2. Black-body Radiation
Consideranisolatedbodyinotherwiseemptyspace.Wedetecttwokindsoflightcomingfrom
ittooureyes.First,thereislightthatisreflectedfromthebody.Thislightmayhaveoriginatedfromthe
sun,reflectedfromthebodyandtravelledtooureyes.Thereisalsolightthatthebodyitselfproduces.
Inthismodelweareinterestedonlyinthelatter,self-radiatedlight.Thisself-radiatedlightisstrongly
dependentonthetemperatureofthebody.Thebestradiatorsarereferredtoasblackbodiesbecause
theyturnouttobegoodabsorbersaswell.Ablackbodyabsorbsanylightthathitsitandradiateslight
witharangeofwavelengthsandintensitiesthatdependsontemperature.
Wecantreatblackbodyradiationasaoneboxproblem,thebodybeingthebox,anditsheat
contentthestuff.Theheatcontentofthebodyisproportionaltoitstemperature𝑇.Inashorttime𝑑𝑡
thebodywillloseenergybyradiationgivenby−𝐹(𝑇) 𝑑𝑡.Weexpressthechangeinthetemperatureof
thebodyas
𝑇(𝑡! + 𝑑𝑡) − 𝑇(𝑡! ) = −𝐹(𝑇) 𝑑𝑡.
Thissaysthefluxofradiantenergyfromabodycausesthebodytocool.
Itturnsoutthattheintensityoftheradiationdependsonwavelengthaswellason
temperature.Wecanverifythisbynotingthatthecolorofanobjectchangesasitstemperature
changes.Thecolorofthefilamentinalightbulbisagoodexample.Theruleforfindingthefluxfor
wavelengthsintherange𝜆to𝜆 + 𝑑𝜆,andtemperature𝑇is
𝐼 𝜆 𝑑𝜆 =
2ℎ𝑐 !
𝜆!
1
!!
𝑒 !!"
𝑑𝜆
−1
where:
𝑐 = 3×10! m/s
ℎ = 6.626×10!!" Js
𝑘 = 1.38×10!!" J/K
𝑇 ≈ 300 K
𝑇 ≈ 6000K
𝜆
Speedoflight
Planck’squantumofaction
Boltzmann’sconstant
Terrestrialobjects
Surfaceofthesun
Wavelength
𝐼(𝜆)𝑑𝜆istheamountofenergyemittedfromasquaremeterofsurfaceinagivendirection(or,
tobeexact,perunitsolidangle),inonesecond,andinthewavelengthrange𝜆to𝜆 + 𝑑𝜆.Thisformula,
calledPlanck’slaw,isoneofthegreatachievementsofhumanunderstanding.ItwasdiscoveredbyMax
PlanckinOctober1900.Wewillrefertoitineachofourtoyclimatemodels.
10
8
6
I( ) (Wm
2
µm
1
Sr
1
)
300K
275K
250K
225K
200K
4
2
0
0
10
20
30
(µm)
40
50
Figure2.1Planck’sequationasafunctionofwavelengthforarangeoftemperaturesthatmightbe
foundonEarth.
Planck’sequationisshowninFigure2.1forseveraldifferenttemperaturesthatmightbe
observedonEarth.Theheightofanyonecurvetellsyouhowbrighttheself-radiationisateach
wavelengthforablackbodyatthattemperature,andtheareaunderthecurvetellsyouthetotalenergy
emitted.Noticethatthe225Kcurveliescompletelyabovethe200Kcurve,andthe250Kcurvelies
completelyabovethe225Kcurve,andsoon.Thismeansthatwarmerblackbodiesemitmoretotal
energy,andmoreenergyforanygivenrangeofwavelengths.Finally,notethatthecurvesarevery
stronglydependentontemperature.InFigure2.1youcanseethatanincreaseintemperaturefrom
200Kto300K,afactorof1.5,resultsinanincreaseinthetotalenergyemittedbytheblackbody(i.e.the
areaunderthecurve)byroughlyafactorof5.Itturnsoutthatthetotalenergyemittedbyablackbody
isproportionaltothefourthpoweroftemperature:
𝐹 𝑇 = 𝜋∫ 𝐼 𝜆 𝑑𝜆 = 𝜎𝑇 ! where𝜎 = 5.67×10!! Wm-2K-4.Thefactorof𝜋comesinasaresultofintegratingoveralldirections
overwhichradiationisemitted.ThisisknownastheStefan-Boltzmannlaw,and𝜎isknownasthe
Stefan-Boltzmannconstant.
