Stefan-Boltzmann and Wien Displacement Laws

The Planck law hands you the entire spectral distribution of blackbody radiation — a curve giving the intensity at every frequency. Almost nobody carries that whole curve around. Two of its summary numbers do the daily work of astrophysics, and both were found by experiment before Planck explained why they held. The Stefan-Boltzmann law reads off the area under the Planck curve: the total power radiated per unit area, which scales as the fourth power of temperature. Wien's displacement law reads off the location of the summit: the wavelength at which the curve peaks, which slides inversely with temperature. Area and peak — one integral and one derivative of the same function. Their reach is enormous precisely because each collapses a spectrum into a single measurable quantity. The steep $F=\sigma T^4$ means a factor-of-two rise in temperature brightens a surface sixteenfold, which is why a star's effective temperature is defined through it and why the luminosities on the Hertzsprung-Russell diagram fan out over so many decades. The color-temperature relation $\lambda_{\max}T=b$ lets you take a source's temperature from its hue alone: a cool red-giant photosphere, the glare of an O star, and the $2.725\,\mathrm{K}$ cosmic microwave background all announce their temperatures by where they peak. And because both $\sigma$ and $b$ are built purely from $k_B$ , $\hbar$ , and $c$ times a pure number, measuring them was one of the first routes to the quantum constants themselves. This article derives both laws from the Planck function, verifies every constant, and shows where they earn their keep.

Key takeaways

The Stefan-Boltzmann law is the area under the Planck curve: a blackbody radiates a flux $F=\sigma T^4$ , with $\sigma=5.670\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ built from $k_B$ , $\hbar$ , $c$ .
Wien's displacement law is the peak of that curve: $\lambda_{\max}T=b=2.898\times10^{-3}\,\mathrm{m\,K}$ , so hotter bodies peak at shorter (bluer) wavelengths.
Both are exact consequences of one spectrum: integrate the Planck function to get $\sigma T^4$ ; differentiate it and set the result to zero to get $\lambda_{\max}T=b$ .
The $T^4$ comes from four powers of $k_BT$ pulled out of the frequency integral; the pure number is $\pi^4/15$ from $\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x$ .
Together they yield effective temperatures, radii, luminosities, and color temperatures — from the Sun ( $5772\,\mathrm{K}$ , peak $502\,\mathrm{nm}$ ) to the CMB ( $2.725\,\mathrm{K}$ , peak $1.06\,\mathrm{mm}$ ).
The frequency peak and the wavelength peak sit at different photon energies — a Jacobian trap that catches everyone once.
Key idea: Integrate the Planck function over frequency and solid angle to get $F=\sigma T^4$ ; differentiate it and set the result to zero to get $\lambda_{\max}T=b$ . Both the $T^4$ scaling and the color-temperature relation are exact properties of one spectrum.

Where this lives today

The two laws are the routine arithmetic of modern observation. Every stellar effective temperature in a survey catalogue — Gaia's billion-plus stars, the SDSS spectroscopic samples — is a Stefan-Boltzmann inversion at heart, and the color indices those surveys report are Wien's law in broadband disguise. Exoplanet science leans on both: a planet's equilibrium temperature balances absorbed stellar flux against its own $\sigma T^4$ re-radiation, the first cut on the habitable zone, and the same $\sigma T^4$ sets Earth's climate budget. At the cosmological frontier, COBE's FIRAS measured the CMB to be a blackbody at $2.72548\pm0.00057\,\mathrm{K}$ with deviations below one part in $10^4$ — the most perfect blackbody ever observed, and direct confirmation that the early Universe was in thermal equilibrium. Current CMB experiments (the Simons Observatory, CMB-S4) work right at the $160\,\mathrm{GHz}$ Wien peak because that is where the signal is strongest against foregrounds. And in stellar interiors, the radiation constant $a=4\sigma/c$ sets the radiation pressure $P=\tfrac13 aT^4$ that holds massive stars up.

