Chapter 7 • Theory & Derivations

Thermal Radiation & Quantum Hypotheses

Kirchhoff's radiation law, blackbody cavity energy density, radiation pressure, Stefan-Boltzmann law, Wien's displacement law, Rayleigh-Jeans formula, Planck's quantum radiation law, and solar astrophysics.

§7.1 Spectral Emissive Powers, Absorptivity & Kirchhoff’s Law

1. Thermal Radiation Fundamentals

All matter at non-zero absolute temperature continuously emits electromagnetic radiation due to thermal fluctuations and accelerating electrical charges within constituent atoms and molecules:

Spectral Emissive Power $e_\lambda$: Radiant energy emitted per unit surface area, per unit time, per unit wavelength interval in all outward directions (in $\text{W/(m}^2\cdot\text{m)}$ or $\text{W/m}^3$).
Total Emissive Power $E$: Integrated radiant flux across all wavelengths:

E = \int_0^\infty e_\lambda d\lambda

Spectral Absorptive Power (Absorptivity) $a_\lambda$: The dimensionless fraction of incident radiation of wavelength $\lambda$ absorbed by the surface ($0 \le a_\lambda \le 1$). An idealized blackbody absorbs 100% of incident radiation across all wavelengths ($a_\lambda \equiv 1$).

2. Kirchhoff's Law of Thermal Radiation

Gustav Kirchhoff (1859) proved through second-law detailed balance that for any body in thermal equilibrium inside an isothermal enclosure at temperature $T$:

\frac{e_\lambda}{a_\lambda} = E_\lambda^{\text{blackbody}}(T)

The ratio of spectral emissive power to absorptive power is an identical universal function of wavelength and temperature for all materials. Physical Consequences:

A good absorber is inevitably a good emitter ($a_\lambda = 1 \implies e_\lambda = E_\lambda^{\text{blackbody}}$). A poor absorber (or shiny mirror, $a_\lambda \ll 1$) is a poor emitter.
A body cannot emit more radiation at any wavelength than an ideal blackbody at the identical temperature.
Cavity Radiator: An isothermal hollow sphere with a small pinhole aperture acts as an experimental realization of a blackbody ($a_\lambda \approx 0.999$), as incoming rays undergo multiple diffuse reflections with internal absorption on every bounce.

§7.2 Energy Density, Radiation Pressure & Stefan-Boltzmann Law

1. Radiation Pressure in an Isotropic Photon Gas

Electromagnetic radiation carries momentum density $\vec{g} = \vec{S}/c^2 = \frac{u}{c} \hat{k}$, where $u$ is the volumetric radiation energy density (in $\text{J/m}^3$). When isotropic radiation impinges on a surface, time-averaged momentum transfer over all angles of incidence yields:

P = \frac{1}{3} u

This isotropic radiation pressure is fundamental to the hydrostatic equilibrium of stars and stellar interiors.

2. Thermodynamic Derivation of the Stefan-Boltzmann Law

Treating the radiation field within an evacuated cavity of volume $V$ as a thermodynamic system:

Internal energy: $U = u(T) V$.
First Law with radiation pressure: $dU = T dS - P dV = T dS - \frac{1}{3} u dV$.
Differential entropy:

dS = \frac{dU + P dV}{T} = \frac{V \frac{du}{dT} dT + u dV + \frac{1}{3} u dV}{T} = \frac{V}{T} \frac{du}{dT} dT + \frac{4 u}{3 T} dV

Since $dS$ is an exact differential, equating cross partial derivatives:

\frac{\partial}{\partial V}\left(\frac{V}{T} \frac{du}{dT}\right) = \frac{\partial}{\partial T}\left(\frac{4 u}{3 T}\right)

\frac{1}{T} \frac{du}{dT} = \frac{4}{3} \left( \frac{1}{T} \frac{du}{dT} - \frac{u}{T^2} \right) = \frac{4}{3 T} \frac{du}{dT} - \frac{4 u}{3 T^2}

\frac{1}{3 T} \frac{du}{dT} = \frac{4 u}{3 T^2} \implies \frac{du}{u} = 4 \frac{dT}{T}

Integrating both sides:

\ln u = 4 \ln T + \text{const} \implies u(T) = a T^4

where $a$ is the radiation density constant. The total emissive power exiting a blackbody surface into a hemisphere is $E = \frac{1}{4} c u$:

E = \sigma T^4

where $\sigma = \frac{a c}{4} = \frac{2\pi^5 k_B^4}{15 c^2 h^3} \approx 5.6704 \times 10^{-8} \text{ W/(m}^2\cdot\text{K}^4)$. This is the Stefan-Boltzmann Law.

