Differentiation of SIE (Approach 2) ¶ Calculation of the amplitudes is computationally expensive,
and a lot of care is needed to make it tenable.
We have tried several approaches which have failed.
The following notes solve only the subproblem of
differentiating ψ \psi ψ
Under Construction
Differentiation of SIE ¶ Using a Cartesian coordinate system aligned with the lens,
the lens potential is given as
ψ R ( x ′ , y ′ ) = C 0 ⋅ f f ′ ⋅ [ y ′ sin − 1 ( f ′ ⋅ y ′ R ) + x ′ sinh − 1 ( f ′ f ⋅ x ′ R ) ] \psi^{\mathrm{R}}(x',y') = C_0\cdot \frac{\sqrt{f}}{f'}\cdot\left[
y'\sin^{-1}\left(f'\cdot\frac{y'}{R}\right)
+ x'\sinh^{-1}\left(\frac{f'}{f}\cdot\frac{x'}{R}\right)
\right] ψ R ( x ′ , y ′ ) = C 0 ⋅ f ′ f ⋅ [ y ′ sin − 1 ( f ′ ⋅ R y ′ ) + x ′ sinh − 1 ( f f ′ ⋅ R x ′ ) ] where
R = x ′ 2 + y ′ 2 R=\sqrt{x'^2+y'^2} R = x ′2 + y ′2 The first order derivatives (see SIE
d d x ′ ψ R ( x ′ , y ′ ) = C 0 f f ′ sinh − 1 ( f ′ f ⋅ x ′ R ) d d y ′ ψ R ( x ′ , y ′ ) = C 0 f f ′ sin − 1 ( f ′ ⋅ y ′ R ) \frac{d}{dx'}\psi^{\mathrm{R}}(x',y') =
C_0\frac{\sqrt{f}}{f'}
\sinh^{-1}\left(\frac{f'}{f}\cdot\frac{x'}{R}\right)
\\
\frac{d}{dy'}\psi^{\mathrm{R}}(x',y') =
C_0\frac{\sqrt{f}}{f'}
\sin^{-1}\left(f'\cdot\frac{y'}{R}\right) d x ′ d ψ R ( x ′ , y ′ ) = C 0 f ′ f sinh − 1 ( f f ′ ⋅ R x ′ ) d y ′ d ψ R ( x ′ , y ′ ) = C 0 f ′ f sin − 1 ( f ′ ⋅ R y ′ ) We want to differentiate in the global coordinate system
( x , y ) (x,y) ( x , y ) ( x , y ) (x,y) ( x , y ) ( x ′ , y ′ ) (x',y') ( x ′ , y ′ ) θ \theta θ
[ x ′ y ′ ] \begin{bmatrix} x' \\\\\\\\ y' \end{bmatrix} ⎣ ⎡ x ′ y ′ ⎦ ⎤ &=
[ cos θ sin θ − sin θ cos θ ] \begin{bmatrix}
\cos\theta & \sin\theta \\\\\\\\
-\sin\theta & \cos\theta
\end{bmatrix} ⎣ ⎡ cos θ − sin θ sin θ cos θ ⎦ ⎤ \cdot
[ x y ] \begin{bmatrix} x \\\\\\\\ y \end{bmatrix} ⎣ ⎡ x y ⎦ ⎤ \end{aligned}
We can differentiate the coordinates, so that
d x ′ d x = cos θ d x ′ d y = sin θ d y ′ d x = − sin θ d y ′ d y = cos θ \begin{align}
\frac{dx'}{dx} & = \cos\theta
\\\\\\\\
\frac{dx'}{dy} & = \sin\theta
\\\\\\\\
\frac{dy'}{dx} & = -\sin\theta
\\\\\\\\
\frac{dy'}{dy} & = \cos\theta
\end{align} d x d x ′ d y d x ′ d x d y ′ d y d y ′ = cos θ = sin θ = − sin θ = cos θ Note that all of these derivatives are constants, so that all
higher-order derivatives vanish.
Differentiation in the global frame ¶ Because ψ \psi ψ
∂ m + n ψ ( ∂ x ) m ( ∂ r ) n = ∑ i = 0 m ∑ j = 0 n ( m + n i + j ) ( ∂ x ′ ∂ x ) m − i ( ∂ y ′ ∂ x ) i ( ∂ x ′ ∂ y ) n − j ( ∂ y ′ ∂ y ) j ∂ m + n ψ ( ∂ x ′ ) m − i + j ( ∂ y ′ ) n − j + i \frac{\partial^{m+n}\psi}{(\partial x)^m(\partial r)^n}
= \sum_{i=0}^m \sum_{j=0}^n \binom{m+n}{i+j}
\big(\frac{\partial x'}{\partial x}\big)^{m-i}
\big(\frac{\partial y'}{\partial x}\big)^{i}
\big(\frac{\partial x'}{\partial y}\big)^{n-j}
\big(\frac{\partial y'}{\partial y}\big)^{j}
\frac{\partial^{m+n}\psi}{(\partial x')^{m-i+j}(\partial y')^{n-j+i}} ( ∂ x ) m ( ∂ r ) n ∂ m + n ψ = i = 0 ∑ m j = 0 ∑ n ( i + j m + n ) ( ∂ x ∂ x ′ ) m − i ( ∂ x ∂ y ′ ) i ( ∂ y ∂ x ′ ) n − j ( ∂ y ∂ y ′ ) j ( ∂ x ′ ) m − i + j ( ∂ y ′ ) n − j + i ∂ m + n ψ Inserting for the derivatives of x ′ x' x ′ y ′ y' y ′
∂ m + n ψ ( ∂ x ) m ( ∂ r ) n = ∑ i = 0 m ∑ j = 0 n ( m + n i + j ) cos m − i + j θ ⋅ sin n − j + i θ ⋅ ( − 1 ) i ∂ m + n ψ ( ∂ x ′ ) m − i + j ( ∂ y ′ ) n − j + i \frac{\partial^{m+n}\psi}{(\partial x)^m(\partial r)^n}
= \sum_{i=0}^m \sum_{j=0}^n \binom{m+n}{i+j}
\cos^{m-i+j}\theta\cdot
\sin^{n-j+i}\theta\cdot
(-1)^i
\frac{\partial^{m+n}\psi}{(\partial x')^{m-i+j}(\partial y')^{n-j+i}} ( ∂ x ) m ( ∂ r ) n ∂ m + n ψ = i = 0 ∑ m j = 0 ∑ n ( i + j m + n ) cos m − i + j θ ⋅ sin n − j + i θ ⋅ ( − 1 ) i ( ∂ x ′ ) m − i + j ( ∂ y ′ ) n − j + i ∂ m + n ψ TODO Check this