# Is the fine structure constant actually a constant or does its value depend on the energy scale?

The value of the fine structure constant is given as $$\alpha = \frac{e^2}{4\pi\varepsilon_0\hbar c} = \frac{1}{137.035\,999..}$$ It's value is only dependent on physical constants (the elementary charge $e$, speed of light $c$, Plancks constant $\hbar$), the vacuum permitivvity $\varepsilon_0$) and the mathematical constant $\pi$, which are considered to be constant under all cirumstances.

However the Wikipedia article Coupling constant states

In particular, at low energies, α ≈ 1/137, whereas at the scale of the Z boson, about 90 GeV, one measures α ≈ 1/127.

I don't understand how this can be possible, except that one of the physical constants above or even $\pi$ are actually not constant, but dependent on the energy scale. But that seems nonsense.

So what do physicists mean when they say that the fine structure constant $\alpha$ increases with energy? Can you perhaps reformulate the quoted sentence above so that it makes more sense?

• The above question (v1), which basically asks What is renormalization?, should not be confused with the following physics.SE question with a similar sounding title physics.stackexchange.com/q/2725/2451 Commented Jun 15, 2011 at 12:47
• I believe this question was already essentially answered here: physics.stackexchange.com/q/9706 Commented Jun 15, 2011 at 14:40
• or perhaps are you asking about general phenomenon of running coupling of quantum field theories that every coupling constant depends on the scale -- en.wikipedia.org/wiki/Coupling_constant#Running_coupling Commented Jun 15, 2011 at 15:04

Expanding on what Vladimir said: the thing that is changing with energy is $e$ (the others are not constants so much as conversion factors between length and time, time and energy, etc.). The reason the charge can vary is that the vacuum is not entirely empty. Sloppily speaking, near a charge, the electric field interacts with virtual (electron/positron) pairs and the effect is that the virtual pairs screens the "raw" electric field. Thus, if you're far away, you see one value, but as you get closer the electric field raises faster than $1/r^2$. With scattering experiments, how close you get to a charge is directly related to the in-going energy of the particles. Now, in modern physics, we account for this by saying that the charge $e$ changes with energy scale; this sounds bizarre in the form I just explained (since you might expect that we just declare the force to be not $1/r^2$), but it turns out that this is the neatest way intellectually to understand it, due to a convergence of issues to do with wanting to preserve observed symmetries in the theory at all scales.

Incidentally, for things like colour charges in QCD, the vacuum anti-screens, which is to say that the observed field strength increases as you get further away. Heuristically, this is what leads to confinement of quarks in the "normal" phase.

• But why can't we say $\varepsilon_0$ changes with energy scale? I could argue that due to the interaction of the virtual particles with the charge the vacuum permittivity changes locally near that charge and that's why the coupling changes with energy. What are the observed symmetries we want to preserve at all scales that force us to vary the elementary charge with energy? Commented Jun 15, 2011 at 13:54
• @asmaier: well, you always have a factor $e^2 / \epsilon_0$ for every interaction with two vertices, so whether you scale $e$ or $\epsilon_0$ is just a matter of convention. In fundamental physics we don't bother writing $\epsilon$s $c$s and $\hbar$s at all, so we can only change $e$ :) The symmetry in QED is a gauge symmetry with $U(1)$ group. Don't get scared by these words, it essentially says that the theory behaves like Maxwell's electrodynamics and in particular there will be familiar Coulomb $1/r^2$ interaction in long distance limit. Commented Jun 15, 2011 at 14:50
• But I don't really subscribe to @genneth's point of view of $1/r^2$ at all distances and forces. At high eneries (and equivalently short distances) it makes no sense to talk about forces. Well, unless you are talking about effective forces due to charge screening but these definitely don't behave like classical Coulomb ones. So I'd like genneth to clarify what he means by preservation of $1/r^2$ at small distances. Commented Jun 15, 2011 at 14:55
• @asmaier: The vacuum permittivity $\epsilon_0=\frac{1}{\mu_0c_0^2}$ is in the current SI convention not a statement about the Vacuum that could be experimentally measured, but rather it is a man-made definition, see en.wikipedia.org/wiki/Vacuum_permittivity#Value In particular, $\epsilon_0$ does not run. Commented Jun 15, 2011 at 18:46
• @asmaier: you should distinguish between fundamental constants of theory and between coupling constants of a concrete model. $\hbar$ and $c$ are very fundamental to QFT (irrespective of the concrete model!) and that's why we work in Planck units and set them equal to one. It is masses and interaction "constants" that you are really interested in when studying some theory and its predictions. Commented Jun 16, 2011 at 11:31

Do not worry, it is a constant.

There is a sloppiness and misunderstanding in physics when the energy dependence of the cross sections is wrongly attributed to the "fundamental constant" whereas it is a cross section feature.

• This is a pet peeve of mine! People invest so much time studying perturbative renormalisation they confuse the couplings that appear in the low energy effective lagrangian with "the" fine structure constant, which if QED doesn't require a UV completion is a perfectly fine thing to define.
– user213887
Commented Nov 26, 2018 at 4:56
• Just to be clear, is the accepted answer wrong? or is it just a matter of language/interpretation? Commented Aug 27, 2022 at 15:18