Deriving the divergence relation in lensing
This note explains the step
\[\nabla \cdot \mathbf{\alpha}(\mathbf{\theta}) = 2\kappa(\mathbf{\theta}),\]
which leads directly to the two-dimensional Poisson equation for the lensing potential,
\[\kappa(\mathbf{\theta}) = \frac{1}{2}\nabla^2\psi(\mathbf{\theta}).\]
1. Start from the scaled deflection angle
The scaled deflection angle can be written as
\[\mathbf{\alpha}(\mathbf{\theta})
=
\frac{1}{\pi}
\int \mathrm{d}^2\theta'\,
\kappa(\mathbf{\theta}')
\frac{\mathbf{\theta}-\mathbf{\theta}'}
{\left|\mathbf{\theta}-\mathbf{\theta}'\right|^2}.\]
This expression comes from the physical thin-lens deflection integral
\[\hat{\mathbf{\alpha}}(\mathbf{\xi})
=
\frac{4G}{c^2}
\int \mathrm{d}^2\xi'\,
\Sigma(\mathbf{\xi}')
\frac{\mathbf{\xi}-\mathbf{\xi}'}
{\left|\mathbf{\xi}-\mathbf{\xi}'\right|^2},\]
after changing from physical coordinates to angular coordinates and using
\[\kappa = \frac{\Sigma}{\Sigma_{\mathrm{cr}}}.\]
2. Take the two-dimensional divergence
The divergence operator is taken with respect to the image-plane coordinate \(\mathbf{\theta}\):
\[\nabla \equiv \nabla_{\mathbf{\theta}}.\]
The integration variable is \(\mathbf{\theta}'\), so \(\kappa(\mathbf{\theta}')\) is treated as fixed when differentiating with respect to \(\mathbf{\theta}\). Therefore,
\[\nabla\cdot\mathbf{\alpha}(\mathbf{\theta})
=
\frac{1}{\pi}
\int \mathrm{d}^2\theta'\,
\kappa(\mathbf{\theta}')
\nabla\cdot
\left[
\frac{\mathbf{\theta}-\mathbf{\theta}'}
{\left|\mathbf{\theta}-\mathbf{\theta}'\right|^2}
\right].\]
3. The key two-dimensional identity
Define
\[\mathbf{x} = \mathbf{\theta}-\mathbf{\theta}'.\]
Then the kernel becomes
\[\frac{\mathbf{x}}{|\mathbf{x}|^2}.\]
In polar coordinates around \(\mathbf{x}=0\), we write \(\mathbf{x}=r\hat{\mathbf{r}}\), where \(r=|\mathbf{x}|\) is the length of \(\mathbf{x}\), and \(\hat{\mathbf{r}}=\mathbf{x}/|\mathbf{x}|\) is the direction of \(\mathbf{x}\).
Away from \(\mathbf{x}=0\), this vector field is
\[\frac{\mathbf{x}}{|\mathbf{x}|^2}
=
\frac{\hat{\mathbf{r}}}{r}.\]
For a purely radial vector field \(\mathbf{A}=A_r(r)\hat{\mathbf{r}}\), the two-dimensional polar-coordinate divergence is
\[\nabla\cdot\mathbf{A}
=
\frac{1}{r}\frac{\partial}{\partial r}\left(rA_r\right).\]
Applying it with \(A_r=1/r\) gives
\[\nabla\cdot\left(\frac{\hat{\mathbf{r}}}{r}\right)
=
\frac{1}{r}\frac{\partial}{\partial r}
\left(r\frac{1}{r}\right)
=
\frac{1}{r}\frac{\partial}{\partial r}(1)
=
0.\]
So the ordinary divergence is zero everywhere except at the singular point \(\mathbf{x}=0\). Since \(\hat{\mathbf{r}}/r\) is not defined at the origin, the missing term has to be understood as a point contribution; the supporting background is in the delta-function note.