§7.3 Wien’s Displacement Law & The Ultraviolet Catastrophe

1. Wien's Displacement Law

Wilhelm Wien (1893) analyzed the adiabatic expansion of an evacuated spherical cavity with perfectly reflecting walls containing blackbody radiation. As the cavity expands, Doppler shifts upon reflection from receding walls stretch each wavelength proportionally to the cavity radius: $\lambda \propto R$. Since isentropic expansion preserves $V T^3 = \text{const} \implies R T = \text{const}$, the product $\lambda T$ remains invariant. Consequently, the spectral energy distribution must take the general form:

u_\lambda d\lambda = \frac{1}{\lambda^5} \phi(\lambda T) d\lambda

Differentiating with respect to $\lambda$ and setting the derivative to zero locates the peak emission wavelength $\lambda_{\max}$:

\lambda_{\max} T = b

where $b \approx 2.8978 \times 10^{-3} \text{ m}\cdot\text{K}$ is Wien's Displacement Constant. As temperature increases, the peak of blackbody emission shifts toward shorter wavelengths (from infrared to visible red, yellow, and blue-white).

2. Rayleigh-Jeans Law & The Ultraviolet Catastrophe

Lord Rayleigh and Sir James Jeans (1900–1905) applied classical electrodynamics and the Maxwell-Boltzmann equipartition theorem to standing electromagnetic waves in a cubical cavity of side $L$.

Number of electromagnetic cavity standing wave modes between $\nu$ and $\nu + d\nu$ (with 2 polarization states):

g(\nu) d\nu = \frac{8\pi V}{c^3} \nu^2 d\nu \implies g(\lambda) d\lambda = \frac{8\pi V}{\lambda^4} d\lambda

By classical equipartition, each harmonic standing wave mode possesses average thermal energy $\langle E \rangle = k_B T$.
The spectral energy density becomes the Rayleigh-Jeans Formula:

u_\lambda d\lambda = \frac{8\pi k_B T}{\lambda^4} d\lambda

While accurate at very long wavelengths (infrared and radio), as $\lambda \to 0$ (ultraviolet, X-ray), the predicted energy density diverges:

\lim_{\lambda \to 0} u_\lambda = \infty \implies \int_0^\infty u_\lambda d\lambda = \infty

This profound classical failure was termed the Ultraviolet Catastrophe by Paul Ehrenfest.

§7.4 Planck’s Quantum Radiation Law & High/Low Asymptotes

1. Max Planck's Quantum Hypothesis

On October 19, 1900, Max Planck resolved the ultraviolet catastrophe by postulating that the material oscillators within the cavity walls emit and absorb electromagnetic radiation only in discrete, indivisible packets (quanta) of energy:

E_n = n h \nu = n \frac{h c}{\lambda}, \quad n = 0, 1, 2, \dots

where $h \approx 6.6261 \times 10^{-34}\text{ J}\cdot\text{s}$ is Planck's constant.

2. Derivation of Average Oscillator Energy

By the Boltzmann distribution, the probability of an oscillator occupying energy level $E_n$ is $P_n \propto e^{-E_n / k_B T} = e^{-n \beta h \nu}$ (where $\beta = 1/k_B T$):

\langle E \rangle = \frac{\sum_{n=0}^\infty n h \nu e^{-n \beta h \nu}}{\sum_{n=0}^\infty e^{-n \beta h \nu}} = -\frac{d}{d\beta} \ln\left( \sum_{n=0}^\infty (e^{-\beta h \nu})^n \right) = -\frac{d}{d\beta} \ln\left(\frac{1}{1 - e^{-\beta h \nu}}\right)

\langle E \rangle = \frac{h \nu e^{-\beta h \nu}}{1 - e^{-\beta h \nu}} = \frac{h \nu}{e^{\frac{h \nu}{k_B T}} - 1}

Multiplying by the density of modes $g(\lambda) d\lambda = \frac{8\pi}{\lambda^4} d\lambda$:

u_\lambda d\lambda = \frac{8\pi h c}{\lambda^5 \left[ e^{\frac{h c}{\lambda k_B T}} - 1 \right]} d\lambda

This is Planck's Radiation Law.

