hi see

and from this reference there is this high-lited alinea on page 360:

Homonuclear diatomic molecules have no dipoleallowed

vibrational-rotational spectra. This

means they do not absorb or emit radiation on

transitions within the same electronic state. They

may have very weak quadrupole transitions.

"Homonuclear diatomic molecules have no dipole allowed vibrational-rotational spectra:"

Understanding that would explain why Nitrogen and Oxygen are not greenhouse gasses. 

They may have very weak quadrupole transitions.

(1)  Is there an equation in the text that could be applied to calculate the maximum energy of a quadrupole transition in a H2 molecule,  and what wavelength of radiation that would be associated with?  That is, when you say "very weak" does that mean "way way weaker than the 21 cm line of hydrogen" or does it mean "very weak but not that weak?"

These may provide a starting point.

http://pages.uoregon.edu/jimbrau/BrauImNew/Chap04/FG04_14.jpg

http://kicp.uchicago.edu/~odom/molecule_database/H2/molecular%20hydrogen%20spectrum%20tabulation.pdf

http://chemistry.tutorvista.com/inorganic-chemistry/hydrogen-absorption-spectrum.html

Yes, George, those lines are associated with the Balmer series, for monatomic hydrogen.  They usually teach about that in first year physics and/or chemistry.

Lyman: E_(n,1)= 13.6 eV (1/n^2 - 1/1^2)  (Ultraviolet)

Balmer: E_(n,2)= 13.6 eV (1/n^2 - 1/2^2)  (Visible)

Paschen: E_(n,3)= 13.6 eV (1/n^2 - 1/3^2) : (infrared)

So long as one has a monatomic version of hydrogen, these equations apply for emission/absorption spectrum.  But I'm envisioning a scenario where no monatomic hydrogen is present.  In other words, where ALL of the hydrogen atoms in the environment has paired up with other hydrogen atoms.

But that leads me to another really important question

(2) When hydrogen atoms form hydrogen molecules, is the interaction endothermic or exothermic?  If the interaction producing H2 from H is exothermic, then we could expect the scenario where NO monatomic hydrogen atoms are present to be the vast majority of cases below 3000 Kelvin.  And then we would not see any Paschen series infrared spectrum from warm (300-3000 Kelvin) Hydrogen gas.

And a third question:

(3) What energy is needed to break the covalent(?) bond between atoms of a hydrogen molecule.  https://socratic.org/questions/how-much-energy-is-needed-to-break-the-bond-in-a-single-molecule-of-hydrogen says it is 7.12x10^-19 Joules, which would be associated with a wavelength of lambda = hc/E = 662x10^-36*3x10^8/7.12x10^-19 = 279 nanometers, based on E = hc/lambda.  And from that, and a rough-estimate application of Wein's Displacement Law, T = b/lambda = (2.9x10^-3 meter Kelvin)/(279x10^-9 meter) = 10432 Kelvin.  

So that means, that in order to create the monatomic hydrogen that would give you the Paschen series, you'd have to have temperatures on the order of 10,000 Kelvin!

JD: Yes, George, those lines are associated with the Balmer series, for monatomic hydrogen.

If you mean the third link Jonathan, the molecular spectrum is at the bottom of the page after the atomic spectra. I think the other two only discuss the molecular spectrum but I've only skimmed them.

Forming molecules is exothermic but UV from stars not only breaks the bonds but ionises the atoms. You need to study where the molecular form would be found.

Sorry about that... I added this third question after you replied, George. 

And a third question:

(3) What energy is needed to break the covalent(?) bond between atoms of a hydrogen molecule.  https://socratic.org/questions/how-much-energy-is-needed-to-break-the-bond-in-a-single-molecule-of-hydrogen says it is 7.12x10^-19 Joules, which would be associated with a wavelength of lambda = hc/E = 662x10^-36*3x10^8/7.12x10^-19 = 279 nanometers, based on E = hc/lambda.  And from that, and a rough-estimate application of Wein's Displacement Law, T = b/lambda = (2.9x10^-3 meter Kelvin)/(279x10^-9 meter) = 10432 Kelvin.  

So that means, that in order to create the monatomic hydrogen that would give you the Paschen series, you'd have to have temperatures on the order of 10,000 Kelvin!

GD: Forming molecules is exothermic but UV from stars not only breaks the bonds but ionises the atoms

Right.  And that is not conflicting with the statement "diatomic hydrogen does not absorb or emit radiation on transitions within the same electronic state."  Because ultraviolet radiation of high enough energy would, of course, put the molecule in a different electronic state.  i.e. breaking the bonds, and ionizing the atoms.

GD: You need to study where the molecular form would be found.

Actually, this was my point.  If you are looking at a region where you find a 21 cm line from monatomic hydrogen, you are probably seeing evidence of a massive production of UV light, from a high energy star, nearby.  Where did this evidence of monatomic hydrogen come from?  It had to come from monatomic hydrogen that exists because of the recent break-up of previously invisible diatomic hydrogen.

But if you're looking at the Cosmic Background Radiation, for instance, you are looking at a light that is estimated to be red at the source, and between there and here, there is not a single source of ultraviolet light, which would interact with diatomic hydrogen.  So, for all we know, the entire space could be composed of perfectly transparent diatomic hydrogen gas.

From the link, the binding energy of the molecule is 4.52 eV while visible light is roughly in the range 1.7 eV to 3.2eV. The Sun is not a hot star but its emission extends to around 200A or about 6 eV.

The first Pop III stars were probably more than 100 times more massive and lasted only around 10 million years so must have been ~100,000 times brighter. Their UV not only split molecules to atoms but also ionised virtually all the atoms to free protons an electrons. Only then did the universe become transparent, before that is called the "Dark Ages" because the medium was opaque. The end is "re-ionisation" and probably peaked around z=11.

The second link is a map of the 21cm emission from neutral, monatomic hydrogen.

JD: It had to come from monatomic hydrogen that exists because of the recent break-up of previously invisible diatomic hydrogen.

No, existing monatomic hydrogen can be excited by other photons and then emits when the atom reverts to the lower energy state, see the third link.

What you are looking for is found in molecular clouds, see the fourth link. The section on "occurrence" describes where they can be found.

http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/hmol.html

http://s3-ap-southeast-2.amazonaws.com/icrar.org/wp-content/uploads/2016/10/12102646/bwinkel_langedit_printer.pdf

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/h21.html

https://en.wikipedia.org/wiki/Molecular_cloud

GD: What you are looking for is found in molecular clouds,

Thanks George.

I see, yes, that would answer the "bold-faced" part of my question.  

In particular, I see here  "Rocket Observation of Interstellar Molecular Hydrogen" by Geourge Carruthers, from August 1970, Astrophysical Journal,

http://adsabs.harvard.edu/abs/1970ApJ...161L..81C

There it is, a spectrum of H2, showing 6 ultraviolet emission bands between 1000 and 1100 angstroms.  Pretty far into the ultraviolet.  

Which pretty well confirms what I was thinking.  Namely that even if diatomic hydrogen were present in high densities,  there would be no evidence of diatomic hydrogen in regions that are not bathed in ultraviolet light, or at temperatures on the order of 10,000 Kelvin.

GD: The end is "re-ionisation" and probably peaked around z=11.

Is that a typo?  Did you mean z = 1100, here?  

quadrupole transitions are very very weak and in the IR-range, but

I do not understand why you ask for transitions of/in hydrogen in IR to UV range; what is your interest

My conjecture is that diatomic hydrogen may be present throughout the universe in much greater quantities than current models currently estimate. 

But I want to get an idea of how to quantify precisely how transparent diatomic hydrogen exactly is.   If it has even a tiny bit of absorbtion of radiation in the infrared... what do they call it... absorption coefficient, and beer's law(?) ... then the cosmic background radiation would have been blocked by a few million light-years of diatomic hydrogen.  

