1.3 Blackbody Spectra and Photon Fluxes
Their technical relevance for the quantitative analysis of incandescent light sources makes it worthwhile to take a closer look at blackbody spectra. Blackbody spectra are also helpful to elucidate the notion of spectra more closely, and to explain that a maximum in a spectrum strongly depends on the choice of independent variable (e.g. wavelength or frequency) and dependent variable (e.g. energy flux or photon flux). In particular, it is sometimes claimed that our sun has maximal radiation output at a wavelengthλmax 500 nm. This statement is actually very misleading if the notion of “radiation output” is not clearly defined, and if no explanation is included that different perfectly suitable notions of radiation output yield very different wavelengths or frequencies of maximal emission. We will see below that the statement above only applies tomaximal power output per unit of wavelength, i.e. if we use a monochromator which slices thewavelengthaxis into intervals of equal length dλ = c|df|/f2, then we find maximal power output in an interval aroundλmax500 nm. However, we will also see that if we use a monochromator which slices the frequency axis into intervals of equal length df = c|dλ|/λ2, then we find maximal power output in an interval around fmax 340 THz, corresponding to a wavelengthc/fmax 880 nm. If we ask for maximal photon counts instead of maximal power output, we find yet other values for peaks in the spectra.
Since Planck’s radiation law (1.25) yields perfect matches to blackbody spectra, it must also imply Stefan’s law and Wien’s law. Stefan’s law is readily derived in the following way. The emitted power per area is
e(T )= ∞
0
df e(f, T )= ∞
0
dλ e(λ, T )=2πkB4T4 h3c2
∞
0
dx x3 exp(x)−1. Evaluation of the integral
∞
0
dx x3 exp(x)−1 =
∞
0
dx x3
∞
n=0
exp[−(n+1)x]
= −
∞
n=1
d3 dn3
∞
0
dx exp(−nx)= −
∞
n=1
d3 dn3
1 n
= ∞
n=1
6
n4 =6ζ (4)= π4
15 (1.26)
implies
e(T )= 2π5k4B
15h3c2T4, (1.27)
i.e. Planck’s law implied a prediction for the Stefan–Boltzmann constant in terms of the Planck constanth, which could be determined previously from a fit to the spectra,
σ = 2π5kB4
15h3c2. (1.28)
An energy fluxe(T ) = 6.33×107W/m2from the Sun yields a remnant energy flux at Earth’s orbit of magnitudee(T )×(R/r⊕)2 = 1.37 kW/m2. HereR = 6.955×108m is the radius of the Sun andr⊕ =1.496×1011m is the radius of Earth’s orbit.
For the derivation of Wien’s law, we set x= hc
λkBT = hf
kBT. (1.29)
Then we have withe(λ, T )=e(f, T )|f=c/λc/λ2,
∂
∂λe(λ, T )= 2π hc2 λ5
1 exp
hc λkBT
−1
⎛
⎝ hc λ2kBT
exp hc
λkBT
exp
hc λkBT
−1
−5 λ
⎞
⎠
= 2π hc2 λ6
1 exp(x)−1
x exp(x) exp(x)−1 −5
, (1.30)
which implies that∂e(λ, T )/∂λ=0 is satisfied if and only if exp(x)= 5
5−x. (1.31)
This condition yieldsx 4.965. The wavelength of maximal spectral emittance e(λ, T )therefore satisfies
λmaxãT hc
4.965kB =2898μmãK. (1.32)
For a heat source of temperatureT =5780 K, like the surface of our sun, this yields λmax=501 nm, c
λmax =598 THz, (1.33)
see Fig.1.3.
