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At fixed $T>0$, substitute $x=hc/(\lambda k_{B}T)>0$ into the Planck function:Here $h$ is the Planck constant, $k_B$ the Boltzmann constant, and $c$ the speed of light. Since $dx/d\lambda =-x/\lambda \ne 0$, an interior maximum in wavelength is also a stationary point in $x$. A logarithmic derivative givesThe required positive root is $x_{\lambda }=4.9651142317\dots$. The rearranged equation also has $x=0$, but that endpoint is not a peak. For $g(x)=5(1-e^{-x})-x$, $g^{\prime }(x)=5e^{-x}-1$ changes sign once, at $x=\log 5$. Thus $g$ first increases and then decreases to negative values, crossing zero exactly once for $x>0$. Since $x^5/(e^x-1)$ tends to zero at both endpoints, this stationary point is the unique global maximum. Solving the definition of $x$ for wavelength givesThe inverse-temperature dependence follows before any numerical root-finding. More generally $B_{\lambda }(T)=T^{5}F(\lambda T)$ for one fixed function $F$, so spectra at all temperatures have the same rescaled shape. This is stronger than a relation between two peak positions. Replacing the denominator by $e^x$ in the Wien approximation would instead give $x=5$ and a peak wavelength about $0.70\%$ too short.

Write $n=n_0+y\mathcal{L}{n}_0+O(y^2)$, with $n_0=(e^x-1{)}^{-1}$ and $\mathcal{L}=x^{-2}{\partial }_x(x^4{\partial }_x)$. Differentiating the Planck photon distribution givesDividing by $n_0$ proves the first-order thermal Sunyaev-Zeldovich effect:In the Rayleigh-Jeans law limit, $x{e}^x/(e^x-1)\to 1$ and $x\coth (x/2)\to 2$, so $\Delta n/n_0\to -2y$. In the Wien approximation, the prefactor approaches $x$ and the bracket approaches $x-4$, so $\Delta n/n_0\sim yx(x-4)\sim y{x}^2$. The last limit is an asymptotic description of the first-order coefficient: a perturbative prediction at large $x$ still requires $y{x}^2\ll 1$.

Hot electrons transfer energy to the cosmic microwave background, removing low-frequency photons and enhancing the high-frequency tail. The first-order distortion crosses zero where $x\coth (x/2)=4$, at $x\simeq 3.830$; for today's CMB this is near $217\mathrm{GHz}$. Photon conservation follows directly from the boundary termIn contrast, integration by parts gives $\int x^3\Delta ndx=4y\int x^3{n}_{0}dx$, so radiation gains a fraction $4y$ of its original energy. The thermal Sunyaev-Zeldovich effect is thus a spectral distortion, not simply a new blackbody temperature.

A galaxy cluster with bulk peculiar velocity along the line of sight additionally produces the kinetic Sunyaev-Zeldovich effect. Define $v_{\parallel }>0$ for recession and $\tau =\int n_e{\sigma }_{T}d\ell$. To first order in velocity and optical depth, $\Delta T/T=-\tau v_{\parallel }/c$: receding gas gives a decrement and approaching gas an increment. Its occupation number perturbation is $(\Delta T/T)x{e}^x/(e^x-1{)}^2$, the spectrum of a small blackbody temperature change. It remains nonzero at the thermal null and therefore complicates measurements near that frequency.

Planck's law gives the spectral radiance of an ideal blackbody in thermal equilibrium at absolute temperature $T>0$. It predicts how much power is carried in each small frequency or wavelength interval, per projected area and per solid angle. The whole curve is determined by $T$: no material-specific parameter enters this ideal Planck function.

Writing $\nu$ for frequency in hertz and $\lambda$ for wavelength in metres, the two forms of the Planck function areHere $h$ is the Planck constant, $k_B$ the Boltzmann constant, and $c$ the speed of light in vacuum. Their exact values in the International System of Units are $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. ^[ref]

The units matter: $B_{\nu }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, whereas $B_{\lambda }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$. Both are spectral radiance, rather than radiant exitance or energy density. To report $B_{\lambda }$ per micrometre, multiply its per-metre value by $10^{-6}$; a micrometre-wide interval is $10^{-6}$ metres wide.

