Temperature perturbation of a Planck spectrum ($\delta B_{\nu }/B_{\nu }=[x/(1-e^{-x})]\delta T/T$)

For a Planck function expressed as spectral radiance per frequency interval,$h$ is the Planck constant, $c$ the speed of light and $k_B$ the Boltzmann constant. Differentiating at fixed frequency givesThe proof uses ${\partial }_{T}x=-x/T$ and the chain rule. At low frequency, the multiplier tends to one, recovering the Rayleigh-Jeans law. At high frequency it grows approximately as $x$: the same fractional temperature change has a larger fractional brightness effect on the declining spectral tail.

For an original example choose $T=2.725\mathrm{K}$ and $\delta T=1\mathrm{mK}$. At $100\mathrm{GHz}$, $x=1.76119$, so the first-order fractional residual is $7.80415\times 10^{-4}$. At $300\mathrm{GHz}$, $x=5.28357$, it is $1.94881\times 10^{-3}$. Both spectra are exact blackbodies: this residual is a temperature mismatch, not evidence of a nonthermal spectrum. Fitting temperature before assessing remaining spectral residuals therefore removes a physically distinct degree of freedom. The figure divides by a fixed peak brightness rather than by the rapidly vanishing high-frequency tail.