The physical picture

Picture the Planck spectrum as a single bell-shaped hill whose whole shape is fixed by one dial: the temperature. Two numbers summarize the hill. Its total area — $B_\nu$ integrated over all frequencies — is the energy the surface pours out each second, the Stefan-Boltzmann quantity. The position of its summit is the Wien quantity. Turn the dial up and two things happen at once: the hill grows dramatically taller (its area climbs as $T^4$ , its peak height even faster, as $T^5$ ) and its summit slides toward shorter wavelengths. The first figure shows this hill for the Sun and marks both numbers on the same curve.

Figure

The Planck spectrum of the Sun at T = 5772 K. Wien's law fixes the peak at λ_max = 502 nm, while the shaded area under the curve is the total flux F = σT⁴ of the Stefan-Boltzmann law — both are read from one temperature.

The second face is the raw numerology of the exponents. Why $T^4$ and not $T^3$ ? Substitute $x=h\nu/k_BT$ in the frequency integral and four factors of $k_BT$ fall out — three from the $\nu^3$ prefactor, one from $\mathrm{d}\nu$ — leaving a pure number behind. (Photon number density alone scales as $T^3$ ; each photon's mean energy $\sim k_BT$ supplies the fourth power.) Wien's law, meanwhile, says the curve is self-similar: written against $x=hc/(\lambda k_BT)$ the Planck shape is universal, so its peak sits at a fixed $x$ and the peak wavelength must scale as $1/T$ . The second figure turns that $1/T$ scaling into a straight line and hangs four real blackbodies on it.

Figure

Wien's displacement law λ_max·T = 2.898×10⁻³ m·K as one straight line on log-log axes, anchored by four blackbodies: the 2.725 K microwave background at 1.06 mm, a 3000 K M dwarf in the near infrared, the 5772 K Sun at 502 nm, and a 30000 K O star in the far ultraviolet.

Because the color-temperature relation is monotonic, you do not even need to resolve the peak to use it: a single flux ratio through two broadband filters pins the temperature of any thermal source — the basis of photometric temperatures for stars too faint to spectrograph.

The core equations

The headline result is the total emergent flux. Integrate the Planck intensity over all frequencies and over the outward hemisphere and every trace of the spectral shape collapses into one number times $T^4$ .

F=\sigma T^4 \qquad (1)

where: $F$ — emergent radiative flux from a blackbody surface (W m⁻²); $\sigma$ — Stefan-Boltzmann constant, $5.670\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ ; $T$ — absolute temperature (K). Double the temperature and the surface brightness rises by $2^4=16$ . The constant $\sigma$ is not fundamental. It is a fixed bundle of the genuinely fundamental constants times the pure number $\pi^4/15$ that came out of the integral, packaged with the hemisphere factor.

\sigma=\frac{2\pi^5 k_B^4}{15\,c^2 h^3}=\frac{\pi^2 k_B^4}{60\,\hbar^3 c^2} \qquad (2)

where: $k_B$ — Boltzmann constant, $1.381\times10^{-23}\,\mathrm{J\,K^{-1}}$ ; $h,\hbar$ — Planck and reduced Planck constant ( $\hbar=h/2\pi$ ); $c$ — speed of light. The two forms are identical: substituting $h=2\pi\hbar$ turns $2\pi^5/15$ into $\pi^2/60$ . The companion headline is the peak location. In wavelength, the product of the peak wavelength and the temperature is a constant.

\lambda_{\max}\,T=b,\qquad b=\frac{hc}{x_5\,k_B}=2.898\times10^{-3}\,\mathrm{m\,K} \qquad (3)

where: $\lambda_{\max}$ — wavelength of the peak of $B_\lambda(T)$ ; $b$ — Wien displacement constant; $x_5\approx4.965$ is the nonzero root of $x=5(1-e^{-x})$ . Reading: at $5772\,\mathrm{K}$ the peak sits at $b/T\approx502\,\mathrm{nm}$ , blue-green. The peak has a different location if you plot per unit frequency instead of per unit wavelength, because the two are densities in different variables. The frequency peak obeys its own transcendental root.

