Page:Über die Energievertheilung im Emissionspectrum eines schwarzen Körpers.pdf/7

 Furthermore $$ \frac{d^2\varphi}{d\lambda^2} = 0$$ for the roots of the equation

$30\lambda^2\vartheta^2 - 12c\lambda\vartheta + c^2 = 0$,

therefore for

$\lambda = \lambda_m \left(1 \pm \sqrt{\tfrac16} \right)$.

For these two points, the curve has inflection points. We will set $$\lambda = \lambda_m(1 + \varepsilon)$$, so that

$ \varphi_\lambda = \frac{C e^{-\frac c{\lambda_m(1 + \varepsilon)\vartheta} } }{\lambda_m^5 (1 + \varepsilon)^5 } = \frac{C e^{-\frac{5}{1 + \varepsilon} }}{\lambda_m^5 (1 + \varepsilon)^5} $,|undefined

therefore

$ \log \frac \varphi{\varphi_m} = - 5\left(\log(1+ \varepsilon) - \frac\varepsilon{1 + \varepsilon} \right) = -5\left(\tfrac12\varepsilon^2 - \tfrac23\varepsilon^3 + \tfrac34\varepsilon^4 \ldots\right) $

We will set $$-\varepsilon$$ for $$\varepsilon$$, so that

$ \log \frac\varphi{\varphi_m} = -5 \left(\tfrac12\varepsilon^2 + \tfrac23 \varepsilon^3 + \tfrac34 \varepsilon^4 \ldots\right) $

Here the absolute value of the series is larger, i.e $$\frac\phi{\phi_m} $$ smaller than with positive $$\varepsilon$$. So long as $$\varepsilon < 1 $$ the ordinates at the same distance from the maximum are smaller on the short wavelength side.

In an earlier work I showed that the energy curves of black bodies at different temperatures cannot intersect. From this it could be further deduced that the curve must drop more slowly on the side of the long waves than the curve

$\frac \mathrm{const.}{\lambda^5}.$

Now this is actually the case with our curve; $$\frac{d\varphi_\lambda}{d\lambda}$$ is always smaller than the absolute value $$\frac{5C}{\lambda^6}$$ and only reaches this limit for $$\vartheta = \infin$$. For infinitely increasing temperature would mean $$\varphi_\lambda = \frac{C}{\lambda^5} $$ and the maximum energy will approach zero wavelength indefinitely.