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

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

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 $\lambda$.