2
$\begingroup$

I'm looking at some dispersion relations for some complex systems and realised I actually don't have a clear understanding of what physics I can get from a dispersion relation from equations that produce complex relations between frequencies and wavevectors. For instance, if you take a diffusion equation

$$\dot u=\nabla^2u$$

for an initial condition $u_0(x)=u(x,t=0)$, then this equation can be simply solved in Fourier space, and the solution is

$$u(x,t) = \int \frac{d^dq}{(2\pi)^d}\tilde u_0(q)e^{-q^2t}e^{i q\cdot r}$$

If the initial equation is transformed to Fourier space also in time we get a complex relation between frequency $\omega$ and wavevector $k$, $$i\omega+q^2=0.$$

Which will be generaly the case for a scalar field $u$ for any overdamped dynamics $\dot u= F[u]$ for any $F$.

My question is what do we make of a relation of that kind? I'm guessing something along the lines of, since this is overdamped and just a scalar field, you don't expect wave-like solutions, so since an initial condition won't propagate like a wave, there's no real correspondence between $\omega $ and $q$. I know this is extremely vague, but I'm looking for some physical insight on this.

To expand the question a bit, if $F[u]$, for instance $F[u]=\nabla^2u +A[u]$ for a nonlinear operator $A$, we can 'solve' for $u$ through its linear propagator $$\tilde u = \frac{1}{i\omega+q^2}\tilde A[\tilde u]$$

How do we interpret this propagator function physically?

Thanks!

$\endgroup$
1
  • $\begingroup$ yes you can solve it in terms of perturbation theory. you can assume $A=0$ then order by order correct the solution. $\endgroup$
    – physshyp
    Commented Nov 5, 2020 at 13:35

1 Answer 1

2
+25
$\begingroup$

Indeed, in wave phenomena the dispersion relation has a clear interpretation in terms of the phase and group velocities. Another place where dispersion frequently comes in play is in discussing non-linear waves: e.g., solitons are often describes as an interplay between the dispersion and the non-linearity.

In Diffusion phenomena speaking of group velocity doesn't make sense, but the Fourier transform still suggests a direct interpretation as the rate of decay/damping of modes with particular wave vectors (or particular spatial structure, if we are not dealing with an open space). In this respect, using Laplace rather than Fourier transform in time makes things clearer.

When it comes to non-linear phenomena, very similar interplays between dispersion and non-linearity are in place: Swift-Hohenberg equation is known to described quite a few types of non-linear phenomena, including some solitary solutions.

Finally, in the quantum domain (since Schrödinger equation is similar to the diffusion equation) the dispersion relation plays the role of the non-relativistic energy-momentum relation. In solid state it may have a rather different and often anisotropic form.

$\endgroup$
1
  • $\begingroup$ Thanks for the answer! I was also looking for some kind of an image of what the correlator means for a perturbative solution of an equation of the kind I wrote in the question though. I however see that for decaying modes that do not represent traveling waves you get said complex dispersion relations which give you the rate of decaying. $\endgroup$ Commented Nov 6, 2020 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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