Radiation laws

Stand near a bonfire and you feel its warmth even with no air moving toward you and no metal touching your skin — that heat crossed empty space as electromagnetic waves. This is radiation, the only mode of heat transfer that needs no medium at all, and it is why the Sun's warmth reaches Earth across 150 million kilometres of vacuum. Every object above absolute zero radiates continuously, and the rate at which it does so obeys a small set of precise, testable laws: the Stefan–Boltzmann law, linking total emitted power to the fourth power of temperature, and Wien's displacement law, linking the colour of the glow to how hot the object is. Together these laws let you read a star's temperature from its colour, or its total power output from a single number.

The core insight

A blackbody radiates total intensity I=σT4I = \sigma T^4 (Stefan–Boltzmann law) with its emission spectrum peaking at a wavelength λmaxT=b\lambda_{max}T = b that shortens as the body gets hotter (Wien's displacement law).

The core idea

A blackbody is an idealised perfect absorber and, correspondingly, the best possible emitter — it radiates the maximum energy physically possible at every wavelength for its temperature. It is modelled as a cavity with a small hole: radiation entering the hole reflects around inside and essentially never re-emerges, so the hole behaves as a perfect absorber, while radiation escaping back out (cavity radiation) is used to study ideal blackbody emission in the laboratory. The Stefan–Boltzmann law gives the total radiant intensity — power emitted per unit surface area, summed over all wavelengths — of a blackbody at absolute temperature TT:

I(T)=σT4I(T) = \sigma T^4

where: II is intensity in watts per square metre, TT is absolute temperature in kelvin, and σ=5.67×10−8 W m−2K−4\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4} is the Stefan–Boltzmann constant. Real surfaces fall short of the ideal, so we scale by the emissivity ϵ\epsilon, a dimensionless number between 0 and 1 with ϵ=1\epsilon = 1 for a perfect blackbody:

I=ϵ σT4I = \epsilon\,\sigma T^4

A polished, reflective surface has ϵ\epsilon close to 0 and radiates poorly; a dull, dark surface has ϵ\epsilon close to 1 and radiates efficiently. This is also why good absorbers are good emitters: a matte black surface absorbs almost all incident radiation and, by the same physics, emits almost as much back out, while a shiny surface reflects most incoming radiation and correspondingly radiates very little of its own — the reasoning behind using a shiny coating (like the silvered wall of a thermos flask) to suppress radiative heat loss. Not all of that emitted power sits at one wavelength. Spectral radiancy R(λ)R(\lambda) describes how power is distributed across wavelength λ\lambda, and the total intensity is the area under its curve:

I(T)=∫0∞R(λ) dλI(T) = \int_0^\infty R(\lambda)\, d\lambda

Plotting R(λ)R(\lambda) for a blackbody at several temperatures produces a family of humped curves: each rises to one peak and falls away, and as TT increases the whole curve lifts (more area, consistent with the T4T^4 growth) while its peak slides toward shorter wavelengths. This shift is quantified by Wien's displacement law:

λmax T=b\lambda_{max}\,T = b

where: λmax\lambda_{max} is the wavelength at which R(λ)R(\lambda) peaks and b≈2.898×10−3 m Kb \approx 2.898\times10^{-3}\ \text{m K} is Wien's constant. This is the physics behind an object glowing dull red, then orange, then white as it is heated further — its peak emission wavelength is sliding out of the infrared, through red, and across the visible spectrum as TT climbs.

Figure
Spectral radiancy against wavelength for three blackbody temperatures showing the peak shifting to shorter wavelength and the curve rising as temperature increases
Hotter blackbodies peak at shorter wavelength and radiate more wavelength (lambda) spectral radiancy R(lambda) 4000 K 5800 K (Sun) 7000 K peak wavelength shortens as temperature rises: lambda-max times T equals b
Derivation AComparing the radiated power of two blackbodies

Given two blackbodies of known radii and surface temperatures, find how many times more total power the hotter or larger one radiates, using only the Stefan–Boltzmann law and the surface area of a sphere — no calculus required, just combining a power law with a ratio.

Step 1. Write total radiated power for a spherical blackbody of radius RR and temperature TT as intensity times surface area: P=I×A=(σT4)(4πR2)P = I \times A = (\sigma T^4)(4\pi R^2). This holds for any two such bodies, labelled 1 and 2.

Step 2. Form the ratio P2/P1P_2/P_1 so the constants σ\sigma and 4π4\pi cancel: P2P1=σT24⋅4πR22σT14⋅4πR12=(R2R1)2(T2T1)4\dfrac{P_2}{P_1} = \dfrac{\sigma T_2^4 \cdot 4\pi R_2^2}{\sigma T_1^4 \cdot 4\pi R_1^2} = \left(\dfrac{R_2}{R_1}\right)^2\left(\dfrac{T_2}{T_1}\right)^4.

