双极坐标系的面积元

本文的推导过程不是用极为简便的主流方法，仅供参考。
参考数学百科_双极坐标系。

本文发布于csdn和知乎，除此之外不做发表。

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标记

记
$d(F_1,F_2)=r$
$d(F_1,P)=r_1$
$d(F_2,P)=r_2$
$\cos \theta (r1,r_2)=\frac{r_1^2+r_2^2-r^2}{2r_1{r}_2}$
$\cos {\theta }_1(r_1,r_2)=\frac{r^2+r_1^2-r_2^2}{2r{r}_1}$
$\cos {\theta }_2(r_1,r_2)=\frac{r^2+r_2^2-r_1^2}{2r{r}_2}$
$x_1(r_1,r_2)=\frac{r^2+r_1^2-r_2^2}{2r}$
$x_2(r_1,r_2)=\frac{r^2+r_2^2-r_1^2}{2r}$
请注意，$-x_2=x_1-r$
若无说明，则状态默认为$(r_1,r_2)$。
另记
$S=\frac{1}{4}\sqrt{r^2(-r^2+r_1^2+r_2^2)+r_1^2(r^2-r_1^2+r_2^2)+r_2^2(r^2+r_1^2-r_2^2)}$

思路

给$r_1$和$r_2$增量，求出如图所示的空出面积即可。

完整的过程

注意到，
$dS$

$={\int }_{x_1(r_1,r_2)}^{x_1(r_1,r_2+d{r}_2)}\sqrt{r_1^2-x^2}dx$
$+{\int }_{x_1(r_1,r_2+d{r}_2)}^{x_1(r_1+d{r}_1,r_2+d{r}_2)}\sqrt{(r_2+d{r}_2{)}^2-(x-r{)}^2}dx$
$+{\int }_{x_1(r_1+d{r}_1,r_2+d{r}_2)}^{x_1(r_1+d{r}_1,r_2)}\sqrt{(r_1+d{r}_1{)}^2-x^2}dx$
$+{\int }_{x_1(r_1+d{r}_1,r_2)}^{x_1(r_1,r_2)}\sqrt{r_2^2-(x-r{)}^2}dx$

$={\int }_{x_1(r_1,r_2)}^{x_1(r_1,r_2+d{r}_2)}\sqrt{r_1^2-x^2}dx$
$+{\int }_{x_1(r_1,r_2+d{r}_2)-r}^{x_1(r_1+d{r}_1,r_2+d{r}_2)-r}\sqrt{(r_2+d{r}_2{)}^2-x^2}dx$
$+{\int }_{x_1(r_1+d{r}_1,r_2+d{r}_2)}^{x_1(r_1+d{r}_1,r_2)}\sqrt{(r_1+d{r}_1{)}^2-x^2}dx$
$+{\int }_{x_1(r_1+d{r}_1,r_2)-r}^{x_1(r_1,r_2)-r}\sqrt{r_2^2-x^2}dx$

$={\int }_{x_1(r_1,r_2)}^{x_1(r_1,r_2+d{r}_2)}\sqrt{r_1^2-x^2}dx$
$-{\int }_{x_2(r_1,r_2+d{r}_2)}^{x_2(r_1+d{r}_1,r_2+d{r}_2)}\sqrt{(r_2+d{r}_2{)}^2-x^2}dx$
$+{\int }_{x_1(r_1+d{r}_1,r_2+d{r}_2)}^{x_1(r_1+d{r}_1,r_2)}\sqrt{(r_1+d{r}_1{)}^2-x^2}dx$
$-{\int }_{x_2(r_1+d{r}_1,r_2)}^{x_2(r_1,r_2)}\sqrt{r_2^2-x^2}dx$

