Blackbody Radiation and the Planck Law

A blackbody is the simplest thermal object in physics and the most consequential in astronomy: a perfect absorber that, by Kirchhoff's law, is also a perfect emitter, radiating a spectrum fixed by one number — its temperature — and nothing else, not its composition, size, or shape. The practical realization is a cavity with opaque walls held at temperature $T$ ; the radiation filling it, sampled through a tiny hole, is blackbody radiation, isotropic and unpolarized. That this spectrum is universal made it the sharpest test of classical physics at the close of the nineteenth century, and classical physics failed it catastrophically. Counting cavity modes and giving each the equipartition energy $k_BT$ gives the Rayleigh–Jeans law $B_\nu = 2\nu^2 k_BT/c^2$ , whose energy density diverges as the mode count climbs with $\nu^2$ — the ultraviolet catastrophe. In 1900 Max Planck cut the divergence by a single, reluctant postulate: the energy of an oscillator of frequency $\nu$ comes in whole quanta $h\nu$ . High-frequency modes, which cost many quanta to excite, are then frozen out, and the mean energy per mode falls exponentially instead of staying pinned at $k_BT$ . The result is the Planck function $B_\nu(T)$ — the cornerstone of astrophysical radiation theory. It sets the source function in local thermodynamic equilibrium, fixes stellar effective temperatures through the Stefan–Boltzmann law, and describes the cosmic microwave background to better than one part in $10^4$ . Its two limiting forms, Rayleigh–Jeans at low frequency and Wien at high, are the everyday workhorses of radio and optical astronomers. This article assembles the full picture: the spectrum, its derivation from mode counting and quantization, its integrals and limits, and where a real source departs from it.

Key takeaways

A blackbody absorbs all incident radiation and re-emits a spectrum set by its temperature $T$ alone; the Planck function $B_\nu(T)=\dfrac{2h\nu^3}{c^2}\dfrac{1}{e^{h\nu/k_BT}-1}$ gives its specific intensity.
Quantization replaces the classical equipartition energy $k_BT$ per mode with the Bose–Einstein mean energy $h\nu/(e^{h\nu/k_BT}-1)$ , curing the ultraviolet catastrophe.
In the Rayleigh–Jeans limit ( $h\nu\ll k_BT$ ), $B_\nu\to 2\nu^2 k_BT/c^2$ ; in the Wien limit ( $h\nu\gg k_BT$ ) it falls exponentially as $e^{-h\nu/k_BT}$ .
Integrating over frequency gives the Stefan–Boltzmann law $F=\sigma T^4$ ; locating the peak gives Wien's displacement law $\lambda_{\max}T=2.898\times10^{-3}\ \mathrm{m\,K}$ .
In local thermodynamic equilibrium the source function equals $B_\nu(T)$ , so an optically thick isothermal medium emerges as a blackbody — the reason stars approximate one.
The cosmic microwave background is the most perfect known blackbody, at $T=2.7255\ \mathrm{K}$ , peaking near 160 GHz.
Key idea: quantizing electromagnetic energy in units of $h\nu$ replaces the classical $k_BT$ per mode with the Bose–Einstein occupation $h\nu/(e^{h\nu/k_BT}-1)$ , giving a spectrum that peaks and then falls exponentially and depends only on $T$ .

The core insight

Quantizing electromagnetic energy in units of $h\nu$ replaces the classical equipartition energy $k_BT$ per mode with the Bose–Einstein mean occupation $h\nu/(e^{h\nu/k_BT}-1)$ , producing a spectrum that peaks and then falls exponentially, and depends only on $T$ .

Where this lives today

The Planck function is not a historical relic; it is the working currency of modern observational astrophysics. Stellar effective temperatures are defined through $L=4\pi R^2\sigma T_{\rm eff}^4$ , and fitting the continuum of a spectrum to $B_\lambda(T)$ yields a colour temperature used to classify every star. In cosmology the CMB is the deepest application: the COBE/FIRAS spectrum measured in the early 1990s remains the most perfect blackbody ever recorded, with departures below one part in $10^4$ , and its near-perfect thermality is direct evidence that the early universe was in thermal equilibrium — the temperature scaling $T\propto(1+z)$ places it near $3000\ \mathrm{K}$ at recombination ( $z\approx1100$ ). The frontier now is the hunt for tiny predicted departures from a pure Planck spectrum: $\mu$ -type and $y$ -type spectral distortions imprinted by energy release before recombination, the target of proposed missions like PIXIE and the ESA Voyage 2050 studies, which would probe physics no anisotropy map can reach. Closer to home, infrared thermography, pyrometry, and the calibration of every space telescope's detectors rest on the same Stefan–Boltzmann and Planck relations, and radio interferometers from the VLA to ALMA report their maps in the brightness-temperature units that the Rayleigh–Jeans limit makes natural.

