totally converted to vibration or electromagnetic waves.

Where is the inertia of the electron located? We again meet the mystery of the "inner field" of the electron. We do not know the inertia distribution close to the classical radius of the electron. The Larmor formula suggests that in the far field of the electron, the inertia is in the mass-energy of the static electric field.

In our previous blog post, the cross section, which we calculated to be 25 millibarn, could be off by a factor 100, depending on the distribution of the inertia of the electron. That is bad news for our claim that the fine structure constant is determined by the geometry of the electron electric field.

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?

Comparison of classical and quantum bremsstrahlung

On May 28, 2021 we wrote about classical bremsstrahlung. If we imagine that the Planck constant goes to zero, we expect quantum bremsstrahlung to approach the corresponding calculation in classical electrodynamics.

Astrophysicists calculate bremsstrahlung with the classical approximation, cutting off impact parameters b where b is smaller than the de Broglie wavelength of the electron.

They motivate their cutoff approximation by claiming that for smaller b, "quantum effects" spoil the classical approximation.

Their motivation is very ad hoc. Why should we ignore the classical contribution of small b? Classically, most of bremsstrahlung energy is emitted with very small b.

Let us try to find a sensible motivation for the cutoff.

The radiated energy in the classical approximation is very small for b > the Compton wavelength

In the earlier blog post we calculated that classically, a "moderately relativistic" electron (1 MeV) which passes a proton at the distance 2.4 * 10^-12 m (the Compton wavelength) will only radiate ~ 10^-7 times of its kinetic energy away, roughly 0.05 eV. Classically, this energy is evenly spread over photons whose energy is up to 500 keV.

If we take the cutoff of astrophysicists literally, we have to assume that through some mysterious mechanism, a classical electron can produce a 500 keV photon even though the electron passes a proton far away, at a distance > 2.4 * 10^-12 m. This does not make much sense.

                                     500 keV photon

                      ~~~~~~~~~~~~~~~~~~~~~

                    /

      e- ------------------------------------------------

                          |

                          | virtual photon

                          |

      p -------------------------------------------------

Let us then think about the Feynman diagram of the process. The momentum transfer in the virtual photon is very large.

Semiclassically, for the moderately relativistic electron to lose a lot of its spatial momentum to the proton, it has to pass the proton at a distance < 2.8 * 10^-15 m (the classical radius of the electron).

The cross section of such a very close encounter is something like

       ~ 10^-7

times the cross section of an encounter at a distance < 2.4 * 10^-12 m.

We see that very close encounters can explain the flux of 500 keV photons which in the astrophysical cutoff approximation were the result of encounters roughly 2.4 * 10^-12 m away.

Now we have a semiclassical approximation which qualitatively explains why the astrophysical cutoff approximation might work for large-energy photons.

What about lower energy photons? A similar argument works for them.

Is there some deep underlying reason why the Feynman diagram method (qualitatively?) reproduces the cutoff approximation of astrophysicists?