$=({\int }_{x_1(r_1+d{r}_1,r_2+d{r}_2)}^{x_1(r_1+d{r}_1,r_2)}\sqrt{(r_1+d{r}_1{)}^2-x^2}dx+{\int }_{x_1(r_1,r_2)}^{x_1(r_1,r_2+d{r}_2)}\sqrt{r_1^2-x^2}dx)$
$-({\int }_{x_2(r_1+d{r}_1,r_2)}^{x_2(r_1,r_2)}\sqrt{r_2^2-x^2}dx+{\int }_{x_2(r_1,r_2+d{r}_2)}^{x_2(r_1+d{r}_1,r_2+d{r}_2)}\sqrt{(r_2+d{r}_2{)}^2-x^2}dx)$

$=(+{\int }_{x_1(r_1+d{r}_1,r_2+d{r}_2)}^{x_1(r_1+d{r}_1,r_2)}\sqrt{(r_1+d{r}_1{)}^2-x^2}dx$
$-{\int }_{x_1(r_1,r_2+d{r}_2)}^{x_1(r_1,r_2)}\sqrt{r_1^2-x^2}dx)$
$+(+{\int }_{x_2(r_1+d{r}_1,r_2+d{r}_2)}^{x_2(r_1,r_2+d{r}_2)}\sqrt{(r_2+d{r}_2{)}^2-x^2}dx$
$-{\int }_{x_2(r_1+d{r}_1,r_2)}^{x_2(r_1,r_2)}\sqrt{r_2^2-x^2}dx)$

$=d{r}_1\cdot \frac{\partial }{\partial r_1}\cdot {\int }_{x_1(r_1,r_2+d{r}_2)}^{x_1(r_1,r_2)}\sqrt{r_1^2-x^2}dx$
$+d{r}_2\cdot \frac{\partial }{\partial r_2}\cdot {\int }_{x_2(r_1+d{r}_1,r_2)}^{x_2(r_1,r_2)}\sqrt{r_2^2-x^2}dx$

这里有两个对称的项，只要算出来一个即可。

${\int }_{x_1(r_1,r_2)}^{x_1(r_1,r_2+d{r}_2)}\sqrt{r_1^2-x^2}dx$

$=\frac{r_1^2}{2}(\arcsin \frac{x_1(r_1,r_2+d{r}_2)}{r_1}-\arcsin \frac{x_2(r_1,r_2)}{r_1})$
$+(\frac{x_1(r_1,r_2+d{r}_2)}{2}\sqrt{r_1^2-x_1^2(r_1,r_2+d{r}_2)}-\frac{x_1(r_1,r_2)}{2}\sqrt{r_1^2-x_1^2(r_1,r_2)})$

$=\frac{r_1^2}{2}($
$(\arcsin \cos {\theta }_1(r_1,r_2+d{r}_2)-\arcsin \cos {\theta }_1(r_1,r_2))$
$+\frac{1}{2}\cdot (\sin 2{\theta }_1(r_1,r_2+d{r}_2)-\sin 2{\theta }_1(r_1,r_2))$
$)$

$=d{r}_2\frac{r_1^2}{2}\frac{\partial }{\partial r_2}(\arcsin \cos {\theta }_1+\frac{1}{2}\cdot (\sin 2{\theta }_1))$

$=d{r}_2\frac{r_1^2}{2}($
$\frac{\partial \arcsin \cos {\theta }_1}{\partial \cos {\theta }_1}\frac{\partial \cos {\theta }_1}{\partial r_2}$
$+\frac{1}{2}\frac{\partial \sin 2{\theta }_1}{\partial 2{\theta }_1}\frac{\partial 2{\theta }_1}{\partial {\theta }_1}(\frac{\partial \cos {\theta }_1}{\partial {\theta }_1}{)}^{-1}\frac{\partial \cos {\theta }_1}{\partial r_2}$
$)$

$=d{r}_2\frac{r_1^2}{2}($
$\frac{\partial \arcsin \cos {\theta }_1}{\partial \cos {\theta }_1}\frac{\partial \frac{r^2+r_1^2-r_2^2}{2r{r}_1}}{\partial r_2}$
$+\frac{1}{2}\frac{\partial \sin 2{\theta }_1}{\partial 2{\theta }_1}\frac{\partial 2{\theta }_1}{\partial {\theta }_1}(\frac{\partial \cos {\theta }_1}{\partial {\theta }_1}{)}^{-1}\frac{\partial \frac{r^2+r_1^2-r_2^2}{2r{r}_1}}{\partial r_2}$
$)$