A linear approximation
Terrestrialtemperaturesarenottoodifferentfromthefreezingpointofwater.For
temperaturesnearthefreezingpointwecanapproximatetheStefan-Boltzmannlawas
𝐹 𝑇 ≈ 𝐴 + 𝐵𝑇
[𝑇]=°C,𝐴 = 320Wm-2,𝐵 = 4.6Wm-2°C-1
𝐴istheenergyradiatedbyasurfaceatthefreezingpoint.𝐵istheadditionalfluxforeach
degreewarmerthanthefreezingpoint.Forexample,ablackbodyat10oCradiates320 + 4.6×10 =
366Wm-2.Wewillencountertheconstant𝐵frequently.Itsetstherateatwhichthefluxofemitted
radiationincreaseswithincreasingtemperature.Roughly,youneedtoincreasethesolarheatfluxby4.6
Wm-2toraisetheEarth’stemperatureonedegree.
Wien’s Law
InFigure2.1,eachblackbodyradiationcurvehasamaximumatsome𝜆!"# .Thepeakinthe
curveshiftstolowerwavelengths(i.e.higherenergyphotons)astemperaturebecomeswarmer.Wien’s
law,whichwasknownbeforePlanckfoundtheblackbodyfunction,isthattheproduct
𝑇 𝜆!"# = constant = 0.003.
ItisusedtodeterminethebrightestwavelengthforabodyoftemperatureT
𝜆!"# = 0.003/𝑇.
Forthesun,surfacetemperaturesareabout6000K.Therefore,thepeakintheradiationcurveis
atabout𝜆!"# = 5×10!! m.Thisisinthevisiblepartofthespectrum.Itmakessense.Wecanseethe
sun.Forterrestrialtemperatures,say300K,thepeakinthespectrumisatabout𝜆!"# = 10!! minthe
infraredpartofthespectrum.Thisisoutsidetheportionofthespectrumthatoureyesaresensitiveto.
WhenIlookatthedaffodilsinourgardenIamseeingsunlightreflectedfromtheblossoms.Iwould
needsomenightvisiongogglesto“see”thethermalemissionfromtheflowers.
Solartemperaturesare20timesterrestrialvalues.Wien’slawtellsusthatthewavelengthsof
theself-radiationfromsolarsourceswillbe1/20timestheradiationfromterrestrialsources.Canwe
makeasimilarstatementaboutthetotalenergy?TheresultfollowsfromtheStefan-Boltzmannlaw.If
thetemperaturesdifferbyafactorof𝑘 thentheradiatedenergiesdifferbyafactorof𝑘 ! .Thuswe
arriveattheconclusionthatasquaremeterofsolarsurfaceradiates160,000timesasmuchenergy
radiatedbyasquaremeterofterrestrialsurface.Forexample,asquaremeterofEarth’ssurfaceatthe
freezingpointradiatesabout320W/m2whileasquaremeterofsolarsurfaceat6000Kreleases
7.3×10! W/m2.
Exercises
2.1Plottheblackbodycurveforafixedtemperature,say𝑇 = 6000K,whichisappropriateforthe
surfaceofthesun.Thinkingahead,youwillwanttodrawthecurveforany𝑇.So𝑇isaparameterwe
willvary,butℎ,𝑘,and𝑐reallyarefixed.Wien’slawsaysthatthemaximumintensityisatawavelength
of𝜆!"# = 3000/𝑇 𝜇m.Soitseemslikelythattaking𝜆intherange𝜆!"# /5to5𝜆!"# willspanthemost
activepartofthespectrum.Isuggesttaking𝑑𝜆 = 𝜆!"# /5.
