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stellar radius and t is the effective temperature this formula can then be rearranged to calculate the temperature t l 4 π r 2 σ 4 displaystyle t sqrt 4 frac l 4 pi r 2 sigma or alternatively the radius r l 4 π σ t 4 displaystyle r sqrt frac l 4 pi sigma t 4 the same formulae can also be simplified to compute the parameters relative to the sun l l r r 2 t t 4 t t l l 1 4 r r 1 2 r r t t 2 l l 1 2 displaystyle begin aligned frac l l_ odot left frac r r_ odot right 2 left frac t t_ odot right 4 1ex frac t t_ odot left frac l l_ odot right 1 4 left frac r_ odot r right 1 2 1ex frac r r_ odot left frac t_ odot t right 2 left frac l l_ odot right 1 2 end aligned where r displaystyle r_ odot is the solar radius and so forth they can also be rewritten in terms of the surface area a and radiant exitance m displaystyle m circ l a m m l a a l m displaystyle begin aligned l am circ 1ex m circ frac l a 1ex a frac l m circ end aligned where a 4 π r 2 displaystyle a 4 pi r 2 and m σ t 4 displaystyle m circ sigma t 4 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 edit 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 earth s own production of energy being small enough to be negligible the luminosity of the sun l is given by l 4 π r 2 σ t 4 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 a 0 the distance between the earth and the sun and the irradiance received power per unit area is given by e l 4 π a 0 2 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 π r 2 displaystyle pi r_ oplus 2 the radiant flux i e solar power absorbed by the earth is thus given by φ abs π r 2 e displaystyle phi _ text abs pi r_ oplus 2 times e_ oplus because the stefan boltzmann law uses a fourth power it has a stabilizing effect on the exchange and the flux emitted by earth tends to be equal to the flux absorbed close to the steady state where 4 π r 2 σ t 4 π r 2 e π r 2 4 π r 2 σ t 4 4 π a 0 2 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 t 4 r 2 t 4 4 a 0 2 t t r 2 a 0 5780 k 6 957 10 8 m 2 1 495 978 707 10 11 m 279 k 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 6 957 times 10 8 rm m over 2 times 1 495 978 707 times 10 11 rm m approx 279 rm k end aligned where t is the temperature of the sun r the radius of the sun and a 0 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 7 1 4 giving 255 k 18 c 1 f 28 29 the above temperature is earth s as seen from space not ground temperature but an average over all emitting bodies of earth from surface to high altitude because of the greenhouse effect the earth s actual average surface temperature is about 288 k 15 c 59 f which is higher than the 255 k 18 c 1 f effective temperature and even higher than the 279 k 6 c 43 f 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 m 2 30 the stefan boltzmann law then gives a temperature of t 1120 w m 2 σ 1 4 375 k displaystyle t left frac 1120 text w m 2 sigma right 1 4 approx 375 text k or 102 c 216 f above the atmosphere the result is even higher 394 k 121 c 250 f 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 edit thermodynamic derivation of the energy density edit the fact that the energy density of the box containing radiation is proportional to t 4 displaystyle t 4 can be derived using thermodynamics 31 15 this derivation uses the relation between the radiation pressure p and the internal energy density u displaystyle u a relation that can be shown using the form of the electromagnetic stress energy tensor this relation is p u 3 displaystyle p frac u 3 now from the fundamental thermodynamic relation d u t d s p d v displaystyle du t ds p dv we obtain the following expression after dividing by d v displaystyle dv and fixing t displaystyle t u v t t s v t p t p t v p 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 s v t p t v 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 u u v displaystyle u uv where the energy density of