# Did Penzias and Wilson use Wein's law (or data based on it) to determine the temperature of CMB?

I'm writing a paper and I used them as an example, but then reconsidered . . . maybe I'm not getting it right! thanks!

At first, I had written that they certainly could have, given that blackbody radiation had been fairly well analyzed since the early 1900's and Wien's law would have been common knowledge among physicists at the time. But it looks like I may have been wrong.

Both the observational paper (Penzias and Wilson) and the theoretical paper (Dicke, Peebles, Roll, Wilkinson) are freely accessible online. (Click on "Full referreed scanned article (GIF)" at the top.) What I didn't realize until I read these papers is that Penzias and Wilson's publication was based upon an observation at a single wavelength, $7.3\text{ cm}$, or equivalently a frequency of $4080\text{ MHz}$. They did not have the ability to measure at a large number of different wavelengths, so they weren't able to determine the full spectrum of their mystery radiation, and certainly they did not have the ability to resolve a peak in that spectrum. So in fact, it seems that they did not use Wien's law. (In fact, if you use Wien's law to calculate the peak wavelength corresponding to their $3.5\text{ K}$ detection, you get only $0.8\text{ mm}$.)

Instead, if I'm understanding correctly, they would have used Planck's law,

$$I = \frac{2hc^3}{\lambda^5}\frac{1}{\exp(\frac{hc}{\lambda kT}) - 1}$$

In this form, the way it's normally written, Planck's law gives the spectral intensity of radiation emitted at any particular wavelength by a blackbody at a particular temperature. But you can rearrange it to this:

$$T = \frac{hc}{\lambda k\ln\Bigl(1 + \frac{2hc^3}{I\lambda^5}\Bigr)}$$

which lets you express a given spectral intensity of radiation at a particular wavelength as the temperature of the blackbody source which would have produced it. If you keep $\lambda$ fixed and avoid small values of $I$, this is an excellent approximation to a linear relationship,

$$T = \frac{\lambda^4}{2kc^2}I\hspace{1cm}\text{if}\quad I \gtrsim \frac{hc^3}{3\lambda^5}$$

This means that if you're making measurements at one specific wavelength, as Penzias and Wilson were, you can report your measured intensities as temperatures instead, because the temperature is proportional to the intensity.

If you look at the paper, you'll see that they report as temperatures the various contributions to the intensity which they identified:

• $2.3\pm 0.3\text{ K}$ from atmospheric absorption
• $0.8\pm 0.4\text{ K}$ from ohmic losses
• $<0.1\text{ K}$ for ground radiation

These add up to a total of $3.2\pm 0.5\text{ K}$ (assuming uncorrelated errors). But the actual measurement yielded a value of $6.7\text{ K}$. The difference, $3.5\text{ K}$, is what they postulated they were detecting from the cosmic microwave background. Now, you could plug that number back into Planck's law to calculate the intensity received from the CMB, but that's not necessary when what you really want to know is the temperature.

• Thanks! Well yesterday I calculated the wavelength they detected and came up with their answer . . . too easy! So I went ahead and wrote in my paper that the were a good example of Wein's Law, even if we cannot prove they used it – Gigi Giles Dec 1 '11 at 17:26
• Excellent answer. For future reference with regard to the strange beast that is radio astronomy, the exact formula gives the "brightness temperature," while the approximation is the definition of "antenna temperature." – user10851 Feb 14 '13 at 9:46