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WayneFrancis
2014-Dec-10, 05:36 AM
In the ATMS forum it has been brought to my attention that my understanding of the CMBR may be flawed and I would like to rectify that.

This is my understanding right now.
In the early universe (< 380,000 years) the universe was a fairly uniform plasma of hydrogen, helium and lithium. At about 380,000 years the expansion of the universe reached a point where that plasma's temperature dipped below the ionization point of hydrogen. This changed the plasma into a neutral gas which allowed the photons to then proceed pretty much unaffected by the medium they where in. This means that a hydrogen spectrum was produced as electrons fell through to the lowest energy state available to them. This hydrogen spectrum has since then been red shifted as the universe expanded by about a factor of 1,000 thus lowing the temperature by about 1,000xs.

What am I getting wrong?
Thanks

Shaula
2014-Dec-10, 06:26 AM
The CMB spectrum is a black body spectrum, It is the emission spectrum of a plasma in thermal equilibrium - much like the underlying basic spectrum of the Sun (which is complicated slightly by absorption lines because we can see into it, so we are not actually looking at just a plasma in thermal equilibrium - in the case of the CMBR we are not seeing into it anywhere near as much).

If it were a hydrogen spectrum it would look like a series of lines rather than the black body curve we see.

StupendousMan
2014-Dec-10, 01:24 PM
And the reason that it's a continuous, blackbody spectrum instead of a emission-line spectrum is that at the time of recombination, the density of the plasma/gas was very high.

ShinAce
2014-Dec-10, 04:18 PM
I say 'everyone's right'!!!

Indeed, it is the settling of hydrogen to its ground state that leads to the current CMBR. In theory, you might be able to detect the H emission lines, but there are difficulties awaiting you:
1) There are on the order of a billion pre-CMBR photons for every one of these hydrogen emission photons. That is why everyone will insist it's a perfect blackbody. The photon gas which existed already is guaranteed to have been in equilibrium.
2) The so called surface of last scattering is not a well defined surface at all. It's not a few seconds thick in the universe's history. It's more like 70,000 years thick. And this is when the universe is only 350,000 years old. I like to call it the 'thick shell of last scattering'.

George
2014-Dec-10, 04:53 PM
I am puzzled about the electrons going to the lowest energy state given a temperature of around 3000K. Wouldn't we see a full range of energy states at Recombination?

ShinAce
2014-Dec-10, 05:48 PM
A free electron being captured by a proton releases 13.6eV.

By comparison, a gas of photons with a mean energy of 1 eV has a temperature of 11,600 Kelvin. When you work out the CMBR, you need to do it in terms of scattering rates compared to the size of the universe(the Hubble parameter). By the time the universe is 3,000 degrees, only a tiny fraction of the photons have enough energy to re-ionize hydrogen.

I'm not sure what you mean by a full range of energy states. Remember that there are already billions of blackbody photons for each proton by the time the CMBR is 'produced'. What we see are the photons which existed before recombination. Not the photons produced by recombination.

I hope that mess makes sense.

edit: If you mean electrons in excited states, but bound, be advised that the population of electrons in excited states follows an exponential law(The boltzmann distribution). What percentage of hydrogen would you expect to be completely ionized on the surface of the sun? It's nearly 6,000 degrees but only 3% ionized.

George
2014-Dec-10, 08:27 PM
...edit: If you mean electrons in excited states, but bound, be advised that the population of electrons in excited states follows an exponential law(The boltzmann distribution). What percentage of hydrogen would you expect to be completely ionized on the surface of the sun? It's nearly 6,000 degrees but only 3% ionized. Perhaps there is a misunderstanding on my part as to what is meant by the "lowest energy state" vs. "excited states".

My simplified picture of Recombination (first combination) is that the expansion cooled things down enough to allow electrons to be captured by the protons (9 out of 10 nuclei being single protons). But I have assumed that the electrons suddenly could be found at multiple energy levels (Bohr shells) in these first atoms, and not necessarily at the "lowest energy state" as mentioned in the OP . The loss off free electrons and their scattering effects allowed photons to start their long paths. Perhaps I am over-playing what was meant by the "lowest energy state" and that Wayne meant a multitude of energy states transpire suddenly, as I expect.

WayneFrancis
2014-Dec-11, 12:03 AM
First thanks for the explanations. :)

So the gas that is at thermal equilibrium wouldn't affect the the emission spectrum?
What I'm hearing is that the emission from hydrogen emission lines, at that the point (surface, thick shell) of recombination, would be swamped by the existing set of photons that have been bouncing about randomly up until that point.

