Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Interstellar medium: Theory and models.

  1. Dec 24, 2004 #1

    hellfire

    User Avatar
    Science Advisor

    I am trying to get a basic understanding of the physics of the interstellar medium and the current models which describe it (FGH and McKee-Ostriker).

    A first question arises regarding the possible assumption of a gas in ionization equilibrium, in which ionizarion and recombination occur at same rates. Does such an assumption imply that a radiation field cannot be a main source of heating of the gas (since the thermal electrons due to ionization will be trapped again)? I wonder that in the FGH model the main sources of heating are cosmic rays and free electrons from dust and metals.
     
  2. jcsd
  3. Dec 28, 2004 #2
    Hi, I have been thinking about it, I'm not sure that I comprehend well the question, but given that nobody answer I will give my 2 cents. According to this page
    http://www-astronomy.mps.ohio-state.edu/~pogge/Ast871/Notes/Introduction.pdf [Broken]
    there are 4 principal kind of equilibria that can be found in the ISM: kinetic equilibrium, excitation equilibrium, ionization equilibrium and pressure equilibrium. The paper says that in the ISM, the radiation field is primarily responsible for ionization
     
    Last edited by a moderator: May 1, 2017
  4. Dec 29, 2004 #3

    hellfire

    User Avatar
    Science Advisor

    Thank you meteor, I do already know these nice lectures. It is not clear for me why only photoionization of dust and metals, as well as cosmic rays, are the only heating sources of the ISM (apart of supernovae shock waves). I assume that the warm phase of the ISM is not heated and remains at a given temperature if it is in equilibrium, unless a supernova blast takes place leading to an evaporation to the hot phase. But radiation from hot stars ionizes hydrogen from the neutral phase leading to the formation of this warm phase. Isn't this process considered a heating?
     
  5. Mar 16, 2005 #4

    hellfire

    User Avatar
    Science Advisor

    Could anyone explain how the terms “Lockman layer” and “Reynolds layer” are used? As far as I know these have something to do with the local environment in the galaxy and probably correspond to a warm neutral medium and to a warm ionized medium, correct? Do these terms refer to some structures around the local bubble?

    Further on, may be someone can answer this question: In the science site of the CHIPS mission I read:

    How can this be empirically differentiated from other cooling mechanisms?
     
  6. Mar 16, 2005 #5

    SpaceTiger

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm not sure I understand the confusion. Is there something else you would expect to heat the ISM? By heating, we're referring to the transfer of energy from something which isn't part of the ISM to something which is. Potential outside sources of heat would be stars, compact objects (like quasars), dark matter, and dark energy. I've never heard of any theories in which those last two can contribute in any major way. Most of the heating comes from stars in one form or another.


    This is a tricky issue. The ISM is certainly not in complete thermodynamic equilibrium, as its constantly moving around and changing states, depending on the stellar activity in its vicinity. However, on human timescales, it can be said to be at a constant temperature with an approximately Maxwell-Boltzmann distribution.


    It is indeed, but this process can be balanced by cooling in some situations. Take the simple case of an HII region. You basically have a massive star ionizing the ISM in its immediate vicinity and adding heat to it in the process. However, by increasing the number of free electrons in the ISM, you also increase the rate of cooling. Eventually, if the flux from the star is constant, you'll reach a balance and the ISM around the star will be in a steady state.
     
  7. Mar 16, 2005 #6

    SpaceTiger

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Evaporative cooling basically refers to the process by which cool clouds embedded in hot gas are "evaporated" by the surrounding hot medium. This alters the density of the medium and, therefore, the cooling rate. Presumably you can distinguish cooling mechanisms by:

    1) Looking at spectra. At what wavelengths is the radiation being emitted? Is it in the form of lines? Which lines?
    2) Looking at where the radiation is coming from. Is it clumpy? This can be resolved by good telescopes.
    3) Looking at the conditions of the gas. This can be derived by examining the relative strengths of absorption and emission lines at various wavelengths.
     
  8. Mar 16, 2005 #7

    hellfire

    User Avatar
    Science Advisor

    Thank you for your clarifying answers.
     
