# Why does the running of the gauge couplings show $\frac{1}{\alpha}$ > $\frac{1}{\alpha_w}$ at low energy?

I thought the coupling constants were something like:

$\alpha \approx 1/137$

$\alpha_w \approx 10^{-6}$

$\alpha_s \approx 1$

And yet if you look at any pictorial representation of the running of the couplings you see something like this (or try google):

which seems to indicate that (at the energies scales we have access to) $\alpha$ is less than $\alpha_w$. Why is the ordering of the size of the couplings messed up in all of these pictures?

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The plots that you are seeing and asking about are plotting the inverse of $\alpha's$. So, nothing is messed up. The larger will become smaller, and the smaller will become larger when you take the inverse of things. –  stupidity Sep 13 '12 at 14:45
Yes, of course the inverse is plotted. That explains why the strong coupling is on the bottom. But why is $\alpha_w$ in the middle and not on the top? –  user1247 Sep 13 '12 at 14:57

I think you're confusing the weak structure constant with the Fermi constant. The Fermi constant is $G_F=1.166\times 10^{-5}\text{ GeV}^{-2}$ and it gives us the effective strength of the four-point interaction of fermions. This four-point interaction is of course mediated by the W boson and by looking at the relevant tree-level Feynman diagrams we have $$\frac{G_F}{\sqrt{2}}=\frac{g_W^2}{8M_W^2},$$ where $g_W$ is the weak coupling and $M_W=80.4\text{ GeV}$ the mass of the W boson. Plugging in numbers, we find $$\alpha_W=\frac{g_W^2}{4\pi}\approx\frac{1}{30}.$$
What is the weak coupling $g_W$ then? –  user1247 Sep 13 '12 at 15:13
They assume that the interaction that causes the decay is the Fermi interaction, with coupling $G_F/\sqrt{2}$. In the UV completion of the Fermi theory, i.e. the weak interaction, this coupling corresponds to $g_W^2/8M_W^2$. –  AndyS Sep 13 '12 at 17:17