WayneFrancis
2014-Dec-11, 01:34 AM
I think I'm getting it.
The photons are being absorbed by the ions within the plasma. New photons are being emitted by the process of thermal radiation. The spectrum of which is simply a function of the temperature of the gas.
this curve (which is different for different at different temperatures). Since the gas is at thermal equilibrium the emission follows a black body in spectrum.


I was getting a little confused by this chart
20162
I was wondering why the peak was shifting to the left as the temperature got higher until I realized that the left is the higher frequency (shorter wave length)

ShinAce
2014-Dec-11, 06:03 AM
What I'm hearing is that the emission from hydrogen emission lines...of recombination, would be swamped by the existing set of photons that have been bouncing about randomly up until that point.

Exactly! Photons themselves bouncing around will 'thermalize' and create the blackbody spectrum. There's just so many of them from before that the actual photons from recombination are outnumbered.

George
2014-Dec-11, 03:00 PM
Exactly! Photons themselves bouncing around will 'thermalize' and create the blackbody spectrum. There's just so many of them from before that the actual photons from recombination are outnumbered. I'm trying to picture the circumstances of that moment. Would the photon/electron ratio just prior to Recombination be fairly low so that the MFP is very short, say centimeters or less?

WayneFrancis
2014-Dec-12, 05:09 AM
Thanks for everyone's help ...I've got another CMB question to clear up now :)

WayneFrancis
2014-Dec-12, 05:48 AM
The fluctuations we see in our image of the CMBR are just do to extremely small temperature differences in temperature. Since the photons we now see where emitted from region a finite distance from us and the photon of the CMBR we would have seen if we were present at this location in space 10 billion years ago are also from a different region, but in the same direction but closer.
I would expect that the temperature fluctuations to have a different pattern. That pattern would have the same ratio of fluctuation, less then 1 part in 100,000 but 'hot' and 'cold' regions would be different am I correct?

Amber Robot
2014-Dec-12, 07:38 AM
Exactly! Photons themselves bouncing around will 'thermalize' and create the blackbody spectrum. There's just so many of them from before that the actual photons from recombination are outnumbered.

But they aren't necessarily outnumbered at a given wavelength. For instance, each recombination will eventually result in a Lyman series photon, with a rest wavelength in the far-ultraviolet. A 3000K blackbody doesn't have many photons in the far-ultraviolet. So, somewhere out there is lyman alpha emission at z=1000, which would have strong contrast against the CMB but most likely completely swamped by nearby sources.

Edited to add: i am just thinking qualitatively. i admit that i have not run the numbers, so i could be wrong.

ngc3314
2014-Dec-12, 02:53 PM
The line photons from recombination have been spread out by the nontrivial time it took for recombination to complete, which is crudely delta z / z =0.1. There are calculations showing the precision it would take to detect them above the blackbody continuum - these results (http://www.mpa-garching.mpg.de/mpa/research/current_research/hl2007-7/hl2007-7-en.html) show that it would be a spectral ripple at the level of a part per million, harder to detect in the spectral direction than similar-amplitude features in the spatial structure.

Amber Robot
2014-Dec-12, 05:12 PM
Thanks. Yeah, good point that the "surface" of last scattering is broad in z. That would smear out the lines tremendously, greatly reducing their observability.

Ken G
2014-Dec-12, 06:01 PM
I'm trying to picture the circumstances of that moment. Would the photon/electron ratio just prior to Recombination be fairly low so that the MFP is very short, say centimeters or less?The photon/electron ratio is a lot different from the mean-free-path, because the latter involves the likelihood of interaction. Photons are vastly more numerous than electrons, by about a factor of a billion. When recombination occurs, each electron adds some relatively small number of photons. Not just one, because like you say, the electrons don't originally enter the neutral atom in the ground state, more often they enter in an excited state. But they eventually end up in the ground state, so by then, they've each caused a few photons to be emitted. Since the photons vastly outnumber the electrons, this is not an important contribution to the photon number. As pointed out above, it might have been an important contribution to the photon number at higher energies, but it turns out that isn't the case either.