  9. Mar 19, 2005 #8
    The heating of atoms/molecules through ionization should in fact be very much negligible. The point is that electrons have a mass only about 1/2000 of that of a proton, so they can only transfer 1/2000 of their energy to a much larger mass in an elastic collision (this follows from the energy and momentum conservation laws). Since the recombination coefficient has about the same order of magnitude as the elastic scattering coefficient, there will be only about one elastic collision before the electron recombines again. Photoelectrons could heat the atoms therefore at best by 100 K or so.
    It is practically always wrong to assume a thermodynamic equilibrium between atoms, molecules and ions on the one hand, and electrons on the other (I have worked on this some years ago in the context of ionospheric physics; see http://www.plasmaphysics.org.uk/research/elspec.htm )

    The situation is however different if you consider photo-dissociation of molecules. Here the energy is imparted to whole atoms and hence it can be fully transferred to other atoms/molecules. Excitation of rotational and vibrational molecular states can also be transferred to translational kinetic energy in a collision.
     
  10. Mar 19, 2005 #9

    SpaceTiger

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It would seem that much of what you learned from your work doesn't apply to the majority of the ISM:

    1. The dominant heating mechanism for most of the interstellar medium is photoionization of neutral atoms.
    2. For most of the ISM, the atoms, molecules, and electrons can be considered to be at the same temperature.
    3. The primary reason that thermodynamic equilibrium fails is the difference in temperature between the ambient radiation field and the particles.

    You have to remember that the ISM has a lot of time to relax. Equapartition is easily achieved, despite the inefficiency of elastic collisions between electrons and protons.
     
  11. Mar 19, 2005 #10
    I am afraid this is as much a myth for the ISM as for the ionosphere (where the same as what you are saying is usually assumed). In neither case do the electrons have enough time to transfer their energies to the ions (the smaller plasma density in the ISM reduces both the recombination and elastic scattering probability by the same amount). Thermal relaxation can only take place through elastic collisions, and if the mass difference of the particles is large, one would need a corresponding large number of collisions for this. However, the electrons will recombine way before this happens. For the ionosphere this is actually confirmed by the recombination continuum that the it emits in the ultraviolet. With an energy distribution relaxed to the ion/neutral temperature this should not be observed at all.
    The increase in temperature due to photoionization will actually be much less than the 100 K that can be transferred in one collision as the degree of ionization is much smaller than 1 (i.e. many neutral atoms/molecules have to be heated by one electron).
     
  12. Mar 19, 2005 #11

    SpaceTiger

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Think of it this way. Imagine you turn on a radiation field in a sea of neutral atoms and you leave this radiation field on (and steady) for millions or billions of years. What happens?

    Well, you will start by ionizing many of your neutral atoms, releasing free electrons which will initially be at a much higher temperature than the heavier atoms and ions. I'll assume that you're right that a given electron will recombine long before it can distribute all of that energy to the heavier particles, but the point is that it will distribute some of it to the protons, heating the gas a little bit before it recombines and radiates away most of the gained energy.

    Since there are no other mechanisms to heat or cool the gas (the low density of the ISM renders collisional excitation negligible most of the time), the net effect is that the temperature of the protons and neutral atoms will increase. There is nothing to stop this increase in temperature until you reach the point at which the electrons that are being freed by ionization are at the same temperature as the rest of the gas.

    It's all about timescales. Your arguments work if the radiation field is changing on timescales less than is required for this process (as is definitely the case for the ionosphere) or if there are other heating/cooling mechanisms that dominate. In the ISM, unless you're in the vicinity of shock waves, photoionization is all there is. I suggest taking a look at chapter 2 of Spitzer's "Physical Processes of the Interstellar Medium".
     
    Last edited: Mar 19, 2005
  13. Mar 20, 2005 #12
    Actually, if you consider a cold interstellar gas cloud, you won't have a sea of neutral atoms but a sea of molecules (i.e. H2) . If you now turn on the UV radiation field of a star, photodissociation of these molecules will lead to a heating of the gas before photoionization becomes even an issue, not only because the heating collisions will be between particles of equal mass, but also because the UV radiation flux in the near-UV (which leads to dissociation) is much stronger than the one in the far-UV (which causes the photoionization).

    It is true that it is only a question of time before an equilibrium situation develops, but the point is that this equilibrium is in most cases not one in Local Thermodynamic Equilibrium (LTE) because one does not have merely elastic collisions affecting the energy distributions. If you look at the results of my model calaculations for the ionosphere, this shows not only the Electron Spectrum , but also the Atomic Level Population to be very much different from a Boltzmann distribution (which is implied by the assumption of LTE) (the corresponding solutions for the interstellar medium should be qualitatively quite similar to this).
    It would be therefore best to actually forget about Spitzer's theoretical treatment and other LTE approaches as these will yield results that are not only quantitatively off the truth by miles, but also prevent a qualitative insight into the physics involved. In this sense, it is likely that the accepted values for HII-regions have in fact little to do with reality and that the temperature is actually much lower than the 10^4 degrees derived under the LTE assumption (one can for instance fully explain the line broadening of radio recombination lines in HII- regions as a a kind of Stark-broadening effect in the plasma rather than as thermal Doppler broadening; see my paper http://www.plasmaphysics.org.uk/papers/radscat2.htm (Chpt.3.4.4. last paragraph)).
     
