# Thread: How can we see the CMB?

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Originally Posted by TooMany
I already answered the black body issue. If you have evidence that the galactic deviation from a black body matters a whole lot in the CMB range (i.e. is a whole lot smaller), can you show it to me?
This has already been answered in this thread. It matters over the entire cpectrum because galaxies do not have anything close to a black body spectrum:
Originally Posted by Tensor
Yeah, that assumption, of Too Many's that galaxies are approximately black body has some holes in it. Here is a graph with two galaxies (M82 and Arp 220) comparing them to fixed temperatures of 30k and 60k. The CMB peak would be over by the 1000 μm, as that is where the 2.7K black body peaks. As you can see, galaxies are nowhere near black bodies.
Originally Posted by TooMany
The only thing I can think of that I might have wrong is this: maybe the shape of the CMB is "like" the shape of a 2.7 K black body curve, but it is actually many orders of magnitude more intense than an actual black body at 2.7 K would be.
You definitely have this wrong since the shape of the CMB is exactly the shape of a 2.7 K black body curve within the measuarable error limits. See the FIRAS CMB Monopole Spectrum

Your "actual black body" is imaginary. It is easy to find a body with an intensity greater than the CMB - the Sun, Earth, planets, Milky Way and many point sources. There are none with a perfect black body spectrum.

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Originally Posted by Nereid
Several posts ago, I asked you this: "Suppose both blackbodies are at a distance of 10 pc from the detector. Suppose the 10K one is a spherical cloud, 1 pc in radius. Suppose the 100 K one is a compact body, 10 km in radius. Which has the greater intensity?"

I'm not sure if you think you addressed the points my question raises.

I'll turn up the contrast.

Imagine a whole slew of objects, each 10 pc from us (and imagine no absorption/extinction, between us and these bodies; no dust, no gas, nothing).

Imagine they all emit as perfect blackbodies.

Suppose they come in triplets; one is tiny (a radius of 10 km, say), one is merely small (a radius of 10^6 km, say), and is huge (a radius of 1 pc, say).

Bodies in the first triplet all have a temperature of 10K; those in the second, 100K; the third, 1,000K; the fourth, 10,000K; and the fifth, 100,000K.

From here, somewhere in space above the Earth's atmosphere, which of these seems the brightest, at ~200 GHz? As in, the sum total of the energy received from the body by a detector, integrated over the whole angular extent of the body (if it's not seen as a point source).

As I understand your argument/reasoning, all three of the 100,000K triplets are equally bright, followed by the 10,000K triplets, and so on.

Of course, I may be wrong; in any case, what's the answer?
Nereid, I'm not playing that game with you, where you throw new questions back at me instead reading my argument and finding the flaw. I'm the person asking the question on this Q&A thread. If you want to contribute, please try to answer the question with mathematics. Show me whatever calculations you like to explain were my argument or calculations are wrong.

Last chance here. Please, carefully read my argument in post #52. Pretend it is a paper submitted to you for review. Tell me exactly what is sufficiently wrong with my argument to be off by many orders of magnitude in assessing the effect of galaxies relative to the CMB in the CMB spectrum. If you cannot, step aside; maybe somebody else in the forum can shed light on it.

I've put many hours of work now into explaining this argument and answering non-quantitative counter arguments. If there is something in my argument that you cannot follow, let me know and I will explain in more detail as best I can.

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Originally Posted by TooMany
Nereid, I'm not playing that game with you, where you throw new questions back at me instead reading my argument and finding the flaw.

I pointed out at least two flaws.

I'm the person asking the question on this Q&A thread. If you want to contribute, please try to answer the question with mathematics. Show me whatever calculations you like to explain were my argument or calculations are wrong.
There are several, apparently quite fundamental, flaws in your approach.

As I have said, several times.

One is that you seem to have conflated "intensity" (or "brightness") with "surface brightness".

Unless and until you grasp the various distinctions in this regard, no "answer with mathematics" will help. This is a conceptual problem.

It is my judgement that you will best be able to grasp the point by working through the "triplets" example I gave.

