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

$$\begin{aligned} E_{\triangle}&\propto S(\nabla\varphi)^2\\ &\propto \frac{bc}{\sin A}\left(\frac{\varphi_{ab}^2}{c^2}+\frac{\varphi_{ac}^2}{b^2}-\frac{2\varphi_{ab}\varphi_{ac}\cos A}{bc}\right)\end{aligned}$$


$$\frac{\pp E_\triangle}{\pp \varphi_a}\propto \left(\frac{b}{c\sin A}-\cot A\right)\varphi_{ab}+\left(\frac{c}{b\sin A}-\cot A\right)\varphi_{ac}\label{a}$$


$$\frac{b}{c\sin A}-\cot A=\frac{b-c\cos A}{c\sin A}=\frac{a\cos C}{a\sin C}=\cot C$$


$$\frac{c}{b\sin A}-\cot A=\cot B$$


$$\frac{\pp E_\triangle}{\pp \varphi_a}\propto \varphi_{ab}\cot C+\varphi_{ac}\cot B \label{comp}$$

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

$$\frac{\pp E}{\pp \varphi_a}=\sum_i\frac{\pp E_i}{\pp \varphi_a} \propto\sum_i \varphi_{ab_i}\cot C_i+\varphi_{ac_i}\cot B_i=0$$


$$\sum_i\left(\cot C_i+\cot B_i\right)\varphi_{a}=\sum_i\left( \varphi_{b_i}\cot C_i+\varphi_{c_i}\cot B_i\right)$$
$$\Rightarrow\varphi_a=\frac{\sum_i \varphi_{b_i}\cot C_i+\varphi_{c_i}\cot B_i}{\sum_i\cot C_i+\cot B_i}$$

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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