    Last edited: Mar 20, 2005
  14. Mar 20, 2005 #13

    SpaceTiger

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What's your point? First of all, the vast majority of the ISM is not molecular. Second, a strong source of UV radiation in a molecular cloud will result in a mostly ionized plasma, just like in the case I presented, only it will go through photodissociation first.


    That's why I referred you to Spitzer. Elastic collisions are orders of magnitude more important than inelastic throughout most of the ISM. The reason your studies were different was that you were dealing with much higher densities.


    I take it you didn't look at the book, nor do you seem to have much respect for ISM physicists. All of these things are presented in great detail, including the fact that the population of energy levels is not in LTE in the ISM. This does not mean, however, that the particles aren't in Maxwellian distributions at the same temperature. Fractional deviations from Maxwellian are typically of order 10^-6. The reason that photoionization can't heat the gases in your case is that the high densities lead to quick cooling by collisional excitation.


    Wow. I don't know if this is stubbornness or ignorance, but you're heading into crackpot territory here. You talk about these concepts as if ISM physicists have no understanding of them.
     
    Last edited: Mar 20, 2005
  15. Mar 22, 2005 #14
    First of all, in order to determine the general awareness regarding this issue, you might want to do a Google search for 'elastic collisions interstellar medium' (without the quotes). What webpage do you think comes up top? It is in fact this forum thread, which I think speaks for itself. Directly behind is a page of Wiley trying to sell Spitzer's book, but I could not find any comprehensive treatment of the relevant issues at all.

    Anyway, I had a look at Spitzer's book now and can't see the mass issue relevant for the electron-proton collisions addressed at all in Chpt.2, i.e. for him the protons would still be heated even if they had infinite mass. He makes an attempt to justify a Maxwellian distribution for the electrons, but even here the argumentation here is less than conclusive. The only element he considers to lead to cooling and hence possible deviations from a Maxwellain is OII. I quote: In HII gas, OII can be one of the more effective coolants. In other words, he just picks some process convenient to him without knowing that is actually the most effective coolant. Why does he not consider recombination as a relevant loss process or for instance collisional excitation of HI? His arbitrary way of justifying his LTE assumptions are also aided by the facts that he uses an elastic collision cross section orders of magnitudes too large due to the Coulomb Logaritm factor (which can be shown to be due to an erroneous treatment of the Coulomb collision; see my page http://www.plasmaphysics.org.uk/#enloss (Energy Loss)) and furthermore because the generally used value for radiative recombination is orders of magnitudes too small (see http://www.plasmaphysics.org.uk/#radrec (Radiative Recombination)).

    In any case, it is clear that Spitzer's theory is not a consistent treatment of the problem, and it is therefore not surprising that his crude ad-hoc assumptions lead to substantial disagreement with observations. I quote from Spitzer's book page 7 regarding the Orion nebula: The root mean square n_e (the electron density) determined from the radiation density is about one-sixth the value computed from the line ratios , but instead of questioning his crude theoretical assumptions, he then concludes suggesting that in each spherical shell the emission comes only from a small fraction (about 1/30 of the volume), a very clumpy radiation source. Well, I think this bold conclusion really speaks for itself regarding the scientific rigour of ISM physicists and does not need further discussion.

    As I said already, the conditions for the ISM are overall not too different to ionospheric condition (where the invalidity of LTE in all respects can well be shown by direct observations), so I am sure that the same physics applies here as well, i.e. photoionization does not contribute significantly to the heating (if there is any heating at all).
     
  16. Mar 22, 2005 #15

    SpaceTiger

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Haha, so you think you're going to get six orders of magnitude from another transition? Do you really expect him to present the treatment for every transition in a textbook? I've done photoionization calculations before (including all of these processes) and, believe it or not, the temperatures do work out to 10^4 K and the collisional processes are negligible.


    You don't seem to understand the situation. The free electrons are ionized by the radiation input, as I said, so every electron that recombines is going to release the same or less energy than was originally put in. Spitzer is looking at collisional processes because they act as an additional source of cooling that can prevent a net heating of the gas by photoionization. For treatment of recombination and a description of what I already said to you, see chapter 5.