\nu_{\max}=\frac{x_3\,k_B}{h}\,T\approx 5.879\times10^{10}\,\mathrm{Hz\,K^{-1}}\;T \qquad (4)

where: $\nu_{\max}$ — frequency of the peak of $B_\nu(T)$ ; $x_3\approx2.821$ is the nonzero root of $x=3(1-e^{-x})$ . Reading: $\nu_{\max}$ and $\lambda_{\max}$ do not satisfy $\lambda_{\max}=c/\nu_{\max}$ — see the warning below. Everything above is extracted from one integrand, the Planck function, here in its per-unit-frequency form.

B_\nu(T)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/k_BT}-1} \qquad (5)

where: $B_\nu$ — specific intensity per unit frequency (W m⁻² Hz⁻¹ sr⁻¹), isotropic for a blackbody. This is the curve whose area gives (1) and whose summit gives (4). The flux out of a surface is $\pi$ times the frequency-integrated intensity, the $\pi$ being the projected solid angle of the outward hemisphere.

F=\pi B=\pi\int_0^\infty B_\nu\,\mathrm{d}\nu \qquad (6)

where: $B$ — frequency-integrated Planck intensity (W m⁻² sr⁻¹); $\pi$ — the value of $\int\cos\theta\,\mathrm{d}\Omega$ over the outward hemisphere. Inside an enclosure the relevant quantity is the isotropic radiation energy density, which carries the same $T^4$ but a different constant.

u=aT^4,\qquad a=\frac{4\sigma}{c}=7.566\times10^{-16}\,\mathrm{J\,m^{-3}\,K^{-4}} \qquad (7)

where: $u$ — radiation energy density; $a$ — radiation constant. The factor $c/4$ separates a one-sided emergent flux from an isotropic interior energy density. Finally, the astrophysical workhorse: multiply the surface flux by the area of a sphere to get a star's bolometric luminosity, which defines its effective temperature.

L=4\pi R^2\,\sigma T_{\mathrm{eff}}^4 \qquad (8)

where: $L$ — bolometric luminosity (W); $R$ — stellar radius; $T_{\mathrm{eff}}$ — effective temperature, the blackbody temperature reproducing the star's output at radius $R$ .

Derivation AStefan-Boltzmann from the Planck function

The goal is to start from nothing but the emergent flux of an isotropic blackbody surface and the Planck intensity, carry out the frequency and angle integrals, and arrive at both the $T^4$ law and the closed form $\sigma=2\pi^5 k_B^4/(15c^2h^3)$ — showing exactly where the fourth power comes from (four factors of $k_BT$ from the substitution) and where the pure number $\pi^4/15$ comes from (a standard Bose-Einstein integral). The constant is built, not quoted.

Step 1. Write the flux emerging from a surface as the intensity weighted by the projection factor $\cos\theta$ and integrated over the outward hemisphere: $F=\int_0^{2\pi}\!\int_0^{\pi/2} B\cos\theta\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi=\pi B$ , where $B$ is the frequency-integrated intensity. Why: a detector on the surface sees each ray foreshortened by $\cos\theta$ , and $\int\cos\theta\,\mathrm{d}\Omega$ over a hemisphere is exactly $\pi$ , not $2\pi$ .

Step 2. Insert the Planck function and non-dimensionalize with $x=h\nu/k_BT$ , so $\nu=(k_BT/h)x$ and $\mathrm{d}\nu=(k_BT/h)\,\mathrm{d}x$ : $B=\int_0^\infty \frac{2h\nu^3}{c^2}\frac{\mathrm{d}\nu}{e^{h\nu/k_BT}-1}=\frac{2h}{c^2}\left(\frac{k_BT}{h}\right)^4\!\int_0^\infty\frac{x^3\,\mathrm{d}x}{e^x-1}$ . Why: the $\nu^3$ contributes three powers of $k_BT/h$ and the $\mathrm{d}\nu$ contributes a fourth, which is the entire origin of the fourth power — the remaining integral is a pure number.

Step 3. Evaluate the dimensionless integral as a Gamma-times-zeta product: $\int_0^\infty\frac{x^3\,\mathrm{d}x}{e^x-1}=\Gamma(4)\,\zeta(4)=6\cdot\frac{\pi^4}{90}=\frac{\pi^4}{15}\approx6.4939$ . Why: expanding $1/(e^x-1)=\sum_{n\ge1}e^{-nx}$ and integrating term by term gives $\Gamma(4)\sum n^{-4}=6\,\zeta(4)$ , and $\zeta(4)=\pi^4/90$ is a standard result.

