Planck's Law from the Photon Gas

Derivation

Planck's Law from the Photon Gas

D-241 Home PU-302 Threads light · chance · energy · matter Depends on Bose-Einstein and Fermi-Dirac Occupation Numbers

Statement

Treating the electromagnetic field inside a cavity in thermal equilibrium at temperature \(T\) as an ideal gas of massless, spin-1 photons with chemical potential \(\mu=0\), the Bose–Einstein occupation of each mode combined with the density of field modes yields the spectral energy density \(u(\nu,T)\,d\nu = \frac{8\pi h \nu^{3}}{c^{3}}\frac{d\nu}{e^{h\nu/k_{B}T}-1}\). Integrating over all frequencies recovers the Stefan–Boltzmann law \(u(T)=aT^{4}\), and extremising \(u(\nu,T)\) recovers Wien's displacement law \(\nu_{\max}\propto T\).

Why it matters

Planck's law is the first result in which the quantisation of energy is forced on us by experiment: no purely classical counting of field modes can reproduce a finite total energy, because equipartition assigns \(k_{B}T\) to each of infinitely many modes (the ultraviolet catastrophe). The photon-gas derivation shows that the single ingredient \(\mu=0\) — photon number is not conserved — together with Bose statistics resolves this completely.

Beyond its historical role, the same integral is the workhorse of thermal radiation, cosmology (the cosmic microwave background is a Planck spectrum to one part in \(10^{5}\)), stellar atmospheres, and radiometry. Stefan–Boltzmann and Wien's law are not independent facts but corollaries of one spectrum, and this derivation exhibits exactly how they descend from it.

Assumptions

The radiation is in thermal equilibrium with cavity walls at a single temperature \(T\). Without equilibrium the mode occupations are not given by a Bose–Einstein distribution at all; \(u(\nu)\) becomes an arbitrary non-thermal spectrum (e.g. synchrotron or laser output) and none of what follows holds. Photon number is not conserved, so the chemical potential vanishes, \(\mu=0\). If \(\mu\neq0\) (as for a photon gas held out of equilibrium, or for conserved massive bosons) the occupation carries an extra factor \(e^{-\mu/k_BT}\); the spectrum shifts and the total energy no longer scales as \(T^{4}\). Each cavity mode is an independent quantum harmonic oscillator whose excitations are non-interacting bosons. Photon–photon interactions are negligible in vacuum QED at these energies; if they were not (strong fields, a nonlinear medium) the modes would not factorise and the grand partition function would not separate into a product over modes. The cavity is large compared with the wavelengths that carry appreciable energy, so the mode spectrum may be treated as a continuum and boundary shape is irrelevant. For a small cavity or at very low \(T\) the discrete mode sum cannot be replaced by an integral; the spectrum becomes cavity-shape dependent and the smooth density of states fails.

Derivation

\[ N(k)\,dk = 2\cdot\frac{V}{(2\pi)^{3}}\,4\pi k^{2}\,dk \]

Count electromagnetic modes in a box of volume \(V=L^{3}\) with periodic boundary conditions: allowed wavevectors sit on a lattice of spacing \(2\pi/L\), so the number per unit \(k\)-volume is \(V/(2\pi)^3\). The shell \(4\pi k^2\,dk\) collects modes of magnitude \(k\), and the factor \(2\) counts the two transverse polarisations of the photon. A

\[ k=\frac{2\pi\nu}{c},\qquad dk=\frac{2\pi}{c}\,d\nu \;\Longrightarrow\; g(\nu)\,d\nu = \frac{8\pi V}{c^{3}}\,\nu^{2}\,d\nu \]

Convert to frequency using the photon dispersion relation \(\omega=ck\), i.e. \(k=2\pi\nu/c\) for massless particles. Substituting \(k^2\,dk=(2\pi\nu/c)^2(2\pi/c)\,d\nu\) into step 1 gives the number of modes per unit volume per unit frequency, \(g(\nu)/V=8\pi\nu^2/c^3\). A

\[ \langle n(\nu)\rangle = \frac{1}{e^{h\nu/k_{B}T}-1} \]

