# Trouble finding the matrix form of potential energy in small oscillations (Goldstein linear triatomic molecule example)

I'm currently trying to learn small oscillations, I kind of comprehend the general theory, but I'm having hard times finding the matrix forms of the potential and kinetic energy. I have been following the Goldstein book, but it doesn't give any explanation about this step, and I'm pretty sure it's obvious, but I can't seem to get it.

The problem is: Starting from the potential energy:

$$\begin{equation} V=\frac{k}{2}\left(x_{2}-x_{1}-b\right)^{2}+\frac{k}{2}\left(x_{3}-x_{2}-b\right)^{2} \end{equation}$$ Coordinate relative to equilibrium position, $$\begin{equation} \eta_{i}=x_{i}-x_{0 i} \end{equation}$$ where, $$\begin{equation} x_{02}-x_{01}=b=x_{03}-x_{02} \end{equation}$$ the potential energy is reduced to: $$\begin{equation} V=\frac{1}{2} k\left(\eta_{2}-\eta_{1}\right)^{2}+\frac{1}{2} k\left(\eta_{3}-\eta_{2}\right)^{2} \end{equation}$$ Developing $$\begin{equation} V=\frac{1}{2} k\left(\eta_{2}^{2}-2 \eta_{2} \eta_{1}+\eta_{2}^{2}+\eta_{3}^{2}-\eta_{2}^{2}+2 \eta_{3} \eta_{2}\right) \end{equation}$$ $$\begin{equation} V=\frac{1}{2}k\left(\eta_{1}^{2}+2 \eta_{2}^{2}+\eta_{3}^{2}-2 \eta_{1} \eta_{2}-2 \eta_{2} \eta_{3}\right) \end{equation}$$ I understand till here, what I don't know is how I'm supposed to pass this to the Matrix form. Specifically: $$\begin{equation} \mathbf{V}=\left(\begin{array}{rrr} k & -k & 0 \\ -k & 2 k & -k \\ 0 & -k & k \end{array}\right) \end{equation}$$

Any specification, help or tip will be very much appreciated.

• the matrix elements are $\mathbf V{ij}=\dfrac{\partial }{\partial x_{i}}\left( \dfrac{\partial V}{\partial x_{j}}\right)$
– Eli
May 25 at 17:03
• Thanks, it works indeed, but I'm really supposed to calculate each element by that formula. Why I'm doing all the previous steps to arrive to $V=\frac{1}{2} k\left(\eta_{1}^{2}+2 \eta_{2}^{2}+\eta_{3}^{2}-2 \eta_{1} \eta_{2}-2 \eta_{2} \eta_{3}\right)$ then? May 25 at 18:03
• $\eta_i$ are the coordinates relative to equilibrium position , replace in the above equation x by $\eta$ you obtain the same matrix. notice that the stiffness matrix is always symmetric
– Eli
May 25 at 18:53

$$\begin{equation} \tag{1} V=\frac{1}{2}k\left(\eta_{1}^{2}+2 \eta_{2}^{2}+\eta_{3}^{2}-2 \eta_{1} \eta_{2}-2 \eta_{2} \eta_{3}\right) \end{equation}$$
We define the vector $$\vec \eta = \begin{pmatrix} \eta_1\\ \eta_2\\ \eta_3 \end{pmatrix}; \,\, \text{ and }\,\, \vec \eta^T = \begin{pmatrix} \eta_1 & \eta_2 & \eta_3 \end{pmatrix}$$
The potential of Eq.(1) is then written as: \begin{align} V =& \begin{pmatrix} \eta_1 & \eta_2 & \eta_3 \end{pmatrix} \left(\begin{array}{rrr} k & -k & 0 \\ -k & 2 k & -k \\ 0 & -k & k \end{array}\right) \begin{pmatrix} \eta_1\\ \eta_2\\ \eta_3 \end{pmatrix}\\ =& \vec \eta^T \mathbf{V}\vec{\eta} \end{align}