3. Asymptotic Reductions

1. Wien Approximation ($\lambda k_B T \ll h c$, high frequencies):

e^{\frac{h c}{\lambda k_B T}} \gg 1 \implies u_\lambda \approx \frac{8\pi h c}{\lambda^5} e^{-\frac{h c}{\lambda k_B T}}

Exponential decay eliminates the ultraviolet catastrophe.

2. Rayleigh-Jeans Limit ($\lambda k_B T \gg h c$, low frequencies):

Expanding $e^x \approx 1 + x$:

e^{\frac{h c}{\lambda k_B T}} - 1 \approx \frac{h c}{\lambda k_B T} \implies u_\lambda \approx \frac{8\pi h c}{\lambda^5 \left(\frac{h c}{\lambda k_B T}\right)} = \frac{8\pi k_B T}{\lambda^4}

recovering the classical Rayleigh-Jeans formula.

§7.5 Solar Constant & Astrophysics of the Sun

1. The Solar Constant

The Solar Constant $S$ is the total radiant energy received from the Sun per unit time, per unit area, on a surface oriented perpendicular to the solar rays at the top of Earth's atmosphere at Earth's mean orbital radius ($1\text{ AU} \approx 1.496 \times 10^{11}\text{ m}$):

S \approx 1361 \text{ W/m}^2

2. Determination of the Surface Temperature of the Sun

Let the Sun have radius $R_\odot \approx 6.963 \times 10^8\text{ m}$ and effective surface photospheric temperature $T_\odot$. By the Stefan-Boltzmann law, the total luminosity $L_\odot$ emitted by the Sun is:

L_\odot = 4\pi R_\odot^2 \sigma T_\odot^4

By energy conservation, this luminosity spreads spherically across space. At the mean Sun-Earth distance $d_{\text{SE}}$:

S = \frac{L_\odot}{4\pi d_{\text{SE}}^2} = \frac{4\pi R_\odot^2 \sigma T_\odot^4}{4\pi d_{\text{SE}}^2} = \sigma T_\odot^4 \left(\frac{R_\odot}{d_{\text{SE}}}\right)^2

Solving for $T_\odot$:

T_\odot = \left[ \frac{S}{\sigma} \left(\frac{d_{\text{SE}}}{R_\odot}\right)^2 \right]^{1/4}

Substituting values:

\frac{d_{\text{SE}}}{R_\odot} = \frac{1.496 \times 10^{11}}{6.963 \times 10^8} \approx 214.85

\left(\frac{d_{\text{SE}}}{R_\odot}\right)^2 = (214.85)^2 \approx 46162

T_\odot = \left[ \frac{1361}{5.6704 \times 10^{-8}} \times 46162 \right]^{1/4} = \left[ (2.4002 \times 10^{10}) \times 46162 \right]^{1/4} = \left[ 1.108 \times 10^{15} \right]^{1/4} \approx 5778\text{ K}

The effective temperature of the solar photosphere is approximately $5780\text{ K}$.

3. Planetary Equilibrium Temperatures

For a planet with planetary radius $R_p$, orbital radius $d$, and Bond albedo $A$ (reflectivity), intercepted solar flux is $\pi R_p^2 S(1 - A)$. Emitting as a sphere of area $4\pi R_p^2$ with emissivity $\epsilon$:

\pi R_p^2 S (1 - A) = 4\pi R_p^2 \epsilon \sigma T_p^4 \implies T_p = \left[ \frac{S (1 - A)}{4 \epsilon \sigma} \right]^{1/4}

For Earth ($A \approx 0.30, \epsilon \approx 1$): $T_p \approx 255\text{ K}$ ($-18^\circ\text{C}$). Greenhouse atmospheric gases warm the actual mean surface temperature to $+15^\circ\text{C}$ ($288\text{ K}$).

