We brought this up with an astronomer friend, and he pointed us to the following graph of predicted absorption of infrared light from outer space by the atmosphere above an infrared telescope on Mauna Kea in Hawaii. We obtained the plot here.
According to the authors, the graph assumes the equivalent of 1 mm of liquid water distributed as water vapor in the air above the telescope. Our graph of absorption length in water gives us an absorption length of around 10 μm for 11-μm infrared. Other peoples graphs show the same thing. If we put a 1-mm layer of liquid water above the telescope, the water would absorb 99.995% of infrared arriving from outer space.
But the graph above appears to predict 98% of 11-μm infrared to reach the telescope from outer space. In other words, the graph suggests that our estimate of absorption by water vapor is off by a factor of a thousand. So we are trying to figure out why the predictions of astronomers (who we trust) are so at variance with our own back-of-the envelope calculations (which we sort of trust).
One possible explanation for the difference is that the behavior of water molecules in water vapor is dramatically different from that of water molecules in liquid water. Sites like this one discuss the difference between water vapor and liquid water, but their graphs appear show water vapor absorbing more than liquid water.
Any advice on this matter would be most welcome. We are looking into it.
UPDATE: The absorption spectrum of water vapor is the spectrum of free water molecules. The molecules vibrate in symmetric and clearly defined ways. The absorption frequencies are sharply-defined. In liquid water, the vibrations are influenced by bonds between molecules. These bonds, called hydrogen bonds occur between the slightly positive hydrogen atoms of one molecule and the slightly negative oxygen atoms of another molecule. The hydrogen bonds serve to spread the absorption wavelengths so that they blend together, giving a black liquid at all infrared frequencies. Water vapor, on the other hand, absorbs in narrow bands that do not always overlap, so that there are bands in which the water is not black at all. Indeed, there are windows in the spectrum, where water vapor is transparent.
For a graph of atmospheric transmission including water vapor and carbon dioxide, take a look at page two of this SOPHIA paper. For a seminal paper on water vapor absorption see Elsasser.
From Lubos Motl at The Reference Frame:
ReplyDeleteDear Kevan, good points. Your question would be even sharper. With your numbers, the concentration of water would be 1 ppm, so 20 microns would be expanded to 20 meters. But I actually think that the amount of water in the atmosphere is actually 1% in average, see
http://en.wikipedia.org/wiki/Earth's_atmosphere
so your 20 microns would become 2 millimeters, much shorter than your 10 meters! This paradox really seems to exist, see the H2O absorption spectrum here:
http://en.wikipedia.org/wiki/File:Water_absorption_spectrum.png
Around 10 micron wavelength, it really seems to wildly contradict this greenhouse graph
http://johnstonanalytics.com/yahoo_site_admin/assets/images/Atmospheric_Transmission.129223740.png
Wikipedia seems to claim that in liquid water, the spectrum is essentially shifted by 60 nm to longer wavelengths only, relatively to vapor:
http://en.wikipedia.org/wiki/Water_absorption
So it shouldn't make much difference.
However, the discussions about the molecular vibrations refer to 1 micron or so. For much longer ones, like 10 microns, I suspect that the hydrogen bond will become absolutely crucial. You know, the fact that this bond is weaker than the hydrogen-oxygen bond doesn't mean that it can always be neglected! It only means that the hydrogen bond-related vibrations affect low-energy i.e. low-frequency spectrum, and I guess that around 10 microns, the influence may be huge.
So I could imagine that the contradiction is resolved if the graph with the huge absorption near 10 microns above refers to liquid water, and most of this absorption is due to the vibrations using the hydrogen bond. Pure water vapor probably ignores these huge wavelengths "almost" altogether.
Yesterday, 2:20:42 PM EST
From my friend and astronomer Teddy Cheung, by e-mail:
ReplyDeleteHi Kevan,
Good to hear from you.
Yes, the absorption of infrared light is strongly altitude dependent,
so as you say, the higher one goes, the more transparent the
atmosphere is to IR light.
Note also that the IR band is a very wide -- typically considered from
~1 micron to ~1000 microns, and the absorption is wavelength
dependent. As you note, most infrared light is indeed almost
completely absorbed by the earth's atmosphere (we say "low
transmission," i.e., close to 0), but there are very "windows" in
which the atmosphere is relatively transparent to IR light (or
transmission >0.5-1).
