Field theoretical origin of mass

The operator form of the relativistic dispersion relation leads to the dynamics of field, yielding the form of the Lagrangian
\begin{equation}
    \mathcal{L} = -\frac{1}{4}F_{\mu\nu} F^{\mu\nu} + \frac{m^2}{2}A_\mu A^\mu
\end{equation}
for a massive (abelian) gauge boson $A_\mu$, and
\begin{equation}\label{lagrangian_fermion}\tag{0}
    \mathcal{L} = i\bar{\psi} \gamma^\mu \partial_\mu \psi - m \bar{\psi} \psi
\end{equation}
for a massive fermion $\psi$. For more generality, let us consider the non-abelian case:
\begin{equation}\label{lie_algebra}\tag{1}
    F_{\mu\nu}^\alpha =\frac{1}{2}\partial_{\left[\mu\right.} A_{\left.\nu\right]}^\alpha - g f^{\alpha \beta \gamma} A_\mu^\beta A_\nu^\gamma .
\end{equation}
A vector boson $A_\mu^\alpha$ requires the form of the mass term to be $-m_{A}^2\mathrm{tr}A_\mu A^\mu$ to satisfy the Klein-Gordon equation. However, if one attempts to impose the gauge invariance to the mass term with respect to the gauge transformation of
\begin{equation}
    A_\mu \rightarrow G A_\mu G^{-1} - \frac{i}{g}\left(\partial_\mu G \right) G^{-1}.
\end{equation}
(where $G\in\mathcal{G}$ is an element of gauge group($\mathcal{G}$) and $g$ is the coupling constant), the mass term then is not invariant. This violation requires a gauge boson to be massless. In other words, the concept of mass confronted an essential skepticism about its ontological status. The basic perspective that inertiality appears as field interactions was formulated by Higgs and explained by the idea called the Higgs mechanism (Higgs 1964). This was actually verified when the Higgs boson was later discovered in 2012 (Aad et al. 2012). That is to say, the concept of mass was fabricated by the Higgs-field interaction for the sake of the symmetry of the nature. It is obvious that imposing gauge symmetry has been an code of today's human intelligence, since laws of physics have evolved following the trend of satisfying gauge in(co)variance, which contains the universal nature of physics. 

In the same manner, the Standard Model, which still has remarkable power to explain our nature, describes mass as an interaction with the Higgs field:
\begin{equation}\label{SM}\tag{2}
    \begin{aligned}
        \mathcal{L}_\text{SM} = & \underbrace{-\frac{1}{4} W_{\mu\nu}^\alpha W^{\alpha\mu\nu} - \frac{1}{4} B_{\mu\nu} B^{\mu\nu}}_{\substack{\text{kinetic terms and self-interactions of} \\ {U(1)\otimes SU(2)\text{ gauge bosons}}}} \\ & + \underbrace{G_\text{Yuk} \bar{\psi}_L \phi \psi_R + G'_\text{Yuk} \bar{\psi}_L \tilde{\phi} \psi_R + h.c.}_\text{Yukawa couplings with fermions and Higgs}\\ 
        & +\underbrace{ i \bar{\psi}_L \gamma^\mu \left( \partial_\mu + i \frac{g}{2} \sigma \cdot W_\mu + i\frac{g'}{2}Y_L B_\mu \right) \psi_L}_\text{kinetic terms and electroweak ...} \\ 
        & \underbrace{+ i \bar{\psi}_R \gamma^\mu \left( \partial_\mu +i \frac{g'}{2} Y_R B_\mu \right)\psi_R }_\text{ ... interactions of quarks and leptons}\\
        & + \underbrace{\left| \left( i\partial_\mu - \frac{g}{2} \sigma \cdot W_\mu - \frac{g'}{2} Y B_\mu \right) \phi \right|^2 - V(\phi)}_{\substack{\text{gauge bosons }W^{\pm}\text{, }Z\text{, }\gamma \text{ and Higgs couplings} \\ {\text{with the potential of }V(\phi)=-\mu^2\phi^\dagger \phi + \lambda \left(\phi^\dagger \phi \right)^2}}}.
    \end{aligned}
\end{equation}
The exact meaning of each term will be discussed later. The purpose of this section is to examine how the concept of mass as an extent to the inertia appears in the interaction field theoretical view.