The physical picture

Picture photons rattling inside a hot oven. Each standing-wave mode of the cavity is an independent electromagnetic oscillator, and classically each should hold the equipartition energy $k_BT$ . But the number of modes with frequency below $\nu$ grows without bound as $\nu^3$ , so the number in a band grows as $\nu^2$ , and if every one carried $k_BT$ the oven would glow infinitely bright in the ultraviolet. Quantization is what saves the picture. To excite a mode of frequency $\nu$ you must pay at least one whole quantum $h\nu$ ; when $h\nu\gg k_BT$ that price is almost never affordable from the thermal bath, so those modes sit dark in their ground state. The Planck curve is the equilibrium compromise between two competing counts — a mode density rising as $\nu^2$ , and a mean energy per mode collapsing exponentially once $h\nu$ exceeds $k_BT$ . Their product peaks and falls off on both sides, a single universal shape locked to the temperature.

Figure

Two competing counts set the Planck curve: the mode density rises as ν² (the Rayleigh–Jeans line, which diverges to the ultraviolet) while the mean energy per mode collapses once hν exceeds k_BT; their product peaks at hν ≈ 2.82 k_BT and falls on both sides

Because $B_\nu(T)$ is the intensity inside an equilibrium cavity, it is the maximum any thermal emitter can achieve at temperature $T$ : a real source whose emissivity is below one at some frequency lies under the Planck curve there, never above it — that is Kirchhoff's law in action. This ceiling is why the spectrum is so useful as a reference. The cosmic microwave background is the sharpest realization nature offers, and its measured spectrum is a textbook Planck curve at a single temperature. The second figure shows that curve on real axes, with the peak carrying genuine numbers.

Figure

The cosmic microwave background is the most perfect blackbody known: its COBE/FIRAS spectrum is a Planck curve at T = 2.7255 K, peaking near 160 GHz (equivalently λ = 1.06 mm), with measured departures below one part in 10⁴

The core equations

The central object is the Planck function in its frequency form: the specific intensity per unit frequency emitted by a blackbody at temperature $T$ . Every other relation in the article is a change of variable, an integral, or a limit of this one line.

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

where: $B_\nu$ — specific intensity per unit frequency, $\mathrm{W\,m^{-2}\,Hz^{-1}\,sr^{-1}}$ ; $h$ — Planck constant, $6.626\times10^{-34}\ \mathrm{J\,s}$ ; $\nu$ — frequency; $c$ — speed of light, $2.998\times10^{8}\ \mathrm{m\,s^{-1}}$ ; $k_B$ — Boltzmann constant, $1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$ ; $T$ — absolute temperature. Reading it: the $\nu^3$ prefactor is the rising mode density times one quantum, and the $1/(e^{h\nu/k_BT}-1)$ factor is the Bose occupation that cuts the high-frequency tail. Spectroscopists work per unit wavelength, and the two are genuinely different functions because a spectral band $\mathrm{d}\nu$ maps to an unequal band $\mathrm{d}\lambda$ . The wavelength form carries an extra pair of $\lambda$ powers from the Jacobian.

B_\lambda(T) = \frac{2hc^2}{\lambda^5}\,\frac{1}{e^{hc/\lambda k_BT}-1} \qquad (2)

where: $B_\lambda$ — specific intensity per unit wavelength, $\mathrm{W\,m^{-2}\,m^{-1}\,sr^{-1}}$ ; $\lambda$ — wavelength. Because the transformation is not linear in the variable, the peak of $B_\lambda$ does not sit at $c/\nu_{\max}$ — a trap developed in the warning below. The factor that does all the physical work is the Bose–Einstein mean occupation number: the average number of photons in a single cavity mode of frequency $\nu$ at temperature $T$ .