Import the mean occupation of a single bosonic mode of energy \(\varepsilon=h\nu\) from Bose–Einstein statistics (prior result: quantum occupation statistics), evaluated at \(\mu=0\). Each mode is an independent oscillator, so its equilibrium excitation number is fixed by the mode energy and \(T\) alone. B

\[ Z_{\text{mode}}=\sum_{n=0}^{\infty} e^{-n\beta h\nu}=\frac{1}{1-e^{-\beta h\nu}},\qquad \langle n\rangle=-\frac{\partial\ln Z_{\text{mode}}}{\partial(\beta h\nu)}=\frac{1}{e^{\beta h\nu}-1} \]

Rigour check on step 3: a single mode is a harmonic oscillator with energies \(nh\nu\) and unrestricted \(n\) (bosons), so its grand partition function is the geometric series above, convergent because \(h\nu>0\). Then \(\langle n\rangle = \sum n\,e^{-n\beta h\nu}/Z = \big(e^{\beta h\nu}-1\big)^{-1}\) with \(\beta=1/k_BT\), confirming the imported occupation from first principles. C

\[ u(\nu,T)\,d\nu = \frac{1}{V}\,g(\nu)\,\langle n(\nu)\rangle\,h\nu\,d\nu \]

Assemble the spectral energy density: (modes per unit volume per unit frequency) \(\times\) (mean photons per mode) \(\times\) (energy \(h\nu\) per photon). Each factor is per-mode independent, so the product is legitimate. A

\[ \boxed{\,u(\nu,T)=\frac{8\pi h\nu^{3}}{c^{3}}\,\frac{1}{e^{h\nu/k_{B}T}-1}\,} \]

Substitute steps 2 and 3 into step 5. This is Planck's spectral energy density (energy per unit volume per unit frequency). A

\[ u(T)=\int_{0}^{\infty} u(\nu,T)\,d\nu=\frac{8\pi h}{c^{3}}\int_{0}^{\infty}\frac{\nu^{3}\,d\nu}{e^{h\nu/k_{B}T}-1} \]

Integrate the spectrum over all frequencies to get the total energy density. The integral converges at both ends: \(\nu^3\) at small \(\nu\), exponential suppression at large \(\nu\). A

\[ x=\frac{h\nu}{k_{B}T}\;\Longrightarrow\; u(T)=\frac{8\pi h}{c^{3}}\left(\frac{k_{B}T}{h}\right)^{4}\int_{0}^{\infty}\frac{x^{3}\,dx}{e^{x}-1} \]

Non-dimensionalise with \(x=h\nu/k_BT\), so \(\nu=(k_BT/h)x\) and \(d\nu=(k_BT/h)dx\); the four powers of \((k_BT/h)\) pull the entire temperature dependence outside the integral. This is why the total energy scales as \(T^{4}\). B

\[ \int_{0}^{\infty}\frac{x^{3}}{e^{x}-1}\,dx=\Gamma(4)\,\zeta(4)=6\cdot\frac{\pi^{4}}{90}=\frac{\pi^{4}}{15} \]

Evaluate the dimensionless integral. Expand \((e^x-1)^{-1}=\sum_{n\ge1}e^{-nx}\) and integrate term by term: \(\int_0^\infty x^3 e^{-nx}dx=6/n^4\), giving \(6\sum_{n\ge1}n^{-4}=6\zeta(4)=\pi^4/15\). C

\[ u(T)=\frac{8\pi^{5}k_{B}^{4}}{15\,h^{3}c^{3}}\,T^{4}\equiv aT^{4} \]

Combine steps 8 and 9. The radiation constant is \(a=8\pi^{5}k_B^4/(15 h^3 c^3)\); the emissive power of a black surface follows from \(j^\star=\tfrac{c}{4}u=\sigma T^4\) with \(\sigma=ac/4\). B

\[ \frac{\partial u}{\partial\nu}=0\;\Longrightarrow\; (3-y)e^{y}=3,\quad y=\frac{h\nu_{\max}}{k_{B}T} \]

