Lensing Jacobian

Jacobian, convergence, shear, and magnification

The lens equation maps an image-plane position \(\mathbf{\theta}\) to a source-plane position \(\mathbf{\beta}\):

\mathbf{\beta}(\mathbf{\theta})=\mathbf{\theta}-\mathbf{\alpha}(\mathbf{\theta}).

For a very small image patch, the lens equation can be linearized. The Jacobian matrix is the local derivative of this map:

A_{ij}(\mathbf{\theta}) = \frac{\partial \beta_i}{\partial \theta_j} = \delta_{ij} - \frac{\partial \alpha_i}{\partial \theta_j}.

Here \(i\) labels the component of the output vector \(\mathbf{\beta}\), or equivalently the row of the Jacobian matrix. The index \(j\) labels which input coordinate \(\theta_j\) is being varied, or equivalently the column of the matrix. In the two-dimensional lens plane, \(i,j\in\{1,2\}\), corresponding to the two angular coordinates on the sky.

So \(A\) answers a local question: if I move a little in the observed image plane, how much does the corresponding source-plane position move?

1. From the lensing potential to the Jacobian

The deflection angle is the gradient of the lensing potential,

\mathbf{\alpha}=\nabla\psi.

Therefore its derivative is the Hessian matrix of \(\psi\):

\frac{\partial \alpha_i}{\partial \theta_j} = \frac{\partial^2\psi}{\partial\theta_i\partial\theta_j} = \psi_{,ij}.

Thus

A = I-\nabla\nabla\psi = \begin{pmatrix} 1-\psi_{,11} & -\psi_{,12}\\ -\psi_{,12} & 1-\psi_{,22} \end{pmatrix}.

The Hessian is symmetric because it comes from second derivatives of one scalar potential. A symmetric \(2\times2\) matrix has three independent numbers. Lensing names those three numbers \(\kappa\), \(\gamma_1\), and \(\gamma_2\).

2. Why this becomes \(\kappa\) and \(\gamma\)

Any symmetric \(2\times2\) matrix can be split into an isotropic trace part plus a traceless anisotropic part. For the lensing potential Hessian, we define

\kappa = \frac{1}{2}\left(\psi_{,11}+\psi_{,22}\right),\qquad \gamma_1 = \frac{1}{2}\left(\psi_{,11}-\psi_{,22}\right),\qquad \gamma_2 = \psi_{,12}.

These definitions mean

\nabla\nabla\psi = \begin{pmatrix} \kappa+\gamma_1 & \gamma_2\\ \gamma_2 & \kappa-\gamma_1 \end{pmatrix}.

The trace part \(\kappa\) is called convergence. It is the locally isotropic focusing term: it changes the size of a small image without choosing a preferred direction. In lensing it is also related to the projected surface mass density by \(\kappa=\Sigma/\Sigma_{\mathrm{cr}}\).

The traceless part is shear. \(\gamma_1\) stretches one coordinate direction while compressing the perpendicular one; \(\gamma_2\) does the same after rotating the axes by \(45^\circ\). Together,

\gamma=\sqrt{\gamma_1^2+\gamma_2^2}

measures the total shear amplitude.

Substituting the decomposition into \(A=I-\nabla\nabla\psi\) gives

A(\mathbf{\theta}) = \frac{\partial\mathbf{\beta}}{\partial\mathbf{\theta}} = \begin{pmatrix} 1-\kappa-\gamma_1 & -\gamma_2\\ -\gamma_2 & 1-\kappa+\gamma_1 \end{pmatrix}.

So the Jacobian is not an extra physical object beyond \(\kappa\) and \(\gamma\). It is the local lens mapping written after decomposing the potential Hessian into isotropic focusing plus anisotropic shear.

3. Why magnification is \(1/\det A\)

The matrix \(A\) maps a small displacement in the image plane to a small displacement in the source plane:

\mathrm{d}\mathbf{\beta}=A\,\mathrm{d}\mathbf{\theta}.

For a linear map in two dimensions, areas are multiplied by the absolute value of the determinant. Therefore

\mathrm{d}^2\beta = |\det A|\,\mathrm{d}^2\theta.

But magnification compares the observed image area with the unlensed source area. Solving the previous relation for the image area gives

\frac{\mathrm{d}^2\theta}{\mathrm{d}^2\beta} = \frac{1}{|\det A|}.

This is why the flux magnification is the inverse absolute determinant. In lensing theory one often keeps the sign and writes the signed magnification as

\mu(\mathbf{\theta}) = \frac{1}{\det A(\mathbf{\theta})}.

The sign records image parity. If \(\det A<0\), the local mapping flips orientation. The observable flux magnification is \(|\mu|\).

Finally, using the matrix above,

\det A = (1-\kappa-\gamma_1)(1-\kappa+\gamma_1)-\gamma_2^2 = (1-\kappa)^2-\gamma_1^2-\gamma_2^2.

With \(\gamma^2=\gamma_1^2+\gamma_2^2\), this becomes

\mu(\mathbf{\theta}) = \frac{1}{(1-\kappa)^2-\gamma^2}.

