Stefan–Boltzmann & Wien's Laws

Thermal imaging and pyrometry. A firefighter's camera, an airport fever scanner and a drone hunting roof heat-leaks are all built on Wien's law: bodies near $300\ \mathrm{K}$ peak around $10\ \mathrm{\mu m}$, so the detector is engineered for the $8$–$14\ \mathrm{\mu m}$ window, and the brightness it records is converted back into a temperature through Stefan's law. The same physics runs inside a blast furnace, where no thermometer survives: an optical pyrometer reads the temperature of the glow itself, and the surface's emissivity is the one number the engineer must get right.

Stars. Astronomers classify stars by colour because colour is temperature: one peaking at $350\ \mathrm{nm}$ is at $8280\ \mathrm{K}$ and blue-white, one peaking at $966\ \mathrm{nm}$ is a $3000\ \mathrm{K}$ red dwarf. Add a measured brightness and $P = \sigma AT^4$ hands you the star's radius — measured, from across the galaxy, with a prism and a fourth power.

Incandescent lighting. The $2500\ \mathrm{K}$ filament peaks at $1.16\ \mathrm{\mu m}$, deep in the infrared — a Wien's-law verdict no engineering can dodge, and the reason the incandescent bulb has been legislated out of existence in most of the world.

Earth's energy balance. The Sun delivers about $1361\ \mathrm{W\,m^{-2}}$ to the top of the atmosphere; the Earth reflects roughly $30\%$, and because a planet intercepts sunlight over $\pi R^2$ but radiates from $4\pi R^2$, it absorbs on average $1361 \times 0.70/4 = 238\ \mathrm{W\,m^{-2}}$. Setting $\sigma T^4 = 238$ gives an equilibrium temperature of $255\ \mathrm{K} = -18\ \mathrm{^\circ C}$ — yet the real mean surface temperature is about $15\ \mathrm{^\circ C}$. That $33\ \mathrm{^\circ C}$ gap is the greenhouse effect, found by taking Stefan's law seriously.

Watch a blacksmith heat an iron bar. Cold, it looks black and does nothing — though it is already radiating, invisibly, in the far infrared. Push it into the fire and it glows a sullen red; hotter, cherry, then orange, then a fierce yellow-white too bright to look at. Two things changed as the temperature climbed, and at wildly different rates. The bar got brighter, savagely so. And it got bluer, steadily. Those are the two laws of this article, and they are two readings of the same graph. A hot body does not emit at a single wavelength; it emits a continuous spectrum, a hump with a definite peak, and that hump's shape is fixed once you know the temperature — nothing else about the body matters. Stefan's law is about the area under the hump: total energy per second per square metre. Wien's law is about the position of its peak: the colour. Area and peak, brightness and colour, Stefan and Wien. Put three such humps on one set of axes and the entire topic becomes visible at once.

Three curves, and every claim of this article is already in the picture. As $T$ rises the peak slides left — the colour walks out of the infrared, through the red, towards the blue. That is Wien. And the whole curve lifts, so violently that the $3000\ \mathrm{K}$ hump must be magnified twenty times before you can even see where its peak sits: its area is sixteen times smaller than the $6000\ \mathrm{K}$ one, its peak height thirty-two times lower. That is Stefan. Notice what the shaded band says about the blacksmith, too: at $3000\ \mathrm{K}$ only the tail of the hump reaches the visible, red end first — which is why hot iron glows red long before it glows white.

In 1879 Josef Stefan noticed something in a table of measurements made by somebody else: the total radiation from a hot body rose not with the temperature but with the temperature raised to the fourth power. Five years later his former student Ludwig Boltzmann proved it from thermodynamics. Their law says a perfect emitter — a black body — pours out, from every square metre of its surface,

where: $E$ — the emissive power, the energy radiated per second from each square metre, in $\mathrm{W\,m^{-2}}$; $T$ — the absolute temperature of the surface, in kelvin, never in Celsius; $\sigma$ — the Stefan–Boltzmann constant, $5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}$, a property of the universe and not of the body. Feel how steep that is. A black body at $1000\ \mathrm{K}$ radiates $E = 5.67\times10^{-8} \times (1000)^4 = 5.67\times10^{4}\ \mathrm{W\,m^{-2}}$ — $56.7$ kilowatts from a single square metre, the output of a small car engine, from a glowing patch the size of a doormat. Equation (1) is per square metre, and for a perfect emitter. Real bodies are neither. Fix both with the surface area $A$ and the emissivity $e$, a number between $0$ and $1$ measuring how black the surface is ($e = 1$ for a black body):

