Radiation

Walk past a blacksmith's forge in Moradabad late at night. The iron bar in the fire is dark. Pump the bellows a few times and the bar turns a dull red. Pump more and it brightens to orange, then yellow, then a fierce white that hurts to look at. The blacksmith has not added any colour — they have only added energy. The iron is sending you photons of different wavelengths, and as it gets hotter it sends both more of them and shifts the mix toward shorter wavelengths. The sky is doing the same thing: the Sun is radiating white because its surface is at 5778 K, the Earth is radiating invisible thermal infrared because it is at 288 K, you are radiating invisible thermal infrared at 310 K, and even a block of ice at 270 K is sending out a faint infrared glow that a thermal camera can see in the dark.

This is thermal radiation, the third of the three heat transfer mechanisms. Conduction carries heat through a solid by molecular jostling. Convection carries it through a fluid by bulk flow. Radiation carries it by electromagnetic waves, and crucially, it needs no medium at all — it is how the Sun reaches you through a hundred and fifty million kilometres of vacuum. Every object above absolute zero radiates. The only question is how much and at what wavelength.

This article builds the three laws that answer those questions — the Stefan-Boltzmann T^4 law, Wien's \lambda_\max T = b law, and Kirchhoff's law relating emission to absorption. You should have read temperature and thermometers so that the T in every formula is firmly an absolute temperature in kelvin. Familiarity with electromagnetic waves is useful but not essential — all you need is that visible light, infrared, and ultraviolet are the same kind of wave, differing only in wavelength.

Every object glows — the idea of thermal radiation

Pick any object in the room. A teacup, your phone, a brick, a glacier. The atoms and molecules inside it are jiggling; if they were not, the object would be at absolute zero, which nothing real ever reaches. Jiggling atoms carry charge (electrons and nuclei) accelerating back and forth. An accelerating charge radiates electromagnetic waves — that is Maxwell's equations at work. So every object, simply by being above 0 K, is emitting electromagnetic waves from its surface.

The wavelengths this radiation contains depend on how hard the atoms are jiggling, which is to say, on temperature. A very cold object (a block of dry ice at 195 K) radiates mostly at wavelengths around 15 microns — deep infrared, invisible. A warm object (you, at 310 K) radiates at roughly 9.3 microns — near-IR, still invisible. A dull red iron bar at 800 K is now radiating with a bit of its spectrum in the visible; most of it is still infrared, but the red tail of the distribution leaks into what your eyes can see. A filament lamp at 2800 K is throwing off visible light plus a lot of waste infrared (which is why incandescent bulbs are terrible fridges). The Sun's surface at 5778 K peaks near 500 nm, smack in the middle of the visible band — which is not a coincidence: your eyes evolved sensitive to precisely the wavelengths that the Sun emits most of.

Three quantitative laws describe this picture. The first, Wien's displacement law, says where on the wavelength axis the peak sits. The second, Stefan-Boltzmann, says how much total power is radiated. The third, Kirchhoff's law, says how emission compares to absorption for a given surface. You can derive all three (or at least motivate them) from the idea of a black body and the equilibrium between a surface and the radiation around it — that is the work of the rest of this article.

The black body — the reference standard

The three laws each involve the notion of a black body: an idealised surface that absorbs every photon of every wavelength that hits it, reflecting and transmitting nothing. Call its absorptivity \alpha = 1. A black body is the cleanest possible emitter as well: because nothing is being reflected back and interfering with its own thermal emission, the spectrum it radiates is purely a function of its temperature, not of what it is made of. This is the key insight — the black body's spectrum is universal.

Real surfaces are not perfectly black. Polished aluminium absorbs only a few percent of the light that hits it and reflects the rest; lampblack soot absorbs about 99%; the inner wall of a hollow cavity with a small hole is almost indistinguishable from a perfect black body because any light that enters the hole rattles around and gets absorbed before it finds its way back out. The small hole in an opaque cavity — an isothermal cavity — is in fact how physicists in the 19th century built black bodies in the lab. Light going in: absorbed. Light coming out of the hole: the thermal radiation of the cavity walls, unpolluted by reflections. The spectrum of that light is what Planck worked out in 1900, and we will use his result below without deriving it (the derivation is quantum mechanics; it belongs in a separate article).

