# What is the intuitution behind a wave with zero wavenumber?

Physics Asked by levente.nas on December 1, 2020

I was recently solving a physics problem that gave you the electric field $$E(x, y, z, t) = E_{0}cos(k_{y}y + psi_{y})sin(k_{x}x-{omega}t)hat{z}$$ and the question asked for a variety of things including $$k_{y}$$, the dispersion relation, and the lowest frequency that will propagate. With the boundary conditions E = 0 at y = 0 and y = b, you can find $$k_{y}$$ to be $$npi/a$$ where n is a number greater than 0. With those boundary conditions, you can also find $$psi_{y}$$ to be $$pi/2$$. With some trig, you can simplify the electric field to $$E(x, y, z, t) = E_{0}sin((npi/a)y)sin(k_{x}x-{omega}t)hat{z}$$ To find the dispersion relation, you can plug this electric field into the wave equation. The result ends up being $$k_{x} = sqrt{(omega^2/c^2) – (npi/a)^2}$$. After this, I got stuck. The answer key says the lowest frequency that will propagate happens at n = 1 (why at n = 1?), and with this they find $$omega_{cutoff}$$ to be $${pi}c/a$$. If this is the case, doesn’t $$k_{x}$$ have to be zero? If that is the case, what does a zero wavenumber mean intuitively?

In general, $$n=0,1,2,...$$. But if $$n=0$$, the field is zero, because the field is proportional to $$sin(npi y/a)$$. So the smallest value of $$n$$ for which the field can propagate is $$n=1$$.

In general, $$k_x=0$$ just means that the field is constant in the $$x$$ direction. A way think about this is that $$k_x=2pi/lambda_x$$, where $$lambda_x$$ is the wavelength in the $$x$$ direction. The limit of 0 wavenumber $$k_xrightarrow 0$$ corresponds to the limit of infinite wavelength $$lambda_x rightarrow infty$$. You can think of a constant as a wave with an infinite wavelength. Similarly, $$|vec{k}|=0$$ means that the field is a constant over all space.

I'm not quite sure why you have written the part of the field varying in the $$x$$ direction as $$sin(k_x x - omega t)$$, as opposed to $$sin(k_x x-omega t +phi_0)$$, for some initial phase $$phi_0$$. If this is allowed by the boundary conditions, then for $$k_x=0$$, we would have $$omega=npi/a$$, so the field would be vary in time but be constant in the $$x$$ direction, explicitly given by by $$sin(- n pi/a + phi_0)$$. For $$phi_0=0$$ (which is the expression you started with), the field would be identically zero.

Correct answer by Andrew on December 1, 2020

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