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peaceharris
2007-Jan-17, 05:12 AM
This website (http://www.phy.duke.edu/~kolena/cmb.htm)claims:

"At some point about 400,000 years after the Bang, the universe had cooled to the point where the matter became neutral, at which point the universe's matter also became transparent to the radiation. (Completely ionized matter can absorb any wavelength radiation; neutral matter can only absorb the relatively few wavelengths that carry the exact energy that match energy differences between electron energy levels.) The temperature at which this transition from ionized to neutral (called the "moment of decoupling") occurred was roughly 3000 K."


What is the evidence to prove that the CMB originates from matter that was at 3000K?

If the CMB originates from hot matter, wouldn't we see transitions from one level to the other? For example, in the spectrum of our sun (http://homepages.wmich.edu/%7Ekorista/sun-images/solar_specbb.jpg), we see hydrogen atoms in the n=2 state absorbing energy and going to the n=3 state. This is the dip at 656nm. So if the CMB originates from hot matter and is redshifted by a factor of 1100, shouldn't we also see all the transitions shifted by the same factor?

If the CMB originates from cold matter, all the atoms will be at the ground state, and we wouldn't see any transitions.

The spectrum of the CMB is a perfect blackbody radiation curve without any of these transitional features.

Have cosmologists verified whether neutral hydrogen at 3000K would produce a perfect blackbody radiation without transitional features?

Ken G
2007-Jan-17, 01:08 PM
Have cosmologists verified whether neutral hydrogen at 3000K would produce a perfect blackbody radiation without transitional features?

Yes, they have. Tiny features due to the transitions you mention (more Lyman lines than Balmer lines) have been calculated, and would be too small to observe. The basic problem is, thermodynamic equilibrium always produces a Planck spectrum, no matter what the lines and continuum opacity is doing. It is simply the most likely distribution of photon energies, if you "shuffle" it enough keeping to a local energy constraint.

Spaceman Spiff
2007-Jan-17, 02:49 PM
This website (http://www.phy.duke.edu/~kolena/cmb.htm)claims:

"At some point about 400,000 years after the Bang, the universe had cooled to the point where the matter became neutral, at which point the universe's matter also became transparent to the radiation. (Completely ionized matter can absorb any wavelength radiation; neutral matter can only absorb the relatively few wavelengths that carry the exact energy that match energy differences between electron energy levels.) The temperature at which this transition from ionized to neutral (called the "moment of decoupling") occurred was roughly 3000 K."



I'd like to point out some "badness" on that part of that information - probably just a matter of using words loosely to keep things simple. Free electrons do not absorb photons; they scatter them (to thus conserve energy and momentum of the system). At low photon energies (compared to the electron rest mass), the scattering is virtually elastic in the frame of the electron (thermal motions of the electrons can smear out the energies of the scattered photons). A 'completely' ionized gas can absorb photons via the free-free process. Neutral H and He opacity comes in many forms. Bound-bound (line) transitions mentioned by the paragraph is just one type. Photoionization (ionizing bound electrons via energetic photons) from ground and excited states is also possible. With far lower cross section is 'Rayleigh' scattering. The gas became neutral as collisional ionization and then photoionization from ground states (mostly) and excited states became negligible as the temperature (and so photon energies and particle kinetic energies) dropped. The primary source of photon opacity in the period near the recombination epoch was free electron scattering. Once the vast majority of these free electrons had attached themselves to nuclei, the opacity dropped to ~zilch for the radiation field present at that time - a thermal blackbody at ~3000K.

Nereid
2007-Jan-17, 03:53 PM
One interesting follow-up question: why did the electrons and protons combine to form neutral hydrogen atoms at ~3000K (and not any other temperature)?

Spaceman Spiff
2007-Jan-17, 04:30 PM
One interesting follow-up question: why did the electrons and protons combine to form neutral hydrogen atoms at ~3000K (and not any other temperature)?

Even in ionized gas, electrons are recombining with nuclei (being that they attract one another) and then being re-ionized. However, the equilibrium ratio of ionized to neutral depends on things like the ratio of thermal energy to ionization potential energy, and potentially other environmental variables. As the temperature drops, the neutral fraction rises. It's nature seeking to minimize its energy ("running downhill"), as usual.

Ken G
2007-Jan-17, 08:43 PM
An interesting followup to that is the issue of density. It actually turns out that you can ionize neutral hydrogen at arbitrarily low temperatures if you simply make the density low enough and wait long enough. In the Big Bang, you have a competition-- dropping temperature and falling density, so they are fighting each other: the lower temperature wants to make neutral hydrogen form while the lower density wants it to ionize. You have to actually look at the rates to find out that the temperature effect wins, and it just turns out that the crossover happens at 3000 K in the model. Stellar atmospheres have higher density than this phase of the Big Bang, so they become neutral at rather higher temperatures around 5000-6000 K, like the Sun.

Spaceman Spiff
2007-Jan-17, 10:28 PM
...and potentially other environmental variables...

(such as density) :). Thanks for elaborating.