For Wien's law, extremise \(u(\nu,T)\) at fixed \(T\). Writing \(u\propto\nu^3/(e^{y}-1)\) with \(y=h\nu/k_BT\), setting \(du/d\nu=0\) gives the transcendental condition \(3(e^y-1)=y\,e^y\), i.e. \((3-y)e^y=3\). B

\[ y=2.8214\ldots\;\Longrightarrow\; \nu_{\max}=\frac{2.8214\,k_{B}}{h}\,T \]

Solve numerically: \(y\approx2.821439\). Hence \(\nu_{\max}\) is directly proportional to \(T\) — Wien's displacement law in frequency. (The wavelength peak differs because \(u_\lambda\,d\lambda\neq u_\nu\,d\nu\); solving \((5-y')e^{y'}=5\) gives \(y'=4.965\) and \(\lambda_{\max}T=b=2.898\times10^{-3}\,\text{m·K}\).) B

Result

\[ u(\nu,T)=\frac{8\pi h\nu^{3}}{c^{3}}\,\frac{1}{e^{h\nu/k_{B}T}-1},\qquad u(T)=\frac{8\pi^{5}k_{B}^{4}}{15\,h^{3}c^{3}}\,T^{4},\qquad \nu_{\max}=2.8214\,\frac{k_{B}T}{h} \]

Reading. The spectral energy density rises as \(\nu^2\) at low frequency (each new mode carries \(\sim k_BT\), the Rayleigh–Jeans regime) and is cut off exponentially once \(h\nu\gtrsim k_BT\), because populating a high-energy mode costs more than the thermal energy available. The peak sits at a frequency proportional to \(T\), and the area under the curve — the total energy — grows as \(T^4\). All three classical laws (Rayleigh–Jeans, Wien, Stefan–Boltzmann) are limits or moments of this one function.

Units check. \([u(\nu,T)]=\dfrac{[\text{J·s}][\text{s}^{-3}]}{[\text{m·s}^{-1}]^3}=\dfrac{\text{J·s}\cdot\text{s}^{-3}}{\text{m}^3\text{s}^{-3}}=\text{J·s·m}^{-3}\), i.e. energy density per unit frequency (J·m\(^{-3}\)·Hz\(^{-1}\)); the Bose factor is dimensionless since \(h\nu/k_BT\) is a pure number. For the total, \([k_B^4 T^4/(h^3c^3)]=\dfrac{(\text{J·K}^{-1})^4\text{K}^4}{(\text{J·s})^3(\text{m·s}^{-1})^3}=\dfrac{\text{J}^4}{\text{J}^3\text{s}^3\cdot\text{m}^3\text{s}^{-3}}=\text{J·m}^{-3}\), an energy density.

Limiting cases

Low frequency \(h\nu\ll k_BT\) (Rayleigh–Jeans): \(e^{h\nu/k_BT}-1\approx h\nu/k_BT\), so \(u\to 8\pi\nu^2 k_BT/c^3\). The \(h\) cancels and each mode carries \(k_BT\) — the classical equipartition result, valid only where it does not diverge.
High frequency \(h\nu\gg k_BT\) (Wien tail): \(e^{h\nu/k_BT}-1\approx e^{h\nu/k_BT}\), so \(u\to (8\pi h\nu^3/c^3)e^{-h\nu/k_BT}\) — Wien's 1896 empirical law, exponentially suppressed.
Classical limit \(h\to0\): the whole spectrum collapses to Rayleigh–Jeans at every \(\nu\); the total energy diverges (ultraviolet catastrophe), showing quantisation is what makes \(u(T)\) finite.
\(T\to0\): \(u(\nu,T)\to0\) for every \(\nu\); the cavity approaches its photon vacuum (zero-point energy excluded from the thermal part).

