Roulette Amplitudes for the SIE Lens

We define a local co-ordinate system aligned with the lens, so that the lens orientation is $\theta=0$.
Let $(x',y')$ be a point in the local frame, corresponding to $(x,y)$ in the global frame.

The calculation of roulette amplitudes takes three steps.

1. Calculate the partial derivatives of $\psi$ with respect to $x'$ and $y'$ up to required order. This is done recursively using a CAS engine.

2. Calculate the partial derivatives of $\psi$ with respect to $x$ and $y$ up to required order, using the chain rule.

3. Calculate the roulette amplitudes $\alpha_s^m$ and $\beta_s^m$.

First-order derivatives¶

In the implementation, we calculate the first order derivatives of $\psi$ as \begin{aligned} \frac{\partial\psi^{\mathrm{R}}}{\partial x} &= C_0\frac{\sqrt{f}}{f’}\cdot \left( \cos\theta\cdot\sinh^{-1}\left(\frac{f’}{f}\cdot\frac{x\cos\theta+y\sin\theta}{R}\right) -\sin\theta\cdot\sin^{-1}\left(f’\cdot\frac{-x\sin\theta+y\cos\theta}{R}\right) \right) \\\\ \frac{\partial\psi^{\mathrm{R}}}{\partial y} &= C_0\frac{\sqrt{f}}{f’}\cdot \left( \sin\theta\cdot\sinh^{-1}\left(\frac{f’}{f}\cdot\frac{x\cos\theta+y\sin\theta}{R}\right) +\cos\theta\cdot\sin^{-1}\left(f’\cdot\frac{-x\sin\theta+y\cos\theta}{R}\right) \right) \end{aligned} where $\theta$ is the orientation of the lens, and \begin{aligned} f’ & =\sqrt{1-f^2}, \\\\ C_0 & = \frac{\xi_0}{D_L^2} \end{aligned} For $\theta=0$, this simplifies to \begin{aligned} \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) \end{aligned}

We take the same approach as we did with the raytrace equation, finding the roulette amplitudes first for $\theta=0$, and then rotating the entire system to find the amplitudes for arbitrary $\theta$.

Inner differentation of the co-ordinate transformation¶

The local co-ordinates are given as the rotation \begin{aligned}

&=

\cdot

\end{aligned} We can differentiate the coordinates, so get

We use this with the chain rule below.

Step 2. Chain rule derivation.¶

\begin{aligned} \frac{\partial^{m+n}\psi}{(\partial x)^m(\partial y)^n} = \sum_{i=0}^m\sum_{j=0}^n \binom{m}{i}\binom{n}{j} \cos^{m-i+j}\theta\cdot \sin^{n-j+i}\theta\cdot (-1)^i\cdot \frac{\partial^{m+n}\psi}{(\partial x’)^{m+n-i-j}(\partial y’)^{i+j}} \end{aligned}

Step 3. Roulette Amplitudes¶

Define \begin{aligned} \Gamma^m_s &= \begin{cases} 0 \quad\text{if $m+s$ is even},\\\\ -\frac{D_L^{m+1}}{2^{m+\delta_{0s}}}\binom{m+1}{\frac{m+1-s}2} \quad \text{otherwise} \end{cases} \\\\ \square &= \frac{\partial^2}{(\partial x)^2} + \frac{\partial^2}{(\partial y)^2} \end{aligned} Note that $\square$ is an operator. Then we have \begin{aligned} \alpha^m_s &= \Gamma^m_s\square^{\frac{m+1-s}2} \sum_{k=0}^{\lfloor s/2 \rfloor} (-1)^k \binom{s}{2k} \frac{\partial^s\psi}{(\partial x)^{s-2k}(\partial y)^{2k}} \\\\ \beta^m_s &=
\Gamma^m_s\square^{\frac{m+1-s}2} \sum_{k=0}^{\lfloor (s-1)/2\rfloor} (-1)^k \binom{s}{2k+1} \frac{\partial^s\psi}{(\partial x)^{s-2k-1}(\partial y)^{2k+1}} \end{aligned}

In practice, we omit the $D_L$ factors, because they cancel against other factors when the amplitudes are used in the deflection formulæ.

For implementation, we need to rewrite $\alpha_s^m$ and $\beta_s^m$ without the $\square$ operator. We get \begin{aligned} \alpha^m_s(\vec{\xi}) &= \Gamma^m_s \sum_{k=0}^{\lfloor s/2 \rfloor} (-1)^k \binom{s}{2k} \sum_{h=0}^H\binom{H}{h} \frac{\partial^{m+1}\psi(\vec{\xi})} {(\partial x)^{m+1-2k-2h}(\partial y)^{2k+2h}} \\\\ \beta^m_s(\vec{\xi}) &=
\Gamma^m_s \sum_{k=0}^{\lfloor (s-1)/2\rfloor} (-1)^k \binom{s}{2k+1} \sum_{h=0}^H\binom{H}{h} \frac{\partial^{m+1}\psi(\vec{\xi})} {(\partial x)^{m-2k-2h}(\partial y)^{2k+2h+1}} \end{aligned} where \begin{aligned} H = \frac{m+1-s}{2} \end{aligned}

Alternative Step 3. Roulette Amplitudes¶

This uses the ECMS paper. \begin{aligned} \alpha^m_s &= -\frac{1}{2^{\delta_{0s}}} D^{m+1}{\textrm{L}} \sum{k=0}^{m}\binom{m}{k} \left({ \mathcal{C}}^{m(k)}s\partial{\xi_1} + \mathcal{C}^{m(k+1)}s\partial{\xi_2} \right) \partial^{m-k}{\xi_1}\partial^k{\xi_2}\psi \\\\ \beta^m_s &=-D^{m+1}\textrm{L}\sum{k=0}^m\binom{m}{k}\left({\mathcal{S}}s^{m(k)}\partial{\xi_1}+{\mathcal{S}}s^{m(k+1)}\partial{\xi_2}\right)\partial_{\xi_1}^{m-k}\partial_{\xi_2}^k\psi \end{aligned} where \begin{aligned} \mathcal{C}^{m(k)}s &= \frac{1}{\pi}\int{-\pi}^{\pi}{\rm d}\phi\sin^k\phi\cos^{m-k+1}\phi\cos s\phi \\\\ \mathcal{S}^{m(k)}s &= \frac{1}{\pi}\int{-\pi}^{\pi}{\rm d}\phi\sin^k\phi\cos^{m-k+1}\phi\sin s\phi \end{aligned} Rearranging, for the sake of implementation, we can write \begin{aligned} \alpha^m_s &= - \frac1{2^{\delta_{0s}}} D_\textrm{L}^{m+1} \sum_{k=0}^m\binom{m}{k}\left({\mathcal{C}}s^{m(k)} \partial{\xi_1}^{m-k+1}\partial_{\xi_2}^k\psi +{\mathcal{C}}s^{m(k+1)} \partial{\xi_1}^{m-k}\partial_{\xi_2}^{k+1}\psi \right) \\\\ \beta^m_s &= -D_\textrm{L}^{m+1}\sum_{k=0}^m\binom{m}{k}\left( {\mathcal{S}}s^{m(k)} \partial{\xi_1}^{m-k+1}\partial_{\xi_2}^k\psi + {\mathcal{S}}s^{m(k+1)} \partial{\xi_1}^{m-k}\partial_{\xi_2}^{k+1}\psi \right) \end{aligned}