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2009-Nov-18, 01:30 PM

This Wiki link (a) shows the Wien displacement for a black-body spectrum that is used to find the peak wavelength and peak frequency of the CMB pressure. The problem is that when one finds the peak wavelength in terms of dP/dw and then finds the peak frequency in terms of dP/df and then finds the peak wavelength using that with w = c/f after changing all of the wavelengths to frequencies in the formula in this manner, then the graph for the same CMB pressure will look slightly different when plotted out each of these two ways and dividing the peak wavelength by the peak frequency will give a constant speed of light that is much less than c. Dividing c by the peak frequency, then, will give a different peak wavelength depending upon which way we do it. For instance, finding the peak wavelength directly for T = 2.72839 K gives around .0019 meters, while finding it in terms of frequency first and then using w = c/f gives .001062 meters.

However, at the very end of the link, it also provides a "compromise" number of 4, that between 5 for the wavelength and 3 for the frequency, which when used, provides the exact same graph for both and we find w * f = c. That gives a single peak wavelength of .001345 meters either way. I had found that using another method as well, and I think it makes the most sense, so wouldn't that be the most likely choice to use as the solution for the peak wavelength and frequency, then, since there is only one graph for one pressure, the same one for both? I know it would depend upon how the peaks are defined, but wouldn't this be the most obvious choice of definition, then, since it provides a single consistent peak wavelength instead of two possibilities depending upon whether we find it directly or through the peak frequency found in this manner?

However, at the very end of the link, it also provides a "compromise" number of 4, that between 5 for the wavelength and 3 for the frequency, which when used, provides the exact same graph for both and we find w * f = c. That gives a single peak wavelength of .001345 meters either way. I had found that using another method as well, and I think it makes the most sense, so wouldn't that be the most likely choice to use as the solution for the peak wavelength and frequency, then, since there is only one graph for one pressure, the same one for both? I know it would depend upon how the peaks are defined, but wouldn't this be the most obvious choice of definition, then, since it provides a single consistent peak wavelength instead of two possibilities depending upon whether we find it directly or through the peak frequency found in this manner?