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...

  • 3
    $\begingroup$ See this table 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. $\endgroup$
    – uhoh
    Commented Aug 10, 2020 at 14:09
  • $\begingroup$ 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. $\endgroup$ Commented Aug 12, 2020 at 15:44
  • $\begingroup$ @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. $\endgroup$
    – uhoh
    Commented Aug 13, 2020 at 3:15
  • $\begingroup$ @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 :-) ) $\endgroup$ Commented Aug 13, 2020 at 11:48
  • $\begingroup$ @CarlWitthoft ya I could not find the vertical bar meaning "evaluated at" in MathJax so I used a subscript on the close parenthesis. $\endgroup$
    – uhoh
    Commented Aug 13, 2020 at 13:02

1 Answer 1


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.

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    $\begingroup$ Looks like maybe you omitted the trailing d nu / d lambda in your final equation? $\endgroup$ Commented Aug 15, 2020 at 14:39
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    $\begingroup$ @ELNJ Not quite, but I did get it wrong. Fixed. $\endgroup$
    – ProfRob
    Commented Aug 15, 2020 at 15:25

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