<p>Because <span class="math-container">$B_\lambda$</span> is not just <span class="math-container">$B_\nu$</span>, with <span class="math-container">$\nu$</span> replaced by <span class="math-container">$c/\lambda$</span>.</p>
<p>The relationship between the two functions is that
<span class="math-container">$$ B_\lambda\ d\lambda = B_\nu\ d\nu$$</span>
since one is defined in terms of flux per unit wavelength, the other as flux per unit frequency.</p>
<p>Thus
<span class="math-container">$$B_\lambda = B_\nu\ \left|\frac{d\nu}{d\lambda}\right| = \frac{c}{\lambda^2} B_\nu, \tag{1}$$</span>
where you can then substitute <span class="math-container">$\nu =c/\lambda$</span> into the expression for <span class="math-container">$B_\nu$</span>.</p>
<p>To find where the maximum of a function, you differentiate and set to zero. Because of the above, the wavelength where <span class="math-container">$B_\lambda$</span> peaks is not <span class="math-container">$c/\nu_{\rm max}$</span>. We can see that by differentiating both sides of eq (1).
<span class="math-container">$$\frac{dB_\lambda}{d\lambda} = -\frac{2c}{\lambda^3}B_\nu + \frac{c}{\lambda^2}\frac{dB_\nu}{d\nu}\frac{d\nu}{d\lambda}$$</span>
<span class="math-container">$$\frac{dB_\lambda}{d\lambda} = -\frac{\nu^3}{c^2}\left(2B_\nu + \nu \frac{dB_\nu}{d\nu}\right).$$</span>
The usefulness of this expression is that you can easily see that if <span class="math-container">$dB_\nu /d\nu$</span> is zero at some frequency, then <span class="math-container">$dB_\lambda/d\lambda$</span> cannot be zero at that same frequency.</p>

CC BY-SA 4.0
2020-08-15T14:39:15.373Z
Looks like maybe you omitted the trailing d nu / d lambda in your final equation?
Eric Jensen
CC BY-SA 4.0
2020-08-15T15:25:49.683Z
@ELNJ Not quite, but I did get it wrong. Fixed.
ProfRob
CC BY-SA 4.0
2020-08-13T07:12:11.157Z
2020-08-15T15:25:02.510Z
2020-08-15T15:25:02.510Z
ProfRob
<p>On Wikipedia, the calculated value of the peak frequency of the cosmic microwave background is 160.23 GHz, but it says if you do the calculating with wavelengths, then convert to frequency, you get a value of 282 GHz.</p>
<p>Why the difference?</p>
<p>There used to be an answer on Quora, I believe, but I can't find it anymore...
Something to do with the summation using integral calculus, I believe...</p>

CC BY-SA 4.0
2020-08-10T14:09:57.693Z
See [this table](https://en.wikipedia.org/wiki/Planck%27s_law#Different_forms) The peaks of $B(\lambda, T)$ and $B(\nu, T)$ where $\lambda$ and $\nu$ will not be the same because one is *per unit wavelength* and the other is *per unit frequency* The bin spacings of the two histograms are different; equally spaced bins for one will be unequally spaced for the other. Near either peak $B$ doesn't change fast since it's near maximum, but the relative bin sizes keep changing.
uhoh
CC BY-SA 4.0
2020-08-12T15:44:01.100Z
Presumably as you allow the discrete "bin size" of unit $\lamba$ and $\omega$ to decrease towards zero, this difference will go away. Probably more confusing than useful to do this.
Carl Witthoft
CC BY-SA 4.0
2020-08-13T03:15:26.933Z
@CarlWitthoft no that's wrong. As bin size decreases to zero this becomes calculus and then one can easily prove that for $$ \left( \frac{\partial B}{\partial \lambda} \right)_{\lambda_0}=0 \ \text{  and  } \left( \frac{\partial B}{\partial \nu} \right)_{\nu_0}=0 $$ $\lambda_0$ and $\nu_0$ are different; i.e.  $\lambda_0\nu_0 \ne c.$ Bin size $\Delta x$ becoming calculus's $dx$ is a [cornerstone of calculus](https://en.wikipedia.org/wiki/Leibniz%27s_notation).
uhoh
CC BY-SA 4.0
2020-08-13T11:48:17.570Z
@uhoh  your  $\lambda_0$ is the location of the peak, right?  Anyway, thanks for the clarification. (yes, I was implying going into a calculus-based analysis :-)  )
Carl Witthoft
CC BY-SA 4.0
2020-08-13T13:02:25.337Z
@CarlWitthoft ya I could not find the vertical bar meaning "evaluated at" in MathJax so I used a subscript on the close parenthesis.
uhoh
CC BY-SA 4.0
2020-08-13T13:39:16.193Z
@uhoh  no problem -   $X_0$ is pretty standard notation for the value of a variable at some "interesting" location
Carl Witthoft
<astrophysics><radio-astronomy><cosmic-microwave-background>
Why is the peak value(s) of the spectral radiance of the CMB different if you calculate with freq. instead of wavelength?
Kurt Hikes
# Why is the peak value(s) of the spectral radiance of the CMB different if you calculate with freq. instead of wavelength?