Step 4. Assemble and identify the constant: $F=\pi B=\frac{2\pi h}{c^2}\left(\frac{k_B}{h}\right)^4\frac{\pi^4}{15}\,T^4=\frac{2\pi^5 k_B^4}{15\,c^2 h^3}\,T^4\equiv\sigma T^4$ . Why: collecting every constant in front of $T^4$ literally defines $\sigma$ ; plugging in CODATA $k_B,h,c$ returns $5.670\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ . *Verified: $\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x=\pi^4/15=6.4939$ , and $2\pi^5 k_B^4/(15c^2h^3)=5.6704\times10^{-8}$ from CODATA constants; matches Carroll & Ostlie Eq. (3.18).*

Go deeper · Derivation B · 1 minWien displacement law by maximizing the Planck curve

Derivation BWien displacement law by maximizing the Planck curve

The goal is to locate the summit of the Planck spectrum rigorously: switch the intensity to a per-unit-wavelength density using the correct Jacobian, differentiate with respect to wavelength, set the derivative to zero, and reduce the condition to a transcendental equation whose root delivers the displacement constant $b=hc/(x_5 k_B)$ . Along the way it becomes clear why the wavelength and frequency peaks obey different equations and sit at different photon energies.

Step 1. Convert $B_\nu$ to $B_\lambda$ with the Jacobian of $\nu=c/\lambda$ , namely $B_\lambda=B_\nu\,|\mathrm{d}\nu/\mathrm{d}\lambda|=B_\nu\,c/\lambda^2$ , giving $B_\lambda(T)=\frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda k_BT}-1}$ . Why: intensity is a density, so it must be transformed with the Jacobian, and it is this $c/\lambda^2$ weighting that will make the wavelength peak differ from the frequency peak.

Step 2. Set $\mathrm{d}B_\lambda/\mathrm{d}\lambda=0$ . Writing $x=hc/(\lambda k_BT)$ and differentiating the product $\lambda^{-5}(e^x-1)^{-1}$ gives, after clearing common factors, the condition $5(e^x-1)=x\,e^x$ , equivalently $x=5(1-e^{-x})$ . Why: at the maximum the two competing trends — the $\lambda^{-5}$ prefactor pushing the peak to short wavelength and the Bose factor cutting off the short-wavelength end — exactly balance, which is what the derivative-equals-zero condition encodes.

Step 3. Solve the transcendental equation numerically. The trivial root $x=0$ is a minimum; the physical root is $x_5=4.965114$ , found by iterating $x\leftarrow5(1-e^{-x})$ to convergence. Why: the equation cannot be solved in closed form, but fixed-point iteration converges quickly to $4.965$ ; the analogous frequency condition $x=3(1-e^{-x})$ has root $x_3=2.821$ , and the two differ because of the Jacobian in Step 1.

Step 4. Extract the constant: $\lambda_{\max}=\frac{hc}{x_5 k_B}\frac{1}{T}$ , so $\lambda_{\max}T=\frac{hc}{x_5 k_B}=2.898\times10^{-3}\,\mathrm{m\,K}$ . Why: with $x_5$ fixed and $h,c,k_B$ known, the product $\lambda_{\max}T$ is a pure constant — the whole content of the displacement law. *Verified: the root of $x=5(1-e^{-x})$ is $4.96511$ , and $hc/(x_5 k_B)=2.8978\times10^{-3}\,\mathrm{m\,K}$ ; the frequency root is $2.82144$ giving $\nu_{\max}/T=5.879\times10^{10}\,\mathrm{Hz\,K^{-1}}$ . Matches Ryden Eq. (2.30)-type values.*

Go deeper · Derivation C · 1 minRadiation energy density and the factor c over 4

Derivation CRadiation energy density and the factor c over 4

The goal is to derive the interior radiation energy density $u=aT^4$ and pin down why its constant $a$ differs from the surface constant $\sigma$ by exactly $4/c$ . This closes the loop between how much energy sits in a radiation field and how much streams out of a hole in it, supplying the $a=4\sigma/c$ that runs through stellar-interior and cosmology calculations.