Alternatively, you may wish to calculate the total area, on the sky (our sky, here on Earth), subtended by all the stars in a galaxy such as M31 or M101. The stars themselves that is. To start, why not assume such a galaxy has 10^11 stars, and that they each have the same radius as our Sun (which subtends an angle of 30', as seen from here). Further assume that no two stars in such a galaxy overlap (as seen from here). Assume Euclidean geometry (the galaxies are close).

(more later, perhaps)

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Originally Posted by Nereid
Um, are you trying to tell me that, in the way you did (or at least presented) your argument/reasoning, it is totally unimportant to distinguish between "brightness" and "surface brightness"?
I stopped arguing using names like "brightness" and "surface brightness". Instead in post #52 I just used physical units of measure concerning radiant power in watts, surface area in m^2, solid angle in steradians, and Hz in bandwidth. Using Planck's law I compared how much radiant power enters the aperture of a detector from the CMB versus a conservative estimate for galaxies in any given area of the sky.

Originally Posted by Nereid
Are you telling me you know this? Or are you asking a question?
OK, so you know, you're telling me.
And this is because a) you've done at least a BoTE calculation (not a hand-waving, piece of verbal fluff)? b) your intuition tells you? c) something else?
This is common knowledge to astronomers. It helps simplify estimates which are complicated as to get exact using LCDM theory. Yes, I know that the non-Euclidean nature of the cosmos does not significantly affect my approximation of the ratio radiant powers emitted by the CMB versus the galaxies. To kill my argument, you need to find an error of several orders of magnitude. If you think you have that kind of error here, can you explain in some more detail? Are you actually arguing that the expansion diminishes the power of galaxies at z = 0.1, z = 1 or even z =2 by several orders of magnitude (by magnitude here I mean powers of 10)?

Originally Posted by Nereid
According to the caption, that is the SED of an intense starburst region in a lensed, z=2.3259 galaxy. Hardly representative of galaxies in general, wouldn't you say?
This is pointless. Show me where typical galaxies are several orders of magnitude dimmer in the CMB range than a black body at the same temperature. That certainly would be the problem, but I can't see anything like that in the SED's that I've looked at. If galaxies don't emit in these bands, then why do astronomers spend so much effort removing the Galactic foreground?

Originally Posted by Nereid
In previous posts, I note that StupendousMan has given you several direct, easily followed suggestions on how you can do searches like this on your own. Do you remember reading his posts? Do you remember the method he suggested?

Again, IIRC, several of the references others have already posted include just such things.

How familiar are you with COBE (in some sense, WMAP's predecessor)?

Do you think COBE would have observed these galaxies, at frequencies relevant to the CMB (if not exactly at 200 GHz)?

May I respectfully suggest that you re-read my post?
This is also pointless. I have chased down links that people have given and found zip about the issue of galaxies except our own. I have calculated that galaxies emit a substantial amount of radiation in the CMB spectrum that would show up in a detector. I have made my presentation and asked what is wrong with it. If you know, show me quantitatively (an approximation is fine), how I ended up several orders of magnitude off, because I don't get it!

Sending me off to read random references is not appropriate as an answer anyway, unless the reference contains a specific explanation. Asking me other questions is not an answer, it's just a distraction.

If that's the way the Q&A forum works here, then it makes no sense to ask a question in the first place because you will just get told to go read some references and do some calculations to figure out the answer yourself.

I have read some references, I have done some calculations!

Can anyone out there help explain this huge error in post #52?

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Originally Posted by Nereid

I pointed out at least two flaws.

There are several, apparently quite fundamental, flaws in your approach.

As I have said, several times.

One is that you seem to have conflated "intensity" (or "brightness") with "surface brightness".
Read post #52, I don't need these terms in the argument. However, if you think it is fantastically important to my argument anyway, then explain why.

Originally Posted by Nereid
Unless and until you grasp the various distinctions in this regard, no "answer with mathematics" will help. This is a conceptual problem.

It is my judgement that you will best be able to grasp the point by working through the "triplets" example I gave.

