2
$\begingroup$

These questions are inspired by the following the paper http://www.soton.ac.uk/~stefano/courses/PHYS2006/chapter7.pdf on 'Normal Modes of a Beaded String'.

Problem Statement

Given a recurrence relation of $-\omega^2A_p=\omega_0^2(A_{p+1}-2A_p+A_{p-1})$ for the displacement amplitude of vibration of the $i^{th}$ bead at frequency $\omega$, we aim to find an expression for the $n^{th}$ mode frequency.

Solution (Taken from the paper)

Suppose that we have already found a mode for the string. If we shift the string, by translational invariance it will be the same. Therefore, if $A_n$ gives a frequency mode $\omega$, the shifted $A_n'$ also gives a frequency mode of $\omega$. We then have that $$A_n'=A_{n+1}$$ Now let’s look for a translation invariant mode, which reproduces itself when we do the shift. Since a mode is arbitrary up to an overall scale, this means $$A_n'=A_{n+1}=hA_n$$ for some constant h, so that the new amplitudes are proportional to the old ones. Applying this relation repeatedly gives $$A_n=h^nA_0$$ After substituting into the initial recurrence relation we have $$-\omega^2h^nA_0=\omega_0^2(h^{n-1}A_{0}-2h^nA_0+h^{n+1}A_{0})$$ $$\omega^2=\omega_0^2\left(2-h-\frac{1}{h}\right) \ \ \ \ (\star)$$

This shows that $h$ and $\frac{1}{h}$ give the same normal mode frequency. Conversely, if the frequency $\omega$ is fixed, the amplitudes $A_n$ must be an arbitrary linear combination of the amplitudes for $h$ and $\frac{1}{h}$ $$A_n=\alpha h^n+\beta h^{-n}$$ Letting $h=e^{i\theta}$ we have $$\omega^2=4\omega_0^2\sin^2(\frac{\theta}{2})$$ My Issues

(1) Since a mode is arbitrary up to an overall scale, this means $A_n'=A_{n+1}=hA_n$. I am not sure what the intuition behind this step actually is? If you shift the beads to left by one why must we scale the amplitude accordingly?

(2)How does $(\star)$ show that $h$ and $\frac{1}{h}$ give the same normal frequency and what does it even mean to say this?

(3)Why does fixing the frequency $\omega$ mean that the amplitudes must be a linear combination of $h$?

$\endgroup$
8
  • 2
    $\begingroup$ This is easy to phrase in the language familiar from quantum mechanics. If the Hamiltonian of your system is invariant under the discrete translations along the string, you can look for the normal modes as simultaneous eigenstates of the Hamiltonian and the translation operator. The relation $A_n'=A_{n+1}=hA_n$ is just the statement that your solution is an eigenstate of the translation operator ($h$ is the eigenvalue). Since the translation operator is unitary, the eigenvalue should be a complex unit, hence $h=e^{i\theta}$. That really summarizes the main idea behind the solution. $\endgroup$ Nov 8 '21 at 22:17
  • $\begingroup$ Is this the only way to describe a system of this sort? I am relatively new to qm and have not studied the Hamiltonian previously. $\endgroup$
    – benmcgloin
    Nov 8 '21 at 22:20
  • 1
    $\begingroup$ You can also view this purely mathematically. Any linear recurrence relation with constant coefficients can be solved by the ansatz $A_n\propto h^n$. Or, if you prefer more physics intuition, remember that you are looking for wave-like solutions. After all, your recurrence relation is a just a discrete approximation to the wave equation. ($A_{n+1}-2A_n+A_{n-1}$ is a discrete approximation to the second derivative.) Then the ansatz $A_n\propto e^{in\theta}$ is natural: it's a discrete version of a wave. $\endgroup$ Nov 9 '21 at 7:36
  • $\begingroup$ That makes sense, I suppose then you could also transform the recurrence relation to a matrix and solve from there. Is it valid to just state the proportionality to $A_0$ or must I make a similar argument with $A_n=hA_{n+1}$? $\endgroup$
    – benmcgloin
    Nov 9 '21 at 7:45
  • 1
    $\begingroup$ Well, once you assume that $A_n=h^nA_0$ then $A_{n+1}=hA_n$ follows automatically, if that's what you mean. $\endgroup$ Nov 9 '21 at 12:41
1
+50
$\begingroup$
  1. The definition of two equivalent normal modes is given in the caption of Figure 7.3: A normal mode remains "the same" if all amplitudes are multiplied by a (complex) constant. You can get an intuition as follows. If one bead configuration exactly coincides with another one after some time evolution (scaling by $h=e^{i \theta}$) or after zooming out vertically (scaling by $h$ real), then they are considered equivalent, or "the same"
  2. The equation is invariant if you exchange $h\leftrightarrow1/h$. It means that if $A_n = h^n A_0$ satisfies the recurrence equation, then $A_n = h^{-n} A_0$ will too. An intuitive way to see this property is by noting that if a translation (to the left) results in an equivalent mode $A_{n+1} = h A_n$, then a translation to the right $A'_n = A_{n-1}$ would also give "the same" normal mode but it holds $A'_n = A_{n-1} = A_{n}/h$ . Moreover, to obtain $\omega>0$ it is required that $h=e^{i \theta}$. This is not a convenience as said in your link, it is a must, because any other real or complex number $h$ will lead to a complex $\omega$, which is not correct
  3. The amplitudes are a linear combination of $h^n$ and $h^{-n}$, not $h$. This is because the wave equation is linear. So, if you found two solutions for the same mode frequency $A_p$ and $B_p$, then their linear combination $\alpha A_p + \beta B_p$ will also be a solution. It is confusing for the author to call $A_n$ both the translation invariant modes and the general superposition, because $\alpha h^n + \beta h^{-n}$ is not translation invariant $h A_n = \alpha h^{n+1} + \beta h^{-n+1} \neq \alpha h^{n+1} + \beta h^{-n-1} = A_{n+1}$. For $h=e^{i \theta}$ this means that the two normal modes with translation invariance $e^{i n \theta}$ and $e^{-i n \theta}$ are two traveling waves in opposite directions. For a finite number of beads this means the solutions are standing waves ($\alpha = -\beta$) with a node at the origin and another node in the last bead

Hope it helps

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.