Say I have a star with radius $R$ and luminosity $L$. I want to find the frequency of its peak emission.

Taking the star as a perfect blackbody, we can use Stefan-Boltzman to see that $$L = 4\pi R^2\sigma T^4 \iff T = \sqrt[4]{ \frac{L}{4\pi R^2\sigma} }$$

Wien's law for wavelength is $\lambda_\text{peak} = \frac{hc}{xkT} = \frac{b}{T}$. Here, $x = 4.965114231744276303...$. We can invert this for frequency, where $\nu_\text{peak} = {cT\over b}$ since $\lambda = \frac{c}{\nu}$.

Wien's law for frequency is $\nu_\text{peak} = \frac{xkT}{h}$, where $x = 2.821439372122078893...$.

They should be equivalent.

Let's say the star in question has a radius of $2R_☉$ and a luminosity of $3L_☉$. Then, it has a temperature of 5371 K.

By the wavelength formula, I get a peak frequency of 555.62 THz. With the frequency formula, I get a peak frequency of 315.86 THz.

Why do I get different things?

Much thanks!

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    $\begingroup$ This has to do with the chain rule when doing changes of variables from the integrations made when deriving Wien's law, and how these results are usually later presented. So the reason is mathematical and relates to how $\mathrm{d}\lambda \neq \mathrm{d}\nu$. I believe it's not the first time this gets asked, e.g., see this and that. $\endgroup$
    – nuwe
    Commented Feb 13 at 19:23
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    $\begingroup$ I agree with @nuwe. There are several physical explanations in the two answers linked to. I kind of like this simple and purely mathematical explanation, though. $\endgroup$
    – pela
    Commented Feb 13 at 21:41
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    $\begingroup$ Also, there is an arxiv paper "The functions Bν and Bλ describe the same physics, but they have different shapes, due to the nonlinear change of variable from wavelength to frequency. As a consequence, the two curves peak at different locations in the spectrum" arxiv.org/abs/1109.3822 $\endgroup$
    – James K
    Commented Feb 13 at 22:29
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    $\begingroup$ Thanks @pela for linking that excellent answer, especially also the one from leftaroundabout! $\endgroup$ Commented Feb 14 at 6:35
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    $\begingroup$ @nuwe yep I plan to do that sooner or later! Thanks for the suggestion :) and happy Valentine’s Day $\endgroup$ Commented Feb 14 at 23:39

1 Answer 1


The frequency at which the flux per unit frequency is maximised and the wavelength at which the flux per unit wavelength is maximised do not agree, in the sense that $$\lambda_{\rm max} f_{\rm max,} \neq c \ . $$

That is because the flux per unit frequency and flux per unit wavelength are different things. They are related by conservation of energy. The flux per unit frequency multiplied by a frequency interval has the same flux as a flux per unit wavelength multiplied by a corresponding wavelength interval. $$B_f\ df = B_\lambda\ d\lambda\ , $$ where $B_f$ and $B_\lambda$ are the expressions for the flux per unit frequency and flux per unit wavelength of a blackbody.

The rest is just maths I'm afraid. The frequency where $B_f$ is maximised is found by differentiating with respect to frequency and equating to zero. We can then replace $B_f$ with $B_\lambda\ d\lambda/df$. i.e. At $f_{\rm max}$ it must be that $$\frac{dB_f}{df}= \frac{d}{df}\left(B_\lambda \frac{d\lambda}{df}\right) = 0 $$ But since $f = c/\lambda$, we can replace $df$ with $df = -c\ d\lambda/\lambda^2$. $$\frac{d}{df}\left(B_\lambda \frac{d\lambda}{df}\right) = \frac{\lambda^2}{c}\frac{d} {d\lambda}\left(B_\lambda \frac{\lambda^2}{c}\right) = 0. $$ Eliminating $\lambda = 0$ as a useless solution, multiplying both sides by $c^2/\lambda^2$ and differentiating the product in the bracket, we get $$ \lambda^2 \frac{dB_\lambda}{d\lambda} + 2\lambda B_\lambda = 0$$ and so $$\frac{dB_\lambda}{d\lambda} = -\frac{2 B_\lambda}{\lambda} \neq 0 \ .$$ What this demonstrates is that, at the frequency which maximises $B_f$, then $dB_\lambda/d\lambda$ is not equal to zero and so $c/f_{\rm max}$ cannot also be the wavelength at which $B_\lambda$ is maximised.

Plenty of other attempts (all of which seem ok to me) to explain this in a variety of ways can be found at https://physics.stackexchange.com/questions/91192/the-strange-thing-about-the-maximum-in-plancks-law .

  • $\begingroup$ Ah I see. So to find the frequency of a star's peak emission, it would be erroneous to find the peak frequency. Instead, I'd need to do peak wavelength -> frequency of peak wavelength. icic $\endgroup$ Commented Feb 14 at 15:26
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    $\begingroup$ @CaptainChicky Wien's law comes in two versions. To find the peak wavelength of $B_\lambda$ you would use one, whilst to find the peak frequency of $B_f$ you would use the other. And multiplying the two results together will NOT give $c$. $\endgroup$
    – ProfRob
    Commented Feb 14 at 16:47

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