雅可比、会聚、剪切与放大率

透镜方程把像平面上的位置 \(\mathbf{\theta}\) 映射到源平面上的位置 \(\mathbf{\beta}\)：

\mathbf{\beta}(\mathbf{\theta})=\mathbf{\theta}-\mathbf{\alpha}(\mathbf{\theta}).

如果只看一小块图像，这个映射可以近似成线性映射。雅可比矩阵就是这个局部映射的导数：

A_{ij}(\mathbf{\theta}) = \frac{\partial \beta_i}{\partial \theta_j} = \delta_{ij} - \frac{\partial \alpha_i}{\partial \theta_j}.

这里 \(i\) 标记输出向量 \(\mathbf{\beta}\) 的第几个分量，也就是雅可比矩阵的第几行；\(j\) 标记正在改变哪个输入坐标 \(\theta_j\)，也就是矩阵的第几列。在二维透镜平面里，\(i,j\in\{1,2\}\)，对应天空上的两个角坐标方向。

所以 \(A\) 回答的是一个局部问题：如果我在观测到的像平面上移动一点，对应的源平面位置会移动多少？

1. 从透镜势到雅可比矩阵

偏折角是透镜势的梯度：

\mathbf{\alpha}=\nabla\psi.

因此偏折角的导数就是 \(\psi\) 的 Hessian 矩阵：

\frac{\partial \alpha_i}{\partial \theta_j} = \frac{\partial^2\psi}{\partial\theta_i\partial\theta_j} = \psi_{,ij}.

于是

A = I-\nabla\nabla\psi = \begin{pmatrix} 1-\psi_{,11} & -\psi_{,12}\\ -\psi_{,12} & 1-\psi_{,22} \end{pmatrix}.

Hessian 是对同一个标量势做二阶导数得到的，所以它是对称矩阵。一个对称的 \(2\times2\) 矩阵有三个独立数。在引力透镜里，这三个数被命名为 \(\kappa\)、\(\gamma_1\)、\(\gamma_2\)。

2. 为什么它会变成 \(\kappa\) 和 \(\gamma\)

任何对称 \(2\times2\) 矩阵都可以拆成一个各向同性的 trace 部分，加上一个没有 trace 的各向异性部分。对透镜势的 Hessian，我们定义

\kappa = \frac{1}{2}\left(\psi_{,11}+\psi_{,22}\right),\qquad \gamma_1 = \frac{1}{2}\left(\psi_{,11}-\psi_{,22}\right),\qquad \gamma_2 = \psi_{,12}.

这些定义等价于

\nabla\nabla\psi = \begin{pmatrix} \kappa+\gamma_1 & \gamma_2\\ \gamma_2 & \kappa-\gamma_1 \end{pmatrix}.

trace 部分 \(\kappa\) 叫会聚。它描述局部各向同性的聚焦：小图像整体变大或变小，但不选择某个特殊方向。在透镜中它还与投影面质量密度有关：\(\kappa=\Sigma/\Sigma_{\mathrm{cr}}\)。

没有 trace 的部分就是剪切。\(\gamma_1\) 表示沿一个坐标方向拉伸、沿垂直方向压缩；\(\gamma_2\) 表示把坐标轴转过 \(45^\circ\) 后的同类变形。二者合起来，

\gamma=\sqrt{\gamma_1^2+\gamma_2^2}

表示总剪切强度。

把这个分解代入 \(A=I-\nabla\nabla\psi\)，就得到

A(\mathbf{\theta}) = \frac{\partial\mathbf{\beta}}{\partial\mathbf{\theta}} = \begin{pmatrix} 1-\kappa-\gamma_1 & -\gamma_2\\ -\gamma_2 & 1-\kappa+\gamma_1 \end{pmatrix}.

所以雅可比矩阵并不是在 \(\kappa\) 和 \(\gamma\) 之外另加的物理量。它只是把局部透镜映射写出来以后，再把势的 Hessian 分解成各向同性聚焦和各向异性剪切。

3. 为什么放大率是 \(1/\det A\)

矩阵 \(A\) 把像平面里的小位移映射到源平面里的小位移：

\mathrm{d}\mathbf{\beta}=A\,\mathrm{d}\mathbf{\theta}.

在二维线性映射中，面积会乘以行列式的绝对值。因此

\mathrm{d}^2\beta = |\det A|\,\mathrm{d}^2\theta.

但是放大率比较的是观测到的像面积和未被透镜放大的源面积。把上式解出像面积，得到

\frac{\mathrm{d}^2\theta}{\mathrm{d}^2\beta} = \frac{1}{|\det A|}.

这就是为什么通量放大率是行列式绝对值的倒数。在透镜理论里，也常保留符号，写成 signed magnification：

\mu(\mathbf{\theta}) = \frac{1}{\det A(\mathbf{\theta})}.

这个符号记录图像 parity。如果 \(\det A<0\)，局部映射会翻转取向。真正的通量放大倍数是 \(|\mu|\)。

最后，对上面的矩阵取行列式：

\det A = (1-\kappa-\gamma_1)(1-\kappa+\gamma_1)-\gamma_2^2 = (1-\kappa)^2-\gamma_1^2-\gamma_2^2.

记 \(\gamma^2=\gamma_1^2+\gamma_2^2\)，就得到

\mu(\mathbf{\theta}) = \frac{1}{(1-\kappa)^2-\gamma^2}.