where: $P$ — the total power radiated by the whole body, in watts; $A$ — its radiating surface area, in $\mathrm{m^2}$ (a sphere has $A = 4\pi r^2$); $e$ — the emissivity, dimensionless. The physical reading: emissivity is a plain scaling knob. A surface with $e = 0.5$ radiates exactly half what a black body of the same size and temperature would, at every wavelength — it does not reshape the curve or shift the peak, only press the hump down. A hot body is also surrounded by things radiating back at it. If the surroundings sit at $T_s$, it absorbs from them while it emits, and what you measure — the rate at which it loses energy — is the difference:

where: $T$ — the body's temperature; $T_s$ — that of the surroundings; both in kelvin. The reading boards test: when $T = T_s$ the net exchange is zero and the body sits in radiative equilibrium — still radiating furiously, but receiving exactly as much as it gives. Nothing has stopped; the books have merely balanced. And note when $T_s$ may be ignored: a filament at $2500\ \mathrm{K}$ in a $300\ \mathrm{K}$ room has $(300/2500)^4 = 2.1\times10^{-4}$, so the room contributes two hundredths of one per cent — for anything glowing, $T_s$ is noise.

The peak does not sit still. Wilhelm Wien found in 1893 that its position moves inversely with the temperature — hence "displacement" — and that the product of the two is a constant of nature:

where: $\lambda_{\max}$ — the wavelength at which emission is strongest, in metres; $T$ — the absolute temperature in kelvin; $b$ — Wien's constant, $2.898\times10^{-3}\ \mathrm{m\,K}$. The physical reading: hotter bodies are bluer, and the relation is inverse, so doubling the temperature halves the peak wavelength. Note what $b$ is not — it is not $\sigma$, it has different units, and it belongs to a different law. Equation (4) is a temperature measuring device, because $T = b/\lambda_{\max}$ works on anything you can take a spectrum of, however distant and however hot. Sunlight peaks at about $500\ \mathrm{nm}$ — green, near the middle of the band your eye evolved to see, which is no coincidence — so the solar surface must sit at $T = 2.898\times10^{-3}/(500\times10^{-9}) = 5796\ \mathrm{K}$, the $5800\ \mathrm{K}$ every textbook quotes. Run it the other way for something cool: your skin at $37\ \mathrm{^\circ C} = 310\ \mathrm{K}$ peaks at $\lambda_{\max} = 2.898\times10^{-3}/310 = 9.35\ \mathrm{\mu m}$, deep in the far infrared — invisible to your eye, blazingly obvious to a camera built to see at $10\ \mathrm{\mu m}$. Every thermal-imaging camera and non-contact fever scanner is engineered around that one number.

That linearisation deserves a number, because it is where two chapters join. Take a black surface at $T_s = 300\ \mathrm{K}$ with an excess $\Delta T = 5\ \mathrm{K}$: the exact Stefan answer is $\sigma(305^4 - 300^4) = 31.4\ \mathrm{W\,m^{-2}}$, while the linear form gives $4\sigma T^3\Delta T = 4 \times 5.67\times10^{-8} \times (300)^3 \times 5 = 30.6\ \mathrm{W\,m^{-2}}$ — agreement to $2.5\%$. That is why a cooling cup of tea obeys Newton's linear law and a cooling star does not: the tea's excess over the room is small, the star's is not.

Board level — the incandescent filament, and why it wastes your money (CBSE/ICSE/NEET pattern)

The tungsten filament of a $60\ \mathrm{W}$ lamp runs at $2500\ \mathrm{K}$. Treating it as a black body, find (a) the power radiated from each square metre, (b) the surface area needed to radiate $60\ \mathrm{W}$, and (c) the wavelength at which its emission peaks. Take $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}$, $b = 2.898\times10^{-3}\ \mathrm{m\,K}$.

Step 1. Check the temperature is already absolute. It is — $2500\ \mathrm{K}$. (Had the question said $2227\ \mathrm{^\circ C}$, the first move would have been to add $273$.)

Step 2. (a) Apply equation (1), fourth power first: $(2500)^4 = 3.906\times10^{13}$, so $E = \sigma T^4 = 5.67\times10^{-8} \times 3.906\times10^{13} = \boxed{2.21\times10^{6}\ \mathrm{W\,m^{-2}}}$ — over two megawatts from every square metre of filament.