Ken G
2014-Dec-12, 06:27 PM
The line photons from recombination have been spread out by the nontrivial time it took for recombination to complete, which is crudely delta z / z =0.1. There are calculations showing the precision it would take to detect them above the blackbody continuum - these results (http://www.mpa-garching.mpg.de/mpa/research/current_research/hl2007-7/hl2007-7-en.html) show that it would be a spectral ripple at the level of a part per million, harder to detect in the spectral direction than similar-amplitude features in the spatial structure.That's a very interesting article, and it points out something I had not realized-- the reason recombination does not produce UV emission lines on top of the relatively dark UV thermal continuum is not just that these lines would be spread out by the expansion, there's another important effect-- they tend to get split into two photons at a range of energies.

The reason the recombination takes so long to complete, and the reason you tend to get two photons in the final step of that recombination, is the same: electrons quickly (in a tiny fraction of a second) fall down to the second level, but it takes a long time for
them to finally settle for the long haul into the ground state, because they need the universe to expand to a point where the mean-free-path of a Lyman alpha photon (the 2->1 transition photon, for those unfamiliar with that term) can cross enough of the expansion to be unlikely to be reabsorbed in some other ground-state H atom somewhere else. For any not following, what I mean is, if you have an H atom go from level 2 to level 1, that's not the end of it-- you still have that Lyman alpha photon, and it wants to re-excite some other ground-state H back to level 2, effectively undoing the original transition. The only time that wouldn't happen is if the density is low enough that the Lyman alpha photon can get far enough before that happens such that the Hubble expansion causes it to arrive too red to re-excite the H atom it would otherwise be absorbed by, and that takes a long time-- if that was all that was going on, the recombination would have taken even longer than it did!

So what actually creates that delta z / z of ~0.1 is not the dropping temperature, that actually has almost nothing to do with it. And it's not the dropping density either-- the temperature and density where the recombination begins to occur in earnest is a fairly precise value. What sets the timescale for the completion of recombination is the likelihood of the 2->1 transition happening via emission of two photons, not one, along with the similar timescale for a Lyman alpha photon to be emitted enough on the redward side of the line for the Hubble expansion to keep it from re-exciting H atoms. The timescale set by these two effects is very long, because they are both quite unlikely, but when they happen, the recombination is finally done, because the photons created will not have enough energy to re-excite H. Also, the spectrum of a two-photon process is continuous, it's not a sharp line, so that's what makes it so hard to see this light-- it's not just spread by the delta z / z as a line would be, it's actually a continuum.

Now, the article points out that only 57% of the electrons reach the ground state via this pathway-- the rest get there by direct emission of a Lyman photon (be it Lyman alpha or from some higher level). So that's why it's a combination of these two effects. That these two pathways are roughly similar strikes me as an odd coincidence, I cannot see any reason for this to necessarily be true. So this strange coincidence means that there should still be something of a line spectrum there, just broadened by that delta z / z effect, but it's only some 43% as strong as it would have been without the two-photon process.

Amber Robot
2014-Dec-12, 06:55 PM
Great post Ken!

George
2014-Dec-21, 12:12 AM
The photon/electron ratio is a lot different from the mean-free-path, because the latter involves the likelihood of interaction. Photons are vastly more numerous than electrons, by about a factor of a billion. When recombination occurs, each electron adds some relatively small number of photons. Not just one, because like you say, the electrons don't originally enter the neutral atom in the ground state, more often they enter in an excited state. But they eventually end up in the ground state, so by then, they've each caused a few photons to be emitted. Since the photons vastly outnumber the electrons, this is not an important contribution to the photon number. As pointed out above, it might have been an important contribution to the photon number at higher energies, but it turns out that isn't the case either.Thanks Ken.

My question is more of what we might see if we were to go there, that moment of Recombination. I was going to spare everyone the story but, for clarity of my question, I've recharged my old wand to power my time machine so that Zeke and I can travel back to that amazing moment. Not that anything could possibly go wrong, but I am hoping to learn what we should see upon our arrival.

I suspect, being there just prior to the beginning of Recombination, we will be able to look out the portal and see a very bright ubiquitous white light with a tint of yellow. Our wand empowered instrumentation would also give us a near perfect blackbody spectrograph that Zeke might not fully appreciate. [I would assume a sp. irr. plot of our light would be closer to an actual sp. irr. data set from a hole in a real furnace, right?] Your last post above, thanks in part to ngc3314, gives it an interesting tweak, but not much given the limited number of atomic photons jumping in an out, not that Zeke would be that interested in this level of detail. He prefers too much the fun of experience over the devilish details. [My wand has not been effective changing this, in case you, ngc3314, or others had ideas of collegiate use of it yourselves.]

