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How do you derive the result for the lagrangian density of a free electromagnetic field

Qmechanic
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Anne
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  • You write down the most general Lagrangian obeying the right symmetries. Then you remove the interaction terms. – Akerai Jan 21 '20 at 15:35
  • 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") – Artem Alexandrov Jan 21 '20 at 15:37
  • Possible duplicates: https://physics.stackexchange.com/q/55291/2451 and links therein. – Qmechanic Jan 21 '20 at 17:19

1 Answers1

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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, $$S=-a\int dtd^3x\,F_{\mu\nu}F^{\mu\nu}.$$

Artem Alexandrov
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