How do you derive the result for the lagrangian density of a free electromagnetic field
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$\begingroup$ You write down the most general Lagrangian obeying the right symmetries. Then you remove the interaction terms. $\endgroup$– AkeraiJan 21, 2020 at 15:35
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$\begingroup$ Lagrangian of free electromagnetic field cannot be derived, it can be deduced by symmetric arguments & invariant properties. See 2-nd volume of Ladau course, paragra 27 ("Action for electromagnetic field") $\endgroup$– Artem AlexandrovJan 21, 2020 at 15:37
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/55291/2451 and links therein. $\endgroup$– Qmechanic ♦Jan 21, 2020 at 17:19
1 Answer
We have following arguments:
- Lagrangian cannot contain potential because they are not defined uniqely due to gauge invariance
- The action should be scalar
These arguments give structure $$S=a\int dtd^3x\,F_{\mu\nu}F^{\mu\nu},$$ where $F_{\mu\nu}F^{\mu\nu}=2({\bf H}^2-{\bf E}^2)$. Electric field contains term $\partial{\bf A}/\partial t$. The term $(\partial{\bf A}/\partial t)^2$ should appears in $S$ with sign "+". If not, for enough fast variation of ${\bf A}$ action $S$ becomes infinite and has not minimum. So, the constant $a$ should be negative. Finally, $$S=-a\int dtd^3x\,F_{\mu\nu}F^{\mu\nu}.$$