Step 3. (b) The filament must radiate the lamp's full $60\ \mathrm{W}$, so from $P = EA$ with $e = 1$: $A = P/E = 60/(2.21\times10^{6}) = 2.71\times10^{-5}\ \mathrm{m^2} = \boxed{0.27\ \mathrm{cm^2}}$. That is why a filament is a hair-thin coiled wire: at two megawatts per square metre, a postage stamp of tungsten would light a small street.

Step 4. (c) Wien, equation (4): $\lambda_{\max} = b/T = 2.898\times10^{-3}/2500 = \boxed{1.16\ \mathrm{\mu m}}$. Now read what you have found. Visible light runs from about $400$ to $700\ \mathrm{nm}$, and the filament's peak sits at $1160\ \mathrm{nm}$ — beyond the red end, in the infrared. The lamp is an excellent heater that leaks a little light. Every LED replacement on the market exists because of that one number.

Competitive level — the filament turned up, and the star turned down (JEE-Main/JEE-Advanced/AP-2 pattern)

(a) A filament is run hotter, from $2000\ \mathrm{K}$ to $2500\ \mathrm{K}$: by what factor does its radiated power rise, and where does its peak wavelength move? (b) Star A has twice the radius of star B but half its surface temperature — find the ratio of their total radiated powers. (c) Star B's spectrum peaks at $350\ \mathrm{nm}$: what is its surface temperature?

Step 1. (a) Do not run two Stefan calculations. Use the ratio: $P_2/P_1 = (T_2/T_1)^4 = (2500/2000)^4 = (1.25)^4 = \boxed{2.44}$. A mere $25\%$ rise in temperature nearly two-and-a-half times the output. (The long way agrees: $E(2000) = 9.07\times10^{5}$, $E(2500) = 2.21\times10^{6}\ \mathrm{W\,m^{-2}}$, ratio $2.44$.)

Step 2. Still in (a), the colour. Wien is inverse, so the peak moves opposite to the temperature: $\lambda_{\max}$ falls from $2.898\times10^{-3}/2000 = 1.45\ \mathrm{\mu m}$ to $\boxed{1.16\ \mathrm{\mu m}}$ — divided by that same $1.25$. Hotter, brighter, bluer; and the brightness moved far faster than the colour, because one is a fourth power and the other a first.

Step 3. (b) The trap: the size changed too, so the area cannot cancel. Use $P = \sigma AT^4$ with $A = 4\pi r^2$, giving $P \propto r^2T^4$. Then $\dfrac{P_A}{P_B} = \left(\dfrac{r_A}{r_B}\right)^{2}\left(\dfrac{T_A}{T_B}\right)^{4} = 2^2 \times \left(\tfrac{1}{2}\right)^{4} = 4 \times \dfrac{1}{16} = \boxed{\dfrac{1}{4}}$. The bigger star is the fainter one, by a factor of four: the fourth power beats the square every time. (The classic wrong answers are $4$ — using only the radius — and $1$ — assuming "twice as big, half as hot" must cancel.)

Step 4. (c) Wien again, on a star: $T_B = b/\lambda_{\max} = 2.898\times10^{-3}/(350\times10^{-9}) = \boxed{8280\ \mathrm{K}}$. A near-ultraviolet peak means a blue-white star far hotter than the Sun — and from (b), star A is at $8280/2 = 4140\ \mathrm{K}$, an orange one. Timing note: ninety seconds for all three parts, and none needs a calculator if you set it up as a ratio — part (a) is $(1.25)^4$, part (b) is $2^2 \times 2^{-4}$, and only part (c) is a genuine division. The question really tests whether you wrote $P \propto r^2T^4$ before substituting: candidates who compute each star's power from scratch lose two minutes and usually drop an arithmetic slip on the way.

Using Celsius. $T$ appears raw in both laws, so the kelvin conversion is not optional: $27\ \mathrm{^\circ C}$ means $T = 300\ \mathrm{K}$. The number-one mark-loser in CBSE, ICSE and NEET radiation questions.

Forgetting the exponent in a factor-of-change question. "Temperature doubles, power doubles" is wrong by a factor of eight. Always write $(T_2/T_1)^4$ first. AP Physics 2 tests both laws this way and expects $16\times$ and $\lambda/2$ in the same breath.