I expect that when our miniature robotic arm (plagiarized from Canadians) extends outward into the photon fog, we will lose sight of it after only xxx (1 cm?). [A blue laser beam shines our way from the end of the arm.] Suddenly, it's the moment of Recombination, and in a matter of milliseconds (?) we can see to the end of other and much longer extended arms. [I may have failed to mention these arms.] Of course, it [the background] is just as bright as it was seconds before, but the distance we can see is advancing at the speed of light. Nevertheless, there is still some tiny amount of fog due to the points in your prior post. [The glare from the glowing atoms (background and foreground) is heavily discounted in this depicition. In about 2 million years after recombination, the foreground glow vanishes from view (1000K), though likely a little sooner and hotter, I suppose.]

Is this close? If so, Zeke may prefer the wand be used for other purposes, especially now that it's the Christmas season.

Ken G
2014-Dec-24, 04:19 AM
My question is more of what we might see if we were to go there, that moment of Recombination. Better bring some reflective gear, you'll be hit by 3000 K radiation from all sides.


I suspect, being there just prior to the beginning of Recombination, we will be able to look out the portal and see a very bright ubiquitous white light with a tint of yellow.Pretty red, I should think.


I expect that when our miniature robotic arm (plagiarized from Canadians) extends outward into the photon fog, we will lose sight of it after only xxx (1 cm?). The density at recombination is a billion times what it is now (1 hydrogen per cubic meter), so it would be 1000 hydrogen per cubic centimeter. The mean free path of free electrons at that density is actually quite large, a few thousand light years. That's about 1/100 of the length of the horizon at that age. (I'm surprised by this, I've never worked it out.)
Suddenly, it's the moment of Recombination, and in a matter of milliseconds (?) we can see to the end of other and much longer extended arms. The recombination is not so sudden-- it takes tens of thousands of years!
In about 2 million years after recombination, the foreground glow vanishes from view (1000K), though likely a little sooner and hotter, I suppose.Yes, it would fade out over millions of years, though I think 1000 K is still fairly bright.

Hornblower
2014-Dec-24, 10:03 AM
I suspect, being there just prior to the beginning of Recombination, we will be able to look out the portal and see a very bright ubiquitous white light with a tint of yellow.

Pretty red, I should think.
Pretty white, I should think. That is my experience with 3000K incandescent lamps providing bright ambient indoor lighting. That same light as a small swatch in contrast with daylight will look pale orange, as does a bright M star in twilight. Remember, our eye-brain system adjusts to ambient incandescent light over a wide range of temperature to perceive it as white.

Jeff Root
2014-Dec-24, 10:03 AM
And the reason that it's a continuous, blackbody spectrum
instead of a emission-line spectrum is that at the time of
recombination, the density of the plasma/gas was very high.
I don't understand this. If ShinAce or Ken explained it in a
later post, I didn't make the connection.



The so called surface of last scattering is not a well defined
surface at all. It's not a few seconds thick in the universe's
history. It's more like 70,000 years thick. And this is when
the universe is only 350,000 years old. I like to call it the
'thick shell of last scattering'.
About fifteen years ago I asked an astronomy professor at
the U of M (who headed the group which later discovered
the largest known cosmic void) about the thickness of the
shell of last scattering. He said it was about 100 parsecs
(300 light-years) thick. I have mentioned this in posts a
couple of times, including here on BAUT, I think. I don't
recall any comments on that figure. It is possible that we
are talking about slightly different things, but I don't recall
the exact wording of my question or the exact wording of
his answer. Alternatively, could his figure possibly have
missed accounting for the two-photon emission that Ken
described, which -- if I understand Ken correctly -- is
responsible for lengthening the moment of decoupling to
70,000 years?

-- Jeff, in Minneapolis

EigenState
2014-Dec-24, 07:50 PM
Greetings,

J.Chluba and R.A.Sunyaev, Induced two-photon decay of the 2s level and the rate of cosmological hydrogen recombination (http://arxiv.org/abs/astro-ph/0508144), Astron.Astrophys.446:39-42 (2006).

It is also not rigorously correct to characterize the HI transitions only in terms of their principle quantum numbers, except perhaps in the case of high Rydberg states. The fine structure splittings within n=2 are only of the order of 0.3 and 0.03 cm-1, yet 2p 2P1/2, 3/2 - 1s 2S1/2 are one photon electric dipole allowed transitions.

