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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?

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    $\begingroup$ 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). $\endgroup$
    – norio
    Mar 10, 2022 at 1:16
  • $\begingroup$ @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)? $\endgroup$
    – agaminon
    Mar 10, 2022 at 1:21
  • $\begingroup$ 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. $\endgroup$
    – norio
    Mar 10, 2022 at 1:54
  • $\begingroup$ Duplicate : Eigenvalue equation for kinetic and potential energy. $\endgroup$
    – Frobenius
    Mar 10, 2022 at 12:56
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    $\begingroup$ Does this answer your question? Eigenvalue equation for kinetic and potential energy $\endgroup$
    – Frobenius
    Mar 10, 2022 at 12:57

1 Answer 1

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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.

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  • $\begingroup$ Sorry if I mix up the notation for $S^T$ and $S^{-1}$. Here, at least, they are the same. $\endgroup$
    – AJ Biffl
    Mar 10, 2022 at 1:43
  • $\begingroup$ 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 $\endgroup$
    – AJ Biffl
    Mar 10, 2022 at 1:43

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