Paper 3, Section II, 36B

Electrodynamics
Part II, 2013

(i) Obtain Maxwell's equations in empty space from the action functional

S[Aμ]=1μ0d4x14FμνFμνS\left[A_{\mu}\right]=-\frac{1}{\mu_{0}} \int d^{4} x \frac{1}{4} F_{\mu \nu} F^{\mu \nu}

where Fμν=μAννAμF_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.

(ii) A modification of Maxwell's equations has the action functional

S~[Aμ]=1μ0d4x{14FμνFμν+12λ2AμAμ},\tilde{S}\left[A_{\mu}\right]=-\frac{1}{\mu_{0}} \int d^{4} x\left\{\frac{1}{4} F_{\mu \nu} F^{\mu \nu}+\frac{1}{2 \lambda^{2}} A_{\mu} A^{\mu}\right\},

where again Fμν=μAννAμF_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu} and λ\lambda is a constant. Obtain the equations of motion of this theory and show that they imply μAμ=0\partial_{\mu} A^{\mu}=0.

(iii) Show that the equations of motion derived from S~\tilde{S} admit solutions of the form

Aμ=A0μeikνxνA^{\mu}=A_{0}^{\mu} e^{i k_{\nu} x^{\nu}}

where A0μA_{0}^{\mu} is a constant 4-vector, and the 4 -vector kμk_{\mu} satisfies A0μkμ=0A_{0}^{\mu} k_{\mu}=0 and kμkμ=1/λ2k_{\mu} k^{\mu}=-1 / \lambda^{2}.

(iv) Show further that the tensor

Tμν=1μ0{FμσFνσ14ημνFαβFαβ12λ2(ημνAαAα2AμAν)}T_{\mu \nu}=\frac{1}{\mu_{0}}\left\{F_{\mu \sigma} F_{\nu}^{\sigma}-\frac{1}{4} \eta_{\mu \nu} F_{\alpha \beta} F^{\alpha \beta}-\frac{1}{2 \lambda^{2}}\left(\eta_{\mu \nu} A_{\alpha} A^{\alpha}-2 A_{\mu} A_{\nu}\right)\right\}

is conserved, that is μTμν=0\partial^{\mu} T_{\mu \nu}=0.