Wavenumber definition in theoretical physics I am trying to understand the physical meaning of the wavenumber, which as explained in wikipedia, is the magnitude of the wave vector, which, if I am not mistaken, the wave vector gives information about the direction of the propagation of an EM/Matter-wave.
The crystallographic def. of the wave number, is pretty clear to me:
$$k=\frac 1 \lambda$$
the number of complete cycles that exist in 1 meter of linear space. A complete cycle translates in space as the distance between two points which have a $2\pi$ phase difference between each other (in other words points of same phase), and this distance is equal to a wavelength $\lambda$. This is how I understand the wave number definition in crystallography. If I am wrong about something, please let me know.
In theoretical physics, the formula is:
$$k=\frac{2\pi}{\lambda}$$
and it's interpreted as the number of radians per unit distance, sometimes called "angular wavenumber". I don't understand this. And I cannot see how the wave numbers of two waves with different wavelengths $\lambda_1$,$\lambda_2$, would be different (considering the above formula).
As I said above, 1 wavelength, is a full cycle, which is $2\pi$ radian. This should be valid for both waves. The only difference, is the number of full cycles per unit of time, in other words the frequency.
What am I getting wrong here?
 A: I expect that you are happy with the equation that describes the displacement of a plane sinusoidal wave propagating in the $x$ direction, when expressed as:
$$y=A\sin2\pi\left(\frac tT -\frac x {\lambda} + \epsilon\right)$$
Every time $t$ changes by a period, $T$, or we change $x$ by a distance $\lambda$, the argument of the sine changes through $2\pi$ so $y$ goes through a complete cycle.
If we put $\omega=\frac{2\pi}T$ and $k=\frac{2\pi}{\lambda}$, the equation can be written more compactly as
$$y=A\sin(\omega t-kx +\epsilon).$$
I think of $k$ as a conversion factor between distance and angle, such that a distance of one wavelength is, through multiplication by $k$, converted to an angle of $2\pi$, and a distance $x$ into an angle of $2\pi\frac x{\lambda}$. This is just as multiplication by  $\omega$ converts a time of one period into an angle of $2\pi$ and a time $t$ into an angle of $2\pi\frac t{T}$.
In short, $k$ defined as $2\pi/\lambda$ gives radians per unit distance in the propagation direction, whereas $k'$ defined as $1/\lambda$ gives cycles per unit distance.
For a wave propagating in any direction we define $\mathbf k$ as a vector in the direction of wave propagation and of magnitude $k=\frac{2\pi}{\lambda}$. Then our wave propagation equation becomes
$$y=A\sin(\omega t-\mathbf k.\mathbf r +\epsilon).$$
Neat, is it not?
A: $k=\frac{2\pi}{\lambda}$ can be understood as phase advance $\Delta\phi$ per length (or if preferred per meter). One cycle corresponds to a phase advance of $2\pi$.
So if one considers the phase advance of one cycle, i.e. $\Delta\phi = 2\pi$ then distance the wave propagated is $x=\lambda$, i.e. one wave length.
We can then write using the concept of the wave vector as phase advance per distance $k=\frac{\Delta\phi}{x}$ with $x$ as distance:
$2\pi =\Delta\phi=\frac{\Delta\phi}{x}x\equiv k\lambda$, therefore we obtain $k = \frac{2\pi}{\lambda}$.
In particular, if the wave length is short, the phase advance on a fixed distance is larger as if the wave length would be large. This fact is nicely expressed in the quantity of the wave vector $k$.
A: You can imagine the wavenumber $k=\frac{2\pi}{\ell}$ of function in space as the spatial counterpart of the relation between angular velocity and period $\omega=\frac{2\pi}{T}$ of functions of time.
The spatial dimension of the wavenumber is $\frac{1}{\text{length}}$, as the dimension of angular velocity is $\frac{1}{\text{time}}$.
To visualize its meaning, try to plot the function
$F(x) = \sin(k x)= \sin\left(\frac{2 \pi x}{\ell}\right)$
