Huygens Principle for a Planar Source

We start off with Maxwell's Equation in the Lorentz gauge:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \square^2A^\mu(\mathbf{r},t) = \square^2A^\mu (r)=\mu j^\mu (r)}

where we use the metric signature (+,+,+,-) and

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A^\mu = (\mathbf{A},\frac{\Phi} {c})}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \square^2=\part_\mu \part^\mu = \nabla^2 - \frac{1}{c^2} \frac{\part^2}{\part t^2}}

$j^{\mu }=(\mathbf {j} ,c\rho ),\partial _{\mu }=(\mathbf {\nabla } ,{\frac {1}{c}}{\frac {\partial }{\partial t}})$

The gauge condition for the Lorentz gauge is

$\partial _{\mu }A^{\mu }=0\Rightarrow \mathbf {\nabla } \cdot \mathbf {A} -{\frac {1}{c^{2}}}{\frac {\partial \Phi }{\partial t}}=0$

Introduce the Green's function at $r=(\mathbf {r} ,t)$ from some impulse source at $r'=(\mathbf {r} ',t')$

$\square _{r}^{2}G(r,r')=\delta ^{4}(r-r')$

and its Fourier transform

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

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

Translational symmetry implies

$G(r-r',0)=G(r,r')\quad$

so that

$G(r,r')={\frac {1}{(2\pi )^{2}}}\int d^{4}q\,e^{iq\cdot (r-r')}{\tilde {G}}(q)$

$\square _{r}^{2}G(r,r')={\frac {1}{(2\pi )^{2}}}\int d^{4}q\,(-q^{2})e^{iq\cdot (r-r')}{\tilde {G}}(q)$

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

where $q=(\mathbf {k} ,{\frac {\omega }{c}})$. But

$\square _{r}^{2}G(r,r')=\delta ^{4}(r-r')={\frac {1}{(2\pi )^{4}}}\int d^{4}q\,e^{iq\cdot (r-r')}$

${\tilde {G}}(q)={\frac {(2\pi )^{2}}{(2\pi )^{4}}}{\frac {1}{-q^{2}}}={\frac {-1}{(2\pi )^{2}q^{2}}}$

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

Chose the "retarded" solution, such that the function is zero unless t>t'.

$G(r,r')={\frac {1}{(2\pi )^{4}}}\int d^{3}k\,e^{i\mathbf {k} \cdot (r-r')}\int d({\frac {\omega }{c}}){\frac {e^{-i\omega (t-t')}}{({\frac {\omega }{c}}-k)({\frac {\omega }{c}}+k)}}\Theta (t-t')$

$={\frac {1}{(2\pi )^{4}}}\int d^{3}k\,e^{i\mathbf {k} \cdot (r-r')}(2\pi i{\frac {e^{ick(t-t')}-e^{-ick(t-t')}}{2k}})\Theta$

$={\frac {-2\pi }{(2\pi )^{4}}}\int _{0}{\frac {k^{2}dk}{k}}\sin \left({ck(t-t')}\right)2\pi \int _{-1}^{1}dze^{ik|\mathbf {r} -\mathbf {r'} |z}\Theta$

$={\frac {-1}{(2\pi )^{2}}}\left({\frac {1}{|\mathbf {r} -\mathbf {r'|} }}\right)2\int _{0}dk\sin(ck(t-t'))\sin(k|\mathbf {r} -\mathbf {r'} |)\Theta$

$={\frac {1}{(2\pi )^{2}}}{\frac {2}{|\mathbf {r} -\mathbf {r} '|}}{\frac {2\pi }{4}}\left[\delta (|\mathbf {r} -\mathbf {r} '|+c(t-t'))-\delta (|\mathbf {r} -\mathbf {r} '|-c(t-t'))\right]\Theta$

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

$F^{\mu }=A(r)\partial _{\mu }G_{1}(r,r')-G_{1}(r,r')\partial _{\mu }A(r)$

$\int \partial _{\mu }F_{\mu }d^{4}r=\int cdt\int d^{3}r[\partial _{\mu }A\partial ^{\mu }G+A\partial _{\mu }\partial ^{\mu }G_{1}-\partial _{\mu }G\partial ^{\mu }A-G_{1}\partial _{\mu }\partial ^{\mu }A]$