\bar n(\nu) = \frac{1}{e^{h\nu/k_BT}-1} \qquad (3)

where: $\bar n$ — mean number of photons per mode; $h\nu$ — energy of one photon of frequency $\nu$ . Multiplying $\bar n$ by the quantum $h\nu$ gives the mean energy per mode, $h\nu/(e^{h\nu/k_BT}-1)$ , which tends to $k_BT$ when $h\nu\ll k_BT$ and to $h\nu\,e^{-h\nu/k_BT}$ when $h\nu\gg k_BT$ . Astronomers frequently need the energy stored in the radiation field itself. For the isotropic cavity field the spectral energy density is $4\pi/c$ times the intensity.

u_\nu(T) = \frac{4\pi}{c}B_\nu(T) = \frac{8\pi h\nu^3}{c^3}\,\frac{1}{e^{h\nu/k_BT}-1} \qquad (4)

where: $u_\nu$ — energy per unit volume per unit frequency, $\mathrm{J\,m^{-3}\,Hz^{-1}}$ ; the factor $4\pi/c$ comes from integrating an isotropic intensity over solid angle. Integrating $u_\nu$ over all frequencies collapses the spectrum to a single power law in temperature — the origin of the $T^4$ scaling.

u(T) = \int_0^\infty u_\nu\,\mathrm{d}\nu = a T^4,\qquad a = \frac{8\pi^5 k_B^4}{15 h^3 c^3} \qquad (5)

where: $u$ — total radiation energy density, $\mathrm{J\,m^{-3}}$ ; $a$ — radiation constant, $7.566\times10^{-16}\ \mathrm{J\,m^{-3}\,K^{-4}}$ . The flux streaming out of a blackbody surface is $c/4$ times this energy density, which is the Stefan–Boltzmann law.

F = \sigma T^4,\qquad \sigma = \frac{2\pi^5 k_B^4}{15 h^3 c^2} = \frac{ac}{4} \qquad (6)

where: $F$ — emergent flux from a blackbody surface, $\mathrm{W\,m^{-2}}$ ; $\sigma$ — Stefan–Boltzmann constant, $5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ . Finally, the bridge to real astrophysical media: in local thermodynamic equilibrium the emission and absorption balance so that the source function — the ratio of emissivity to opacity — is exactly the Planck function.

S_\nu = B_\nu(T) \quad \text{(in LTE)} \qquad (7)

where: $S_\nu$ — source function, $j_\nu/\alpha_\nu$ ; the equality holds wherever collisions maintain a common local temperature for matter and radiation. This is why solving the transfer equation for an optically thick isothermal slab returns a blackbody spectrum.

Derivation APlanck's law from mode counting and quantized energy

The goal is to build $B_\nu(T)$ from first principles by combining two independent ingredients — how many electromagnetic modes a cavity supports in a frequency band, and how much energy each mode holds on average once its energy is quantized — and to show that the quantization is the one and only step that separates the correct spectrum from the classical Rayleigh–Jeans divergence. We count standing waves in a conducting box, apply the Boltzmann distribution to a ladder of energy levels spaced by $h\nu$ , take the product, and convert the resulting energy density to a specific intensity.

Step 1. Count the modes. Standing waves in a cubic cavity of side $L$ have $\nu=(c/2L)\sqrt{n_x^2+n_y^2+n_z^2}$ for non-negative integers $n_i$ ; counting lattice points in the positive octant of a spherical shell of radius $\propto\nu$ and multiplying by two transverse polarizations gives a mode density per unit volume $\mathrm{d}N/(V\,\mathrm{d}\nu)=8\pi\nu^2/c^3$ . Why: this factor is purely geometric and classical — Rayleigh and Jeans had it right; the error was never in the counting.

Step 2. Compute the mean energy per mode with quantized levels. Restricting a mode's energy to integer multiples of $h\nu$ and weighting each level by the Boltzmann factor gives $\langle E\rangle=\dfrac{\sum_{n=0}^\infty n h\nu\, e^{-nh\nu/k_BT}}{\sum_{n=0}^\infty e^{-nh\nu/k_BT}}=\dfrac{h\nu}{e^{h\nu/k_BT}-1}$ , since both sums are geometric series. Why: this is Planck's decisive break — the continuous equipartition result $\langle E\rangle=k_BT$ is replaced by a quantity that decays exponentially once $h\nu>k_BT$ , freezing out the high-frequency modes.

