Small oscillations: How to find normal modes?

[bznm]

Hi,
I'm studying Small Oscillations and I'm having a problem with normal modes.
In some texts, there is written that normal modes are the eigenvectors of the matrix $V- \omega^2 V$ where V is the matrix of potential energy and T is the matrix of kinetic energy.
Some of them normalize the eigenvector, other don't do it.

In other texts, there is written that normal modes are the coordinates that uncouple the equation of motion and that I can find them as ζ=$B^-1$ η where ζ is the column vector of these normal modes, η is the column vector of initial coordinates and $B^-1$ is the modal matrix (but... for the modal matrix, do I have to normalize eigenvectors?)

Which is the most correct way to find normal modes?
Thank you

 

[Jano L.]

The normalization is just for convenience. Let's say we have the Hamiltoninan

$$
H = \sum_a \frac{p_a^2}{2m} + \sum_a \frac{1}{2}m \omega_a^2 r_a^2 + \sum_a \sum_b {}^{'} \frac{1}{2} V_{ab} r_a r_b
$$
Any set of eigenvectors of the matrix $A_{ab} = V_{ab}/m + \omega_a^2 \delta_{ab}$ can be used to define new, "normal" basis.

The elements of the transformation matrix C will depend on the choice of the eigenvectors, but otherwise everything is the same for any set. The choice in which all the eigenvectors have unit length is particularly useful since then the transformation matrix C is orthogonal and its inverse, which is also needed, can be obtained easily by simple transposition

$$
C^{-1} = C^T.
$$

 

[bznm]

The choice in which all the eigenvectors have unit length is particularly useful since then the transformation matrix C is orthogonal and its inverse, which is also needed, can be obtained easily by simple transposition

$$
C^{-1} = C^T.
$$

You are right! I haven't thought about it!

So if I have to find normal modes, I have to find eigenvectors, then I normalize them (using $$Tu \cdot u=1$$) and find modal matrix C. Then, I find $$C^T$$ and post-moltiplicate it for the vector column of the coordinates that at the beginning of the exercise I have choosen to describe the situation: I'll obtain normal modes. Is it correct?

 

[Jano L.]

I think the normalization we speak about does not involve any T. I will try to explain how I would do the transformation.

Let $\mathbf v_n$ be the eigenvector of the matrix A with eigenvalues $\lambda_n$. Since the matrix is symmetric, these eigenvectors can be chosen to be orthogonal. Then they can be used to form a new basis (basis of normal modes), so any original oscillation can be expressed as linear combination of them.

The vector in the old coordinates can be expressed in the new coordinates $R_n$ in this way:

$$
\mathbf r = \sum_n R_n \mathbf v_n.
$$

so in coordinates we have
$$
r_a = C_{an} R_n,
$$

where $C_{an}$ is the $a-$th component of the $n$-the vector $v_{n,a}$.
This is a special case of the coordinate transformation

$$
r_a = \frac{\partial r_a }{\partial R_n} R_n
$$

The matrix C of transformation is directly given by the old coordinates of the eigenvectors. So far the normalization was not important.

However, we still have to transform the momenta in the kinetic energy. We want them transformed in such a way so that the equations of motion will hold in the new variables $R_n, P_n$, and this implies that the momenta have to transform according to

$$
p_a = \frac{\partial R_n }{\partial r_a} P_n
$$
which is
$$
p_a = (C^{-1})_{na} P_n.
$$

If we did not normalize the vectors $\mathbf v$, we would have to keep in consideration this inverse matrix $C^{-1}$. But if we normalize them so that
$$
\sum_a v_{an} v_{an} = \delta_{nm},
$$
the matrix C will be orthogonal (columns will be orthogonal unit vectors) and the inverse is calculated easily just by transposition. Then the transformation of momenta is written

$$
p_a = C_{an} P_n,
$$

which is the same as for the coordinates.