Step 3. Read the result. The power ratio depends only on the ratios of radius and temperature, not on their absolute values — a body twice the radius radiates four times the power at equal temperature, while a body with double the absolute temperature radiates sixteen times the power at equal radius, because power depends on the fourth power of TT but only the square of RR.

Step 4. This ratio form is the standard tool for comparing two stars, two filaments, or two furnace walls without recomputing each intensity separately — plug in the two radius ratios and temperature ratios and multiply.

Worked example

A comparison star, Sirius A, has surface temperature T2≈9940 KT_2 \approx 9940\ \text{K} and radius R2≈1.71 R⊙R_2 \approx 1.71\,R_\odot, where R⊙R_\odot is the Sun's radius. Using the Sun's known surface temperature T1=5800 KT_1 = 5800\ \text{K} and total power P1≈3.91×1026 WP_1 \approx 3.91\times10^{26}\ \text{W} (from the solar luminosity calculation), find Sirius A's total radiated power.

Step 1. Compute the radius ratio squared: (R2R1)2=(1.71)2≈2.92\left(\dfrac{R_2}{R_1}\right)^2 = (1.71)^2 \approx 2.92.

Step 2. Compute the temperature ratio to the fourth power: T2T1=99405800≈1.714\dfrac{T_2}{T_1} = \dfrac{9940}{5800} \approx 1.714, so (T2T1)4≈(1.714)4≈8.63\left(\dfrac{T_2}{T_1}\right)^4 \approx (1.714)^4 \approx 8.63.

Step 3. Multiply the two factors to get the power ratio: P2P1≈2.92×8.63≈25.2\dfrac{P_2}{P_1} \approx 2.92 \times 8.63 \approx 25.2.

Step 4. Scale up the Sun's known power: P2≈25.2×3.91×1026≈9.85×1027 WP_2 \approx 25.2 \times 3.91\times10^{26} \approx 9.85\times10^{27}\ \text{W}. Sirius A radiates about 25 times the Sun's total power, roughly 9.9×1027 W9.9\times10^{27}\ \text{W}, dominated by its higher surface temperature rather than its modestly larger size.

Common traps

  • Using Celsius temperature in I=σT4I = \sigma T^4 or in Wien's law — both require absolute (kelvin) temperature, and a fourth-power law is extremely sensitive to this mistake.
  • Treating I=σT4I = \sigma T^4 as total power directly — it gives intensity (power per unit area); multiply by surface area to get total power.
  • Forgetting the emissivity factor ϵ\epsilon for real, non-blackbody objects — a shiny object radiates far less than σT4\sigma T^4 predicts.
  • Assuming a hotter body's spectral peak shifts to a longer wavelength — it is the opposite: hotter bodies peak at shorter wavelengths, λmax∝1/T\lambda_{max} \propto 1/T.

Ask the tutor: want a walkthrough of how these radiation laws connect to Newton's law of cooling, or a two-body Stefan's-law comparison problem to practise on?

Why does the Stefan-Boltzmann law use kelvin and not Celsius?

Because I=σT4I = \sigma T^4 is a genuine fourth-power relationship in absolute temperature, and Celsius has an arbitrary zero point that does not correspond to zero thermal energy. Using a Celsius value would make the fourth power meaningless — doubling a Celsius reading does not double the actual thermal energy the way doubling a kelvin reading does. Always convert to kelvin first, add 273 to the Celsius value, before applying either the Stefan-Boltzmann law or Wien's displacement law.

What is the practical difference between intensity and total power in these problems?

Intensity I=σT4I = \sigma T^4 (or ϵσT4\epsilon\sigma T^4) is power radiated per square metre of surface, so it depends only on temperature and emissivity, never on the object's size. Total power PP is intensity multiplied by the entire radiating surface area, so a small hot object and a large cool object can radiate the same total power. Exam questions that ask for total power almost always require you to compute the surface area separately and multiply, as in the solar luminosity calculation.

Does a hotter blackbody stop emitting the cooler wavelengths it used to emit?

No. Raising the temperature shifts where the peak of the spectral radiancy curve sits and raises the height of the whole curve at every wavelength, but the object still emits some power across a broad range of wavelengths on both sides of the peak. A red-hot poker continues to emit infrared even as it turns orange and then white; it is simply emitting far more at every wavelength, with the balance now weighted toward shorter wavelengths, which is why its apparent colour shifts.

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