The best documents to look at are for actual telescope observatories
that rely on knowing the precisely the "transmission curve" of a
specific site. Take a look at the Erickson (2004) pdf from here:
http://www.sofia.usra.edu/Science/publications/sci_publications.html
Fig 2 shows the transmission curve for Mauna Kea, HI, in comparison to
their proposed mission to fly a IR telescope on a retrofitted Boeing
747 (SOFIA), which actually just began observations these last few
months. You can see the obvious strong advantage of going up to the
stratosphere (that's what the "S" means).
Note around the 2 bands you mentioned -- 2 micron and (just above) 10
micron -- are two of the relatively open windows in the transmission
curve.
For a better view of the 2 micron window from the ground, see:
http://web.ipac.caltech.edu/staff/waw/2mass/opt_cal/index.html#s19
Indeed, in this page, you see clearly 3 windows of high transmission
and infrared telescopes on the ground operate quite well observing in
these bands. We even have historical names for these three bands:
J 1.0 - 1.3 μ
H 1.4 - 1.8 μ
K 2.0 - 2.4 μ
See here for more details (copy of someone else's page):
http://www.slac.stanford.edu/~teddy/spectrum.html
Are you thinking of buildling something new?
best,
Teddy
From my friend and astronomer Teddy Cheung:
ReplyDeleteKevan,
Here's a nice page showing the transmission for Mauna Kea for a wider
wavelength range. Indeed, ~10 micron is quite transparent.
http://www2.keck.hawaii.edu/inst/lws/atran.html
Just remember that the light being let through is just half the
problem. Even if you have high transmission in a band, the infrared
atmosphere can be quite turbulent (or lots of 'twinkling' if you're
talking to a layman about observing with an optical telescope) so IR
telescopes typically have to have (lots of) short exposures, and also
there is a huge zenith angle dependence (best to look straight up, and
impossible to look near the horizon).
ciao,
Teddy
From Teddy by e-mail:
ReplyDeleteHi Kevan,
OK, first, I want to check whether 1 mm of water vapor was a
reasonable value since I don't have a good sense of relative scales. I
found this doc from googling describing the best/driest sites on earth
(South pole, Atacama desert, then Mauna Kea=MK):
www.iac.es/site-testing/images/stories/espasmaunakea.pdf
As seen in Fig 4, 1mm is indeed a reasonable value in MK for the dry
observing months (Jan - June) but for only ~25% of the nights there.
This corroborates my understanding of observing at MK as there are
only a few 'prime' nights for IR observing and for the rest of the
nights, they are typically pursuing optical observations instead.
The MK conditions are confirmed in a primary reference in a refereed
journal (see Fig 13) so I'm satisfied with this number:
http://adsabs.harvard.edu/abs/1987PASP...99..560B
OK, I'll wait to see if you find something on absorption of water
vapour. I know that we / they astronomers do observe at these select
IR bands, however inefficiently.
The blog looks fun.
ciao,
Teddy
Question: is it possible that the energy arriving to the telescope is already filtered by the atmosphere, thus containing no absorbable frequencies by the water?
ReplyDeleteHi Mario,
ReplyDeleteYes, if there is a cloud, no long-wave radiation will get to the telescope. But that's not what you are asking, I expect. Clouds are made of water. If the clear sky is clear, it contains only water vapor, which is the gaseous phase of water. You could call it steam. Gaseous water absorbs far less long-wave radiation than does liquid water for the same mass per square centimeter. That is: a 1 mm thick layer of water absorbs all long-wave radiation, while the same water evaporated into a 100-m thick layer of air absorbs only some long-wave radiation. We look at the absorption by water vapor here:
http://homeclimateanalysis.blogspot.com/2010/06/continuum-absorption-length.html
and here:
http://homeclimateanalysis.blogspot.com/2010/07/earths-atmosphere.html
With humidity 1% by mass, water vapor in the first 3,000 m of the Earth's atmosphere absorbs almost all radiation in the range 5-8 um. So a telescope on the ground in a temperate climate will see no 5-8 um infra-red from the stars. Nor will it see >13 um because that's absorbed by CO2 and other things.
Yours, Kevan