Saturday, 24 June 2017

lagrangian formalism - Why do we require local gauge invariance



My thought on this are somewhat scattered so I apologise in advance.


Maxwell's equations are gauge invariant. The physical Electric and Magnetic fields don't depend on whether we use $A_\mu$ or $A_\mu+\partial_\mu\Lambda$. We have the free EM Lagrangian $\mathcal{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ from which we can derive Maxwell's equations. It thus makes sense that the action stemming from this Lagrangian also be gauge invariant.


My issue arises when we couple to matter.


Let's take the Lagrangian for EM fields coupled to matter;


$$\mathcal{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+j^\mu A_\mu +\bar{\Psi}(i\gamma^\mu\partial_\mu-m)\Psi$$


Under $A_\mu\rightarrow A_\mu+\partial_\mu\Lambda$ the interaction term transforms as


$$\delta(j^\mu A_\mu)=j^\mu \partial_\mu\Lambda=\partial_\mu(\Lambda j^\mu)-\partial_\mu j^\mu\Lambda$$


Ignoring the total derivative term the invariance of the action then requires $\partial_\mu j^\mu=0$.


This is generally the point where textbooks say that we can find a $j^\mu$ made up of the matter fields using Noether's theorem; for spinor fields under global $U(1)$ transformations we would have $j^\mu=\bar{\Psi}\gamma^\mu\Psi$.


However, this conserved current is only obtained when the matter fields are on-shell, but we (presumably) require gauge invariance of the action even when the fields are off-shell. (Please correct me if this is incorrect). If the fields are off-shell we can cancel out the extra term we obtain by requiring that the matter fields transform in an appropriate way under local $U(1)$ transformation. We can then update our definition of gauge invariance to a local $U(1)$ transformation of the matter fields and the usual transformation of the EM fields.



My question; Why should we want to do this anyway? Even if we have the extra term $j^\mu\partial_\mu\Lambda$ Maxwell's equations are unaffected which was the reason we wanted gauge invarance in the first place. From this perspective, the extra term is not an issue.


Now I must admit my classical EM knowledge is a bit rusty so bear with me on this next point. Obviously the spinor equations of motion will not be invariant under the transformation of $A_\mu$ but is there any particular reason why the should be? Just because the 4-potential is not an observable in the pure EM sector is it necessarily like that in the spinor sector?


A thought has also occurred to me when writing this post; if we allow the term $j^\mu\partial_\mu\Lambda$ to appear in the Lagrangian then $\partial_\mu\Lambda$ itself becomes a dynamical field. I can imagine this could produce issues with regard to renormalizability (guessing) but a more immediate consequence I can see of this is the impossibility of interactions; the field equation for $\partial_\mu\Lambda$ is $j^\mu=0$, so we have no interaction. We could though claim that the field is $\Lambda$ instead of $\partial_\mu\Lambda$, in which case the field equation is $\partial_\mu j^\mu=0$.


I admit this post is a bit all over the place. My questions essentially boil down to;




  • Since the current is only conserved on-shell, why do we use Neother's theorem to create the current which we couple to the EM potential?




  • Even if we don't banish the extra term, how screwed are we? Does the extra term produce issues with regards to renormalizability or the resulting spinor equations have properties which disagree with experiments? We could just go from global to local transformations of the spinors and everything works out nicely, but why would we want to do this, apart from local transformations seeming like a more general alternative to global ones?







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