<p>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
<span class="math-container">$$\lambda_{\rm max} f_{\rm max,} \neq c \ . $$</span></p>
<p>That is because the flux per unit frequency and flux per unit wavelength are different things. They are <em>related</em> 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.
<span class="math-container">$$B_f\ df = B_\lambda\ d\lambda\ , $$</span>
where <span class="math-container">$B_f$</span> and <span class="math-container">$B_\lambda$</span> are the expressions for the flux per unit frequency and flux per unit wavelength of a blackbody.</p>
<p>The rest is just maths I'm afraid. The frequency where <span class="math-container">$B_f$</span> is maximised is found by differentiating with respect to frequency and equating to zero. We can then replace <span class="math-container">$B_f$</span> with <span class="math-container">$B_\lambda\ d\lambda/df$</span>. i.e. At <span class="math-container">$f_{\rm max}$</span> it must be that
<span class="math-container">$$\frac{dB_f}{df}= \frac{d}{df}\left(B_\lambda \frac{d\lambda}{df}\right) = 0 $$</span>
But since <span class="math-container">$f = c/\lambda$</span>, we can replace <span class="math-container">$df$</span> with <span class="math-container">$df = -c\  d\lambda/\lambda^2$</span>.
<span class="math-container">$$\frac{d}{df}\left(B_\lambda \frac{d\lambda}{df}\right) = \frac{\lambda^2}{c}\frac{d} {d\lambda}\left(B_\lambda \frac{\lambda^2}{c}\right) =  0. $$</span>
Eliminating <span class="math-container">$\lambda = 0$</span> as a useless solution, multiplying both sides by <span class="math-container">$c^2/\lambda^2$</span> and differentiating the product in the bracket, we get
<span class="math-container">$$ \lambda^2 \frac{dB_\lambda}{d\lambda} + 2\lambda B_\lambda = 0$$</span>
and so
<span class="math-container">$$\frac{dB_\lambda}{d\lambda} = -\frac{2
B_\lambda}{\lambda} \neq 0 \ .$$</span>
What this demonstrates is that, at the frequency which maximises <span class="math-container">$B_f$</span>, then <span class="math-container">$dB_\lambda/d\lambda$</span> is <em>not</em> equal to zero and so <span class="math-container">$c/f_{\rm max}$</span> cannot also be the wavelength at which <span class="math-container">$B_\lambda$</span> is maximised.</p>
<p>Plenty of other attempts (all of which seem ok to me) to explain this in a variety of ways can be found at <a href="https://physics.stackexchange.com/questions/91192/the-strange-thing-about-the-maximum-in-plancks-law">https://physics.stackexchange.com/questions/91192/the-strange-thing-about-the-maximum-in-plancks-law</a> .</p>

CC BY-SA 4.0
2024-02-14T15:26:32.520Z
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
Captain Chicky
CC BY-SA 4.0
2024-02-14T16:47:36.943Z
@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 $c$.
ProfRob
CC BY-SA 4.0
2024-02-14T08:26:16.463Z
2024-02-14T10:41:20.547Z
2024-02-14T10:41:20.547Z
ProfRob
<p>Say I have a star with radius <span class="math-container">$R$</span> and luminosity <span class="math-container">$L$</span>. I want to find the frequency of its peak emission.</p>
<p>Taking the star as a perfect blackbody, we can use Stefan-Boltzman to see that
<span class="math-container">$$L = 4\pi R^2\sigma T^4 \iff T = \sqrt[4]{ \frac{L}{4\pi R^2\sigma} }$$</span></p>
<p>Wien's law for wavelength is <span class="math-container">$\lambda_\text{peak} = \frac{hc}{xkT} = \frac{b}{T}$</span>. Here, <span class="math-container">$x = 4.965114231744276303...$</span>. We can invert this for frequency, where <span class="math-container">$\nu_\text{peak} = {cT\over b}$</span> since <span class="math-container">$\lambda = \frac{c}{\nu}$</span>.</p>
<p>Wien's law for frequency is <span class="math-container">$\nu_\text{peak} = \frac{xkT}{h}$</span>, where <span class="math-container">$x = 2.821439372122078893...$</span>.</p>
<p>They should be equivalent.</p>
<hr />
<p>Let's say the star in question has a radius of <span class="math-container">$2R_☉$</span> and a luminosity of <span class="math-container">$3L_☉$</span>. Then, it has a temperature of <a href="https://www.wolframalpha.com/input?i=%28%283%20solar%20luminosities%29%2F%284pi%20*%20%282%20solar%20radii%29%5E2%20*%20stefan-boltzman%20constant%29%29%5E%281%2F4%29" rel="nofollow noreferrer">5371 K</a>.</p>
<p>By the wavelength formula, I get a peak frequency of <a href="https://www.wolframalpha.com/input?i=%28speed%20of%20light%20*%205371%2F%20%280.002898%20meter%29%29%20to%20hertz" rel="nofollow noreferrer">555.62 THz</a>. With the frequency formula, I get a peak frequency of <a href="https://www.wolframalpha.com/input?i=2.821439372122078893%20*%20boltzman%20constant%20*%205371%20kelvins%20%2F%20%28planck%20constant%29" rel="nofollow noreferrer">315.86 THz</a>.</p>
<p>Why do I get different things?</p>
<p>Much thanks!</p>