Alternatively, you may wish to calculate the total area, on the sky (our sky, here on Earth), subtended by all the stars in a galaxy such as M31 or M101. The stars themselves that is. To start, why not assume such a galaxy has 10^11 stars, and that they each have the same radius as our Sun (which subtends an angle of 30', as seen from here). Further assume that no two stars in such a galaxy overlap (as seen from here). Assume Euclidean geometry (the galaxies are close).

(more later, perhaps)
I ignored the stars, I just took the temperature of dust/gas which is diffuse but has considerable optical depth.
Your just distracting from the specific issue without really trying to explain what is wrong. I don't wish to do your calculations, I'm asking you to identify what is wrong with mine. If you cannot do that I'm done responding to you but still looking for someone who can please help truly explain the problem with my argument.

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Originally Posted by TooMany
Can anyone out there help explain this huge error in post #52?
The huge error in post #52 is its conclusion!
If we look at the WMAP data before foreground removal we do not see every galaxy in the universe as implied by your conclusion (just do your calculation for a single galaxy).
What we see is the Milky Way and a relatively small number of point sources. We do not see 100 billion point sources.

ETA: Another huge error in post #52 is that galaxies do not have a temperature of 5K. If you wanted to assign a temperature to a galaxy it would that of its stars which are emitting the light that we see, i.e. ~5000K.

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Originally Posted by TooMany
If that's the way the Q&A forum works here, then it makes no sense to ask a question in the first place because you will just get told to go read some references and do some calculations to figure out the answer yourself.

I have read some references, I have done some calculations!

Can anyone out there help explain this huge error in post #52?
The errors have been explained to you, several times, by several different posters, in several different ways. You've rejected those explanations. It's not the fora's fault if you don't like the answers or explanations.

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Originally Posted by Tensor
The errors have been explained to you, several times, by several different posters, in several different ways. You've rejected those explanations. It's not the fora's fault if you don't like the answers or explanations.
I rejected those explanations because they amount to hand waving (non-quantitative) and no one has bothered to explain or demonstrate how they make a difference of several orders of magnitude. If there is a kindly real astronomer out there who has read this thread, I still want an explanation that makes sense.

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Originally Posted by Reality Check
The huge error in post #52 is its conclusion!
If we look at the WMAP data before foreground removal we do not see every galaxy in the universe as implied by your conclusion (just do your calculation for a single galaxy).
What we see is the Milky Way and a relatively small number of point sources. We do not see 100 billion point sources.

ETA: Another huge error in post #52 is that galaxies do not have a temperature of 5K. If you wanted to assign a temperature to a galaxy it would that of its stars which are emitting the light that we see, i.e. ~5000K.
Sorry "Reality Check" but you can only check reality if you know what your talking about. Stars are tiny and widely spaced, you cannot assign a stellar temperature to a galaxy over it's entire area. If there were such a galaxy we wouldn't need a sun.

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Originally Posted by Reality Check
ETA: Another huge error in post #52 is that galaxies do
not have a temperature of 5K. If you wanted to assign a
temperature to a galaxy it would that of its stars which
are emitting the light that we see, i.e. ~5000K.
In that case TooMany's point would be very greatly
strengthened. Microwave radiation from distant galaxies
would vastly outshine the CMBR.

-- Jeff, in Minneapolis

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TooMany,

As far as I can tell, the only major flaw in your analysis
is in how the light from stars, nebulae, and galaxies is
reduced by distance. That one flaw can explain why the
light is faint enough to be ignored.

-- Jeff, in Minneapolis

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Originally Posted by TooMany
Sorry "Reality Check" but you can only check reality if you know what your talking about. Stars are tiny and widely spaced, you cannot assign a stellar temperature to a galaxy over it's entire area. If there were such a galaxy we wouldn't need a sun.
Sorry "TooMany" but you do not get the point since you need to know what you are talking about.
The point is that galaxies do not have a temperature. Stars, interstellar dust, etc have temperature.
But if you wanted to give them an effective temperature then it could be the temperature of the light that we receive from them and that light comes mainly from the stars.
Another effective temperature is that of Eddington's 1926 estimate which was 3.18 K and very wrong: Eddington's Temperature of Space.

