Integrating the Blackbody Curve, Part 1

Brian M. Sutin
Posted 2014-05-17 00:00

When designing an infrared system, one major source of noise (and signal) that the detector sees is blackbody radiation from thermal sources. The distribution of blackbody radiation with respect to wavelength and temperature follows Planck's law,

B_{T} (λ) = \frac{2 h c^{2}}{λ^{5}} \frac{1}{e^{\frac{hc}{λ k_{B} T}} - 1} .

(1.1)

What is needed for thermal noise calculations is the integral of this over the wavelengths that are able to pass the filters and be absorbed by the detector. The normalized integral is

F_{T} (λ) = \frac{\int_{0}^{λ} B_{T} (λ) d λ}{\int_{0}^{\infty} B_{T} (λ) d λ} .

(1.2)

The normalization factor in the denominator is given by the Stefan-Boltzmann law as

\int_{0}^{\infty} B_{T} (λ) d λ = \frac{2 π^{5} k_{B}^{4} T^{4}}{15 h^{3} c^{2}} .

(1.3)

plot of integrated blackbody approximation errors

If we define a new variable v as

v = \frac{hc}{λ k_{B} T},

(1.4)

then all of the physical constants drop out, giving a purely mathematical formula which can be expanded as an infinite summation,

\begin{matrix} F (v) & = & \frac{15}{π^{4}} \int_{0}^{v} \frac{x^{3} dx}{e^{x} - 1} \\ = & \frac{15}{π^{4}} \int_{0}^{v} \frac{x^{3} e^{- x} dx}{1 - e^{- x}} \\ = & \frac{15}{π^{4}} \int_{0}^{v} x^{3} e^{- x} \sum_{m = 0}^{\infty} e^{- mx} dx \\ = & \frac{15}{π^{4}} \int_{0}^{v} x^{3} \sum_{m = 1}^{\infty} e^{- mx} dx \\ = & \frac{15}{π^{4}} \sum_{m = 1}^{\infty} \int_{0}^{v} x^{3} e^{- mx} dx \\ = & \frac{15}{π^{4}} \sum_{m = 1}^{\infty} e^{- mv} [\frac{v^{3}}{m^{}} + \frac{3 v^{2}}{m^{2}} + \frac{6 v}{m^{3}} + \frac{6}{m^{4}}] . \end{matrix}

(1.5)

What we would like to do is have an expression for F(v) that is easy to compute accurately. The problem is that each of the m^-n series in (1.5) is logarithmically convergent, and thus converges very slowly, especially for small v, or large λ. See Figure 1.

Figure 1: plot of series approximation to the blackbody distribution

Can this be improved?

If the series is terminated at some finite number of terms, then the m^-4 term adds the most error at small v. In the limit as v goes to zero F(v) must go to unity, so the first three terms in the summation drop out, giving

\frac{15}{π^{4}} \sum_{m = 0}^{\infty} \frac{6}{m^{4}} = 1 .

(1.6)

Adding both sides of this equation to the formula for F(v) gives

F (v) = 1 + \frac{15}{π^{4}} \sum_{m = 1}^{\infty} e^{- mv} [\frac{v^{3}}{m^{}} + \frac{3 v^{2}}{m^{2}} + \frac{6 v}{m^{3}} + 6 \frac{1 - e^{mv}}{m^{4}}] .

(1.7)

Now we have a similar formula to (1.5) that is accurate for small v. How well does this work? As shown in Figure 2, the terms are still converging on the left hand side, albeit very slowly.

Figure 2: plot of alternate series approximation to the blackbody distribution

If the e^mv term is expanded out as a Taylor series, then

\frac{6 e^{mv}}{m^{4}} \approx \frac{6}{m^{4}} + \frac{6 v}{m^{3}} + \frac{3 v^{2}}{m^{2}} + \frac{v^{3}}{m^{}} + \frac{v^{4}}{4^{}} + .. .

(1.8)

so all of the other terms cancel, leaving the v⁴ term as the dominant term, giving better convergence for small v. Oddly the published Taylor series expansion by Wiebelt (1966) for 1-F(v) goes as v³.

The story continues in Integrating the Blackbody Curve, Part 2.

Categories Optics, Physics

The shortest path between two points might be a corkscrew.

Integrating the Blackbody Curve, Part 1

Search