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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?

One Answer

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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