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Originally Posted by TooMany
I rejected those explanations because they amount to hand waving (non-quantitative) and no one has bothered to explain or demonstrate how they make a difference of several orders of magnitude.
Of course, this is according to you. But, as Reality Check has pointed out, the largest proof is that they don't show up in the maps. So, rejecting those explanations appears to be nothing more than you don't like it.

However, you have been given the links to the one, three, five and seven year data releases. Since, according to you, there are all those galaxies out there contaminating the map, you should be able to simply demonstrate where in the maps, from the WMAP probe, those galaxies are, right? After all, according to you, those galaxies are shining brighter than CMB, right? So it should be a simple matter for you to show how your calculations and assumptions clearly and exactly show where those brightly shining galaxies are on the map or invalidate the assumptions and calculations IN THE PAPERS. Feel free to show us.

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This paper has a nice overview of the different statistical methods used to isolate the CMB spectrum from the signals received. Astronomers working on the WMAP data know they cannot hope to precisely find and characterise every source at all distances - however they can take a range of methods and use them to extract their signal without requiring a priori knowledge of the source of every single foreground signal.

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Originally Posted by Shaula
This paper has a nice overview of the different statistical methods used to isolate the CMB spectrum from the signals received. Astronomers working on the WMAP data know they cannot hope to precisely find and characterise every source at all distances - however they can take a range of methods and use them to extract their signal without requiring a priori knowledge of the source of every single foreground signal.
Thanks Shaula. I'll have to read that paper in the morning when my eyelids aren't falling. Primarily it addresses our Galaxy, but I did notice that extragalactic sources (beyond our Galaxy) are mention about 5 times, so there might be info about removal of other galaxies.

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Originally Posted by Jeff Root
TooMany,

As far as I can tell, the only major flaw in your analysis
is in how the light from stars, nebulae, and galaxies is
reduced by distance. That one flaw can explain why the
light is faint enough to be ignored.

-- Jeff, in Minneapolis
Hi Jeff,

I'll try to explain this tomorrow. Basically I agree with you that for a given object, the power received from that object is reduced by distance. The power received by a detector from a given galaxy is proportional to 1/d^2, where d is the distance of the detector from the galaxy.

But that's not the all that needs to be considered I'll finish manjana.

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Primarily it addresses our Galaxy, but I did notice that extragalactic sources (beyond our Galaxy) are mention about 5 times, so there might be info about removal of other galaxies.
It details a range of methods which are agnostic to the actual source of the signal. When they say galactic foreground they usually mean all foreground that is of galactic origin, not just our galaxy. It also has links to papers which have highlighted some of the problems with the methods used and how they are fixed, and some papers which try to do the reverse calculations and remove the CMB to derive a galactic foreground map. It is a good summary and and there is no 'might' about it addressing distant non-resolved sources. It does. That is the beauty of a statistical approach rather than one based on what will always be incomplete observational maps.

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Originally Posted by Shaula
It details a range of methods which are agnostic to the actual source of the signal. When they say galactic foreground they usually mean all foreground that is of galactic origin, not just our galaxy. It also has links to papers which have highlighted some of the problems with the methods used and how they are fixed, and some papers which try to do the reverse calculations and remove the CMB to derive a galactic foreground map. It is a good summary and and there is no 'might' about it addressing distant non-resolved sources. It does. That is the beauty of a statistical approach rather than one based on what will always be incomplete observational maps.
My ears pricked up when I saw this sentence in Section 5 (Quote, emphasis mine):

A random CMB realization was created, starting from the publicly available best-fit cosmological parameters of a ΛCDM model to the combination of five-year WMAP data with supernovae and baryon acoustic oscillations.

However I don't think it is all that serious, because they are just using this synthetic universe to test their other procedures on?

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As the paragraph right before it states - this is just a way to test the methods they are using and get a measure of their retrieval accuracy and any artefacts that they might introduce.

