advanced32 min studyLesson 329 in Physics

Count thermal photons with zero chemical potential

Derive Planck energy density from mode counting and connect cavity energy, isotropic pressure, and surface emission.

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What you’ll understand

  • Explain why equilibrium blackbody photons have μ=0.
  • Keep angular-frequency state density distinct from ordinary-frequency density.
  • Integrate the spectrum to obtain temperature scaling and distinguish energy density from emitted flux.

Thermal radiation is an equilibrium gas whose particle number is not conserved. Cavity walls can absorb and emit photons until temperature determines their occupation, giving zero photon chemical potential. Combining that occupation with the number of electromagnetic modes produces Planck's distribution. We will use angular frequency consistently, count both polarizations, and separate the energy stored per volume from pressure and the flux escaping a small hole. Those quantities differ by geometric and speed factors.

A mode has many possible photon occupations

Each electromagnetic mode is a harmonic oscillator with energy spacing hbar times angular frequency. Its thermal state permits occupation zero, one, two, and so on. Because photon number can adjust through equilibrium interaction with the walls, no independent conserved-number multiplier is imposed: the photon chemical potential is zero. Summing the geometric occupation weights gives a Bose occupation. This applies to equilibrium blackbody light; driven lasers and approximately number-conserving photon arrangements can require a different effective description.

The thermal excitation energy per mode is its photon energy times mean occupation. A temperature-independent zero-point term is omitted from this thermal spectrum; questions involving changes in vacuum energy or boundary stresses need their own treatment. At low photon energy compared with thermal energy the occupation becomes large and the mode approaches the classical equipartition value. At high energy the exponential suppresses occupation. This high-frequency suppression prevents the ultraviolet divergence obtained by assigning thermal energy to infinitely many classical modes.

μγ=0,Zω=∑n=0∞e−nβℏω=11−e−βℏω,nˉω=1eβℏω−1\mu_\gamma=0,\qquad Z_\omega=\sum_{n=0}^{\infty}e^{-n\beta\hbar\omega}=\frac1{1-e^{-\beta\hbar\omega}},\qquad \bar n_\omega=\frac1{e^{\beta\hbar\omega}-1}
ω is angular frequency in rad/s, ℏω energy in joules, and β=1/(kBT). Zω omits the zero-point energy factor and describes one equilibrium bosonic mode at positive T.↔ Scroll wide equations horizontally.

Mode density supplies the missing frequency powers

In a large vacuum cavity, shell counting in wavevector space gives mode density proportional to angular frequency squared. There are two independent transverse polarizations. Counting them once produces the density used here. Multiplying mode density by energy per photon and Bose occupation gives thermal spectral energy density per angular-frequency interval. Density per ordinary frequency differs because angular frequency is two pi times ordinary frequency; the interval Jacobian must accompany any change of variable.

Substitute hbar times angular frequency divided by thermal energy into the total-energy integral. Every frequency factor and the integration interval contributes a temperature factor, yielding total energy density proportional to temperature to the fourth power. The remaining dimensionless integral is pi to the fourth divided by fifteen. A spectral maximum depends on whether density is expressed per frequency or per wavelength, so a peak cannot be converted between those plots simply by inverting its horizontal coordinate.

g(ω)V=ω2π2c3,uω(T)=ℏω3π2c3(eℏω/(kBT)−1),u(T)=∫0∞uω dω=π2kB415ℏ3c3T4\frac{g(\omega)}V=\frac{\omega^2}{\pi^2c^3},\qquad u_\omega(T)=\frac{\hbar\omega^3}{\pi^2c^3(e^{\hbar\omega/(k_BT)}-1)},\qquad u(T)=\int_0^\infty u_\omega\,d\omega=\frac{\pi^2k_B^4}{15\hbar^3c^3}T^4
g(ω)dω counts modes including two polarizations; u is thermal energy density in J/m³ and uω is energy density per rad/s. The continuum approximation assumes a cavity large compared with relevant thermal wavelengths.↔ Scroll wide equations horizontally.

