# Question about the frequency of normal modes in coupled oscillators and their derivation [closed]

Consider a two-mass system that is coupled by three springs, such that: $$m_1=m_2=m$$; $$k_1=k_3=k; k_2=k_{12}$$. It can be written in terms of the following coordinates: $$\eta_1=x_1-l_1$$ and $$\eta_2=x_2-(l_1+l_2)$$. It's relatively simple to find the motion equations for such a system using lagrangians, we can write them in matrix form:

$$\ddot{\vec{\eta}}+\mathrm{W}\vec{\eta}=\vec{0} \rightarrow \vec{\eta}=\begin{pmatrix}\eta_1\\\eta_2\end{pmatrix};\mathrm{W}=\frac{1}{m}\begin{pmatrix}k+k_{12}&&-k_{12}\\-k_{12}&&k+k_{12}\end{pmatrix}$$

Now my notes state that we have to decouple these equations by changing basis, diagonalizing as $$\mathrm{S}^{-1}\mathrm{W}\mathrm{S}=\mathrm{W}_D$$. This much makes sense. However it states that, therefore, $$\vec{\eta}=\mathrm{S}\vec{Q}$$. I believe $$\vec{Q}$$ is an arbitrary vector though it is not stated explicitly. The solutions for $$Q_{1,2}(t)$$ are simple harmonic oscillators with frecuencies $$\omega_{1,2}$$.

Now the problem is that it also states that the eigenvalues in $$\mathrm{W}_D$$ are equal to $$\omega_{1,2}^2$$. But why is this? Is it, too, arbitrary, coming from our at-choice-chosen $$\vec{Q}$$? That's my current understanding. If not, what is going on?

• Substitute $\eta = S Q$ in $\ddot{\eta} + W\eta = 0$. Then, you get the equation of motion in the new coordinates $Q$. This far is true for any transformation $S$, i.e., for any coordinates $Q$. Now multiply $S^{-1}$ from the left and impose the condition $S^{-1} W S = W_D$. Then, the equation of motion for $Q$ is decoupled for its components. This special $Q$ is the normal mode coordinates (except some scaling factor or other details that I didn't check). Mar 10, 2022 at 1:16
• @norio That appears too in my notes, but why is it the case that the frequencies $\omega_{1,2}$, squared, are equal to the eigenvalues? That's what I don't see a reason for, is it just what the definition of "normal mode" entails (and is, therefore, arbitrary)? Mar 10, 2022 at 1:21
• Let one of the decoupled equations for $Q$ be $\ddot{q}= -a q$. Here $a$ is a diagonal element of $W_D$. You said you get a simple harmonic oscillator solution, $q(t) = A \cos(\omega t +\phi)$, where $A$ and $\phi$ are constants to be determined by initial conditions. If you substitute this solution into the first equation of this comment, you get $a = \omega^2$. In other words, if you have derived the formula $\omega = \sqrt{k/m}$ for a single harmonic oscillator, you can derive the same relation for a diagonal element of $W_D$ and the frequency of a normal mode in the same way. Mar 10, 2022 at 1:54
• Mar 10, 2022 at 12:56
• Does this answer your question? Eigenvalue equation for kinetic and potential energy Mar 10, 2022 at 12:57

Firstly, $$\vec{Q}$$ is not an arbitrary vector. It's right there in your question: $$\vec{\eta} = S\vec{Q}$$, or equivalently, $$\vec{Q} = S^{-1}\vec{\eta}$$. This is a physical problem - for a given initial condition, only one thing happens, and that one unique thing is represented mathematically by the vector $$\vec{\eta}(t)$$. The transformation lets you simply change the coordinates to write your equations of motion more simply, as two separate equations for $$Q_1$$ and $$Q_2$$ instead of all the components of $$\vec{\eta}$$ mixed together.

Now, when you perform the transformation $$W_D = S^{-1}WS$$, you should be using an orthogonal transformation matrix (so that $$S^{-1} = S^T$$). When you use an orthogonal transformation matrix, the eigenvalues of $$W_D$$ are the same as the eigenvalues of $$W$$! In fact, finding the eigenvalues and eigenvectors of $$W$$ is an integral part of calculating the transformation matrices!

Let's say $$W$$ has two eigenvalues, $$\omega_1^2$$ and $$\omega_2^2$$, with corresponding eigenvectors $$\vec{v}_1 = [v_{11}\hspace{1ex}v_{12}]$$ and $$\vec{v}_2 = [v_{21}\hspace{1ex}v_{22}]$$ (so $$W\vec{v}_1 = \omega_1^2\vec{v}_1$$ and $$W\vec{v}_1 = \omega_1^2\vec{v}_1$$). We should also make sure that $$\vec{v}_1$$ and $$\vec{v}_2$$ are both unit vectors - both should have a magnitude of 1.