To find its normalization, integrate over a disk enclosing the origin and use the two-dimensional divergence theorem:
\[\int_{\mathrm{disk}} \mathrm{d}^2x\,
\nabla\cdot
\left(
\frac{\mathbf{x}}{|\mathbf{x}|^2}
\right)
=
\oint_{\partial\mathrm{disk}}
\frac{\mathbf{x}}{|\mathbf{x}|^2}
\cdot \hat{\mathbf{n}}\,\mathrm{d}s.\]
On a circle of radius \(R\),
\[\frac{\mathbf{x}}{|\mathbf{x}|^2}
=
\frac{\hat{\mathbf{r}}}{R},
\qquad
\mathrm{d}s = R\,\mathrm{d}\phi.\]
Thus the boundary flux is
\[\oint
\frac{\hat{\mathbf{r}}}{R}
\cdot
\hat{\mathbf{r}}\,
R\,\mathrm{d}\phi
=
\int_0^{2\pi}\mathrm{d}\phi
=
2\pi.\]
Therefore, as a distribution,
\[\nabla\cdot
\left[
\frac{\mathbf{x}}{|\mathbf{x}|^2}
\right]
=
2\pi\delta^{(2)}(\mathbf{x}).\]
Substituting back \(\mathbf{x}=\mathbf{\theta}-\mathbf{\theta}'\), we obtain
\[\nabla\cdot
\left[
\frac{\mathbf{\theta}-\mathbf{\theta}'}
{\left|\mathbf{\theta}-\mathbf{\theta}'\right|^2}
\right]
=
2\pi\delta^{(2)}(\mathbf{\theta}-\mathbf{\theta}').\]
4. Apply the identity to the deflection integral
Substituting this identity into the divergence of the deflection field gives
\[\nabla\cdot\mathbf{\alpha}(\mathbf{\theta})
=
\frac{1}{\pi}
\int \mathrm{d}^2\theta'\,
\kappa(\mathbf{\theta}')
2\pi\delta^{(2)}(\mathbf{\theta}-\mathbf{\theta}').\]
The delta function selects \(\mathbf{\theta}'=\mathbf{\theta}\), so
\[\nabla\cdot\mathbf{\alpha}(\mathbf{\theta})
=
2\kappa(\mathbf{\theta}).\]
5. Connect to the lensing potential
The lensing potential is defined so that
\[\mathbf{\alpha}(\mathbf{\theta})
=
\nabla\psi(\mathbf{\theta}).\]
Therefore,
\[\nabla\cdot\mathbf{\alpha}
=
\nabla\cdot\nabla\psi
=
\nabla^2\psi.\]
Combining this with
\[\nabla\cdot\mathbf{\alpha}=2\kappa\]
gives
\[\nabla^2\psi=2\kappa,\]
or equivalently
\[\kappa(\mathbf{\theta})
=
\frac{1}{2}\nabla^2\psi(\mathbf{\theta}).\]
推导引力透镜中的散度关系
这篇笔记解释下面这一步：
\[\nabla \cdot \mathbf{\alpha}(\mathbf{\theta}) = 2\kappa(\mathbf{\theta}),\]
它会直接导向透镜势的二维 Poisson 方程：
\[\kappa(\mathbf{\theta}) = \frac{1}{2}\nabla^2\psi(\mathbf{\theta}).\]
1. 从缩放偏折角出发
缩放后的偏折角可以写成
\[\mathbf{\alpha}(\mathbf{\theta})
=
\frac{1}{\pi}
\int \mathrm{d}^2\theta'\,
\kappa(\mathbf{\theta}')
\frac{\mathbf{\theta}-\mathbf{\theta}'}
{\left|\mathbf{\theta}-\mathbf{\theta}'\right|^2}.\]
这个表达式来自物理薄透镜偏折积分：
\[\hat{\mathbf{\alpha}}(\mathbf{\xi})
=
\frac{4G}{c^2}
\int \mathrm{d}^2\xi'\,
\Sigma(\mathbf{\xi}')
\frac{\mathbf{\xi}-\mathbf{\xi}'}
{\left|\mathbf{\xi}-\mathbf{\xi}'\right|^2},\]
经过从物理坐标到角坐标的变换，并使用
\[\kappa = \frac{\Sigma}{\Sigma_{\mathrm{cr}}}.\]
2. 取二维散度
散度算符是对像平面坐标 \(\mathbf{\theta}\) 取的：
\[\nabla \equiv \nabla_{\mathbf{\theta}}.\]