Standard University Exam Examination Problems

Rigorous Analytical & Numerical Solved Problems

Comprehensive step-by-step mathematical proofs, dimensional evaluations, and calculations matching B.Sc. Honors university examinations.

Standard Example 7.1: Solar Surface Temperature & Terrestrial Radiation Budget

The measured solar constant outside Earth's atmosphere is $S = 1361.0\text{ W/m}^2$. The mean Earth-Sun distance is $d_{\text{SE}} = 1.496\times 10^{11}\text{ m}$, and the solar radius is $R_\odot = 6.963\times 10^8\text{ m}$. (a) Calculate the total electromagnetic luminosity $L_\odot$ of the Sun. (b) Calculate the effective photospheric surface temperature $T_\odot$ of the Sun assuming blackbody emission ($\sigma = 5.6704\times 10^{-8}\text{ W/(m}^2\cdot\text{K}^4)$). (c) Calculate the peak emission wavelength $\lambda_{\max}$ of the solar spectrum using Wien's displacement law ($b = 2.8978\times 10^{-3}\text{ m}\cdot\text{K}$).

Step 1: Calculate Total Solar Luminosity $L_\odot$

The solar flux spreads over a sphere of radius $d_{\text{SE}}$:

L_\odot = 4\pi d_{\text{SE}}^2 S = 4\pi (1.496 \times 10^{11}\text{ m})^2 (1361.0\text{ W/m}^2)

4\pi (2.2380 \times 10^{22}) = 2.8124 \times 10^{23}\text{ m}^2

L_\odot = (2.8124 \times 10^{23})(1361.0) = 3.8277 \times 10^{26}\text{ W}

The Sun radiates nearly $3.83 \times 10^{26}\text{ Joules}$ of energy every second.

Step 2: Calculate Effective Surface Temperature $T_\odot$

Total solar surface area: $A_\odot = 4\pi R_\odot^2 = 4\pi (6.963 \times 10^8)^2 = 4\pi (4.8483 \times 10^{17}) = 6.0927 \times 10^{18}\text{ m}^2$. By the Stefan-Boltzmann law $L_\odot = A_\odot \sigma T_\odot^4$:

T_\odot^4 = \frac{L_\odot}{A_\odot \sigma} = \frac{3.8277 \times 10^{26}}{(6.0927 \times 10^{18})(5.6704 \times 10^{-8})} = \frac{3.8277 \times 10^{26}}{3.4548 \times 10^{11}} = 1.1079 \times 10^{15}\text{ K}^4

T_\odot = (1.1079 \times 10^{15})^{1/4} = 5777.6\text{ K} \approx 5778\text{ K}

Step 3: Calculate Peak Wavelength $\lambda_{\max}$

By Wien's displacement law:

\lambda_{\max} = \frac{b}{T_\odot} = \frac{2.8978 \times 10^{-3}\text{ m}\cdot\text{K}}{5777.6\text{ K}} = 5.0156 \times 10^{-7}\text{ m} = 501.6\text{ nm}

This peak wavelength corresponds to green visible light ($502\text{ nm}$), explaining why human vision evolved its highest sensitivity in the green region of the spectrum.

Final Answer & Physical Insight

(a) Total solar luminosity $L_\odot = 3.83\times 10^{26}\text{ W}$. (b) Effective solar temperature $T_\odot = 5778\text{ K}$. (c) Peak wavelength $\lambda_{\max} = 501.6\text{ nm}$ (green visible light).

Standard Example 7.2: Cryogenic Radiation Shielding Between Concentric Spheres

A spherical liquid helium storage dewar of radius $r_1 = 0.250\text{ m}$ is maintained at $T_1 = 4.20\text{ K}$ and has an emissivity of $\epsilon_1 = 0.040$. It is enclosed within an outer concentric spherical vacuum jacket of radius $r_2 = 0.350\text{ m}$ at ambient room temperature $T_2 = 300.0\text{ K}$ with emissivity $\epsilon_2 = 0.040$. (a) Calculate the net radiative heat influx without any intermediate radiation shields. (b) If a thin polished copper radiation shield ($\epsilon_s = 0.030$) of radius $r_s = 0.300\text{ m}$ is inserted in the vacuum space between them, calculate the equilibrium temperature $T_s$ of the shield. (c) Calculate the new net heat influx and the percentage reduction in cryogenic boil-off achieved by the shield.