Step 3. Multiply mode density by mean energy per mode to get the spectral energy density: $u_\nu=\dfrac{8\pi\nu^2}{c^3}\cdot\dfrac{h\nu}{e^{h\nu/k_BT}-1}=\dfrac{8\pi h\nu^3}{c^3}\dfrac{1}{e^{h\nu/k_BT}-1}$ . Why: the $\nu^2$ from counting and the $\nu$ from the quantum combine to the $\nu^3$ prefactor, while the Bose denominator supplies the cutoff that makes the frequency integral converge.

Step 4. Convert energy density to specific intensity. For an isotropic radiation field the energy density and intensity are related by $u_\nu=(4\pi/c)B_\nu$ , so $B_\nu=\dfrac{c}{4\pi}u_\nu=\dfrac{2h\nu^3}{c^2}\dfrac{1}{e^{h\nu/k_BT}-1}$ , the boxed Planck law. Why: intensity is the observable an astronomer measures along a line of sight, so this last conversion is what makes the cavity result usable for stars and the CMB. *Verified: reduces to Rayleigh–Jeans $2\nu^2k_BT/c^2$ for $h\nu\ll k_BT$ and to Wien $2h\nu^3e^{-h\nu/k_BT}/c^2$ for $h\nu\gg k_BT$ ; matches Carroll & Ostlie Eq. 3.22 and Rybicki & Lightman Ch. 1.*

Go deeper · Derivation B · 1 minThe Stefan–Boltzmann law by integration

Derivation BThe Stefan–Boltzmann law by integration

The goal is to collapse the full Planck spectrum into a single number — the total energy radiated per unit area — and to show that the temperature dependence is exactly the fourth power, with a constant built only from $h$ , $c$ , and $k_B$ . The strategy is to integrate the spectral energy density over all frequencies, extract the temperature scaling by a dimensionless substitution, evaluate the remaining pure number as a standard Bose integral, and convert energy density to emergent flux.

Step 1. Set up the integral. The total energy density is $u=\displaystyle\int_0^\infty\frac{8\pi h\nu^3}{c^3}\frac{\mathrm{d}\nu}{e^{h\nu/k_BT}-1}$ . Why: the ultraviolet catastrophe was precisely the statement that this integral diverges for the classical integrand; here the Bose denominator guarantees convergence.

Step 2. Substitute $x=h\nu/k_BT$ , so $\nu=k_BTx/h$ and $\mathrm{d}\nu=(k_BT/h)\,\mathrm{d}x$ , giving $u=\dfrac{8\pi k_B^4 T^4}{h^3 c^3}\displaystyle\int_0^\infty\frac{x^3}{e^x-1}\,\mathrm{d}x$ . Why: every factor of temperature is pulled outside the integral in one stroke, so the $T^4$ law is visible before any number is computed — the integral is a pure constant.

Step 3. Evaluate the Bose integral. Expanding $1/(e^x-1)=\sum_{n\ge1}e^{-nx}$ and integrating term by term gives $\displaystyle\int_0^\infty\frac{x^3}{e^x-1}\,\mathrm{d}x=\Gamma(4)\zeta(4)=6\cdot\frac{\pi^4}{90}=\frac{\pi^4}{15}$ . Why: this is the single transcendental input; $\Gamma(4)=3!=6$ and $\zeta(4)=\pi^4/90$ are standard, so the whole constant reduces to elementary factors and powers of $\pi$ .

Step 4. Collect and convert to flux. Then $u=\dfrac{8\pi^5 k_B^4}{15 h^3 c^3}T^4=aT^4$ , and the flux leaving a surface is $F=\dfrac{c}{4}u=\sigma T^4$ with $\sigma=\dfrac{ac}{4}=\dfrac{2\pi^5 k_B^4}{15 h^3 c^2}$ . Why: the factor $c/4$ is the standard angular average converting an isotropic interior energy density to a one-sided emergent flux; putting numbers in gives $\sigma=5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ , the measured value. *Verified: numerically $\sigma=5.670\times10^{-8}$ and $a=4\sigma/c=7.566\times10^{-16}$ , both matching CODATA; consistency $\sigma=ac/4$ holds exactly.*

Go deeper · Derivation C · 1 minWien's displacement law by locating the peak

Derivation CWien's displacement law by locating the peak

The goal is to find where the wavelength-form Planck spectrum is brightest and to show that the product of peak wavelength and temperature is a universal constant — the reason a hotter body is bluer. The method is to differentiate $B_\lambda(T)$ with respect to $\lambda$ at fixed $T$ , set the derivative to zero, reduce the condition to a transcendental equation in the single dimensionless variable $x=hc/\lambda k_BT$ , and solve it numerically.