Ken G
2007-Jan-17, 11:28 PM
I have to correct something I had backwards in our earlier discussion. I said that photons cool faster than nonrelativistic matter when they are decoupled, but that's not right, it is photon's energy density that drops faster than matter, but not their temperature. (It's because of the rest mass of the matter that keeps the matter energy higher.) As I see it, the temperature depends only on the kinetic energy, and that actually drops slower for photons than for nonrelativistic matter. So what this means is, while photons are coupled to the gas temperature prior to recombination, the photons are being forced to cool at the faster rate of the adiabatic gas expansion, than they would cool if left to themselves. If left to themselves they'd conserve photon number, but as they are cooling faster than that, they are losing photon number! And quite substantially, prior to recombination. So even though photon scattering off electrons is not an "absorption" process, and that is mostly what is happening while electrons are free, it still must be true that considerable true absorption is going on during that time (free-free absorption presumably), because you have to get rid of the vast majority of the photons prior to releasing the CMB into free space.

This also means that after decoupling to the gas, the photons cool at a slower rate than adiabatic gas expansion, so the cosmological gas temperature must by now be much lower than 2.7 K. Given all the gravitational heating of gas going on, this is pretty meaningless, but it's interesting nevertheless.

peaceharris
2007-Jan-18, 12:07 PM
Yes, they have. Tiny features due to the transitions you mention (more Lyman lines than Balmer lines) have been calculated, and would be too small to observe. The basic problem is, thermodynamic equilibrium always produces a Planck spectrum, no matter what the lines and continuum opacity is doing. It is simply the most likely distribution of photon energies, if you "shuffle" it enough keeping to a local energy constraint.

If you claim that tiny features are expected, but not observed, then we can claim that the CMB is atleast in approximate thermodynamic equilibrium. So let me use the word 'approximate thermodynamic equilibrium'

If you had hydrogen at 3000K in the universe, would it be in approximate thermodynamic equilibrium with its surrounding? How do cosmologists verify this?

We see absorbtion lines in the sun because the sun's atmosphere is at a cooler temperature. Only if the sun's atmosphere is at the same temperature as deeper inside, only then we would not see any absorption lines.

Are there Lyman alpha forests receding with z=1000, z=900 etc in between the CMB and an observer here on earth? If yes, are these forests also in approximate thermodynamic equilibrium with the CMB such that we don't see any absorption lines?

Do you know the reason why cosmologists do not believe that this radiation is from cold matter? Do you know the reason why cosmologists do not accept CMBR as a verification of Eddington's temperature of space?

Ken G
2007-Jan-19, 02:44 AM
If you had hydrogen at 3000K in the universe, would it be in approximate thermodynamic equilibrium with its surrounding? How do cosmologists verify this?We understand the equilibration rates, and how much time was available in the Big Bang model, and it comes out with the clear expectation that is was in near equilibrium. Then the observations show a nearly perfect equilibrium CMB spectrum, so all is well.


We see absorbtion lines in the sun because the sun's atmosphere is at a cooler temperature. Only if the sun's atmosphere is at the same temperature as deeper inside, only then we would not see any absorption lines.
That's not always correct, because there is an effect called scattering that would produce absorption lines anyway, but at high enough density, you are correct.


Are there Lyman alpha forests receding with z=1000, z=900 etc in between the CMB and an observer here on earth? If yes, are these forests also in approximate thermodynamic equilibrium with the CMB such that we don't see any absorption lines?They are not in thermodynamic equilibrium because they are scattering lines, but they are illuminated from all sides and are in what's called radiative equilibrium, so the answer is more or less the same-- no lines.



Do you know the reason why cosmologists do not believe that this radiation is from cold matter? Do you know the reason why cosmologists do not accept CMBR as a verification of Eddington's temperature of space?Are you asking me why cosmologists think the CMB is redshifted? Where should I begin?

Amber Robot
2007-Jan-19, 03:10 AM
Are there Lyman alpha forests receding with z=1000, z=900 etc in between the CMB and an observer here on earth? If yes, are these forests also in approximate thermodynamic equilibrium with the CMB such that we don't see any absorption lines?

Well, the spectrum of a 3000K blackbody has essentially zero flux at 1216 Angstroms, so no there is no z=900-1000 lyman alpha forest in the CMB.

Ken G
2007-Jan-19, 03:42 AM
That is so, and there would also be little signal in lines if they were lower energy because it is radiative equilibrium. Put the two together, and it's really a tiny signal to look for.

peaceharris
2007-Jan-19, 06:00 AM
Well, the spectrum of a 3000K blackbody has essentially zero flux at 1216 Angstroms, so no there is no z=900-1000 lyman alpha forest in the CMB.

I overlooked that.


Are you asking me why cosmologists think the CMB is redshifted? Where should I begin?

Yeah. What evidence is there to prove that the CMB is redshifted radiation and not radiation from cold matter? Is there any reason to assume that particles in space should not be at 2.7K, considering the fact that Eddington predicted a value very close?

You can begin with telling me what's the strongest evidence in favor of the current interpretation of the CMBR over the interpretation that the CMBR is a local effect caused by cold matter in space.


We understand the equilibration rates, and how much time was available in the Big Bang model, and it comes out with the clear expectation that is was in near equilibrium.

