B4.21

Electrodynamics
Part II, 2004

Using Lorentz gauge, A,aa=0A_{, a}^{a}=0, Maxwell's equations for a current distribution JaJ^{a} can be reduced to Aa(x)=μ0Ja(x)\square A^{a}(x)=\mu_{0} J^{a}(x). The retarded solution is

Aa(x)=μ02πd4yθ(z0)δ(zczc)Ja(y)A^{a}(x)=\frac{\mu_{0}}{2 \pi} \int d^{4} y \theta\left(z^{0}\right) \delta\left(z_{c} z^{c}\right) J^{a}(y)

where za=xayaz^{a}=x^{a}-y^{a}. Explain, heuristically, the rôle of the δ\delta-function and Heaviside step function θ\theta in this formula.

The current distribution is produced by a point particle of charge qq moving on a world line ra(s)r^{a}(s), where ss is the particle's proper time, so that

Ja(y)=qdsVa(s)δ(4)(yr(s))J^{a}(y)=q \int d s V^{a}(s) \delta^{(4)}(y-r(s))

where Va=r˙a(s)=dra/dsV^{a}=\dot{r}^{a}(s)=d r^{a} / d s. Show that

Aa(x)=μ0q2πdsθ(X0)δ(XcXc)Va(s),A^{a}(x)=\frac{\mu_{0} q}{2 \pi} \int d s \theta\left(X^{0}\right) \delta\left(X_{c} X^{c}\right) V^{a}(s),

where Xa=xara(s)X^{a}=x^{a}-r^{a}(s), and further that, setting α=XcVc\alpha=X_{c} V^{c},

Aa(x)=μ0q4π[Vaα]s=sA^{a}(x)=\frac{\mu_{0} q}{4 \pi}\left[\frac{V^{a}}{\alpha}\right]_{s=s^{*}}

where ss^{*} should be defined. Verify that

s,a=[Xaα]s=s.s^{*}, a=\left[\frac{X_{a}}{\alpha}\right]_{s=s^{*}} .

Evaluating quantities at s=ss=s^{*} show that

[Vaα],b=1α2[VaVb+SaXb]\left[\frac{V^{a}}{\alpha}\right]_{, b}=\frac{1}{\alpha^{2}}\left[-V^{a} V_{b}+S^{a} X_{b}\right]

where Sa=V˙a+Va(1XcV˙c)/αS^{a}=\dot{V}^{a}+V^{a}\left(1-X_{c} \dot{V}^{c}\right) / \alpha. Hence verify that Aa,a(x)=0A^{a}{ }_{, a}(x)=0 and

Fab=μ0q4πα2(SaXbSbXa).F_{a b}=\frac{\mu_{0} q}{4 \pi \alpha^{2}}\left(S_{a} X_{b}-S_{b} X_{a}\right) .

Verify this formula for a stationary point charge at the origin.

[Hint: If f(s)f(s) has simple zeros at si,i=1,2,s_{i}, i=1,2, \ldots then

δ(f(s))=iδ(si)f(si)\delta(f(s))=\sum_{i} \frac{\delta\left(s_{i}\right)}{\left|f^{\prime}\left(s_{i}\right)\right|}