# How do you change Planck's law from frequency to wavelength? [duplicate]

I have to derive Wien's displacement law by using Planck's law. I've tried but I come to a unsolvable equation (well I can't solve it) anywhere I look online it comes to the same conclusion, you need to solve an equation involving transcendentals which I have no idea what they are nor any of the math classes required for this physics course teach it. Once that equation is solved, the next steps are quite simple, by using the solution.

Maybe my professor just wants us to google the solution to the equation and used it to keep solving the problem?

anyway, what really bothers me is this:

My book shows Planck's law in terms of frequency

$$B_\nu (T) = \frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/(k_B T)} - 1}$$

but to do the problem I need it in terms of wavelength

$$B_\lambda (T) = \frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/(\lambda k_B T)} - 1}$$

And $v = c/\lambda$, so by that then it can be said that the exponent of $\lambda$ would be 3 when it is expressed in terms of lambda (plus $c$ would have an exponent of 1). But Wikipedia says it is 5 and not 3. It multiplies by the derivative of $\nu$, but I don't get why. I'm just substituting, not differentiating, no need to use the chain rule.

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## marked as duplicate by Qmechanic♦Sep 29 '13 at 21:43

Maybe you're not differentiating anything but you should. The functions $B(T)$ express the density of energy per unit interval of frequencies – or unit interval of wavelengths. But the width of the unit intervals aren't the same.
In particular, the total energy in an interval $(\nu,\nu+d\nu)$ is $$E(\nu,\nu+d\nu) = B_\nu(T)\cdot d\nu \sim \frac{\nu^3}{c^2} \cdot d\nu$$ That's what the formula for the density per unit frequency means. When you want to translate it to the language of the wavelength, you must realize that $$\nu = \frac{c}{\lambda}$$ and therefore (the line below is obtained just by taking a derivative) $$d\nu = -\frac{c}{\lambda^2} \cdot d\lambda$$ The minus sign should be ignored because it just means that the graphs should be left-right-reflected. However, this $d\nu$ may be substituted to the first equation to get $$E(\nu,\nu+d\nu)\sim \left(\frac{c}{\lambda}\right)^3\cdot \frac{1}{c^2} \cdot \frac{c}{\lambda^2}\cdot d\lambda \sim \frac{c^2}{\lambda^5}d\lambda$$ which gives you the right dependence on $\lambda$ and $c$, namely $c^2/\lambda^5$. All the other factors are the same and there is no change in the numerical prefactor. Note that I can call this energy in the interval $E(\lambda,\lambda+d\lambda)$ because throughout the text, I was assuming it was physically the same interval, just parameterized either via $\nu$ or by $\lambda$.