Units of a wavenumber spectrum

Short definition of Units of a wavenumber spectrum
There are two commonly used units for a wavenumber and it is important to be clear about which one you are using because the level of a spectrum depends on the unit.

This is the common definition for Units of a wavenumber spectrum, but other definitions may be discussed within the wiki.

Mathematicians and theoreticians usually use ‘angular’ units expressed in radians and this should be indicated by $\mathrm{r}\mathrm{a}\mathrm{d}{\mathrm{m}}^{\mathrm{-}\mathrm{1}}$ -– radians per meter. It is the counterpart to frequency expressed in $\mathrm{r}\mathrm{a}\mathrm{d}{\mathrm{s}}^{\mathrm{-}\mathrm{1}}$ -– radians per second. Never express the units as ${\mathrm{m}}^{\mathrm{-}\mathrm{1}}$ just because an angle technically has no units. This usage is ambiguous. The other unit, which is preferred by investigational scientists because it is derived naturally by a Fourier transform, among other reasons, is $\mathrm{c}\mathrm{p}\mathrm{m}$ -– cycles per meter. It is the counterpart of $\mathrm{H}\mathrm{z}$ -– cycles per second. The two measures of wavenumber differ by a factor of $2\pi$ which is not small compared to one.

Here we use the symbol $\hat{k}$ to indicate the angular wavenumber expressed in units of $\mathrm{r}\mathrm{a}\mathrm{d}{\mathrm{m}}^{\mathrm{-}\mathrm{1}}$, and we use the symbol $k$ to indicate the cyclic wavenumber in units of $\mathrm{c}\mathrm{p}\mathrm{m}$. Their relationship is

$\hat{k}=2\pi k$

Regardless of the unit of wavenumber that you employ, the integral over a wavenumber band gives the variance within that band and this variance must be wavenumber-unit independent. Here are some examples that apply in the inertial subrange. For the velocity spectrum, we must have

$F_{22}({\hat{k}}_1)\mathrm{d}{\hat{k}}_1=F_{22}(k_1)\mathrm{d}{k}_1$

and substituting ( ) gives

Failed to parse (unknown function "\begin{equation}"): {\displaystyle \begin{equation} \begin{split} \tilde{F}_{22} (\hat{k}_1) \, \mathrm{d}\hat{k}_1 &= \frac{4}{3} C_1 \left(2\pi k_1 \right)^{-5/3} \mathrm{d} (2\pi k_1 ) \\ &= \left(2\pi\right)^{-2/3}\, \frac{4}{3} C_1 \,k_1^{-5/3}\, \mathrm{d}k_1 \end{split} \end{equation} }

which means that, in the inertial subrange, the cross-profile spectrum of velocity, ${\tilde{F}}_{22}(k_1)$, expressed in units of $\mathrm{c}\mathrm{p}\mathrm{m}$, is smaller than the same spectrum, ${\tilde{F}}_{22}({\hat{k}}_1)$, expressed in units of $\mathrm{r}\mathrm{a}\mathrm{d}{\mathrm{m}}^{\mathrm{-}\mathrm{1}}$.

Similarly, the universal shear spectrum is

Failed to parse (unknown function "\begin{equation}"): {\displaystyle \begin{equation} \begin{split} \tilde{G}_{22} (\hat{k}_1) \, \mathrm{d}\hat{k}_1 &= \frac{4}{3} C_1 \left(2\pi k_1 \right)^{1/3} \mathrm{d} (2\pi k_1 ) \\ &= \left(2\pi\right)^{4/3}\, \frac{4}{3} C_1 \,k_1^{1/3}\, \mathrm{d}k_1 \end{split} \end{equation} }

which means that the shear spectrum, expressed in units of $\mathrm{c}\mathrm{p}\mathrm{m}$, is larger by a factor of $(2\pi {)}^{4/3}$ in the inertial subrange than the shear spectrum expressed in units of $\mathrm{r}\mathrm{a}\mathrm{d}{\mathrm{m}}^{\mathrm{-}\mathrm{1}}$. Finally, the complete shear spectrum must integrate to 2/15 over all wavenumbers and, therefore, the peak of the shear spectrum expressed in units of $\mathrm{c}\mathrm{p}\mathrm{m}$ is larger than the shear spectrum expressed in units of $\mathrm{r}\mathrm{a}\mathrm{d}{\mathrm{m}}^{\mathrm{-}\mathrm{1}}$ by a factor of $2\pi$.