But if it is perfectly transparent, then it is possible that when we see the CMBR through billions of light-years of transparent diatomic hydrogen gas.

JD: In particular, I see here  "Rocket Observation of Interstellar Molecular Hydrogen" by Geourge Carruthers, from August 1970, Astrophysical Journal, http://adsabs.harvard.edu/abs/1970ApJ...161L..81C There it is, a spectrum of H2, showing 6 ultraviolet emission bands between 1000 and 1100 angstroms.  Pretty far into the ultraviolet.

That would just be the band observed by their instrument.

JD: .. if diatomic hydrogen were present in high densities,  there would be no evidence of diatomic hydrogen in regions that are not bathed in ultraviolet light, or at temperatures on the order of 10,000 Kelvin.

No, it would mean that we will not see emission in those lines other than from clouds at that temperature. Those bands would also only absorb energy from high energy photons but the spectrum extends all the way down through the visible. Any higher energy photon can lose a tiny amount of energy and just be deflected (so that energy and momentum are conserved).

For the visible molecular spectrum of hydrogen and its isotopes, see the attached. It also gives you tables listing the known lines (it's a bit old). Just note how many there are!

GD: The end is "re-ionisation" and probably peaked around z=11.

JD: Is that a typo?  Did you mean z = 1100, here?

No, 1089 is when the CMB was emitted because the dense gas cooled and became transparent, that transition is similar to that at the photosphere of the Sun, from the opaque surface to the transparent corona. That was at an age of ~380k years.

After that you had the "dark ages" when there were no stars. The first stars formed we think at around z=25 or perhaps a bit higher (Naoz suggests up to z=65). However, even a small amount of neutral gas in the mix makes it effectively opaque over astronomical distances so it only became transparent at around z=11. The Lyman Alpha line is in the UV, but the redshift means it blocked what we see now at much lower frequencies.

JD: But I want to get an idea of how to quantify precisely how transparent diatomic hydrogen exactly is.

It isn't. It only forms in denser clouds and those are completely opaque. That was the point of the previous link, objects like the Horsehead Nebula show what they are like.

http://kicp.uchicago.edu/~odom/molecule_database/H2/molecular%20hydrogen%20spectrum%20tabulation.pdf

So, George, you have some idea of the "dark ages"

GD: After that you had the "dark ages" when there were no stars.  The first stars formed we think at around z=25 or perhaps a bit higher 

and you understand the surface of dense gas with redshift of 1089

GD: 1089 is when the CMB was emitted because the dense gas cooled and became transparent, that transition is similar to that at the photosphere of the Sun, from the opaque surface to the transparent corona. That was at an age of ~380k years.

So, when I tell you that I am talking about this region where 25

JD: So, when I tell you that I am talking about this region where 25

GD: OK, I had the impression you were talking about present day conditions

Maybe the issue here is that I can't literally show you what I'm talking about.  I have no examples of objects that have been observed with redshifts between 12

GD: OK, I had the impression you were talking about present day conditions

JD: Maybe the issue here is that I can't literally show you what I'm talking about.  I have no examples of objects that have been observed with redshifts between 12

I'm simply saying that what you said is entirely correct and is well known in the present cosmological model

So then, you would agree that there are vast regions of transparent hydrogen gas, in the region between z=25 and z=1100?  I had the impression you were arguing with me.

No, Jonathan, GC only studied that region. I've given you a paper which lists spectral lines for H2 twice already, this is the third time.

I see you posted "The molecular spectrum of Molecular Hydrogen and its Isotopes" by G. H. Dieke 

http://kicp.uchicago.edu/~odom/molecule_database/H2/molecular%20hydrogen%20spectrum%20tabulation.pdf

I must have missed this article earlier, because my response only discussed the Lyman, Paschen, Balmer series.  But more likely, I may have glanced through the article and actually believed it was ostensibly a discussion of emission spectra of monatomic hydrogen.  The text is actually  a discussion of the emission spectra of diatomic hydrogen, and isotopes of diatomic hydrogen, in a discharge tube.

Your question was "Where can I find the emission and absorption spectra of diatomic hydrogen gas?" and I've given you a paper that does that, you can't dismiss it just because you don't like the result.

If you had presented the Dieke paper with "Here.  This is the emission spectrum of diatomic hydrogen and its isotopes", then I would like your result.  But instead, you presented the article as though it was an argument against the proposition that diatomic hydrogen is transparent.  

It would be incorrect, though, to call Dieke's result the "absorption spectrum" of diatomic hydrogen, and that has little to do with like, or dislike on my part, but more to do with the meaning of the word "absorption spectrum" which is "a spectrum of electromagnetic radiation transmitted through a substance, showing dark lines or bands due to absorption of specific wavelengths."

If we wanted the absorption spectrum of diatomic hydrogen, we would have to fill a chamber with cold hydrogen gas, and shine a continuous spectrum light through it and measure the absorption lines on the other side of the gas.  

What Dieke's (et al) experiment did was to put the hydrogen gas in a discharge tube, and send electrical pulses through it to ionize the gas, and then measure the resulting emission spectrum.

The emission spectrum from a plasma created from diatomic hydrogen is not the same as the absorption spectrum of cold diatomic hydrogen.

Actually, I think that in the case of monatomic hydrogen, in fact, the emission and absorption spectra are the same...  But these results from Dieke are more indicative of a phosphorescent-like behavior.

GD: Your question was "Where can I find the emission and absorption spectra of diatomic hydrogen gas?" and I've given you a paper that does that, you can't dismiss it just because you don't like the result.

You know, I guess I can confess to a little bit of that.  I claim that Diatomic hydrogen gas is transparent.  I'm sure enough of that that I will work pretty hard to figure out a reason to dismiss evidence against that conclusion.

If you give me really rock-solid evidence that I'm wrong--that diatomic hydrogen has an infrared absorption spectrum, I might believe it.  But if I think it's not sufficient evidence, then I'm going to tell you...

a...  That's not diatomic hydrogen, that's monatomic hydrogen.

b...  Those are not frequencies in the infrared, radio, and microwave spectrum

c...  That's not cold diatomic hydrogen, that's hot diatomic hydrogen.

d... That's not cold diatomic hydrogen, that's hot hydrogen in a discharge tube.

e.  That's not an absorption spectrum, that's an emission spectrum.

You might see my argument evolve as you present more information, and you might see that as somehow "dismissing the evidence because I don't like the result."  But no, I don't think I'm really dismissing any of the actual evidence.  

I'm just dismissing certain interpretations of the evidence that suggest that cold diatomic hydrogen has an absorption spectrum.

JD: e. That's not an absorption spectrum, that's an emission spectrum.

JD: I'm just dismissing certain interpretations of the evidence that suggest that cold diatomic hydrogen has an absorption spectrum.

Previously:

JD: Actually, I think that in the case of monatomic hydrogen, in fact, the emission and absorption spectra are the same.

That is always the case. Spectral lines represent changes within the molecule that have specific energy differences. If the gas is cold and illuminated, it will preferentially absorb at the line resulting in molecules in the lower energy state being boosted to the higher state and we see an absorption line. If the gas is hot and in a cold environment, the molecules will naturally decay from the high energy state to the low and emit a photon, we then see the line in emission. If there is a resonance, whether it is seen in absorption or emission depends only on the environment and which mode is used to make the measurement is only a matter of convenience. Dieke et al used a discharge to excite the gas which produced both the atomic and molecular spectra, but the various series of the atomic lines are extermely well known (Lyman, Balmer etc.) and as they note they were able to use silver coating to catalyse the recombination reducing the intensity of those.

These are extracts from the paper:

• The spectra of all six known isotopic species of molecular hydrogen, H2, DH, D2, TH, TD, T2 have been obtained with high dispersion under several excitation conditions. This has led to a complete wavelength table of these spectra containing about 100,000 different lines.

While there will be some commonality, that's going to be about 16000 lines per species.

