I know this might be a silly question, but is it necessary to know Planck's Law in order to show that $\lambda_{max}\propto\frac{1}{T}$? If you set \begin{equation} u(\lambda,T)=\frac{f(\lambda T)}{\lambda^5} \end{equation} then \begin{equation} \frac{\partial u}{\partial \lambda} = \frac{1}{\lambda^5}\frac{\partial f}{\partial \lambda}-\frac{5}{\lambda ^4}f=0 \end{equation}

\begin{equation} \frac{\frac{\partial f}{\partial \lambda}}{f} = 5\lambda \end{equation}

But I am stuck here because if I integrate the L.H.S. \begin{equation} \int \frac{\frac{\partial f}{\partial \lambda}}{f} d\lambda = \log (f(\lambda T)) = \frac{5}{2}\lambda^2 \end{equation}

  • $\begingroup$ The notation $\partial f/\partial \lambda$ does not make sense if $f=f(\lambda T)$ is only a function of the product. This should be notated as $f'(\lambda T)$, or (if you absolutely must have partial-derivative notation) as $\frac{\partial f}{\partial (\lambda T)}$. $\endgroup$ – Emilio Pisanty Feb 2 '18 at 11:01

You've made a mistake in your derivation.

\begin{equation} \frac{\partial u}{\partial \lambda} = \frac{\color{red} T}{\lambda^5}\frac{\partial f}{\partial \lambda}-\frac{5}{\lambda^{\color{red} 6}}f=0 \end{equation}

  • $\begingroup$ So $5\frac{f}{\frac{\partial f}{\partial \lambda}}$ must be a constant. However I don't see why this should be true for any $f$ $\endgroup$ – math4everyone Feb 3 '18 at 4:44

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