In a model, are there not many free variables that we can play with? None of these are observable, in the sense you are free to assume the temperature, density and distribution of matter. Matter in today's universe is not in equilibrium, but you have the right to assume that matter in the universe billions of years ago was.

Eddington did not have any free variables to play with.

Let's say you have 5 points that fall on a straight line. You can say that it is a linear equation. But we also can find a higher order polynomial that fits these points, since we have more coefficients to play with. If a straight line can explain the 5 points, should we not accept the simpler model?

Spaceman Spiff
2007-Jan-19, 06:30 AM
I overlooked that.



Yeah. What evidence is there to prove that the CMB is redshifted radiation and not radiation from cold matter? Is there any reason to assume that particles in space should not be at 2.7K, considering the fact that Eddington predicted a value very close?

You can begin with telling me what's the strongest evidence in favor of the current interpretation of the CMBR over the interpretation that the CMBR is a local effect caused by cold matter in space.



In a model, are there not many free variables that we can play with? None of these are observable, in the sense you are free to assume the temperature, density and distribution of matter. Matter in today's universe is not in equilibrium, but you have the right to assume that matter in the universe billions of years ago was.

Eddington did not have any free variables to play with.

Let's say you have 5 points that fall on a straight line. You can say that it is a linear equation. But we also can find a higher order polynomial that fits these points, since we have more coefficients to play with. If a straight line can explain the 5 points, should we not accept the simpler model?

Try here (http://www.astro.ucla.edu/%7Ewright/Eddington-T0.html), here (http://www.astro.ucla.edu/%7Ewright/tiredlit.htm), and here (http://www.astro.ucla.edu/%7Ewright/stars_vs_cmb.html).

Whatever else you say the cmb is, it's gotta be an opaque blackbody emitter that is everywhere isothermal to about 1 part in 100,000. This sums it up well: "The local Universe is transparent and has a wide range of temperatures, so it does not produce a blackbody, which requires an isothermal absorbing situation."

In a nutshell, this is a non-starter.

Ken G
2007-Jan-19, 06:40 AM
Yeah. What evidence is there to prove that the CMB is redshifted radiation and not radiation from cold matter? Is there any reason to assume that particles in space should not be at 2.7K, considering the fact that Eddington predicted a value very close?
Tons and tons. The Eddington calculation is not to be taken seriously, there are many problems with it. To name a few, galactic starlight is not going to be a perfect blackbody, the 3 K result is merely an equivalent temperature. It only applies if there is no dust absorption, and it does not come from all directions. It's a totally bogus comparison to the CMB.



You can begin with telling me what's the strongest evidence in favor of the current interpretation of the CMBR over the interpretation that the CMBR is a local effect caused by cold matter in space.Covered.



In a model, are there not many free variables that we can play with? None of these are observable, in the sense you are free to assume the temperature, density and distribution of matter. Matter in today's universe is not in equilibrium, but you have the right to assume that matter in the universe billions of years ago was.Is there a question there? The model is based on valid physical assumptions, and works about as well as any model in the history of science, with the one caveat that we need to understand dark matter to understand the universe's past, and dark energy to understand its future. But at least this proves that the Big Bang is not a model with unlimited adjustable parameters, or we wouldn't be looking for either of those physical effects.


Eddington did not have any free variables to play with.


Eddington himself would never have used that calculation in comparison to CMB observations, for all the reasons I've given.

peaceharris
2007-Jan-19, 09:55 AM
Tons and tons. The Eddington calculation is not to be taken seriously, there are many problems with it. To name a few, galactic starlight is not going to be a perfect blackbody, the 3 K result is merely an equivalent temperature. It only applies if there is no dust absorption, and it does not come from all directions. It's a totally bogus comparison to the CMB.


What do you mean by 'It only applies if there is no dust absorption'?

Dust will get heat as it absorbs radiation from all its surrounding stars, and radiate it outward. As some temperature, the heat loss due to dust radiating heat outward will be the same as the heat it gets from the surrounding stars.

Eventhough starlight is not a perfect blackbody, does that mean particles in space also are not a perfect blackbody? What kind of radiation will you get from particles that radiate out all the heat that it gets?

Why do you say that radiation from space cannot come from all directions. I think what you say would only be true, if in some direction, there isn't any dust particles.

Nereid
2007-Jan-19, 01:26 PM
[snip]

Eventhough starlight is not a perfect blackbody, does that mean particles in space also are not a perfect blackbody? What kind of radiation will you get from particles that radiate out all the heat that it gets?

[snip]Indeed, warm dust does radiatiate much like a blackbody, as do warm rocks, warm mountains, warm asteroids, warm ....

However, none of them radiate like a blackbody to 1 part in 100,000 (as Spaceman Spiff noted above).

Results from ISO (http://www.iso.vilspa.esa.es/) and Spitzer (http://www.spitzer.caltech.edu/) (to name just two IR missions) have given us some good insights into what warm dust (etc) looks like, in the nearby universe, and the deviations of the observed SEDs (spectral energy distribution) of such dust from that of a blackbody is typically in the few% range - such deviations would be many hundreds, or thousands, of times the 1 sigma error bars of measurements of the CMB's SED.