Consider the simplest example, an abelian gauge boson:
\begin{equation}\label{abelian_gauge}\tag{3}
    \mathcal{L}=-\frac{1}{4} F_{\mu\nu}F^{\mu\nu} + \left| (\partial_\mu -igA_\mu) \phi \right|^2 + \mu^2\phi^\dagger \phi - \lambda \left(\phi^\dagger \phi \right)^2,
\end{equation}
where $g(=e)$ is a coupling constant. It is nothing but the scalar QED I have examined. Under the gauge transformation of 
\begin{equation}
    A_\mu \to A_\mu - \partial_\mu \Lambda, \;\;\;\; \phi \to e^{ig\Lambda}\phi
\end{equation}
our Lagrangian is invariant. But one can easily show that if $\mu^2>0$, the theory has non-zero vacuum, generating the expectation value of $\left< \phi \right> = \mu/\sqrt{2\lambda}$. Spontaneously $U(1)$ symmetry will be broken to require the vacuum refinement. Since the original $\phi$ was a complex scalar, having two degrees of freedom, one can replace it with two real scalars: the direction of argument $\chi$ and the radial direction near the true vacuum $\phi_H$ 
\begin{equation}
    \phi = \left[ \left<\phi\right>+\frac{\phi_H}{\sqrt{2}} \right] e^{\frac{i\chi}{\sqrt{2}\left<\phi\right>}} .
\end{equation}
Changing variables gives our refined Lagrangian,
\begin{align}
    \mathcal{L} =& \underbrace{-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}  +\frac{g^2 \mu^2}{4} A_\mu A^\mu}_{\text{kinetic term of massive bosons}} \; + \underbrace{\frac{1}{2} \partial_\mu \chi \partial^\mu \chi}_\text{massless Goldstone} \\
    &+ \underbrace{\frac{1}{2}\left(\partial_\mu \phi_H \partial^\mu \phi_H - 2\mu^2\phi^2_H\right)}_{\text{Higgs with mass }\sqrt{2}\mu} + \;\text{(interactions)}.
\end{align}
Thus the gauge boson $A_\mu$ has acquired the mass $g\mu/\sqrt{2}$. After the symmetry has been broken, the gauge boson $A_\mu$ has gained a degree of freedom due to its massiveness and the scalar field $\phi$ has lost a degree of freedom to grant a mass to the field $A_\mu$, preserving the total four degrees of freedom.

Now, consider a gauge theory with $U(1) \otimes SU(2)$ electro-weak interactions, the second and third lines in (2). Here, the left and right chiral projections of fermions are 
\begin{equation}
    \psi_L = \frac{1-\gamma^5}{2}\psi,\;\;\;\; \psi_R = \frac{1+\gamma^5}{2} \psi.
\end{equation}
The left-handed quarks and leptons $\psi_L$ are chiral doublets, and the right-handed quarks and leptons $\psi_R$ are chiral singlets. Each doublets and singlets transforms as
\begin{equation}
    \psi_L \to \psi_L'=e^{iY_L\theta} e^{i\sigma^i\eta^i/2} \psi_L,\;\;\;\; \psi_R \to \psi_R'=e^{iY_R\theta}\psi_R
\end{equation}
where $\sigma^i/2$ denote the generators of $SU(2)_L$ Lie algebra whose structure constants are levi-civita $f^{\alpha\beta\gamma}=\epsilon^{\alpha\beta\gamma}$ in (2). The transformation of $W_\mu$ and $B_\mu$ are determined by the above gauge transformation.
\begin{equation}
    \sigma \cdot W_\mu \to \sigma\cdot W_\mu' = e^{i \sigma \cdot \eta / 2} \sigma \cdot W_\mu e^{-i \sigma \cdot \eta / 2} + \frac{1}{g}\left( \partial_\mu e^{i \sigma \cdot \eta(x) / 2} \right) e^{-i \sigma \cdot \eta / 2}
\end{equation}
\begin{equation}
    B_\mu \to B_\mu'=B_\mu -\frac{1}{g'}\partial_\mu \theta(x).
\end{equation}
The four gauge parameters $\eta^i$ and $\theta$ contain the $SU(2)$ gauge bosons $W^i$ coupling with weak-isospin $T=\sigma/2$ and a $U(1)$ gauge boson $B$ coupling with hypercharge $Y$. A given fermion's electric charge $Q$ is determined by adding its weak-isospin $T^3$ and hypercharge $Y$, which have fixed values of $Q=T^3+Y/2$. Since for $\psi_L$ the eigenvalue of $T^3$ is $\pm 1/2$ and for $\psi_R$ is $0$, the lepton's eigenvalues are fully determined as $Y_L^\text{leptons}=-1$, $Y_R^\text{leptons}=-2$, $Y_L^\text{quarks}=1/3$, $Y_R^\text{(u)}=4/3$, $Y_R^\text{(d)}=-2/3$. 

