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.
 A: 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.
Hope this helps, if needed I will give some more examples.
A: 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.
