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这是一份sydney悉尼大学PHYS3035/PHYS3935 的成功案例
$$
G(q, \omega)=\frac{1}{q^{2}-q_{0}^{2}}\left(U-\frac{q q}{q_{0}^{2}}\right)
$$
which sometimes is useful. Like the Huygens propagator, the dyadic Green function is also singular for $q=q_{0}$.
The plane-wave representation of the dyadic Green function for the magnetic field is obtained by inserting the Fourier integral transformation
$$
\boldsymbol{G}{\mathrm{M}}(\boldsymbol{R} ; \omega)=(2 \pi)^{-3} \int{-\infty}^{\infty} \boldsymbol{G}_{\mathrm{M}}(q, \omega) \mathrm{e}^{i \boldsymbol{q} \cdot \boldsymbol{R}} \mathrm{d}^{3} q
$$
Calculations analogous to those carried out to determine $\boldsymbol{G}(\boldsymbol{R} ; \omega)$ result in
$$
\boldsymbol{G}{\mathrm{M}}(\boldsymbol{q}, \omega)=\frac{q}{q{0}} \frac{1}{q^{2}-q_{0}^{2}} \boldsymbol{U} \times \hat{\boldsymbol{q}}
$$
an expression which also is singular for $q=q_{0}$. The folding theorem in $\boldsymbol{r}$-space gives when applied to
$$
\boldsymbol{B}(q, \omega)=\frac{i \mu_{0} \omega}{c_{0}} \boldsymbol{G}{\mathrm{M}}(q, \omega) \cdot J(q, \omega) $$ A current density parallel to the $q$-direction does not give rise to a magnetic field, and the magnetic field generated by a current density perpendicular to $\hat{q}$ always lies in a plane perpendicular to $\hat{q}$. To prove the above-mentioned claims we expand the unit dyad after a triple set of orthogonal unit vectors $\hat{q}{\perp}^{(1)}, \hat{q}{\perp}^{(2)}$, and $\hat{q}$, with $\hat{\boldsymbol{q}}{1}^{(1)} \times \hat{\boldsymbol{q}}{2}^{(2)}=\hat{\boldsymbol{q}}$, i.e., $$ \boldsymbol{U}=\hat{\boldsymbol{q}}{\perp}^{(1)} \hat{q}{\perp}^{(1)}+\hat{\boldsymbol{q}}{\perp}^{(2)} \hat{\boldsymbol{q}}_{\perp}^{(2)}+\hat{q} \hat{q}
$$
PHYS3035/PHYS3935 COURSE NOTES ：
In the mixed representation, the electric field, $\boldsymbol{E}(z ; q |, \omega)$, from a current density distribution, $\boldsymbol{J}(z: q |, \omega)$, is outside the distribution given by
$$
E\left(z ; q_{|}, \omega\right)=i \mu_{0} \omega \int_{-\infty}^{\infty} G\left(z-z^{\prime} ; q_{1}, \omega\right) \cdot J\left(z^{\prime} ; q_{|}, \omega\right) \mathrm{d} z^{\prime}
$$
where
$$
G\left(Z ; q_{|}, \omega\right)=\Gamma\left(s g n Z ; q_{1}, \omega\right) \mathrm{e}^{i x_{\perp}|Z|}
$$
with
$$
\Gamma\left(\operatorname{sgn} Z: q_{|}, \omega\right)=\frac{i}{2 q_{0}^{2} \kappa \perp}\left[q_{0}^{2} U-q_{|} q_{|}-\kappa_{\perp}^{2} \hat{z} \bar{z}-\left(q_{|}^{\hat{z}}+\hat{z} q_{|}\right) \kappa_{\perp} \operatorname{sgn} Z\right]
$$
Let us now assume that the source current density is nonvanishing only on a plane sheet located at $z^{\prime}=z_{0}$. Thus,
$$
J\left(z^{\prime} ; q_{|}, \omega\right)=J_{0}\left(q_{|}, \omega\right) \delta\left(z^{\prime}-z_{0}\right)
$$