Using Wien's Law to show spectral distruibution function of one temperature represents all temperatures

Question

This is a exercise question from Quantum Mechanic textbook by Bransden:

Using Wien's Law to show that if the spectral distribution function of black body radiation, $\rho(\lambda,T)$ is known at one temperature then it can be obtained at any temperature (so that a single curve can be used to represent black body radiation at all temperatures.

Wien's law being: $\rho(\lambda, T)=\lambda^{-5}f(\lambda T)$

I understand this question as that temperature $T$ only acts as a scalar to the $\lambda$ variable, so that the differentiation of Wien's Law with respect to $T$ will give a constant multiple to the whole function, but in fact it is not clear that this is true by examining only on the Wien's Law.

Any help would be appreciated.

Answer 1

Suppose that, for a temperature $T_1$, you know $$ \rho(\lambda,T_1) = \lambda^{-5}f(\lambda T_1) $$ for every value of $\lambda$. Now, for a temperature $T_2$, let's introduce a variable $$\bar{\lambda} = \lambda T_2/T_1. $$ Then $$ \begin{align} \rho(\lambda,T_2) &= \lambda^{-5}f(\lambda T_2)\\ &= (T_2/T_1)^5 \,\bar{\lambda}^{-5}f(\bar{\lambda} T_1)\\ &= (T_2/T_1)^5\rho(\bar{\lambda},T_1)\\ &= (T_2/T_1)^5\rho(\lambda T_2/T_1, T_1) \end{align} $$ so that you also know $\rho(\lambda,T_2)$ for every value of $\lambda$.