Real surfaces are described by an emissivity \varepsilon between 0 and 1, defined so that the power they emit at temperature T is

Some typical values: fresh snow \varepsilon \approx 0.8–0.9 (surprisingly close to black in the IR), skin \varepsilon \approx 0.98 (almost perfectly black in the IR — this is why you show up clearly on a thermal camera regardless of skin colour, which only affects visible light), polished aluminium \varepsilon \approx 0.05, oxidised copper \varepsilon \approx 0.6, a diya's clay pot \varepsilon \approx 0.9. The numbers are wavelength-dependent in general — a surface can be black to visible light and shiny to IR (snow is one example: white to your eyes, nearly black to a thermal camera). When you see \varepsilon without a subscript, it usually means the "total" emissivity averaged over the spectrum the body is actually emitting at that temperature.

Assumption that runs through this article: unless stated otherwise, we talk about surfaces where \varepsilon is roughly constant across the relevant wavelengths. Real surfaces are often more interesting — the "selective surface" coatings on solar hot-water panels are engineered to have high \varepsilon at visible wavelengths (to absorb sunlight) and low \varepsilon at IR wavelengths (to suppress their own thermal losses). We flag these cases when they come up.

The Planck spectrum — what a black body actually radiates

Before you can derive Stefan-Boltzmann or Wien, you need to know what a black body radiates at each wavelength. That is the Planck spectrum, the spectral radiance of a black body at temperature T:

where h = 6.626\times 10^{-34} J·s is Planck's constant, c = 3\times 10^{8} m/s, k_B = 1.38\times 10^{-23} J/K is Boltzmann's constant, and u(\lambda,T)\,d\lambda is the power radiated per unit surface area of the body in the wavelength interval \lambda to \lambda + d\lambda. Deriving this formula from first principles requires quantum mechanics (Planck had to assume that an oscillator could only take discrete energies n h\nu to make the integral finite — that assumption was the beginning of quantum physics). We will not re-derive it here; for this article, take it as the experimentally verified spectrum and use it to derive Wien and Stefan-Boltzmann.

Two features of this curve are what the two laws capture:

1. It has a peak. The peak wavelength \lambda_\max depends on T.
2. The area under it is the total radiated power. The total rises steeply with T.

Wien says where the peak is. Stefan-Boltzmann says what the area is. Both come out of the Planck spectrum by calculus.

Wien's displacement law — from the peak of the Planck curve

Claim: at temperature T the wavelength at which a black body radiates most strongly is

Triple the temperature, the peak moves to one-third the wavelength.

Derivation. You find the peak of u(\lambda,T) by \frac{\partial u}{\partial \lambda} = 0. Let x = hc/(\lambda k_B T) so that the \lambda-dependence of the Planck expression becomes tidy.

Step 1. Write the spectrum in terms of x. Since \lambda = hc/(x k_B T), the factor 1/\lambda^5 becomes (x k_B T/(hc))^5, and the spectrum becomes

Why: pulling the T-dependence and the constants out of the way reduces the problem to finding the maximum of a single function of a dimensionless variable x. The answer for \lambda_\max will follow by untwisting the substitution.

Step 2. Differentiate f(x) = x^5/(e^x - 1) and set the derivative to zero.

Set the numerator to zero and divide by x^4:

Why: the quotient rule plus the fact that e^x - 1 > 0 for x > 0 means only the numerator needs to vanish. The x^4 factor gets cancelled because x = 0 is a minimum (the spectrum vanishes at very short wavelengths), not a maximum.

Step 3. Solve the transcendental equation 5 - x = 5 e^{-x} numerically. Starting from x = 5 and iterating, you get x \approx 4.965.

Why: this equation has no closed-form solution, but one iteration of Newton's method converges quickly: x_{n+1} = 5 - 5 e^{-x_n}, with x_0 = 5 giving x_1 = 4.966, x_2 = 4.965. That is as far as you need to go.

Step 4. Untwist the substitution. x = hc/(\lambda_\max k_B T) = 4.965, so

Plug in h, c, k_B:

Why: the constants gather into a single universal number b — the Wien displacement constant. The equation \lambda_\max T = b is a statement about a fundamental combination of h, c, k_B, not about any particular body.

Wien's law is a stunningly useful diagnostic. Point a spectrometer at a distant star, measure its peak wavelength, and you have its surface temperature without ever touching it. Bring the same reasoning indoors: a diya flame glowing dull red must be at roughly 2.898\times 10^{-3}/(700\times 10^{-9}) \approx 4140 K if 700 nm is really where its peak lies — but careful observation suggests the flame peak is actually in the near-IR around 1600 nm, which gives T \approx 1810 K, a far more realistic diya temperature. The red your eye sees is only the short-wavelength tail of a much broader IR emission.

