In statistical mechanics, we have formulas like $$ n = \frac{g}{2\pi^2} \int_m^\infty \frac{E(E^2 - m^2)^{1/2}}{e^{(E - \mu)/T} \pm 1}\ dE . $$ where $g$ is the internal degrees of freedom. What is the value of $g$ for a Weyl spinor? For context, I am looking at the first part of eq. (6) in this paper. For the charge, we can write more explicitly $$ Q = 3 \cdot N \cdot \frac{2}{3} (\mu_{\rm uL} + \mu_{\rm uR}) + 3 \cdot N \cdot \left( -\frac{1}{3} \right) (\mu_{\rm dL} + \mu_{\rm dR}) + (-1)\cdot \sum_i(\mu_{iL} + \mu_{iR}) - 2\cdot 2 \mu_W - 2m\mu_- $$ The factor of 3 on the quark chemical potentials is due to color, $N$ is the number of generations and the fractional powers are their charge. The bosons $W$ and $\phi^-$ get an extra factor of two since they are bosons (there are $m$ Higgs doublets in this paper). However, the additional factor of two for $\mu_W$ only makes sense if the leptons have $g = 1$ and the W boson has $g = 2$ (before electroweak symmetry breaking).
Am I missing something? Thanks!