Step 1. The isotropic energy density is the intensity integrated over all $4\pi$ steradians and divided by $c$ : $u=\frac{1}{c}\int B\,\mathrm{d}\Omega=\frac{4\pi}{c}B$ . Why: energy density counts photons flowing in every direction, so it uses the full solid angle $4\pi$ and converts the streaming intensity to a stored density by dividing by the propagation speed $c$ .

Step 2. Substitute the frequency-integrated Planck intensity $B=\frac{2\pi^4 k_B^4}{15c^2h^3}T^4$ from the previous derivation (note $F=\pi B=\sigma T^4$ so $B=\sigma T^4/\pi$ ): $u=\frac{4\pi}{c}\cdot\frac{\sigma T^4}{\pi}=\frac{4\sigma}{c}T^4\equiv aT^4$ . Why: the same spectral integral that built $\sigma$ builds $a$ ; only the geometric prefactor changes from $\pi$ (one-sided flux) to $4\pi/c$ (all-sky density).

Step 3. Compare the two constants: $a=4\sigma/c=7.566\times10^{-16}\,\mathrm{J\,m^{-3}\,K^{-4}}$ , and equivalently the emergent flux from a cavity is $F=\frac{c}{4}u$ . Why: the $c/4$ is the standard effusion factor — averaging $c\cos\theta$ over the outward hemisphere and normalizing over all directions gives $c/4$ , the same factor that relates gas pressure to flux through a hole. *Verified: $a=4\sigma/c=7.5657\times10^{-16}\,\mathrm{J\,m^{-3}\,K^{-4}}$ and $F=(c/4)u$ reproduce $\sigma$ ; consistent with Rybicki & Lightman Ch. 1.*

Limiting cases

The two laws are the integrated and extremal summaries of one spectrum, so their behavior in each corner of the Planck curve is worth reading off directly.

Case	Result	Meaning
Rayleigh-Jeans (low frequency, $h\nu\ll k_BT$ )	$B_\nu\approx\dfrac{2\nu^2 k_B T}{c^2}$	Classical limit; integrating this alone over all $\nu$ diverges (the ultraviolet catastrophe), so the finite $T^4$ demands the full Planck cutoff.
Wien tail (high frequency, $h\nu\gg k_BT$ )	$B_\nu\approx\dfrac{2h\nu^3}{c^2}e^{-h\nu/k_BT}$	The spectrum falls exponentially, which is what makes the peak sharp and $\lambda_{\max}T$ finite; historically Wien's own 1896 formula.
Effective temperature	$L=4\pi R^2\sigma T_{\mathrm{eff}}^4$	Defines $T_{\mathrm{eff}}$ for any star as the blackbody temperature giving its observed luminosity at radius $R$ , even when the true spectrum is only approximately Planckian.
Energy density vs flux	$u=aT^4,\ a=\dfrac{4\sigma}{c}$	The radiation constant relates interior energy density to $\sigma$ ; a factor $c/4$ separates isotropic density from one-sided flux.
Wavelength peak vs frequency peak	$x_5=4.965$ vs $x_3=2.821$	Different transcendental roots, so $\lambda_{\max}\neq c/\nu_{\max}$ ; the mismatch is the Jacobian $c/\lambda^2$ , not an error.

Common trapThe wavelength peak and the frequency peak are not the same photon

It is tempting to convert Wien's wavelength law into a frequency with $\nu=c/\lambda_{\max}$ and expect the frequency peak. It fails, and not by a little. Because $B_\nu\,\mathrm{d}\nu=B_\lambda\,\mathrm{d}\lambda$ relates the two densities through a Jacobian $c/\lambda^2$ , maximizing $B_\lambda$ and maximizing $B_\nu$ solve different transcendental equations — $x=5(1-e^{-x})$ giving $x_5=4.965$ for wavelength, versus $x=3(1-e^{-x})$ giving $x_3=2.821$ for frequency. For the Sun the wavelength peak is $502\,\mathrm{nm}$ but the frequency peak corresponds to about $850\,\mathrm{nm}$ , in the near infrared. Neither is "the" color of the blackbody; each is the peak of a particular density. Always state which variable you are peaking in, and never convert one peak into the other with $c=\lambda\nu$ .