Interactive: drag the temperature, watch the peak shift

The Stefan-Boltzmann law — total radiated power

Claim: the total power per unit surface area radiated by a black body at temperature T is

For a real surface with emissivity \varepsilon, multiply by \varepsilon: P/A = \varepsilon \sigma T^4.

Derivation. The total radiated power per unit area is the integral of the Planck spectrum over all wavelengths:

Step 1. Substitute x = hc/(\lambda k_B T), so \lambda = hc/(x k_B T) and d\lambda = -hc/(x^2 k_B T)\,dx. Reversing the limits (as \lambda goes 0\to\infty, x goes \infty \to 0) absorbs the minus sign:

Tidying the powers of h, c, k_B, T:

Why: this substitution is the standard move — it separates the T-dependence cleanly, leaving a dimensionless integral that is a pure number. The T^4 is already visible in front of the integral; that is the origin of the fourth-power law.

Step 2. Evaluate the dimensionless integral. It is a standard result in mathematical physics:

Why: expand 1/(e^x - 1) = \sum_{n=1}^\infty e^{-nx} (a geometric series, valid for x > 0) and integrate term by term to get \sum_{n=1}^\infty 6/n^4 = 6\,\zeta(4) = \pi^4/15. Knowing \zeta(4) = \pi^4/90 from the theory of the Riemann zeta function closes the calculation. If this detail is new, read it as "the integral evaluates to about 6.49" and move on; the rest of the derivation does not depend on knowing why.

Step 3. Put it together:

With numerical values plugged in:

Why: the Stefan-Boltzmann constant \sigma is a combination of fundamental constants alone. The T^4 is the key physics — it makes the power rise with extraordinary steepness. Go from 300 K to 600 K, the power rises by a factor of 2^4 = 16. Go from 300 K to 6000 K (as the Sun does), it rises by 20^4 = 160{,}000.

For a real surface, replace \sigma by \varepsilon\sigma. For a body at temperature T sitting in surroundings at temperature T_s, the net rate at which it loses energy to radiation is

because the body emits \varepsilon\sigma T^4 and absorbs \varepsilon\sigma T_s^4 from the surroundings (by Kirchhoff's law, its absorptivity equals its emissivity — we prove this in the next section). This is the form that matters for heat-loss calculations: a cup of chai losing heat to a room, a satellite losing heat to the 2.7 K cosmic background, the Earth losing heat to space.

Assumption check: equation (6) assumes the surroundings are themselves approximately a black body at temperature T_s (the walls of a room, the sky as a whole, etc.). If the surroundings have a different emissivity, you need a more general "form factor" analysis, which belongs in an engineering heat-transfer text.

Kirchhoff's law — good absorbers are good emitters

Claim: at any given wavelength and temperature, a body's emissivity equals its absorptivity:

Derivation. Place any object inside an isothermal cavity at temperature T. The cavity is full of black-body radiation at temperature T with spectrum u(\lambda, T). In thermal equilibrium the object must be at the same temperature T — any other temperature would imply a net energy flow that would change its temperature, contradicting equilibrium.

Thermal equilibrium requires detailed balance at every wavelength. The power the object absorbs from the radiation in wavelength interval d\lambda is

The power it emits in the same wavelength interval is

because \varepsilon_\lambda u(\lambda, T) is by definition the body's emission compared to a black body's.

Why: a black body would emit u(\lambda, T)\, A\, d\lambda in that wavelength interval. A real body emits \varepsilon_\lambda times that. The absorptivity \alpha_\lambda is defined as the fraction of incident radiation the body absorbs at wavelength \lambda.

In equilibrium, the two must be equal (otherwise energy at wavelength \lambda would accumulate or deplete, contradicting the steady state):

Since this must hold for every wavelength independently (radiation at different wavelengths does not interconvert in equilibrium), it is a point-wise equality, not a spectrum-averaged one.

Why the detail balance? If the absorption rate at some \lambda exceeded the emission rate, energy at \lambda would decrease in the cavity until that wavelength's radiation dropped and equilibrium was restored — but then the cavity radiation would no longer match the universal u(\lambda, T), which contradicts the known universality of black-body radiation. So emission and absorption have to match independently at every wavelength.

