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.


1 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 $\lambda$.

  • 1
    $\begingroup$ Never imagine you can solve this problem this way, thanks $\endgroup$ Commented Sep 10, 2013 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.