Advanced Classical Mechanics/Small Oscillations and Perturbed Motion

From testwiki
Jump to navigation Jump to search

In Linear Motion, we argued that all sufficiently small oscillations are harmonic. In this section we will exploit this result in several ways to understand

  1. The motion of systems with many degrees of freedom near equilibrium,
  2. The motion of systems perturbed from known solutions, and
  3. The motion of systems with Lagrangians perturbed from systems with known solutions.

All three of these points are applications of perturbation theory, and they all start with the harmonic oscillator.

Normal Modes

The modes of oscillation of systems near equilibrium are called the normal modes of the system. Understanding the frequencies of the normal modes of the system is crucial to design a system that can move (even if it isn't meant to). Let's look at a system with many degrees of freedom; we have

L=12i,jTijq˙iq˙jV(q1,qn).

Let q0,i be an equilibrium position and expand about this point qi=q0,i+ηi so q˙i=η˙i.

We can expand the potential energy to give

V(q1,qn)=V(q0,1,q0,n)+i(Vqi)q0,iηi+12i,j(2Vqiqj)q0,iηiηj+.

The first term is a constant with respect to ηi and constant terms do not affect the motion. The second term is zero, because q0,i is a point of equilibrium. If we drop all terms higher than the third term shown, we are left with

L=12i,j(Tijη˙iη˙jVijηiηj)

where

Tij=Tij(q0,1,q0,n) and Vij=(2Vqiqj)q0,i,

yielding the equations of motion

j(Tijη¨jVijηj)=0

This is a linear differential equation with constant coefficients. We can try the solution

ηi=Caieiωt

so we have

j(Vijajω2Tijaj)=0.

This is a matrix equation such that

Aa=0 with

a=[a1a2aj]

and

A=[V11ω2T11V12ω2T12V21ω2T21V22ω2T22]

This equation only has a solution is detA=0. This gives a nth-degree polynomial to solve for ω2. We will get n solutions for ω2 that we can substitute into the matrix equation and solve for aj.

Is this guaranteed to work? Yes, it turns out. Look at the equation in terms of matrices we have

Va=ω2Ta.

The matrix V is symmetric and real. The matrix T should be positive definite (because a negative kinetic energy doesn't make sense). Technical issue: If T has a null space, the degrees of freedom corresponding to the null space are massless and cannot be excited unless they are in the null space of V. Either way, you can drop the null space from both sides of the equation.

Assuming that T is invertable we have

(T)1Va=ω2a

and we have a standard eigenvalue equation. In most examples, the kinetic energy matrix will be diagonal, so it is straightforward to construct the quotient matrix and diagonize it.

Perturbations about Steady Motion

Let's say I have some solution to the equations of motion and I would like to look at small deviations from the solution. Let's q0,i(t) satisfy

ddtLq˙iLqi=0,

and let's look at

qi(t)=q0,i(t)+ηi(t)

where ηi is small. Let's expand the entire Lagrangian to find the equations of motion for the deviations ηi. We have

L(q1,qn)=L(q0,1,q0,n;q˙0,1,q˙0,n)+i[(Lqi)q0,iηi+(Lq˙i)q0,iη˙i]+

                12i,j[(2Lqiqj)q0,iηiηj+(2Lq˙iq˙j)q0,iη˙iη˙j+2(2Lqiq˙j)q0,iηiη˙j]+.

Now let's apply Lagrange's equations for the deviations

ddtLη˙iLηi=0

to give

ddt{(Lq˙j)q0,i+i[(2Lq˙iq˙j)q0,iη˙i+(2Lqiq˙j)q0,iηi]}

        (Lqj)q0,ii[(2Lqiqj)q0,iηi+(2Lqiq˙j)q0,iη˙i]=0

The two terms without η actually cancel each other out, leaving the following equations of motion.

ddt{i[(2Lq˙iq˙j)q0,iη˙i+(2Lqiq˙j)q0,iηi]}

        i[(2Lqiqj)q0,iηi+(2Lqiq˙j)q0,iη˙i]=0.

In steady motion, the partial derivatives are taken to be constant in time yielding the even simpler result

i[(2Lq˙iq˙j)q0,iη¨i(2Lqiqj)q0,iηi]=0.

Again we have a linear differential equation with constant coefficients, and all of the results from the previous section carry over.

Perturbed Lagrangians

What about finding solutions to Lagrangians that are almost like ones that we have already solved? Let's say we have

L=L0+L1

where L1 is considered to be small compared to L0 Let's say I have some solution to the equations of motion for L0 and I would like to look at small deviations from the solution induced by the change in the Lagrangian. Let's say q0,i(t) satisfy

ddtL0q˙iL0qi=0,

and let's look at

qi(t)=q0,i(t)+ηi(t)

where ηi is small. Let's expand the entire Lagrangian to find the equations of motion for the deviations ηi. We have

L(q1,qn)=L(q0,1,q0,n;q˙0,1,q˙0,n)+i[(Lqi)q0,iηi+(Lq˙i)q0,iη˙i]+

                12i,j[(2Lqiqj)q0,iηiηj+(2Lq˙iq˙j)q0,iη˙iη˙j+2(2Lqiq˙j)q0,iηiη˙j]+

                L1(q0,1,q0,n;q˙0,1,q˙0,n)+i[(L1qj)q0,iηi+(L1q˙j)q0,iη˙i]+.

Now let's apply Lagrange's equations for the deviations

ddtLη˙iLηi=0

to give

ddt{(L0q˙j)q0,i+(L1q˙j)q0,i+i[(2L0q˙iq˙j)q0,iη˙i+(2L0qiq˙j)q0,iηi]}

        (L0qj)q0,i+(L1qj)q0,i+i[(2L0qiqj)q0,iηi+(2L0qiq˙j)q0,iη˙i]=0

The two lowest orders terms without η actually cancel each other out, leaving the following equations of motion.

ddt{i[(2L0q˙iq˙j)q0,iη˙i+(2L0qiq˙j)q0,iηi]}

        i[(2L0qiqj)q0,iηi+(2L0qiq˙j)q0,iη˙i]=(L1qj)q0,iddt(L1q˙j)q0,i.

Let's specialize and assume that the unperturbed motion is steady so the partial derivatives of the unperturbed Lagrangian are constant in time, to obtain

i[(2L0q˙iq˙j)q0,iη¨i(2L0qiqj)q0,iηi]=(L1qj)q0,iddt(L1q˙j)q0,i.

which is the equation of a coupled set of driven harmonic oscillators. Template:Subpage navbar