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Gottfried and Weisskopf write in their book, in the chapter on electroweak theory, that

$\frac{m_W}{m_Z} + \left(\frac{e}{g}\right)^2 = 1$

In this expression, $m_W$ and $m_Z$ are the masses of the $W$ and $Z$ bosons, and $e$ and $g$ are the electromagnetic and the weak coupling constants. I want to understand the equation; so I have a question. Why exactly is the boson mass ratio related to the coupling ratio?

After all, this expressions seems to tell that mass is due to coupling strength. But why is mass ratio related in this "elliptical" way to coupling ratio?

I also have another, minor question. I suppose that by $e^2$ the authors mean what is called the fine structure constant $\alpha$, and by $g^2$ what is called $\alpha_w$. Is this correct? Yes, it is, within a factor of $4 \pi$ for both of them.

Thank you in advance for all your help to make this clearer. Also any book or article on that equation would be helpful. I will improve the question with all comments that I get.

Despite the beautiful answer below, to me it remains a deep fascination why the mass of a particle should depend on the coupling, and why a charged particle (the W) is less massive than a similar neutral one (the Z) - naively, I would expect that electric charge adds to mass. Maybe I can ask this in other terms: why is the Higgs mechanism the way it is? Where does it come from? Since nobody knows yet, a more realistic question is: what is the simplest way to explain mass generation in weak bosons? The rotation in the abstract space is so hard to picture in concrete terms. Is there a simpler way?

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After all, this expressions seems to tell that mass is due to coupling strength. But why is mass ratio related in this "elliptical" way to coupling ratio?

The relation between the couplings and the masses follows from the Higgs mechanism. It generates the masses for $W^{\pm}-,Z-$bosons. The masses depend on the $SU_{L}(2)$ gauge coupling $g_{1}$ and the $U_{Y}(1)$ gauge coupling $g_{2}$; precisely, the $W$-boson mass depends on the weak coupling $g \equiv g_{1}$ linearly, while the $Z$-boson mass is proportional to $\sqrt{g_{1}^{2}+g_{2}^{2}}$. From the other side, the EM coupling is proportional to $g_{2}/\sqrt{g_{1}^{2}+g_{2}^{2}}$. The ellipticity then follows from the simple trigonometric relation between the ratio $m_{Z}/m_{W}$ and $e/g_{1}$.

Such proportionality is the result of non-trivial "symmetry breaking" $G_{\text{electroweak}} \simeq SU_{L}(2)\times U_{Y}(1) \to U_{\text{EM}}(1)$, where $U_{\text{EM}}(1)$ (which is actually the EM gauge group) is the subgroup of $G_{\text{electroweak}}$ which isn't contained in $U_{Y}(1)$ or $SU_{L}(2)$ separately. Let's trace it explicitly.

The interaction of the $SU_{L}(2)\times U_{Y}(1)$ gauge bosons with the Higgs doublet $H$ is given by the lagrangian $$ \tag 1 L_{H} = |D_{\mu}H|^{2}, \quad D_{\mu} \equiv \partial_{\mu} - i\frac{T_{H}}{2}g_{1}t_{a}W_{\mu}^{a}-i\frac{Y_{H}}{2}g_{2}B_{\mu}, $$ where $W_{\mu,a}, a = 1,2,3$ are the $SU_{L}(2)$ gauge fields, $B_{\mu}$ is the $U_{Y}(1)$ gauge field, $t_{a}$ are the Pauli matrices, $T_{H} = 1$ is the weak isospin number for the Higgs doublet, and $Y_{H} = 1$ is the hypercharge number for the Higgs doublet.

After the Higgs driven electroweak crossover the $H$ acquires non-zero VEV, $H \to H + v\begin{pmatrix} 0 \\ 1\end{pmatrix}$, which generates the mass term $$ \tag 2 L_{\text{mass}}= v^{2}g_{1}^{2}|W_{\mu}|^{2} + v^{2}(g_{2}B_{\mu}-g_{1}W_{\mu 3})^{2} $$ where $W_{\mu} \equiv \frac{1}{\sqrt{2}}(W_{\mu,1} -iW_{\mu,2})$.

One can introduce two linear combinations of $W_{\mu, 3}$ and $B_{\mu}$ called $Z$-boson and photon $A$, $$ \tag 3 \begin{pmatrix} W_{\mu 3} \\ B_{\mu}\end{pmatrix} = \begin{pmatrix} \cos(\theta_{W}) & \sin(\theta_{W})\\ -\sin(\theta_{W}) & \cos(\theta_{W})\end{pmatrix}\begin{pmatrix} Z_{\mu} \\ A_{\mu}\end{pmatrix}, \quad \cos(\theta_{W}) = \frac{g_{1}}{\sqrt{g_{1}^{2}+g_{2}^{2}}} $$ which diagonalize $(2)$: $$ L_{\text{mass}} = \frac{v^{2}g_{1}^{2}}{2}|W_{\mu}|^{2} + \frac{v^{2}(g_{1}^{2}+g_{2}^{2})}{2}Z^{2} $$ Note that the photon is massless, as it must be.

Therefore $m_{W} = \frac{vg_{1}}{\sqrt{2}}$ and $m_{Z} = \frac{m_{W}}{\cos(\theta_{W})}$, and $$ \frac{m_{W}}{m_{Z}} = \cos(\theta_{W}) $$ The only thing which is remained is the calculation of the coupling $e$ to the photon $A$ in terms of the constants $g_{1}, g_{2}$. By using $(3)$ and the explicit form of the covariant derivative $(1)$, one obtains $$ e =g_{2}\cos(\theta_{W})= g_{1}\sin(\theta_{W}) $$ Therefore, one obtains the desired result $$ \left( \frac{m_{W}}{m_{Z}}\right)^{2}+\left(\frac{e}{g_{1}}\right)^{2} = \cos^{2}(\theta_{W}) +\sin^{2}(\theta_{W})= 1 $$

I also have another, minor question. I suppose that by e2 the authors mean what is called the fine structure constant $\alpha$, and by $g^{2}$ what is called $\alpha_{W}$. Is this correct?

It depends on the units which you use. For the case of natural units, the relations read $\alpha = \frac{e^{2}}{4\pi}$ and $\alpha_{W} = \frac{g^{2}}{4\pi}$.

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  • $\begingroup$ Thank you very much for this beautiful explanation; I still dream of simplifying it so much that I am able to tell it to my grandparents. But this is not an easy task. $\endgroup$ – Giulia Tozzi Oct 1 '17 at 18:01
  • $\begingroup$ I added some more questions above. $\endgroup$ – Giulia Tozzi Oct 1 '17 at 20:02
  • $\begingroup$ @GiuliaTozzi : the most of the added questions are likely to open as another question. $\endgroup$ – Name YYY Oct 1 '17 at 20:14

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