Given that the fine structure constant is a dimensionless ratio quantifying the strength of the coupling between a charged particle and the electromagnetic field, is there a symmetric set of such irreducible, dimensionless constants for the coupling between the strong and the weak nuclear, and the higgs fields and their respective particles?

The definition of the gravitational coupling constant seems to imply that. Are their values known? Are they somehow related to the fine structure constant, or depending on it?


1 Answer 1



I apologize in advance if some of these things might not be clear to you. I started to kind of ramble at some point. Hopefully you can get something out of it.


The existence of a dimensionless coupling constant is a feature of all gauge theories in four dimensions. And, more generally, it is a feature of any theory that has a marginal coupling.

The gauge interactions are the electroweak (electromagnetic + weak) and the strong. The Higgs is not a new force but rather a particle that interacts with the electroweak field. It thus does not give any new gauge couplings. It does give a lot of couplings called "Yukawa" couplings. However they depend on the expectation value of the Higgs field in the vacuum we live in, and are not as "fundamental" as the other constants.${}^1$

The gravitational constant

The constant you quote is not the same thing though. The gravitational constant is dimensionful (it is an irrelevant coupling in the renormalization group sense, see the Wikipedia link). The quantity you linked instead is the ratio of the gravitational coupling constant and the mass of the electron. While this is an ok thing to do, it is an arbitrary choice. It's the same as saying that we decide to measure that constant with a "ruler" whose steps are set by the mass of the electron.

So where are the couplings?

On the other hand, the strong and weak force do have couplings in the same fashion as the fine-structure constant $\alpha$. See the PDG (the one for the strong force is given under the entry "strong coupling constant". While the one for the weak force is given in a rather cryptic way through the Weinberg angle).

The way these results are quoted seems confusing: one depends on the mass of the $Z$ boson $m_Z$ and the other on some weird symbol $\overline{\mathrm{MS}}$. And also there is a strange looking comment in the $\dagger$ note of the fine-structure constant...

So, how do we understand this?

The reason is that these constants are not constants at all! They depend on the energy scale at which the measurement is made. So the value quoted is the value at some convenient energy range. The $\overline{\mathrm{MS}}$ is a whole another story since it is related to renormalization, but the idea is the same (only that we do not compute the coupling at a specific energy value, but rather we compute it in a convenient scheme, from which one can infer the value at any other energy).

This behaviour, namely that a coupling constant values depends on the energy, is refereed to as running.

For $\alpha$ the result typically quoted is at zero energy since the coupling changes rather smoothly and in most cases we are interested in interactions at low energy.

Are there any really constant constants?

Not in the standard model. But we physicists have a lot of creativity and we came up with models where the marginal couplings do not run at all. To show this rigorously we need additional weapons in our arsenal, namely supersymmetry. The mother example is $\mathcal{N}=4$ Super Yang Mills, where the coupling constant $g_{\mathrm{YM}}$ does not run at all. There are other examples without supersymmetry in four dimensions, but we still struggle to pinpoint those theories precisely.

The big difference between the two links I gave you is that in the first (supersymmetric) case, there exists a theory for every value of the coupling. Whereas for the second case the value of the coupling would be fixed.

$\;{}^1\;$This is not entirely correct: the values of the Yukawa couplings are fundamentals. But what I'm trying to say there is that the main consequence that they have on nature heavily depends on the vacuum expectation value of the Higgs $v$. That is because they give masses to the fermions as $m \sim g_{\mathrm{Yukawa}} v$. But other than that, yes, they can be regarded as fundamental constants.

  • $\begingroup$ That there are no "constant constants" is a misconception. Even in the Standard Model with its running couplings it is possible to define "constant constants" by taking combinations of couplings and the energy scale that are invariant under the RG flow. Of course you can only do this up to a fixed order in perturbation theory, but the theory itself only exists in the perturbative form, so it is good enough. There are always "constant constants", otherwise the theory simply doesn't make sense. Its just that these are not the bare couplings that appear in the action. $\endgroup$ Mar 26, 2020 at 2:19
  • $\begingroup$ If you are talking about something like $\Lambda_{\mathrm{QCD}}$, however constant, it doesn't have an interpretation as a coupling. Otherwise, other quantities that I can think of which are constant, are trivially so. Namely, if $g_0$ is the coupling at a certain energy, then the running coupling is $g \equiv g(g_0,\mu)$, $\mu$ being the scale. Then $g^{-1}(g,\mu) = g_0$ is obviously constant as it is the initial condition of the RG equations. Am I misunderstanding your comment? Do you have other examples? $\endgroup$
    – MannyC
    Mar 26, 2020 at 2:37
  • $\begingroup$ What is a coupling but a free constant that parametrizes your theory/model? Yes, what you've described is an example. $\endgroup$ Mar 26, 2020 at 2:53

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