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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?

Qmechanic
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imbAF
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3 Answers3

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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?

Philip Wood
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  • I understand what you mean. My question has to do mostly with this definition of it:"it's interpreted as the number of radians per unit distance, sometimes called "angular wavenumber"." If we have two waves with two different wavelengths, the division above, gives us their respective k values, which should be different, and can be interpreted as two circles with 2 different radiuses. – imbAF Oct 09 '22 at 18:16
  • But there's nothing here, that has two do with nr. of radians per unit length. if you divide $w\pi$ with the respective wavelengths, you'd get the respective wave number. I don't see the involvement of x amount of radians. Basically the description of the formula doesn't look as is describing the formula. That's my issue – imbAF Oct 09 '22 at 18:19
  • Don't follow you, I'm afraid. I've added a little to my answer to try and make it clearer. – Philip Wood Oct 09 '22 at 18:33
  • $\frac{k}{2\pi}$ equals number of wave cycles per cm. $k$ is number of radians per cm. A radian is simply a subdivision of a cycle, and there are $2\pi$ radians in a cycle. You can also say there are 360° in a cycle. – RC_23 Oct 09 '22 at 20:31
  • As a side note, you'll see wavenumber $k$ used quite a bit in quantum mechanics, because the energy of a photon is directly proportional to it, $E=(hc)k$. Measurements made using interferometry are most conveniently done by counting "fringes per cm," which directly measures $k$. So partly for historical reasons it is used in that field. – RC_23 Oct 09 '22 at 20:35
  • @RC_23 I've just understood: your second comment ("As a side note...") refers to $k$ defined as $1/\lambda$. Using the notation $\hbar=h/2\pi$ we can also write the photon energy formula neatly in terms of $k=2\pi/\lambda$ thus: $E=\hbar c k$. – Philip Wood Oct 12 '22 at 08:49
  • Yes, perhaps I was careless with the two versions of $k$. Good catch. – RC_23 Oct 12 '22 at 15:32
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$k=\frac{2\pi}{\lambda}$ 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 $k$.

Frederic Thomas
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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

$F(x) = \sin(k x)= \sin\left(\frac{2 \pi x}{\ell}\right)$

basics
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