# Stefan–Boltzmann law

Graph of a function of total emitted energy of a black body ${\displaystyle j^{\star }}$ proportional to its thermodynamic temperature ${\displaystyle T\,}$. In blue is a total energy according to the Wien approximation, ${\displaystyle j_{W}^{\star }=j^{\star }/\zeta (4)\approx 0.924\,\sigma T^{4}\!\,}$

The Stefan–Boltzmann law describes the power radiated from a black body in terms of its temperature. Specifically, the Stefan–Boltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black-body radiant emittance or radiant exitance), ${\displaystyle j^{\star }}$, is directly proportional to the fourth power of the black body's thermodynamic temperature T:

${\displaystyle j^{\star }=\sigma T^{4}.}$

The constant of proportionality σ, called the Stefan–Boltzmann constant derives from other known constants of nature. The value of the constant is

${\displaystyle \sigma ={\frac {2\pi ^{5}k^{4}}{15c^{2}h^{3}}}=5.670373\times 10^{-8}\,\mathrm {W\,m^{-2}K^{-4}} ,}$

where k is the Boltzmann constant, h is Planck's constant, and c is the speed of light in a vacuum. Thus at 100 K the energy flux is 5.67 W/m2, at 1000 K 56,700 W/m2, etc. The radiance (watts per square metre per steradian) is given by

${\displaystyle L={\frac {j^{\star }}{\pi }}={\frac {\sigma }{\pi }}T^{4}.}$

A body that does not absorb all incident radiation (sometimes known as a grey body) emits less total energy than a black body and is characterized by an emissivity, ${\displaystyle \varepsilon <1}$:

${\displaystyle j^{\star }=\varepsilon \sigma T^{4}.}$

The irradiance ${\displaystyle j^{\star }}$ has dimensions of energy flux (energy per time per area), and the SI units of measure are joules per second per square metre, or equivalently, watts per square metre. The SI unit for absolute temperature T is the kelvin. ${\displaystyle \varepsilon }$ is the emissivity of the grey body; if it is a perfect blackbody, ${\displaystyle \varepsilon =1}$. In the still more general (and realistic) case, the emissivity depends on the wavelength, ${\displaystyle \varepsilon =\varepsilon (\lambda )}$.

To find the total power radiated from an object, multiply by its surface area, ${\displaystyle A}$:

${\displaystyle P=Aj^{\star }=A\varepsilon \sigma T^{4}.}$

Wavelength- and subwavelength-scale particles,[1] metamaterials,[2] and other nanostructures are not subject to ray-optical limits and may be designed to exceed the Stefan–Boltzmann law.

## History

The law was deduced by Josef Stefan (1835–1893) in 1879 on the basis of experimental measurements made by John Tyndall and was derived from theoretical considerations, using thermodynamics, by Ludwig Boltzmann (1844–1906) in 1884. Boltzmann considered a certain ideal heat engine with light as a working matter instead of a gas. The law is highly accurate only for ideal black objects, the perfect radiators, called black bodies; it works as a good approximation for most "grey" bodies. Stefan published this law in the article Über die Beziehung zwischen der Wärmestrahlung und der Temperatur (On the relationship between thermal radiation and temperature) in the Bulletins from the sessions of the Vienna Academy of Sciences.

## Examples

### Temperature of the Sun

With his law Stefan also determined the temperature of the Sun's surface. He learned from the data of Charles Soret (1854–1904) that the energy flux density from the Sun is 29 times greater than the energy flux density of a certain warmed metal lamella (a thin plate). A round lamella was placed at such a distance from the measuring device that it would be seen at the same angle as the Sun. Soret estimated the temperature of the lamella to be approximately 1900 °C to 2000 °C. Stefan surmised that ⅓ of the energy flux from the Sun is absorbed by the Earth's atmosphere, so he took for the correct Sun's energy flux a value 3/2 times greater than Soret's value, namely 29 × 3/2 = 43.5.

