According to definition of normal modes, which says if all the different independent parts of a system vibrate at same frequency and their amplitude preserve a fixed ratio then such a motion is a normal mode of that system then since in sinusoidal travelling waves also different parts move with same frequency and different parts preserve a ratio, shouldn't they too be normal modes?

So are sinusoidal traveling waves normal modes?

  • $\begingroup$ Coincidentally, I spent the last week studying the mathematical similarity between two coupled oscillators and two coupled waveguides. It turns out that there's a very strong mathematical similarity between those two cases wherein you can see clearly that the "normal modes" in the coupled oscillator case correspond exactly with the waves in the coupled waveguide case. The calculation is not hard, but it involves defining a bit of notation. Are you interested in that? Would a discussion of the mathematical similarities between coupled oscillators and coupled waveguides answer your question/ $\endgroup$ – DanielSank Mar 25 '19 at 0:27
  • $\begingroup$ Yeah sure I would love to see it, but I did play around with the idea a bit and it turns out that the ratio pf amplitudes of different moving parts is not preserved during a travelling wave so it shouldn't a normal mode, right? $\endgroup$ – Lucifer Mar 25 '19 at 3:56
  • $\begingroup$ Physical systems supporting travelling waves absolutely have normal modes. You just have to think about what "normal modes" means. The fact that the ratios of amplitudes of moving parts can change really isn't that important. It's not the amplitudes themselves that you should look at, but rather the amplitudes normalized by the impedance of the mode. I can explain/show what that means in a full answer, but I need to take some time to write it. $\endgroup$ – DanielSank Mar 25 '19 at 4:55
  • $\begingroup$ Alright, let me know what you think of the answer. I'm happy to add more detail and clarify whatever you need. $\endgroup$ – DanielSank Mar 25 '19 at 7:05

Yes, travelling wave systems have normal modes. In fact, the physics and mathematics of two coupled oscillators is strikingly similar to that of coupled waveguides. I'm most familiar with electrical oscillators, but everything written below applies to any coupled harmonic oscillators.

Coupled oscillators

Consider two electrical oscillators "$a$" and "$b$". Oscillator $a$ has capacitance $C_a$ and inductance $L_a$, and similarly for oscillator $b$. The oscillators are coupled through a capacitance $C_g$ and mutual inductance $L_g$. Each oscillator has a magnetic flux $\Phi$ and an electric charge $Q$.$^{[a]}$

We could study this system using Kirchhoff's laws, but it's a lot easier to convert everything to the Hamiltonian formalism. The Hamiltonian for the system is $$H = \frac{\Phi_a^2}{2 L_a'} + \frac{\Phi_b^2}{2 L_b'} + \frac{Q_a^2}{2 C_a'} + \frac{Q_b^2}{2 C_b'} + \frac{Q_a Q_b}{C_g'} - \frac{\Phi_a \Phi_b}{L_g'} $$ where all those primes on the various constants have to do with the fact that the coupling renormalizes each oscillator's capacitance and inductance. Now we introduce the variables \begin{align} a &= \frac{1}{\sqrt{2}}\left( \frac{\Phi_a}{\sqrt{Z_a'}} + i \sqrt{Z_a'} Q_a \right) \\ b &= \frac{1}{\sqrt{2}}\left( \frac{\Phi_b}{\sqrt{Z_b'}} + i \sqrt{Z_b'} Q_b \right) \end{align} where the impedance $Z$ is defined by $Z \equiv \sqrt{L/C}$. With these variables, the Hamiltonian becomes \begin{align} H &= \omega_a' a^* a + \omega_b' b^* b \\ &-\left( ab + a^* b^* \right) \underbrace{\frac{1}{2} \left( \frac{1}{C_g' \sqrt{Z_a' Z_b'}} + \frac{\sqrt{Z_a' Z_b'}}{L_g'} \right)}_\chi \\ &+\left( a b^* + a^* b \right) \underbrace{\frac{1}{2} \left( \frac{1}{C_g' \sqrt{Z_a' Z_b'}} - \frac{\sqrt{Z_a' Z_b'}}{L_g'} \right)}_g \, . \end{align} Let's remember for a moment what the Hamiltonian means: it provides a way to get the time evolution of the system. In the present case the time dependences come from $$ \dot a(t) = -i \frac{\partial H}{\partial a^*} \qquad \dot b(t) = -i \frac{\partial H}{\partial b^*} \, . $$ Using these equations, we can write a matrix equation for the whole system: $$ \frac{d}{dt} \left( \begin{array}{c} a \\ b \\ a^* \\ b^* \end{array} \right) = -i \left( \begin{array}{cc} \omega_a' & g & 0 & - \chi \\ g & \omega_b' & -\chi & 0 \\ 0 & \chi & -\omega_a' & -g \\ \chi & 0 & -g & -\omega_b' \end{array} \right) \left( \begin{array}{c} a \\ b \\ a^* \\ b^* \end{array} \right) \, . $$ Ok now here's the point: the normal modes and frequencies of the system are precisely the eigenvectors and eigenvalues of that matrix. If the coupling is turned off (i.e. $C_g=0$ and $L_g=0$), then $g = \chi = 0$ and the eigenvalues are $\pm \omega_a'$ and $\pm \omega_b'$, which makes complete sense.