    Haha, yeah, that 13.6 eV energy gap will lead to a great coolant. You do realize that virtually all of the neutral hydrogen is in its ground state, right?



    Ah, I see, you actually are a crackpot. My apologies for wasting your time.
     
  17. Mar 22, 2005 #16
    I know next to nothing about the ISM but have read this thread. Only part of Thomas2's agruement that makes sense to me is the bit about the population levels being far from Boltzman. One must agree with ST that there is a lot of time available so fact that each elastic collision on more massive particle transfers less than 1% of the electron's energy does not imply much about the commonality of temperatures between them and the more massive particles.
    In the case of an inelastic collision with HI I believe the elcetron is very likely to first be dropped in a very high n level, then most likely rapidly fall to n-1 etc, effectively "down shifting" in energy the UV that liberated the electron in the first place to far IR. In fact, I don't think there is any rapid change in the transition probabilities as it cascades down thru the "n level" giving off many IR photons each time. Thus, seem quite plausible to me that if one were to look at the nearly uniform populations in the high n levels only one could claim the ISM was very hot. Thus these levels are not Boltzman distributred at either the much lower electron or massive particle temperature, evenif you do not want them to be essentially the same. For LTE to exist, this non-Boltzman population distribution in the high n levels and the incident uv would need to be absent. (The radiation field is part of LTE also and based on above argument, I suspect it has peak in IR as well as the UV.)

    Even though LTE it is not, I bet the Saha equations, with heavy particle temperature (probably same as electron also) still gets the fraction ionized just about right, unless the UV flux is really intense.

    As I stated initially, I don't know much about the ISM, but if any of the above is drasticaly wrong, please someone tell me. thanks.
     
    Last edited: Mar 22, 2005
  18. Mar 22, 2005 #17

    SpaceTiger

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    They are far from Boltzmann, I don't disagree with that. It doesn't imply that protons and electrons are then at different temperatures.
     
    Last edited: Mar 22, 2005
  19. Mar 22, 2005 #18
    What's the problem? A little table of half a dozen transitions might well do the job. But then again, how should he know the relevant data without making already the LTE assumptions when deriving them from observations? The point is, that without being able to make separate in situ measurements of the relevant constituents (like it is for the ionosphere) your model is either consistently right or consistently wrong, and I am afraid with the LTE assumption it is likely to be the latter.

    I think it is you who does not understand the situation. As indicated earlier by me already, if recombination occurs before the electron can transfer their kinetic energy to protons in elastic collisions (which is more than likely in view of the large mass difference), then there won't be any significant heating of the protons through photoionization. Recombination is the only true loss process for the electrons and hence the most important one for the energy balance of the electrons (and as a consequence also for the heating of the protons).

    Collisional excitation of HI transitions (e.g. Lyα) does not take 13.6 eV, but even if you are referring to collisional ionization, you are in fact right , it is a very effective way of energy degradation. If you had done any computations in this respect, then you would know that (for the sun) the maximum of the primary photoelectron production is around 25-30 eV (see the corresponding peak near 2 Ry in the curves for the http://www.plasmaphysics.org.uk/research/elspec.htm" [Broken]). For the stars responsible for HII regions the average photoelectron energy might even be higher due to their much stronger UV emission. This means that the majority of electrons will lose 13.6 eV in one go for each ionizing collisions, i.e. their energy decreases very rapidly to smaller eV values where they will either recombine (the "[URL [Broken]
    recombination cross section[/url] increases very strongly towards smaller energies) or can excite other transitions.
     
    Last edited by a moderator: May 1, 2017
  20. Mar 22, 2005 #19

    SpaceTiger

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This actually was a mistake, as I meant 10 eV, but the argument stands. I'm not going to waste my time with the rest, as I'm likely to get as frustrated with you as the referees of your "papers".
     
    Last edited: Mar 22, 2005
  21. Mar 23, 2005 #20
    After having a look at Spitzer's book it seems that the level population is indeed computed for non-LTE. This actually would have to be so as the atomic decay times are so much shorter than the collision times. It is not really a big deal either to do non-LTE calculations of the level populations, especially if it is restricted to a few levels only. It is much more more difficult to do a consistent non-LTE calculation of the free electron spectrum as this is contiunuous.

    It seems you are speaking of recombination here because direct excitation from the ground state is very unlikely to be into higher levels as the excitation cross section decreases very sharply with increasing quantum number for the upper level (less than 10% will be excited from the ground state into a level higher than n=2).