Precise measurements of atmospheric absorption were not made until 1888 and 1904. The temperature Stefan obtained was a median value of previous ones, 1950 °C and the absolute thermodynamic one 2200 K. As 2.574 = 43.5, it follows from the law that the temperature of the Sun is 2.57 times greater than the temperature of the lamella, so Stefan got a value of 5430 °C or 5700 K (the modern value is 5778 K[3]). This was the first sensible[clarification needed] value for the temperature of the Sun. Before this, values ranging from as low as 1800 °C to as high as 13,000,000 °C[citation needed] were claimed. The lower value of 1800 °C was determined by Claude Servais Mathias Pouillet (1790–1868) in 1838 using the Dulong-Petit law. Pouillet also took just half the value of the Sun's correct energy flux.

### Temperature of stars

The temperature of stars other than the Sun can be approximated using a similar means by treating the emitted energy as a black body radiation.[4] So:

${\displaystyle L=4\pi R^{2}\sigma {T_{e}}^{4}}$

where L is the luminosity, σ is the Stefan–Boltzmann constant, R is the stellar radius and T is the effective temperature. This same formula can be used to compute the approximate radius of a main sequence star relative to the sun:

${\displaystyle {\frac {R}{R_{\odot }}}\approx \left({\frac {T_{\odot }}{T}}\right)^{2}\cdot {\sqrt {\frac {L}{L_{\odot }}}}}$

where R, is the solar radius, L is the solar luminosity, and so forth.

With the Stefan–Boltzmann law, astronomers can easily infer the radii of stars. The law is also met in the thermodynamics of black holes in so-called Hawking radiation.

### Effective Temperature of the Earth

Similarly we can calculate the effective temperature of the Earth T by equating the energy received from the Sun and the energy radiated by the Earth, under the black-body approximation. The luminosity of the Sun, L, is given by:

${\displaystyle L_{\odot }=4\pi R_{\odot }^{2}\sigma T_{\odot }^{4}}$

At Earth, this energy is passing through a sphere with a radius of a0, the distance between the Earth and the Sun, and the irradiance (received power per unit area) is given by

${\displaystyle E_{\oplus }={\frac {L_{\odot }}{4\pi a_{0}^{2}}}}$

The Earth has a radius of R, and therefore has a cross-section of ${\displaystyle \pi R_{\oplus }^{2}}$. The radiant flux (i.e. solar power) absorbed by the Earth is thus given by:

${\displaystyle \Phi _{\text{abs}}=\pi R_{\oplus }^{2}\times E_{\oplus }:}$

Assuming the exchange is in a steady state, the flux emitted by Earth must equal the flux absorbed, and so:

{\displaystyle {\begin{aligned}4\pi R_{\oplus }^{2}\sigma T_{\oplus }^{4}&=\pi R_{\oplus }^{2}\times E_{\oplus }\\&=\pi R_{\oplus }^{2}\times {\frac {4\pi R_{\odot }^{2}\sigma T_{\odot }^{4}}{4\pi a_{0}^{2}}}\\\end{aligned}}}

T can then be found:

{\displaystyle {\begin{aligned}T_{\oplus }^{4}&={\frac {R_{\odot }^{2}T_{\odot }^{4}}{4a_{0}^{2}}}\\T_{\oplus }&=T_{\odot }\times {\sqrt {\frac {R_{\odot }}{2a_{0}}}}\\&=5780\;{\rm {K}}\times {\sqrt {696\times 10^{6}\;{\rm {m}} \over 2\times 149.598\times 10^{9}\;{\rm {m}}}}\\&\approx 279\;{\rm {K}}\end{aligned}}}

where T is the temperature of the Sun, R the radius of the Sun, and a0 is the distance between the Earth and the Sun. This gives an effective temperature of 6 °C on the surface of the Earth, assuming that it perfectly absorbs all emission falling on it and has no atmosphere.

The Earth has an albedo of 0.3, meaning that 30% of the solar radiation that hits the planet gets scattered back into space without absorption. The effect of albedo on temperature can be approximated by assuming that the energy absorbed is multiplied by 0.7, but that the planet still radiates as a black body (the latter by definition of effective temperature, which is what we are calculating). This approximation reduces the temperature by a factor of 0.71/4, giving 255 K (−18 °C).[5][6]

However, long-wave radiation from the surface of the earth is partially absorbed and re-radiated back down by greenhouse gases, namely water vapor, carbon dioxide and methane.[7][8] Since the emissivity with greenhouse effect (weighted more in the longer wavelengths where the Earth radiates) is reduced more than the absorptivity (weighted more in the shorter wavelengths of the Sun's radiation) is reduced, the equilibrium temperature is higher than the simple black-body calculation estimates. As a result, the Earth's actual average surface temperature is about 288 K (15 °C), which is higher than the 255 K effective temperature, and even higher than the 279 K temperature that a black body would have.

