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Using the usual notation for rectangular (Cartesian), cylindrical, and spherical coordinates, which of the following expressions represent a plane wave propagating in free space? $$e^{-jk_0R\sin(\theta)} $$ $$e^{-jk_0R\cos(\theta)} $$ $$e^{-jk_0R}$$ $$ e^{-jk_0r}$$ where $j$ is the imaginary unit, and $k_0$ is the wavenumber in free space.

I found the real notation of each expression, because I thought it would reveal an expression that stood out from the others. $$ \cos(\omega t-k_0R\sin(\theta))$$ $$ \cos(\omega t-k_0R\cos(\theta))$$ $$ \cos(\omega t-k_0R)$$ $$ \cos(\omega t-k_0r)$$

But as can be seen, this is not the case. In fact, the only difference between the third and fourth expression, is that one is in spherical coordinates, while the other is in cylindrical.

So how do I distinguish a plane wave from a non-plane wave. I feel like I'm missing some fundamental reasoning.

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1 Answer 1

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The following may be useful.

The usual form of a plane wave will be something that involves a vector dot product between the $\vec{k}$-vector and the position vector, call it $\vec{x}$. Something of the form $\vec{k} \cdot \vec{x}$ will appear in the argument of the sinusoid.

I hope this helps.

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  • $\begingroup$ Oh, so because there has to be a position vector in the argument of the sinusoid, it must be expression 2, right? Actually, $e^{-jk_0R\cos(\theta)} $ in spherical coordinates is the same as $e^{-jk_0z}$ in rectangular coordinates, right? $\endgroup$
    – Carl
    Nov 22, 2020 at 17:33

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