Step 1. Differentiate. Writing $B_\lambda\propto\lambda^{-5}(e^{hc/\lambda k_BT}-1)^{-1}$ and setting $\mathrm{d}B_\lambda/\mathrm{d}\lambda=0$ produces, after clearing common factors, the condition $5(e^{x}-1)=x\,e^{x}$ with $x=hc/\lambda k_BT$ . Why: the two competing $\lambda$ -dependences — the $\lambda^{-5}$ prefactor pushing the peak blue and the exponential pushing it red — balance exactly at the maximum, and that balance is the whole content of the displacement law.

Step 2. Reduce and solve. Rearranged, the condition is $x=5\left(1-e^{-x}\right)$ , a transcendental equation whose nonzero root is $x=4.9651$ . Why: no closed form exists, but the root is a fixed pure number, so it converts directly into a dimensional constant that is the same for every blackbody.

Step 3. Assemble the constant. Then $\lambda_{\max}=\dfrac{hc}{4.9651\,k_B\,T}$ , i.e. $\lambda_{\max}T=\dfrac{hc}{4.9651\,k_B}=2.898\times10^{-3}\ \mathrm{m\,K}$ . Why: this is Wien's displacement law; repeating the same steps on $B_\nu$ instead gives the different root $x=2.8214$ , which is exactly why the frequency and wavelength peaks disagree. *Verified: $x=4.9651$ solves $x=5(1-e^{-x})$ and gives $\lambda_{\max}T=2.898\times10^{-3}\ \mathrm{m\,K}$ ; the $B_\nu$ peak root $2.8214$ gives $h\nu_{\max}=2.82\,k_BT$ .*

Limiting cases

The behaviour of $B_\nu(T)$ is governed entirely by the dimensionless ratio $x\equiv h\nu/k_BT$ : whether a photon quantum is small or large compared with the thermal energy.

Case	Result	Meaning
Rayleigh–Jeans ( $x\ll1$ )	$B_\nu(T)\approx\dfrac{2\nu^2 k_BT}{c^2}$	Low-frequency / radio regime; linear in $T$ , independent of $h$ — the classical result, valid where $h\nu\ll k_BT$ .
Wien ( $x\gg1$ )	$B_\nu(T)\approx\dfrac{2h\nu^3}{c^2}e^{-h\nu/k_BT}$	High-frequency tail; exponential cutoff. Historically Wien's fit before Planck; accurate on the blue side of the peak.
Frequency peak	$h\nu_{\max}=2.821\,k_BT$	Maximum of $B_\nu$ ; root of $x=3(1-e^{-x})$ .
Wavelength peak	$\lambda_{\max}T=2.898\times10^{-3}\ \mathrm{m\,K}$	Maximum of $B_\lambda$ ; root of $x=5(1-e^{-x})$ . Hotter bodies peak bluer.
Total flux	$F=\sigma T^4$	Integral over all frequencies; $\sigma=5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ .

Common trapThe νₘₐₓ and λₘₐₓ peaks are not the same photon energy

It is tempting to convert the frequency peak to a wavelength by $\lambda=c/\nu_{\max}$ and expect the wavelength peak — but $B_\nu(T)$ and $B_\lambda(T)$ are different functions, related by $B_\lambda\,\mathrm{d}\lambda=B_\nu\,\mathrm{d}\nu$ with the Jacobian $|\mathrm{d}\nu/\mathrm{d}\lambda|=c/\lambda^2$ . That extra factor of $\lambda^{-2}$ shifts the maximum: the $B_\nu$ peak sits at $h\nu_{\max}=2.82\,k_BT$ , while the $B_\lambda$ peak corresponds to $hc/\lambda_{\max}=4.97\,k_BT$ . So $\nu_{\max}\neq c/\lambda_{\max}$ , and quoting "the peak" without saying which form you mean is ambiguous by nearly a factor of two in photon energy. A second, related trap: blackbody emission at temperature $T$ is the upper bound on thermal intensity at that $T$ — it is a ceiling real spectra approach from below, not a curve they straddle.

Worked examples

1 · The Sun as a blackbody (fully worked)

Treat the Sun as a blackbody at its effective temperature $T_{\rm eff}=5772\ \mathrm{K}$ (radius $R_\odot=6.957\times10^8\ \mathrm{m}$ ). Find the wavelength of peak $B_\lambda$ , the wavelength at which $B_\nu$ peaks, the emergent surface flux, and the total luminosity.

