Cosmic microwave confusion

[trekkiee]

i'm confused about how COBE (the Cosmic Background Explorer) measured the frequency peak of the Cosmic Microwave Background (CMB).

 

the following seems clear to me and relatively straightforward:

for black body radiation, i.e., for an ideal photon gas in local thermodynamic equilibrium with matter, e.g., the surface of last scattering of the CMB, the spectral radiance, I_nu (T) = [2h/c^2][nu^3/(exp(h nu/[k T])-1)] or I'_lamba (T) = [2hc^2][1/lamba^5(exp(h c/[lambda k T])-1)] <forgive my clumsy notation, i don't know how to do LaTeX notation>, peaks at nu_max or lambda_max, respectively, such that c/lambda_max = 1.76*nu_max.

 

although the lambda_max peak is the actual energy peak at which a black body radiates the maximum energy [photons at lambda_max are 1.76 times as energetic as photons at nu_max], radio astronomers seem to prefer the frequency peak. in fact, doing a google image search of "cosmic microwave background," then picking out spectral graphs [spectral radiance vs frequency or wavelength], will find almost exclusively graphs depicting the frequency peak.

 

examples include:

http://map.gsfc.nasa.gov/media/ContentMedia/990015b.jpg

http://www.phy.duke.edu/~kolena/cmbspectrum1.gif

which show the frequency peak of 384 MJy/sr = 384 megaJanskys per steradian = 3.84e-18 W/(m^2-Hz-sr) = I_nu (2.725 K) at 1.87mm (160 GHz).

 

File:Firas spectrum.jpg - Wikipedia, the free encyclopedia

which shows the frequency peak of 1.15e-4 erg/(s-cm^2-cm^-1-sr) = 3.84e-18 W/(m^2-Hz-sr) = I_nu (2.725 K) at 1.87mm = 5.34 cm^-1 (160 GHz).

 

my understanding is that the FIRAS interferometer on board COBE compared the spectral radiance of the CMB to an on-board black body.

 

what is NOT clear to me is exactly how these measurements were made. i would think that if u measured the CMB's [or any black body's] spectral radiance, u would measure the wavelength peak, not the frequency peak, since the wavelength peak is the physical maximum, that is, the point on the EM spectrum where the black body radiates the maximum energy. the frequency peak seems to me to have no physical significance and to only be a mathematical tool. i don't see how physical measurements can measure any energy peak other than the wavelength peak. i would very much appreciate it if some1 could clear this up for me. thx in advance.

 

ö¿ö¬ E=mc²

~

[Tormod]

Hello and welcome to Hypography.

 

We kindly ask that you take care when writing. Do it legibly, like capitalizing the first letter in each sentence for readability. Also please refrain from writing in txt language ("u", "some1"). A lot of our members are foreigners and it can be hard to read that kind of text.

[modest]

Welcome, Trekkiee.

 

I think which spectrum you use shouldn't matter in this case. The purpose as I understand was to compare the curve to a blackbody and find the temperature. That can easily be done either with the frequency plot or wavelength. Aside from that, I do agree that wavelength is usually used, wikipedia talks a little about it in the Planck's law article:

 

The wavelength and frequency peaks are in bold and occur at 25.0% and 64.6% respectively. The 41.8% point is the wavelength-frequency-neutral peak. These are the points at which the quotients by e^{hν / kT} − 1 of the respective Planck-law functions 1 / λ⁵, ν³, and (ν / λ)² attain their maxima.

 

Which peak to use depends on the application. The conventional choice is the wavelength peak at 25.0% given by Wien's displacement law. For some purposes the median or 50% point dividing the total radiation into two halves may be more suitable. The latter is closer to the frequency peak than to the wavelength peak because the radiance drops exponentially at short wavelengths and only polynomially at long. The neutral peak occurs at a shorter wavelength than the median for the same reason.

 

So, I think in this particular application it didn't really matter which curve was used.

 

~modest