2.10: Derivation of Wien's and Stefan's Laws
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Wien's and Stefan's Laws are found, respectively, by differentiation and integration of Planck's equation. Neither of these is particularly easy, and they are not found in every textbook. Therefore, I derive them here.
Wien's Law
Planck's equation for the exitance per unit wavelength interval (equation 2.6.1) is
MC=1λ5(eK/λT−1),
in which I have omitted some subscripts. Differentiation gives
1CdMdλ=−1(eK/λT−1)2⋅[5λ4⋅(eK/λT−1)+λ5⋅(−Kλ2T)eK/λT].
M is greatest when this is zero; that is, when
x=5(1−e−x),
where x=KλT.
Hence, with equation 2.6.9, the wavelength at which M is a maximum, is given by
λ=hckxT.
The maximum value of M is found be substituting this vale of λ back into Planck's equation, to arrive at equation 2.7.16. The corresponding versions of Wien's Law appropriate to the other version's of Planck's equation are found similarly.
Stefan's Law
Integration of Planck's equation to arrive at Stefan's law is a bit more tricky.
It should be clear that ∫∞0Mλdλ=∫∞0Mνdν, and therefore I choose to integrate the easier of the functions, namely Mν. To integrate Mλ, the first thing we would do anyway would be to make the substitution ν=c/λ.
Planck's equation for the blackbody exitance per unit frequency interval is
Mν=C3∫∞0ν3dνeK2ν/T−1.
Let x=K2ν/T; then Mν=2πk4T4c2h3∫∞0x3dxex−1,
And, except for the numerical value of the integral, we already have Stefan's law. The integral can be evaluated numerically, but not without difficulty, and there is an analytical solution for it.
Consider the indefinite integral and integrate it by parts:
∫x3dxex−1=x3ln(1−e−x)−3∫x2ln(1−e−x)dx+const.
Now put the limits in:
∫∞0x3dxex−1=−3∫∞0x2ln(1−e−x)dx.
Write down the Maclaurin expansion of the integrand:
∫∞0x3dxex−1=3∫∞0x2(e−x+12e−2x+13e−3x+...)dx
and integrate term by term to obtain
∫∞0x3dxex−1=6(1+124+134+...).
We must now evaluate 1+124+134+...
The series ∞∑11nm is the Riemann ζ-function. For m=1, it diverges. For m=3,5,7, etc., it has to be evaluated numerically. For m=2,4,6, etc., the sums can be written explicitly in terms of π. For example:
ζ(2)=π26,
ζ(4)=π490,
ζ(6)=π6945.
One of the stages necessary in evaluating the ζ-function is to derive the infinite product
sinαπαπ=[1−α2][1−(12α)2][1−(13α)2]...
If we can do that, we are more than halfway there.
Let's start by considering the Fourier expansion of cosθx:
cosθx=∞∑0ancosnx
In Equation 2.11.10 n is an integer, θ not necessarily so; we shall suppose that θ is some number between 0 and 1. There is no need to consider any sine terms, because cosθx is an even function of x. We work out what the Fourier coefficients are in the usual way, to get
an=(−1)n2θsinθπθ2−n2,n=1,2,3,...
As usual, and for the usual reason, a0 is an exception:
a0=sinθπθπ.
We have therefore arrived at the Fourier expansion of cosθx:
cosθx=2θsinθππ(12θ2−cosxθ2−12+cos2xθ2−22−cos3xθ2−32+...).
Put x=π and rearrange slightly:
πcotθπ−1θ=2θ(1θ2−12+1θ2−22+...).
Since we are assuming that θ is some number between 0 and 1, we shall re-write this so that the denominators are all positive:
πcotθπ−1θ=−2θ12−θ2−2θ22−θ2−...
Now multiply both sides by dθ and integrate from θ=0 to θ=α. The integration must be done with care. The indefinite integral of the left hand side is lnsinθπ−lnθ+constant, i.e. ln(sinθπθ)+constant. The definite integral between 0 and α is ln(sinαπα)−limθ→ 0ln(sinθπθ).
The limit of the second term is lnπ, so the definite integral is ln(sinαπαπ). Integrating the right hand side is a bit easier, so we arrive at
ln(sinαπαπ)=ln(12−α212)+ln(22−α222)+...
On taking the antilogarithm, we arrive at the required infinite product:
sinαπαπ=[1−α2][1−(12α)2][1−(13α)2]...
Now expand this as a power series in α2:
sinαπαπ=1+()α2+()α4+()α6+...
The first one is easy, but subsequent ones rapidly get more difficult, but you do have to get at least as far as α4.
Now compare this expansion with the ordinary Maclaurin expansion:
sinαπαπ=1−π23!α2+π45!α4−...
and we arrive at the correct expressions for the Riemann ζ-functions. We then get for Stefan's law:
M=2π5k415h3c2T4=σT4,
where σ=5.6705×10−8 W m−2K−4.
Questions
Finally, now that you have struggled through Riemann’s zeta-function, let’s just make sure that you have understood the really simple stuff, so here are a couple of easy questions – and you won’t have to bother with zeta-functions.
1. By what factor should the temperature of a black body be increased so that
a) The integrated radiance (over all frequencies) is doubled?
b) The frequency at which its radiance is greatest is doubled?
c) The spectral radiance per unit wavelength interval at its wavelength of maximum spectral radiance is doubled?
2. A block of shiny silver (absorptance = 0.23) has a bubble inside it of radius 2.2cm, and it is held at a temperature of 1200K.
A block of dull black carbon (absorptance = 0.86) has a bubble inside it of radius 4.3cm, and it is held at a temperature of 2300K,
Calculate the ratio Integrated radiation energy density inside the carbon bubbleIntegrated radiation energy density insdie the silver bubble.
Answers. 1. a) 1.189 b) 2.000 c) 1.149
2. 13.5