Huygens Principle for a Planar Source

We start off with Maxwell's Equation in the Lorentz gauge:
 * $$\square^2A^\mu(\mathbf{r},t) = \square^2A^\mu (r)=\mu j^\mu (r)$$

where we use the metric signature (+,+,+,-) and
 * $$A^\mu = (\mathbf{A},\frac{\Phi} {c})$$
 * $$\square^2=\part_\mu \part^\mu = \nabla^2 - \frac{1}{c^2} \frac{\part^2}{\part t^2}$$
 * $$j^\mu = (\mathbf{j},c\rho), \part_\mu= (\mathbf{\nabla}, \frac{1}{c} \frac{\part}{\part t})$$

The gauge condition for the Lorentz gauge is
 * $$\part_\mu A^\mu = 0 \rArr \mathbf{\nabla} \cdot \mathbf{A}-\frac{1}{c^2} \frac{\part\Phi}{\part t}=0$$

Introduce the Green's function at $$ r=(\mathbf{r},t)$$ from some impulse source at $$ r'=(\mathbf{r}',t')$$
 * $$\square^2_rG(r,r')=\delta^4(r-r')$$

and its Fourier transform
 * $$ \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4r\, e^{-iq\cdot r} G(r,0)$$
 * $$ G(r,0)=\frac{1}{(2\pi)^2} \int d^4q\, e^{iq\cdot r} \tilde{G}(q)$$

Translational symmetry implies
 * $$G(r-r',0)=G(r,r')\quad$$

so that
 * $$ G(r,r')=\frac{1}{(2\pi)^2}\int d^4q\, e^{iq\cdot (r-r')} \tilde{G} (q)$$
 * $$\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^4q\,(-q^2)e^{iq\cdot (r-r')}\tilde{G}(q)$$
 * $$\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^4q\, e^{iq\cdot (r-r')}(-k^2+\frac{\omega^2}{c^2})$$

where $$q=(\mathbf{k},\frac{\omega}{c})$$. But
 * $$\square^2_rG(r,r')=\delta^4(r-r')=\frac{1}{(2\pi)^4}\int d^4q\, e^{iq\cdot (r-r')}$$
 * $$\tilde{G}(q)=\frac{(2\pi)^2}{(2\pi)^4}\frac{1}{-q^2}= \frac{-1}{(2\pi)^2q^2}$$
 * $$G(r,r')=\frac{-1}{(2\pi)^4} \int d^4q\, e^{iq\cdot (r-r')} \frac{1}{(k+\frac{\omega}{c})(k-\frac{\omega}{c})}$$

Chose the "retarded" solution, such that the function is zero unless t>t'.
 * $$G(r,r')=\frac{1}{(2\pi)^4}\int d^3k\, e^{i\mathbf{k}\cdot (r-r')}\int d(\frac{\omega}{c}) \frac{e^{-i\omega(t-t')}}{(\frac{\omega}{c}-k)(\frac{\omega}{c}+k)}\Theta(t-t')$$
 * $$=\frac{1}{(2\pi)^4}\int d^3k\, e^{i\mathbf{k}\cdot (r-r')}(2\pi i \frac{e^{ick(t-t')}-e^{-ick(t-t')}}{2k})\Theta$$
 * $$=\frac{-2\pi}{(2\pi)^4}\int_0 \frac{k^2dk}{k} \sin\left({ck(t-t')}\right) 2\pi\int_{-1}^1 dze^{ik|\mathbf{r}-\mathbf{r'}|z}\Theta$$
 * $$=\frac{-1}{(2\pi)^2}\left(\frac{1}{|\mathbf{r}-\mathbf{r'|}}\right)2\int_0 dk \sin(ck(t-t')) \sin(k|\mathbf{r}-\mathbf{r'}|)\Theta$$
 * $$=\frac{1}{(2\pi)^2}\frac{2}{|\mathbf{r}-\mathbf{r}'|}\frac{2\pi}{4} \left[\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))-\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))\right]\Theta$$

But the term $$\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'$$ so that
 * $$  G(r,r')=\frac{-1}{4\pi}\quad \frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}$$

Now to get the $$G_1(r,r')\quad $$ in the half-space with z>0 with the boundary condition $$G_1\quad $$ at$$ r_3=z=0 \quad$$   we take the difference:
 * $$G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}'+2z'\hat{e_3}|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'+2z'\hat{e_3}|}\right)$$

