It has been recently pointed out to me that the solution of the heat equation in a semi-infinite material with an oscillating boundary condition at the surface is not an evanescent wave. The argument was that in the solution to the equation (which is a temperature field that oscillates in an exponentially decaying envelope), the oscillating part is a propagating wave. Do you agree with this argument? Is the solution an evanescent wave or not?

To make things clear, I am talking about waves of this form:

$T(z,t) = T_0 + A \exp(-z/\delta) \cos (\omega t - z/\delta)$

I am myself a bit concerned about the argument since Wikipedia simply states:

An evanescent wave is a near-field wave with an intensity that exhibits exponential decay without absorption as a function of the distance from the boundary at which the wave was formed.

Also, another definition I found is that of a plane wave which amplitude decays exponentially with the distance to the source.

From both definitions, it seems to me that the solution to the heat problem described above is an evanescent wave.

While investigating the question, I found that it is widely accepted that the two following examples are evanescent waves:

  • The field behind an interface in the case of a plane light wave hitting the interface in the normal direction in conditions of total reflection. In that case, the field is a standing wave in an exponentially decaying envelope.

  • Same as above, but with the light wave hitting the interface with an incidence other than normal. Here, the evanescent field has a propagating part parallel to the interface.

However, I did not find an example that has a propagating part in the direction normal to the interface and that is clearly called an evanescent wave.

  • $\begingroup$ I have the equation of the electric field of an evanescent electromagnetic wave and it is a wave propagating in direction parallel and decaying in direction normal to the interface, so this seems to match what you've been told. The equation you've written represents a decaying wave, such as an electromagnetic wave entering a conductor. I've seen evanescent waves called "standing evanescent waves", but the equation clearly shows that they do propagate. $\endgroup$
    – auxsvr
    Jun 5, 2014 at 16:27

1 Answer 1


Such waves are not evanescent waves. Evanescent waves are waves which are not absorbed. They always appear as a remaining part of a totally reflected wave — be it a total internal reflection or reflection from frequency band gap (as e.g. in photonic crystals) or whatever else.

If you try to impose an oscillating boundary condition with a frequency which would normally reflect from given medium, you'll not get evanescent wave — instead you'll get a sum of the waves which can propagate in the medium, with ever changing amplitudes and constant average energy.

What your wave given by the equation you've written resembles more is a damped wave. Consider a damped wave equation:


where $c$ is speed of wave and $k$ is friction coefficient. Imposing a condition $u(0,t)=\sin(t)$ will give you something like this result:

enter image description here

Resembles your wave, doesn't it? Although it seems precisely your one can't be a solution of $(1)$, but maybe changing type of damping will be able to give your wave. Still, the main conclusion is that this is not an evanescent wave, but a damped wave.

  • $\begingroup$ I agree with the fact that there is no "interface" like in the examples of the electromagnetic wave. About the absorption, it seems more tricky, since in heat diffusion, the medium does not absorb heat (in average), or "temperature". It looks more like a kind of dispersion where a wave of a single frequency can "disperse" itself... So to answer the initial question, that is still a little unclear to me. $\endgroup$
    – Roan
    Jun 9, 2014 at 16:39
  • $\begingroup$ I'd argue that the "energy" ($l^2$-norm of the function) does get absorbed, in a way. Suppose the oscillation stops at value of zero. Then norm of the function will monotonically decrease with time. It's easy to see, solving heat equation on a finite domain. $\endgroup$
    – Ruslan
    Jun 9, 2014 at 19:27
  • $\begingroup$ " Evanescent waves are waves which are not absorbed." That's the key thing! +1 $\endgroup$ Dec 22, 2014 at 11:45

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.