spring representation of graphs Suppose we have a finite graph $G$ which we want to embed in ${\bf R}^d$; fix the positions of some nodes and connect all the nodes of the graphs with ideal springs of varying strength; (i.e. there is a force between every two nodes bringing them together where the magnitude is constant times the euclidean distnance between the two nodes), and consider the equilibrium position (i.e. all non-fixed nodes have net force 0).  
Now suppose that we take two adjacent nodes $a$ and $b$ whose positions were not fixed, and increase the strength of the spring that connects them, and consider the new equilibrium.  How do I show that the length of the edge between $a$ and $b$ necessrarily decreases?
I know that the "Energy" (sum of (strength * square of length of edge)) is a convex function; the energy always increases when the edge weights increase, and that if $x$ and $y$ are the equilibria before and after increasing the edge length, the energy in $x$ is greater than the energy in $y$. 
 A: Here is the sketch of a low tech solution:
Suppose that the graph is embedded with $ \vec{x}_i $ for $ i = 1 \cdots n $ the coordinates of the $i$th vertex. Let $ k_{ij} $ be the spring constant connecting vertices $ i $ and $ j$. Suppose the first $ m $ vertices are free, and the remaining $ n-m $ are fixed. You can check that the equilibrium position $ \vec{x} $ solves the matrix equation
$$
\begin{pmatrix} 
-\sum_{j = 1}^{n} k_{1j}  & k_{12} & \cdots & k_{1m} \\
k_{21} & -\sum_j k_{2j} & \cdots & k_{2m}  \\
\vdots & & \ddots  \\
 k_{m1}& k_{m2} & \cdots & -\sum_j k_{mj}
\end{pmatrix}
\begin{pmatrix}
\vec{x}_1 \\
\vec{x}_2 \\
\vdots \\
\vec{x}_m
\end{pmatrix}
= -\begin{pmatrix} \sum_{j = m+1}^n  k_{1j} \vec{x}_j \\ \sum_j k_{2j} \vec{x}_j \\ \vdots \\ \sum_j k_{mj} \vec{x}_j
\end{pmatrix}
$$
Write the above equation as $ M x = v $. Suppose you are increasing $ k_{12} $. Differentiate the above equation with respect to $k_{12} $ to find:
$$ 
\frac{dM}{d  k_{12} } x + M \frac{d x}{dk_{12}} =  0 
$$
Now since $ \frac{dM}{d  k_{12} } $ looks like: $$
\begin{pmatrix}
-1 & 1 & 0 & \cdots \\
1 & -1 & 0 & \cdots \\
0 & 0 & 0 & \cdots \\
\vdots & & \ddots
\end{pmatrix} $$
You can interpret the equation:
$$ 
 M \frac{d x}{dk_{12}} =  -\frac{dM}{d  k_{12} } x
$$
as solving for the equilibrium position of the spring system with the same spring constants, with only $ \frac{d x_1}{dk_{12}} $ and $ \frac{d x_2}{dk_{12}} $ attached by springs to fixed external points at $ \pm (x_2 - x_1) $ . You can finish up to argue that it follows that $ \frac{d x}{dk_{12}} $ decreases the distance between $ x_1 $ and $ x_2 $. 