2.2Whenplottedasafunctionofwavelength,thePlanckfunctionindicateshowmuchenergyis
radiatedatanygivenrangeofwavelengths.Tobeprecise,theareaunderthecurvefrom𝜆! to𝜆! equals
theradiatedenergyatwavelengthsbetween𝜆! and𝜆! .Foranarrowrangeofwavelengths,𝑑𝜆,thearea
issimplytheareaofarectangleofheight𝐼(𝜆)andwidth𝑑𝜆.Forthetotalenergywewouldhavetoadd
uptheareasofmanysuchrectangles.
totalenergy= 𝐼(𝜆! ) (𝜆! − 𝜆! ) + 𝐼(𝜆! ) (𝜆! − 𝜆! ) + ⋯
Inthespreadsheetcalculationof𝐼(𝜆),the𝜆! areevenlyspaced,sothetotalenergyradiatedis
totalenergy= (𝐼(𝜆! ) + … + 𝐼(𝜆! )) 𝑑𝜆.
Modifyyourspreadsheettoincludethiscalculation.Makeadiagramthatillustrateswhatismeantby
theareaunderthecurvefrom𝜆! to𝜆! .
2.3Theblackbodyradiationcurvehasamaximumatsome𝜆!"# .Byinspectingtheradiationcurvesfor
differenttemperatures,makeatableof𝜆!"# (𝑇).Foreachtemperature,calculatetheproduct
𝑇𝜆!"# (𝑇).DoesWien’slaw,
𝑇 𝜆!"# = constant = 0.003,
holdtrue?
2.4TheconstantσintheStefan-Boltzmannlawcanbeshowntoequal
𝜎=
!!! ! !
!"! ! ! !
.
Don’taskmehow.Wecangainsomeinsightbynotingthat𝜎 musthavedimensionsWm-2K-4.
Boltzmann’sconstant𝑘hasdimensionsJK-1andistheonlysymbolaroundinvolvingK.Therefore𝜎
mustbeproportionalto𝑘 ! .ThisintroducesJ4ontherighthandside.Itwilltake3factorsofℎinthe
denominatortobalancetheenergyunits,andtwofactorsof𝑐toputthingsright.Writeoutthe
dimensionsofeachquantityandverifydimensionaluniformityfortheequation.
2.5Let𝑢 = 𝜆𝑘𝑇/ℎ𝑐,or𝑇 = ℎ𝑐𝑢/𝜆𝑘.ThenthePlanckformulais
𝑐! 𝐼 𝜆 =
1
𝑢 ! exp
1
−1
𝑢
ℎ! 𝑐 !
𝑐! =
2𝜋𝑘 ! 𝑇 !
Showthat𝑢isdimensionless.Therighthandsideisafunctiononlyof𝑢.Useaspreadsheettocalculate
therighthandsidefor0 < 𝑢 < 1,andverifythatithasamaximumat𝑢 = 0.2014andarea6.3.Show
thatthefirstresultisequivalenttoWien’slawandthesecondresultisequivalenttotheStefanBoltzmannlaw.
Plotthefunction
𝑐! 𝐼(𝑢) =
1
𝑢 ! exp
1
−1
𝑢
Therighthandsidedependsonlyon𝑢,andvaluesof𝑢aredisplayedonthehorizontalaxis.Ifweare
interestedinaparticular𝑇,wecanrelabelthehorizontalaxisintermsofwavelength
𝜆 = ℎ𝑐𝑢/𝑘𝑇.
Fromthispointofview,theshapeoftheblackbodycurveisalwaysthesame.Fordifferent𝑇wemust
labelthehorizontalaxisdifferently.