Step 1: Radiative Heat Influx Without Radiation Shield

For concentric spheres, net radiative heat exchange is:

Q_{12} = \frac{\sigma (A_1)(T_2^4 - T_1^4)}{\frac{1}{\epsilon_1} + \frac{A_1}{A_2} \left(\frac{1}{\epsilon_2} - 1\right)}

Inner area $A_1 = 4\pi r_1^2 = 4\pi (0.25)^2 = 0.7854\text{ m}^2$. Outer area $A_2 = 4\pi r_2^2 = 4\pi (0.35)^2 = 1.5394\text{ m}^2$. Area ratio: $A_1 / A_2 = (0.25 / 0.35)^2 = 0.5102$. Denominator: $\frac{1}{0.04} + 0.5102 \left(\frac{1}{0.04} - 1\right) = 25.0 + 0.5102 (24.0) = 25.0 + 12.245 = 37.245$. Temperature terms: $T_2^4 = (300)^4 = 8.10 \times 10^9\text{ K}^4$, $T_1^4 = (4.2)^4 \approx 311 \approx 0$.

Q_{12} = \frac{(5.6704 \times 10^{-8})(0.7854)(8.10 \times 10^9)}{37.245} = \frac{360.75}{37.245} = 9.686\text{ W}

Step 2: Determine Equilibrium Shield Temperature $T_s$

In steady state, heat flowing from outer jacket (2) to shield (s) equals heat from shield (s) to inner dewar (1): $Q_{2s} = Q_{s1}$. Shield area $A_s = 4\pi (0.30)^2 = 1.1310\text{ m}^2$.

Q_{2s} = \frac{\sigma A_s (T_2^4 - T_s^4)}{\frac{1}{\epsilon_s} + \frac{A_s}{A_2} \left(\frac{1}{\epsilon_2} - 1\right)}, \quad Q_{s1} = \frac{\sigma A_1 (T_s^4 - T_1^4)}{\frac{1}{\epsilon_1} + \frac{A_1}{A_s} \left(\frac{1}{\epsilon_s} - 1\right)}

Denominator 1: $\frac{1}{0.03} + \left(\frac{0.30}{0.35}\right)^2 (24) = 33.33 + (0.7347)(24) = 33.33 + 17.63 = 50.96$. Denominator 2: $\frac{1}{0.04} + \left(\frac{0.25}{0.30}\right)^2 (32.33) = 25.0 + (0.6944)(32.33) = 25.0 + 22.45 = 47.45$. Equating fluxes and solving yields: $T_s^4 \approx 0.528 T_2^4 \implies T_s \approx (0.528)^{1/4} (300\text{ K}) = (0.852)(300) = 255.6\text{ K}$.

Step 3: New Heat Influx and Boil-off Reduction

The new heat reaching the helium bath is:

Q_{\text{shielded}} = \frac{(5.6704 \times 10^{-8})(0.7854)(255.6^4)}{47.45} = \frac{(4.4535 \times 10^{-8})(4.270 \times 10^9)}{47.45} = \frac{190.16}{47.45} = 4.008\text{ W}

Percentage reduction:

\text{Reduction} = \frac{9.686 - 4.008}{9.686} \times 100\\% = 58.62\\%

A single radiation shield cuts cryogenic heat leak and helium boil-off by nearly $59\\%$.

Final Answer & Physical Insight

(a) Unshielded heat influx $Q = 9.69\text{ W}$. (b) Equilibrium shield temperature $T_s = 255.6\text{ K}$ ($-17.5^\circ\text{C}$). (c) Shielded heat influx $Q_{\text{shielded}} = 4.01\text{ W}$, achieving a $58.6\\%$ reduction in helium boil-off.

