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.


1 Answer 1


Expanding on Ron's comment:

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

  • 5
    $\begingroup$ Why can't you just use $\nu \to \frac{c}{\lambda }$ to change variables? Neither $d \lambda$ or $d \nu$ appear in the equations of the original question, so why is it necessary to calculate $\frac{d \nu}{d \lambda }$ at all? $\endgroup$
    – iwantmyphd
    Commented Jan 30, 2021 at 21:13
  • $\begingroup$ @iwantmyphd because $I(\nu, T) d\nu = I(\lambda, T) d\lambda$ $\endgroup$
    – Cheng
    Commented Oct 14, 2021 at 4:55

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