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

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

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

Translational symmetry implies

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

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

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

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

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

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

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

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 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'.

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 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')}

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}{(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}

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{-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}

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}{(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}

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}{(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 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 \delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'} 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 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 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 G_1(r,r')\quad } in the half-space with z>0 with the boundary condition $G_{1}\quad$ atFailed 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_3=z=0 \quad}   we take the difference:

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 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

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 F^\mu=A(r)\part_\mu G_1(r,r')-G_1(r,r')\part_\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 $z=kr\sin {\theta }'\quad$. So

$J_{0}(kr\sin {\theta }')=\left({\frac {1}{kr\sin {\theta }'}}{\frac {\partial }{\partial (kr\sin {\theta }')}}\right)(kr\sin {\theta }')J_{1}(kr\sin {\theta }')={\frac {d}{k\sin {\theta }'dr}}J_{1}(kr\sin {\theta }')$

This gives us the equation${\mathcal {J}}(r')=2\pi {\frac {e^{ikr'}}{r'^{2}}}\int _{0}^{a}rdr{\frac {d}{k\sin {\theta }'dr}}J_{1}(kr\sin {\theta }')$

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

${\mathcal {J}}(r')=2\pi {\frac {e^{ikr'}}{k\sin {\theta }'r'^{2}}}\left[aJ_{1}(ka\sin {\theta }')-0J_{1}(0k\sin {\theta }')\right]=2\pi {\frac {e^{ikr'}}{k\sin {\theta }'r'^{2}}}aJ_{1}(ka\sin {\theta }')$

and $A(r')={\frac {z'{\dot {A_{0}}}}{2\pi c}}e^{-ikct'}2\pi {\frac {e^{ikr'}}{k\sin {\theta }'r'^{2}}}aJ_{1}(ka\sin {\theta }')={\frac {z'{\dot {A_{0}}}a}{c}}{\frac {e^{ikr'-ikct'}}{k\sin {\theta }'r'^{2}}}J_{1}(ka\sin {\theta }')$

To find the angle to the diffraction minimum, we must find the zeroes of this amplitude function. This will occur when $J_{1}(ka\sin {\theta }')=0\quad$

Plot of Bessel function of the first kind, J_α(x), for integer orders α=0,1,2.

To the right is a graph of three Bessel functions of the first order, specifically $J_{0}(x),J_{1}(x),andJ_{2}(x)\quad$. It 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 $A(r')\quad$ sould 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 tamplitude 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 \theta=0 \quad} . The next zero of 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(x) \quad} corresponds to the first minumum of the diffraction pattern. In this case, this zero occurs at x=3.832. 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 ka\sin{\theta}'=3.832 \quad} . SinceFailed 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 k=\frac{2\pi}{\lambda}\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 a=\frac{D}{2}\quad}

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