(Asked by: username_0 on 2020-08-10)

On Wikipedia, the calculated value of the peak frequency of the cosmic microwave background is 160.23 GHz, but it says if you do the calculating with wavelengths, then convert to frequency, you get a value of 282 GHz.

Why the difference?

There used to be an answer on Quora, I believe, but I can't find it anymore...
Something to do with the summation using integral calculus, I believe...

**Tags:** astrophysics, radio-astronomy, cosmic-microwave-background

> **Comments:**
> **username_1**: See [this table](https://en.wikipedia.org/wiki/Planck%27s_law#Different_forms) The peaks of $B(\lambda, T)$ and $B(\nu, T)$ where $\lambda$ and $\nu$ will not be the same because one is *per unit wavelength* and the other is *per unit frequency* The bin spacings of the two histograms are different; equally spaced bins for one will be unequally spaced for the other. Near either peak $B$ doesn't change fast since it's near maximum, but the relative bin sizes keep changing.  
> **username_2**: Presumably as you allow the discrete "bin size" of unit $\lamba$ and $\omega$ to decrease towards zero, this difference will go away. Probably more confusing than useful to do this.  
> **username_1**: @username_2 no that's wrong. As bin size decreases to zero this becomes calculus and then one can easily prove that for $$ \left( \frac{\partial B}{\partial \lambda} \right)_{\lambda_0}=0 \ \text{  and  } \left( \frac{\partial B}{\partial \nu} \right)_{\nu_0}=0 $$ $\lambda_0$ and $\nu_0$ are different; i.e.  $\lambda_0\nu_0 \ne c.$ Bin size $\Delta x$ becoming calculus's $dx$ is a [cornerstone of calculus](https://en.wikipedia.org/wiki/Leibniz%27s_notation).  
> **username_2**: @username_1  your  $\lambda_0$ is the location of the peak, right?  Anyway, thanks for the clarification. (yes, I was implying going into a calculus-based analysis :-)  )  
> **username_1**: @username_2 ya I could not find the vertical bar meaning "evaluated at" in MathJax so I used a subscript on the close parenthesis.  
> **username_2**: @username_1  no problem -   $X_0$ is pretty standard notation for the value of a variable at some "interesting" location  

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## 1 answers

### Answer by username_3 on 2020-08-13. Score: 5

Because $B\_\lambda$ is not just $B\_\nu$, with $\nu$ replaced by $c/\lambda$.

The relationship between the two functions is that
$$ B\_\lambda\ d\lambda = B\_\nu\ d\nu$$
since one is defined in terms of flux per unit wavelength, the other as flux per unit frequency.

Thus
$$B\_\lambda = B\_\nu\ \left|\frac{d\nu}{d\lambda}\right| = \frac{c}{\lambda^2} B\_\nu, \tag{1}$$
where you can then substitute $\nu =c/\lambda$ into the expression for $B\_\nu$.

To find where the maximum of a function, you differentiate and set to zero. Because of the above, the wavelength where $B\_\lambda$ peaks is not $c/\nu\_{\rm max}$. We can see that by differentiating both sides of eq (1).
$$\frac{dB\_\lambda}{d\lambda} = -\frac{2c}{\lambda^3}B\_\nu + \frac{c}{\lambda^2}\frac{dB\_\nu}{d\nu}\frac{d\nu}{d\lambda}$$
$$\frac{dB\_\lambda}{d\lambda} = -\frac{\nu^3}{c^2}\left(2B\_\nu + \nu \frac{dB\_\nu}{d\nu}\right).$$
The usefulness of this expression is that you can easily see that if $dB\_\nu /d\nu$ is zero at some frequency, then $dB\_\lambda/d\lambda$ cannot be zero at that same frequency.

> **Comments:**
> **username_4**: Looks like maybe you omitted the trailing d nu / d lambda in your final equation?  
> **username_3**: @ELNJ Not quite, but I did get it wrong. Fixed.  

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