Huygens Principle for a Planar Source

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

$\square ^{2}A^{\mu }(\mathbf {r} ,t)=\square ^{2}A^{\mu }(r)=\mu j^{\mu }(r)$

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}}

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 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

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 \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 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=(\mathbf{r},t)} from some impulse source at 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'=(\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

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

In cylindrical coordinates, $d^{2}r=rdrd\phi \quad$. Without loss of generality, we consider a harmonic solution with a particular frequency ω = 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 $|\mathbf {r} -\mathbf {r} '|={\sqrt {(x-x')^{2}+(y-y')^{2}+z'^{2}}}$

$={\sqrt {r^{2}+r'^{2}+2r\rho ^{2}\cos \phi }}$

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}'}

so that 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}'|=r'-r\cos{\phi}\sin{\theta}'} 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 \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 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{2r\sin\theta'\cos\phi}{r'}\rightarrow 0 } 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 \frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}}

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 \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 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 \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 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 kr\sin{\theta}' \quad} as the argument. This gives us the equation:

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 \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

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 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 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 v=k=1 \quad} 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 z=kr\sin{\theta}' \quad} . 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 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}')=\frac{d}{k\sin{\theta}'dr} J_1(kr\sin{\theta}')}

This gives us the equationFailed 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')=2\pi\frac{e^{ikr'}}{r'^2}\int_0^a rdr \frac{d}{k\sin{\theta}'dr} J_1(kr\sin{\theta}')}

We invoke the principle that the integral of a derivative is the function evaluated at the end points to give us the equation

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 \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 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'} 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'-ikct'}}{k\sin{\theta}'r'^2}J_1(ka\sin{\theta}'}