Understanding solution to equations of motion of n coupled oscillators

[zenterix]

Homework Statement

I'd like to understand a section of Ch. 3 "Normal Modes" of Georgi's "The Physics of Waves".

Relevant Equations

Below I show the reasoning shown in the section and I will directly quote the portions that generate my doubts.

Consider a system of $n$ coupled oscillators.

$n$ particles are constrained to move in the $x$ direction.

$x_j$ measures the displacement of the $j$th particle from equilibrium.

From Newton's 2nd law

$$m_j\ddot{x_j}=F_j\tag{1}$$

"Because the system is linear, we expect that we can write the force as follows"

$$F_j=-\sum\limits_{k=1}^n K_{jk}x_k\tag{2}$$

For $j=1$ to $n$.

The equations of motion are thus

$$m_j\ddot{x_j}=-\sum\limits_{k=1}^n K_{jk}x_k\tag{3}$$

for $j=1$ to $n$.

We can determine the coefficient $K_{jk}$ by considering a small displacement $x_k$ of the $k$th particle keeping all others fixed at equilibrium positions. We measure the force on the $j$th particle with only the $k$th particle displaced.

$$F_{jk}=-K_{jk}x_k\tag{4}$$

Because the system is linear, the total force on particle $j$ is the sum of the contributions from the displacements of each of each mass.

$$F_j=\sum_k F_{jk}=-\sum_k K_{jk}x_k\tag{5}$$

$$K_{jk}=-\frac{F_{jk}}{x_k}\tag{6}$$

Since the system has no damping, the forces can be derived from potential energy

$$F_j=-\frac{\partial V}{\partial x_j}\tag{7}$$

If we differentiate (5) we get

$$\frac{\partial F_j}{\partial x_k}=-\frac{\partial^2 V}{\partial x_j\partial x_k}=-K_{jk}\tag{8}$$

Similarly, we have

$$\frac{\partial F_k}{\partial x_j}=-\frac{\partial^2 V}{\partial x_k\partial x_j}=-K_{kj}\tag{9}$$

Thus

$$K_{jk}=K_{kj}\tag{10}$$

We can write the equations of motion in (3) in matrix form

$$M\ddot{X}=-KX\tag{11}$$

$$\ddot{X}=-M^{-1}KX\tag{12}$$

(10) tells us that $K$ is symmetric.

(12) represents a system of $n$ linear time-invariant 2nd order differential equations.

It can be shown that we can always find so-called irreducible solutions to these equations.

To find such solutions, we first complexify (12)

$$\ddot{Z}=-M^{-1}KZ\tag{13}$$

Note that $X=\text{Re}(Z)$.

Irreducible solutions have the form

$$Z(t)=Ae^{-i\omega t}\tag{14}$$

where $A$ is some constant $n$-vector and $\omega$ is a number, the angular frequency.

If we calculate the second derivative of (14) and plug it into (13), together with (14), then we end up with an eigenvalue equation

$$\omega^2 A=M^{-1}KA\tag{15}$$

$$(M^{-1}K-\omega^2 I)A=0\tag{16}$$

$$\det{(M^{-1}K-\omega^2 I)}=0\tag{17}$$

The latter is an $n$th order polynomial equation for $\omega^2$.

So far so good. My question arise from the next steps. I have included the relevant snippets from Georgi's "The Physics of Waves" that I don't quite understand.

Physically, we expect all the solutions for $\omega^2$ to be real and positive whenever the system is in stable equilibrium because we expect such systems to oscillate.

Negative $\omega^2$ are associated with unstable equilibrium.

How can $\omega^2$ be negative?

I can imagine it being complex, but how can it be real and negative?

The vector $A$ is called the "normal mode" of the system associated with frequency $\omega$. Because $A$ is real, in the absence of friction, the complex solutions (14) can be put together into real solutions. The general solution is of the form

$$X(t)=\text{Re}[(b+ic)Z(t)]\tag{18}$$

$$=bA\cos{\omega t}+cA\sin{\omega t}=dA\cos{(\omega t-\theta)}\tag{19}$$

where $b$ and $c$ (or $d$ and $\theta$ are real numbers.

Where did $b+ic$ come from?

We can now construct the complete solution to the equation of motion. Because of linearity, we get it by adding together all the normal mode solutions with arbitrary coefficients that must be set by the initial conditions.