A full simulation of the WMAP instrument was used to test the MCMC and template cleaning methods, in order to investigate the interaction of systematics in both time-domain data and sky maps. Starting with a set of synthetic sky inputs for the CMB and foregrounds (described below), the scanning of the instrument was applied to the inputs to produce a time stream of data, which was then put through the same entire calibration and map-making pipeline as used for real data.

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Originally Posted by TooMany
Read post #52, I don't need these terms in the argument. However, if you think it is fantastically important to my argument anyway, then explain why.
OK.

I ignored the stars, I just took the temperature of dust/gas which is diffuse but has considerable optical depth.

Let's see ...

Your just distracting from the specific issue without really trying to explain what is wrong. I don't wish to do your calculations, I'm asking you to identify what is wrong with mine.
Several specific things have been pointed out, that are wrong with your calculation.

If you'd like to re-do your calculation, fixing the flaws already identified, please do so. Whether these reduce the estimated ~200 GHz emission received from galaxies other than our own, to levels well below the CMB signal, well, it's your calculation ...

To post #52 ...
Originally Posted by TooMany
Yes I really want to know where my reasoning is wrong.

The word intensity unfortunately has many different meanings so to keep things straight, we would have to use terminology very carefully, so let's use units of measure and avoid labels. Planck's law is about the power of radiation emitted by a surface at a given temperature at a given frequency. More specifically, the "intensity" in this case is called the "spectral radiance". It is measured as W m−2·sr−1·Hz−1 when stated in SI units. In other words, it is the power in watts radiated by one square meter of surface at temperature T into one steradian of solid angle (measure of angular area over which the radiation is spread) per Hertz of bandwidth. In order to get the total power emitted per square meter per steradian, you must integrate over the entire spectrum. The surprisingly simple integral is this:

u = a*T^4/pi

The units of u are Watts per square meter (of radiating surface) per steradian (of angular area into which it radiates). The constant a is the Stefan-Boltzmann constant. This power however is not directly relevant as described below.

We are only interested in the radiation in the CMB range within the entire spectrum. Planck's law can give us the peak power per Hertz of a black body at 2.7 K. The peak occurs at approximately 200 GHz. The area under that entire black body spectral curve (total power) is roughly proportional to this peak.

Consider a surface that is warmer than 2.7 K. According to Planck's law, the spectral radiance of the warmer surface is greater than the cooler surface at every frequency in the entire spectrum, including the CMB range. I calculated that at 5 K the spectral radiance is about 6 times that of the CMB at the peak CMB frequency. That means that the spectral power within the CMB range radiated by a surface a 5 K is even more than 6 times the power radiated by the same surface at 2.7 K. (Look at some black body curves at different temps to see that the integral over the CMB frequency range must be even more than 6 times that of the CMB.)
If the 2.7K and 5K surfaces have the same area.

If the 5K surface has an area 6 times smaller than the 2.7K one, then the power radiated by each, at ~200 GHz, will be the same (approximately). Right?

Let's figure out what a detector sees when it looks at a fixed area of surface at some temperature. The detector itself has a fixed sized aperture. At a given distance from the emitting surface, the solid angle of the aperture as viewed from the emitting surface depends inversely on the square of the distance. The power entering the aperature from the surface follows this proportionality:

p ~ S(T,H)/d^2

Where S(T,H) is the total radiant power emitted from the fixed area of surface S at temperature T at frequency H. So the power detected from this fixed area surface declines with the square of the distance d. Now from the point of view of the detector, the radiation is coming from is coming from a surface of fixed area, but the solid angle s (in steradians) occupied by that surface from the detector's viewpoint depends on it's distance d from the detector and follows this proportionality:

s ~ 1/d^2

That is, the solid angle of a fixed area surface declines as the square of it's distance from the detector. Now let's ask how much power the detector receives per steradian occupied by that emitting surface at distance d from the detector. This is just p/s so

p/s ~ (S(T,H)/d^2)/(1/d^2) = S(T,H)

So the power received by a aperture from a surface at temperature T at frequency H only depends on the temperature of the surface and the solid angle occupied by the surface; it does not depend on the distance.
In a static, Euclidean universe.