Best regards,
ES

George
2014-Dec-24, 09:08 PM
Better bring some reflective gear, you'll be hit by 3000 K radiation from all sides.The wand's Adiabatic button will be set on before we depart. Thanks for worrying about us. :)


Pretty red, I should think.Hornblower makes the argument I like to use. A tungsten filament heats to between 2000K and 3000K (melting point is 3695K), so white would be the likely result, except when compared to a hotter and bright light source, like the Sun. [Also, our eyes tend to make the bright source whiter than it would normally be, as do cameras for white balance. [Heliochromologist(s) call this white-shifting. Too often there is a redshift mentality when temperatures are below solar temperatures (5850K for a BB fit, 5777K for a thermal fit). This seems to be how the yellow Sun story began, thanks mainly, or partly, to Sechii.] With a hotter light source in the mix, 3000K would become white with a tint of yellow.


The density at recombination is a billion times what it is now (1 hydrogen per cubic meter), so it would be 1000 hydrogen per cubic centimeter. The mean free path of free electrons at that density is actually quite large, a few thousand light years. That's about 1/100 of the length of the horizon at that age. (I'm surprised by this, I've never worked it out.)The recombination is not so sudden-- it takes tens of thousands of years! Yes, it would fade out over millions of years, though I think 1000 K is still fairly bright. Ok, so more than a cm. :) Wow, I thought, obviously, density would have been much greater. So does this mean the CMBR is more of a shell a few thousand light years thick; that's a heck of a photosphere! The implications of that will make for some headscratchin and probably give me even greater respect for those who are mapping it. [My head or the CMBR]

Ken G
2014-Dec-25, 12:51 AM
Pretty white, I should think. That is my experience with 3000K incandescent lamps providing bright ambient indoor lighting. That same light as a small swatch in contrast with daylight will look pale orange, as does a bright M star in twilight. Remember, our eye-brain system adjusts to ambient incandescent light over a wide range of temperature to perceive it as white.Yes, I'm confused about this, because a 3000 K star looks quite red (for example, Betelgeuse is even a little hotter than that, at 3500 K, and noticeably red). But light bulb filaments don't usually get much hotter than that, you're right. So it may have to do with how our eyes work and what the brightness level is, etc.

Ken G
2014-Dec-25, 01:00 AM
About fifteen years ago I asked an astronomy professor at
the U of M (who headed the group which later discovered
the largest known cosmic void) about the thickness of the
shell of last scattering. He said it was about 100 parsecs
(300 light-years) thick. There are two separate issues here. One is, how long did recombination take to complete-- that is some tens of thousands of years, due to the need to wait for those 2-photon processes. The other is, what was the mean-free-path of a photon just before recombination, which is about 103light years. The "thickness of the surface of last scattering" can involve either of those numbers, because even if recombination is instantaneous, that mean-free path is still the thickness of last scattering. But since recombination is not instantaneous, there is a lot of different lengths that photons could have as the mean free path gets longer over tens of thousands of years.

Hornblower
2014-Dec-26, 02:32 AM
Yes, I'm confused about this, because a 3000 K star looks quite red (for example, Betelgeuse is even a little hotter than that, at 3500 K, and noticeably red). But light bulb filaments don't usually get much hotter than that, you're right. So it may have to do with how our eyes work and what the brightness level is, etc.
Yes, the reddish tint we see with Betelgeuse is that of a small spot surrounded by darkness, or by a bluish background during twilight. My eyes tend to have a set point of seeing an F star as approximately neutral white, while cooler ones look reddish or yellowish and hotter ones bluish. When any of those tints become bright ambient light my sensation shifts to neutral white. Only when an incandescent source is cool enough to have practically no blue light in its spectrum do I see it as reddish.

Ken G
2014-Dec-26, 03:00 AM
Yes, that must be typical of how the eye responds.

weltevredenkaroo
2014-Dec-28, 08:57 PM
How do CMBR analysts prune out 21cm noise from intervening IGM and young galaxies? For that matter, what signal-to-noise ratios are typical in COBE and related signal processors?

EigenState
2014-Dec-28, 09:26 PM
Greetings,


How do CMBR analysts prune out 21cm noise from intervening IGM and young galaxies? For that matter, what signal-to-noise ratios are typical in COBE and related signal processors?

A search on Google Scholar for "CMB data analysis" yields over 83 000 results. I suggest that you begin there, or try the sites associated with the various individual missions. Any published paper will also provide details of their data analysis procedures.