For a computed example, at $T=6000\mathrm{K}$ and $\lambda =0.500\mu \mathrm{m}$, the exponent is $hc/(\lambda k_{B}T)\approx 4.79592$. Substitution into Planck's law gives $B_{\lambda }\approx 3.17569\times 10^{13}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$, or $3.17569\times 10^7\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ per micrometre. At the same wavelength, the illustrative temperatures $3000\mathrm{K}$ and $4500\mathrm{K}$ give $2.60268\times 10^5$ and $6.37793\times 10^6$ in the latter units. These are ideal predictions for stipulated temperatures.

The key change of variables is $\nu =c/\lambda$. Corresponding positive intervals carry the same power, so $B_{\lambda }d\lambda =B_{\nu }|d\nu |$ andEqual-width wavelength intervals have unequal widths in frequency; the factor $c/{\lambda }^2$ compensates for this. Merely replacing the horizontal-axis labels of $B_{\nu }$ by $c/\nu$ does not produce $B_{\lambda }$. This explains the frequency and wavelength peaks of Planck's law.

A short derivation uses the Boltzmann distribution for a single electromagnetic mode with thermal energies $nh\nu$, $n=0,1,\dots$. Put $q=e^{-h\nu /(k_{B}T)}$. The geometric series gives $Z={\sum }_{n\ge 0}{q}^n=(1-q{)}^{-1}$ and mean thermal energy $h\nu q/(1-q)$. The electromagnetic mode density in a box, including both independent polarization of an electromagnetic wave states of a photon, is $8\pi {\nu }^2/c^3$ per volume per unit frequency. Multiplication gives spectral energy density $u_{\nu }=8\pi h{\nu }^3/[c^3(e^{h\nu /(k_{B}T)}-1)]$; isotropy gives $B_{\nu }=c{u}_{\nu }/(4\pi )$. This counts thermal excitation energy, excluding zero-point energy. ^[ref]

Planck's law connects quantum mechanics to a measurable radiation curve. Its small-$h\nu /(k_{B}T)$ limit is the Rayleigh-Jeans law, $B_{\nu }\approx 2k_{B}T{\nu }^2/c^2$; its large-$h\nu /(k_{B}T)$ limit is the Wien approximation, $B_{\nu }\approx (2h{\nu }^3/c^2)e^{-h\nu /(k_{B}T)}$. The suppression at high frequency makes the total energy density finite. Integrating the curve gives the Stefan–Boltzmann law, while locating its peak gives Wien's displacement law. Thus one formula predicts both the spectral shape and the total emission.

Wien's displacement law says that the wavelength maximizing the Planck function per unit wavelength, $B_{\lambda }(T)$, moves inversely with absolute temperature:Here $b$ is the Wien displacement constant. Heating an ideal blackbody therefore shifts this peak toward shorter wavelengths, while also increasing its emission at every fixed wavelength. The peak identifies where equal narrow wavelength intervals carry the most power; it is not a single emitted wavelength. Planck's law supplies the full curve, and the Stefan–Boltzmann law supplies its integral.

Specify the spectral coordinate before finding a peak. The frequency form of Wien's displacement law maximizes $B_{\nu }$, which is spectral radiance per unit frequency. Its maximum is not $c/{\lambda }_{\max }$. The frequency and wavelength spectral radiance conversion changes the density as well as the horizontal coordinate; the existing frequency and wavelength peaks of Planck's law give a complementary numerical comparison.

In his 11 December 1911 Nobel lecture, Wilhelm Wien explained the displacement principle through a thought experiment: compress black-body radiation between reflecting moving walls. Work against radiation pressure raises the energy density, while reflection changes each frequency by the Doppler effect. The Second law of thermodynamics connects the resulting spectrum to one at a higher temperature. This determines how the spectrum scales between temperatures, without determining its complete shape. The Wien approximation is a separate, restricted spectral formula.^[ref]