Step 1. Wavelength peak from Wien's law: $\lambda_{\max}=\dfrac{2.898\times10^{-3}}{5772}=5.02\times10^{-7}\ \mathrm{m}=\mathbf{502\ nm}$ , in the green — which is why the solar spectrum peaks in the middle of the visible band our eyes evolved to use.

Step 2. Frequency peak, to expose the trap. Here $\nu_{\max}=2.821\,k_BT/h=(2.821)(1.381\times10^{-23})(5772)/(6.626\times10^{-34})=3.39\times10^{14}\ \mathrm{Hz}$ , which corresponds to $c/\nu_{\max}=\mathbf{883\ nm}$ in the near-infrared. The two "peaks" differ by nearly a factor of two — a direct demonstration of the warning above.

Step 3. Surface flux from Stefan–Boltzmann: $F=\sigma T_{\rm eff}^4=(5.670\times10^{-8})(5772)^4=\mathbf{6.29\times10^{7}\ W\,m^{-2}}$ . Every square metre of the photosphere radiates about 63 megawatts.

Step 4. Luminosity by multiplying by the surface area: $L=4\pi R_\odot^2\,F=4\pi(6.957\times10^8)^2(6.29\times10^7)=\mathbf{3.83\times10^{26}\ W}$ , matching the measured solar luminosity $L_\odot=3.828\times10^{26}\ \mathrm{W}$ — a clean confirmation that a single-temperature blackbody captures the Sun's bulk output.

2 · The cosmic microwave background (fully worked)

The CMB is a near-perfect blackbody at $T=2.7255\ \mathrm{K}$ . Find the frequency of peak $B_\nu$ , the wavelength of peak $B_\lambda$ , and the total radiation energy density.

Step 1. Frequency peak: $\nu_{\max}=2.821\,k_BT/h=(2.821)(1.381\times10^{-23})(2.7255)/(6.626\times10^{-34})=\mathbf{160\ GHz}$ . This is the microwave band, and it is the frequency the COBE/FIRAS instrument was tuned around.

Step 2. Wavelength peak: $\lambda_{\max}=\dfrac{2.898\times10^{-3}}{2.7255}=\mathbf{1.06\ mm}$ . Note that $c/\lambda_{\max}=283\ \mathrm{GHz}\neq160\ \mathrm{GHz}$ — the same $B_\nu$ -versus- $B_\lambda$ shift as for the Sun.

Step 3. Energy density from $u=aT^4$ : $u=(7.566\times10^{-16})(2.7255)^4=\mathbf{4.17\times10^{-14}\ J\,m^{-3}}$ . Dividing by the electron-volt gives $\approx0.26\ \mathrm{eV\,cm^{-3}}$ — the CMB photons carry a diffuse but perfectly uniform energy density filling all of space, and the corresponding photon number density is about $411\ \mathrm{cm^{-3}}$ .

3 · Rayleigh–Jeans and brightness temperature (radio)

A thermal radio source at $T=100\ \mathrm{K}$ is observed at $\nu=1.4\ \mathrm{GHz}$ . Confirm the Rayleigh–Jeans limit applies and compute the intensity.

Step 1. Check the regime: $x=h\nu/k_BT=(6.626\times10^{-34})(1.4\times10^9)/[(1.381\times10^{-23})(100)]=6.7\times10^{-4}\ll1$ , so the Rayleigh–Jeans form holds to better than a part in a thousand.

Step 2. Apply $B_\nu\approx2\nu^2k_BT/c^2=2(1.4\times10^9)^2(1.381\times10^{-23})(100)/(2.998\times10^8)^2=\mathbf{6.0\times10^{-20}\ W\,m^{-2}\,Hz^{-1}\,sr^{-1}}$ . Because intensity is linear in $T$ here, radio astronomers simply report the brightness temperature $T_b$ defined by $I_\nu\equiv2\nu^2k_BT_b/c^2$ ; for this optically thick thermal source $T_b=100\ \mathrm{K}$ equals the physical temperature, but for a non-thermal source $T_b$ is just a convenient linear stand-in for intensity.

Physics of the portrait

The Planck function is the equation that opened the twentieth century. It is worth remembering that its author did not believe the physics he had introduced — he called the quantum "an act of desperation," a bookkeeping trick to fit a curve — and spent years trying to derive it away before conceding that energy really does come in lumps.