Isotropic radiation has pressure and an escape flux

Photons carry momentum equal to energy divided by c. In an isotropic cavity, the average squared directional cosine is one third, so radiation pressure is energy density divided by three. A small hole emits only outward-going photons; weighting their crossing rate by the directional cosine gives flux equal to c times energy density divided by four. The one-third and one-quarter factors describe different angular averages and cannot be interchanged.

The escaping energy flux is the Stefan–Boltzmann exitance of an ideal black surface at that temperature. A real object's emissivity, spectral response, surrounding radiation, and geometry alter its net heat exchange. A single directed beam does not obey the isotropic pressure formula. The quadratic lab can verify the mode-density frequency-square factor alone. Its parabola has neither Bose occupation nor a thermal spectrum, so the Planck curve and total temperature-fourth-power law remain analytical deductions.

Pγ=u3,M=cu4=σT4,σ=π2kB460ℏ3c2P_\gamma=\frac u3,\qquad M=\frac{cu}{4}=\sigma T^4,\qquad \sigma=\frac{\pi^2k_B^4}{60\hbar^3c^2}
Pγ is isotropic cavity pressure in pascals, M is outward radiant exitance in W/m², and σ in W/(m²·K⁴). The hole is small enough to leave cavity equilibrium undisturbed; exterior incident radiation is excluded from this outward flux.↔ Scroll wide equations horizontally.

Convert cavity energy density into surface emission

At temperature T₀ a blackbody cavity has thermal energy density 4.0×10⁻⁶ J/m³. Increase its temperature to 2T₀. Using c=3.0×10⁸ m/s, find the new energy density, isotropic pressure, and outward small-hole flux.

  1. Energy density follows T⁴, so doubling temperature multiplies it by sixteen. The new value is 6.4×10⁻⁵ J/m³.

  2. Isotropic pressure is u/3, giving approximately 2.13×10⁻⁵ Pa. This uses the isotropic momentum distribution rather than the pressure formula for a single directed beam.

  3. Outward exitance is cu/4=(3.0×10⁸)(6.4×10⁻⁵)/4=4800 W/m². The one-quarter factor accounts for angular crossing of outward rays; it is not simply all energy moving normally at c.

The result

At 2T₀ the cavity has u=6.4×10⁻⁵ J/m³, pressure about 2.13×10⁻⁵ Pa, and outward flux 4800 W/m². These are thermal equilibrium quantities with distinct units.

Explore the model

Interactive lab

Interactive lab · function

Test the photon mode-density factor

For each fresh run Reset and set BOTH coefficient a=1, position x=1; then Reset and set a=1, x=2. Regard x as normalized angular frequency and y as normalized mode density. Predict a factor four. The model shows only y=ax² and does not include Planck occupation, energy per photon, or thermal emission.

Test the photon mode-density factorThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forMode count grows quadratically, while exponential occupation suppresses the high-frequency energy spectrum.

Check your understanding

Why is blackbody photon chemical potential zero in equilibrium with ordinary emitting and absorbing walls?

Try it yourself

For one equilibrium photon mode suppose ℏω=kBT ln3. Find its mean occupation and thermal excitation energy. If the same frequency is observed at twice the temperature, what is the new mean occupation?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the exponential occupation directly; the mode frequency remains fixed while β changes.

Compare with the solution

Initially exp(ℏω/kBT)=3, so mean occupation is 1/2 and thermal energy is ℏω/2. At twice the temperature the exponent halves, giving exponential √3 and mean 1/(√3−1)=(√3+1)/2≈1.366. This is a fixed-frequency occupation comparison, not the sixteenfold integrated energy-density scaling across all frequencies.

Take it with you

  • Equilibrium blackbody photon number is unconstrained, giving μγ=0.
  • Planck density combines two-polarization mode counting with Bose occupation and photon energy.
  • Isotropic pressure u/3 and outward flux cu/4 use different directional averages.

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