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I have read that the mass term appearing in the electroweak Lagrangian stops it (the Lagrangian) from becoming gauge invariance. Can someone explain where and why this term is creating the problem?

Qmechanic
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user56751
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    I could probably guess it, but: Which mass term? Which Lagrangian? Which gauge? – ACuriousMind Aug 06 '14 at 15:05
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    Welcome to Physics! As it stands, this question is unclear/does not contain enough information to be useful to this site. Please improve the question by adding more information, or it is likely to be closed. – Danu Aug 06 '14 at 15:30
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    Come on chaps, engage the brain cells. If it wasn't obvious the OP is asking about the Higgs mechanism a brief look at their other answer should make it obvious this is the case. – John Rennie Aug 06 '14 at 16:23
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    @John Rennie: While you certainly make a compelling argument, it is still preferred if OP tells us rather than we have to guess. For all we know, OP could e.g. in principle be talking about abelian or non-abelian gauge groups, mass terms for boson or mass terms for fermions, for starters. user56751: Could you please look over and approve the edits to your post? – Qmechanic Aug 06 '14 at 18:15
  • @JohnRennie Fair point, but as a user on a mobile phone (at the moment) it isn't a very nice experience to have to try and pick through someone's physics.SE history. Additionally, I concur with Qmechanic and think every asker should at least have the decency to provide some context, or cannot reasonably expect an answer. – Danu Aug 06 '14 at 22:35

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Let's assume a typical fermionic mass-term (interacting leptons and quarks are spin 1/2-particles):

$$ \tag 1 \bar{\Psi}\Psi = \bar{\Psi}\left(\frac{1 + \gamma_{5}}{2} + \frac{1 - \gamma_{5}}{2}\right)\Psi = \left| \bar{\Psi}\left( 1 \pm \gamma_{5} \right) = \left( (1 \mp \gamma_{5})\Psi\right)^{\dagger}\gamma_{0} \right| = $$ $$ =\bar{\Psi}_{L}\Psi_{R} + \bar{\Psi}_{R}\Psi_{L}. $$ Then let's assume $SU(2)\otimes U(1)$ gauge-invariant, realistic theory (the electroweak part of the SM). According to this theory, the left representation $\Psi_{L}$ transforms as the doublet part under the gauge transformations, while $\Psi_{R}$ transforms as the singlet. So of course, the mass term isn't gauge invariant.

But if we assume only $U(1)$ gauge theory, there isn't doublets, so the mass term is indeed gauge invariant (except Majorana case, when $\Psi = \hat{C} \Psi$, where $\hat{C}$ refers to the charge conjugation).

This is the reason why we must include (gauge-invariant) interaction of the Yukawa-type with scalar doublets. For example, I will illustrate my statement by describing the mechanism of appearance of mass of charged leptons into the Standard model. We "replace" the mass term $(1)$ by $$ L_{int} = -G\bar{\Phi}_{L}\varphi \Psi_{R} + h.c. $$ Here $\varphi = \begin{pmatrix} \varphi_{1} & \varphi_{2} \end{pmatrix}^{T}$ refers to the doublet of the scalar complex field, and $\Phi_{L} = \begin{pmatrix} \nu_{L} & \Psi_{L}\end{pmatrix}^{T}$. After using unitary gauge (\varphi \to $\begin{pmatrix} 0 & \sigma \end{pmatrix}^{T}$) and shifting the vacuum ($\sigma \to \sigma + \eta$) we will give the mass term and interaction with Higgs boson:

$$ L_{\int} = -G\eta (\bar{\Psi}_{L}\Psi_{R} + h.c.) - G\sigma (\bar{\Psi}_{L}\Psi_{R} + h.c.). $$ So we have the gauge-invariant mass term. But the payment for this is the appearance of Yukawa-interaction with massive real scalar field.

Andrew McAddams
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    thank for responding. actually i was trying to indicate about that mass term which appears in lagrangian of quantum electrodynamics...., but there was some writing tool problem in my laptop so that word was missed..hence i am sorry to all for inconvenience..... – user56751 Aug 07 '14 at 05:28