Standard Example 7.3: Planck Radiation Integrated Power & Wien Limit Verification

Starting from Planck's spectral energy density $u_\lambda = \frac{8\pi h c}{\lambda^5 (e^{h c / \lambda k_B T} - 1)}$: (a) derive the Stefan-Boltzmann law by performing the definite integral $u = \int_0^\infty u_\lambda d\lambda$, using the Riemann zeta identity $\int_0^\infty \frac{x^3}{e^x - 1} dx = \frac{\pi^4}{15}$. (b) Derive Wien's displacement law and find the transcendental equation for $x = \frac{h c}{\lambda_{\max} k_B T}$. (c) Solve the transcendental equation numerically to find Wien's displacement constant $b$.

Step 1: Integrate Planck Formula to Derive Stefan-Boltzmann Law

Let $x = \frac{h c}{\lambda k_B T} \implies \lambda = \frac{h c}{k_B T x}$ and $d\lambda = -\frac{h c}{k_B T x^2} dx$.

u = \int_0^\infty \frac{8\pi h c}{\lambda^5 (e^{h c / \lambda k_B T} - 1)} d\lambda = 8\pi h c \int_0^\infty \frac{1}{\left(\frac{h c}{k_B T x}\right)^5 (e^x - 1)} \left( \frac{h c}{k_B T x^2} dx \right)

u = 8\pi h c \left(\frac{k_B T}{h c}\right)^4 \int_0^\infty \frac{x^3}{e^x - 1} dx = \frac{8\pi k_B^4 T^4}{h^3 c^3} \left(\frac{\pi^4}{15}\right) = \frac{8\pi^5 k_B^4}{15 c^3 h^3} T^4

Setting $a = \frac{8\pi^5 k_B^4}{15 c^3 h^3}$, the radiant emissive power is $E = \frac{c}{4} u = \sigma T^4$, where:

\sigma = \frac{2\pi^5 k_B^4}{15 c^2 h^3}

This confirms the Stefan-Boltzmann law directly from quantum first principles.

Step 2: Derive Wien's Transcendental Equation

To locate $\lambda_{\max}$, set $\frac{d u_\lambda}{d\lambda} = 0$. Write $u_\lambda = C \lambda^{-5} (e^{h c / \lambda k_B T} - 1)^{-1}$ with $x = \frac{h c}{\lambda k_B T}$:

\frac{d u_\lambda}{d\lambda} = C \left[ -5\lambda^{-6} (e^x - 1)^{-1} - \lambda^{-5} (e^x - 1)^{-2} e^x \left(-\frac{h c}{k_B T \lambda^2}\right) \right] = 0

-5 (e^x - 1) + x e^x = 0 \implies x e^x = 5 (e^x - 1)

Dividing by $e^x$:

x = 5 (1 - e^{-x}) \implies 5 - x = 5 e^{-x}

This is Wien's classic transcendental equation.

Step 3: Numerical Root Finding for $x$ and Evaluation of $b$

Solving $f(x) = x + 5 e^{-x} - 5 = 0$ via Newton-Raphson iteration:

$x_0 = 5 \implies f(5) = 5 + 5 e^{-5} - 5 = 5(0.006738) = 0.03369$.
$f'(x) = 1 - 5 e^{-x} \implies f'(5) = 1 - 0.03369 = 0.9663$.
$x_1 = 5 - \frac{0.03369}{0.9663} = 5 - 0.03487 = 4.9651$.

Converged root: $x = 4.965114$. Since $x = \frac{h c}{\lambda_{\max} k_B T} \implies \lambda_{\max} T = \frac{h c}{x k_B} = b$:

b = \frac{(6.62607 \times 10^{-34}\text{ J}\cdot\text{s})(2.99792 \times 10^8\text{ m/s})}{(4.965114)(1.38065 \times 10^{-23}\text{ J/K})} = \frac{1.986445 \times 10^{-25}}{6.855085 \times 10^{-23}} = 2.89777 \times 10^{-3}\text{ m}\cdot\text{K}

Final Answer & Physical Insight

(a) Exact Stefan-Boltzmann constant derived: $\sigma = \frac{2\pi^5 k_B^4}{15 c^2 h^3}$. (b) Transcendental equation is $x = 5(1 - e^{-x})$ with root $x = 4.9651$. (c) Wien's displacement constant is $b = 2.898\times 10^{-3}\text{ m}\cdot\text{K}$.