Help finding the vibrational frequencies and normal modes

[1v1Dota2RightMeow]

Homework Statement

Let's say that I have a potential $U(x) = \beta (x^2-\alpha ^2)^2$ with minima at $x=\pm \alpha$. I need to find the normal modes and vibrational frequencies. How do I do this?

Homework Equations

$U(x) = \beta (x^2-\alpha ^2)^2$
$F=-kx=-m\omega ^2 x$
$\omega = \sqrt{\frac{k}{m}}$
$U(x)=\frac{1}{2}m\omega ^2 x^2$

The Lagrangian for this system is given by:

$L=(1/2)m(\dot{x}_1^2 + \dot{x}_2^2)-4\alpha^2 \beta (x_1 + \alpha)^2 -4\alpha^2 \beta (x_2 + \alpha)^2 - (1/2) k (x_2 - x_1 -2\alpha)^2$

The Attempt at a Solution

I found an example that uses matrices. The general idea seems to be to put the Lagrangian into matrix form, set up the characteristic equation, and then solve for eigenvalues and eigenvectors. But how do I put this Lagrangian into matrix form?? Note - I have Mathematica at my disposal.

 

[JoePhysics]

It isn't writing the Lagrangian in matrix form. Rather, one would apply the Euler-Lagrange equations to the Lagrangian to get the equations of motion. This will result in, in essence*, an equation that looks like
$$\frac{d^2\mathbf{x}}{dt^2} = -K \mathbf{x}$$
where $\mathbf{x}$ is a column vector with your two generalized coordinates ($x_1$ and $x_2$ in your case) and $K$ is a 2x2 matrix. The trick then is to find a pair of coordinates such that the equations of motion are decoupled. Do you know how to do that? Think of what you can do to the matrix $K$, that you may have covered in a linear algebra course.

*There could be constant terms here, but they can be dealt with by translating the coordinates appropriately.

Edit: I noticed your potential isn't quadratic, but that's OK; the trick is to work out what the system looks like for small enough oscillations.

 

[1v1Dota2RightMeow]

Edit: I noticed your potential isn't quadratic, but that's OK; the trick is to work out what the system looks like for small enough oscillations.

Yup, in the Lagrangian I reduced using Taylor series.

 

[JoePhysics]

Yup, in the Lagrangian I reduced using Taylor series.

Excellent, then you're all set. Reduce the system to that format (which if you notice, is analogous to that of a harmonic oscillator but with variables that are vectors) and see what else you can do.

 

[1v1Dota2RightMeow]

It isn't writing the Lagrangian in matrix form. Rather, one would apply the Euler-Lagrange equations to the Lagrangian to get the equations of motion. This will result in, in essence*, an equation that looks like
$$\frac{d^2\mathbf{x}}{dt^2} = -K \mathbf{x}$$
where $\mathbf{x}$ is a column vector with your two generalized coordinates ($x_1$ and $x_2$ in your case) and $K$ is a 2x2 matrix. The trick then is to find a pair of coordinates such that the equations of motion are decoupled. Do you know how to do that? Think of what you can do to the matrix $K$, that you may have covered in a linear algebra course.

*There could be constant terms here, but they can be dealt with by translating the coordinates appropriately.

Edit: I noticed your potential isn't quadratic, but that's OK; the trick is to work out what the system looks like for small enough oscillations.

Ok so I found the coupled equations to be

$\ddot{x_1} =(\frac{-8\alpha^2 \beta -k}{m})x_1 + (\frac{-8\alpha^3 \beta + kx_2 -2k \alpha}{m})$
$\ddot{x_2} =(\frac{-8\alpha^2 \beta -k}{m})x_2 +( \frac{8\alpha^3 \beta + kx_1 +2k \alpha}{m})$

 

[JoePhysics]

Excellent, now get it in the form
$$\begin{pmatrix}
\displaystyle\frac{d^2x_1}{dt^2} \\
\displaystyle\frac{d^2x_2}{dt^2}
\end{pmatrix} = -K
\begin{pmatrix}
x_1 \\
x_2
\end{pmatrix} + \text{whatever}
$$
where $K$ is a 2x2 matrix. Mind you, I am not checking your algebra!

