11
$\begingroup$

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."

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – TwoBs Sep 6 '14 at 12:24
  • $\begingroup$ My question is more about fundamental particles though. $\endgroup$ – Konstantin Schubert Sep 6 '14 at 12:30
  • 1
    $\begingroup$ 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. $\endgroup$ – TwoBs Sep 6 '14 at 12:52
15
$\begingroup$

It is possible for particles to get masses at loop level while they are absent at tree level as long as there is no (non-anomalous) symmetry that forbids it. However, in most models if there is a particle that doesn't have a bare mass then its due to a symmetry which then protects it from getting masses at loop level.

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.

This topic is often discussed in the context of neutrinos.

$\endgroup$
6
$\begingroup$

In addition to JeffDror answer :

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 $m_\Lambda^2<0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.