Light as Information

Section 4 of 9

Blackbody Radiation

Part 4: Blackbody Radiation

What Is a Blackbody?

A blackbody is an idealized object that absorbs all radiation falling on it — no reflection, no transmission. When heated, it re-emits radiation with a spectrum that depends only on its temperature. The name is misleading: a cold blackbody looks black, but a hot one glows — the hotter, the brighter and bluer.

Blackbody

An idealized perfect absorber and emitter: it absorbs all incident radiation and, when heated, emits a spectrum (the Planck curve) that depends only on its temperature $T$ . Stars are approximate blackbodies.

Stars are approximate blackbodies — their spectra deviate due to absorption lines, but the overall shape follows the blackbody curve. Other astronomical blackbodies: planets (Earth emits as a $\sim 255$ K blackbody in the infrared), the Cosmic Microwave Background (the most perfect blackbody ever measured, at 2.725 K), and accretion disks around black holes.

The Planck Spectrum: Qualitative Picture

When you heat an object it glows, and the color depends on temperature: a stovetop burner dull red ( $\sim$ 700 K), an incandescent filament yellow-white ( $\sim$ 2700 K), the Sun yellow ( $\sim$ 5800 K), a welding arc blue-white ( $\sim$ 6000+ K).

Graph showing three Planck curves for stars at 8000 K (blue, peaks in the near-UV around 360 nm), 5000 K (yellow, peaks near 580 nm), and 3000 K (red, peaks in the near-IR around 970 nm). The visible band (about 400-700 nm) is marked. Y-axis is brightness; X-axis is wavelength. — Figure 8Hotter stars peak at shorter (bluer) wavelengths and emit more total light: an 8000 K star peaks near 360 nm, a 3000 K star near 970 nm. Wien's law: lambda_peak = b/T with b = 0.2898 cm K.JWST/STScI

The blackbody spectra (Planck curves) show two patterns: (1) the peak shifts to shorter wavelengths as temperature increases — hot objects peak in the blue/UV, cool objects in the red/IR; (2) the total energy emitted increases dramatically with temperature — the area under the curve grows as $T^4$ (the Stefan-Boltzmann law, quantified in Module 2).

Pause & Predict #1

Think about the shape of the Planck curve, not just the peak location.

The blue star (8000 K) peaks in the ultraviolet. Why does it still appear blue to our eyes, rather than invisible?

The red star (3000 K) peaks in the infrared. Why does it still glow visibly red?

Check your reasoning against the answers below.

Multiple choice

An 8000 K star peaks in the ultraviolet, yet it still looks blue to your eyes (not invisible). Why?

Its Planck curve is broad — it emits plenty of visible light, and more blue than redIts peak is actually in the visible, not the UVUV light reddens to blue in our atmosphereAll hot stars look white

The Ultraviolet Catastrophe

Before 1900, classical physics gave the Rayleigh-Jeans law, $B_\lambda \propto T\lambda^{-4}$ . It works at long wavelengths, but as $\lambda \to 0$ the intensity diverges — classical physics predicted any warm object should emit infinite energy in the ultraviolet. This ultraviolet catastrophe was a fundamental failure: ovens don’t emit infinite UV.

Planck’s Quantum Solution

In 1900 Max Planck found the fix — “an act of desperation.” He proposed that energy is emitted in discrete packets (quanta) with energy $E = h\nu$ . To emit a short-wavelength (high-frequency) photon, an oscillator needs energy $h\nu$ ; if the available thermal energy ( $\sim k_B T$ ) is less than $h\nu$ , emission is suppressed — you can’t emit a fraction of a photon. This quantum suppression cuts off the catastrophe. It took Einstein (1905) to take quanta seriously, showing light itself comes in photons.

The Planck Function: Meeting a Real Equation

The complete blackbody spectrum is the Planck function. It looks intimidating — don’t panic; we read it piece by piece.

$B_\lambda(T)$ is the spectral radiance — power emitted per unit area, per unit solid angle, per unit wavelength. The $\lambda^{-5}$ factor would drive short wavelengths up, but the exponential term in the denominator suppresses them; the competition between the two creates the peak. You don’t need to compute it from scratch — you need to recognize its shape and know which term controls which regime.

Spectral radiance

The power a surface emits per unit area, per unit solid angle (“per direction”), per unit wavelength interval — what the Planck function $B_\lambda(T)$ gives. CGS units: $\mathrm{erg\,s^{-1}\,cm^{-2}\,sr^{-1}\,cm^{-1}}$ .

The spectral radiance tells you how bright a surface looks in a given direction at each wavelength. A unit check confirms $[B_\lambda] = \mathrm{erg\,s^{-1}\,cm^{-2}\,sr^{-1}\,cm^{-1}}$ — power per area per solid angle per wavelength interval, exactly what “spectral radiance” should be.

Limiting Cases: Taming Complex Equations

The way to not be intimidated by a scary equation is to take its limiting cases — what happens when one term dominates.

Case 1 — long wavelengths (Rayleigh-Jeans). When $hc/\lambda k_B T \ll 1$ , expand $e^x - 1 \approx x$ , so

$B_\lambda \approx \frac{2hc^2}{\lambda^5}\cdot\frac{\lambda k_B T}{hc} = \frac{2ck_B T}{\lambda^4}.$

Classical physics works here because quantum effects are negligible when $h\nu \ll k_B T$ .

Case 2 — short wavelengths (Wien tail). When $hc/\lambda k_B T \gg 1$ , the $-1$ is negligible and

$B_\lambda \approx \frac{2hc^2}{\lambda^5}\,e^{-hc/\lambda k_B T}.$

The exponential cuts off short-wavelength emission — the quantum suppression that prevents the catastrophe. You can’t emit photons whose energy exceeds the available thermal energy.

Regime	Condition	Behavior	Physical reason
Long $\lambda$	$hc/\lambda \ll k_B T$	$B_\lambda \propto T\lambda^{-4}$	Classical limit; photons are “cheap”
Short $\lambda$	$hc/\lambda \gg k_B T$	$B_\lambda \to 0$ exponentially	Quantum suppression; photons are “expensive”

Problem

Using $B_\lambda \propto T\lambda^{-4}$ : (1) if $T$ doubles (fixed $\lambda$ ), intensity does what? (2) If $\lambda$ doubles (fixed $T$ )? (3) Two blackbodies at the same $T$ , one at $\lambda = 1$ cm and one at $\lambda = 0.5$ cm — how do their Rayleigh-Jeans intensities compare?