积分变量是 \(\mathbf{\theta}'\)，所以当我们对 \(\mathbf{\theta}\) 求导时，\(\kappa(\mathbf{\theta}')\) 被视为固定。因此，
\[\nabla\cdot\mathbf{\alpha}(\mathbf{\theta})
=
\frac{1}{\pi}
\int \mathrm{d}^2\theta'\,
\kappa(\mathbf{\theta}')
\nabla\cdot
\left[
\frac{\mathbf{\theta}-\mathbf{\theta}'}
{\left|\mathbf{\theta}-\mathbf{\theta}'\right|^2}
\right].\]
3. 关键的二维恒等式
定义
\[\mathbf{x} = \mathbf{\theta}-\mathbf{\theta}'.\]
于是核函数变成
\[\frac{\mathbf{x}}{|\mathbf{x}|^2}.\]
在以 \(\mathbf{x}=0\) 为中心的极坐标中，我们写作 \(\mathbf{x}=r\hat{\mathbf{r}}\)，其中 \(r=|\mathbf{x}|\) 是 \(\mathbf{x}\) 的长度，而 \(\hat{\mathbf{r}}=\mathbf{x}/|\mathbf{x}|\) 是 \(\mathbf{x}\) 的方向。
在 \(\mathbf{x}=0\) 之外，这个矢量场为
\[\frac{\mathbf{x}}{|\mathbf{x}|^2}
=
\frac{\hat{\mathbf{r}}}{r}.\]
对于纯径向矢量场 \(\mathbf{A}=A_r(r)\hat{\mathbf{r}}\)，二维极坐标中的散度公式是
\[\nabla\cdot\mathbf{A}
=
\frac{1}{r}\frac{\partial}{\partial r}\left(rA_r\right).\]
这里代入 \(A_r=1/r\)，因此
\[\nabla\cdot\left(\frac{\hat{\mathbf{r}}}{r}\right)
=
\frac{1}{r}\frac{\partial}{\partial r}
\left(r\frac{1}{r}\right)
=
\frac{1}{r}\frac{\partial}{\partial r}(1)
=
0.\]
所以普通意义下的散度在除奇点 \(\mathbf{x}=0\) 之外处处为零。由于 \(\hat{\mathbf{r}}/r\) 在原点没有定义，缺失项必须理解成一个点贡献；相关背景见二维 delta 函数解释。
为了确定它的归一化，在包围原点的圆盘上积分，并使用二维散度定理：
\[\int_{\mathrm{disk}} \mathrm{d}^2x\,
\nabla\cdot
\left(
\frac{\mathbf{x}}{|\mathbf{x}|^2}
\right)
=
\oint_{\partial\mathrm{disk}}
\frac{\mathbf{x}}{|\mathbf{x}|^2}
\cdot \hat{\mathbf{n}}\,\mathrm{d}s.\]
在半径为 \(R\) 的圆上，
\[\frac{\mathbf{x}}{|\mathbf{x}|^2}
=
\frac{\hat{\mathbf{r}}}{R},
\qquad
\mathrm{d}s = R\,\mathrm{d}\phi.\]
因此边界通量为
\[\oint
\frac{\hat{\mathbf{r}}}{R}
\cdot
\hat{\mathbf{r}}\,
R\,\mathrm{d}\phi
=
\int_0^{2\pi}\mathrm{d}\phi
=
2\pi.\]
因此，作为分布，
\[\nabla\cdot
\left[
\frac{\mathbf{x}}{|\mathbf{x}|^2}
\right]
=
2\pi\delta^{(2)}(\mathbf{x}).\]
代回 \(\mathbf{x}=\mathbf{\theta}-\mathbf{\theta}'\)，得到
\[\nabla\cdot
\left[
\frac{\mathbf{\theta}-\mathbf{\theta}'}
{\left|\mathbf{\theta}-\mathbf{\theta}'\right|^2}
\right]
=
2\pi\delta^{(2)}(\mathbf{\theta}-\mathbf{\theta}').\]
4. 将恒等式用于偏折积分
把这个恒等式代入偏折场的散度，得到
\[\nabla\cdot\mathbf{\alpha}(\mathbf{\theta})
=
\frac{1}{\pi}
\int \mathrm{d}^2\theta'\,
\kappa(\mathbf{\theta}')
2\pi\delta^{(2)}(\mathbf{\theta}-\mathbf{\theta}').\]
delta 函数选出 \(\mathbf{\theta}'=\mathbf{\theta}\)，因此
\[\nabla\cdot\mathbf{\alpha}(\mathbf{\theta})
=
2\kappa(\mathbf{\theta}).\]
5. 与透镜势联系起来
透镜势定义为满足
\[\mathbf{\alpha}(\mathbf{\theta})
=
\nabla\psi(\mathbf{\theta}).\]
因此，
\[\nabla\cdot\mathbf{\alpha}
=
\nabla\cdot\nabla\psi
=
\nabla^2\psi.\]
将它与
\[\nabla\cdot\mathbf{\alpha}=2\kappa\]
结合，得到
\[\nabla^2\psi=2\kappa,\]
或等价地，
\[\kappa(\mathbf{\theta})
=
\frac{1}{2}\nabla^2\psi(\mathbf{\theta}).\]