CosmoSim is a simulator for gravitational lensing.
We build on the notation defined in Lens Potential. The derivation is based on Principles of Gravitational Lensing by Congdon and Keeton 2018.
Working with angular co-ordinates, $\beta$ denotes the source and $\theta$ the image. We consider only a single deflected point.
The amplification tensor $A$ is given in terms of its inverse which
is the Jacobian of the lens mapping.
\begin{equation}
\mathcal{A}(\boldsymbol{\theta})
= \frac{\partial\boldsymbol{\beta}}{\partial\boldsymbol{\theta}}
=
\begin{bmatrix}
\frac{\partial u}{\partial \theta_1} &
\frac{\partial u}{\partial \theta_2} \\\
\frac{\partial v}{\partial \theta_1} &
\frac{\partial v}{\partial \theta_2}
\end{bmatrix}
\end{equation}
where $\boldsymbol{\beta}=(u,v)$ and $\boldsymbol{\theta}=(\theta_1,y)$.
The raytrace equation is given as
\begin{equation}
\boldsymbol{\beta} = \boldsymbol{\theta} - \nabla\psi
\end{equation}
which gives
\begin{equation}
\mathcal{A}(\boldsymbol{\theta})
=
\begin{bmatrix}
1 - \psi_{\theta_1\theta_1} & -\psi_{\theta_1\theta_2} \\\
-\psi_{\theta_1\theta_2} & 1 - \psi_{\theta_2\theta_2}
\end{bmatrix}
\end{equation}
We can equivalently write it in normalised screen space co-cordinates
\begin{equation}
\mathcal{A}(\mathbf{x})
=
\begin{bmatrix}
1 - \psi_{x_1x_1} & -\psi_{x_1y_2} \\\
-\psi_{x_1y_2} & 1 - \psi_{y_2y_2}
\end{bmatrix}
\end{equation}
Since $\mathcal{A}$ is real and symmetric, we can decompose it using the identity matrix
and the Pauli spin matrices
\begin{equation}
R_- =
\begin{bmatrix}
1 & 0 \\ 0 & -1
\end{bmatrix}
\quad
R_/ =
\begin{bmatrix}
0 & 1 \\ 1 & 0
\end{bmatrix}
\end{equation}
We get
\begin{equation}
\mathcal{A}(\mathbf{x})
=
(1-\kappa)I - \gamma_+R_- - \gamma_\times R_/
\end{equation}
where this convergence (or mass distribution) is
\begin{equation}
\kappa(\theta) =
\frac12( \psi_{\theta_1\theta_1} + \psi_{\theta_2\theta_2} )
\end{equation}
and
\begin{aligned}
\gamma_+(\theta) &=
\frac12( \psi_{\theta_1\theta_1} - \psi_{\theta_2\theta_2} ) \\\
\gamma_\times(\theta) &= \psi_{\theta_1\theta_2}
\end{aligned}
We can then write
\begin{equation}
\mathcal{A}(\boldsymbol{\theta})
=
\begin{bmatrix}
1 - \kappa - \gamma_+ & - \gamma_\times \\\
- \gamma_\times & 1 - \kappa + \gamma_+
\end{bmatrix}
\end{equation}
It is then straight forward to prove that the eigenvalues of
$\mathcal{A}(\boldsymbol{\theta})$ are
\begin{equation}
\lambda_\pm =
(1 - \kappa) \pm \sqrt{\gamma_+^2 + \gamma_\times^2}
\end{equation}
and we can verify that the inverse is given as
\begin{equation}
\mathcal{A}^{-1}(\boldsymbol{\theta})
=
\frac1{(1-\kappa)^2 - \gamma_+^2 - \gamma_\times^2}
\begin{bmatrix}
1 - \kappa + \gamma_+ & \gamma_\times \\\
\gamma_\times & 1 - \kappa - \gamma_+
\end{bmatrix}
\end{equation}
The magnification of an image is given as \begin{equation} \mu(\boldsymbol{\theta}) = \det \mathcal{A}^{-1}(\boldsymbol{\theta}) = \frac1{\det\mathcal{A}} = \lambda_-\lambda_+ = \frac1{(1-\kappa)^2 - \gamma_+^2 - \gamma_\times^2} \end{equation}
TODO What is the significance of this?
TODO Is this consistent with the inverse magnification given in some sources? The inverse magnification is \begin{equation} \mu^{-1}(\theta) = \frac{\beta}{\theta}\cdot\frac{d\beta}{d\theta} \end{equation}
The critical curve is defined as the set of points where the map from source to image plane breaks down, i.e. when the magnification tends to infinity. Thus the critical curve is the solution of the equation $\det\mathcal{A}=0$, or \begin{equation} 0 = \lambda_-\lambda_+ = (1-\kappa)^2 - \gamma_+^2 - \gamma_\times^2 \end{equation}
TODO Define caustic