# What happens to the $2\pi$ factor when calculating Raman-shifts in units of wavenumbers?

So from the classical theory, you find a formula for a dipole in a planar electromagnetic wave, where there will be two cosine terms with a frequency (actually angular velocity in $[rad/s]$, as the argument of a trigonometric function should be dimensionless and it is multiplied by $t[s]$)

$\omega_i \pm \omega_n [rad/s]$

corresponding with the Stokes & anti-Stokes Raman shifts. ($i$ for impinging and $n$ for denoting the $n$th normal vibrational mode)

$\Delta \omega_{\pm} = \frac{1}{\lambda_i}\pm \frac{1}{\lambda_n}$

Now here is something that alarms me: this is not the same $\omega$ (though it is proportional to it), it can't be. The units are all wrong. On top of that, even if you would just accept that assignment to $\Delta \omega$, then it's not a wavenumber, as the wavenumber $k$ is defined as

$k=\frac{2\pi}{\lambda} [rad/m]$

I understand the cm$^{-1}$ part (magnitudes are then usually between 0-2000, which is perfect), but what happened to the $2\pi [rad]$ factor? Is the "Raman shift" actually in terms of "inverse wavelength" ? Do you have to multiply $\Delta \omega$ by $2\pi [rad]$ to get the $\text{actual}$ wavenumber?

Extra remark: nowhere in the wikipedia-article is implied what $\omega$ should be (which is something I (possibly wrongly) assume), though I find it a bit confusing to use the symbol commonly known as the angular velocity ($\propto$ frequence) for denoting a difference in wavenumbers (usually $k$)

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There are two definitions for the wavenumber, one is $k=\frac{2\pi}{\lambda}$, the number of wavelengths per $2\pi$ units of distance (i.e. on the circle), the other one is $k=\frac{1}{\lambda}$, which is the number of wavelengths per unit distance. The formula for the Raman shift is refering to the latter definition.
Just looked it up on Wikipedia, apparently that is the case. Though I'm still a bit confused why they use $\omega$ Thanks! – PatronBernard Apr 13 '13 at 15:32
There is no unique convention for naming the frequency. Usually, $\omega$ denotes the angular frequency, but in this case it seems to be the "normal" frequency, which is usually denoted by $\nu$. – AGP Apr 13 '13 at 16:28