• With most molecules only a few electronic levels can be studied with sufficient detail. In the molecular hydrogen spectrum there is a great variety of electronic levels each one with its superimposed vibrational and rotational structures.
• The work on the present program is concerned primarily with the spectrum between 3000A and 3 microns.

That covers the visible (from about 4000A to 7000A) through into the near infra-red and there are thousands of lines just in that region.

http://kicp.uchicago.edu/~odom/molecule_database/H2/molecular%20hydrogen%20spectrum%20tabulation.pdf

JD: Actually, I think that in the case of monatomic hydrogen, in fact, the emission and absorption spectra are the same.

GD: That is always the case.

If it is always the case, that would be a serious blow to my conjecture that diatomic hydrogen could be transparent, with no absorption spectra.  

1.  Would you agree with me, though, that a diatomic hydrogen gas will not absorb light in the Balmer, Paschen, Lyman series?  For instance, if I shine a light corresponding to the Hydrogen alpha line, into a bottle of cold diatomic hydrogen, it won't absorb it right?

2.  Would you agree that in the paper, the diagram for H_2 looks much less complicated than the other diagrams?  If I can narrow the search, I am looking for emission lines in the H2 spectrum here, that are NOT in the Balmer, Paschen, Lyman series. 

3.  Would you agree that the DH, D2, TH, TD, T2 consist, in part, of deuterium and tritium, which have miniscule half-lives relative to the age of the universe, and should not be found in significant quantites?

What remains of relevance here to the question?  Here is something that I will not dismiss.  If you look at the H2 emission spectra (image attached), there seem to be some large emissions that are numbered.  My impression is that those numbers are associated with their order in the Paschen, Balmer, or Lyman series.  So I would not consider these to be emissions from the diatomic hydrogen, but rather from monatomic hydrogen that was produced by the electric current.

However, what remains here are several very small peaks.  If the diatomic hydrogen can be heated by shining light of those frequencies into the gas, then it is an absorption spectra.

GD: They were able to use silver coating to catalyse the recombination reducing the intensity of those.

Is this a form of Light amplified by stimulated emission of radiation?  That is, did they find a way to do exactly what I was saying:  heat the diatomic hydrogen by reflecting those particular wavelengths back on it?  

They say "The atomic spectrum can never be completely suppressed

but its intensity can be greatly reduced by silvering the walls of the discharge

tube lightly in order to render them good catalyzers for the recombination of the

atoms formed by the discharge."

They're running 1.5 amps of electricity through the gas at very high voltages, which produces large numbers of monatomic hydrogen atoms.  Then, somehow, by means I'm sure I don't entirely understand, coating the tube with silver made it possible for many of the protons to recombine with the electrons without emitting any light... I guess, by traveling through the conducting surface.  So what remains in the record are hydrogen atoms combining with hydrogen atoms.  

...and that's where to look for the absorption spectra of diatomic hydrogen.  How many semi-stable "orbitals" do two hydrogen atoms have within a hydrogen molecule, and what are their levels?  

http://kicp.uchicago.edu/~odom/mole...ydrogen%20spectrum%20tabulation.pdf

JD: ...and that's where to look for the absorption spectra of diatomic hydrogen.  How many semi-stable "orbitals" do two hydrogen atoms have within a hydrogen molecule, and what are their levels?  

By the way, I considered this question was already answered in the first two replies in this thread by the attached pdf document, and the emphasis made by Osamah Nawfal Oudah and Harry ten Brink

"and from this reference there is this high-lited alinea on page 360: 'Homonuclear diatomic molecules have no dipole allowed vibrational-rotational spectra. This means they do not absorb or emit radiation on transitions within the same electronic state. They may have very weak quadrupole transitions.'

My interpretation here, of "same electronic state" was "any state that can be described as diatomic hydrogen"  and my interpretation of quadrupole transitions was "huh?"

Harry ten Brink has said those quadrupole transitions are very weak, and in the IR range, but I still don't know for sure whether these are "emission only" transitions, or "emission/absorption" transitions.  If they are "emission/absorption" transitions, then my hypothesis would need modification.

When speaking of a "quadrupole" interaction in a symmetric diatomic molecule, is this really an interaction with a single molecule, or is it an interaction with a pair of molecules--(that is, four atoms.)  

I would like to posit that even if the absorption spectrum occurs with regularity in a dense cold diatomic hydrogen gas, the probability of it occurring might tend to zero as the density of the gas decreases below some critical level.

I see in https://en.wikipedia.org/wiki/Spin_isomers_of_hydrogen that the equation for the energies is

E_J = {J(J+1) hbar^2}/{2I} 

 I calculated the Delta E for transitions from 1-0, 2-1, and 2-0, and the expected wavelengths of resulting photons, using the formula and got wavelengths of 41, 13 and 20 micrometers, respectively.  

Are these transitions regularly observed?

Where and when is the rotation of the axis of quantization allowed? Is it system dependent or planet dependent (or something more complicated)?. Available from: https://www.researchgate.net/post/Where_and_when_is_the_rotation_of_the_axis_of_quantization_allowed_Is_it_system_dependent_or_planet_dependent_or_something_more_complicated#5900b35296b7e4d0666c27d2 [accessed Apr 26, 2017].

Sorry, we've been away for a while.

GD: Actually, I think that in the case of monatomic hydrogen, in fact, the emission and absorption spectra are the same.

GD: That is always the case.

JD: If it is always the case, that would be a serious blow to my conjecture that diatomic hydrogen could be transparent, with no absorption spectra.

It is certainly the case though there are some subtleties. If you pump an atom from a low level to a higher energy, say A to C, it may decay in two steps via an intermediate level B. Of course it is also possible to pump from A to B and then B to C so the emission/absorption process still shows all the possible lines.

JD: 1.  Would you agree with me, though, that a diatomic hydrogen gas will not absorb light in the Balmer, Paschen, Lyman series?

I would agree since the levels change, but there is the possibility that one of the di-atomic molecule lines coincides with a line in the series, they get very close towards the low frequency end.

JD: For instance, if I shine a light corresponding to the Hydrogen alpha line, into a bottle of cold diatomic hydrogen, it won't absorb it right?

It might but only by coincidence.

JD: 2.  Would you agree that in the paper, the diagram for H_2 looks much less complicated than the other diagrams?  If I can narrow the search, I am looking for emission lines in the H2 spectrum here, that are NOT in the Balmer, Paschen, Lyman series.

I believe the tables do not include the monatomic lines, they would have been removed but only the mass of the nucleus differs so the same patterns will exist just moved in frequency by a constant ratio.

3.  Would you agree that the DH, D2, TH, TD, T2 consist, in part, of deuterium and tritium, which have miniscule half-lives relative to the age of the universe, and should not be found in significant quantites?

Deuterium is stable, it still exists from primordial nucleogenesis and is a key measurement in cosmology. Tritium would have decayed, it's half life is  about 12.3 years.

JD: What remains of relevance here to the question?

The document tabulates over 27,000 lines for D2. All the same lines will exist for hydrogen but shifted due to the change in the mass of the nuclei. They say they expect a larger number for the H2 table but it was not complete at time of publication.

JD: Here is something that I will not dismiss.  If you look at the H2 emission spectra (image attached), there seem to be some large emissions that are numbered.  My impression is that those numbers are associated with their order in the Paschen, Balmer, or Lyman series.  So I would not consider these to be emissions from the diatomic hydrogen, but rather from monatomic hydrogen that was produced by the electric current.

You could be right, or some may be due to contaminants. the tables contain just the relevant molecular lines.

JD: However, what remains here are several very small peaks.  If the diatomic hydrogen can be heated by shining light of those frequencies into the gas, then it is an absorption spectra.

Exactly. That is the answer to your original question.

JD: Is this a form of Light amplified by stimulated emission of radiation?

No, just adsorption of the plasma giving it a chance to cool I expect.

JD: They say "The atomic spectrum can never be completely suppressed but its intensity can be greatly reduced by silvering the walls of the discharge tube lightly in order to render them good catalyzers for the recombination of the atoms formed by the discharge."

