 Frequency and wavelength spectral radiance conversion

In vacuum, $\nu =c/\lambda$, where $c$ is the speed of light. Increasing wavelength decreases frequency, so the ordinary signed differentials satisfy $d\nu =-c{\lambda }^{-2}d\lambda$. A spectral slice's positive contribution to radiance must be unchanged. The change of variables formula therefore gives the absolute Jacobian determinantThese formulas use $L_{\lambda }$ per metre and $L_{\nu }$ per hertz. The factor $c/{\nu }^2$ has units $\mathrm{m}\mathrm{s}=\mathrm{m}\mathrm{H}{\mathrm{z}}^{-1}$, which converts the former density into the latter. If ${\ell }_{\mathrm{nm}}$ denotes the numerical density per nanometre in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{n}{\mathrm{m}}^{-1}$, its per-metre value is $10^9{\ell }_{\mathrm{nm}}$. Conversely, the per-nanometre value is $10^{-9}$ times the per-metre value. Likewise, a density per terahertz is $10^{12}$ times its per-hertz value because $1\mathrm{THz}=10^{12}\mathrm{Hz}$.

For a concrete stipulated spectrum, let $L_{\lambda }=C=10^9\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$ between ${\lambda }_1=500\mathrm{nm}$ and ${\lambda }_2=600\mathrm{nm}$, and zero outside. This is exactly $1\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{n}{\mathrm{m}}^{-1}$ throughout the band. Using the exact International System of Units value $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, the corresponding frequency bounds areThe transformed spectral radiance is $L_{\nu }=Cc/{\nu }^2$ on this interval. At ${\nu }_1$ it is approximately $1.200830743\times 10^{-12}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, and at ${\nu }_2$ it is $8.339102380\times 10^{-13}$ in the same units. These are respectively $1.200830743$ and $0.833910238$ per terahertz. Thus a flat wavelength density becomes a decreasing frequency density; relabelling the horizontal axis alone would miss both the shape and the units. ^[ref].

The equality of band radiances has a short exact proof:The reversed bounds and the Jacobian determinant preserve the same spectral measure. Each half-band, $500$–$550\mathrm{nm}$ and $550$–$600\mathrm{nm}$, carries $50\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, although the corresponding frequency widths are about $54.508$ and $45.423\mathrm{THz}$.

A spectrum's peak is consequently a statement about a particular density and coordinate. The frequency and wavelength peaks of Planck's law differ for precisely this reason. Integrated radiance over a specified physical band supplies a comparison that survives the change of variables.