It's also safe to assume $$W$$ is Hermitian (such matrices in physics always are. For a real-valued matrix, like here, that means $$W$$ is symmetric. The coupling between 1 and 2 is the same as between 2 and 1). This means $$\vec{v}_1$$ and $$\vec{v}_2$$ have the nice property that $$\vec{v}_1\cdot\vec{v}_2 = 0$$.

So now, if we construct a matrix $$S$$ like this:

$$S = [\vec{v}_1 | \vec{v}_2] = \begin{bmatrix}v_{11} & v_{21} \\ v_{12} & v_{22}\end{bmatrix}$$

notice what happens: the first column of $$WS$$ will be $$W\vec{v}_1$$, which is just $$\omega_1^2\vec{v}_1$$. The second column will similarly be $$\omega_2^2\vec{v}_2$$. So:

$$WS = [\omega_1^2\vec{v}_1 | \omega_2^2\vec{v}_2] = \begin{bmatrix}\omega_1^2v_{11} & \omega_2^2v_{21} \\ \omega_1^2v_{12} & \omega_2^2v_{22}\end{bmatrix}$$

(if you don't believe me you should try it yourself. It has to do with how eigenvectors are defined - this is their whole thing)

Now, let's multiply this whole thing by $$S^T$$:

$$S^T = \left[\frac{\vec{v}_1}{\vec{v}_2}\right] = \begin{bmatrix}v_{11} & v_{12} \\ v_{21} & v_{22}\end{bmatrix}$$

and compute $$S^TWS$$. But notice what happens when we do so: when we multiply the first row by the first column, we're really taking the dot product of $$\vec{v}_1$$ and $$\omega_1^2\vec{v}_1$$. When we multiply the first row and the second column, we are really taking the dot product $$\vec{v}_1\cdot\omega_2^2\vec{v}_2$$. In all:

$$S^TWS = \begin{bmatrix} \omega_1^2(\vec{v}_1\cdot\vec{v}_2) & \omega_2^2(\vec{v}_1\cdot\vec{v}_2) \\ \omega_1^2(\vec{v}_2\cdot\vec{v}_1) & \omega_2^2(\vec{v}_2\cdot\vec{v}_2) \end{bmatrix}$$

BUT remember: $$\vec{v}_1$$ and $$\vec{v}_2$$ are unit vectors, so $$\vec{v}_1\cdot\vec{v}_1 = \vec{v}_2\cdot\vec{v}_2 = 1$$, and we also saw that $$\vec{v}_1\cdot\vec{v}_2 = \vec{v}_2\cdot\vec{v}_1 = 0$$ because $$W$$ is symmetric. So:

$$S^TWS = \begin{bmatrix} \omega_1^2 & 0 \\ 0 & \omega_2^2 \end{bmatrix} = W_D$$

And voila! Diagonalized! We can immediately see that $$W_D$$ has eigenvectors $$[1\hspace{1ex}0]$$ and $$[0\hspace{1ex}1]$$, with corresponding eigenvalues $$\omega_1^2$$ and $$\omega_2^2$$, the same as the eigenvalues of $$W$$, just on a new basis.

Now here's the kicker.

We take the original equation of motion:

$$\ddot{\vec{\eta}} = -W\vec{\eta}$$

and multiply both sides by $$S^T$$:

$$S^T\ddot{\vec{\eta}} = -S^TW\vec{\eta}$$

We can also sneak in $$SS^T$$ in between $$W$$ and $$\vec{\eta}$$ on the right hand side, because $$SS^T = SS^{-1}$$ is just the identity matrix:

$$S^T\ddot{\vec{\eta}} = -S^TWSS^T\vec{\eta}$$

Inserting some parentheses:

$$S^T\ddot{\vec{\eta}} = -(S^TWS)(S^T\vec{\eta})$$

and we recover (given $$\vec{Q} = S^T\vec{\eta}$$):

$$\ddot{\vec{Q}} = -W_D\vec{Q}$$

which of course decouples into

$$\ddot{Q}_1 = -\omega_1^2 Q_1$$

and

$$\ddot{Q}_2 = -\omega_2^2 Q_2$$

So the solutions $$Q_{1,2}$$ are oscillators with frequencies $$\omega_{1,2}^2$$ because $$\omega_{1,2}^2$$ are the eigenvalues of $$W$$ (and $$W_D$$). The transformation doesn't change the eigenvalues of your matrix, it only mixes up your coordinates so that the math is easier to deal with.

• Sorry if I mix up the notation for $S^T$ and $S^{-1}$. Here, at least, they are the same. Mar 10, 2022 at 1:43
• I should also highlight that the frequencies of the normal modes are SET by the eigenvalues of your original $W$ matrix - at no point was anything defined arbitrarily Mar 10, 2022 at 1:43