In our universe the distance-angular size relationship is not quite the same.

Further, there's redshift. If redshift doesn't change the shape of a blackbody SED (merely moves it 'to the right' - you might like to check this), then what is a 5K blackbody in the frame of the emitter will be less than 5K in the frame of the observer. A redshift of 1 will make the 5K emitter look like a 2.5K one, for example.

This is why we have Obler's Paradox in an infinite universe. No matter how far away a star is, the power we receive from it per steradian of the sky area that it occupies is the same (without considering dust etc.).
It's actually a bit more complicated than that (and dust doesn't make any difference to the paradox). Oh, and the guy's name is Olbers, with an s, so it's Olbers' paradox.

Do you know which is more important, in resolving Olbers' paradox, the fact that universe is not infinite, or the fact that it's expanding?

The radiation from the CMB is said to have a temperature of 2.7 K. It can be thought of as a surface at 2.7 T that covers the entire sky. Distance of the CMB matters not all here, only the measured temperature and the fact that it occupies 4*pi steradians matters.

Now what about galaxies? First, overall they are warmer than 2.7 K. I used a conservative estimate of 5 K over the extent of the galaxy.
Several BAUTians have challenged this particular assumption of yours.

RC, for example, pointed to some work Eddington did, showing that if you converted all the EMR (electromagnetic radiation) emitted by all the stars of our galaxy into a blackbody with the same total energy (density), the temperature of said bb would be a mere 3.18K. IIRC, much more recently Narlikar did a similar calculation, using more recent data, and came up with a similar number. In any case, your 'conservative estimate' is clearly wrong.

Perhaps you could revisit your estimate?

A surface at 5 K radiates at least 6 times more power in the CMB spectrum than the CMB does (using Planck's law).
In the rest frame, yes; in the observer frame it does not.

I used the HUDF image to argue for a conservative estimate of 4% sky coverage by galaxies. From that 4% of the sky we receive at least 6 times as much power (over it's solid angle which is 4% of 4*pi) than from the CMB over the same solid angle. Since 4%*6 = 24%, 24% of detected radiation in the CMB spectrum must come from galaxies (conservatively).
Assume 10 billion stars. Assume each star is square (as seen from a distance). Assume the stars comprise a galaxy. Assume the galaxy covers an area of sky 1' by 1'. Assume Euclidean geometry. Assume the galaxy is flat. Assume the stars are evenly spaced, and cover the galaxy with no gaps or overlaps. The stars would then be 0.6 mas (milli-arcsecs) square.

With me so far?

If the stars are, physically, 1 million km on each side, how far away is the galaxy? The back of my small envelope says ~3 x 10^14 km, barely 10 pc! (please check my arithmetic).

At 10 Mpc - just around the corner, cosmologically speaking - 1 million km stars would cover a quite trivial fraction of the 1' x 1' area the galaxy subtends on the sky (1 part per million ... or is it 1 part per 10^12? What do you think?)

Very few sightlines towards such a galaxy would hit a star.

Now I have made my argument more detailed by explaining how surfaces in the sky affect the detector according to their temperature and the solid angle they occupy on the sky. I've shown how radiation received from a given solid angle is not affected by distance, but only by temperature of the source.
And I've now pointed out - again - at least two major flaws in your argument.

Let me summarize the argument: There is a background of radiation from the entire sky (the CMB) that has the temperature 2.7 K. However, about 4% of entire sky is covered with galaxies at a very conservative temperature of 5 K.
Let's revisit this when you've redone your estimate/calculation.

Wherever there is a galaxy in the sky it delivers six times as much power in the CBM range of the spectrum as the CMB delivers from the same angular area (solid angle) in the sky as occupied by the galaxy. Since we receive from 4% of the total solid angle of the sky six times as much power per unit solid angle than we receive from the CMB, in total at least 24% of the power received in the CMB spectral range is from galaxies.