Breaks when

Chemical potential is non-zero. If the photon gas is driven out of equilibrium so that photon number is effectively conserved on the relevant timescale (e.g. a photon Bose–Einstein condensate in a dye microcavity, or Compton-frozen plasma), \(\mu\neq0\) and the occupation gains a factor \(e^{-\mu/k_BT}\); the spectrum is no longer Planckian and \(u\not\propto T^4\).
Cavity dimensions approach the thermal wavelength. When \(L\sim hc/k_BT\), the discrete mode sum cannot be replaced by \(\int g(\nu)d\nu\); the density of states develops gaps, near-field and shape effects dominate, and the smooth \(\nu^2\) counting fails (relevant to micro/nano-cavities and Casimir physics).
Strong-field or medium nonlinearity. In matter with a frequency-dependent refractive index \(n(\nu)\), or in fields strong enough for photon–photon scattering, the dispersion relation \(\nu=ck/2\pi\) and the independence of modes both break, so \(g(\nu)\) and the product form of the partition function are invalid.
Relativistic/finite-particle number fluctuations. For a system with only a few photons the grand-canonical average \(\langle n\rangle\) mis-describes the actual fluctuating occupation; the mean spectrum is not the observed one.

Failure modes

Forgetting the polarisation factor of 2. Dropping it halves \(g(\nu)\), Stefan–Boltzmann, and \(\sigma\) — a factor-of-two error that survives to the final constant.
Using \(\mu\neq0\) out of habit from massive-boson problems. Photons are not conserved; inserting a chemical potential is the single most common conceptual slip and destroys the \(T^4\) scaling.
Conflating the frequency peak with the wavelength peak. \(\nu_{\max}\) solves \((3-y)e^y=3\) while \(\lambda_{\max}\) solves \((5-y')e^{y'}=5\); they are not related by \(\lambda_{\max}=c/\nu_{\max}\). Students often quote one condition for the other.
Applying equipartition to all modes. Assigning \(k_BT\) per mode everywhere reproduces the ultraviolet catastrophe; the Bose factor is essential precisely where the classical argument fails.
Confusing energy density \(u\) with emissive flux \(j^\star\). They differ by \(c/4\) (from the angular integral of an isotropic photon gas leaking through a hole); forgetting this factor mis-scales \(\sigma\).
Mismatched spectral variable. Writing \(u_\lambda=u_\nu\) instead of \(u_\lambda=u_\nu\,|d\nu/d\lambda|=u_\nu\,c/\lambda^2\); the Jacobian is mandatory when changing between \(\nu\) and \(\lambda\).

Discussion

The physical crux is the vanishing chemical potential. In the grand canonical ensemble a species has \(\mu=0\) precisely when its number is not conserved — the walls of the cavity freely emit and absorb photons until \(u\) reaches the value that minimises the free energy, which is exactly where \((\partial F/\partial N)_{T,V}=\mu=0\). This single fact is the whole difference between a photon gas and, say, a gas of massive atoms, for which \(N\) is fixed and \(\mu\) is a genuine variable. It is why radiation has an equation of state with no independent particle-number knob: everything is set by \(T\).

The mode-counting factor \(g(\nu)\propto\nu^2\) is purely geometric and classical — it is the same density of states that appears in acoustics and in the Debye model of solids. What quantum mechanics supplies is the occupation: replacing the classical equipartition value \(k_BT\) with \(h\nu\,\langle n\rangle\). The two ingredients are cleanly separable, which is why the Debye heat capacity of a solid and the Planck spectrum of radiation share the same \(x^3/(e^x-1)\) integral and the same low-temperature power laws. Recognising that shared structure is one of the deep unifications in statistical physics.

At the level of quantum field theory, each cavity mode is a mode of the free electromagnetic field, and its excitations are genuine field quanta obeying \([\hat a,\hat a^\dagger]=1\). The grand partition function factorises into a product over modes precisely because the free field is a sum of decoupled oscillators; interactions (QED vertices) would spoil this, but at optical energies the fine-structure suppression makes photon–photon scattering utterly negligible, so the ideal-gas treatment is essentially exact. The zero-point energy \(\tfrac12 h\nu\) per mode is present but \(T\)-independent and is subtracted as an additive constant in the thermal energy; it reappears physically only through differences, as in the Casimir effect. That the same field, same modes, and same statistics give both the blackbody spectrum and the Casimir force is a striking consistency check on the whole framework.