Exactly, it catalyses the normal recombination and keeps the light generated out of the main beam if it is released on the walls.

They're running 1.5 amps of electricity through the gas at very high voltages, which produces large numbers of monatomic hydrogen atoms.  Then, somehow, by means I'm sure I don't entirely understand, coating the tube with silver made it possible for many of the protons to recombine with the electrons without emitting any light... I guess, by traveling through the conducting surface.  So what remains in the record are hydrogen atoms combining with hydrogen atoms.  

JD: ...and that's where to look for the absorption spectra of diatomic hydrogen.  How many semi-stable "orbitals" do two hydrogen atoms have within a hydrogen molecule, and what are their levels?

Tens of thousands in just that narrow visible and near infra-red band, and it is easier to measure them than calculate them.

JD: When speaking of a "quadrupole" interaction in a symmetric diatomic molecule, is this really an interaction with a single molecule, or is it an interaction with a pair of molecules--(that is, four atoms.)

It's a single molecule, two nuclei and two electrons giving four 'poles' whereas a single atom has just one positive and one negative making a dipole. See the diagrams in the link.

http://encyclopedia2.thefreedictionary.com/Electric+Quadrupole+Radiation

Resolved issues.

1.  In general, the lyman, alpha, paschen series photons are not part of the absorption spectrum of pure diatomic hydrogen.  

2.  Deuterium is indeed stable.  (I had been fooled by a google search of "Deuterium Half-Life," and a careless reading of the result.  Try it, and you'll probably see what I mean.)

Issues where I think you may be right:

3.  I counted the emission lines in the spectrum for H2 (43 lines), D2 (73 lines) and DH (110 lines).  I thought this was because the H2 spectrum was going to be a more simple spectrum.  You said (GD) "All the same lines will exist for hydrogen but shifted due to the change in the mass of the nuclei."

4.  The quadrupole interactions in the discharge tube are coming from dipole pairs within the same hydrogen molecule.  (I'm still wondering whether you couldn't have quadrupole interactions between dipoles in two neighboring molecules, though)

Outstanding questions

5.  Just what do those numbered emission lines in figure 1 represent?

6.  Does adsorption cause any emission spectra?  

7.  I especially need to learn the notations here... things like "2a=2s^3Sigma" "K=0" "v=0" "v=1" etc.

Why I'm still not quite convinced that diatomic hydrogen has an absorption spectrum:

 8.  I still think that an absorption spectrum is not always the same as the emission spectrum.  Let's not forget that what is being measured here is an emission spectrum from a gas in a high electric field.  To measure an absorption spectrum, you would have to pass a continuous blackbody spectrum through the gas, and then measure how much each wavelength is dimmed.  To say "It is easier to measure than to calculate", I would say we need both.  Measuring will probably give some false positives.  If there are frequencies not predicted by theoretical models diatomic hydrogen, they may be caused by interactions with the silver, or trace gasses in the tube that weren't meant to be there.

9.  The electric field of the discharge tube contorts the molecules into conditions of asymmetry in the quadrupoles so that it can emit a photon in that wavelength.  Can these same contortions occur in a region of zero electric field, so that a stable molecule, with no quadrupole, can absorb a photon which puts it into the quadrupole configuration?   Common knowledge here... (Kirchhoff's second Law of radiation: "For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.") ... suggests that of course it can absorb whatever wavelength it can emit.  I am suggesting, to the contrary, that there may be some wavelengths that an electrically stimulated hydrogen molecule can emit, but a non-stimulated hydrogen atom  cannot absorb.

All that being said, I need to learn the formulas, and derivations of the formulas that Deike uses to calculate the emission spectra of diatomic hydrogen.  Deike refers to Richardson and Fulcher's theoretical work.  I wonder if there might be any introductory treatments of the theory that I could follow.

JD: Common knowledge here... (Kirchhoff's second Law of radiation: "For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.") ... suggests that of course it can absorb whatever wavelength it can emit.

That's right. If you understand that then imagine an experiment where you have a clear glass container filled with hydrogen on one side of a dichroic mirror and a black body on the other, all three being inside a perfectly reflective container. The mirror is designed to reflect all wavelengths other than a narrow band at which it is transparent. Kirchoff's Law must apply to emission and absorption in that band in isolation otherwise, if the hydrogen emitted but didn't absorb, the temperature of the gas would fall while that of the black body would rise. Photons emitted by the gas would then pass through it until they encountered the black body and were absorbed. Photons from the black body would do the same. The temperature difference that would arise could then be used to drive a heat engine.

Either hydrogen molecules absorb at the same spectral lines as they emit or you have a simple method of building a "perpetual motion" or "over unity" machine (of the second kind). In fact the law means that if you know the probability of an atom in an excited state emitting a photon to drop to a lower energy state, you can also calculate the probability that a photon passing an atom in the low energy state will absorb it.

Interesting setup.  This reminds me of a quora answer that I put up a couple years ago:

https://www.quora.com/What-will-happen-if-an-electric-bulb-is-lit-inside-a-room-full-of-perfect-mirrors/answer/Jonathan-Doolin

At the time, I was thinking, like you, that the absorption spectrum of air would be the same as the emission spectrum. And I still think that's the case for greenhouse gasses, like carbon-dioxide, methane, and water-vapor.  

However, non-green-house gasses, like diatomic nitrogen, diatomic oxygen, and diatomic hydrogen will neither absorb, nor emit light in such a situation.  While they could still gain or lose heat by convection, they would neither gain nor lose heat by radiation.  

These non-greenhouse: diatomic gasses,  have an emission spectrum when they're put into a discharge tube because thousands of volts are applied across the gas, to push them past their dielectric strength.  But the cold version of the same molecules doesn't have absorption spectrum in those same frequencies.  On the other hand, cold greenhouse gasses, can absorb energy in the same bandwidths as their emission spectra, and they can pass that energy on to the other transparent molecules via convection.

GD:  if the hydrogen emitted but didn't absorb, the temperature of the gas would fall while that of the black body would rise. 

You're imagining an environment where the diatomic gasses emit light but they don't absorb, in which case, yes, they would have heat passing from a cooler to a hotter body, thus violating a law of thermodynamics.   But what I have been proposing is that the diatomic hydrogen is transparent in such situations, neither emitting, nor absorbing.  It could only transfer heat via convection.

JD: You're imagining an environment where the diatomic gasses emit light but they don't absorb, in which case, yes, they would have heat passing from a cooler to a hotter body, thus violating a law of thermodynamics.

Right, that was in response to your hypothesis that hydrogen gas could emit but not absorb at some particular spectral line, it can't for exactly that reason.

JD: But what I have been proposing is that the diatomic hydrogen is transparent in such situations, neither emitting, nor absorbing.

The article we have been discussing states that they have tabulated more than 27000 spectral lines for D2 in the band they observed and expect more for H2. Each of those lines must also be able to absorb in accordance with Kirchoff's Law. I got the impression you were saying the lines might be present in emission but not in absorption, please re-read your own posts if that was not the case as I am sure you gave that impression.

GD:  got the impression you were saying the lines might be present in emission but not in absorption,

That's partially correct.  

I was saying the 27,000 lines should be present in the emission spectrum from hydrogen gas put into a discharge tube and high voltages which break down the dielectric.  

However, I was also saying, the 27,000 lines should not be present in either the absorption spectrum or emission spectrum of cold diatomic hydrogen in an electrically neutral environment.

And I was saying if the 27,000 lines were present in the absorption spectrum of cold diatomic hydrogen, then diatomic hydrogen should be categorized as a greenhouse gas.

The paper is specifically looking at the molecular form so the atomic series, Lyman, Balmer, etc. are excluded from the tables. Those thousands of lines therefore exist as a result of the structure of the molecule. If the gas is hot, they will be seen in emission, if it is cold they will absorb. Kirchoff's Law requires that symmetry.