The gauge kinetic term is the two terms in the first line in (1), where
\begin{equation}
    W_{\mu\nu}^\alpha = \frac{1}{2}\partial_{[\mu} W_{\nu]}^\alpha - g\epsilon^{\alpha\beta\gamma} W_\mu^\beta W_\nu^\gamma, \;\; B_{\mu\nu} = \frac{1}{2}\partial_{[\mu} B_{\nu]}.
\end{equation}

The last two terms in (2) are the Higgs terms, where $\phi$ is a complex scalar field. In $SU(2)_L$ spinor representation, $\phi=(\phi^{+}, \phi^0)^T$ has the $U(1)$ weak hypercharge $Y_\phi =1$. The $U(1)_Y$ symmetry is henceforth required to be a system with massless gauge boson. In order for the theory to be renormalizable and $SU(2)_L\otimes U(1)_Y$ invariant, the Higgs potential has to be the same potential form as (3). Like the abelian example above, if $\mu^2>0$, our theory has non-zero vacuum, leading to a spontaneous symmetry breaking. Since there are an infinitely degenerate vacuum with $\phi^\dagger \phi=\left<\phi\right>^2/2=-\lambda^2/2\mu$, one can take $\left(\phi^{+},\phi^0\right)=\left(0,\left(\left<\phi\right> + \phi_H\right)/\sqrt{2}\right)$ by specifying the direction. Conservation of charge leads $\phi^0$ to be interpreted as an electrically neutral and the total charge to become neutral, meaning the $U(1)_Q$ symmetry stays unbroken. In the unitary gauge, the Goldstone boson does not exist. Thus the only physical one is Higgs after symmetry breaking. Calculating kinetic terms of Higgs in (2) gives (after lengthy algebra)

where I have only considered the contribution for gauge boson masses. For the charged boson $W_\mu^{\pm}$ and neutral boson $Z_\mu$ and $A_\mu$, they are defined as
\begin{align}
    W_\mu^{\pm}:=\frac{1}{\sqrt{2}} \left( W^1 \mp W^2 \right)_\mu & \;\;\;\; \text{with mass}\;\;\;\; m_W=\frac{g\left<\phi\right>}{2} \\
    Z_\mu:=\frac{gW_\mu^3 - g'B_\mu}{\sqrt{g^2+g'^2}} \;\;\;\;\;\;\;\;\;\;& \;\;\;\;\text{with mass}\;\;\;\; m_Z=\frac{\left<\phi\right>}{2}\sqrt{g^2+g'^2} \\
    A_\mu:=\frac{g'W_\mu^3 + gB_\mu}{\sqrt{g^2+g'^2}} \;\;\;\;\;\;\;\;\;\;& \;\;\;\;\text{with mass}\;\;\;\; m_A=0.
\end{align}
Counting the degrees of freedom gives $4(\phi)+2(B_\mu)+6(W_\mu^\alpha)=12$ before spontaneous symmetry breaking, and $1(\phi_H)+3(Z_\mu)+6(W^{\pm}_\mu)+2(A_\mu)=12$.