Or look at it this way, the blackness of space is very cold, it has a temperature of 2.7 K, the part covered by galaxies is warmer and occupies 4% of the sky. Because galaxies are warmer (5 K) we receive 6 times as much power from them than the from the same 4% of the blackness. Thus it seems almost intuitive that galaxies must be removed to find the true CMB.
Due to a major flaw earlier in the logic chain, the conclusion has an unknown truth value.

Sorry, no can do.

21. To cut to the chase - the Planck blackbody function specifies not only the spectral shape but its surface brightness. (It is thus nontrivial that the CMB has the appropriate surface brightness for its temperature, something invariant to redshift because multiplying wavelength in the equation by (1+z) gives a new blackbody with temperature lower by that factor). Something with a blackbody spectral shape but lower surface brightness (such as the average of a galaxy taken crudely) is a graybody. A galaxy with, say, a 20 K dust component has surface brightness orders of magnitude lower than an equivalent blackbody, diluted by the covering fraction of optically thick material at the relevant wavelength. To take Nereid's example in an earlier post, if we approximate the Milky Way as 100 billion sunlike stars, their collective cross-section is about 1011 x (106 km)2 = 1023 km2 spread over (100,000 light-years)2 = 1036 km2, so the collective spectrum would be an approximate 5500 K blackbody distribution, but with surface brightness lower by a factor 1013 than predicted by the Planck blackbody function.

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Originally Posted by ngc3314
To cut to the chase - the Planck blackbody function specifies not only the spectral shape but its surface brightness. (It is thus nontrivial that the CMB has the appropriate surface brightness for its temperature, something invariant to redshift because multiplying wavelength in the equation by (1+z) gives a new blackbody with temperature lower by that factor). Something with a blackbody spectral shape but lower surface brightness (such as the average of a galaxy taken crudely) is a graybody. A galaxy with, say, a 20 K dust component has surface brightness orders of magnitude lower than an equivalent blackbody, diluted by the covering fraction of optically thick material at the relevant wavelength. To take Nereid's example in an earlier post, if we approximate the Milky Way as 100 billion sunlike stars, their collective cross-section is about 1011 x (106 km)2 = 1023 km2 spread over (100,000 light-years)2 = 1036 km2, so the collective spectrum would be an approximate 5500 K blackbody distribution, but with surface brightness lower by a factor 1013 than predicted by the Planck blackbody function.
OK that is beginning to make sense to me. I did check that the measured CMB radiation is in fact the same as a solid black body at 2.7 K and not just similar in shape to such a black body.

I believe your calculation that overall surface brightness of a galaxy from stars is a factor of 10^13 less than the actual surface brightness of a single star. This is fairly easy to understand because stars are so tiny in comparison to the distance between them. The coverage of the galaxy by stars is ultra sparse.

The shape of the SED of a galaxy in the CMD spectral area is fairly similar to the shape of a black body at 20 K (I think). (I'm not sure how much energy is contributed by stars in this part of the spectrum versus the dust and gas that may actually have an average temperature of about 20 K.)

Here's what I still don't understand and perhaps you can explain it just a bit more:

A galaxy with, say, a 20 K dust component has surface brightness orders of magnitude lower than an equivalent blackbody, diluted by the covering fraction of optically thick material at the relevant wavelength.
When I look at galactic images, the dust appears to be fairly opaque. I was assuming that this opacity was an indication that it radiates nearly like solid surface. It certainly is somewhat transparent but it looks substantial, like black smoke. This made me think that it could be treated as a solid black body, or at least as some substantial percentage of a black body radiator.

But this may be my huge error concerning the dust! The dust may be close to a blackbody at optical/NIR wavelengths, but it's almost completely transparent at CMB wavelengths so it has virtually no optical depth at these wavelengths. Is that it?

You have really cut to the chase (for me)! I apologize to Nereid because Nereid was saying that galaxies are nothing like blackbodies. Their SEDs are similar to the shape of a bb, but they 12 or 13 orders or magnitude "grayer" than a black body at CMB wavelenghts.