Molecular hydrogen is classed as an indirect greenhouse gas similar to CO2 but there is a surfeit of oxygen in the atmosphere and ozone in the upper reaches so it reacts to produce water very quickly keeping the fraction quite low. The figure I've found is 0.5ppm for hydrogen compared to 379ppm for CO2 in the first link. It's consequences are discussed in the second link.

Bear in mind also that evidence for the existence of dark matter comes from structure formation which requires that DM was collapsing into halos prior to the CMB being released, a time when we know that the hydrogen/helium mix was very hot, dense and completely opaque. That follows from the fact that the CMB is an exact match to the Planck curve. Kirchoff's Law again is important, if the emission is 100% of a black body, its absorption must also be 100%.

http://www.kayelaby.npl.co.uk/chemistry/3_1/3_1_4.html

http://www.ghgonline.org/otherhydrogen.htm

To say diatomic hydrogen is an "indirect" greenhouse gas is a reasonable and well intended classification.  It means that while diatomic hydrogen, on its own is transparent in the infrared, the emissions are still going to contribute to the greenhose effect, because these light molecules will immediately clime to the thermosphere where they will find lone oxygen atoms, and combine, forming hydroxide, which is not transparent to infrared light.

However, you seem to suggest here that carbon dioxide is an indirect greenhouse gas.  That's not accurate.  Carbon Dioxide is generally quite stable, and is not transparent in the infrared.  It's contribution to the greenhouse effect is quite direct.

GD: That follows from the fact that the CMB is an exact match to the Planck curve.

It's funny.  The last person I was discussing this with insisted that the CMB should consist of emissions from positron-electron interactions.

JD: It means that while diatomic hydrogen, on its own is transparent in the infrared

It isn't transparent in the infra-red, haven't we just gone over the fact that in one decade of the visible to infra-red region, there are over 27000 spectral lines?

JD: However, you seem to suggest here that carbon dioxide is an indirect greenhouse gas.

Not me, I know little about the topic, see the page I cited for the source of that.

JD: Carbon Dioxide is generally quite stable, and is not transparent in the infrared.

The same is true of hydrogen but I believe both get involved in other reactions.

JD: It's funny. The last person I was discussing this with insisted that the CMB should consist of emissions from positron-electron interactions.

That is correct, the majority of the photons were produced by that process when the universe was around a few seconds old, but they bounced around in the cooling thermal bath of free protons and electrons for ~380 thousand years so remained in thermal equilibrium until released, hence they are seen with the thermal spectrum of the time when they decoupled.

GD: It isn't transparent in the infra-red, haven't we just gone over the fact that in one decade of the visible to infra-red region, there are over 27000 spectral lines?

We have repeated ad nauseam that there are 27000 emission lines when you put hydrogen gas in a discharge tube that runs 1.5 amps of current through the gas.   

GD: Not me, I know little about the topic, see the page I cited for the source of that.

Okay,  well, neither of the pages you cited claimed that diatomic hydrogen was a greenhouse gas, nor did either one claim that carbon dioxide's contribution to the greenhouse effect was "indirect".  And if you think they did, you need to do more research.

GD: they bounced around in the cooling thermal bath of free protons and electrons for ~380 thousand years so remained in thermal equilibrium until released

So what are you saying here?   Are you saying that when free protons and electrons combine, they will produce a blackbody spectrum because they are in thermal equilibrium when they do it?  

I don't think it matters what is the temperature of a proton and an electron, if they are separate, they will come together in a chemical reaction that produces an emission spectrum including balmer, lyman, paschen series.  

The table would look like this

Temperature              Composition and Spectrum

>10,000 Kelvin          Bath of free protons and electrons:  

                                     Lyman, Balmer, Paschen series dominant

3000-10,000 Kelvin   Mostly diatomic hydrogen; Blackbody radiation

                                     if surface is dense enough

0-3000 Kelvin           Almost entirely diatomic hydrogen.  

                                    Completely transparent

JD: We have repeated ad nauseam that there are 27000 emission lines when you put hydrogen gas in a discharge tube that runs 1.5 amps of current through the gas.

We have also repeated ad nauseam that neutral single atoms produce the Lyman, Balmer, etc. series and that all other lines are from the diatomic molecule.

We have also repeated several times that Kirchoff's Law requires that any line that exists in the emission spectrum must also exist in absorption.

The above three points therefore answer your original question.

GD: they bounced around in the cooling thermal bath of free protons and electrons for ~380 thousand years so remained in thermal equilibrium until released

JD: So what are you saying here?   Are you saying that when free protons and electrons combine, they will produce a blackbody spectrum because they are in thermal equilibrium when they do it?

No, perhaps I wasn't clear.

When positrons annihilate with electrons, it produces pairs of gamma rays. At the end of that phase in the early universe when it was a few seconds old, for every free proton, there was one free electron and about a billion photons, and all three had such high energy that they behaved like the classical "billiard ball" model of particles.

If you shoot one fast ball into a table of slowly moving balls on a frictionless table, the energy will be shared out through impacts until all are moving at roughly the same mean speed, and the actual speeds will have a well defined statistical spread about the mean.

Similarly, every time a high energy photon hit a low energy proton or electron, the photon would lose some energy while the particle would gain kinetic energy. As a result, the bath of photon, electrons and protons would quickly reach equilibrium. That would then be maintained until the electrons combined with the protons and the mean free path of photons with energy below the Lyman Alpha line suddenly increased. The source of the photons was electron-positron annihilation but the energy distribution reflects the thermal spectrum of the last free charged particles they bounced off. That's the source of the technical term "surface of last scattering".

Note I am glossing over the effect of neutrino decoupling and Helium in the gas mix.

JD: I don't think it matters what is the temperature of a proton and an electron, if they are separate, they will come together in a chemical reaction that produces an emission spectrum including balmer, lyman, paschen series.

That's basically right but the temperature needs to be low enough for them to stick, at higher temperatures they have too much energy and bounce of each other.

JD: The table would look like this

JD: Temperature              Composition and Spectrum

JD: >10,000 Kelvin          Bath of free protons and electrons:  

                                     Lyman, Balmer, Paschen series dominant

Those series are characteristic of the transitions between energy levels in the neutral atome, the free particles have no intrinsic energy levels.

JD: 3000-10,000 Kelvin   Mostly diatomic hydrogen; Blackbody radiation

                                     if surface is dense enough

Above about 5000K, almost entirely free protons and electrons, thermal equilibrium, the gas is opaque

Above about 3000K, mostly neutral gas (monatomic) but a small fraction of free protons and electrons, enough to keep the free path quite short, photons still in thermal equilibrium.

JD: 0-3000 Kelvin Almost entirely diatomic hydrogen.

                                    Completely transparent

Below 2970K, virtually all neutral atomic hydrogen  Possibly Lyman Alpha emission from denser patches which future telescopes hope to see at high redshift.

Strong absorption in residual patches of neutral gas produce the "Lyman Alpha Forest" and denser regions produce "Damped Lyman Alpha" regions where the spectral line is broadened.

This lasted until the end of the "Dark Ages" at about z=11 when UV light started re-ionising the gas into free protons and electrons at very low density leaving it completely transparent other than the effects of dispersion due to free electrons. The end of the re-ionisation was around z=6 to 7 and is indicated by the start of the Gunn-Peterson Trough.

Luckily molecular hydrogen only occurs in denser nebulae because it is not transparent, unlike the atomic version.

Obviously in all these discussions, the density of whatever form the hydrogen is in is also a critical factor.

JD: Okay, well, neither of the pages you cited claimed that diatomic hydrogen was a greenhouse gas, nor did either one claim that carbon dioxide's contribution to the greenhouse effect was "indirect". And if you think they did, you need to do more research.

