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Suppose that there exists a medium whose dispersion relation has more than one root in $k$ for a certian $\omega$. One such medium (out of the many possibilities) might have a dispersion relation like,

$$ (\omega^2-c_a^2k^2)(\omega^2-c_b^2k^2) = 0 $$

Which would support one wave with phase velocity $c_a$ and another with phase velocity $c_b$.

Now, suppose that this medium is put next to a "regular" one that obeys the wave equation. Then you'd get the "reflection at normal incidence" problem, but instead of having one transmitted wave you would have two.

Continuity and derivative continuity don't constraint the problem enough. What do I need to do in order to solve this reflection problem in the general case?

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  • $\begingroup$ How many polarization states does your wave have? $\endgroup$ Commented Jul 26, 2018 at 20:53
  • $\begingroup$ You have to switch to microscopic equations. Think about a big bunch of masses on springs which you drive at a given frequency. If there are multiple modes resonant at that frequency, where does the energy go? Clearly you can’t answer this knowing the frequencies alone, but it’s trivial using $F=ma$. $\endgroup$
    – knzhou
    Commented Jul 26, 2018 at 21:09
  • $\begingroup$ The reason I ask my question, BTW, is that it makes a difference for the character of the PDE that underlies the dispersion relation, which is why the usual boundary condition matching may not be working for you. $\endgroup$ Commented Jul 27, 2018 at 12:55

2 Answers 2

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Summary: As a general rule, if the dispersion relation is an $n$th-order polynomial in $k$, then you need to specify $n$ spatial boundary conditions. In particular, you would need to specify four spatial boundary conditions at the interface between the media. If we're talking about a scalar wave, without any polarization states, this implies that you would need to specify continuity of the function and its first through third derivatives to solve for the transmitted wave.


Let's assume that we are talking about a scalar field propagating in 1-D. In other words, we are talking about the behavior of a scalar function $u(x,t): \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ that obeys some kind of PDE. We know that $u(x,t)$ will admit a Fourier representation: $$ u(x,t) = \frac{1}{2 \pi} \iint \tilde{u}(k,\omega) e^{-i k x} e^{-i \omega t} \, dk \, d \omega $$ The dispersion relation given implies that waves can only propagate when $(\omega^2 - c_a^2 k^2)(\omega^2 - c_b^2 k^2) = 0$; this implies that the support of $\tilde{u}$ is confined to this surface, and that therefore $$ (\omega^2 - c_a^2 k^2)(\omega^2 - c_b^2 k^2) \tilde{u}(k, \omega) = 0. $$ Taking the Fourier transform of this equation, we have that $$ \frac{1}{2 \pi} \iint (\omega^2 - c_a^2 k^2)(\omega^2 - c_b^2 k^2) \tilde{u}(k,\omega) e^{-i k x} e^{-i \omega t} \, dk \, d \omega = 0 \\ \frac{1}{2 \pi} \iint \left[ - \frac{\partial^2}{\partial t^2} + c_a^2 \frac{\partial^2}{\partial x^2} \right]\left[ - \frac{\partial^2}{\partial t^2} + c_b^2 \frac{\partial^2}{\partial x^2} \right] \tilde{u}(k,\omega) e^{-i k x} e^{-i \omega t} \, dk \, d \omega = 0 \\ \left[ - \frac{\partial^2}{\partial t^2} + c_a^2 \frac{\partial^2}{\partial x^2} \right]\left[ - \frac{\partial^2}{\partial t^2} + c_b^2 \frac{\partial^2}{\partial x^2} \right] \left\{ \frac{1}{2 \pi} \iint \tilde{u}(k,\omega) e^{-i k x} e^{-i \omega t} \, dk \, d \omega \right\}= 0 \\ \left[ - \frac{\partial^2}{\partial t^2} + c_a^2 \frac{\partial^2}{\partial x^2} \right]\left[ - \frac{\partial^2}{\partial t^2} + c_b^2 \frac{\partial^2}{\partial x^2} \right] u(x,t) = 0 $$ Thus, $u(x,t)$ must satisfy the fourth-order PDE $$ \frac{\partial^4 u}{\partial t^4} - (c_a^2 + c_b^2) \frac{\partial^4 u}{\partial t^2 \partial x^2} + c_a^2 c_b^2 \frac{\partial^4 u}{\partial x^4} = 0. $$

From this equation, we can see why specifying two conditions at the boundary between the interfaces is insufficient. As a general rule, to find a unique solution to a PDE, one must specify as many boundary conditions in a given direction as there are derivatives in that direction.

The problem of wave transmission corresponds to saying that the function and all of its time derivatives are zero at some "early" time, and that we specify (via the shape of the incoming wave) the function at the spatial "boundary" of the medium at all times. Since there are fourth spatial derivatives present in the PDE above, we would need to specify four pieces of data at the spatial boundary. In other words, if the boundary is at $x = 0$, we would need to specify $$ u(0,t), \qquad u'(0,t), \qquad u''(0,t), \qquad u'''(0,t), $$ where a prime $'$ denotes $\partial/\partial x$. This, along with the initial value conditions $$ u(x,0) = \dot{u}(x,0) = \ddot{u}(x,0) = \dddot{u}(x,0) = 0 $$ (dots representing time derivatives) will generically lead to a unique solution for $u$.

This method can be extended to any polynomial dispersion relation in $\omega$ and $k$; in general, if the dispersion relation is of the form $f(\omega, k) = 0$, then the wave equation must be $$ f \left( i \frac{\partial}{\partial t}, i \frac{\partial}{\partial x} \right) \left[ u(x,t) \right] = 0. $$ The number of boundary conditions required will then be determined by the order of this PDE, which will in turn be determined by the order of the polynomial $f(\omega, k)$.

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There is a phenomenon called optical birefringence, which resembles what you describe. In low symmetry crystals the dielectric response is different for the two polarisation directions. A famous example is calcium carbonate.

It is also quite possible that the refractive index depends on the direction of propagation of light in a medium.

Analogous phenomena should be possible for sound waves, which in solids have three polarisation directions.

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  • $\begingroup$ This requires that there be two different polarization states, though; it's not clear from the OP whether this is the case, or whether they have in mind a single polarization state (e.g., longitudinal wave in a string.) $\endgroup$ Commented Jul 26, 2018 at 20:53
  • $\begingroup$ @Michael Seifert Of course, light has two polarisation states but sound has three. So for sound I guess media can exist with trirefringence. $\endgroup$
    – my2cts
    Commented Jul 26, 2018 at 23:12

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