$=d{r}_2\frac{r_1^2}{2}($
$\frac{1}{\sin {\theta }_1}\frac{-2r_2}{2r{r}_1}$
$+\frac{1}{2}\cos 2{\theta }_{1}2\frac{1}{\sin {\theta }_1}\frac{-2r_2}{2r{r}_1}$
$)$

$=d{r}_2-\frac{r_1^2}{2}\frac{r_2}{r{r}_1}\frac{1}{\sin {\theta }_1}(1+\cos 2{\theta }_1)$

$=d{r}_2-\frac{r_1^2}{2}\frac{r_2}{r{r}_1}\frac{1}{\sin {\theta }_1}2{\sin }^2{\theta }_1$

$=d{r}_2-\frac{r_1{r}_2}{r}\sin {\theta }_1$

则

$\frac{\partial }{\partial r_1}\cdot {\int }_{x_1(r_1,r_2+d{r}_2)}^{x_1(r_1,r_2)}\sqrt{r_1^2-x^2}dx$

$=d{r}_2\frac{\partial }{\partial r_1}\cdot \frac{r_1{r}_2}{r}\sin {\theta }_1$

$=d{r}_2\frac{r_2}{r}\frac{\partial r_1\sin {\theta }_1}{\partial r_1}$

$=d{r}_2\frac{r_2}{r}(\sin {\theta }_1+r_1\frac{\partial \sin {\theta }_1}{\partial {\theta }_1}(\frac{\partial \cos {\theta }_1}{\partial {\theta }_1}{)}^{-1}\frac{\partial \cos {\theta }_1}{\partial r_1})$

$=d{r}_2\frac{r_2}{r}(\sin {\theta }_1+r_1\frac{\partial \sin {\theta }_1}{\partial {\theta }_1}(\frac{\partial \cos {\theta }_1}{\partial {\theta }_1}{)}^{-1}\frac{\partial \frac{r^2+r_1^2-r_2^2}{2r{r}_1}}{\partial r_1})$

$=d{r}_2\frac{r_2}{r}(\sin {\theta }_1+r_1\frac{\partial \sin {\theta }_1}{\partial {\theta }_1}(\frac{\partial \cos {\theta }_1}{\partial {\theta }_1}{)}^{-1}\frac{(2r{r}_1)\cdot \frac{\partial }{\partial r_1}(r^2+r_1^2-r_2^2)-(r^2+r_1^2-r_2^2)\cdot \frac{\partial }{\partial r_1}(2r{r}_1)}{(2r{r}_1{)}^2}$

$=d{r}_2\frac{r_2}{r}(\sin {\theta }_1+r_1\cos {\theta }_1\frac{1}{\sin {\theta }_1}\frac{r_1^2+r_2^2-r^2}{2r{r}_1^2})$

$=d{r}_2\frac{r_2}{r}(\sin {\theta }_1+\cot {\theta }_1\cos \theta \frac{r_2}{r})$

代入到原式。

$dS$
$=d{r}_{1}d{r}_2(\frac{r_2\sin {\theta }_1+r_1\sin {\theta }_2}{r}+\frac{\cos \theta }{r^2}(r_2^2\cot {\theta }_1+r_1^2\cot {\theta }_2))$
$=\frac{d{r}_{1}d{r}_2}{r^2}(\frac{S}{2}(\frac{r_2}{r_1}+\frac{r_1}{r_2})+\frac{r(r_1^2+r_2^2-r^2)}{4S}(r_1+r_2))$
这与学术界的结论不一致，也与之前自己推的结果不一致，但暂时不知道哪里算错了。
在之前计算过这样一个结果，可惜草稿纸找不到了。
记
$F=r^2+r_1^2+r_2^2$
$H=r^4+r_1^4+r_2^4$
有
$dS=d{r}_{1}d{r}_2\frac{(F-r^2)(\sqrt{F^2-2H}+F)-2(H-r4)}{2r^2{r}_1{r}_2}$