radiation only depends on the temperature therefore u v t u v v t u displaystyle left frac partial u partial v right _ t u left frac partial v partial v right _ t u now the equality is u t p t v p displaystyle u t left frac partial p partial t right _ v p after substitution of u v t displaystyle left frac partial u partial v right _ t meanwhile the pressure is the rate of momentum change per unit area since the momentum of a photon is the same as the energy divided by the speed of light u t 3 u t v u 3 displaystyle u frac t 3 left frac partial u partial t right _ v frac u 3 where the factor 1 3 comes from the projection of the momentum transfer onto the normal to the wall of the container since the partial derivative u t v displaystyle left frac partial u partial t right _ v can be expressed as a relationship between only u displaystyle u and t 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 d u 4 u d t t displaystyle frac du 4u frac dt t which leads immediately to u a t 4 displaystyle u at 4 with a displaystyle a as some constant of integration derivation from planck s law edit deriving the stefan boltzmann law using 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 θ as the zenith angle and φ as the azimuthal angle and the small flat blackbody surface lies on the xy plane where θ π 2 the intensity of the light emitted from the blackbody surface is given by planck s law i ν t 2 h ν 3 c 2 1 e h ν k t 1 displaystyle i nu t frac 2h nu 3 c 2 frac 1 e h nu kt 1 where i ν t displaystyle i nu t is the amount of power per unit surface area per unit solid angle per unit frequency emitted at a frequency ν displaystyle nu by a black body at temperature t h displaystyle h is the planck constant c displaystyle c is the speed of light and k displaystyle k is the boltzmann constant the quantity i ν t a cos θ d ν d ω displaystyle i nu t a cos theta d nu d omega is the power radiated by a surface of area a through a solid angle d ω in the frequency range between ν and ν dν the stefan boltzmann law gives the power emitted per unit area of the emitting body p a 0 i ν t d ν cos θ d ω displaystyle frac p a int _ 0 infty i nu t d nu int cos theta d omega note that the cosine appears because black bodies are lambertian i e they obey lambert s cosine law meaning that the intensity observed along the sphere will be the actual intensity times the cosine of the zenith angle to derive the stefan boltzmann law we must integrate d ω sin θ d θ d φ textstyle d omega sin theta d theta d varphi over the half sphere and integrate ν displaystyle nu from 0 to p a 0 i ν t d ν 0 2 π d φ 0 π 2 cos θ sin θ d θ π 0 i ν t d ν displaystyle begin aligned frac p a int _ 0 infty i nu t d nu int _ 0 2 pi d varphi int _ 0 pi 2 cos theta sin theta d theta pi int _ 0 infty i nu t d nu end aligned then we plug in for i p a 2 π h c 2 0 ν 3 e h ν k t 1 d ν 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 evaluate this integral do a substitution u h ν k t d u h k t d ν displaystyle begin aligned u frac h nu kt 6pt du frac h kt d nu end aligned which gives p a 2 π h c 2 k t h 4 0 u 3 e u 1 d u 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 the polylogarithm or the riemann zeta function ζ s displaystyle zeta s the value of the integral is γ 4 ζ 4 π 4 15 displaystyle gamma 4 zeta 4 frac pi 4 15 where γ s displaystyle gamma s is the gamma function giving the result that for a perfect blackbody surface m σ t 4 σ 2 π 5 k 4 15 c 2 h 3 π 2 k 4 60 ℏ 3 c 2 displaystyle m circ 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 collection 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 energy density edit the total energy density u can be similarly calculated except the integration is over the whole sphere and there is no cosine and the energy flux u c should be divided by the velocity c to give the energy density u u 1 c 0 i ν t d ν d ω displaystyle u frac 1 c int _ 0 infty i nu t d nu int d omega thus 0 π 2 cos θ sin