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Show that the action integral for the electromagnetic field in a matter with dielectric constant [tex]\epsilon[/tex] is given by [tex]S_m=\frac{1}{8 \pi} \int dt d^3 \bf x \left[ \epsilon \bf E^2 - \bf B^2 \right][/tex]

where the standard expressions for the E field and B field [tex]E=-\frac{1}{c}\partial_t \bf A - \nabla \phi[/tex] and [tex]\bf \nabla \times \bf A[/tex] are used.

I'm familiar with the Lagrangian principle L = T - V, where T and V are kinetic and potential energy respectively, and I'm familar with the concept that minimising the action integral of a system leads us to finding the equations of motion for that system. In this case [tex]d^3 \bf x \left[ \epsilon \bf E^2 - \bf B^2 \right][/tex] is the Lagrangian density of the system. I can see that the expression for Lagrangian density has units of energy per unit volume as one would expect.. But why does the Lagrangian density take that particular form?

Thank you for taking the time to read this; any help would be much appreciated.

where the standard expressions for the E field and B field [tex]E=-\frac{1}{c}\partial_t \bf A - \nabla \phi[/tex] and [tex]\bf \nabla \times \bf A[/tex] are used.

I'm familiar with the Lagrangian principle L = T - V, where T and V are kinetic and potential energy respectively, and I'm familar with the concept that minimising the action integral of a system leads us to finding the equations of motion for that system. In this case [tex]d^3 \bf x \left[ \epsilon \bf E^2 - \bf B^2 \right][/tex] is the Lagrangian density of the system. I can see that the expression for Lagrangian density has units of energy per unit volume as one would expect.. But why does the Lagrangian density take that particular form?

Thank you for taking the time to read this; any help would be much appreciated.

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