Worked examples

1 · Effective temperature and color of the Sun (fully worked)

Given the solar luminosity $L_\odot=3.83\times10^{26}\,\mathrm{W}$ and radius $R_\odot=6.96\times10^{8}\,\mathrm{m}$ , find the effective temperature from the Stefan-Boltzmann law, then predict the peak wavelength from Wien's law.

Step 1. Invert equation (8) for temperature: $T_{\mathrm{eff}}=\left(\dfrac{L}{4\pi R^2\sigma}\right)^{1/4}$ . Assemble the denominator first: $4\pi R_\odot^2=4\pi(6.96\times10^{8})^2=6.09\times10^{18}\,\mathrm{m^2}$ , and multiplying by $\sigma=5.670\times10^{-8}$ gives $3.45\times10^{11}\,\mathrm{W\,K^{-4}}$ .

Step 2. Divide and take the fourth root: $L/(4\pi R^2\sigma)=3.83\times10^{26}/3.45\times10^{11}=1.11\times10^{15}\,\mathrm{K^4}$ , so $T_{\mathrm{eff}}=(1.11\times10^{15})^{1/4}=\mathbf{5772\,K}$ . This is the defined effective temperature of the Sun — the standard IAU value is $5772\,\mathrm{K}$ , an exact match.

Step 3. Feed that temperature into Wien's wavelength law (3): $\lambda_{\max}=b/T=2.898\times10^{-3}/5772=5.02\times10^{-7}\,\mathrm{m}=\mathbf{502\,nm}$ . That lands in the blue-green, entirely consistent with the Sun's G2 spectral class. T_eff = 5772 K and λ_max = 502 nm. Note the Sun looks white, not blue-green, because the eye integrates the whole broad curve — the peak is only where the density is largest.

2 · Temperature of the cosmic microwave background (fully worked)

The CMB is the most perfect blackbody known, with $T=2.725\,\mathrm{K}$ . Find its peak in frequency and in wavelength, and confirm they disagree.

Step 1. Frequency peak from equation (4): $\nu_{\max}=(5.879\times10^{10}\,\mathrm{Hz\,K^{-1}})\times2.725=\mathbf{1.60\times10^{11}\,Hz}=160\,\mathrm{GHz}$ , squarely in the microwave band that COBE/FIRAS measured.

Step 2. Wavelength peak from equation (3): $\lambda_{\max}=b/T=2.898\times10^{-3}/2.725=\mathbf{1.06\times10^{-3}\,m}=1.06\,\mathrm{mm}$ .

Step 3. Check the Jacobian trap: naively converting the frequency peak gives $c/\nu_{\max}=1.87\,\mathrm{mm}$ , not the $1.06\,\mathrm{mm}$ wavelength peak — the ratio is $x_5/x_3=4.965/2.821=1.76$ , exactly as the two roots predict. 160 GHz and 1.06 mm are both correct; they are peaks of different densities.

3 · Radius of a red supergiant from both laws

Betelgeuse has bolometric luminosity $L\approx1.0\times10^{5}\,L_\odot$ and photospheric temperature $T\approx3600\,\mathrm{K}$ (peak at $b/T\approx805\,\mathrm{nm}$ , deep red). Estimate its radius.

Step 1. Solve equation (8) for radius: $R=\sqrt{L/(4\pi\sigma T^4)}$ , with $L=1.0\times10^{5}\times3.83\times10^{26}=3.83\times10^{31}\,\mathrm{W}$ and $T^4=(3600)^4=1.68\times10^{14}\,\mathrm{K^4}$ .

Step 2. Denominator $4\pi\sigma T^4=1.20\times10^{8}\,\mathrm{W\,m^{-2}}$ , so $R=\sqrt{3.83\times10^{31}/1.20\times10^{8}}=\mathbf{5.6\times10^{11}\,m}$ , i.e. $R/R_\odot\approx\mathbf{800}$ .