This leads me to one additional question. How the heck can you even measure the dust emissions in the CMB range if the CMB outshines it by 10^13. I have seen papers that talk about measuring it.

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Originally Posted by Reality Check
Sorry "TooMany" but you do not get the point since you need to know what you are talking about.
The point is that galaxies do not have a temperature. Stars, interstellar dust, etc have temperature.
But if you wanted to give them an effective temperature then it could be the temperature of the light that we receive from them and that light comes mainly from the stars.
Another effective temperature is that of Eddington's 1926 estimate which was 3.18 K and very wrong: Eddington's Temperature of Space.
Eddington was not wrong about what he actually calculated and stated. He estimated the energy density in deep space of starlight. He then calculated that this energy density was equivalent to the energy density of a blackbody sky at a temperature of 3.18 K. It turned out much later (which Wright states is a coincidence) that the energy density of the CMB is close to average energy density of stars . You cannot interpret the stellar energy density as the CMB directly however, because the spectrum is entirely different.

24. Order of Kilopi
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Originally Posted by Jeff Root
TooMany,

As far as I can tell, the only major flaw in your analysis
is in how the light from stars, nebulae, and galaxies is
reduced by distance. That one flaw can explain why the
light is faint enough to be ignored.

-- Jeff, in Minneapolis
May I ask, in light of the recent posts in this thread, what do you now consider the major flaw(s) to be?

In particular, what do you consider the role of "surface brightness" in any flaw to be?

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Originally Posted by Shaula
When they say galactic foreground they usually mean all foreground that is of galactic origin, not just our galaxy.
I don't think that is true. I checked all instances of "galactic" in the paper. They are all capitalized as "Galactic". This is done deliberately to designate our galaxy.
There is mention of "extragalactic" in the paper, but I haven't really understood those yet.
Last edited by TooMany; 2012-Jun-07 at 10:55 PM. Reason: Fix quote tag

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Nereid, I understand you are trying to help me resolve this issue, but I'm still a little puzzled. I certainly understand that one cannot say that a galaxy is a blackbody shining with stars at 5000 K. The intensity of stars is greatly diluted over the area of the galaxy because stars are so very tiny in comparison. So yes the overall surface brightness is tiny compared to that of a star. I'll try to figure out how much stars in a galaxy dump into the CMB spectrum.

Now the dust is another thing. Dust seems to extend over a large area in some galaxies and has a typical temperature of 20 K. But the output of this dust in the CMB spectrum must be about a million times less than that of a blackbody at 2.7 K in order to be negligible in CMB differential measurements. Why is surface brightness of the dust so extremely small even though it does cover substantial area of the galaxy (unlike the stars)? The dust appears to be nearly opaque in the optical range.

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Dust in our galaxy seems to have a surface brightness that is equivalent in effect to about 20 micro K (at least in this reference). Using my guess of 4% sky coverage by other galaxies, the contamination from dust could amount to fluctuations of about 1 micro K in the measured CMB. This is roughly the magnitude of the SZ effect for clusters.

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You are wrong. You continue to ignore the simple fact that the intenstity varies as 1/r^2. So the contamination from the dust in a normal galaxy will be lower than that of the Milky Way.

29. Three pages. Time to move to S&T.

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Originally Posted by TooMany
I don't think that is true. I checked all instances of "galactic" in the paper. They are all capitalized as "Galactic". This is done deliberately to designate our galaxy.
There is mention of "extragalactic" in the paper, but I haven't really understood those yet.
The paper is about improving the removal of all sources of foreground contamination including extragalactic point sources:
We provide an updated map of the cosmic microwave background (CMB) using the internal linear combination method, updated foreground masks, and updates to point source catalogs using two different techniques. With additional years of data, we now detect 471 point sources using a five-band technique and 417 sources using a three-band CMB-free technique. In total there are 62 newly detected point sources, a 12% increase over the five-year release.
...
These expanded masks are then combined with the point source mask as in previous releases, which has been updated with newly detected sources.
I agree that the capitalization of Galactic refers to the Milky Way.

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