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I would like some help understanding absorption spectra, for example, by a cloud of gas in space. Photons might be absorbed if their energy matches a transition in one of the atoms or molecules in the gas. I am happy with that part but surely the atoms or molecules drop back to their ground state at some point and emit another photon of the same energy. So, does not the gas reach a steady state with no net absorption? I have done some reading and I think that I found that the answer is as simple as the emitted photon may be in a different direction.

If I have not gone wrong so far then please confirm or correct this thought experiment.

  1. An object is detected in space emitting a beam of light. This is analysed and found to be a pure black body spectrum with no absorption lines.

  2. Three space stations A, B, and C are set up. Initially, they are in line and within the beam. A is nearest and C is furthest. They all observe the same spectrum (maybe slightly weaker for B and C if the beam spreads a bit).

  3. Station B is moved perpendicularly to the line between them out of the beam. So, now B cannot observe the beam.

  4. A cloud of gas floats to where B was. A observes no change. C observes an spectrum with absorption lines caused by the cloud. B now sees a faint glow from the cloud as some of the re-emitted photons are in its direction.

Clarification on point 4. I don't mean that only station B sees the glow from the cloud. I expect that it goes in all directions. This is why I said "faint glow", station B just gets a small portion of it. "A observes no change" - well if it looked in the other direction then it should also see the glow.

Edit: correction made in 4.

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  • $\begingroup$ Related? physics.stackexchange.com/q/327960/104696 $\endgroup$
    – Farcher
    Aug 23, 2018 at 17:26
  • $\begingroup$ Close though I don't quite follow it. I think that my question is simpler so I was hoping for a confirmation or correction of my understanding. Then I might reread that and other related questions. $\endgroup$
    – badjohn
    Aug 23, 2018 at 17:58
  • $\begingroup$ Is absorption a suitable tag for this? $\endgroup$ Aug 23, 2018 at 18:23
  • $\begingroup$ Is that the wrong sort of absorption? I expected to see an absorption spectrum tag and didn't find it. $\endgroup$
    – badjohn
    Aug 23, 2018 at 18:46

4 Answers 4

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In your first paragraph it appears your assuming the cloud will reach an equilibrium and then let light pass through without an absorption spectrum. This is not correct because the new photons emitted from the cloud will be in random directions and not on toward the screen.

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  • $\begingroup$ By equilibrium i meant that the same quantity and frequency of photons would be emitted as absorbed but not necessarily in the same directions. That is what I was trying to illustrate in my thought experiment. However, since then my understanding has advanced thanks to the other answers. $\endgroup$
    – badjohn
    Aug 23, 2018 at 21:38
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Your thought experiment is not quite right. If I understand correctly you put a cloud of (cold?) gas in between A and C, with A closest to the source of (collimated?) light.

C will observe a blackbody spectrum modified by whatever has happened to it in the cloud of gas. Let us assume that the cloud is in thermal equilibrium, so as you say, it must emit as much as it absorbs. Further to that, the principle of detailed balance tells us that, in thermal equilibrium, all microscopic processes are balanced by their reverse processes.

Let us further assume that we are only talking about radiative processes. If that is so, then photons from the incoming beam may be absorbed or travel through the cloud. However, for every photon absorbed another must be emitted. There are basically two emission processes - spontaneous emission, where an atom/molecule de-excites by emitting a photon, and stimulated emission, which is where an incoming photon stimulates the de-excitation of the atom/molecule and results in the emission of a further photon with the same wavelength and direction as the incoming photon.

Thus spontaneous emission effectively takes photons out of the beam, whereas stimulated emission actually amplifies the beam (and is the principle behind laser/maser action). The balance between spontaneous and stimulated emission is frequency dependent - high (optical, infrared) frequencies favour spontaneous emission, low frequencies (radiowaves) favour stimulated emission. If the beam is being observed in optical light it is therefore likely that the cloud superimposes absorption lines on the blackbody spectrum as observed by C (not emission lines as you claimed in #4).

Whether this is the case depends on its optical depth as a function of wavelength. If the cloud is optically thin in the absorption line it will also be optically thin at other wavelengths. This means that almost none of the incoming photons are absorbed (and thus there is also very little spontaneous emission if the cloud is in thermal equilibrium and only heated by incoming radiation) and C will observe a blackbody spectrum with very weak absorption lines. In this case B will observe a very faint emission spectrum dominated by lines and with no continuum.