One can also derive an analogue of Wien’s law for the frequencyfmaxof maximal spectral emittancee(f, T ). We have
1.3 Blackbody Spectra and Photon Fluxes 11
Fig. 1.3 The spectral emittancee(λ, T )for a heat source of temperatureT =5780 K
∂
∂fe(f, T )= 2π hf2 c2
1 exp
hf kBT
−1
⎛
⎝3− hf kBT
exp hc
λkBT
exp
hc λkBT
−1
⎞
⎠
= 2π hf2 c2
1 exp(x)−1
3−x exp(x) exp(x)−1
, (1.34)
which implies that∂e(f, T )/∂f =0 is satisfied if and only if exp(x)= 3
3−x, (1.35)
with solutionx 2.821. The frequency of maximal spectral emittance e(f, T ) therefore satisfies
fmax
T 2.821kB
h =58.79GHz
K . (1.36)
This yields for a heat source of temperatureT =5780 K, as in Fig.1.1, fmax=340 THz, c
fmax =882 nm. (1.37)
The photon fluxes in the wavelength scale and in the frequency scale,j (λ, T )and j (f, T ), are defined below. The spectral emittance per unit of frequency,e(f, T ),
is directly related to the photon flux per fractional wavelength or frequency interval dlnf =df/f = −dlnλ= −dλ/λ. We have with the notations used in (1.9) for spectral densities and integrated fluxes the relations
e(f, T )=hfj (f, T )=hf ∂
∂fj[0,f](T )=h ∂
∂ln(f/f0)j[0,f](T )
=hj (ln(f/f0), T )=hλj (λ, T )=hj (ln(λ/λ0), T ). (1.38) Optimization of the energy flux of a light source for given frequency bandwidthdf is therefore equivalent to optimization of photon flux for fixed fractional bandwidth df/f = |dλ/λ|.
The number of photons per area, per second, and per unit of wavelength emitted from a heat source of temperatureT is
j (λ, T )= λ
hce(λ, T )=2π c λ4
1 exp
hc λkBT
−1
. (1.39)
This satisfies
∂
∂λj (λ, T )=j (λ, T ) λ
x exp(x) exp(x)−1−4
=0 (1.40)
if
exp(x)= 4
4−x. (1.41)
This has the solutionx 3.921. The wavelength of maximal spectral photon flux j (λ, T )therefore satisfies
λmaxãT hc
3.921kB =3670μmãK. (1.42)
This yields for a heat source of temperatureT =5780 K λmax=635 nm, c
λmax =472 THz, (1.43)
see Fig.1.4.
The photon flux in the wavelength scale,j (λ, T ), is also related to the energy fluxes per fractional wavelength or frequency intervaldlnλ=dλ/λ= −dlnf =
−df/f, j (λ, T )= λ
hce(λ, T )= 1
hce(ln(λ/λ0), T )= f
hce(f, T )= 1
hce(ln(f/f0), T ).
1.3 Blackbody Spectra and Photon Fluxes 13
Fig. 1.4 The spectral photon fluxj (λ, T )for a heat source of temperatureT =5780 K
Therefore optimization of photon flux for fixed wavelength bandwidth dλ is equivalent to optimization of energy flux for fixed fractional bandwidthdλ/λ =
|df/f|.
Finally, the number of photons per area, per second, and per unit of frequency emitted from a heat source of temperatureT is
j (f, T )= e(f, T )
hf =2πf2 c2
1 exp
hf kBT
−1
. (1.44)
This satisfies
∂
∂fj (f, T )=j (f, T ) f
2−x exp(x) exp(x)−1
=0 (1.45)
if
exp(x)= 2
2−x. (1.46)
This condition is solved byx 1.594. Therefore the frequency of maximal spectral photon fluxj (f, T )in the frequency scale satisfies
fmax
T 1.594kB
h =33.21GHz
K . (1.47)
Fig. 1.5 The spectral photon fluxj (f, T )for a heat source of temperatureT =5780 K
This yields for a heat source of temperatureT =5780 K fmax=192 THz, c
fmax =1.56μm, (1.48)
see Fig.1.5.
The flux of emitted photons is j (T )=
∞
0
df j (f, T )=2πkB3T3 h3c2
∞
0
dx x2
exp(x)−1. (1.49) Evaluation of the integral
∞
0
dx x2 exp(x)−1 =
∞
0
dx x2
∞
n=0
exp[−(n+1)x]
= ∞
n=1
d2 dn2
∞
0
dx exp(−nx)= ∞
n=1
d2 dn2
1 n
=
∞
n=1
2
n3 =2ζ (3) (1.50)
1.5 Wave-Particle Duality 15 yields
j (T )= 4π ζ (3)kB3
h3c2 T3=1.5205×1015 T3
m2ãsãK3. (1.51) A surface temperatureT = 5780 K for our sun yields a photon flux at the solar surface 2.94×1026m−2s−1and a resulting photon flux at Earth’s orbit of 6.35× 1021m−2s−1. The average photon energye(T )/j (T )=1.35 eV is in the infrared.