Now use Green's theorem, with the generating function
 * $$F^\mu=A(r)\part_\mu G_1(r,r')-G_1(r,r')\part_\mu A(r)$$
 * $$\int \part_\mu F_\mu d^4r= \int cdt \int d^3r[\part_\mu A \part^\mu G+A\part_\mu \part^\mu G_1-\part_\mu G \part^\mu A -G_1\part_\mu \part^\mu A]$$
 * $$\part_\mu \part^\mu G_1(r,r')=\delta^4(r-r')$$
 * $$\part_\mu \part^\mu A(r)= \mu j(r)$$, let $$j(r)=0 \quad$$
 * $$\int \part_\mu F_\mu d^4r=A(r')$$

Now invoke the divergence theorem on the half space $$z>0 \quad$$:
 * $$A(r')=-\int d^2r\int cdt\left[A(r)\frac{\part}{\part z}G_1(r,r')-G_1(r,r')\frac{\part}{\part z}A(r)\right]$$, where the last term is zero by the condition of$$G_1(z=0,r')=0 \quad$$
 * $$A(r')=-c\int dt\int d^2rA(r)\frac{\part}{\part z}G_1(r,r')$$

To do the t integral, I need to bring out the z derivative. To do this, I first turn it into a z' derivative, using the relation
 * $$G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}|-c(t-t'))}{|\mathbf{r}-\mathbf{r}|}\right)$$

where $$\mathbf{r}''=\mathbf{r}'-2z'\hat{e_3}$$
 * $$\frac{\part}{\part z}G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\part}{\part z}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}|-c(t-t'))}{|\mathbf{r}-\mathbf{r}|}\right)\right)$$

&there4; $$A(r')=\frac{-1}{4\pi}\frac{\part}{\part z'}\int_{z=0} d^2r\left(2\frac{A(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c})}{|\mathbf{r}-\mathbf{r}'|}\right)$$ At $$z=0 \quad $$, $$|\mathbf{r}-\mathbf{r}'|=\sqrt{r^2+z'^2}=S, dS=\frac{rdr}{\sqrt{r^2+z'^2}}$$ If$$A(\mathbf{r},t) \quad$$ is independent of position, as in a plane wave propagating along the z axis, then:
 * $$A(r')=\frac{-\part}{\part z'}\int_{z'}^\infin dS A\left(\mathbf{0},t-\frac{S}{c}\right)=A\left(\mathbf{\mathbf{0}},t'-\frac{z'}{c}\right)$$

This gives us uniform translation of waves at velocity c. More generally:
 * $$A(r')=\frac{-1}{2\pi}\int_{z=0} d^2r\frac{\part}{\part z'}\left(\frac{A\left(\mathbf{r}, t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|}\right)$$
 * $$=\frac{-1}{2\pi}\int_{z=0} d^2r\left(\frac{A\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^3}(-z')+\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|c}\frac{-z'}{|\mathbf{r}-\mathbf{r}'|}\right)$$
 * $$A(r')=\frac{-1}{2\pi}\int_{z=0} d^2r\left(\frac{A\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^3}(-z')+\frac{1}{c}\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^2}(-z')\right)$$

In our case, we consider only those waves which drop off as $$\frac{1}{r'} \quad$$, so
 * $$A(r')=\frac{z'}{2\pi c}\int_{z=0} d^2r\left(\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^2}\right)$$

In cylindrical coordinates, $$d^2r=rdrd\phi \quad$$. Without loss of generality, we consider a harmonic solution with a particular frequency &omega; = kc.
 * $$\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)=\dot{A}(\mathbf{r},0)e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}$$
 * $$A(r')=\frac{ikz'}{2\pi}\,e^{-i\omega t'} \int_{z=0} rdrd\phi\, \frac{e^{ik|\mathbf{r}-\mathbf{r}'|)}}{|\mathbf{r}-\mathbf{r}'|^2}A_0(\mathbf{r},0)$$

