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 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}')}

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}

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