# For a particle to have physical mass, is it always necessary to have a mass term in the lagrangian?

Since the self-energy adds to the bare mass defined in the Lagrangian, is it possible to create a physical particle mass from the self-energy alone, with no mass terms occuring in the Lagrangian?

On a possibly related note, Wikipedia says: "The photon and gluon do not get a mass through renormalization because gauge symmetry protects them from getting a mass."

• Notice that particles may have mass without even appearing in the lagrangian. Take e.g. QCD where the lagrangian contains quarks and gluons and yet the proton is massive. Sep 6 '14 at 12:24
• My question is more about fundamental particles though. Sep 6 '14 at 12:30
• ah, ok. You didn't say it and the title says 'always necessary' that I interpreted in full generality. Then JeffDror's answer is OK. Sep 6 '14 at 12:52

This is often a subtle topic due to chiral symmetry, \begin{equation} \psi \rightarrow e ^{i \gamma_5 \alpha}\psi \end{equation} which can protect fermions from getting masses under loop corrections. This symmetry is broken by a fermionic mass term, $\bar{\psi} \psi$, but can be conserved by the rest of the Lagrangian. In this case if the mass term doesn't appear at tree level (due to some imposed symmetry) it can't appear at higher orders since the chiral symmetry will protect it.
In the case of scalar fields, this is definitely the case. In order to get a massless field, you need to fine-tune one parameter of the Lagrangian. In the case of the a $\varphi^4$ theory defined by the parameters ($m_\Lambda$, $g_\Lambda$, $\Lambda$), i.e. the bare mass, interaction and the UV cut-off, one need to fine-tune one of the parameter (usually one chooses $m_\Lambda$) to insure that the renormalized mass $m_R=0$.
Note that generically the correction to the mass is positive, and one needs $m_\Lambda^2<0$.