Wien's law gives two different results?

Question

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.

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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!

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 .