Derivation Notes on the transverse vibration modes of carbon ring

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Derivation Notes on the transverse vibration modes of carbon ring

Last updated: April, 1, 2014

Vectorial formalism for harmonic oscillator

Two massive points, namely <math>m_1</math> and <math>m_2</math> are connected by a spring of force constant <math>k</math> and unstretched length <math>l_0</math>. Now, if the displacements of the two massive points from their equilibrium positions are <math>\vec{u}_1</math> and <math>\vec{u}_2</math>. The force exerted on massive point 1 by the spring simply reads

<math>\begin{align}

f_{2\rightarrow 1}=-k{(\vec{u}_1-\vec{u}_2)} \end{align}</math>

which is slightly different from the form in the dissertation.

Equation of motion and parity symmetry

Now let us consider a carbon ring of 8 atoms, and use the convention of the plot on P.62 of the dissertation, one may write down explicitly four equations of motion for the massive points 0 - 3.

<math>\begin{align}

&m\ddot{\vec{u}}_0=-(\kappa+\lambda_1)({\vec{u}}_0-{\vec{u}}_1)-\kappa({\vec{u}}_0-{\vec{u}}_7)\\ &m\ddot{\vec{u}}_1=-(\kappa+\lambda_1)({\vec{u}}_1-{\vec{u}}_0)-\kappa({\vec{u}}_1-{\vec{u}}_2)\\ &m\ddot{\vec{u}}_2=-(\kappa-\lambda_2)({\vec{u}}_2-{\vec{u}}_3)-\kappa({\vec{u}}_2-{\vec{u}}_1)\\ &m\ddot{\vec{u}}_3=-(\kappa-\lambda_2)({\vec{u}}_3-{\vec{u}}_2)-\kappa({\vec{u}}_3-{\vec{u}}_4) \end{align}</math>

Obviously, the above equations do not close, since it involves <math>\vec{u}_4</math> and <math>\vec{u}_7</math>. To close these equations, one may assume the following parity symmetry (reflection with respect to the origin).

<math>\begin{align}

&\vec{u}_4=-\vec{u}_0\\ &\vec{u}_5=-\vec{u}_1\\ &\vec{u}_6=-\vec{u}_2\\ &\vec{u}_7=-\vec{u}_3 \end{align}</math>

It is straightforward to verify that the above assumption guarantees that the center of the mass of the system remains motionless. As a results, one obtains the following set of four equations:

<math>\begin{align}

&m\ddot{\vec{u}}_0=-(\kappa+\lambda_1)({\vec{u}}_0-{\vec{u}}_1)-\kappa({\vec{u}}_0+{\vec{u}}_3)\\ &m\ddot{\vec{u}}_1=-(\kappa+\lambda_1)({\vec{u}}_1-{\vec{u}}_0)-\kappa({\vec{u}}_1-{\vec{u}}_2)\\ &m\ddot{\vec{u}}_2=-(\kappa-\lambda_2)({\vec{u}}_2-{\vec{u}}_3)-\kappa({\vec{u}}_2-{\vec{u}}_1)\\ &m\ddot{\vec{u}}_3=-(\kappa-\lambda_2)({\vec{u}}_3-{\vec{u}}_2)-\kappa({\vec{u}}_3+{\vec{u}}_0) \end{align}</math>

Solution of mirror symmetry

Some solution of the above equations can be obtained by further assuming the following mirror symmetry (reflection with respect to <math>x</math> or <math>y</math> axis), namely,

<math>\begin{align}

&\vec{u}_1=P_y \vec{u}_0\\ &\vec{u}_2=P_x \vec{u}_3 \end{align}</math>

where <math>P_x</math> <math>P_y</math> are the operations which mirror the displacement <math>\vec{u}</math> with respect to <math>x</math> and <math>y</math> axis. For instance, <math>P_x (u_x, u_y) = (u_x, -u_y)</math>. It is necessarily and straightforwardly to verify that, the above four equations reduces to the following two equations.

<math>\begin{align}

&m\ddot{\vec{u}}_1=-(\kappa+\lambda_1)((1-P_y){\vec{u}}_1)-\kappa({\vec{u}}_1-{\vec{u}}_2)\\ &m\ddot{\vec{u}}_2=-(\kappa-\lambda_2)((1-P_x){\vec{u}}_2)-\kappa({\vec{u}}_2-{\vec{u}}_1) \end{align}</math>

At this point, it is worth noting that one can introduce "any symmetry" as one likes, for instance <math>\vec{u}_1=\pm\vec{u}_0</math> and <math>\vec{u}_2=\pm\vec{u}_3</math>, but usually the four equations will become conflict against one another if the guessed symmetry is not "good" enough. The example guess <math>\vec{u}_1=\pm\vec{u}_0</math> and <math>\vec{u}_2=\pm\vec{u}_3</math> in fact lead to conflict between the equations (instead of reducing to two pairs of identical equations).

By assuming the form <math>{\vec{u}}={\vec{u}}_0 e^{i\omega t}</math>, one gets

<math>\begin{align}

&-m\omega^2{\vec{u}}_1=-(\kappa+\lambda_1)((1-P_y){\vec{u}}_1)-\kappa({\vec{u}}_1-{\vec{u}}_2)\\ &-m\omega^2{\vec{u}}_2=-(\kappa-\lambda_2)((1-P_x){\vec{u}}_2)-\kappa({\vec{u}}_2-{\vec{u}}_1) \end{align}</math>

which implies

<math>\begin{align}

{\vec{u}}_1=\frac{1}{\kappa^2}[(\kappa+\lambda_1)(1-P_y)+\kappa+m\omega^2][(\kappa-\lambda_2)(1-P_x)+\kappa+m\omega^2]{\vec{u}}_1 \end{align}</math>

To solve it, one may write the above equation in terms of components and leads to two equations

<math>\begin{align}

&Template:U_{1x}=\frac{1}{\kappa^2}[2(\kappa+\lambda_1)+\kappa+m\omega^2][\kappa+m\omega^2]Template:U_{1x}\\ &Template:U_{1y}=\frac{1}{\kappa^2}[\kappa+m\omega^2][2(\kappa-\lambda_2)+\kappa+m\omega^2]Template:U_{1y} \end{align}</math>

Since the two equations do not attain the non-trivial solution simultaneously. This implies the following two sets of isolated modes

<math>\begin{align}

&Template:U_{1x}\ne 0, Template:U_{1y}=0\\ &Template:U_{1y}\ne 0, Template:U_{1x}=0 \end{align}</math>

Each of which gives two different solutions for <math>\omega</math>.

Solutions with less symmetry