Energy of a triangle
For a triangle \(ABC\), assume the points have potential \(\varphi_a,\varphi_b,\varphi_c\). As shown in Fig. ???, \(AB'\perp AC, AC'\perp AB\), combine the components of \(\nabla\varphi\), we have
$$(\nabla\varphi)^2=\frac{1}{\sin^2 A}\left(\frac{\varphi_{ab}^2}{c^2}+\frac{\varphi_{ac}^2}{b^2}-\frac{2\varphi_{ab}\varphi_{ac}\cos A}{bc}\right)$$
The area of triangle is
$$S=\frac{bc\sin A}{2}$$
The energy of this triangle is
$$\begin{aligned}
E_{\triangle}&\propto S(\nabla\varphi)^2\\
&\propto \frac{bc}{\sin A}\left(\frac{\varphi_{ab}^2}{c^2}+\frac{\varphi_{ac}^2}{b^2}-\frac{2\varphi_{ab}\varphi_{ac}\cos A}{bc}\right)\end{aligned}$$
对\(\varphi_a\)求导得到：
$$\frac{\pp E_\triangle}{\pp \varphi_a}\propto \left(\frac{b}{c\sin A}-\cot A\right)\varphi_{ab}+\left(\frac{c}{b\sin A}-\cot A\right)\varphi_{ac}\label{a}$$
其中：
$$\frac{b}{c\sin A}-\cot A=\frac{b-c\cos A}{c\sin A}=\frac{a\cos C}{a\sin C}=\cot C$$
同理
$$\frac{c}{b\sin A}-\cot A=\cot B$$
故偏导化为
$$\frac{\pp E_\triangle}{\pp \varphi_a}\propto \varphi_{ab}\cot C+\varphi_{ac}\cot B
\label{comp}$$
Stationary point equation for energy minimum
Let \(E\) be the total energy and use \(i\) to mark all triangles containing \(A\). \(B_i, C_i\) are other points in triangle \(i\). In the stationary point, we have
$$\frac{\pp E}{\pp \varphi_a}=\sum_i\frac{\pp E_i}{\pp \varphi_a}
\propto\sum_i \varphi_{ab_i}\cot C_i+\varphi_{ac_i}\cot B_i=0$$
So,
$$\sum_i\left(\cot C_i+\cot B_i\right)\varphi_{a}=\sum_i\left( \varphi_{b_i}\cot C_i+\varphi_{c_i}\cot B_i\right)$$
$$\Rightarrow\varphi_a=\frac{\sum_i \varphi_{b_i}\cot C_i+\varphi_{c_i}\cot B_i}{\sum_i\cot C_i+\cot B_i}$$
Assume we know coordinates of a triangle \(ABC\). To calculate \(\cot A\), we define \(\bm b=\overrightarrow{AB}, \bm c=\overrightarrow{AC}\).
$$\cot A=\frac{bc\cos A}{bc\sin A}=\frac{\bm{b\cdot c}}{|\bm{b\times c}|}=\frac{\bm{b\cdot c}}{2S}$$
The denominator \(2S\) is the same for a triangle.
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