A recent piece of major news in the physics world is that the proton's weak-force charge was measured to be .0719.

Is that a ratio? A dimensionless number, with no units? The articles I read didn't say.....


2 Answers 2


You haven't linked to any of the articles that you've read, but the value that you've quoted seems to come from this Nature paper that was published this week. Note that the news there is the precision of the measurement, $q_\text{weak}^\text{p} = 0.0719 \pm 0.0045$; the central value is consistent with the preliminary result from 2013 of $ 0.064 \pm 0.012$, but the new uncertainty is much improved. Those papers use a definition given by e.g. Erler, Kurylov, and Ramsey-Musolf (2003), and also discussed by the Particle Data Group in section 10.3 of their Review of Particle Properties.

The short answer to your question is that the result isn't quite dimensionless, but doesn't neatly correspond to a macroscopic unit, either. This makes the literature on the subject somewhat harder to follow than necessary.

Compare to the electric charge, which we historically measure in coulombs. However, we have discovered in the 20th century that electric charge in nature occurs only in lumps, where each of those lumps is about one-sixth of an attocoulomb. And later in 2018, we expect to redefine what we mean by a coulomb to be a certain (large) number of these fundamental charges.

One might even be tempted to refer to those unit charges as "dimensionless," but isn't quite right. In the electromagnetic part of the electroweak Lagrangian, the fundamental charge $e$ appears as a coupling constant,

$$ \mathcal L_\text{em} = e J_\mu^\text{em} A^\mu \tag1 $$

in the interaction between the electromagnetic field $A^\mu$ and the electromagnetic currents

$$ J_\mu^\text{em} = \sum_f q_f \overline f \gamma_\mu f \tag2 $$

depend on the charge quantum numbers $q$ of the various fermion fields $f$. Those quantum numbers are the "unit" charges, $+1$ for the positron, $-1/3$ for the down quark, et cetera. The unit for the fundamental charge $e$ in this Lagrangian depends on choices made elsewhere, but is essentially never the coulomb. In the usual unit convention where $\hbar = c = 1$, and the Lagrangian density $\mathcal L$ has units of $\mathrm{GeV}^4$, the fundamental electric charge $e$ winds up being dimensionless. It is related to the (also dimensionless) electroweak coupling constants $g,g'$ by $e = g\sin\theta_W = g'\cos\theta_W$, where $\theta_W$ is the weak mixing angle, and to the dimensionful Fermi coupling constant, $G_F$, and the mass $m_W$ of the charged weak boson by

$$ \frac{G_F}{(\hbar c)^3} = \frac{\sqrt2 g^2}{8 (m_W c^2)^2} \approx 1.17\times10^{-5}\,\rm GeV^{-2} \tag3 $$

Why all this stuff? It's because the weak charge comes from the other neutral term in the electroweak Lagrangian, giving the coupling between the matter particles and the neutral weak gauge field $Z^\mu$:

$$ \mathcal L_N = \frac g{\cos\theta_W} \left( J_\mu^3 + J_\mu^\text{em} \sin^2\theta_W \right) Z^\mu \tag4 $$

Here the weak neutral current is $$ J_\mu^3 = \sum_f I^3 \overline f \gamma_\mu \frac{1-\gamma^5}2 f \tag5 $$

with $I^3$ the weak isospin, and the electromagnetic current $J_\mu^\text{em}$ we've seen already.

What the Erler et al. approach does (at the risk of oversimplifying) is to compress this mess down into a single effective coupling between the fermion fields and the $Z$ boson, and to refer to that coupling constant as the "weak charge." Via some algebra that is opaque to me, the coupling constants seem to wind up proportional to

$$ Q_W \approx 2 I_3 - 4 q_e \sin^2\theta_W $$

This normalization is nice, because it means that the neutron and the neutrino, the electrically neutral ($q_e=0$) members of their respective isospin doublets ($|I_3|=\frac12$), end up with approximately unit weak charge. More importantly, because the weak mixing angle obeys $\sin^2\theta_W\approx\frac14$, the weak charge for the electrically-charged particles (the electron and proton) nearly vanishes, making those weak charge quite sensitive to the weak mixing angle. This normalization also gives us a nice duality between electric charge and weak charge:

$$ \begin{array}{cccc} \text{particle} & q_e & 2I_3 - 4q_e\sin^2\theta_W & \text{measured } q_W \\\hline \text{neutrino} & 0 & +1 \\ \text{electron} & -1 & \text{small} \\ \hline \text{up quark} & +2/3 & +1/3 & +0.375 \pm 0.004 \\ \text{down quark} & -1/3 & -2/3 & -0.678 \pm 0.005 \\ \hline \text{proton} & +1 & \text{small} & +0.072\pm0.004 \\ \text{neutron} & 0 & -1 & -0.981\pm0.006 \\ \end{array} $$

The "measured" values here are taken from Table 1 of the recent Nature paper. (Disclaimer: I'm a co-author on that paper, and the 2013 experimental paper linked earlier.)

So the long answer to your question is that the proton's weak charge is $+0.072$ in a system of units where the corresponding charge on the neutron is approximately $-1$, and the too-long answer is an attempt to clarify just what that unit means.

  • 1
    $\begingroup$ A previous version of this answer on another question. $\endgroup$
    – rob
    May 12, 2018 at 21:54

The charges in particle physics are mostly dimensionless along with the corresponding coupling constants.

In particle physics, as you may notice, some universal constants are set to unity: $$ \hbar=c=1 $$ This implies that any renormalizable term in the Lagrangian would have the charge and the coupling constant to be dimensionless, since for example in QED, you write the interaction term of electrons and photons, $$ e \; Q_\psi \, \bar\psi \gamma ^\mu \psi \; A_\mu $$ where $Q_\psi$ is the electric charge of electron field $\psi$, $e$ is the electromagnetic coupling constant (the unit charge), $A_\mu$ is the photon field, and the rest is the current density for electrons. Here $\psi$'s have the dimensions of $GeV^{3/2}$ while the photon has dimensions of $GeV$, thus in total becomes $GeV^4$ which is the Lagrangian dimensions for 4D. Therefore, $e$ and $Q_\psi$ needs to be dimensionless.

Sometimes the charge is defined together with the coupling constant, instead of being a fraction to it. But in each case the charges are dimensionless.

There could be dimensionful constants or charges, as well. For example in gravity the coupling constant is the Newton's gravitational constant, $G_N$, which is in dimensions $GeV^{-2}$. The charges in gravity are the 4-momentum vector, $P_\mu$, which has dimensions $GeV$.

In early years of particle physics, there was Fermi's theory which has the Fermi constant, $G_F$, for a four-fermion interaction and has dimensions $GeV^{-2}$ similar to the gravity but the (electric) charge was still dimensionless because of the dimensions of the fermion as I have stated above.

Of course, you may find the metric dimensions of these constants and charges by setting back $\hbar$ and $c$ to their corresponding metric values. It could be a good exercise to internalize the relation between the natural scales and human scale, if you haven't done yet. You start to really feel the difference between physics as a geometry and the actual physics. After that you set also $G_N=1$ and then you lose it again :)


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