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It seems to me that nearly all the common linear dispersive equations have dispersion relations which vanish at the zero spatial frequency. For example:

  • The Schrodinger dispersion relation is $\omega(\xi)=|\xi|^2$
  • The KdV/Airy dispersion relation is $\omega(\xi)=-|\xi|^3$
  • The wave equation dispersion relation is $\omega(\xi)=|\xi|$ (maybe not technically dispersive since particles travel at the same speed)

But the Klein-Gordon equation's dispersion relation $\omega(\xi)=\sqrt{1+|\xi|^2}$ seems to be the only one for which $\omega(0)\neq 0$.

Is there a good physical/mathematical interpretation for dispersion relations which vanish at zero vs. ones that don't?

The best answer I can come up with is that the zeros of $\omega(\xi)$ seems relevant to understanding oscillations in time. More precisely, consider a solution to a linear dispersive equation

$$ u(t,x)=\int e^{ix\cdot\xi+it\omega(\xi)}\hat{u}_0(\xi)d\xi$$

Then to understand time oscillations of $u$, one could consider integrating it in time against some other function and then integrating by parts in time, using $\frac{1}{i\omega(\xi)}\partial_t e^{ix\cdot\xi+it\omega(\xi)}=e^{ix\cdot\xi+it\omega(\xi)}$:

$$\int_0^T u(t,x)f(t) dt = -\int \int_0^T f'(t)\frac{e^{ix\cdot\xi+it\omega(\xi)}}{i\omega(\xi)}\hat{u}_0(\xi)d\xi dt + \text{boundary terms}$$

But if one is after $L^\infty$ estimates, this only works if $\omega(\xi)$ is nonzero on the support of $\hat{u}_0$. Is time oscillation behavior unique to Klein-Gordon?

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    $\begingroup$ Physically, the $+1$ for the KG dispersion describes a relativistic particle with some (generally nonzero) mass. Reinserting all physical constants, it reads $E^2 = (m \, c^2)^2 + (c\,p)^2$ with $E$ the energy (your $\omega$), $m$ the rest mass, $c$ the speed of light, and $p$ (your $\xi$) the momentum. The first term on the RHS (your $+1$) is the particle's rest energy, cf Einstein's famous $E=m\,c^2$. For massless particles $m = 0$, and one is left with $E = |\xi|$ like in your other examples. See en.wikipedia.org/wiki/Klein%E2%80%93Gordon_equation for more. Does that help? $\endgroup$ Commented Nov 29, 2023 at 18:04
  • $\begingroup$ That helps! Although I'm still wondering why the other equations have $\omega(0)=0$, or what the physical interpretation is there. I'd imagine it's context-dependent. $\endgroup$ Commented Nov 29, 2023 at 20:16
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    $\begingroup$ It means that the energy is purely kinetic (no potential energy): if the particle (excitation; or wave front, perhaps) doesn't move, it has zero energy $\endgroup$ Commented Nov 29, 2023 at 20:43

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