If the cloud is optically thick (at the wavelength of an absorption line and at the continuum wavelengths too), then all incoming photons are absorbed and the cloud will emit a blackbody spectrum at the temperature required by its thermal equilibrium (power absorbed = power radiated). As a result C will observe a blackbody spectrum at that temperature and B will also observe the same blackbody spectrum from the cloud. (Note it cannot be a perfect blackbody spectrum unless the cloud can be made magically isothermal, so there will be some very weak absorption lines).

The in-between cases are the interesting ones - where the cloud is optically thick at the absorption line wavelengths, but optically thin in the continuum. C will observe the original blackbody spectrum with deep absorption lines superimposed. Meanwhile B will observe the cloud to be a strong emission line object with a very weak continuum.

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  • $\begingroup$ Thank you very much for that detailed explanation. I think that I now understand. The "emission" in 4 was a mistake that I have corrected. My understanding was not that far off, just my typing. $\endgroup$
    – badjohn
    Aug 23, 2018 at 20:44
  • $\begingroup$ @badjohn But please note, the optical depth as a function of wavelength must be part of your thought experiment. I think you were assuming the latter "interesting "case, but that depends specifically on the density, temperature and size of the cloud. $\endgroup$
    – ProfRob
    Aug 23, 2018 at 20:48
  • $\begingroup$ You are right. Prior to your explanation, I was unknowingly assuming that. $\endgroup$
    – badjohn
    Aug 23, 2018 at 20:54
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I don't fully get your 5 points, but to your initial assumption: yes you can excite the gas into transparency but you will need very much power for that. Probably you will not reach it. The major point you might not have considered is that the the emissions which relaxe the excited atoms will be emitted at a random direction, so most likely not in the direction of the initial beam. So every gas atom will constantly absorb and emit light, thus efficiently reduce the energy of your primary beam.

So you with increasing gas debsity you will get stronger absorption lines in the primary beam and any other observer which are not in the beam path will see a nice nebular glowing in the colors of the previously absorbed spectral lines.

If your light beam is very intense though stimulated emissions will also play a role. But I don't think that this is what you had mind when asking the question.

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  • $\begingroup$ 5 points? I told a little story of 3 space stations observing a source. I was hoping for comments on whether it was plausible or not. Was the thought experiment unclear? $\endgroup$
    – badjohn
    Aug 23, 2018 at 17:48
  • $\begingroup$ Ok, I got you question a little bit wrong and yes in addition there are only 4 points, sorry. So 1,2,3 are correct. But at 4 of cause all observers see the glow of the gas cloud. $\endgroup$
    – user_na
    Aug 23, 2018 at 17:54
  • $\begingroup$ Thanks. I did not mean to suggest that only station B observed the emission from the cloud. I meant that it could now see the light which it could not before. $\endgroup$
    – badjohn
    Aug 23, 2018 at 18:09
  • $\begingroup$ Then you got everything right. $\endgroup$
    – user_na
    Aug 23, 2018 at 18:22
  • $\begingroup$ Absorption by the electron can/will get converted into electron vibration, then molecular vibration and likely the eventual emission of lower energy IR photons. Your experiment above is better suited to when photons are scattered and not absorbed. $\endgroup$ Aug 23, 2018 at 18:39
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Absorption by the electron can/will get converted into electron vibration, then molecular vibration and likely the eventual emission of lower energy IR photons. Your experiment above is better suited to when photons are scattered and not absorbed.

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  • $\begingroup$ Thanks. That's what I needed. So, the emission from the cloud may have the same total energy (once equilibrium has been reached) but its spectrum will probably not match the absorption. It will be at a lower frequency and less discrete. $\endgroup$
    – badjohn
    Aug 23, 2018 at 19:12
  • $\begingroup$ No problem. There may be a temperature increase of the cloud, i..e more vibration but in the long long term I think you are mostly correct. $\endgroup$ Aug 23, 2018 at 19:28
  • $\begingroup$ I would expect the cloud to heat up initially but not indefinitely unless the source was incredibly bright. The Sun heats the Earth but not without limit. $\endgroup$
    – badjohn
    Aug 23, 2018 at 19:38

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