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:

enter image description here

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.

  • $\begingroup$ the matrix elements are $\mathbf V{ij}=\dfrac{\partial }{\partial x_{i}}\left( \dfrac{\partial V}{\partial x_{j}}\right) $ $\endgroup$
    – Eli
    May 25 at 17:03
  • $\begingroup$ 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? $\endgroup$
    – James
    May 25 at 18:03
  • $\begingroup$ $\eta_i$ 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 $\endgroup$
    – Eli
    May 25 at 18:53

Starting from your potential form:

\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}

This is the definition of matix potential.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.