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The Planck distribution law for blackbody radiation is stated in terms of frequency as:

$$ d\rho (\nu, T) = \rho_{\nu}(T)d\nu = \dfrac{8\pi h}{c^3}\dfrac{\nu ^3}{e^{\left(\dfrac{h\nu}{k_B T}\right)}-1}d\nu $$

Using the substitutions $\lambda \nu = c$ and $d\nu = \dfrac{-c}{\lambda^2}d\lambda$, the law has also been stated in terms of wavelength:

$$ d\rho (\lambda, T) = \rho_{\lambda}(T)d\lambda = \dfrac{8\pi hc}{\lambda^5}\dfrac{1}{e^{\left(\dfrac{hc}{\lambda k_B T}\right)}-1}d\lambda $$

I tried the substitutions, and I don't understand what to do with the extra negative sign? How did it just disappear like that?

$$ \begin{align} & \dfrac{8\pi h}{c^3}\dfrac{\nu ^3}{e^{\left(\dfrac{h\nu}{k_B T}\right)}-1}d\nu \\ =&\dfrac{8\pi h}{c^3}\dfrac{\left(\dfrac{c}{\lambda}\right) ^3}{e^{\left(\dfrac{h\left(\dfrac{c}{\lambda}\right)}{k_B T}\right)}-1}\left( \dfrac{-c}{\lambda ^2} d\lambda \right) \\ =& -\dfrac{8\pi hc}{\lambda^5}\dfrac{1}{e^{\left(\dfrac{hc}{\lambda k_B T}\right)}-1}d\lambda \end{align} $$

Reference

  1. Physical Chemistry: A Molecular Approach
    McQuarrie, D. A., Simon, J. D.
    Viva Books Private Limited (2011 edition, 2015 reprint)
ananta
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The integral operator$$\int_0^{+\infty} dk \rightarrow -\int_{+\infty}^0 \frac{2 \pi}{\lambda^2} d\lambda=\int_{0}^{\infty} \frac{2 \pi}{\lambda^2} d\lambda$$

J.G.
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John
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