Spaceman Spiff
2007-Jan-19, 04:05 PM
What do you mean by 'It only applies if there is no dust absorption'?

Dust will get heat as it absorbs radiation from all its surrounding stars, and radiate it outward. As some temperature, the heat loss due to dust radiating heat outward will be the same as the heat it gets from the surrounding stars.

Eventhough starlight is not a perfect blackbody, does that mean particles in space also are not a perfect blackbody? What kind of radiation will you get from particles that radiate out all the heat that it gets?

Why do you say that radiation from space cannot come from all directions. I think what you say would only be true, if in some direction, there isn't any dust particles.

So all the dust grains must each radiate as a mathematical blackbody (Planck function) to roughly a few parts in 100,000 AND they must all be at the same temperature to a similar precision because a sum of blackbodies is not a Planck function unless they all are at the same temperature.

Next - typical grains do not emit precisely as a Planck spectrum - those who have made similar claims have had to resort to contrived large "iron whiskers". And then how is it that they are all at the same temperature throughout the universe and distributed so homogeneously throughout the universe?

In other words, there are extremely strong quantified observational constraints on the properties of the cmb. Have you read the links I supplied you?

As I said, this is a non-starter.

Ken G
2007-Jan-19, 10:45 PM
What do you mean by 'It only applies if there is no dust absorption'?Our galaxy is a disk. That means that if you have dust absorption of starlight, the re-radiation will be predominantly out of the plane. It's the same reason that a glow-in-the-dark frisbee will look brighter from the top than from the side. Eddington did not include that important effect.



Dust will get heat as it absorbs radiation from all its surrounding stars, and radiate it outward. As some temperature, the heat loss due to dust radiating heat outward will be the same as the heat it gets from the surrounding stars.
I'm well aware of that, see my answer above.


Eventhough starlight is not a perfect blackbody, does that mean particles in space also are not a perfect blackbody? What kind of radiation will you get from particles that radiate out all the heat that it gets?You are saying that dust can convert starlight into a blackbody temperature, and that is true, locally and to some imperfect degree (the wavelength of 2.7 K radiation is larger than dust), but a much larger problem is that the CMB is isotropic. We do see IR radiation from dust, and not surprisingly, it traces the galactic disk. This is a complete "non-starter", indeed. The interactions of light and matter are among the best understood physical processes in the universe. It's when matter doesn't interact with light that we know much less about it!



Why do you say that radiation from space cannot come from all directions. I think what you say would only be true, if in some direction, there isn't any dust particles.
Yes, and that's just the case. The galaxy is not optically thick in dust in all directions, it's a disk of dust. Were that not true, we wouldn't even see other galaxies! Check out Andromeda some time.

peaceharris
2007-Jan-20, 01:53 AM
... Next - typical grains do not emit precisely as a Planck spectrum - those who have made similar claims have had to resort to contrived large "iron whiskers". And then how is it that they are all at the same temperature throughout the universe and distributed so homogeneously throughout the universe?


... but a much larger problem is that the CMB is isotropic. We do see IR radiation from dust, and not surprisingly, it traces the galactic disk

If you superpose IR radiation from dust that is hot and and microwave radiation from dust that is cold, and observe the radiation using an instrument that observes microwave wavelengths, will we see 2 planck curves or just 1?

For example, if there are 100 hot dust particles from the Large Magellanic Cloud, and 100 cold dust particles at 3K in a region 10 light years away, what will an instrument that is capable of receiving wavelengths of the order of 0.1cm register? If the cold dust particles at 3K are all around in a radius of 100 light years, and the hot particles are much further away what will you observe using an instrument that is capable of receiving wavelengths of the order of 0.1cm?



Results from ISO and Spitzer (to name just two IR missions) have given us some good insights into what warm dust (etc) looks like, in the nearby universe, and the deviations of the observed SEDs (spectral energy distribution) of such dust from that of a blackbody is typically in the few% range - such deviations would be many hundreds, or thousands, of times the 1 sigma error bars of measurements of the CMB's SED.

If you observe hydrogen gas at 3000K, you will observe something that is much much much further away from an ideal blackbody. People who claim that the CMB is due to hot hydrogen are assuming equilibrium conditions just after the hypothesized big bang. If you want an 'apple to apple' comparison, try releasing hydrogen gas at 3000K in the universe and verify whether it is in thermal equilibrium with its surrounding. Please don't compare the non-Planck curves we observe with the fictitious Planck curves that is imagined. The universe as we observe, is not in equilibrium.

Secondly, using an IR telescope, we cannot detect a perfect blackbody from hot dust far away since cold dust close to us will absorb the IR radiation from the hot dust far away. Definitely hot dust far away is not in thermal equilibrium with cold dust near to us. Again it's not an 'apple to apple' comparison since you are comparing hot dust that is not in equilibrium with cold dust that may be in equilibrium.

The question is: Is the cold dust close to us in thermal equilibrium with its surrounding?



Have you read the links I supplied you?

Yes, but I disagree and don't understand the views in those links. To discuss the views in those links, please ask the author of that website to join this forum. Otherwise, let's just discuss the views that members of this forum subscribe to.