Fermion masses comes from Yukawa coupling terms in the second line of (2). For each quark and lepton, writing down all the Yukawa terms including all spinor indices reads
\begin{equation}\begin{aligned}[b]
    \mathcal{L}_\text{Yuk} = & G^\text{(q,d)}_{\alpha\beta} \bar{\psi}^\text{(q)}_{\alpha,L}\phi \psi^\text{(d)}_{\beta,R} + G^\text{(q,u)}_{\alpha\beta} \bar{\psi}^\text{(q)}_{\alpha,L}\tilde{\phi}\psi^\text{(u)}_{\beta,R} \\
    & + G^\text{(l,e)}_{\alpha\beta} \bar{\psi}^\text{(l)}_{\alpha,L} \phi \psi^\text{(e)}_{\beta,R} + G^\text{(l,$\nu$)}_{\alpha\beta} \bar{\psi}^\text{(l)}_{\alpha,L}\tilde{\phi}\psi^\text{($\nu$)}_{\beta,R} + h.c.
\end{aligned}\end{equation}
where $\tilde{\phi}_\alpha = \epsilon_{\alpha\beta} \phi_\beta^*$. Definitely, Yukawa terms have no gauge symmetry violator since they consist of $SU(2)_L$ singlet. Mass terms are required to be a trivial hypercharge. The up quarks $\psi^\text{(u)}$ and electrons $\psi^\text{(e)}$ can have mass with proper mix up of $\phi$ and $\tilde{\phi}$ due to the eigenvalue $Y_\phi=1/2$ and $Y_{\tilde{\phi}}=-1/2$. But, in case of neutrino $\psi^\text{($\nu$)}$, there is no right-handed partner in the Standard Model. Neutrinos, thus, were expected to be massless (Cottingham and Greenwood
2007). (However, neutrinos should have nonzero masses in order to exhibit the empirically proven phenomena of neutrino oscillation, which combines neutrino flavour states with neutrino mass states. For more information, see Schechter and Valle (1980) and Grossman and Lipkin (1997).) Under $U \in SU(2)$, each representation transforms ($\tilde{\phi}_\alpha'=\epsilon_{\alpha\beta} \phi_j'^*=\epsilon_{\alpha\beta}U_{\beta\gamma}^*\phi_\gamma^*=(U^\dagger)_{\gamma\beta}\epsilon_{\alpha\beta}\phi_\gamma^*=U_{\alpha\beta}\epsilon_{\beta\gamma}\phi_\gamma^*$ since $\epsilon_{\alpha\beta}(U^\dagger)_{\gamma\alpha}(U^\dagger)_{\delta\beta}=\epsilon_{\gamma\delta}\det{(U^\dagger)}$) as
\begin{equation}
    \phi \to \phi'=U \phi, \;\; \tilde{\phi} \to \tilde{\phi}'=U \tilde{\phi}.
\end{equation}
The same procedure, putting $\phi=\left(0, \left<\phi\right>+\phi_H \right)^T/\sqrt{2}$, gives
\begin{equation}\begin{aligned}[b]
    \mathcal{L}_\text{Yuk} &= G^\text{(q,d)} \bar{\psi}^\text{(q)}_{L}\phi \psi^\text{(d)}_{R} + G^\text{(q,u)} \bar{\psi}^\text{(q)}_{L}\tilde{\phi} \psi^\text{(u)}_{R} + G^\text{(l,e)} \bar{\psi}^\text{(l)}_{L}\phi \psi^\text{(e)}_{R} + h.c. \\
    &=\frac{G^\text{(d)} \left<\phi\right>}{\sqrt{2}} \left( \bar{\psi}^\text{(d)}_{L} \psi^\text{(d)}_{R} + \bar{\psi}^\text{(d)}_{R} \psi^\text{(d)}_{L} \right) + \frac{G^\text{(u)} \left<\phi\right>}{\sqrt{2}} \left( \bar{\psi}^\text{(u)}_{L} \psi^\text{(u)}_{R} + \bar{\psi}^\text{(u)}_{R} \psi^\text{(u)}_{L} \right) \\ &\;\;\;\; + \frac{G^\text{(e)} \left<\phi\right>}{\sqrt{2}} \left( \bar{\psi}^\text{(e)}_{L} \psi^\text{(e)}_{R} + \bar{\psi}^\text{(e)}_{R} \psi^\text{(e)}_{L} \right) + h.c.
\end{aligned}\end{equation}
Comparing with the fermionic mass term in (0), masses for the fermions are determined as
\begin{equation}
    m_\text{d}=\frac{G^\text{(d)} \left<\phi\right>}{\sqrt{2}}, \;\; m_\text{u}=\frac{G^\text{(u)} \left<\phi\right>}{\sqrt{2}}, \;\; \text{and}\;\; m_\text{e}=\frac{G^\text{(e)} \left<\phi\right>}{\sqrt{2}}.
\end{equation}