Origins · 1 minMax Planck and how this was discovered

Max Planck

1858–1947 · the birth of the quantum

Max Planck was a conservative Berlin thermodynamicist, temperamentally the last person one would expect to overturn classical physics. Through the 1890s he worked on the blackbody spectrum as a problem in irreversibility and entropy, and in October 1900, prompted by new far-infrared measurements from Lummer, Pringsheim, Rubens, and Kurlbaum that broke Wien's exponential fit, he guessed an interpolation formula between the Rayleigh–Jeans and Wien limits that matched the data perfectly. Forced to justify it, in December 1900 he found he could derive it only by assuming the oscillators in the cavity walls exchanged energy in discrete elements $\varepsilon=h\nu$ , introducing a new constant of nature $h=6.63\times10^{-34}\ \mathrm{J\,s}$ . He regarded the quantization as a formal device and hoped to recover the continuum; instead Einstein in 1905 took the quanta literally to explain the photoelectric effect, and Bose and Einstein in 1924 showed the statistics were those of a gas of indistinguishable photons — recovering Planck's occupation number $1/(e^{h\nu/k_BT}-1)$ from pure counting. Planck received the 1918 Nobel Prize; the constant, the spectrum, and the units all now bear his name, and the satellite that measured the CMB to exquisite precision was called Planck in his honour.

History

1859

— Kirchhoff states that a blackbody's emission is a universal function of temperature and frequency

1879

— Stefan finds empirically that total radiated power scales as T⁴

1884

— Boltzmann derives the T⁴ law thermodynamically

1893

— Wien states the displacement law λ_max T = const

1900

— Planck introduces the quantum h and the full spectrum

1924

— Bose and Einstein derive the occupation number from photon statistics

Going further

Texts. Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., Ch. 3 · Rybicki & Lightman, Radiative Processes in Astrophysics, Ch. 1 · Ryden, Introduction to Cosmology, 2nd ed., Ch. 2 & 8 · Dodelson, Modern Cosmology, Ch. 1–2 · Mo, van den Bosch & White, Galaxy Formation and Evolution, Ch. 3 · Binney & Tremaine, Galactic Dynamics, App. on radiative quantities.

Provenance. Equations from Carroll & Ostlie Ch. 3, Rybicki & Lightman Ch. 1, and Ryden Ch. 2. A verifier should recompute: (1) the Bose integral $\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x=\Gamma(4)\zeta(4)=\pi^4/15$ ; (2) $\sigma=2\pi^5k_B^4/(15h^3c^2)=5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ and $a=4\sigma/c=7.566\times10^{-16}\ \mathrm{J\,m^{-3}\,K^{-4}}$ ; (3) the Wien peaks — $\lambda_{\max}T=2.898\times10^{-3}\ \mathrm{m\,K}$ from $x=5(1-e^{-x})$ giving $x=4.9651$ , and $h\nu_{\max}=2.821\,k_BT$ from $x=3(1-e^{-x})$ ; (4) the Rayleigh–Jeans and Wien limits by expanding $e^x-1$ ; (5) worked-example numbers — solar $\lambda_{\max}=502\ \mathrm{nm}$ , $F=6.29\times10^7\ \mathrm{W\,m^{-2}}$ , $L=3.83\times10^{26}\ \mathrm{W}$ at $T_{\rm eff}=5772\ \mathrm{K}$ , and CMB $\nu_{\max}=160\ \mathrm{GHz}$ , $u=4.17\times10^{-14}\ \mathrm{J\,m^{-3}}$ at $T=2.7255\ \mathrm{K}$ . Constants are CODATA/IAU standard values.

Prerequisites. Specific Intensity and Radiative Flux · Local Thermodynamic Equilibrium · Kirchhoff's Laws and Thermal Emission. Leads to. Stefan–Boltzmann and Wien Displacement Laws · Emission and the Source Function · The Equation of Radiative Transfer.

Ask the tutor: Ask me to derive Wien's displacement constant from scratch, or to work out why the CMB stays a perfect blackbody even as the universe expands.

Why is a blackbody spectrum independent of the material?

