# Waveguide cut off frequency derivation - Wave equation to Helmholtz equation

I'm trying to derive the cut off frequency for a wave guide. I found a derivation on wikipedia, but I don't understand the first step where we go from the wave equation to the helmholtz equation. Why does only considering $$\psi(x,y,z,t)=\psi(x,y,z)e^{i\omega t}$$ mean you can go from the wave equation to the helmholtz equation? ## 1 Answer

By guessing a time dependence of the form $$e^{i\omega t}$$ you remove the time dependence from the equation. specifically, it transforms to the Helmholz equation.

I recommend you set $$\psi(x,y,z,t)=u(x,y,z)e^{i\omega t}$$ and substitute into the wave equation and do the algebra and see for yourself. You'll get the Helmholz equation on the spatial part $$u(x,y,z)$$.

I suggest you read on separation of variables

• I think I see it now, let $\psi (x,y,z,t)e^{i\omega t}$ in the wave equation, then expand the brackets. I see where the $\omega^2$ comes from now in the helmholtz equation, but where does the $-e^{i\omega t}$ go when you take the second derivative wrt t? Oh, does it cancel out with the other $-e^{i\omega t}$ from the lapace? – David Mar 16 at 13:34
• sorry for the delay. the factor $e^{i\omega t}$ indeed cancels out with the one left over from applying the Laplacian . @David – Tomka Mar 17 at 5:37