You got me on that one, I mis-read the page, it refers to carbon monoxide as being indirect. From the page I linked previously:

• "Hydrogen (H2) is similar to carbon monoxide in that it acts as an indirect greenhouse gas through its effect on hydroxyl (OH) radicals. By reducing the levels of OH in the atmosphere, hydrogen increases the lifetime of some direct greenhouse gases, such as methane."

https://en.wikipedia.org/wiki/Lyman-alpha_forest

https://en.wikipedia.org/wiki/Damped_Lyman-alpha_system

https://en.wikipedia.org/wiki/Gunn%E2%80%93Peterson_trough

http://www.ghgonline.org/otherhydrogen.htm

http://www.ghgonline.org/otherco.htm

JD: We have repeated ad nauseam that there are 27000 emission lines when you put hydrogen gas in a discharge tube that runs 1.5 amps of current through the gas.

GD: We have also repeated ad nauseam that neutral single atoms produce the Lyman, Balmer, etc. series and that all other lines are from the diatomic molecule.

Neutral, but excited single atoms of hydrogen gas, at sufficient temperatures will, indeed, produce the Lyman, Balmer, Paschen series. Then, they will produce further radiation when two neutral single atoms combine to form the covalent bond of diatomic hydrogen. However, once that covalent bond has been formed, there will be no further emissions unless the molecules are separated, either by bumping into other molecules thermally, or being hit by ultraviolet or higher light.

GD: We have also repeated several times that Kirchoff's Law requires that any line that exists in the emission spectrum must also exist in absorption.

That depends on whether the emission spectra comes from stable excited states in the neutral particle, or if it comes from the formation of the particle. For greenhouse gasses, such as carbon dioxide, and water vapor, and methane, and OH, they have modes, such as asymmetric stretch, symmetric stretch, bending, wagging, twisting, scissoring,

http://www.scienceinschool.org/sites/default/files/articleContentImages/12/climate/issue12climate1_large.jpg

http://courses.chem.psu.edu/chem210/mol-gallery/methane-vib/pictures/vibrational-modes.jpg

Hydroxide can interact with a photon because of it's polarity

https://chem.libretexts.org/@api/deki/files/33967/image001.png?revision=1

Monatomic hydrogen, can also interact with a photon because of it's polarity, which shows up in the time-dependent schrodinger wave function

http://vergil.chemistry.gatech.edu/notes/quantrev/node9.html

in the e^(-iE_i t/hbar) term.... once you plug in the boundary conditions that turn it into a real-valued function.

GD: The above three points therefore answer your original question.

No, because you are conflating the emission of light caused by the formation of a covalent bond in a molecule, with the emission of light from vibrational and rotational modes of a molecule.

GD: When positrons annihilate with electrons, it produces pairs of gamma rays.

Okay.

GD: At the end of that phase in the early universe when it was a few seconds old, for every free proton, there was one free electron and about a billion photons, and all three had such high energy that they behaved like the classical "billiard ball" model of particles.

I would agree that every region of the observable universe went through a stage of this description. I think, more likely a few hundred or thousand years, rather than a few seconds. But yes, you've got Weinberg's "First Three Minutes" theory here.

GD: The source of the photons was electron-positron annihilation but the energy distribution reflects the thermal spectrum of the last free charged particles they bounced off. That's the source of the technical term "surface of last scattering".

But regardless of the handwaving, the actual surface from which the CMBR emanates is a dense region filled, largely with diatomic hydrogen, at 3000 Kelvin.

GD: Note I am glossing over the effect of neutrino decoupling and Helium in the gas mix.

I'm fine with rough estimates for now. The vast majority of the gas will be hydrogen.

JD: I don't think it matters what is the temperature of a proton and an electron, if they are separate, they will come together in a chemical reaction that produces an emission spectrum including balmer, lyman, paschen series.

GD: That's basically right but the temperature needs to be low enough for them to stick, at higher temperatures they have too much energy and bounce of each other.

Okay, your proposition seems to be that the reason that we don't see CMBR emissions from high temperature regions of the big bang is that high temperature regions won't produce atomic hydrogen that "sticks". My proposition is, OF COURSE IT STICKS, It sticks, and then, shortly after it sticks, it will be hit either by a photon that unsticks it, or it will be hit by another particle that unsticks it. The reason that we don't see it is because the layer of diatomic gas between us and it is essentially an opaque blackbody surface around 3000 Kelvin.

GD: Those series are characteristic of the transitions between energy levels in the neutral atome, the free particles have no intrinsic energy levels.

Yes, the free particles have have relative energy levels, which I've given above (Oops... editing snafu.  I deleted several paragraphs, so this is the first I've mentioned it in this thread.  See the attached diagram, and following link.) as zero eV per pair, -13.6 eV per pair, and -16 eV per pair. This is the binding energy per pair for a monatomic hydrogen atom, and roughly the binding energy per pair of diatomic hydrogen. (http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/hmol.html Here I'm taking half of the 4.52 eV binding energy for the whole molecule, and adding it to the 13.6 eV binding energy of each atom.)

GD: Above about 5000K, almost entirely free protons and electrons, thermal equilibrium, the gas is opaque. Above about 3000K, mostly neutral gas (monatomic) but a small fraction of free protons and electrons, enough to keep the free path quite short, photons still in thermal equilibrium.

What theory are you using to estimate the proportions of forms for the gas?

Applying the use of the canonical partition function, (Image Attached) I found:

At 5000 Kelvin: Diatomic Hydrogen:~100%

Monatomic Hydrogen: 2.0321*10^-14

Free Protons and Electrons:1.07807*10^-16

I'm not 100% sure that the canonical partition function is exactly the right way to go here, but I do feel, that the ratio between the chemical potentials of the various forms and the value of k_B*T is relevant.

Hi Jonathan,

I just came across a thesis paper which is mainly about formation of H2 in astrophysical conditions. It huge, 216 pages, and I have only skimmed it, but there are relevant short paragraphs on formation in the early universe from 1.2 to 1.4 and there are spectra in section 4.4 starting on page 163 of the PDF. Note these are computer generated based on rotational models and the aspect of interest is the relative amplitudes rather than the frequencies.

The theory is related to actual experiments using HD formation.

http://discovery.ucl.ac.uk/19475/1/19475.pdf

By the way, thanks, George, for bringing that Gunn Peterson trough to my attention. 

http://adsabs.harvard.edu/abs/1965ApJ...142.1633G 

I may have more comments on this, later.

JD: We have repeated ad nauseam that there are 27000 emission lines when you put hydrogen gas in a discharge tube that runs 1.5 amps of current through the gas.

GD: We have also repeated ad nauseam that neutral single atoms produce the Lyman, Balmer, etc. series and that all other lines are from the diatomic molecule.

JD: Neutral, but excited single atoms of hydrogen gas, at sufficient temperatures will, indeed, produce the Lyman, Balmer, Paschen series. Then, they will produce further radiation when two neutral single atoms combine to form the covalent bond of diatomic hydrogen.

The binding energy is ~4.5eV so if the molecule formed in a single step, that would be the minimum energy of the photon emitted, and that is in the UV range. However the atoms would also have some kinetic energy as they approached which would add to that so the spectrum would have a higher frequency tail. However, if you read the parts of the large paper I indicated, formation is a two-step process. Those would explain at most half a dozen lines.

JD: However, once that covalent bond has been formed, there will be no further emissions unless the molecules are separated, either by bumping into other molecules thermally, or being hit by ultraviolet or higher light.

The ~27,000 lines are ro-vibrational modes of the neutral molecule.

GD: We have also repeated several times that Kirchoff's Law requires that any line that exists in the emission spectrum must also exist in absorption.

JD: That depends on whether the emission spectra comes from stable excited states in the neutral particle, or if it comes from the formation of the particle.

No, the 4.5eV energy emitted on formation would correspond to the minimum energy at which the gas would absorb by disassociating the molecule into neutral atoms, with an excess energy resulting in kinetic energy of the atoms.