Coupled waveguides

Alright now suppose we have two waveguides "$a$" and "$b$" that are coupled to each other through some mutual capacitance and inductance per length of the waveguide. Denote the rightward and leftward moving amplitudes in waveguide $a$ as $a_\pm$, and similarly for waveguide $b$. If you work it all out, you find that $$ \frac{d}{dx} \left( \begin{array}{c} a_+ \\ b_- \\ a_- \\ b_+ \end{array} \right) = i \left( \begin{array}{cc} k_a' & -g & 0 & \chi \\ g & -k_b' & -\chi & 0 \\ 0 & -\chi & -k_a' & g \\ \chi & 0 & -g & k_b' \end{array} \right) \left( \begin{array}{c} a_+ \\ b_- \\ a_- \\ b_+ \end{array} \right) $$ where the $k$'s are the wave numbers associated with each waveguide (and I should say that the meanings of $g$ and $\chi$ are slightly different than they were for the coupled oscillator case).

Comparison between the problems

Thus the coupled waveguide problem has the same form as the coupled oscillator problem (the signs are different, but that's just because of how we ordered the variables). In both cases we have a system of first-order differential equations, and in both cases the eigenvectors and eigenvalues of the matrix in the differential equation tell us what the normal modes and frequencies (for the oscillator case) or wave numbers (for the waveguide case) of the system are.

$[a]$: There's a direct correspondence between electrical and mechanical oscillators. The electrical flux and charge correspond to position and momentum. Capacitance corresponds to mass, and inductance corresponds to one over the spring constant. Where we use Kirchhoff's laws for the electrical case, we use Newton's law ($F = ma$) in the mechanical case.

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  • $\begingroup$ Thanks it helped a lot. The only question I have is that should travelling waves themselves be considered normal modes of a system? $\endgroup$ – Lucifer Mar 25 '19 at 15:14
  • $\begingroup$ @Lucifer Well, yeah, why not? Are you trying to understand what "normal mode" means in the travelling wave context? $\endgroup$ – DanielSank Mar 25 '19 at 18:25

If the equations of motion of the vibrating system are equivalent to real and symmetric mass and stiffness terms, the normal modes will be real vectors, which means that all parts of the system move in the same phase.

That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave.

There is a special case, when two (or more) vibration modes have identical frequencies. In that situation, a combination of the different mode shapes with different phases may "look like" a travelling wave. However this may only be a theoretical possibility, because the tolerances in a real-life structures often separate the two theoretically-identical frequencies.

However there are mechanical systems which do have "travelling" normal vibration modes. A simple example is a gyroscope, where the vibration modes include precession and nutation.

In general, the equations of motion of a system rotating with constant angular velocity will include Coriolis terms, if it is modelled in a rotating coordinate system fixed to the undeformed shape of the body. The equations of motion are then Hermitian matrices rather than real symmetric matrices. The eigenvalues (natural frequencies) are still real, but the mode shapes are now complex vectors which can be interpreted as travelling waves.

In general, the speed at which the "mode shape" rotates around the object is different from the rotation speed of the object itself. If the two speeds coincide for some particular rotation speeds, that can have severe consequences for the design of real rotating machinery - for example the so-called "critical speeds" of rotating shafts.

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  • $\begingroup$ "That excludes travelling waves, where there is a phase difference between points in the direction of travel of the wave."- but even in a standing wave, which is a normal mode for a string fixed at both ends, there is phase difference between points separated by a node or in two adjacent "loops". Isn't the requirement of a normal mode is that all moving parts have same frequency and not necessarily same phase? $\endgroup$ – Lucifer Mar 24 '19 at 17:37
  • $\begingroup$ Tell me if I am wrong, but I think that the reason sinusoidal travelling wave is not a normal mode because the ratio of amplitude of different parts don't remain constant. Like say the ratio between amplitude at two points A and B is say k but as time passes there will be a time when the same ratio is (1/k) or may be some other value which doesn't happen in case of normal modes like standing waves on a string. $\endgroup$ – Lucifer Mar 24 '19 at 17:57

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