Common misconceptions. Planck did not "assume light is made of particles" — his 1900 quantisation was of the oscillator energies of the wall material, \(E=nh\nu\); the photon as a particle came later (Einstein, 1905). The modern photon-gas derivation given here is logically cleaner but historically anachronistic. A second misconception is that the ultraviolet catastrophe was an observed divergence; it was a theoretical prediction of the classical spectrum that experiment never showed, and Planck's law was fitted to the data before it was derived. Finally, "\(\mu=0\)" does not mean photons have zero energy or that there are zero of them — it means adding a photon at fixed \(T,V\) costs no free energy at equilibrium.

Worked examples

Peak frequency and Wien check for the Sun's photosphere, \(T=5772\ \text{K}\).

Use the frequency-peak form \(\nu_{\max}=2.8214\,k_BT/h\). Symbols first, then numbers. A

\[ \nu_{\max}=\frac{2.8214\,(1.381\times10^{-23}\,\text{J·K}^{-1})(5772\,\text{K})}{6.626\times10^{-34}\,\text{J·s}} \]

Insert \(k_B,\,T,\,h\) in SI. A

\[ \nu_{\max}=\frac{2.8214\times1.381\times10^{-23}\times5772}{6.626\times10^{-34}}=3.39\times10^{14}\ \text{Hz} \]

Arithmetic: numerator \(=2.249\times10^{-19}\,\text{J}\); divide by \(h\). A

\[ \lambda_{\max}^{(\lambda)}=\frac{b}{T}=\frac{2.898\times10^{-3}\,\text{m·K}}{5772\,\text{K}}=502\ \text{nm} \]

The wavelength-peak law (separate condition) for comparison — green-yellow, consistent with sunlight. Note \(c/\nu_{\max}=884\,\text{nm}\neq\lambda_{\max}\), illustrating the peak-mismatch point. B

\[ \nu_{\max}\approx 3.4\times10^{14}\ \text{Hz},\qquad \lambda_{\max}\approx 502\ \text{nm} \]

Reading. The Sun peaks in the visible, near green; the frequency and wavelength peaks are genuinely different points on the spectrum, not reciprocals.

Units check. \(\big[k_BT/h\big]=\dfrac{\text{J·K}^{-1}\cdot\text{K}}{\text{J·s}}=\text{s}^{-1}=\text{Hz}\). Correct.

Total radiated power of a spherical blackbody: a tungsten filament of radius \(r=25\ \mu\text{m}\), length \(\ell=5.0\ \text{cm}\), at \(T=2800\ \text{K}\), treated as an ideal emitter.

Use \(P=\sigma A T^4\) with \(\sigma=ac/4=5.670\times10^{-8}\,\text{W·m}^{-2}\text{K}^{-4}\), derived above. Symbols first. A

\[ A=2\pi r\ell=2\pi(2.5\times10^{-5}\,\text{m})(5.0\times10^{-2}\,\text{m})=7.85\times10^{-6}\ \text{m}^2 \]

Lateral area of the cylindrical filament (end caps negligible). A

\[ T^4=(2800)^4=6.147\times10^{13}\ \text{K}^4 \]

Fourth power of the temperature. A

\[ P=\sigma A T^4=(5.670\times10^{-8})(7.85\times10^{-6})(6.147\times10^{13}) \]

Assemble Stefan–Boltzmann. A

\[ P\approx 27\ \text{W} \]

Reading. A pure blackbody of this size at 2800 K radiates about 27 W; a real tungsten filament has emissivity \(\varepsilon\approx0.35\), giving \(\sim\)9 W, the right order for a small bulb.

Units check. \(\text{W·m}^{-2}\text{K}^{-4}\cdot\text{m}^2\cdot\text{K}^4=\text{W}\). Correct.

