Paper 1, Section II, C

Electrodynamics
Part II, 2009

The action for a modified version of electrodynamics is given by

I=d4x(14FabFab12m2AaAa+μ0JaAa)I=\int d^{4} x\left(-\frac{1}{4} F_{a b} F^{a b}-\frac{1}{2} m^{2} A_{a} A^{a}+\mu_{0} J^{a} A_{a}\right)

where mm is an arbitrary constant, Fab=aAbbAaF_{a b}=\partial_{a} A_{b}-\partial_{b} A_{a} and JaJ^{a} is a conserved current.

(i) By varying AaA_{a}, derive the field equations analogous to Maxwell's equations by demanding that δI=0\delta I=0 for an arbitrary variation δAa\delta A_{a}.

(ii) Show that aAa=0\partial_{a} A^{a}=0.

(iii) Suppose that the current Ja(x)J^{a}(x) is a function of position only. Show that

Aa(x)=μ04πd3xJa(x)xxemxxA^{a}(\mathbf{x})=\frac{\mu_{0}}{4 \pi} \int d^{3} x^{\prime} \frac{J^{a}\left(\mathbf{x}^{\prime}\right)}{\left|\mathbf{x}-\mathbf{x}^{\prime}\right|} e^{-m\left|\mathbf{x}-\mathbf{x}^{\prime}\right|}