 

[1v1Dota2RightMeow]

Excellent, now get it in the form
$$\begin{pmatrix}
\displaystyle\frac{d^2x_1}{dt^2} \\
\displaystyle\frac{d^2x_2}{dt^2}
\end{pmatrix} = -K
\begin{pmatrix}
x_1 \\
x_2
\end{pmatrix} + \text{whatever}
$$
where $K$ is a 2x2 matrix. Mind you, I am not checking your algebra!

Ok so then $K=\begin{bmatrix}\frac{8\alpha^2\beta + k}{m} & -k/m\\-k/m & \frac{8\alpha^2\beta + k}{m}\end{bmatrix}$. I removed a negative sign from inside the matrix because it looks like you pulled it out to be in front of the $K$.

 

[1v1Dota2RightMeow]

But now I have absolutely no clue about how to decouple these equations. Linear algebra wasn't my best math course... hehe...

 

[1v1Dota2RightMeow]

Can I just completely throw out those constants that you've labeled "whatever"??

 

[JoePhysics]

I was reviewing a bit (I also have my graduate qualifier in a bit ), and it seems there may be a less "systematic" way to solve these kinds of problems. Here's how it goes. First, the "whatever" terms need to go away; this can be done by setting $\ddot{x}_1 = \ddot{x}_2 = 0$ and seeing for what $x_1$ and $x_2$ there is equilibrium; we call these positions $x_{1\text{eq}}$ and $x_{2\text{eq}}$ respectively. You then define a new pair of variables $x_1' = x_1 - x_{1\text{eq}}$ and $x_2' = x_2 - x_{2\text{eq}}$, and if you plug these new variables into your differential equations, you should get rid of any constant terms.

And this is where the actual process begins; let us assume a solution of the form $x_1'(t) = A_1 \exp{(i \omega t)}$ and $x_2'(t) = A_2 \exp{(i \omega t)}$. If you plug this into your differential equations and cancel out the complex exponential, you end up with an expression that looks like
$$B \begin{pmatrix}
A_1 \\
A_2
\end{pmatrix} =
\begin{pmatrix}
0 \\
0
\end{pmatrix}
$$
where $B$ is a 2x2 matrix, where you'll have somewhere in it the $\omega$ that we introduced before in our ansatz. Now here is the most important thing about these problems: we are interested in finding nontrivial solutions to this system of equations in $A_1$ and $A_2$. Thus we are interested in seeing when the inverse of the matrix $B$ does not exist. This will happen when its determinant vanishes, and this is what we need to do to find the eigenfrequencies. From here on it's mostly an algebra problem. Do you know how to continue, and then later find the normal modes?

 

[1v1Dota2RightMeow]

First, the "whatever" terms need to go away; this can be done by setting $\ddot{x}_1 = \ddot{x}_2 = 0$ and seeing for what $x_1$ and $x_2$ there is equilibrium; we call these positions $x_{1\text{eq}}$ and $x_{2\text{eq}}$ respectively. You then define a new pair of variables $x_1' = x_1 - x_{1\text{eq}}$ and $x_2' = x_2 - x_{2\text{eq}}$, and if you plug these new variables into your differential equations, you should get rid of any constant terms.

Going in order of what you've written, this is my first question: are the equilibirum points the points where the potential of the wells is at a minimum? I found those to be at $x=\pm \alpha$. But then again this does not include the spring potential...

 

[1v1Dota2RightMeow]

Oh gosh oh gosh - I just realized that I forgot to explicitly mention that there is a spring connecting the two particles. The scenario is that there is a double well potential with one particle in each well. The particles are connected to each other by a spring. We are considering small oscillations about the minima.