Getting Wien's direction backwards. Hotter means a shorter peak wavelength, since $\lambda_{\max} = b/T$ is inverse; "displacement" means the peak moves towards the blue as $T$ rises. If your answer makes a hotter star redder, you divided the wrong way round.

Mixing up $\sigma$ and $b$. Different constants, different laws, different units: $\sigma$ for power, $b$ for wavelength. Swapping them gives an answer that is not merely wrong but dimensionally absurd.

Forgetting that $E = \sigma T^4$ is per unit area. $E$ is a flux, $\mathrm{W\,m^{-2}}$. For a real object's total power multiply by area and emissivity, $P = e\sigma AT^4$ — and a sphere's area is $4\pi r^2$, not $\pi r^2$.

Using $T^4$ where the question wants $T^4 - T_s^4$. A body cooling into its surroundings needs both terms. Drop $T_s$ only when the body is far hotter: fine for a filament, nonsense for a cup of tea in a warm room.

A board note, honestly stated. AP Physics 2 assesses both laws together with factor-of-change reasoning, and JEE-Advanced's published syllabus names "Stefan's law" and "Wien's displacement law" explicitly. The 2024 NTA rationalisation removed the literal string "Stefan's law" from the JEE-Main syllabus PDF — but NCERT Class 11 still teaches it, previous papers still use it, and coaching sheets still drill it. Know it regardless: the document dropped a phrase, not the physics.

Texts. NCERT Physics Class 11, Ch. 11 (Thermal Properties of Matter) — Stefan's and Wien's laws as CBSE and JEE teach them · Halliday, Resnick and Walker, "Thermal Radiation" · Serway and Jewett, "Blackbody Radiation and Planck's Hypothesis" (the AP-2 treatment).

Provenance. Standard results; $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}$, $b = 2.898\times10^{-3}\ \mathrm{m\,K}$. Every number here is recomputed in verify_hs2-stefan-boltzmann-wien.py: the Stefan values $E(T)$ at $T = 300, 600, 900, 1000, 2000, 2500, 3000, 4500, 5796$ and $6000\ \mathrm{K}$ ($459$, $7348$, $3.72\times10^{4}$, $5.67\times10^{4}$, $9.07\times10^{5}$, $2.21\times10^{6}$, $4.59\times10^{6}$, $2.33\times10^{7}$, $6.40\times10^{7}$, $7.35\times10^{7}\ \mathrm{W\,m^{-2}}$); the scalings $2^4 = 16$, $3^4 = 81$, $(1.25)^4 = 2.44$, $(1.01)^4 = 1.041$, $(300/2500)^4 = 2.1\times10^{-4}$ and the star ratio $2^2\times(1/2)^4 = 1/4$; the Wien values $9.35\ \mathrm{\mu m}$ at $310\ \mathrm{K}$, $2.90\ \mathrm{\mu m}$ at $1000\ \mathrm{K}$, $1.45$ and $1.16\ \mathrm{\mu m}$ at $2000$ and $2500\ \mathrm{K}$, $966/644/483\ \mathrm{nm}$ at $3000/4500/6000\ \mathrm{K}$, with the inversions $5796\ \mathrm{K}$ from $500\ \mathrm{nm}$ and $8280\ \mathrm{K}$ from $350\ \mathrm{nm}$; the filament area $0.27\ \mathrm{cm^2}$; the solar area $6.09\times10^{18}\ \mathrm{m^2}$ and luminosity $3.90\times10^{26}\ \mathrm{W}$ against the measured $3.85\times10^{26}$; the Planck peak-height ratio $32$; the linearisation $30.6$ against the exact $31.4\ \mathrm{W\,m^{-2}}$; and Earth's balance $238\ \mathrm{W\,m^{-2}} \Rightarrow 255\ \mathrm{K}$.

Prerequisites. Thermal Radiation and the Black Body (emissivity, Kirchhoff's law, why a black body is the reference). Leads to. Newton's Law of Cooling — which, as the limiting-cases table shows, is this article's $T^4$ law linearised for a small temperature excess · ARCHETYPE: Cooling and Radiation Numericals. Go deeper (undergraduate). "Stefan-Boltzmann Law and Radiation Pressure" derives the fourth power the way Boltzmann did — by giving the radiation in a cavity a pressure of its own and doing thermodynamics on it — and shows that the photon gas which carries the energy also pushes.