Our galaxy is a disk. That means that if you have dust absorption of starlight, the re-radiation will be predominantly out of the plane. It's the same reason that a glow-in-the-dark frisbee will look brighter from the top than from the side. Eddington did not include that important effect.

Are you saying that matter on side of of the disk closer to the center of the milky way is hotter than matter on the other side? If we are just considering dust particles within a radius of 100 light years, can we neglect the effect caused by certains regions of space which have a higher density of stars?

Ken G
2007-Jan-20, 03:35 AM
If you superpose IR radiation from dust that is hot and and microwave radiation from dust that is cold, and observe the radiation using an instrument that observes microwave wavelengths, will we see 2 planck curves or just 1?You'll see any number of Planck curves, depending on the number of temperatures present, but the short answer is, yes.

If the cold dust particles at 3K are all around in a radius of 100 light years, and the hot particles are much further away what will you observe using an instrument that is capable of receiving wavelengths of the order of 0.1cm?This is a very simple calculation to determine, but it depends quite sensitively on a property known as optical depth. How "optically thick" are these dust contributions? You need to know that. If they are optically thin (and inside 100 LY or away from the disk, they are), then you do not see a Planck spectrum even from that dust (forget the other dust)-- at the very least you'd see what's known as a diluted Planck spectrum. That's not what the CMB is.





If you observe hydrogen gas at 3000K, you will observe something that is much much much further away from an ideal blackbody. And do you know any physics to make this claim, or do you feel you should simply be allowed to have an opinion on this, like it was a flavor of jelly bean?


People who claim that the CMB is due to hot hydrogen are assuming equilibrium conditions just after the hypothesized big bang. And here's the funny thing about that-- they do know some physics, it's not just their "opinion" on this.


If you want an 'apple to apple' comparison, try releasing hydrogen gas at 3000K in the universe and verify whether it is in thermal equilibrium with its surrounding.You will need to also specify a density and a timescale. And this will be one of the easiest calculations in physics to determine whether or not it will be in thermal equilibrium. Physics has been applied to astronomy successfully once or twice, you know.


The universe as we observe, is not in equilibrium.News flash: the density is a lot lower now, and the conditions far less homogeneous, than what we observe in the CMB. Hey-- the universe has evolved!


Secondly, using an IR telescope, we cannot detect a perfect blackbody from hot dust far away since cold dust close to us will absorb the IR radiation from the hot dust far away. Assuming it's optically thick. Again, look at Andromeda sometime-- the dust isn't optically thick. Are you planning on continuing to ignore this point? Is it that you don't want the truth?

If we are just considering dust particles within a radius of 100 light years, can we neglect the effect caused by certains regions of space which have a higher density of stars?I'm not sure what you are asking, but again I am sure that you need to understand the concept of optical depth. It's the difference between a window and a wall, and in this case, between your window to enlightenment and a wall of misconception.

Spaceman Spiff
2007-Jan-20, 03:58 AM
So here we are again, debating an ATM with someone who already has the answers to his own question but isn't interested in understanding the physics. I thought this was the Q&A forum...(enter forum moderator...?)

IMO, it's time to move on.

Nereid
2007-Jan-20, 10:38 AM
Indeed.

peaceharris, the Q&A section of BAUT is not intended to be forum for presenting and arguing an idea which is against the mainstream, nor for presentation of your own beefs with mainstream astronomy, astrophysics, cosmology, or space science.

BAUT has a section set up specifically for that - Against The Mainstream (http://www.bautforum.com/forumdisplay.php?f=17) (a.k.a. ATM).

If you'd like, I will move this thread to that section.

Or you may start a new one in that section.

If you still have questions about the CMB, please ask them.

peaceharris
2007-Jan-20, 11:48 AM
You are saying that dust can convert starlight into a blackbody temperature, and that is true, locally and to some imperfect degree (the wavelength of 2.7 K radiation is larger than dust)

If you go through the derivation of Planck's blackbody curve, you will realize that the size of dust doesn't play any role.


This is a very simple calculation to determine, but it depends quite sensitively on a property known as optical depth. How "optically thick" are these dust contributions? You need to know that. If they are optically thin (and inside 100 LY or away from the disk, they are), then you do not see a Planck spectrum even from that dust (forget the other dust)-- at the very least you'd see what's known as a diluted Planck spectrum.

Again, if you go through the derivation of Planck's blackbody curve, you will realize that the density of dust doesn't change the blackbody curve.


BAUT has a section set up specifically for that - Against The Mainstream (a.k.a. ATM).

If you'd like, I will move this thread to that section.

Or you may start a new one in that section.

OK, let's stop discussing what causes the CMB here. Maybe later we can continue this discussion at ATM. Can we just finish discussing whether or not size and density play a role in the determination of Planck's curve?

Discussing Planck's curve is not 'against the mainstream', since the textbooks which I have don't claim that size and density can change the blackbody curve and produce what Ken call's 'a diluted Planck spectrum'


Assuming it's optically thick. Again, look at Andromeda sometime-- the dust isn't optically thick. Are you planning on continuing to ignore this point? Is it that you don't want the truth?