θ d θ textstyle int _ 0 pi 2 cos theta sin theta d theta is replaced by 0 π sin θ d θ textstyle int _ 0 pi sin theta d theta giving an extra factor of 4 thus in total u 4 c σ t 4 displaystyle u frac 4 c sigma t 4 the product 4 c σ displaystyle frac 4 c sigma is sometimes known as the radiation constant or radiation density constant 32 33 decomposition in terms of photons edit the stefan boltzmann law can be expressed as 34 m σ t 4 n p h o t e p h o t displaystyle m circ sigma t 4 n_ mathrm phot langle e_ mathrm phot rangle where the flux of photons n p h o t displaystyle n_ mathrm phot is given by n p h o t π 0 b ν h ν d ν displaystyle n_ mathrm phot pi int _ 0 infty frac b_ nu h nu mathrm d nu n p h o t 1 5205 10 15 photons s 1 m 2 k 3 t 3 displaystyle n_ mathrm phot left 1 5205 times 10 15 textrm photons cdot textrm s 1 cdot textrm m 2 cdot mathrm k 3 right cdot t 3 and the average energy per photon e phot displaystyle langle e_ textrm phot rangle is given by e phot π 4 30 ζ 3 k t 3 7294 10 23 j k 1 t displaystyle langle e_ textrm phot rangle frac pi 4 30 zeta 3 k t left 3 7294 times 10 23 mathrm j cdot mathrm k 1 right cdot t marr and wilkin 2012 recommend that students be taught about e phot displaystyle langle e_ textrm phot rangle instead of being taught wien s displacement law and that the above decomposition be taught when the stefan boltzmann law is taught 34 see also edit black body radiation rayleigh jeans law sakuma hattori equation notes edit 2022 codata value stefan boltzmann constant the nist reference on constants units and uncertainty nist may 2024 retrieved 2024 05 18 1 2 thermal insulation heat transfer by radiation vocabulary iso_9288 2022 international organization for standardization 2022 retrieved 2023 06 17 1 2 3 4 siegel robert howell john r 1992 thermal radiation heat transfer 3 ed taylor francis isbn 0 89116 271 2 1 2 reif f 1965 fundamentals of statistical and thermal physics waveland press isbn 978 1 57766 612 7 bohren craig f huffman donald r 1998 absorption and scattering of light by small particles wiley pp 123 126 isbn 978 0 471 29340 8 narimanov evgenii e smolyaninov igor i 2012 beyond stefan boltzmann law thermal hyper conductivity conference on lasers and electro optics 2012 osa technical digest art qm2e 1 optical society of america arxiv 1109 5444 citeseerx 10 1 1 764 846 doi 10 1364 qels 2012 qm2e 1 isbn 978 1 55752 943 5 s2cid 36550833 cite conference cite uses deprecated parameter citeseerx help golyk v a krüger m kardar m 2012 heat radiation from long cylindrical objects phys rev e 85 4 046603 arxiv 1109 1769 bibcode 2012phrve 85d6603g doi 10 1103 physreve 85 046603 hdl 1721 1 71630 pmid 22680594 s2cid 27489038 radiant exitance electropedia the world s online electrotechnical vocabulary international electrotechnical commission retrieved 20 june 2023 goody r m yung y l 1989 atmospheric radiation theoretical basis oxford university press isbn 0 19 505134 3 grainger r g 2020 a primer on atmospheric radiative transfer chapter 3 radiometric basics pdf earth observation data group department of physics university of oxford retrieved 15 june 2023 radiation energy density hyperphysics retrieved 20 june 2023 tyndall john 1864 on luminous i e visible and obscure i e infrared radiation philosophical magazine 4th series 28 329 341 see p 333 in his physics textbook of 1875 adolph wüllner quoted tyndall s results and then added estimates of the temperature that corresponded to the platinum filament s color wüllner adolph 1875 lehrbuch der experimentalphysik textbook of experimental physics in german vol 3 leipzig germany b g teubner p 215 from wüllner 1875 p 215 wie aus gleich zu besprechenden versuchen von draper hervorgeht also fast um das 12fache zu as follows from the experiments of draper which will be discussed shortly a temperature of about 525 c corresponds to the weak red glow a temperature of about 1200 c to the full white glow thus while the t...