Step 3. A star with the same $T_{\mathrm{eff}}$ as a cool dwarf but $10^5$ times its luminosity must be enormous — which is exactly why the coolest, most luminous stars occupy the supergiant branch of the Hertzsprung-Russell diagram. R ≈ 800 R_⊙.

Physics of the portrait

Behind the two tidy laws stand four physicists and two decades of argument. The empirical relations came first — a measured slope, a measured shift — and the man who turned the peak into a theorem, and set the stage Planck would later fill, was Wilhelm Wien.

Origins · 1 minWilhelm Wien and how this was discovered

Wilhelm Wien

1864-1928 · thermal radiation and the quantum threshold

Wilhelm Wien worked at the Physikalisch-Technische Reichsanstalt in Berlin, whose precise blackbody measurements would soon force the quantum revolution. In 1893 he proved the displacement law by a thermodynamic argument: subject cavity radiation to a slow adiabatic compression, track how each wavelength Doppler-shifts off the moving walls and how the temperature rises, and the whole spectral shape must be a universal function of the single combination $\lambda T$ — so the peak wavelength has to scale as $1/T$ , with no microscopic model. This "displacement" of the peak is the law that bears his name. In 1896 he went further and guessed a full spectral form, $B_\lambda\propto\lambda^{-5}e^{-a/\lambda T}$ , which fit the short-wavelength data but failed in the infrared — the failure that Planck patched in 1900 by interpolating between Wien's exponential tail and the Rayleigh-Jeans slope, forcing him to quantize energy. Wien's law thus sits exactly at the hinge of classical and quantum physics: it is derivable from thermodynamics alone, yet its constant $b=hc/(x_5 k_B)$ secretly contains $h$ . He received the 1911 Nobel Prize "for his discoveries regarding the laws governing the radiation of heat." The total-power side is older: Josef Stefan inferred $F\propto T^4$ empirically in 1879, and his student Ludwig Boltzmann derived it thermodynamically in 1884 by treating radiation as a gas with pressure $P=u/3$ — which is why that law carries both names.

History

1859

— Kirchhoff poses the universal blackbody problem and names the ideal radiator

1879

— Stefan infers the total-power law F ∝ T⁴ from measurements

1884

— Boltzmann derives F = σT⁴ thermodynamically from radiation pressure

1893

— Wien proves the displacement law λ_max·T = constant

1900

— Planck fits the full spectrum and quantizes energy, fixing σ and b in terms of h

1911

— Wien receives the Nobel Prize for the laws of heat radiation

Going further

Texts. Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., Ch. 3 (blackbody radiation, effective temperature) · Ryden, Introduction to Cosmology, 2nd ed., Ch. 2 and Ch. 8 (CMB blackbody) · Rybicki & Lightman, Radiative Processes in Astrophysics, Ch. 1 (thermal radiation, the $c/4$ factor) · Dodelson, Modern Cosmology, Ch. 1-2 · Mo, van den Bosch & White, Galaxy Formation and Evolution, Ch. 10 (stellar photometry) for the color-temperature machinery in a galaxy-survey context.

Provenance. Equations and constants are grounded in Carroll & Ostlie (Ch. 3), Ryden (Ch. 2), and Rybicki & Lightman (Ch. 1). A verifier should recompute $\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x=\pi^4/15=6.4939$ ; confirm the transcendental roots $x_5=4.96511$ from $x=5(1-e^{-x})$ and $x_3=2.82144$ from $x=3(1-e^{-x})$ ; reproduce $\sigma=2\pi^5 k_B^4/(15c^2h^3)=5.6704\times10^{-8}$ , $b=hc/(x_5 k_B)=2.8978\times10^{-3}\,\mathrm{m\,K}$ , and $a=4\sigma/c=7.5657\times10^{-16}$ from CODATA $k_B,h,c$ ; and check the worked numbers $T_{\mathrm{eff},\odot}=5772\,\mathrm{K}$ , $\lambda_{\max,\odot}=502\,\mathrm{nm}$ , $\nu_{\max,\mathrm{CMB}}=160\,\mathrm{GHz}$ , $\lambda_{\max,\mathrm{CMB}}=1.06\,\mathrm{mm}$ .