Dust that is optically will absorb much more radiation and reach the temperature of space. Dust that is optically thin, need to absorb much less radiation to reach the temperature of space. Everyone knows that 2kg of water needs more heat to boil than 1kg of water.

Optically thin dust will absorb IR radiation, but much less than optically thick dust. I do know that dust locally is optically thin. Is optically thin dust capable of producing a blackbody curve?

Nereid
2007-Jan-20, 12:34 PM
[snip]
BAUT has a section set up specifically for that - Against The Mainstream (a.k.a. ATM).

If you'd like, I will move this thread to that section.

Or you may start a new one in that section.OK, let's stop discussing what causes the CMB here. Maybe later we can continue this discussion at ATM. Can we just finish discussing whether or not size and density play a role in the determination of Planck's curve?

Discussing Planck's curve is not 'against the mainstream', since the textbooks which I have don't claim that size and density can change the blackbody curve and produce what Ken call's 'a diluted Planck spectrum'Sure.

Assuming it's optically thick. Again, look at Andromeda sometime-- the dust isn't optically thick. Are you planning on continuing to ignore this point? Is it that you don't want the truth?Dust that is optically will absorb much more radiation and reach the temperature of space. Dust that is optically thin, need to absorb much less radiation to reach the temperature of space. Everyone knows that 2kg of water needs more heat to boil than 1kg of water.

Optically thin dust will absorb IR radiation, but much less than optically thick dust. I do know that dust locally is optically thin. Is optically thin dust capable of producing a blackbody curve?I think you're missing a word - 'thin'? - between the two words I bolded.

It seems you are mixing up several different factors ... optical depth depends upon many things - size distribution of the absorbing particles, wavelength (when it comes to absorption, there are several different wavelength dependencies), total path length through the absorbing material, ... And the equilibrium temperature of any absorbers in space depends on their location (among other things) - the dust which produces zodiacal light is considerably warmer than that in the local ISM, which in turn would be warmer than any in the depths of inter-cluster voids.

In any case, real dust doesn't radiate as a blackbody (a point I made earlier), although the deviations may be relatively small (a few percent).

peaceharris
2007-Jan-20, 02:17 PM
Sure.I think you're missing a word - 'thin'? - between the two words I bolded.

It seems you are mixing up several different factors ... optical depth depends upon many things - size distribution of the absorbing particles, wavelength (when it comes to absorption, there are several different wavelength dependencies), total path length through the absorbing material, ... And the equilibrium temperature of any absorbers in space depends on their location (among other things) - the dust which produces zodiacal light is considerably warmer than that in the local ISM, which in turn would be warmer than any in the depths of inter-cluster voids.

In any case, real dust doesn't radiate as a blackbody (a point I made earlier), although the deviations may be relatively small (a few percent).


Yeah, I thought that optically thick meant high density. I will have to edit that post. Sorry.

peaceharris
2007-Jan-20, 02:50 PM
Please ignore the earlier post: This is the corrected version.



You are saying that dust can convert starlight into a blackbody temperature, and that is true, locally and to some imperfect degree (the wavelength of 2.7 K radiation is larger than dust)

If you go through the derivation of Planck's blackbody curve, you will realize that the size of dust doesn't play any role.


This is a very simple calculation to determine, but it depends quite sensitively on a property known as optical depth. How "optically thick" are these dust contributions? You need to know that. If they are optically thin (and inside 100 LY or away from the disk, they are), then you do not see a Planck spectrum even from that dust (forget the other dust)-- at the very least you'd see what's known as a diluted Planck spectrum.

Again, if you go through the derivation of Planck's blackbody curve, you will realize that the optical thickness of dust doesn't change the blackbody curve.


BAUT has a section set up specifically for that - Against The Mainstream (a.k.a. ATM).

If you'd like, I will move this thread to that section.

Or you may start a new one in that section.

OK, let's stop discussing what causes the CMB here. Maybe later we can continue this discussion at ATM. Can we just finish discussing whether or not size and density play a role in the determination of Planck's curve?

Discussing Planck's curve is not 'against the mainstream', since the textbooks which I have don't claim that size and density can change the blackbody curve and produce what Ken call's 'a diluted Planck spectrum'


Assuming it's optically thick. Again, look at Andromeda sometime-- the dust isn't optically thick. Are you planning on continuing to ignore this point? Is it that you don't want the truth?

So you are saying, other than absorption by cold dust near us, there are other reasons for an imperfect blackbody curve from dust? Would I be right in saying that the main reason it doesn't produce a perfect blackbody is because it is not in equilibrium with its surrounding? Perhaps the dust is continuously expanding because it keeps getting heat from the stars inside. So the heat received is used to expand the gas. As far as I know, the condition for blackbody radiation is heat received should be equal to heat radiated. So again its not an apple to apple comparison. You are comparing hot particles that are expanding against cold particles that have reached an equilibrium state.

Ken G
2007-Jan-20, 08:04 PM
If you go through the derivation of Planck's blackbody curve, you will realize that the size of dust doesn't play any role.

Your knowledge of physics is understandably incomplete, but it is incomplete nevertheless. Learn some more about blackbodies, and try that derivation again, you will find that it does play a role when the dust is not small compared to the wavelength, and the dust is optically thin, as is the case here.


