# Role of Fine Structure Constant in Q.E.D.?

• I
Gold Member
TL;DR Summary
Why are BOTH the FSC and the fundamental electric charge, e, both described as coupling constants for the EM field?
Why, in a nutshell, are there two coupling constants for the EM field, the fundamental electric charge, e, and the Fine Structure Constant? If a brief answer is not possible, can you recommend good literature on this subject?

## Answers and Replies

Mentor
Why, in a nutshell, are there two coupling constants for the EM field, the fundamental electric charge, e, and the Fine Structure Constant?
There aren't. There is just one, but which numerical value you use depends on what specific version of the relevant equations you are using.

From an "in principle" standpoint, the fine structure constant is "the" coupling constant for the EM field, since it is the one that is independent of any choice of units; it's a dimensionless number. The fundamental electric charge is not a dimensionless number (or at least not in all systems of units, but it is in at least one--see below); it depends on your choice of units, and from an "in principle" standpoint is best viewed as a "translation" of the "true" coupling constant, the fine structure constant, into systems of equations that, for both historical and practical reasons, are more frequently used by many people who want to solve problems involving the EM field.

Note that in "natural" units, in which ##\hbar = c = 1##, we have ##\alpha = e^2 / 4 \pi##, so both numbers are dimensionless and which one you pick as "the coupling constant" is just a matter of preference and doesn't change any of the physics.

LarryS, protonsarecool, aaroman and 4 others
Homework Helper
Gold Member
"The fundamental electric charge, e" cannot be fundamental because it can have many numerical values, depending on which which conference defines it. Defining e^2 in terms of alpha works, but removes e as a 'fundamental' unit in SI.

LarryS
Gold Member
2022 Award
In the newest SI ##e##, the elementary charge, has been defined to take a fixed value defining the base unit of electric current, the Ampere. Also Plancks quantum of action, ##h##, and also the vacuum speed of light, ##c##, take defined values in the SI, but the permittivity of the vacuum, ##\epsilon_0##, must be measured, ##\alpha=e^/4\pi \epsilon_0\hbar c## is a derived quantity within the SI.

LarryS
Mentor
the permittivity of the vacuum, ##\epsilon_0##, must be measured, ##\alpha=e^/4\pi \epsilon_0\hbar c## is a derived quantity within the SI.
I think it would be more accurate to say that ##\alpha## in SI units is not treated as a "physical constant" whose value is fixed by the system of units; it is treated as a measured physical constant, whose measured value determines the value of ##\epsilon_0## and ##\mu_0##.

vanhees71
Homework Helper
Gold Member
I used to joke about doing that. Nature can be fooled. The only joke I have left is measuring pi, but we did that in 7th grade.

vanhees71
WernerQH
Also Plancks quantum of action, ##h##, and also the vacuum speed of light, ##c##, take defined values in the SI, but the permittivity of the vacuum, ##\epsilon_0##, must be measured, ##\alpha=e^/4\pi \epsilon_0\hbar c^2## is a derived quantity within the SI.
I don't think this is correct. The permittivity of the vacuum must still satisfy the relation ## \epsilon_0 \mu_0 c^2 = 1 ##, and also ## \mu_0 = 4 \pi × 10^{-7} {\rm Vs/Am} ## is still exact. With the values from the Wikipedia page on the SI system I calculated the inverse fine structure constant ## 1 / \alpha ## as ## 137.035999 ##. I suppose that the definition of the electron charge (i.e. the Ampere) will be revised when a more accurate determination of the fine structure constant becomes available.

Gold Member
2022 Award
No, that was due to the "old SI" valid till 2019. There they fixed indeed ##\mu_0=4 \pi \cdot 10^{-7} \text{Vs}/\text{Am}## due to the old definition of the Ampere. In this version, indeed since also ##c## has been fixed in the way it still is, i.e., via ##\nu_{\text{Cs}}## and the definition of the m via fixing the value of ##c##.

This is now changed by fixing the elementary charge, ##e##, now. BTW the changes of the units by this change of the SI are largest for the elecromagnetic units (order of ##10^{-9}##, if I remember right).

https://en.wikipedia.org/wiki/Vacuum_permeability

The idea of the "new SI" is to change the fundamental constants not anymore. They are defined once and for all at fixed values to be considered exact. These are ##\nu_{\text{Cs}}## (the hyperfine transition frequency of Cs-133), ##c## (speed of light in vacuo, or rather the "limiting speed" of relativity), ##h## (Planck's quantum of action), ##N_{\text{A}}## (Avogadro constant), ##k_{\text{B}}## (Boltzmann constant), together defining the 6 physical base units of the SI, i.e., the second (s), the metre (m), the mass (kg), the mole (mol), and the Kelvin (K).

https://en.wikipedia.org/wiki/SI_base_unit

What may improve with is the accuracy of the realization of the units. For sure, alreadt now time can be more accurately measured than via the hyperfine transition of Cesium, using optical atomic clocks or, maybe soon, nuclear clocks (Thorium).

https://en.wikipedia.org/wiki/Atomic_clock
https://en.wikipedia.org/wiki/Nuclear_clock

WernerQH
No, that was due to the "old SI" valid till 2019. There they fixed indeed ##\mu_0=4 \pi \cdot 10^{-7} \text{Vs}/\text{Am}## due to the old definition of the Ampere.
Thanks for the clarification. I'm still surprised that this relation is no longer exact (with the volt suitably defined). But there's only a finite number of constraints that can be satisfied.

https://arxiv.org/pdf/1610.02910
In the revised SI the value of ## \mu_0 / 4 \pi ## will still be ## 10^{-7} {\rm N/A^2} ## at the time of adoption (see Table II), but subsequently the value of ## \mu_0 / \alpha ## will be fixed. The recommended value and uncertainty of ## \mu_0 / \alpha ## must then evolve in perfect correlation with improved experimental determinations of ## \alpha ##.
## \alpha = \mu_0 c e^2 / 2h ##

vanhees71