The SIE Lens Hans Georg Schaathun
NTNU---Norwegian University of Science and Technology
Ben David Normann
NTNU---Norwegian University of Science and Technology
SIE (Singular isothermal ellipsoid) ¶ The lens potential of SIE is given as (see SIE
ψ SIE R ( θ E , f , λ L ; θ , ϕ ) = θ E f 1 − f 2 θ ⋅ ( [ sin ( ϕ − λ L ) ] ⋅ sin − 1 ( 1 − f 2 ⋅ sin ( ϕ − λ L ) ) + [ cos ( ϕ − λ L ) ] ⋅ sinh − 1 ( 1 − f 2 f cos ( ϕ − λ L ) ) ) . \begin{aligned}
\begin{split}
\psi^R_\textrm{SIE}(\theta_E,f,\lambda_L;\theta,\phi) & =
\theta_E\sqrt{\frac{f}{1-f^2}}\theta
\\
& \cdot \Bigg([\sin(\phi-\lambda_L)]\cdot\sin^{-1}\left(\sqrt{1-f^2}\cdot \sin{(\phi-\lambda_L)}\right)
\\&
+[\cos(\phi-\lambda_L)]\cdot\sinh^{-1}\left(\frac{\sqrt{1-f^2}}{f}\cos(\phi-\lambda_L)\right)\Bigg).
\end{split}
\end{aligned} ψ SIE R ( θ E , f , λ L ; θ , ϕ ) = θ E 1 − f 2 f θ ⋅ ( [ sin ( ϕ − λ L )] ⋅ sin − 1 ( 1 − f 2 ⋅ sin ( ϕ − λ L ) ) + [ cos ( ϕ − λ L )] ⋅ sinh − 1 ( f 1 − f 2 cos ( ϕ − λ L ) ) ) . where ( θ , ϕ ) (\theta,\phi) ( θ , ϕ ) λ L \lambda_L λ L θ \theta θ x x x λ L \lambda_L λ L
In the following, we use the following shorthands:
f ′ = 1 − f 2 , C 0 = ξ 0 D L 2 \begin{aligned}
f' & =\sqrt{1-f^2},
\\
C_0 & = \frac{\xi_0}{D_L^2}
\end{aligned} f ′ C 0 = 1 − f 2 , = D L 2 ξ 0 Derivation at canonical orientation ¶ In this section we assume λ L = 0 \lambda_L=0 λ L = 0 ψ R \psi^{\mathrm{R}} ψ R
ψ R ( r , ϕ ) = C 0 ⋅ f f ′ R ⋅ [ sin ϕ sin − 1 ( f ′ ⋅ sin ϕ ) + cos ϕ sinh − 1 ( f ′ f ⋅ cos ϕ ) ] \psi^{\mathrm{R}}(r,\phi) =
C_0\cdot \frac{\sqrt{f}}{f'}R\cdot\left[ \sin\phi\sin^{-1}(f'\cdot\sin\phi) + \cos\phi\sinh^{-1}\left(\frac{f'}{f}\cdot\cos\phi\right)\right] ψ R ( r , ϕ ) = C 0 ⋅ f ′ f R ⋅ [ sin ϕ sin − 1 ( f ′ ⋅ sin ϕ ) + cos ϕ sinh − 1 ( f f ′ ⋅ cos ϕ ) ] In Cartesian coordinates, we get
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 deflection is given by the gradient ∇ ψ R \nabla\psi^{\mathrm{R}} ∇ ψ R
d d x sin − 1 x = 1 1 − x 2 \frac{d}{dx}\sin^{-1} x = \frac{1}{\sqrt{1-x^2}} d x d sin − 1 x = 1 − x 2 1 d d x sinh − 1 x = 1 1 + x 2 \frac{d}{dx}\sinh^{-1} x = \frac{1}{\sqrt{1+x^2}} d x d sinh − 1 x = 1 + x 2 1 Differentiation with respect to x x x ¶ Firstly, we differentiate the sin − 1 \sin^{-1} sin − 1
t 1 = d d x y sin − 1 ( f ′ ⋅ y R ) = y 1 − ( f ′ y R ) 2 ( f ′ y ) d d x 1 R = y 1 − ( f ′ y R ) 2 ( f ′ y ) − 1 R 2 d R d x = y 1 − ( f ′ y R ) 2 ( f ′ y ) − 1 R 2 x R = − f ′ x y 2 1 − ( f ′ y R ) 2 1 R 3 = − f ′ x y 2 R 3 ⋅ 1 1 − ( f ′ y R ) 2 \begin{aligned}
\begin{split}
