# Cosmic microwave confusion

1. Jun 27, 2009

### trekkiee

hi, i'm new to this forum and i would like to pose a question that have been bothering me for awhile.

the peak wavelength of the Cosmic Microwave Background (CMB) is lambda_max=[2.898x10^-3 m K]/[2.725 K]=1.06 mm [or 282 GHz], where lambda_max is the maximum, or peak, wavelength, i.e., the wavelength at which the object [the CMB] emits the most energy. lambda_max=[2.898x10^-3 m K]/T is Wien's Law [T is temperature], which applies to blackbody radiation, of which the CMB is about the best example in nature. my confusion arises from trying to find a graph of the spectrum of the CMB which displays this peak wavelength. every graph i find on the web shows the peak at about 2mm, not 1mm. do a google image search of "cosmic microwave background" and u find this image a lot http://en.wikipedia.org/wiki/File:Firas_spectrum.jpg, which clearly peaks close to 2mm [5 waves/cm], not 1mm [10 waves/cm], and u find these images: http://upload.wikimedia.org/wikipedia/commons/5/5c/Cmb_intensity.gif [Broken], http://www.rpi.edu/dept/phys/Courses/Astro_F96/lnCMB/cmbSpecData.gif, http://www.symmetrymagazine.org/images/200611/logSM.jpg, http://map.gsfc.nasa.gov/media/ContentMedia/990015b.jpg, http://www.astro.ubc.ca/people/scott/ispectrum.gif, etc. what i find on the net is data, originating almost exclusively from COBE [the cosmic background explorer], i think, displaying peaks near 2mm, 5 waves/cm, or 150 GHz. what i can't find are graphs displaying peaks at 1.06 mm. clearly, i'm missing something relatively simple and obvious here, and i would very much appreciate it if some1 could explain it to me. thx in advance.

ö¿ö¬ E=mc²
~

Last edited by a moderator: May 4, 2017
2. Jun 28, 2009

### Chalnoth

Basically, where the peak is depends upon whether you are looking at the maximum intensity per wavelength interval versus intensity per frequency interval. The law you quoted above cites the maximum per wavelength interval. The maximum per frequency interval will be different. To find the maximum per frequency interval you'd have to find the maximum of the following function:

$$I(\nu, T) = \frac{2h\nu^3}{c^2} \frac{1}{e^{\frac{h\nu}{kT}}-1}$$

This Wikipedia article talks about finding the peak to this function, if you don't want to do it yourself:
http://en.wikipedia.org/wiki/Planck's_law

3. Jul 6, 2009

### trekkiee

i'm STILL 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).

http://en.wikipedia.org/wiki/File:Firas_spectrum.jpg
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
~

4. Jul 6, 2009

### Chalnoth

The instrument itself has no bearing whatsoever on whether the frequency or wavelength peak is measured: they're just two different ways of representing the information provided by the instrument.

5. Jul 6, 2009

### Chronos

'sigh' i tire too easily these days. Chalnoth has captured the essence of this argument.

6. Jul 6, 2009

### trekkiee

2 questions:
would u happen to know what information was provided by the FIRAS interferometer on board COBE which is then converted into MJy/sr and then represented in the NASA graph i referenced?

can W/(m^2-Hz-sr) be converted into W/(m^2-m-sr) w/o multiplying I_nu (T) by |d{nu}/d{lamda}|, which just seems to convert the frequency peak spectrum into the wavelength peak spectrum?

ö¿ö¬ E=mc²
~

7. Jul 6, 2009

### Chalnoth

Well, I don't know the specifics of the instrument off hand, but the way all these things work is that you have some sort of way of detecting the photons that strike the telescope, and the energy of the photons is converted into a voltage that is detected by the instrument. Said voltage is recorded by the electronics, and send to the ground. The scientists on the ground then make use of knowledge of the instrument as well as observations of "known" sources to determine how those voltages relate to the amount of energy deposited in the detector. The energy deposited in the detector is frequently quoted in units of MJy/sr.

8. Jul 7, 2009

### marcus

Trekkiee, could it be that the information FIRAS provided was the fourier transform of the power spectrum?

Suppose you have a Michelson interferometer with two arms and there is an electric motor on one of the two arms that can move the mirror steadily and lengthen the arm continuously and suppose you record the continuuous intensity of what comes out, after interference. What kind of signal will that be?

Each individual wavelength component will be running thru its interference "fringe" pattern at a different rate. And it is all mixed together so you measure the combined power of every wavelength component, each if which is experiencing blips. Their powers are all fluctuating at different rates because of interference. That time varying power signal has information in it.

You get a timevarying signal which contains information about every wavelength or every wavelength bucket, if you discretize. Can you reconstruct the power spectrum? Can you fourier transform back and somehow tell how much power is in each wavelength band?

Tell me if this is a stupid idea, or out of touch with reality. I don't know this topic.
What I want you to explain to me is a fourier spectrometer.

What I'm picturing is a two-arm Michelson with a motor driving one smooth pass of one arm, from short to long. And a horn antenna feeding the Michelson. And a second horn antenna with a black body radiator in the mouth of the horn to provide a comparison spectrum. My intuition is the information is there all right. Fourier transform goes from time domain to frequency, or from frequency domain to time domain. The motor steadily sweeping the length of the arm from short to long makes a kind of artificial time varying signal. I'm vague about this, sorry. Can you clarify? I've almost not studied this at all, and you have, so you need to do about 3/4 or more of the work here. Explain very simply and plainly as if I didn't know hardly anything. Best way.

Very interesting question how did COBE work. COBE was like the greatest experiment of the 20th century. brought us face to face with the universe. so close we could see her freckles and count her eyelashes

9. Jul 7, 2009

### trekkiee

marcus & chalnoth:
thx for ur responses, i found them both to be very helpful. u pointed me in the right direction, i think. i don't think i ever studied fourier spectrometry [and 5000 other subjects], but
http://www.astro.ljmu.ac.uk/courses/phys362/notes/ [Broken]
has an excellent analysis. i had to look up a lot of stuff as i worked thru the calculations, lol, cus it's been 16yrs since i did any serious calculations and i sometimes feel that i've forgotten everything i ever learned! [plz direct me to any other (free) online resources that u may know of that cover the same subject

my initial confusion was that i didn't understand the physical meaning of the black body frequency peak, but now i think i'm getting my mind around it. because the frequency and wavelength peaks of black body radiation occur at different points on the spectrum, i somehow got it in my head that they were physically different maxima. but there are not 2 different peaks to black body radiation. there is only 1, and whether or not that peak is the frequency peak depends on how the data is measured and displayed. the peak in black body radiation displayed as the frequency peak and the corresponding frequency spectrum has physical meaning in that it allows the observer to verify the black body nature of the radiation by verifying the black body shape of the measured frequency distribution. but the wavelength peak has the added physical meaning of being the point on the EM spectrum where the black body radiates the maximum power. correct me if i'm wrong, plz.

P.S. can any1 verify that there is no physically meaningful way to convert W/(m^2-Hz-sr) into W/(m^2-m-sr)? i initially thought multiplying I_nu (T) by |d{nu}/d{lamda}| would convert the units properly, but this just converts the frequency spectrum into the wavelength spectrum. thx in advance :)

ö¿ö¬ E=mc²
~

Last edited by a moderator: May 4, 2017