JD: For greenhouse gasses, such as carbon dioxide, and water vapor, and methane, and OH, they have modes, such as asymmetric stretch, symmetric stretch, bending, wagging, twisting, scissoring,

The same is true of H2 but the symmetry means there is no electric dipole, the modes still exist though.

GD: The above three points therefore answer your original question.

JD: No, because you are conflating the emission of light caused by the formation of a covalent bond in a molecule, with the emission of light from vibrational and rotational modes of a molecule.

No, the formation energy of 4.5eV is in the UV, the lines we've been discussing are rotational and vibrational modes resulting in lines in the visible to IR band.

GD: When positrons annihilate with electrons, it produces pairs of gamma rays.

JD: Okay.

GD: At the end of that phase in the early universe when it was a few seconds old, for every free proton, there was one free electron and about a billion photons, and all three had such high energy that they behaved like the classical "billiard ball" model of particles.

JD: I would agree that every region of the observable universe went through a stage of this description. I think, more likely a few hundred or thousand years, rather than a few seconds. But yes, you've got Weinberg's "First Three Minutes" theory here.

The positron annihilation was in the first few seconds, the "billiard ball" behaviour of the resulting photons is what lasted for ~380k years.

JD: But regardless of the handwaving, the actual surface from which the CMBR emanates is a dense region filled, largely with diatomic hydrogen, at 3000 Kelvin.

No, it is a dense gas at the transition from free protons and electrons to neutral atoms, much hotter than where molecules can start to form.

JD: Okay, your proposition seems to be that the reason that we don't see CMBR emissions from high temperature regions of the big bang is that high temperature regions won't produce atomic hydrogen that "sticks".

I said nothing like that. The reason we don't see higher temperature regions is the same as why we don't see the core of the Sun, the free electron/proton plasma is opaque. The neutral atom gas that forms from it can only absorb via the Lyman, Balmer, etc. series.

JD: I'm taking half of the 4.52 eV binding energy for the whole molecule, and adding it to the 13.6 eV binding energy of each atom.)

You have the key there, the plasma cools until the tail energies fall below ~13.6eV at which point neutral atoms form and the gas becomes transparent. It has to continue to cool below ~4.5eV before molecules can form so they play no part in the transition.

JD: What theory are you using to estimate the proportions of forms for the gas?

I'm giving values from memory of lecture notes I saw some weeks ago so I'll try to dig them up. The calculation I think was similar to yours but the lecturer than noted that while it gave a result accurate to an order of magnitude, the actual temperatures had to be lower because a small number of free particles in the tail of the Gaussian distribution was enough to keep the mixture opaque a bit longer.

http://discovery.ucl.ac.uk/19475/1/19475.pdf by FARAHJABEEN ISLAM

Islam makes some really surprising claims made in equations 1.1, 1.2, 1.3 here.

Equation 1.1 is saying that a hydrogen atom with two electrons is more stable than a neutral hydrogen atom, and will release a photon on formation. (no source cited)

Equation 1.2 is saying that a diatomic hydrogen with one electron is more stable than a neutral hydrogen, and a proton, and will release a photon on formation. (no source cited)

Are those well-known chemical interactions?

Then Islam goes on to say, in section 1.4, that the process that creates diatomic hydrogen from monatomic hydrogen is extremely unlikely, and he cites [19] Desorption From Interstellar Ices https://arxiv.org/pdf/0708.3374.pdf as support for this argument.

However, I didn't find an argument in the cited article that supports this claim.   If you see something I overlooked, let me know. 

Islam's argument in section 1.4, though, says that if two hydrogen atoms meet, they will not have sufficient contact time to release 4.5 eV.  But I'm not sure of what accepted principle of physics or chemistry requires a certain amount of contact-time between particles for an energy to be released.  Heisenberg's uncertainty principle came to mind, but if it applies as Delta E Delta t >= hbar/2, then the necessary contact time would be Deltat t >=7.3x10^-17 seconds, which would become more and more likely with colder and colder gasses.

JD: Islam makes some really surprising claims made in equations 1.1, 1.2, 1.3 here.

JD: Are those well-known chemical interactions?

JD: .. he cites [19] Desorption From Interstellar Ices https://arxiv.org/pdf/0708.3374.pdf as support for this argument.

JD: However, I didn't find an argument in the cited article that supports this claim.   If you see something I overlooked, let me know.

I think you've misread the citations, what he says in 1.1 to 1.4 is standard knowledge for which he cites [3], "Dyson, J.E. and D.A. Williams,The Physics of the Interstellar Medium. 2nd edition, 1997, Bristol: Institute of Physics.". Looking at the index on Google, I think he's referring to section 3.4 starting on page 34 but there's no preview for that page. Ref [19] is about an alternative formation mechanism.

JD: Islam's argument in section 1.4, though, says that if two hydrogen atoms meet, they will not have sufficient contact time to release 4.5 eV.  But I'm not sure of what accepted principle of physics or chemistry requires a certain amount of contact-time between particles for an energy to be released.

Again, all that is not "Islam's argument", it is taken from a standard textbook.I haven't studied the physics in that level of detail so you'd need to research it yourself but I would think you'd find some explanations in on-line lecture material.

His thesis was on adding one new mode extending existing simulation software for predicting the spectrum.

https://books.google.co.uk/books?id=k7x3Es3OrV8C&printsec=frontcover#v=onepage&q&f=false

I'm not sure if it matters where the argument comes from.  (Except in some cases, it might be interesting to see exactly where the miscommunication or misconception arose.) What matters is that it is false.  

To say  as a general rule  "There is not sufficient contact time for the excited hydrogen molecule to radiate away its binding energy of 4.5 eV before it is likely to dissociate" , is a false generality, regardless of who made the error.

The only time that would be true is if the temperature was high enough to make collisions of 4.5 eV common-place, and density of the gas were so high that the average time between collisions was well under 7.3*10^-17 seconds.

JD: What matters is that it is false.

If that's what you think, I suggest you start looking at the textbooks and examining the source of the information. This clearly has been what is taught in undergraduate classes for at least two decades and numerous astronomers have worked for many years on finding mechanisms by which the higher-than-expected fraction of molecular hydrogen could be explained, hence the concentration and studies of the effects of dust grains and adsorption. You could start with the one cited, it's available second hand for around $12. If you can show the calculations are in error and solve a problem nobody else has cracked, you'll be famous.

That is what I think.  I'd be very surprised if there were any calculation in the referenced text regarding the matter, since it can't be supported by any basic principle.  But I'll see what I can find.

Scientists don't just pluck numbers out of thin air. Of course calculations must have been done and when the observed density was much higher than the result, they will have been checked before anyone spent time looking for alternative formation routes.

The relevant part of "The Physics of the Interstellar Medium" is attached.  Dyson and Williams say "At the temperatures, densities, and radiation intensities in the interstellar medium molecules do not form readily nor do they persist indefinitely."

And they also say "The problem of formation of molecules is enhanced by the fact that they are rapidly destroyed"

However, Dyson and Williams do not take any care to make any distinction here between different molecules, and different properties of different interstellar regions.  

D&W refer to a largest transition probability "usually about 10^8/second".  Above (page 4 of thread) I thought that the transition probability per second ought to be tied to the Heisenberg Uncertainty Principle.  But I've also been told that the "Fermi Golden Rule" might apply here...  Something I'll have to examine further. So that I don't forget: https://farside.ph.utexas.edu/teaching/qm/Quantum/node87.html#e8.196 

I gather, this notation is something I need to understand better than I currently do.

The other thing here is that D&W seem to assume (as the mean interstellar UV flux) here that there are 10 billion ultraviolet photons per square meter per second in every nanometer wavelength of ultraviolet light (e.g. 10 billion per second at 130 nanometers, 10 billion per second at 131 nanometers, 10 billion per second at 132 nanometers, etc; all the way from 120 to 190 nanometers)  I did a calculation, and this seems to be around 10 microWatts of energy in these wavelengths.  (attached below)

I will have to do some further research to know which interstellar regions that "mean interstellar UV flux" would be common.  Is that what we are currently receiving from the sun?  Or is that what we are receiving from all the other stars, after we remove the contribution of the sun?  Or is that what we would be receiving, if we removed all the stars and the sun, and only considered the contribution of UV from distant galaxies?  

