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But it does not explain why $m^2 A_\mu A^\mu$ represent the mass term. … Therefore, one can identify $m$ as the mass of the quanta of the theory. The reason is same for a massive Proca field ("massive photon"). …

Light does not have mass and Newton's theory of gravity doesn't work here. Any massive object, like the Sun, locally deforms the flat fabric of spacetime into a curved one. …

This is why only absolute value of the atmospheric mass squared difference can be inferred from oscillation experiments. This is explained in PDG and Lecture-2 by F. Feruglio. …

How is $\theta_{12}$ identified with the Solar mixing angle and $\Delta m^2_{21}$ the Solar mass-squared difference? …

In the early universe, the Lagrangian Higgs field was such that $\mu^2<0$, and hence cannot be directly associated with the mass of the particle. …

In chiral basis, $\psi=\begin{pmatrix} \psi_L\\ \psi_R \end{pmatrix}$ and therefore, $\overline\psi=\psi^\dagger\gamma^0=\begin{pmatrix} \psi^\dagger_L & \psi^\dagger_R \end{pmatrix}\gamma^0=\begin{pm …

This contribution modifies the pole of the propagator from $$m_0^2\to m^2= m_0^2+\Sigma=m_0^2\Big(1-\frac{\lambda_0}{16\pi^2\epsilon}\Big)+\text{finite}\tag{2}$$ where $m^2$ is the physical mass. …

If not, then both the masses must arise by $SU(2)\times U(1)$ symmetry breaking and both the mass scales are fixed by electroweak symmetry breaking scale. Right? …

How can neutrinos have a magnetic moment in spite of being elementary (as opposed to a neutron) and electrically neutral (as opposed to a proton)? How can it even be defined, and measured?

In an interacting quantum field theory, for example, QED, the Dirac mass $m\bar{\psi}\psi$ is a piece of the free Dirac Lagrangian. … However, the answer here by Lubos Motl states "...the Dirac mass term destroys a particle and creates a new one, or destroys/creates a particle-antiparticle pair, or destroys an anti-particle and creates …

Now the question is, if the interaction vanishes, the bare mass $m_0$ change to renormalized or physical mass. But one always assumes the asymptotic free states to have physical mass $m$. …

In the type-I seesaw, the expression for the effective light neutrino mass matrix is given by $$M_\nu=-m_D^TM_R^{-1}m_D$$ where $m_D$ is the Dirac mass and $M_R$ is Majorana mass for the right-handed electroweak …

I should also be able to verify that in the limit the mass $m\to 0$, a Lorentz transformation doesn't change the helicity. …

This is what happens for a Proca field. The corresponding "electric field" will not remain divergenceless i.e., $\boldsymbol{\nabla}\cdot\textbf{E}\neq 0$ but the corresonding "magnetic field" will re …

$$\newcommand{\slashed}[1]{#1\!\!\!/}$$ Since $\Gamma_\mu$ has a Lorentz index $\mu$, it must involve $\gamma_\mu$, $p_\mu, p'_\mu$ (or equivalently, the linear combinations $p_\mu\pm p'_\mu$) such …