In the above discussion, we have assumed that the whole surface of the earth is at one temperature. Another interesting question is to ask what the temperature of a blackbody surface on the earth would be assuming that it reaches equilibrium with the sunlight falling on it. This of course depends on the angle of the sun on the surface and on how much air the sunlight has gone through. When the sun is at the zenith and the surface is horizontal, the irradiance can be as high as 1120 W/m2.[9] The Stefan–Boltzmann law then gives a temperature of

${\displaystyle T=\left({\frac {1120{\text{ W/m}}^{2}}{\sigma }}\right)^{1/4}\approx 375{\text{ K}}}$

or 102 °C. (Above the atmosphere, the result is even higher: 394 K.) We can think of the earth's surface as "trying" to reach equilibrium temperature during the day, but being cooled by the atmosphere, and "trying" to reach equilibrium with starlight and possibly moonlight at night, but being warmed by the atmosphere.

## Origination

### Thermodynamic derivation of the energy density

[10]

The fact that the energy density of the box containing radiation is proportional to ${\displaystyle T^{4}}$ can be derived using thermodynamics. It follows from the Maxwell stress tensor of classical electrodynamics that the radiation pressure ${\displaystyle p}$ is related to the internal energy density ${\displaystyle u}$:

${\displaystyle p={\frac {u}{3}}}$.

${\displaystyle dU=TdS-pdV}$,

we obtain the following expression, after dividing by ${\displaystyle dV}$ and fixing ${\displaystyle T}$ :

${\displaystyle \left({\frac {\partial U}{\partial V}}\right)_{T}=T\left({\frac {\partial S}{\partial V}}\right)_{T}-p=T\left({\frac {\partial p}{\partial T}}\right)_{V}-p}$.

The last equality comes from the following Maxwell relation:

${\displaystyle \left({\frac {\partial S}{\partial V}}\right)_{T}=\left({\frac {\partial p}{\partial T}}\right)_{V}}$.

From the definition of energy density it follows that

${\displaystyle U=uV}$

where the energy density of radiation only depends on the temperature, therefore

${\displaystyle \left({\frac {\partial U}{\partial V}}\right)_{T}=u\left({\frac {\partial V}{\partial V}}\right)_{T}=u}$.

Now, the equality

${\displaystyle \left({\frac {\partial U}{\partial V}}\right)_{T}=T\left({\frac {\partial p}{\partial T}}\right)_{V}-p}$,

after substitution of ${\displaystyle \left({\frac {\partial U}{\partial V}}\right)_{T}}$ and ${\displaystyle p}$ for the corresponding expressions, can be written as

${\displaystyle u={\frac {T}{3}}\left({\frac {\partial u}{\partial T}}\right)_{V}-{\frac {u}{3}}}$.

Since the partial derivative ${\displaystyle \left({\frac {\partial u}{\partial T}}\right)_{V}}$ can be expressed as a relationship between only ${\displaystyle u}$ and ${\displaystyle T}$ (if one isolates it on one side of the equality), the partial derivative can be replaced by the ordinary derivative. After separating the differentials the equality becomes

${\displaystyle {\frac {du}{4u}}={\frac {dT}{T}}}$,

which leads immediately to ${\displaystyle u=AT^{4}}$, with ${\displaystyle A}$ as some constant of integration.

### Derivation from Planck's law

The law can be derived by considering a small flat black body surface radiating out into a half-sphere. This derivation uses spherical coordinates, with φ as the zenith angle and θ as the azimuthal angle; and the small flat blackbody surface lies on the xy-plane, where φ = π/2.

The intensity of the light emitted from the blackbody surface is given by Planck's law :

${\displaystyle I(\nu ,T)={\frac {2h\nu ^{3}}{c^{2}}}{\frac {1}{e^{\frac {h\nu }{kT}}-1}}.}$
where
• ${\displaystyle I(\nu ,T)\,}$ is the amount of power per unit surface area per unit solid angle per unit frequency emitted at a frequency ${\displaystyle \nu \,}$ by a black body at temperature T.
• ${\displaystyle h\,}$ is Planck's constant
• ${\displaystyle c\,}$ is the speed of light, and
• ${\displaystyle k\,}$ is Boltzmann's constant.