Again, if you go through the derivation of Planck's blackbody curve, you will realize that the density of dust doesn't change the blackbody curve.
What it changes is the optical depth, and the dilution of the Planck spectrum. That's all about the optical depth issue.

Discussing Planck's curve is not 'against the mainstream', since the textbooks which I have don't claim that size and density can change the blackbody curve and produce what Ken call's 'a diluted Planck spectrum'
Then either they are quite poorly written, or you should read them quite a bit more carefully. You have yet to comment on the concept of "optical depth"-- can it be that you feel this is not relevant? At some point, your textbook will define a blackbody as an object that absorbs all light that tries to pass through it. Have you ever looked at dust in a sunbeam? Does that seem like a blackbody to you? But that's what you are talking about looking out of the plane of our galaxy.


Dust that is optically thin, need to absorb much less radiation to reach the temperature of space. The issue is not the temperature achieved by the dust, it is the brightness of the radiation you will see when you look at it. That is what you are trying to compare to the CMB, yes or no?


Optically thin dust will absorb IR radiation, but much less than optically thick dust. I do know that dust locally is optically thin. Is optically thin dust capable of producing a blackbody curve?

Once more, it is capable of producing a diluted (i.e., less bright) blackbody curve, and only at wavelengths much smaller than the dust.

Ken G
2007-Jan-20, 08:11 PM
So you are saying, other than absorption by cold dust near us, there are other reasons for an imperfect blackbody curve from dust?Far away the big reason you won't get anything like a Planck spectrum from dust is that it is optically thin. It also won't be isotropic if you look far enough away-- you'll see many different temperatures locally. Furthermore, galactic dust emission is seen all the time, and presents no exceptional particular mysteries, and certainly would not look like the CMB in anyone's wildest dreams.

peaceharris
2007-Jan-22, 01:17 PM
Your knowledge of physics is understandably incomplete, but it is incomplete nevertheless. Learn some more about blackbodies, and try that derivation again, you will find that it does play a role when the dust is not small compared to the wavelength, ...

Please explain why large sized dust inside an ideal black body cavity doesn't produce an ideal black body curve.

Ken G
2007-Jan-22, 03:53 PM
That dust will come to the temperature of the cavity, and emit a blackbody spectrum except at the longer wavelengths. Your point being?

peaceharris
2007-Jan-23, 04:57 AM
That dust will come to the temperature of the cavity, and emit a blackbody spectrum except at the longer wavelengths. Your point being?

I did not ask you to reassert your statement. I asked you to explain why.

If you are not going to answer my questions, why should I answer yours?

Ken G
2007-Jan-23, 05:36 AM
I was telling you that you asked me to explain something that isn't true. So I'll instead assume you are asking me to explain something that is true, such as why a dust particle in thermal equilibrium with a cavity does not emit a Planck spectrum at all wavelengths. It goes like this: in thermal equilibrium, all processes are balanced by their inverse process. That means that every erg of energy at every wavelength that is absorbed by the dust particle is equal, erg for erg and wavelength by wavelength, to what it emits. So here's the kicker: the cross section of the dust particle (and the rate it absorbs at any wavelength equals its cross section at that wavelength times the incident perfect Planck function, right?) is not constant with wavelength. Therefore, the rate that it emits is not a perfect Planck function. I'm terribly sorry if your reference material did not make that clear, presumably they were not talking about dust. You see, when the wavelength is comparable to the dust particle size, that's when you get into complicated dependences of the cross section on the wavelength. Complicated, but well understood. And that is just one of the reasons that the Eddington calculation you mention is irrelevant to the CMB, and furthermore, Eddington would have known that.

Tim Thompson
2007-Jan-24, 12:54 AM
... and furthermore, Eddington would have known that.
Eddington did know that, and said so explicitly. He said he was calculating the temperature of space, in the sense that if you held a thermometer out there, that's what it would read. But he also explicity said that the radiation field was far from equilibrium and would therefore not have a thermal (Planck Law) spectrum. We need to have a page about this somewhere, I get tired of repeating all this everytime somebody comes along and claims that Eddington "predicted" the CMB, which he certainly did not do.

Ken G
2007-Jan-24, 01:04 AM
Hopefully, that settles the issue for the present company!

peaceharris
2007-Jan-24, 05:23 AM
That means that every erg of energy at every wavelength that is absorbed by the dust particle is equal, erg for erg and wavelength by wavelength, to what it emits. So here's the kicker: the cross section of the dust particle (and the rate it absorbs at any wavelength equals its cross section at that wavelength times the incident perfect Planck function, right?) is not constant with wavelength. Therefore, the rate that it emits is not a perfect Planck function.

By 'equilibrium', I meant that the heat energy absorbed is not used for expanding the dust cloud. By 'equilibrium' I mean that the total heat energy absorbed is radiated. In other words, if the dust absorbs 1 photon of wavelength 1m, and if it radiates 10 photons of wavelength 10m, it is in 'equilibrium'.

Using my definition of 'equilibrium', a body at 'equilibrium' will emit what kind of spectrum? What's the mathematical formula of this spectrum?

