Page:Über die Energievertheilung im Emissionspectrum eines schwarzen Körpers.pdf/5

 where $$F$$ and $$f$$ are two unknown functions and $$\vartheta$$ is the absolute temperature.

Now the change of radiation with temperature, according to the theory given by Boltzmann and myself consists of an increase of the total energy in proportion to the fourth power of the absolute temperature and a change in wavelength for each quantum of energy enclosed between $$\lambda$$ and $$\lambda + d\lambda $$, in the sense that the associated wavelength changes in inverse proportion to the absolute temperature. If one then imagines the energy at a temperature to be plotted as a function of wavelength, this curve would remain unchanged, with changing temperature, if the scale of the graph were changed so that the ordinates were proportionally diminished and the abscissas proportionally increased. The latter is for our value of $$\varphi_\lambda $$ only possible if $$\lambda$$ and $$\vartheta$$ occur in the exponent and only if they do so as the product $$\lambda\vartheta$$

If $$c$$ is a constant, then

$ \frac{f(\lambda)}{\vartheta} = \frac c{\lambda\vartheta} $

The increase in the total energy determines the value of $$ F(\lambda) $$. Because it has to be

$ \int\limits_0^{\infin} F(\lambda)e^{-\frac c{\vartheta\lambda}} d\lambda = \mathrm{const.}\, \vartheta^4 $.


 * undefined

$$ F(\lambda) $$ can be determined using the method of undetermined coefficients.

We imagine $$ F(\lambda) $$ expanded as a series and set $$\lambda = \frac c{y\theta}$$, so that

$\begin{align} F(\lambda) = F\left(\frac c {y\vartheta}\right) = a_0 & + a_{+1} \frac{\vartheta y}c + a_{+2} \frac{\vartheta^2 y^2}{c^2} + \ldots + a_{n}\frac{\vartheta^n y^n}{c^n} + \ldots \\ & +a_{-1}\frac c{\vartheta y} + a_{-2} \frac{c^2}{\vartheta^2 y^2} + \ldots + a_{-n} \frac{\vartheta^{-n}y^{-n}}{c^{-n}}. \end{align}$
 * undefined