    The situation is in general not as drastic for recombination, but this depends very strongly on the energy that the plasma electron has before recombining:

    if the recombination coefficient (cross section x velocity) into the ground state (n=1) for an electron with an energy of 1 Rydberg (13.6 eV) is set to 1, then the recombination coefficients into the first few levels are

    n=1 : 1
    n=2 : 0.43
    n=3 : 0.14
    n=4 : 5.4*10^-3

    i.e. 95% of all electrons with an energy of 13.6 eV will recombine into the levels n=1-3 ;

    However, if the electron energy is only 0.1 Rydberg (1.36 eV) then the same percentage of electrons is distributed between levels n=1-10. The recombination coefficients (which can be compared directly to the above values) are

    n=1 : 0.95
    n=2 : 5.6
    n=3 : 7.9
    n=4 : 6.7
    n=5 : 5.0
    n=6 : 3.4
    ..
    n=10 : 0.76

    (As a rule of thumb, if an electron has an energy E (normalized to 1 Rydberg = 13.6 eV), then the maximum of the recombination will be into level n_max=1/√E with a significant spread of +-n_max around this).

    As is evident from these values, the recombination coefficients for lower energy electrons tends to be much higher (the recombination coefficient into level n=3 differs for instance by a factor 56) , which demonstrates how important it is to know the exact electron spectrum for calculating the atomic level populations. The discrepancy regarding the determination of the electron density in the Orion nebuly that Spitzer mentions in his book on page 7 could well be related to this circumstance.

    It should be noted that one can get these numerical figures only by computing the recombination cross section using the exact quantum mechanical wave functions both for the free electron energy as well the energy of the bound level in the http://www.plasmaphysics.org.uk/#overlap" that I have referenced above already (and if multiplied with the velocities to the above relative recombination coefficients).

    This basically applies to all plasmas not just the ISM.
     
    Last edited by a moderator: Apr 21, 2017
  22. Mar 23, 2005 #21
    OK, T2, ST and I all agree that there are few excited HI atoms and those that are have non-thermal population distribution. So I am beating a “dead horse” to say more, but in view of the relative to electron capture rates, the electron that is captured (inelastic collision) will just fall down thru the levels quickly. Hence in a large ensemble of HI atoms the population of the upper levels will just be proportional to the radiative life times of the levels (neglecting the rare reionization by a second collision or there being intense UV near the excited HI).

    Of course, Thomas2 I was speaking of recombination - that was implied by “the electron being “dropped” into a high level.” But I want to add a few words more on this (just from intuition, not calculation): I think that the free protons will be traveling in many directions with many different speeds. I bet (because the orbital for n =50 is so much larger than for n=5) that some electron, relative to some proton, is “nearly co-traveling” and easily falls into a high level (weakly bound and now truly “co- traveling”) giving any energy difference away as IR photon continuum. Once “bound” it is going to fall down thru the level (I think one of the radiative selection rules is delta n =+/_ 1, but that is from memory many years old. Perhaps it is just much more rapid than greater change in n.) In any case, I expect as I said before that the inelastic recombine collisions plus the cascade down thru the levels will effective effectively “downshift” the stellar UV. Further more, this IR is not going to excite any HI atom out of ground state (might spin or shake some H2 molecules - just guessing - I am too lazy to calculate. Or even read Spitzer -30 years ago I read great little book by a Spitzer on plasma - perhaps new book is his son/daughter‘s?)

    TO Thomas2: In post 18, you said:
    “Recombination is the only true loss process for the electrons and hence the most important one for the energy balance of the electrons”
    I am not knowledgeable enough to argue (or support) your idea that recombination is the dominate means of taking energy out of the free electrons; however, it would seem to me, that bremstrallum radiation might be significant, especially if there are any higher Z ions around. C++++ would be very effective, (might be there from an old star that fused 3 Heliums before “going nova“) as the remaining two electrons would be “tight in” and it would look to a passing electron as a point with four plus charges and scatter the passing electron, producing a little bremstrallum radiation. As protons are probably much more numerous, they may be more important, despite individually being at least four times less effective. (Four charges would have same electric field at twice the distance, so cross section is four times larger, but transit time thru field and effect on electron trajectory would also be larger so this is very conservative lower limit) I also suspect that C++++ would capture electrons of higher energy well - the levels would be like HI but with four times the spacing. BTW thanks for infro about what level matches with what energy electron. -It is pretty much as I would guess from the “resonate” nature of many capture processes, even nuclear ones.

