# Planck's Law in terms of wavelength

I am drawing a blank when it comes to equation transformation. Wikipedia gives two equations for the spectral radiance of black body:

• First as a function of frequency $\nu$: $$I(\nu, T) = \frac{2 h \nu^3}{c^2}\cdot\frac{1}{e^\frac{h \nu}{k T} - 1}$$
• Then as a function of wavelength $\lambda$ : $$I'(\lambda, T) = \frac{2hc^2}{\lambda^5}\cdot\frac{1}{e^\frac{h c}{\lambda k T}-1}$$

And I don't see how they get $\lambda^5$ term. I'm assuming that the transformation is just $\nu \rightarrow c/\lambda$, but that gives $$\frac{2 h \nu^3}{c^2} \Rightarrow \frac{2 hc}{\lambda^3} \neq \frac{2hc^2}{\lambda^5}$$

Similar transformation happens at other parts in the article also. I'm obviously missing something, likely completely trivial.

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Sometimes wikipedia is wrong .. although in this case $I$ may not represent the same thing as $I'$. I wish I knew more about black body radiation. – ja72 Aug 16 '11 at 1:53
Wikipedia is not wrong, the spectrum is a density and contains a differential factor $d\lambda$ which you need to transform to get the correct density. – Ron Maimon Aug 16 '11 at 2:34
@Ron: you might want to post that as an answer :-P – David Z Aug 16 '11 at 2:43

## 1 Answer

Expanding on Ron's comment:

$$I(\nu ,T)d\nu =\frac{2h\nu ^3}{c^2}\frac{d\nu }{e^{\frac{h\nu }{kT}}-1}$$ $$\nu \to \frac{c}{\lambda },\quad d\nu \to c\frac{d\lambda }{\lambda ^2}$$ $$I(\lambda ,T)d\lambda =\frac{2h}{c^2}\left(\frac{c}{\lambda }\right)^3\frac{1}{e^{\frac{hc}{\lambda kT}}-1}c\frac{d\lambda }{\lambda ^2}=\frac{2hc^2}{\lambda ^5}\frac{d\lambda }{e^{\frac{hc}{\lambda kT}}-1}$$

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