This is from Hobson, Riley, Bence Mathematical Methods, p 322. A spring system is described as follows (they are floating in air like molecules):

enter image description here The equilibrium positions of four equal masses M of a square with sides 2L are $R_n=\pm L_i\pm L_j$ and displacements from equilibrium are $q_n=x_ni+y_nj$.

In the following equation for the potential energy, the assumption is made that displacement is much less than equilibrium position $|R_m-T_n|>>|q_m-q_n|$

enter image description here Plugging in each mass to (2) above gives the following:

enter image description here In the squared middle terms 1/2 is the normalization factor of the L's in $R_m-R_n$ and the reversal of minus signs for the Y's reflects the direction of the vectors in the image.

The text states that the Potential Matrix is thus constructed from the above but I don't see how: Where did these numbers come from? What does each row and column represent?

enter image description here


The energy of the system is conserved and this can be expressed in a compact form with vectors and a matrix:

$(M/2)\dot{X^2}+(k/2)X^TBX=E_0$, with $X^T=(x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4)$ and $B$ the matrix you draw. It's a huge sum of terms for the kinetic energy and the potential energy. Each element of the matrix comes from the development of the squared sums you have for the potential energy coming from the compresion/lengthen of the springs, so is, you are adding all the elements of the column tagged as $2b_{mn}/k$. E.g.



  • $\begingroup$ Should $b_{13}$ be $b_{12}$ and $y_1-y_4$?. This is very good. I understand now though. $\endgroup$ – user5389726598465 Jun 10 '17 at 8:07
  • $\begingroup$ No, they are not differences, they are products of coordinates. e.g. $b_{13}=b_{31}=-2$, added result in $-4$, corresponding to the coefficient of $x_1x_2$ $\endgroup$ – Rafa Budría Jun 10 '17 at 8:12
  • $\begingroup$ I see. The author infact did alternate $(x_1, y_1, x_2 \dots)$ in this matrix. $\endgroup$ – user5389726598465 Jun 10 '17 at 8:13
  • $\begingroup$ Yes, so is. Consider this too: $\dot{X^2}=\dot X^T\dot X=x_1^2+y_1^2+x_2^2\dots$ $\endgroup$ – Rafa Budría Jun 10 '17 at 8:18

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