The quantity ${\displaystyle I(\nu ,T)~A~d\nu ~d\Omega }$ is the power radiated by a surface of area A through a solid angle in the frequency range between ν and ν + .

The Stefan–Boltzmann law gives the power emitted per unit area of the emitting body,

${\displaystyle {\frac {P}{A}}=\int _{0}^{\infty }I(\nu ,T)d\nu \int d\Omega \,}$

To derive the Stefan–Boltzmann law, we must integrate Ω over the half-sphere and integrate ν from 0 to ∞. Furthermore, because black bodies are Lambertian (i.e. they obey Lambert's cosine law), the intensity observed along the sphere will be the actual intensity times the cosine of the zenith angle φ, and in spherical coordinates, = sin(φ) dφ dθ.

{\displaystyle {\begin{aligned}{\frac {P}{A}}&=\int _{0}^{\infty }I(\nu ,T)\,d\nu \int _{0}^{2\pi }\,d\theta \int _{0}^{\pi /2}\cos \phi \sin \phi \,d\phi \\&=\pi \int _{0}^{\infty }I(\nu ,T)\,d\nu \end{aligned}}}

Then we plug in for I:

${\displaystyle {\frac {P}{A}}={\frac {2\pi h}{c^{2}}}\int _{0}^{\infty }{\frac {\nu ^{3}}{e^{\frac {h\nu }{kT}}-1}}d\nu \,}$

To do this integral, do a substitution,

${\displaystyle u={\frac {h\nu }{kT}}\,}$

${\displaystyle du={\frac {h}{kT}}\,d\nu }$

which gives:

${\displaystyle {\frac {P}{A}}={\frac {2\pi h}{c^{2}}}\left({\frac {kT}{h}}\right)^{4}\int _{0}^{\infty }{\frac {u^{3}}{e^{u}-1}}\,du.}$

The integral on the right is standard and goes by many names: it is a particular case of a Bose-Einstein integral, or the Riemann zeta function, ${\displaystyle \zeta (4)}$, or the Polylogarithm. The value of the integral is ${\displaystyle {\frac {\pi ^{4}}{15}}}$, giving the result that, for a perfect blackbody surface:

${\displaystyle j^{\star }=\sigma T^{4}~,~~\sigma ={\frac {2\pi ^{5}k^{4}}{15c^{2}h^{3}}}={\frac {\pi ^{2}k^{4}}{60\hbar ^{3}c^{2}}}.}$

Finally, this proof started out only considering a small flat surface. However, any differentiable surface can be approximated by a bunch of small flat surfaces. So long as the geometry of the surface does not cause the blackbody to reabsorb its own radiation, the total energy radiated is just the sum of the energies radiated by each surface; and the total surface area is just the sum of the areas of each surface—so this law holds for all convex blackbodies, too, so long as the surface has the same temperature throughout. The law extends to radiation from non-convex bodies by using the fact that the convex hull of a black body radiates as though it were itself a black body.

## Notes

1. ^ Bohren, Craig F.; Huffman, Donald R. (1998). Absorption and scattering of light by small particles. Wiley. pp. 123–126. ISBN 0-471-29340-7.
2. ^ "Beyond Stefan-Boltzmann Law: Thermal Hyper-Conductivity." 26 September 2011.
3. ^ http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html
4. ^ "Luminosity of Stars". Australian Telescope Outreach and Education. Retrieved 2006-08-13.
5. ^
6. ^ Solar Radiation and the Earth's Energy Balance
7. ^ Das, P. K. (1996), "The Earth's Changing Climate" (PDF), Resonance, 1 (3): 54–65, doi:10.1007/bf02835622
8. ^ Cole, George H. A.; Woolfson, Michael M. (2002). Planetary Science: The Science of Planets Around Stars (1st ed.). Institute of Physics Publishing. pp. 36–37, 380–382. ISBN 0-7503-0815-X.
9. ^ "Introduction to Solar Radiation". Newport Corporation. Archived from the original on Oct 29, 2013.
10. ^ http://www.pha.jhu.edu/~kknizhni/StatMech/Derivation_of_Stefan_Boltzmann_Law.pdf