t_1 = \frac{d}{dx}
y\sin^{-1}\left(f'\cdot\frac{y}{R}\right)
&= \frac{y}{\sqrt{1 - (f'\frac{y}{R})^2}}(f'y)
\frac{d}{dx}\frac{1}{R}
\\\\&
= \frac{y}{\sqrt{1 - (f'\frac{y}{R})^2}}(f'y)
\frac{-1}{R^2}
\frac{dR}{dx}
\\\\&
= \frac{y}{\sqrt{1 - (f'\frac{y}{R})^2}}(f'y)
\frac{-1}{R^2}
\frac{x}{R}
\\\\&
= -\frac{f'xy^2}{\sqrt{1 - (f'\frac{y}{R})^2}}
\frac{1}{R^3}
\\\\&
= -\frac{f'xy^2}{R^3}\cdot
\frac{1}{\sqrt{1 - (f'\frac{y}{R})^2}}
\end{split}
\end{aligned} t 1 = d x d y sin − 1 ( f ′ ⋅ R y ) = 1 − ( f ′ R y ) 2 y ( f ′ y ) d x d R 1 = 1 − ( f ′ R y ) 2 y ( f ′ y ) R 2 − 1 d x d R = 1 − ( f ′ R y ) 2 y ( f ′ y ) R 2 − 1 R x = − 1 − ( f ′ R y ) 2 f ′ x y 2 R 3 1 = − R 3 f ′ x y 2 ⋅ 1 − ( f ′ R y ) 2 1 Secondly, we differentiate the sinh − 1 \sinh^{-1} sinh − 1
d d x x sinh − 1 ( f ′ f ⋅ x R ) = sinh − 1 ( f ′ f ⋅ x R ) + x 1 + ( f ′ f x R ) 2 f ′ f ⋅ d d x x R = sinh − 1 ( f ′ f ⋅ x R ) + t 2 \begin{aligned}
\begin{split}
\frac{d}{dx}
x\sinh^{-1}\left(\frac{f'}{f}\cdot\frac{x}{R}\right)
&= \sinh^{-1}\left(\frac{f'}{f}\cdot\frac{x}{R}\right)
+ \frac{x}{\sqrt{1 + (\frac{f'}{f}\frac{x}{R})^2}}\frac{f'}{f}
\cdot \frac{d}{dx}\frac{x}{R}
\\&
= \sinh^{-1}\left(\frac{f'}{f}\cdot\frac{x}{R}\right) + t_2
\end{split}
\end{aligned} d x d x sinh − 1 ( f f ′ ⋅ R x ) = sinh − 1 ( f f ′ ⋅ R x ) + 1 + ( f f ′ R x ) 2 x f f ′ ⋅ d x d R x = sinh − 1 ( f f ′ ⋅ R x ) + t 2 where
t 2 = x 1 + ( f ′ f x R ) 2 f ′ f ⋅ d d x x R = x 1 + ( f ′ f x R ) 2 f ′ f ⋅ [ 1 R − x 1 R 2 x R ] = x 1 + ( f ′ f x R ) 2 f ′ f ⋅ x 2 + y 2 − x 2 R 3 = f ′ f ⋅ x y 2 1 + ( f ′ f x R ) 2 ⋅ 1 R 3 = f ′ x y 2 R 3 ⋅ 1 f 2 + ( f ′ x R ) 2 \begin{aligned}
\begin{split}
t_2 &= \frac{x}{\sqrt{1 + (\frac{f'}{f}\frac{x}{R})^2}}\frac{f'}{f}
\cdot \frac{d}{dx}\frac{x}{R}
\\&
= \frac{x}{\sqrt{1 + (\frac{f'}{f}\frac{x}{R})^2}}\frac{f'}{f}
\cdot \left[ \frac{1}{R} - x\frac{1}{R^2}\frac{x}{R} \right]
\\&
= \frac{x}{\sqrt{1 + (\frac{f'}{f}\frac{x}{R})^2}}\frac{f'}{f}
\cdot \frac{x^2+y^2 - x^2}{R^3}
\\&
= \frac{f'}{f}\cdot\frac{xy^2}{\sqrt{1 + (\frac{f'}{f}\frac{x}{R})^2}}
\cdot \frac{1}{R^3}
\\&
= \frac{f'xy^2}{R^3} \cdot \frac{1}{\sqrt{f^2 + (f'\frac{x}{R})^2}}
\end{split}
\end{aligned} t 2 = 1 + ( f f ′ R x ) 2 x f f ′ ⋅ d x d R x = 1 + ( f f ′ R x ) 2 x f f ′ ⋅ [ R 1 − x R 2 1 R x ] = 1 + ( f f ′ R x ) 2 x f f ′ ⋅ R 3 x 2 + y 2 − x 2 = f f ′ ⋅ 1 + ( f f ′ R x ) 2 x y 2 ⋅ R 3 1 = R 3 f ′ x y 2 ⋅ f 2 + ( f ′ R x ) 2 1 The partial derivative of the lens potential is then
