Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Feynman says in his book "QED" that the square root of the fine structure constant is the probability for a charged particle to emit a photon. But for which wavelength? Or is it an average over all wavelengths?

Note: I meant virtual photon, and I meant a stable charged particle, like the electron. One way to rephrase it would be: how many virtual photons (per unit volume) are there in the Coulomb field around an electron?

share|cite|improve this question
You'll have to provide more context. The full quote from the book would be helpful. – user346 Feb 19 '11 at 8:13
"how many virtual photons are there in the Coulomb field around an electron?" You know that the coulomb field extends to infinity? – Georg Feb 19 '11 at 13:30

3 Answers 3

by Hans de Vries: a numerical elegant approach can be found here:

The fine structure constant: A radiative series leading to it’s exact value

Vries use both 'pi' and 'e' in the formula and reminds me Feynman (wikipedia - Fine-structure_constant ) ' it related to pi or perhaps to the base of natural logarithms? ...' .

I do not know the meaning of this serie.

share|cite|improve this answer

The fine structure constant $\alpha$ describes the fine structure of the Hydrogen levels of energy.

"I meant a stable charged particle, like the electron."

Is the electron really "stable" if any push makes radiation (inelastic "deformation")? I am afraid it is not. Such radiation occurs always ( probability = 1). The "Coulomb center" picture is obtained by summation of all inelastic pictures. It is called an "inclusive" picture. It is not an "elastic" (non-destructive) one.

share|cite|improve this answer

It's neither - or everything. Feynman means that the "vertex in the Feynman diagram" which has 2 external lines corresponding to a charged particle and 1 external line corresponding to a photon (all three particles have any energy/frequency and momentum/inverse_wavelength you want) is proportional to $e\approx \sqrt{\alpha}$. So he really means the probability amplitude.

One needs lots of (or at least several) other calculations to calculate the probability. In particular, a stable charged particle can never emit a photon because it would violate the energy or momentum conservation: in its initial rest frame, the energy is minimized, so one can't afford to increase the energy by changing the state of the motion (plus emitting a photon, which would make the final energy even higher). An unstable particle can decay into another particle and a photon. The decay rate will be proportional to the fine-structure constant (without the square root) but the detailed decay rate depends on the mass of the decaying particle as well as the other final decay product.

Also, $\alpha$ slowly depends on a scale - logarithmically. It's the value at $E=0$, or - approximately - anywhere at masses $m<m_e$ where $m_e$ is the electron mass where $\alpha$ has the familiar value $1/137.03604$. Even though I didn't tell you a particular process that is linked to $\alpha$, it's actually a good idea in the process of "renormalization" to define $\alpha$ operationally exactly as a particular probability or amplitude at particular energies, just like you suggested. However, there are many choices how to do that.

share|cite|improve this answer
Ok, so the root of alpha is the probability amplitude. How does this allow to answer the note I added to the question? What is the virtual photon density around an electron? – user2107 Feb 19 '11 at 17:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.