Kirchhoff's law has three immediate practical consequences.

1. A good absorber is a good emitter. A black surface (high \alpha) at temperature T radiates more than a shiny surface at the same temperature. That is why solar collectors are painted black — they absorb more sunlight and, as a bonus, emit their own infrared well (but this IR loss is a problem that selective surfaces solve).

2. Polished metal radiates very little. An aluminised surface at 300 K emits at maybe 5% of the black-body rate. That is why Dewar flasks (thermos bottles) have silvered inner walls — radiation loss from the hot liquid is cut by a factor of 20 compared to a black bottle.

3. "Thermally transparent" means cannot radiate. A window pane that transmits visible light is not radiating visible light. But glass is nearly black in the thermal IR (\varepsilon \approx 0.95 at 10 μm) and radiates like a black body at window temperature — which is why your thermal camera sees you through a glass pane only partially. It sees the pane's IR emission superimposed on the room's IR reflections.

Worked examples

Common confusions

• "Radiation needs a medium." No — this is exactly the opposite. Radiation is the only heat transfer mechanism that works in vacuum. That is why the Sun can warm the Earth across empty space: conduction and convection cannot cross 150 million km of nothing, but electromagnetic waves can.

• "Hot objects emit radiation; cold objects do not." Every object above absolute zero radiates. An ice cube at 270 K is sending out more than 300 W/m² of thermal infrared. What is true is that net heat transfer goes from hotter to colder, because hotter bodies emit more intensely — but both bodies are always radiating.

• "A black body is just a very dark object." A black body is an idealisation: something that absorbs every photon at every wavelength. Real surfaces that look black to the eye (lampblack, a deep hole in a box) approach this, but even they reflect some UV or some radio. The important use of the "black body" concept is that its emission spectrum depends only on temperature, not on material — which is why Wien's and Stefan-Boltzmann's constants are universal.

• "Stefan's law uses Celsius." Never. The Stefan-Boltzmann law requires the absolute temperature in kelvin. Using Celsius (e.g. raising 100 to the fourth power instead of 373) will give an answer off by an enormous factor. The T^4 only makes physical sense when zero really does mean zero emission.

• "Emissivity equals absorptivity means the object looks the same colour as what it absorbs." The equality \varepsilon_\lambda = \alpha_\lambda is wavelength-by-wavelength, not averaged. A white surface (like dry snow) has low \alpha for visible light but high \alpha for thermal IR — so it is a poor absorber of sunlight (visible) but an excellent emitter of its own IR. The colour your eye sees tells you about visible-wavelength properties; it says nothing about the IR.

• "Cooking food by radiation is slow because radiation is weak." Actually, microwaves and tandoor ovens cook fast precisely by radiation — but the radiation is at different wavelengths. A 2.45 GHz microwave penetrates several centimetres into food because water absorbs strongly at that frequency but only near the surface. A tandoor's 700 K walls radiate at about 4 μm (Wien), which water absorbs strongly, so the naan is browned on the surface in seconds. Radiation is not inherently slow; it depends on the match between emission spectrum and absorption spectrum.

• "Newton's law of cooling and Stefan's law are unrelated." Newton's law of cooling (\text{rate} \propto \Delta T) is in fact the linearisation of Stefan's law for small temperature differences. Expand T^4 - T_s^4 = (T - T_s)(T^3 + T^2 T_s + T T_s^2 + T_s^3) \approx 4 T_s^3\,(T - T_s) when T - T_s \ll T_s. So for small \Delta T, the T^4 law reduces to a first-order law linear in \Delta T with slope 4\sigma\varepsilon T_s^3. That is why your cup of tea, radiating to a room not much colder, cools exponentially in the Newton-law regime.

Where this leads next

• Newton's Law of Cooling — the linearisation of Stefan's T^4 law for small temperature differences, and why your cup of chai cools exponentially.
• Photoelectric Effect — the other experiment that forced quantum mechanics on physics (after black-body radiation), and where the photon picture first landed.
• Electromagnetic Spectrum — the full catalogue of wavelengths, which thermal radiation samples according to the body's temperature.
• Wave Nature of Electromagnetic Waves — the Maxwell-equations story for why accelerating charges radiate in the first place.
• Kinetic Theory of Gases — the temperature in Stefan-Boltzmann is the same kinetic-theory temperature, and the two chapters together tell you how to compute the emission of any hot gas.