# Finding the components of the tensor for potential and kinetic energy

I have a rather poor understanding of what a tensor is, but enough to apply it to the biggest part of the classical mechanics I'm studying.
However, I've run into a small problem while studying "Free vibrations of a linear triatomic molecule". (We're assuming all 3 atoms are in 1 straight line and the forces they exert on each other are represented by springs with constant values for $k$).

My potential energy is described as:

$$V=\frac{k}{2}(\eta_1^2+2\eta_2^2+\eta_3^2-2\eta_1\eta_2-2\eta_2\eta_3)$$
Where $\eta$ is a coordinate relative to the equilibrium position.

And 'hence' the tensor has the form:

$$\vec V=\begin{bmatrix} k & -k & 0\\ -k & 2k & -k\\ 0 & -k & k \end{bmatrix}$$

How do I go from 1 to the other and vice versa?
I've tried googling terms like "Equation to tensor", "Potential energy tensor", "Absolute value of tensor" etc. but they didn't yield anything usable for me.

• The potential energy is $\frac{1}{2} \eta^T V \eta$, with $\eta^T = (\eta_1, \eta_2, \eta_3)$. – auxsvr Jun 11 '14 at 14:29

I believe this is the "missing link", stated in a less abstract fashion than in the above comment: https://en.wikipedia.org/wiki/Quadratic_form . Some programs in physics cover that in undergraduate algebra courses, some leave it for later. Notably, this method doesn't apply just to tensors, it's a general connection between symmetric matrices (of spaces $\mathcal{M}_{n\times n}({ℝ})$, since they are quadratic) and equations.
If you have an expression like $Q\left({\xi }_{1},\dots ,{\xi }_{n}\right)=\sum _{{i}_{i}j=1}^{n}{\alpha }_{ij}{\xi }_{i}{\xi }_{j}$ , which you do since the lagrangian of small oscillations has quadratic terms of both $x$ and $\stackrel{.}{x}$, you may transform the quadratic form of the potential energy (your first expression) into a symmetric matrix by putting values of ${\alpha }_{ii}$ on the diagonal (in this case the coefficients are ${\alpha }_{11}=k$, ${\alpha }_{22}=2k$, and ${\alpha }_{33}=k$, and, because the coefficients are symmetric (see definition), the matrix is symmetric as well. Hence, other elements are put in their appropriate position ( in this case ${\alpha }_{12}={\alpha }_{21}=-k$, ${\alpha }_{23}={\alpha }_{32}=-k$, ${\alpha }_{13}={\alpha }_{31}=0$). Note also that the matrix coefficients of nondiagonal elements are half of the ${\alpha }_{ij}$ coefficients from the quadratic form, since they are represented twice in the matrix.
Your squared general coordinates correspond with the tensors diagonal axis. The others are the cross terms. Since we are dealing with commuting terms, it will always be of the form $2\eta_{i}\eta_{j}$ although it is truly $\eta_{i}\eta_{j}+\eta_{j}\eta_{i}$. Obviously, the $\frac{1}{2}$ is absorbed by the energy equation.