Using Lorentz gauge, A,aa=0, Maxwell's equations for a current distribution Ja can be reduced to □Aa(x)=μ0Ja(x). The retarded solution is
Aa(x)=2πμ0∫d4yθ(z0)δ(zczc)Ja(y)
where za=xa−ya. Explain, heuristically, the rôle of the δ-function and Heaviside step function θ in this formula.
The current distribution is produced by a point particle of charge q moving on a world line ra(s), where s is the particle's proper time, so that
Ja(y)=q∫dsVa(s)δ(4)(y−r(s))
where Va=r˙a(s)=dra/ds. Show that
Aa(x)=2πμ0q∫dsθ(X0)δ(XcXc)Va(s),
where Xa=xa−ra(s), and further that, setting α=XcVc,
Aa(x)=4πμ0q[αVa]s=s∗
where s∗ should be defined. Verify that
s∗,a=[αXa]s=s∗.
Evaluating quantities at s=s∗ show that
[αVa],b=α21[−VaVb+SaXb]
where Sa=V˙a+Va(1−XcV˙c)/α. Hence verify that Aa,a(x)=0 and
Fab=4πα2μ0q(SaXb−SbXa).
Verify this formula for a stationary point charge at the origin.
[Hint: If f(s) has simple zeros at si,i=1,2,… then
δ(f(s))=i∑∣f′(si)∣δ(si)