Problems

(A) Rayleigh–Jeans limit. Show that for \(h\nu\ll k_BT\) the Planck spectrum reduces to \(u(\nu,T)=8\pi\nu^2 k_BT/c^3\), and state why Planck's constant disappears.
Solution
For small \(x=h\nu/k_BT\), expand \(e^{x}-1\approx x+\tfrac{x^2}{2}\approx x\). Then \(u=\dfrac{8\pi h\nu^3}{c^3}\cdot\dfrac{1}{x}=\dfrac{8\pi h\nu^3}{c^3}\cdot\dfrac{k_BT}{h\nu}=\dfrac{8\pi\nu^2 k_BT}{c^3}\). The factor \(h\) cancels because in this limit every mode holds the classical equipartition energy \(k_BT\), independent of quantisation. This spectrum diverges as \(\nu\to\infty\), which is the ultraviolet catastrophe.
(A) CMB peak. The cosmic microwave background is a blackbody at \(T=2.725\ \text{K}\). Find its wavelength peak using \(\lambda_{\max}T=2.898\times10^{-3}\,\text{m·K}\).
Solution
\(\lambda_{\max}=\dfrac{2.898\times10^{-3}}{2.725}=1.06\times10^{-3}\ \text{m}=1.06\ \text{mm}\), in the microwave band. The corresponding frequency-peak is \(\nu_{\max}=2.8214\,k_BT/h=2.8214\times(1.381\times10^{-23}\times2.725)/(6.626\times10^{-34})=1.60\times10^{11}\ \text{Hz}=160\ \text{GHz}\).
(B) Temperature scaling of total energy. A cavity's total radiant energy density doubles. By what factor did its temperature change? If instead the peak frequency doubled, by what factor did \(T\) change?
Solution
Since \(u\propto T^4\), doubling \(u\) means \(T\) scales by \(2^{1/4}=1.189\), a 19% rise. Since \(\nu_{\max}\propto T\), doubling the peak frequency doubles \(T\) (factor 2), which would raise \(u\) by \(2^4=16\).
(B) Stefan–Boltzmann constant from fundamentals. Compute \(\sigma=\dfrac{2\pi^5 k_B^4}{15 h^3 c^2}\) numerically and confirm it matches the tabulated value.
Solution
Note \(\sigma=ac/4\) with \(a=8\pi^5 k_B^4/(15h^3c^3)\), so \(\sigma=2\pi^5 k_B^4/(15 h^3 c^2)\). Insert \(k_B=1.381\times10^{-23}\), \(h=6.626\times10^{-34}\), \(c=2.998\times10^{8}\): \(k_B^4=3.637\times10^{-92}\); \(h^3=2.909\times10^{-100}\); \(c^2=8.988\times10^{16}\). Numerator \(2\pi^5\times3.637\times10^{-92}=2\times306.0\times3.637\times10^{-92}=2.226\times10^{-89}\). Denominator \(15\times2.909\times10^{-100}\times8.988\times10^{16}=3.922\times10^{-82}\). Ratio \(=5.67\times10^{-8}\ \text{W·m}^{-2}\text{K}^{-4}\), matching the accepted \(5.670\times10^{-8}\).
(C) Photon number density and mean energy. Show that the photon number density is \(n=\dfrac{8\pi}{c^3}\left(\dfrac{k_BT}{h}\right)^3\!\int_0^\infty\dfrac{x^2\,dx}{e^x-1}\), evaluate the integral in terms of \(\zeta(3)\), and hence find the mean energy per photon as a multiple of \(k_BT\).
Solution
Number density is energy density with the \(h\nu\) factor removed: \(n=\int_0^\infty \dfrac{g(\nu)/V}{e^{h\nu/k_BT}-1}\,d\nu=\dfrac{8\pi}{c^3}\int_0^\infty\dfrac{\nu^2\,d\nu}{e^{h\nu/k_BT}-1}\). Substituting \(x=h\nu/k_BT\) gives the stated form with \(\int_0^\infty\dfrac{x^2}{e^x-1}dx=\Gamma(3)\zeta(3)=2\zeta(3)=2(1.20206)=2.404\). Thus \(n=\dfrac{8\pi}{c^3}\Big(\dfrac{k_BT}{h}\Big)^3(2.404)\). The mean energy per photon is \(\dfrac{u}{n\,}=\dfrac{(8\pi h/c^3)(k_BT/h)^4(\pi^4/15)}{(8\pi/c^3)(k_BT/h)^3(2\zeta(3))}=\dfrac{\pi^4/15}{2\zeta(3)}\,k_BT=\dfrac{6.494}{2.404}\,k_BT=2.701\,k_BT\). So a typical blackbody photon carries about \(2.7\,k_BT\).