Because it is the spectrum of radiation in thermodynamic equilibrium, and equilibrium erases the memory of how the radiation was made. Inside an opaque cavity a photon is absorbed and re-emitted enormously many times before it escapes, so the emergent spectrum reflects only the wall temperature, not the wall's chemistry. Kirchhoff proved this in 1859 with a thermodynamic argument: if two cavities at the same temperature but of different materials had different spectra, you could connect them and drive a net heat flow from cold to hot, violating the second law. So the spectrum must be a universal function of $T$ and $\nu$ alone — the function Planck later wrote down.

What exactly went wrong in the ultraviolet catastrophe?

Classical statistical mechanics assigns every quadratic degree of freedom the energy $k_BT$ (equipartition), and each cavity mode is such a degree of freedom. But the number of modes in a frequency band grows as $\nu^2$ without limit, so the total energy $\int\nu^2 k_BT\,\mathrm{d}\nu$ diverges — the field should carry infinite energy, concentrated at high (ultraviolet) frequencies. The Rayleigh–Jeans law captures this and matches experiment only at low frequency, then runs away. Quantization fixes it by making the mean energy per mode $h\nu/(e^{h\nu/k_BT}-1)$ rather than $k_BT$ : for $h\nu\gg k_BT$ this is exponentially small, so high-frequency modes contribute almost nothing and the integral converges.

Are real stars actually blackbodies?

Only approximately, and it is worth being precise about the sense in which they are. A star has no solid surface; its photosphere is a layer where the gas becomes transparent, and the different wavelengths escape from different depths at slightly different temperatures. Where the gas is optically thick and near local thermodynamic equilibrium the emergent continuum is close to $B_\lambda(T)$ , which is why a single effective temperature describes the bulk output well. But superimposed on that continuum are absorption lines from atoms in the cooler outer layers, and in the ultraviolet and in the cores of strong lines the assumption of LTE fails, so the true spectrum departs from any single Planck curve. The blackbody is the zeroth-order model, refined by stellar-atmosphere theory.

Why does the frequency-form peak differ from the wavelength-form peak?

Because "per unit frequency" and "per unit wavelength" slice the spectrum into unequal bins. A fixed frequency interval $\mathrm{d}\nu$ maps to a wavelength interval $\mathrm{d}\lambda=(c/\nu^2)\,\mathrm{d}\nu$ , which is wide at long wavelengths and narrow at short ones. When you re-bin $B_\nu$ into $B_\lambda$ you multiply by that stretching factor $c/\lambda^2$ , and multiplying a function by a monotonic factor moves its maximum. The upshot: $B_\nu$ peaks at $h\nu_{\max}=2.82\,k_BT$ while $B_\lambda$ peaks at $hc/\lambda_{\max}=4.97\,k_BT$ . Both are correct; they simply answer different questions ("most energy per Hz" versus "most energy per nm"), so you must always state which form you mean.

Where do the Stefan–Boltzmann and radiation constants come from?

They are not independent constants of nature but combinations of $h$ , $c$ , and $k_B$ fixed once you integrate the Planck spectrum. Carrying out $\int_0^\infty u_\nu\,\mathrm{d}\nu$ pulls out $T^4$ and leaves the pure number $\pi^4/15$ from the Bose integral, giving the radiation constant $a=8\pi^5k_B^4/(15h^3c^3)=7.566\times10^{-16}\ \mathrm{J\,m^{-3}\,K^{-4}}$ . The flux leaving a surface is $c/4$ times the interior energy density, so $\sigma=ac/4=2\pi^5k_B^4/(15h^3c^2)=5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ . Historically Stefan (1879) and Boltzmann (1884) found the $T^4$ law before anyone could compute the constant; Planck's spectrum was the first theory to predict its value.

What is brightness temperature, and is it a real temperature?

Brightness temperature $T_b$ is defined by inverting the Rayleigh–Jeans law: $T_b\equiv c^2 I_\nu/(2\nu^2 k_B)$ , so it is simply the intensity re-expressed in kelvin. It is a genuine physical temperature only for an optically thick thermal source in LTE, where $I_\nu=B_\nu(T)$ and $T_b=T$ . For an optically thin source, a non-thermal synchrotron source, or a maser, $T_b$ is just a convenient linear label for intensity and can be wildly larger than any real temperature — pulsar and quasar cores reach $T_b>10^{12}\ \mathrm{K}$ , which no thermal population could produce. Radio astronomers use it because at their frequencies $h\nu\ll k_BT$ , so intensity and temperature are proportional and detector noise is naturally quoted the same way.