$\partial _{\mu }\partial ^{\mu }G_{1}(r,r')=\delta ^{4}(r-r')$

$\partial _{\mu }\partial ^{\mu }A(r)=\mu j(r)$, let $j(r)=0\quad$

$\int \partial _{\mu }F_{\mu }d^{4}r=A(r')$

Now invoke the divergence theorem on the half space $z>0\quad$:

$A(r')=-\int d^{2}r\int cdt\left[A(r){\frac {\partial }{\partial z}}G_{1}(r,r')-G_{1}(r,r'){\frac {\partial }{\partial z}}A(r)\right]$, where the last term is zero by the condition of$G_{1}(z=0,r')=0\quad$

$A(r')=-c\int dt\int d^{2}rA(r){\frac {\partial }{\partial z}}G_{1}(r,r')$

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 :$\mathbf {r} ''=\mathbf {r} '-2z'{\hat {e_{3}}}$

${\frac {\partial }{\partial z}}G_{1}(r,r')={\frac {-1}{4\pi }}\left({\frac {\partial }{\partial z}}\left({\frac {\delta (|\mathbf {r} -\mathbf {r} '|-c(t-t'))}{|\mathbf {r} -\mathbf {r} '|}}-{\frac {\delta (|\mathbf {r} -\mathbf {r} ''|-c(t-t'))}{|\mathbf {r} -\mathbf {r} ''|}}\right)\right)$

∴ $A(r')={\frac {-1}{4\pi }}{\frac {\partial }{\partial z'}}\int _{z=0}d^{2}r\left(2{\frac {A(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}})}{|\mathbf {r} -\mathbf {r} '|}}\right)$

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:

$A(r')={\frac {-\partial }{\partial z'}}\int _{z'}^{\infty }dSA\left(\mathbf {0} ,t-{\frac {S}{c}}\right)=A\left(\mathbf {\mathbf {0} } ,t'-{\frac {z'}{c}}\right)$

This gives us uniform translation of waves at velocity c. More generally:

$A(r')={\frac {-1}{2\pi }}\int _{z=0}d^{2}r{\frac {\partial }{\partial z'}}\left({\frac {A\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|}}\right)$

$={\frac {-1}{2\pi }}\int _{z=0}d^{2}r\left({\frac {A\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|^{3}}}(-z')+{\frac {{\dot {A}}\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|c}}{\frac {-z'}{|\mathbf {r} -\mathbf {r} '|}}\right)$

$A(r')={\frac {-1}{2\pi }}\int _{z=0}d^{2}r\left({\frac {A\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|^{3}}}(-z')+{\frac {1}{c}}{\frac {{\dot {A}}\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|^{2}}}(-z')\right)$

In our case, we consider only those waves which drop off as ${\frac {1}{r'}}\quad$, so:

$A(r')={\frac {1}{2\pi }}\int _{z=0}d^{2}r\left({\frac {1}{c}}{\frac {{\dot {A}}\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|^{2}}}(z')\right)$

$A(r')={\frac {z'}{2\pi c}}\int _{z=0}d^{2}r\left({\frac {{\dot {A}}\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|^{2}}}\right)$

In cylindrical coordinates, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d^2r=rdrd\phi \quad} . Also, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}'|)}} . So:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A(r')=\frac{z'\dot{A_0}}{2\pi c}\int_{z=0} rdrd\phi \frac{e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}}{|\mathbf{r}-\mathbf{r}'|^2}}

Special Case

Picture an opaque screen with a circular aperture of radius a.

LetFailed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{J}(r')=\int_0^a rdr\int_0^{2\pi} d\phi \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}}

Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =r'-\frac{2r\rho'\cos\phi}{2r'}, \frac{\rho'}{r'}=\sin\theta'}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}\left(1+\frac{2r\sin\theta'\cos\phi}{r'}\right)}