AnexcelspreadsheettogeneratePlanck’sdistributionforanytemperatureisgivenas
excel2_blackbody_radiation.Afteryouhaveyourspreadsheetworking,minormodificationswillallow
youtocompletetheexercises.
Hints/Solutions
2.1Seecm2ex2.1Pickafixedtemperature𝑇andthenplot𝐼versus𝜆.Because𝐼issuchastrong
functionoftemperatureitishardtoplot𝐼(𝜆, 𝑇 = 300 K)and𝐼(𝜆, 𝑇 = 6000 K)onthesameaxes.Use
themastermodeltoplottheradiationcurvesfor𝑇 = 6000 Kandfor300K.Modifythemasterprogram
togetthetotalradiationsummedoverwavelength,asafunctionof𝑇.
T=6000K
3.50E-01
3.00E-01
2.50E-01
2.00E-01
1.50E-01
1.00E-01
5.00E-02
0.00E+00
0.00E+00 5.00E-07 1.00E-06 1.50E-06 2.00E-06 2.50E-06 3.00E-06
Series2
T=300K
3.50E-01
3.00E-01
2.50E-01
2.00E-01
1.50E-01
1.00E-01
5.00E-02
0.00E+00
0.00E+00 1.00E-05 2.00E-05 3.00E-05 4.00E-05 5.00E-05 6.00E-05
Series1
Comparingthesetwocurves,onerepresentingtheradiationfromthesunandtheotherrepresenting
theradiationfromtheEarth,weseethatthecurveshavethesameshape.Theonlydifferenceisinhow
theaxesarelabelled.
Tosketchthesolarandterrestrialspectraonthesameaxes,firstdrawthecurveforthesun.It
willhaveitsmaximumat𝜆,say,andanareaof𝐴,say.Nowit’seasytosketchtheterrestrialspectrum.It
hasexactlythesameshapeasthesolarspectrumbutitsmaximumwillbeat20𝜆anditsareawillbe
𝐴/160000.Ifyoustartedwithahandsomecurveforthesolarspectrum,youmustplacetheterrestrial
spectrumfartotherightandwithanamplitudethatistoosmalltobeseenonyourgraph.Butdon’t
thinktheterrestrialradiationisunimportant.
2.2
Thedarkshadedareaequalsthe 1.20E+14
energyforthatrangeofwavelengths.
1.00E+14
8.00E+13
6.00E+13
4.00E+13
2.00E+13
0.00E+00 2.3
Wien'slaw
0.000035
0.00003
0.000025
0.00002
0.000015
0.00001
0.000005
0
0
2000
4000
6000
8000
10000
Figure2.2Plotof𝜆!"# (verticalaxis)asafunctionof𝑇(horizontalaxis).
12000
2.4Use[]todenotedimensionsofasin[k]=J/K,[h]=Js,[c]=L/s,and[T]=K.
Then[k4h-3c-2T4]=J4-3K-4+4s-3+2L-2=Js-1L-2=W/m2.
2.5Theshapeoftheblackbodycurveisthesameforalltemperatures.
1/[u^5(e^1/u-1)]u=lambdakT/hc
25
20
15
10
5
0
0
0.2
0.4
0.6
0.8
1
1.2
0
0.2
0.4
0.6
0.8
1
1.2
00.511.52T=6000K
010203040T=300K
Figure2.3(top)Theblack-bodyradiationcurve,plottedasafunctionofthedimensionless𝑢.Forfixed𝑇
thehorizontalaxisisproportionaltowavelengthheregivenin𝜇m.(bottom)Thefigureillustratesthat
theshapeoftheblack-bodycurveisthesameforalltemperatures.Ifyouplottheblack-bodyspectrum
fortwotemperaturestheonlythingthatwillbedifferentbetweenthetwoplotsistheaxislabels(see
horizontalaxis).Notethatthepeakradiationoccursat10𝜇mforterrestrialtemperaturesandatabout
0.5𝜇mforsolartemperatures.
.