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

$A^{\mu }=(\mathbf {A} ,{\frac {\Phi }{c}})$

$\square ^{2}=\partial _{\mu }\partial ^{\mu }=\nabla ^{2}-{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}$

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

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

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

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 $|\mathbf {r} -\mathbf {r} '|={\sqrt {(x-x')^{2}+(y-y')^{2}+z'^{2}}}$

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

$=r'-{\frac {2r\rho '\cos \phi }{2r'}},{\frac {\rho '}{r'}}=\sin {\theta }'$

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:

${\mathcal {J}}(r')={\frac {e^{ikr'}}{r'^{2}}}\int _{0}^{a}rdr2\pi J_{0}(kr\sin {\theta }')$

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

$z^{v-k}J_{v-k}(z)=\left({\frac {1}{z}}{\frac {\partial }{\partial z}}\right)^{k}z^{v}J_{v}(z)$

In this case, we take $v=k=1\quad$ 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}')}

To find the angle to the first diffraction minimum, we must find the first zero of this amplitude function. This will occur when 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_1(ka\sin{\theta}')=0 \quad} File:Bessels J0J1J2.svg