Bremsstrahlung always contains an infinite number of soft photons but that does not matter

In the previous blog post we presented a conjecture that even a single electron always emits an infinite number of soft photons when it passes a nucleus. Let us analyze this in more detail.

          superheavy electron

         e-  -------------------------->

                          ● Z+

                          macroscopic charge

Let us make the electron very heavy, so that it becomes an almost classical particle. We replace the nucleus Z+ with a macroscopic heavy body with a macroscopic charge.

Then it is obvious that the superheavy electron draws a slowly bending classical path. It receives momentum from the macroscopic charge in many small packets as it draws its path.

The Feynman diagram for the process does not depend on the masses or the charge of the nucleus. The diagram shows the nucleus sending just one packet of momentum to the electron.

Overlapping probabilities

         sectors

                   \   |   /

                     \ | /

                       ●    Z+

Let us divide the space around Z+ into many sectors. There might be a million of such sectors. The electron goes through each sector as it passes the nucleus.

Let us use a separate Feynman diagram to analyze the journey through each sector.

In each sector, there is a tiny probability P that the electron emits a soft photon of a certain energy range ΔE.

If we sum all the probabilities for a very wide range of energies for every sector, we will get a sum which is much larger than 1. Is this an infrared divergence?

No. We forgot that if we sum probabilities, the cases have to be separate. If we want to know the probability of emitting exactly one soft photon, we must demand that one sector emits a single photon and other sectors emit zero photons.

Feynman integral formulas seem to lack the sophistication to ban emission of other photons besides the one which is emitted in the diagram. Feynman integrals then sum overlapping probabilities. That is the reason for the infrared divergence. The integral seems to claim that the probability of emitting a single soft photon is infinite.

Regularization of the infrared divergence

Classically, the electron emits a wave which is like a rounded Dirac delta distribution. There is very little total energy in soft photons. There may be a few soft photons which might be observable, but the rest of the soft photons contain infinitesimal total energy. (See the boldface "Classical number of photons" in the previous blog post.)

We can simply ignore photons whose energy is below some tiny threshold. Putting a cutoff or assigning a tiny "mass" to a photon is acceptable.

If we are interested in a cross section of some event whose probability is much less than 1, then the overlap of probabilities has a very small effect and we can ignore the overlap.

Regularization does not remove the overlap. Regularization only sweeps under the rug the ugly fact that some calculated probabilities would be > 1 because of the flawed method of summing probabilities.