As the title states, is it possible to derive a hyperbolic relationship in the form of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ between frequency $\omega$ and wavenumber $k$

I have tried to start this from the general phase-group velocity relationship $v_pv_g = c^2$ where $v_p$ = $\frac{w}{k}$ and $v_g = \frac{dw}{dk}$ and upon substitution arrived at the obvious conclusion that $w^2/k^2 = c^2$. Can someone help me carry this train of thought further?



The freespace dispersion equation is $\omega^2 = k^2\,c^2$ and this cannot change: this simply follows from considering plane wave components of propagating fields, which all fulfil the Helmholtz equation

$$\nabla^2 A_j + \frac{\omega^2}{c^2} A_j = 0\tag{1}$$

which is fulfilled by all Cartesian components of the moncrhomatic EM field vectors and, for a plane wave, $\nabla^2 = -|k|^2$, whence the dispersion relationship.

The dispersion of a massive particle fulfilling the Klein Gordon equations yields what you want. All particles fulfilling the Dirac equation in their first quantised (semiclassical) description fulfill the KG equation (but the converse is not true).

The Klein Gordon equation is:

$$\frac {1}{c^2} \frac{\partial^2}{\partial t^2} \psi - \nabla^2 \psi + \frac {m^2 c^2}{\hbar^2} \psi = 0\tag{2}$$

where $m$ is the particle's restmass and, on writing $\nabla^2 \mapsto -|k|^2$, $\partial_t\mapsto-i\,\omega$ for a plane, monochromatic wave we find the dispersion relationship is:

$$\frac{\omega^2}{c^2} - k^2 = \frac{m^2\,c^2}{\hbar^2}\tag{3}$$

i.e. the equation for a hyperbola, as you wished for. This hyperbolic relationship actually yields a really interesting insight into mass at the quantum level: from classical mechanics and general relativity, the notions of inertial mass and gravitational mass are wonted to us, but a third interpretation is that rest mass measures a particles "stay-putability": the group velocity is:

$$v_g = \frac{\mathrm{d}\,\omega}{\mathrm{d}\,k} = \frac{c}{\sqrt{1+\frac{m^2\,c^2}{\hbar^2\,k^2}}}\tag{4}$$

Massless particles must always be observed to be travelling at speed $c$, as shown by (4) (which becomes $v_g=c$ if $m=0$). They are always dispersionless. However, if $m$ is nonzero in (4) you can slow a particle down, or "make it stay put" by making the momentum $\hbar\,k$ very small. You can see now from (4) what I mean by mass measures a particle's "stay puttability". You can make a massive particle stay put, at least as far as the Heisenberg uncertainty principle lets you.

So now, can we simulate this dispersion for light? On elimination of the magnetic field vectors from Maxwell's equations, we find, for each of the Cartesian components of the electric field:

$$-\nabla^2 E + \frac{1}{c^2}\frac{\partial^2}{\partial\,t^2} E +\mu\,\frac{\partial}{\partial\,t} J = 0\tag{5}$$

where $J$ is the corresponding Cartesian component of the current density. So, to simulate your hyperbolic relationship, we need to reproduce the Klein-Gordon equation, which in turn requires that:

$$E = \mu\,\frac{\partial}{\partial\,t} J\tag{6}$$

This is a highly unusual relationship: Ohm's law is $J = \sigma\, E$, so (6) implies a nondissipative current density in phase quadrature with the electric field. I can't imagine a physical material that would do this.


Technically, $\omega^2/1^2-k^2/c^2=0$ is a degenerate hyperbola if that counts. But I don't think you can derive an equation of the form $\omega^2/a^2 - k^2/b^2 = 1$ for waves propagating in free space. You may however find something of the kind if you consider materials with fancier dispersion relations than $\omega = ck$, like e.g. plasmas.


Wavenumber k is the number of waves per metre. Frequency w is number of waves per second.

The number w/k is the speed of the wave.


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