We can now see that the number of different modes is always equal to $n$, the number of degrees of freedom. Label the normal modes $A^\alpha$, where $\alpha$ is a label that (we will argue below) goes from $1$ to $n$. Label the corresponding frequencies $\omega_\alpha$. Then the most general possible motion of the system is a sum of all the normal modes

$$Z(t)=\sum\limits_{\alpha=1}^n\omega_\alpha A^\alpha e^{-i\omega_\alpha t}\tag{20}$$

or in real form (with $\omega=b+ic$)

$$X(t)=\sum\limits_{\alpha=1}^n [b_\alpha A^\alpha \cos{(\omega_\alpha t)}+c_\alpha A^\alpha\sin{(\omega_\alpha t)}]\tag{21}$$

$$=\sum\limits_{\alpha=1}^n d_\alpha A^\alpha \cos{(\omega_\alpha t-\theta_\alpha)}\tag{22}$$

where $b_\alpha$ and $c_\alpha$ are real numbers that must be determined from initial conditions.

Why are $\omega_\alpha$ being used as "arbitrary coefficients" in (20)?

 

[zenterix]

Here are my thoughts on what the answers to the last two questions in the OP might be.

The trial irreducible solution is given by (14), ie $Z(t)=Ae^{-i\omega t}$.

Suppose we've found $n$ different eigenvalues $\omega_\alpha^2$ and we chose an associated eigenvector $A^\alpha$.

Each associated complex solution is $Z_\alpha(t)=A^\alpha e^{-i\omega_\alpha t}$.

The associated real solution is $X_\alpha (t)=\text{Re}(Z_\alpha (t))$.

A linear combination of the complex solutions is

$$Z(t)=\sum\limits_{\alpha=1}^n a_\alpha A^\alpha e^{-i\omega_\alpha t}$$

where $a_\alpha$ are arbitrary complex constants.

If $a_\alpha=b_\alpha+ic_\alpha$ then we get

$$Z(t)=\sum\limits_{\alpha=1}^n (b_\alpha+ic_\alpha) A^\alpha e^{-i\omega_\alpha t}$$

$$=\sum\limits_{\alpha=1}^n (b_\alpha+ic_\alpha)A^\alpha (\cos{\omega_\alpha t}-i\sin{\omega_\alpha t})$$

$$=\sum\limits_{\alpha=1}^n A^\alpha(b_\alpha\cos{\omega_\alpha t}+c_\alpha\sin{\omega_\alpha t}+i(c_\alpha\cos{\omega_\alpha t}-b_\alpha\sin{\omega_\alpha t}))$$

$$X(t)=\text{Re}(Z(t))=\sum\limits_{\alpha=1}^n A^\alpha(b_\alpha\cos{\omega_\alpha t}+c_\alpha\sin{\omega_\alpha t})$$

If these calculations are correct then it seems that the answer to my last question in the OP is that $\omega_\alpha$ used in the book is a typo.

 

[TSny]

A linear combination of the complex solutions is

$$Z(t)=\sum\limits_{\alpha=1}^n a_\alpha A^\alpha e^{-i\omega_\alpha t}$$
where $a_\alpha$ are arbitrary complex constants.

If $a_\alpha=b_\alpha+ic_\alpha$ then . . .

If these calculations are correct then it seems that the answer to my last question in the OP is that $\omega_\alpha$ used in the book is a typo.

This looks good to me.

The textbook writes

If you look closely, the symbol that I circled in blue is the script letter for w rather than the Greek letter $\omega$. Unfortunately, the font chosen for the book's printing makes these look almost identical.

 

[zenterix]

@TSny Do you understand what the book means when it says

Physically, we expect all the solutions for $\omega^2$ to be real and positive whenever the system is in stable equilibrium because we expect such systems to oscillate.

Negative $\omega^2$ are associated with unstable equilibrium.

How can $\omega^2$ be negative?

 

[TSny]

How can $\omega^2$ be negative?

I can imagine it being complex, but how can it be real and negative?

$\omega^2$ is found by solving equation (3.73) in the textbook: $$\textrm{det}\left[ M^{-1}K - \omega^2 I \right] = 0$$
The left side is a polynomial in the unknown $\omega^2$. Generally, the roots of a polynomial can be real or complex. However, the particular polynomial in this case can be shown to have only real roots. See Georgi's comments just below equation (3.73). Thus, $\omega^2$ is real.

If one of the roots of the polynomial is negative, then $\omega^2$ is negative. In this case, $\omega$ is an imaginary number: $\pm ia$, where $a$ is a positive number. Then the equation $Z(t) = A e^{-i \omega t}$ (equation 3.66) implies exponential growth or decay of the motion. See Georgi's discussion of the inverted pendulum (Fig. 3.5).

So, for systems that oscillate about stable equilibrium with no friction, we expect $\omega^2$ to be positive.