Best regards,
ES

Shaula
2014-Dec-28, 09:37 PM
How do CMBR analysts prune out 21cm noise from intervening IGM and young galaxies? For that matter, what signal-to-noise ratios are typical in COBE and related signal processors?
The fairly recently published Planck results paper has pretty extensive discussion (or citations for) the signal processing done. You could try: http://planck.caltech.edu/publications2013Results.html - it seems to have a fairly comprehensive overview.

George
2014-Dec-29, 04:55 PM
Yes, I'm confused about this, because a 3000 K star looks quite red (for example, Betelgeuse is even a little hotter than that, at 3500 K, and noticeably red). But light bulb filaments don't usually get much hotter than that, you're right. So it may have to do with how our eyes work and what the brightness level is, etc. Don't forget the CLV (center to limb variation) of red giants. A 3000K temp. is an effective temperature, like the Sun's 5777K, but the Sun's clv ranges from 5000K to 6390K, so the orangish-red, or red, color you see in red giants may be more a result of their limb color.

weltevredenkaroo
2014-Dec-29, 08:45 PM
The fairly recently published Planck results paper has pretty extensive discussion (or citations for) the signal processing done. You could try: http://planck.caltech.edu/publications2013Results.html - it seems to have a fairly comprehensive overview.

Thanks, Shaula, this paper doesn't address indiv wavebands but sect 3.2.1 pretty much answers the point if not the question:

http://planck.caltech.edu/pub/2013results/Planck_2013_results_09.pdf

EigenState: Searching CMB data analysis 21 cm H1 pulls in 1600 cites but none directly relevant. I didn't calculate what a 21 cm signal becomes at < z = 1089 > z = 1000. Back to the homework.

Ken G
2014-Dec-30, 02:12 AM
Don't forget the CLV (center to limb variation) of red giants. A 3000K temp. is an effective temperature, like the Sun's 5777K, but the Sun's clv ranges from 5000K to 6390K, so the orangish-red, or red, color you see in red giants may be more a result of their limb color.That's an interesting point, I wonder if it explains why Capella looks so orange? I've always wondered that, given that one of the stars is the same spectral type as the Sun. The red giant is a little cooler, but it didn't seem enough to make it look so orange, unless perhaps there's an important contribution from the limb darkening you are talking about.

ngc3314
2014-Dec-30, 09:26 AM
Here's a paper (http://www.sciencedirect.com/science/article/pii/S0370157306002730) discussing the challenges and payoff of detecting the 21-cm hydrogen signal from the epoch of reionization and before. There are different tradeoffs to using angular versus frequency behavior to filter foreground contributions. The ands of interest are tens of MHz, with people working on approaches from very radio-quiet locations on Earth and proposals for missions to take data awhile orbiting over the lunar farside [1] (the best place within several AU to reduce low-frequency interference, since auroral emission and the Van Allen belts are powerful at these frequencies).

[1] There was in fact a precursor, Radio Astronomy Explorer B orbiting the Moon in the 1970s with pairs of 230-meter dipole antennae, generating a low-resolution sky map at a few Mhz (not far above the cutoff where the interstellar medium and solar wind soak up longer wavelengths).

EigenState
2014-Dec-30, 05:40 PM
Greetings,


Here's a paper (http://www.sciencedirect.com/science/article/pii/S0370157306002730) discussing the challenges and payoff of detecting the 21-cm hydrogen signal from the epoch of reionization and before.

Interesting. Unfortunately I could not get the paper to display properly.


The [b]ands of interest are tens of MHz, with people working on approaches from very radio-quiet locations on Earth and proposals for missions to take data awhile orbiting over the lunar farside [1] (the best place within several AU to reduce low-frequency interference, since auroral emission and the Van Allen belts are powerful at these frequencies).

Tough frequency domain to work in especially without being able to utilize phase sensitive detection.

Best regards,
ES

StupendousMan
2014-Dec-31, 12:00 AM
You can find a copy of Furlanetto's paper mentioned by ngc3314 here:

http://arxiv.org/abs/astro-ph/0608032

Perhaps the PDF copy available at this URL will be easier for you to read.

EigenState
2014-Dec-31, 12:03 AM
Greetings,


You can find a copy of Furlanetto's paper mentioned by ngc3314 here:

http://arxiv.org/abs/astro-ph/0608032

Perhaps the PDF copy available at this URL will be easier for you to read.

Terrific! Thank you!

Best regards,
ES