Why is the peak value(s) of the spectral radiance of the CMB different if you calculate with freq. instead of wavelength? 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...
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.