It occurred to me you could use an integral of Planck's distribution to figure out how much of the 1300 Watts we receive from the sun ought to theoretically be in the range from 130-190 nanometers, at 1 Astronomical Unit from a class G star.  Assuming that at 1AU from the sun, we receive about 1300 Watts per square meter, I calculated, ("How-Far-From-Sun.PNG) from integrating the Planck Distribution this would provide about 1 Watt per square meter in wavelengths between 120-190 nanometers.  

Then, assuming that luminosity varied with distance squared,  L/L_0 = (1/D^2)/(1/D_0^2) I found that the distance from the sun at which you would receive 10 microWatts per square meter, would be about 0.008 light-years from the sun.  That's ten times the distance of pluto, but I think it still seems too close to the sun to be considered "the mean interstellar UV flux", seeing as the average distance between stars is close to 5 light years, 

https://www.youtube.com/watch?v=jspPcFgq04Y&t=9s

George's last comment here, a little less than a year ago, was to claim that "Scientists don't just pluck numbers from thin air". I demonstrated in my last two posts, in fact, Dyson and Williams did, in fact, pluck numbers from thin air. Then Farahjabeen, took those numbers, applied a factor of 10^16 to one of them, misused them.

George decided not to respond, so I didn't press it, but now he's responded in another thread.

https://www.researchgate.net/post/Is_the_use_of_relativity_of_simultaneity_in_standard_cosmology_appropriate

GD: " The CMB comes from 380k years after the bang in the standard model, I don't know where it could come from ... as there is no matter to produce it.... "

(I removed a few words that are irrelevant to the context here.)

So, since George seems unaware of the prevalent theory of where the CMB comes from, George, the CMB comes from the thermal interaction of Hydrogen gas at about 3000 Kelvin. I'm not making that up. It's on Wikipedia, even!

https://en.wikipedia.org/wiki/Cosmic_microwave_background.

" As the universe expanded, adiabatic cooling caused the energy density of the plasma to decrease until it became favorable for electrons to combine with protons, forming hydrogen atoms. This recombination event happened when the temperature was around 3000 K or when the universe was approximately 379,000 years old.[12] As photons did not interact with these electrically neutral atoms, the former began to travel freely through space, resulting in the decoupling of matter and radiation.[13] "

I thought it might be useful to revisit this question. To do so, I re-watched the video in my last post, and am giving time-stamped descriptions of what I'm figuring. In this way, I provide a way that others may respond, without having to work quite so hard to describe something from the video.

0:00 I believe that diatomic hydrogen is transparent and there is a lot more than the expert's think.

0:25 The formation of interstellar hydrogen...

0:45 Hydrogen atom with two electrons? Surprising.

0:58 Ionized diatomic hydrogen in a stable state? That thing will bust apart.

1:30 An interaction that should be commonplace: Two hydrogen atoms combine to form diatomic hydrogen... Fareed argues is "inefficient".... "There is not sufficient cotact time for the excited hydrogen molecule to radiate away its binding energy of 4.5 eV before it is likely to dissociate."

2:15 It sounds like he thinks it takes a longer amount of time to get rid of a large amount of energy, which is backwards

2:40 Delta E Delta t >= hbar, so the minimum amount of time to get rid of 4.5 eV of energy is on the order of 146x10^-18 seconds. That is all the time it takes for two hydrogen atoms to emit the photon and be forever bound. Only H atoms moving at extremely high velocities could avoid combining.

4:20 One radian of the photon is at 146x10^-18 seconds. The period of the photon is 918x10^-18 seconds. More than enough time for 4.5 electron volts to dissipate.

5:30 Fahrajabeen says there is a 10^-8 per second probability of the two atoms combining. If Delta t had come out to be on the order of 10^8 seconds, I would agree with him. But as it is, Delta t is on the order of 10^-17 per second, so the probability is about 100% per second of being in close proximity.

6:40 The period of vibrational oscillation is an absurd thing to be concerned about, since we want to know how long the atoms are in proximity.

7:20 GD "This is not Islam's argument. It is taken from a standard textbook" The Physics of the INterstellar Medium... (Dyson Williams) Many of the words and numbers from Dyson Williams appear out of context in Farajabeen's text. Dyson is clearly using some of the same numbers, and cites his source as Dyson Williams, but the only thing you can clearly conclude from this is that Farajabeen has no clue what Dyson Williams were saying, and neither did the peer-review, or graduate advisor who allowed Farajabeen's text to be published.

8:20 Dyson and Williams do talk about the time of contact being very short, and they compare the time of contact to the period of vibrational oscillation. They say they are comparable, which is true. You can compare any two numbers.

8:55 Dyson and Williams use 10^8 per second. Farahjabeen uses 10^-8 per second. There is a sixteen order of magnitude difference here.

9:30 I said "It might be interesting to see where the miscommunication or misconception arose." To say "there is not sufficient contact time to radiate away 4.5 eV of energy" is false, regardless of who made the error. In a very very hot plasma, it might be true. But in a cold gas, it is false.

10:15 Dyson Williams didn't make any mistake, but they did introduce this irrelevant topic: the period of oscillation.

11:20 George Dishman says: "This clearly has been what is taught in undergraduate classes for at least two decades" Egad! I hope not.

11:40 JD: "I'd be very surprised if I can find any calculation in [Dyson/Williams] regarding the matter, since it can't be supported by any basic principle."

11:51 GD: "Scientists don't just pluck numbers from thin air"

12:00 Dyson-Williams: "Consider two atoms, A and B" without making a distinction of what those atoms are... saying "contact is very short" without giving a description of contact radius, or velocity, then saying "it's comparable to the period of a vibrational oscillation" (which it is, because two numbers with the same units are always comparable) then saying "the largest transition probabilities are around 10^8 per second"

12:35 Dyson and Williams were very clearly "plucking numbers from thin air." I don't object to that, because they're just kind of thinking through some things without any specific examples. But when you take that "general" method and "general" reasoning, you need to solve the problems that Dyson-Williams have worked out... Scientists don't OFTEN pluck numbers

13:40 I don't fault Dyson/Williams for plucking numbers from thin air, but Farajabeen got the probability off by 16 orders of magnitude, and he misunderstood their argument.

14:30 However, in Dyson Williams, on the next page, I think I do see an error. "the mean interstellar UV flux in these wavelengths" At first I thought, 10 billion photons in each wavelength from 120 nm to 190 nm? That's got to be a huge amount of energy" It turned out, though to be about 10 microWatts.

16:08 Dyson Williams claimed this was the mean amount of ultraviolet radiation to which all the interstellar medium was exposed.

16:20 How much ultraviolet radiation does the earth get, then? I build a Planck Spectrum sketch to figure out how much intensity we get from 0 to 190 nanometers.

20:45 The flux of ultraviolet photons with wavelength under 190 nanometers is about 1.2606 Watts per square meter

21:15 Earth's position is not "average interstellar space" Earth's thermosphere is exposed to 1.26 watts... Where do you get 10 microwatts per square meter? Luminosity has an inverse square of distance pattern.

23:15 Setting 10 microWatts/1.26 Watts to (1AU)^2/(D)^2 and solving for D, is about 369 Astronomical Units, or about 0.5% of a light-year. The Oort Cloud is 50,000 Astronomical units out. Most of the interstellar gas would be outside the Oort Cloud.

25:25 Using 1AU^2, compared to the distance to the inner Oort Cloud. The amount of Intensity of ultraviolet radiation at the Oort Cloud would be 50.8 nanoWatts per square meter MAXIMUM... This is a HIGH estimate for the amount of ultraviolet radiation that interstellar gas would be exposed to.