Prerequisites. Blackbody Radiation and the Planck Law · Specific Intensity and Radiative Flux · Kirchhoff's Laws and Thermal Emission. Leads to. Local Thermodynamic Equilibrium · The Saha Ionization Equation.

Ask the tutor: Ask me why the Sun's frequency peak lands in the infrared while its wavelength peak is green — the Jacobian that trips up everyone.

Is the Stefan-Boltzmann law only true for a perfect blackbody?

Strictly, $F=\sigma T^4$ is the flux from an ideal blackbody surface. A real ("grey") body radiates $F=\varepsilon\sigma T^4$ , where the emissivity $\varepsilon\le1$ measures how close it comes to ideal. For stars this is folded into the definition of effective temperature: rather than track a wavelength-dependent emissivity, one defines $T_{\mathrm{eff}}$ as the temperature of the blackbody emitting the star's actual total flux from its actual radius. So $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$ holds exactly, by construction, with all the non-ideal detail absorbed into the single number $T_{\mathrm{eff}}$ .

Where does the pure number π⁴/15 actually come from?

It is the value of the dimensionless integral $\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x$ left after the temperature and physical constants are factored out of the Planck integral. Expand the Bose factor as a geometric series $1/(e^x-1)=\sum_{n\ge1}e^{-nx}$ and integrate term by term: each term gives $\Gamma(4)/n^4=6/n^4$ , and summing over $n$ gives $6\,\zeta(4)=6\cdot\pi^4/90=\pi^4/15\approx6.494$ . So the $\pi$ in $\sigma$ traces back to the zeta value $\zeta(4)=\pi^4/90$ , not to any geometric circle.

Why is there a factor of π in F = πB but a factor of 4π/c in the energy density?

Both come from integrating over solid angle, but they answer different questions. Flux asks how much energy crosses the surface per second, so each ray is weighted by its projection $\cos\theta$ and integrated over the outward hemisphere: $\int\cos\theta\,\mathrm{d}\Omega=\pi$ . Energy density asks how much energy is stored per unit volume, weighting every direction equally over the full $4\pi$ steradians and dividing by the propagation speed: $u=(1/c)\int B\,\mathrm{d}\Omega=4\pi B/c$ . Combining the two gives the effusion relation $F=(c/4)u$ , hence $a=4\sigma/c$ .

The Sun peaks at 502 nm (green) — so why isn't the Sun green?

Because your eye responds to the whole broad spectrum, not to the single wavelength of maximum spectral density. The Planck curve at $5772\,\mathrm{K}$ is wide and only gently peaked; it delivers copious red, green, and blue, and the brain integrates all of it to "white" (sunlight defines white). The peak wavelength is a real feature of the density curve, but it is not "the color" of the source — and it depends on whether you plot per wavelength or per frequency, the Sun's frequency peak sitting near $850\,\mathrm{nm}$ in the infrared.

Can I use Wien's law on a source that isn't a perfect blackbody?

With care. Wien's law locates the peak of the Planck function; a source with strong emission or absorption lines, a non-thermal component, or significant dust reddening will have a peak displaced from $b/T$ . For a stellar photosphere the continuum is close enough to Planckian that the peak (or, more robustly, a broadband color index) gives a good color temperature, though it can differ modestly from the effective temperature. The safe practice is to fit a color index calibrated against models rather than read a single peak off a low-resolution spectrum.

How were σ and b used to measure the fundamental constants?

Turn the logic around. Measure $\sigma$ in the laboratory (total radiated power versus temperature) and independently measure $b$ (peak position versus temperature), and you have two equations, $\sigma=2\pi^5 k_B^4/(15c^2h^3)$ and $b=hc/(x_5 k_B)$ , in the unknowns $h$ and $k_B$ (with $c$ known). Planck did exactly this in 1900: from the fitted radiation data he extracted the first reasonable values of both $h$ and $k_B$ , and from $k_B$ with the gas constant $R$ he even obtained Avogadro's number. The blackbody spectrum was historically one of the first precision routes to the quantum of action.