Tim Thompson
2007-Jan-24, 06:05 AM
Using my definition of 'equilibrium', a body at 'equilibrium' will emit what kind of spectrum? What's the mathematical formula of this spectrum?
There is no specific formula. In the case you have in mind, the emitted spectrum is very dependent on the emissivity as a function of wavelength, for the specific material. In the case of interstellar dust, there is usually a fair amount of silicon, which results in an obvious spectral feature at about 10 microns wavelength (in abosrption if the dust is cool, emission if it is warm). Gas spectra are more complicated. But in any case, it would certainly not be a Planck Law spectrum. You need to appeal to the general tools of radiative transfer theory to construct a mathematical model for your specific spectrum.

There are two extremely strong reasons for ruling out local, cool material as the source for the CMB, either being in its own right fatal to the claim.

1) The CMB spectrum, as measured by the FIRAS instrument is a Planck Law, thermal spectrum (the most precise such spectrum ever measured, in fact). We know that we are not surrounded by anything that is optically thick; if we were, we would be unable to see through it. Only an optically thick medium can emit with such a precise Planck Law spectrum. But we are surrounded by optically thin material, so it cannot be the source for the CMB.

2) The CMB temperature is the same in all directions, over all 4pi steradians of the sky, except for the anisotropies that exist at the few milli Kelvin level. There is no mechanism known that could bring everything we see to such a precisely defined temperature that is the same everywhere, in every direction. The light travel time is too long to allow such a large volume to reach the thermal equilibrium required to emit a Planck Law spectrum (which we aleady know it could not do anyway).

These two points are strong evidence that the CMB, whatever it is, must be cosmological in origin. In the context of big bang cosmology, it is thermal emission, redshifted from a time when the universe was in causal contact and able to reach thermal equilibrium throughout its volume.

Ken G
2007-Jan-24, 07:47 AM
By 'equilibrium' I mean that the total heat energy absorbed is radiated.You are talking about what is called "radiative equilibrium", which is quite a bit different from "thermodynamic equilibrium". Google them, or take some physics.


Using my definition of 'equilibrium', a body at 'equilibrium' will emit what kind of spectrum? What's the mathematical formula of this spectrum?As Tim said, there is no answer to this question. I can easily give you examples where radiative equilibrium can give you any spectrum you like. If you are fishing for "a Planck spectrum", you are simply wrong, for all the reasons that have been explained to you by now.

peaceharris
2007-Jan-25, 05:27 AM
Only an optically thick medium can emit with such a precise Planck Law spectrum. But we are surrounded by optically thin material, so it cannot be the source for the CMB.


Once the vast majority of these free electrons had attached themselves to nuclei, the opacity dropped to ~zilch for the radiation field present at that time - a thermal blackbody at ~3000K

How am I supposed to intepret this: Someone says 'Only an optically thick medium can emit with such a precise Planck Law spectrum' and another one says 'the opacity dropped to ~zilch for the radiation field present at that time - a thermal blackbody at ~3000K'

Isn't 'optically thick' and 'opacity dropped to ~zilch' opposite concepts?


Have you ever looked at dust in a sunbeam? Does that seem like a blackbody to you?

No. It is reflecting light in the visible range. If I observed this using an instrument capable of viewing microwaves, I won't see the reflected visible light.


In the case of interstellar dust, there is usually a fair amount of silicon, which results in an obvious spectral feature at about 10 microns wavelength (in abosrption if the dust is cool, emission if it is warm).

I agree with this. To prove that the CMB is due to dust, we need to show that spectral features due to dust exist.

Tim Thompson
2007-Jan-25, 07:25 AM
How am I supposed to intepret this: Someone says 'Only an optically thick medium can emit with such a precise Planck Law spectrum' and another one says 'the opacity dropped to ~zilch for the radiation field present at that time - a thermal blackbody at ~3000K'

Isn't 'optically thick' and 'opacity dropped to ~zilch' opposite concepts?

The key point here is that the optically thin universe is not emitting the CMB. Rather, the CMB is simply propagating through it. When light goes through a glass window, we do not argue that the window is the source of the light. Likewise, the optically thin universe around us is not the source of the CMB, it's just the "window" through which it propagates.

The CMB was emitted when the universe was optically thick, and was in causal contact. The optically thick part allows the spectrum to be thermal, while the causal contact part allows the temperature to be nearly the same everywhere. As the universe expands, the CMB is just along for the ride. It was emitted when the universe was about 300,000 years old, and has been propagating ever since.

Ken G
2007-Jan-25, 02:54 PM
Isn't 'optically thick' and 'opacity dropped to ~zilch' opposite concepts?
Only if you completely ignore the word "dropped". If it dropped to zilch, what do you think it was before that? Well, that's when the CMB was in thermal equilibrium.



No. It is reflecting light in the visible range. If I observed this using an instrument capable of viewing microwaves, I won't see the reflected visible light.
But it would be optically thin-- you can see through it. That's the point of my analogy, not the wavelength.



I agree with this. To prove that the CMB is due to dust, we need to show that spectral features due to dust exist.

The other arguments given clearly rule out the CMB as being due to dust. You can waste your time looking for dust spectral features if you like-- knock yourself out.