CC BY-SA 4.0
2024-02-13T19:23:12.717Z
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](https://physics.stackexchange.com/questions/211049/frequency-and-wavelength-peak-for-wiens-displaement-law-of-a-blackbody) and [that](https://physics.stackexchange.com/questions/91192/the-strange-thing-about-the-maximum-in-plancks-law).
nuwe
CC BY-SA 4.0
2024-02-13T21:41:01.287Z
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](https://physics.stackexchange.com/a/315994/70207), though.
pela
CC BY-SA 4.0
2024-02-13T22:29:09.093Z
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"  https://arxiv.org/abs/1109.3822
James K
CC BY-SA 4.0
2024-02-14T06:35:27.400Z
Thanks @pela for linking that excellent answer, especially also the one from leftaroundabout!
planetmaker
CC BY-SA 4.0
2024-02-14T17:52:14.460Z
I wanted to add as a little sidenote that deriving Wien's result can be very useful to grasp this
nuwe
CC BY-SA 4.0
2024-02-14T23:39:05.307Z
@nuwe yep I plan to do that sooner or later! Thanks for the suggestion :) and happy Valentine’s Day
Captain Chicky
<temperature><luminosity><wavelength>
Wien's law gives two different results?
Captain Chicky
# Wien's law gives two different results?

(Asked by: username_0 on 2024-02-13)

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](https://www.wolframalpha.com/input?i=%28%283%20solar%20luminosities%29%2F%284pi%20*%20%282%20solar%20radii%29%5E2%20*%20stefan-boltzman%20constant%29%29%5E%281%2F4%29).

By the wavelength formula, I get a peak frequency of [555.62 THz](https://www.wolframalpha.com/input?i=%28speed%20of%20light%20*%205371%2F%20%280.002898%20meter%29%29%20to%20hertz). With the frequency formula, I get a peak frequency of [315.86 THz](https://www.wolframalpha.com/input?i=2.821439372122078893%20*%20boltzman%20constant%20*%205371%20kelvins%20%2F%20%28planck%20constant%29).

Why do I get different things?

Much thanks!

**Tags:** temperature, luminosity, wavelength
> **Comments:**
> **username_1**: 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](https://physics.stackexchange.com/questions/211049/frequency-and-wavelength-peak-for-wiens-displaement-law-of-a-blackbody) and [that](https://physics.stackexchange.com/questions/91192/the-strange-thing-about-the-maximum-in-plancks-law).  
> **username_2**: 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](https://physics.stackexchange.com/a/315994/70207), though.  
> **username_3**: 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"  https://arxiv.org/abs/1109.3822  
> **username_4**: Thanks @pela for linking that excellent answer, especially also the one from leftaroundabout!  
> **username_1**: I wanted to add as a little sidenote that deriving Wien's result can be very useful to grasp this  
> **username_0**: @nuwe yep I plan to do that sooner or later! Thanks for the suggestion :) and happy Valentine’s Day  

---

## 1 answers

### Answer by username_5 on 2024-02-14. Score: 3 | Accepted answer

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> .

> **Comments:**
> **username_0**: 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  
> **username_5**: @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 $c$.  

---