    TO SpaceTiger: In post 17 you seem tobe over reacting - I expected that form T2 because In post 16 I said:
    Only part of Thomas2's agruement that makes sense to me is the bit about the population levels…One must agree with ST that there is a lot of time available so fact that each elastic collision on more massive particle transfers less than 1% of the electron's energy does not imply much about the commonality of temperatures between them and the more massive particles.”
    and also later that:
    “I bet the Saha equations, with heavy particle temperature (probably same as electron also) still gets the fraction ionized just about right
     
  23. Mar 24, 2005 #22

    Chronos

    User Avatar
    Science Advisor
    Gold Member

    Interesting, I was thinking more along the lines of remarkable restraint.
     
  24. Mar 24, 2005 #23
    Just in order to avoid any misunderstanding in case you are referring to my diagram showing the http://www.plasmaphysics.org.uk/research/levpop.htm" [Broken] in an atom when mentioning the quantum number n=50 i.e. the level with the highest population apart from the ground state (which for HII regions would actually even be higher): you shouldn't mix up the population with the rate of electrons falling through the levels. The population only increases for higher levels because the lifetime of the levels becomes so long i.e. the electrons accumulate in these levels (later it decreases again because elastic collions interfere with the recombination). Because of these long lifetimes actually only very few of these electrons are going to lower levels. The rate at which electrons fall into the levels is in fact given by the thin solid curve in this plot and this decreases all the way for increasing n here. Although there is an increase if the initial energy of the electron is smaller than 10 eV or so (as is evident from the figures in my post above) this extends at best to n=10 if the electron energy is about 1 eV.
     
    Last edited by a moderator: May 1, 2017
  25. Mar 24, 2005 #24
    No, I had not looked at your figure before now. On my display the powers of 10 are not legible, but as I now know the solid line peak is near 50, I can deduce them. On the selection rule I knew that for the lower levels the delta n could be any value as we do speak of the "lyman" and "Balmer" series, but thought that it might be much stronger, essentailly -1 for cascade down as the adjacent orbital must overlap much better than levels that differ by a few n - as you confirm of the "level-by-Level" cascade. I did not consider that the high levels might have such long life times that they would often be collisionally excited before radiative decay. Does this imply there is a set of well separated IR lines, not essentially a continuum of IR from these upper levels? My only experience has been at the opposite extreme, dense lab plasmas
     
  26. Mar 25, 2005 #25
    I am sorry, I made actually a mistake here. The cascading probability is in fact smallest for neighbouring levels and largest into the ground state. Although it is true that the orbitals overlap better for adjacent states, the decay probability is proportional to the frequency cubed, and since the frequencies of transitions between neigbouring levels are much smaller, the overall probability is smaller too. For a given upper level n, the decay probability is in fact proportional to m^-1.8 if m is the lower level (I have given the full formula on my webpage http://www.plasmaphysics.org.uk/#atdecay (Atomic Decay Probability; this formula has in principle been derived for large n and m, but it is also a reasonable approximation for lower levels). This means for instance that if an electron has recombined into n=5, it will go with a probability of 66% into the ground state m=1, with 19% into m=2, with 9% into m=3 and only with 6% into the neigbouring level m=4. So there is actually only a minor amount of downshifting from the UV region. Still there should be of course some infrared emission as well and these would then be sharp emission lines unless the lines are collisionally broadened (which however is only an issue for very high plasma densities like in the solar photosphere). If you go to much higher levels however (i.e. lower frequencies), the spectrum becomes also quasi-continuous for much lower plasma densities. You can see this for a plasma density of 10^14 cm^-3 for my calculation regarding the Solar Recombination Spectrum.

    I am sorry by the way that some of the annotations to the plots are not very well legible. These are all scanned in from A4 sized plots that I did at the time and the size of the lettering was OK for this size but not for further reduction (I had to reduce the size of the plots considerably to fit onto all computer screens, so some of the lettering has become quite small; maybe with most people having now larger computer screens I should re-scan some of the plots to a larger size). For the time being, if you print out the print versions, you should (just about) be able to read everything (the resolution is twice as good; you have to print them though because they size is reduced on the page i.e. not all pixels show on screen). In most cases you can anyway deduce the powers from the context (as you did) and where it changes from single to double figures for instance.

    P.S.: I have re-edited my previous post regarding the cascading probabilities (i.e. I have deleted my original statement)
     
    Last edited: Mar 25, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook