Sometimes, I see that wavenumber units are $\text{m}^{-1}$, but on the other hand, (and by definition of $k = \omega/v = 2\pi/\lambda$), it is $\text{rad/m}$. What is correct?
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Wavenumber has units of inverse length. – Arturo don Juan Jul 14 '21 at 01:21
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@ArturodonJuan Ok, but why? From k=ω/v=2π/λ it is rad/m. – Yoni Berkovitch Jul 14 '21 at 01:28
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Both are correct. It's just an issue of angular versus linear units. The linear wavenumber $\tilde{v}$ is $$\tilde{v}=\frac{1}{\lambda}$$ while the angular wavenumber $k$ is $$k=2\pi\tilde{v}=\frac{2\pi}{\lambda}$$ The linear wavenumber is seldom used compared to the angular wavenumber, so most of the time, by "wavenumber", we mean the angular one $k$. Remember that the whole issue of angular and linear arises because $$1 \;\text{cycle} = 2\pi \;\text{radians}$$ As for the radian itself, it is a dimensionless quantity because it is defined as the ratio of lengths. For more information, see this post and this post.
Vincent Thacker
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Radians are dimensionless as they are the ratio of an arclength (which has units of length) to the radius (also units of length).
ZeroTheHero
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