1. hi, i'm new to this forum and i would like to pose a question that has 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/wikipedi..._intensity.gif, http://www.rpi.edu/dept/phys/Courses...mbSpecData.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
~  2.

3. Welcome and thanks for this interesting question. It puzzled me a while myself, but now I found the answer.

You have to do the calculations based on the Planck law of the frequency, not the wavelength. The measurements were done in the frequency space, i.e. the intensity was measured per unit frequency, not per unit wavelength. Both representations are not entirely identical. This equation is correct, not . But the frequency plot only displays . In order to have the law of Wien's displacement, you have to derive it in this picture, i.e. it is: For a temperature of 2.725 K, this is Hz which is equivalent to a wavelength of 1.87 mm.

So, the spectra distributed are based on the Planck formula of the form (B is replaced by I in this equation):   4. thank u for ur response, it was very helpfull. i reread the wikipedia articles on planck's law and wien's displacement law, went thru and calculated the integrals, etc, even got out landau & lifshitz (1980) and shu (radiation, 1991) which i haven't studied in 16yrs just to refresh. a lot of work, lol, but i needed the review.

its clear to me that graphing I (nu,T) as W/m^2-Hz-sr vs Hz will peak at nu_max and graphing I'(lambda,T) as W/m^2-m-sr vs m will peak at lambda_max and graphing (nu) I(nu,T) which equals (lambda) I'(lambda,T) as W/m^2-sr will peak at the so-called "neutral peak."

what still is NOT clear to me is at which point on the electromagnetic spectrum a blackbody emits the maximum energy. thx in advance for ur response.

ö¿ö¬ E=mc²
~  5. Originally Posted by trekkiee
what still is NOT clear to me is at which point on the electromagnetic spectrum a blackbody emits the maximum energy. thx in advance for ur response.
I'd suppose that it should be simply , because E and frequency are linearly dependent. In order to be sure, you will have to substitute and , find and then find the maximum of the function . But this is not trivial. I will try to check this as soon as possible.  6. Originally Posted by Dishmaster Originally Posted by trekkiee
what still is NOT clear to me is at which point on the electromagnetic spectrum a blackbody emits the maximum energy. thx in advance for ur response.
I'd suppose that it should be simply , because E and frequency are linearly dependent. In order to be sure, you will have to substitute and , find and then find the maximum of the function . But this is not trivial. I will try to check this as soon as possible.
ur reply was very helpfull again. it seems to me now that a blackbody radiates max energy at lambda_max, which emits photons 1.76 times a energetic as the photons emitted at nu_max [since both distributions equipartition the energy].

btw, ur suggestion to substitute E=h nu, dE=h d{nu} into I(nu,T) and then finding the peak of I(E,T) by dI/dE=0 will
give u E_max=h nu_max, but no new insight, i think, same w/I'(lambda,T).

but it seems that the "real" energy peak is the wavelength peak. correct me if i'm wrong. thx in advance.

o-o- E=mc^2
~  7. 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
~  8. I think, the answer is that it is impossible to strictly measure either or . You always have an interval in wavelength or frequency, how small it may be. It cannot be infinitely small. So, in fact, you always measure . The calibration of the instrument allows you then to make the transition to the flux density measured in Jy, just as you described it. So, finally, it is more a definition to what the incoming flux is translated. The latter has historical regions, since radio astronomy emerged from high frequency physics.  9. Originally Posted by Dishmaster
I think, the answer is that it is impossible to strictly measure either or . You always have an interval in wavelength or frequency, how small it may be. It cannot be infinitely small. So, in fact, you always measure . The calibration of the instrument allows you then to make the transition to the flux density measured in Jy, just as you described it. So, finally, it is more a definition to what the incoming flux is translated. The latter has historical regions, since radio astronomy emerged from high frequency physics.
ur response is very helpful, as always so i guess a certain amount of energy is measured over a frequency or wavelength range, delta E = I_nu (T) delta nu = I_lambda (T) delta lambda (per unit surface area per unit solid angle). it seems to me then that there is no physical significance to the frequency peak - it's just a mathematically different representation of the wavelength peak, which is the true energy peak. correct me if i'm wrong, plz.

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?

thx in advance ö¿ö¬ E=mc²
~  10. The maximum of the energy output can be derived in a different way. The energy content is given by: If you derive the peak of this distribution, you will see that it is the same, regardless of choosing the wavelength or the frequency.  So, it is slightly off of the wavelength peak of .  Bookmarks
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