d d x ψ R ( x , y ) = C 0 f f ′ ⋅ [ sinh − 1 ( f ′ f ⋅ x R ) + t 1 + t 2 ] \frac{d}{dx}\psi^{\mathrm{R}}(x,y) =
C_0\frac{\sqrt{f}}{f'}\cdot\left[
\sinh^{-1}\left(\frac{f'}{f}\cdot\frac{x}{R}\right)
+ t_1 + t_2
\right] d x d ψ R ( x , y ) = C 0 f ′ f ⋅ [ sinh − 1 ( f f ′ ⋅ R x ) + t 1 + t 2 ] We can prove that t 1 + t 2 = 0 t_1+t_2=0 t 1 + t 2 = 0
t 1 + t 2 = f ′ x y 2 R 3 ⋅ [ 1 f 2 + ( f ′ x R ) 2 − 1 1 − ( f ′ y R ) 2 ] t_1 + t_2
= \frac{f'xy^2}{R^3} \cdot \left[ \frac{1}{\sqrt{f^2 + (f'\frac{x}{R})^2}}
- \frac{1}{\sqrt{1 - (f'\frac{y}{R})^2}} \right] t 1 + t 2 = R 3 f ′ x y 2 ⋅ [ f 2 + ( f ′ R x ) 2 1 − 1 − ( f ′ R y ) 2 1 ] It suffices to prove that T = 0 T=0 T = 0
T = 1 f 2 + ( f ′ x R ) 2 − 1 1 − ( f ′ y R ) 2 T = \frac{1}{\sqrt{f^2 + (f'\frac{x}{R})^2}}
- \frac{1}{\sqrt{1 - (f'\frac{y}{R})^2}} T = f 2 + ( f ′ R x ) 2 1 − 1 − ( f ′ R y ) 2 1 We have
T = R f 2 R 2 + ( 1 − f 2 ) x 2 − R R 2 − ( 1 − f 2 ) y 2 T = \frac{R}{\sqrt{f^2R^2 + (1-f^2)x^2}}
- \frac{R}{\sqrt{R^2 - (1-f^2)y^2}} T = f 2 R 2 + ( 1 − f 2 ) x 2 R − R 2 − ( 1 − f 2 ) y 2 R Using the fact that R 2 = x 2 + y 2 R^2=x^2+y^2 R 2 = x 2 + y 2
T = = R f 2 y 2 + x 2 − R x 2 + f 2 y 2 = 0 , T =
= \frac{R}{\sqrt{f^2y^2+x^2}}
- \frac{R}{\sqrt{x^2 + f^2y^2}} = 0, T == f 2 y 2 + x 2 R − x 2 + f 2 y 2 R = 0 , as required.
We conclude that
d d x ψ R ( x , y ) = C 0 f f ′ sinh − 1 ( f ′ f ⋅ x 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) d x d ψ R ( x , y ) = C 0 f ′ f sinh − 1 ( f f ′ ⋅ R x ) Differentiation with respect to y y y ¶ The differentiation with respect to y y y
d d y ψ R ( x , y ) = C 0 f f ′ sin − 1 ( f ′ ⋅ y R ) \frac{d}{dy}\psi^{\mathrm{R}}(x,y) =
C_0\frac{\sqrt{f}}{f'}
\sin^{-1}\left(f'\cdot\frac{y}{R}\right) d y d ψ R ( x , y ) = C 0 f ′ f sin − 1 ( f ′ ⋅ R y ) Deflection for arbitrary orientation ¶ For Raytrace simulation, we only require the first order derivatives.
We find it easiest to calculate the deflection vector in a local
co-ordinate system where θ = 0 \theta=0 θ = 0 Approach 1
In Roulette simulation, we require higher order derivatives.
The recursive formulæ for roulette amplitudes proved intractible,
but the non-recursive formulæ are straight forward if higher-order
partial derivatives of ψ \psi ψ
The rotation trick which worked for first-order derivativs does not
easily generalise to higher orders.
We ended up using a co-ordinate substitution and chain rule
differentiation. This is elaborated as
Approach 2
A possible Approach 3 would differentiate the general formula for
ψ R \psi^{\mathrm{R}} ψ R