Special Case
Picture an opaque screen with a circular aperture of radius a. Let$$\mathcal{J}(r')=\int_0^a rdr\int_0^{2\pi} d\phi\, \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|^2}$$ Then $$A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')$$ But $$|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}$$
 * $$=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}$$
 * $$=r'-\frac{2r\rho'\cos\phi}{2r'}, \frac{\rho'}{r'}=\sin{\theta}'$$

so that $$|\mathbf{r}-\mathbf{r}'|=r'-r\cos{\phi}\sin{\theta}'$$ and $$ \frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}\left(1+\frac{2r\sin\theta'\cos\phi}{r'}\right)$$ In this particular case, we are dealing with far-field effects only, so $$\frac{2r\sin\theta'\cos\phi}{r'}\rightarrow 0 $$ and $$\frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}$$ So, $$ \mathcal{J}(r')=\int_0^a rdr\int_0^{2\pi} d\phi\, \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}=\frac{e^{ikr'}}{r'^2}\int_0^a rdr\int_0^{2\pi} d\phi\, e^{-ikr\sin{\theta}'\cos{\phi}}$$ The integral $$\int_0^{2\pi} d\phi\, e^{-ikr\sin{\theta}'\cos{\phi}}$$ is the integral representation of the zero order Bessel function of the first kind with $$ kr\sin{\theta}' \quad$$ as the argument. This gives us the equation:
 * $$\mathcal{J}(r')=\frac{e^{ikr'}}{r'^2}\int_0^a rdr 2\pi J_0(kr \sin{\theta}') $$

To simplify the math, we make use of the fact that we can represent this Bessel functions as the derivative of a Bessel function of a different order. In general, the formula to compute this derivative is
 * $$z^{v-k}J_{v-k}(z)=\left(\frac{1}{z}\frac{\part}{\part z}\right)^kz^vJ_v(z)$$

In this case, we take $$v=k=1 \quad$$ and $$z=kr\sin{\theta}' \quad$$. So
 * $$J_0(kr\sin{\theta}')=\left(\frac{1}{kr\sin{\theta}'}\frac{\part}{\part (kr\sin{\theta}')}\right)(kr\sin{\theta}')J_1(kr\sin{\theta}')$$

This gives us the equation
 * $$\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{r'^2}\int_0^a rdr \left(\frac{1}{kr\sin{\theta}'}\frac{\part}{\part (kr\sin{\theta}')}\right)(kr\sin{\theta}')J_1(kr\sin{\theta}')$$

Let $$x=kr\sin{\theta'} \quad $$ so that
 * $$\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{k^2\sin^2{\theta}'r'^2}\int_0^{ka\sin{\theta'}} dx \frac{d}{dx}xJ_1(x)$$

$$\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}\left[aJ_1(ka\sin{\theta}')-0J_1(0k\sin{\theta}')\right]=2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}aJ_1(ka\sin{\theta}')$$ and $$A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-i\omega t'} 2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}aJ_1(ka\sin{\theta}')=\frac{z'\dot{A_0}a}{c}\frac{e^{ikr'-i\omega t'}}{k\sin{\theta}'r'^2}J_1(ka\sin{\theta}')$$ To find the angle to the diffraction minimum, we must find the zeroes of this amplitude function. This will occur when $$J_1(ka\sin{\theta}')=0 \quad$$

To the right is a graph of three Bessel functions of the first order, specifically $$ J_0(x), J_1(x), and J_2(x) \quad$$. As it is shown, the first zero of $$J_1(x) \quad$$ will occur at $$x=0 \quad$$. This will correspond to the center of the pattern, at $$\theta=0 \quad$$. Here, we would expect a bright spot, so $$A(r') \quad$$ should be positive and finite. At $$\theta=0 \quad$$ the term $$\frac{J_1(ka\sin{\theta}')}{\sin{\theta}'}$$ is positive and finite, so this expression gives the correct amplitude at $$\theta=0 \quad$$. The next zero of $$ J_1(x) \quad$$ corresponds to the first minumum of the diffraction pattern. In this case, this zero occurs at x=3.832. So, $$ka\sin{\theta}'=3.832 \quad$$. Since$$ k=\frac{2\pi}{\lambda}\quad$$ and $$ a=\frac{D}{2}\quad$$

$$\frac{2\pi D\sin{\theta}'}{2\lambda}=3.832\rightarrow \sin{\theta}'= \frac{1.22\lambda}{D}$$