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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.

For a non-scattering atmosphere above a blackbody surface, the difference between the two curves is the transmitted surface term $B_{\lambda }(T_s){\mathcal{T}}_{\lambda ,s}$. The atmospheric-only curve is an integral of local thermal emission weighted by transmission to space. Molecular absorption blocks some wavelengths and permits narrow windows at others. In a transparent window, adding the hot surface substantially increases the emerging radiance; in an opaque interval, the two curves nearly coincide because the surface contribution is extinguished.

The original graph shows the clearest surface-dominated window around $0.99$–$1.04\mu \mathrm{m}$, with the blue curve near $5\mu \mathrm{W}\mathrm{c}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mu {\mathrm{m}}^{-1}$ at its maximum and the atmospheric-only contribution small. Weaker possibilities are around $0.85$–$0.92\mu \mathrm{m}$ and near $0.8\mu \mathrm{m}$, with less available signal. The $1.17$–$1.20\mu \mathrm{m}$ peak contains a measurable surface contribution but a larger atmospheric background. The near-equality around $1.27$–$1.29\mu \mathrm{m}$ makes that peak a poor surface thermometer despite its appreciable total radiance. These are approximate readings, not precisely tabulated band edges.

At $T_s\simeq 766\mathrm{K}$, the Wien displacement law puts the thermal wavelength peak near $3.8\mu \mathrm{m}$. A $1\mu \mathrm{m}$ window is still in the Wien approximation regime, but its radiance is exponentially larger than for an Earth-like $300\mathrm{K}$ or colder Martian surface. For example, at $1\mu \mathrm{m}$ the Planck function ratio $B_{\lambda }(766)/B_{\lambda }(300)$ is of order $10^{12}$. A modest hot-spot temperature increase likewise gives a relatively large fractional radiance change. Inverting this signal for surface temperature requires an atmospheric transmission and emission model, not just identifying a peak.

Under an additional thermalization assumption, an opaque star emits approximately black-body radiation characterized by the temperature near the layer from which thermal photons escape. Planck's law gives a Rayleigh-Jeans law low-frequency tail $B_{\nu }\propto T{\nu }^2$ and a Wien approximation high-frequency tail $B_{\nu }\propto {\nu }^3{e}^{-h\nu /(k_{B}T)}$.

Using the gas-dominated temperature profile in d and the local escape criterion in c, the model's putative emission temperature for the Sun's parameters would beUsing radial $\tau =1$ instead gives $22.8\mathrm{K}$. The frequency form of Wien's displacement law for the maximum of $B_{\nu }$ is $h{\nu }_{\max }/(k_{B}T)=2.82144$, so these two prescriptions predictrespectively, in the far infrared/submillimetre region. These are conditional estimates for this model, not observed solar peaks. A realistic solar effective temperature near $5.8\times 10^3\mathrm{K}$ gives ${\nu }_{\max }\simeq 3.4\times 10^{14}\mathrm{Hz}$ in $B_{\nu }$. The maximum of $B_{\lambda }$ occurs near $0.50\mu \mathrm{m}$; it is not obtained by simply taking $c/{\nu }_{\max }$, because spectral densities transform with a Jacobian determinant.

The model is unsuitable for the solar photosphere: fully ionized hydrogen at $16$–$23\mathrm{K}$ is inconsistent with thermal ionization, and the assumed opacity and near-surface temperature gradient do not resemble the Sun's. Even its central temperature is substantially below the actual solar value, about $1.6\times 10^7\mathrm{K}$.

There is a further logical limitation: the question retains only Thomson scattering. In its coherent, cold-electron limit, this redistributes photon directions without generating a Planck photon distribution. For a concrete counterexample, monochromatic injected radiation remains monochromatic after any number of such elastic scatterings; large optical depth cannot turn it into black-body radiation. True absorption and emission, or a specified pre-existing thermal radiation field with appropriate energy exchange, are missing. Thus an approximately blackbody emergent spectrum is the usual additional physical assumption, while the stated scattering-only model does not uniquely predict an emergent spectrum or its peak.

For $x=h\nu /(k_{B}T)\ll 1$, the Taylor expansion $e^x-1=x+O(x^2)$ gives the Rayleigh-Jeans law for equilibrium radiation,This is the classical limit: each electromagnetic mode has mean thermal energy approximately $k_{B}T$. Extending it to arbitrarily large frequencies would give a divergent integrated energy density, the ultraviolet catastrophe.

For $x\gg 1$, $e^x-1\simeq e^x$, giving the Wien approximationPlanck's quantum hypothesis assigns equally spaced mode energies $E_n=nh\nu$. When the spacing is small relative to $k_{B}T$, many levels are occupied and the classical approximation works. When $h\nu \gg k_{B}T$, exciting even one quantum is exponentially unlikely, suppressing the high-frequency tail. The single expression $⟨E⟩=h\nu /(e^x-1)$ contains both limits. Comparing temperature directly to $h\nu$ would be dimensionally incorrect; the low-temperature condition means $k_{B}T\ll h\nu$.

Put $x=h\nu /(k_{B}T)=hc/(\lambda k_{B}T)$. In the long-wavelength limit $x\ll 1$, the Taylor expansion $e^x-1=x+O(x^2)$ in Planck's law gives the Rayleigh-Jeans law,This is the classical equipartition theorem result: each electromagnetic mode carries mean energy $k_{B}T$. Extending it to arbitrarily high frequency would make the total energy density diverge, the ultraviolet catastrophe.

In the short-wavelength limit $x\gg 1$, $e^x-1\simeq e^x$, giving the Wien approximation,Large photon energies are thermally suppressed, removing the classical high-frequency divergence.

If the requested “form” is expressed per unit wavelength, the change of variables includes its Jacobian determinant: $U_{\lambda }=U_{\nu }|d\nu /d\lambda |=U_{\nu }c/{\lambda }^2$. ThusSubstituting $\nu =c/\lambda$ without the Jacobian determinant would give an incorrect spectral density.

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]