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In another question I tried to answer what a sample of the Sun's photosphere or core would look like, if it could be brought into the lab.

Here is a broader question - if I have a small inert container of gas in local thermodynamic equilibrium, that has the solar photospheric composition and photospheric density ($\sim 10^{-9}$ g/cm$^3$ -i.e. it will be optically thin for a small lab sample), what would it look like (to the human eye) and how would it change as I cranked up the temperature from say 1000K through the solar photospheric temperature to say as high as 100,000K?

EDIT: Just as a steer - I know it will look nothing like a blackbody - that is why I am interested in it and why I emphasize the optically thin nature of the problem.

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ProfRob
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  • Wikipedia says photospheric density is approximately 2E-4 kg/m^3. I think 100g/cc of stuff at 4500-6000K would be relatively opaque due to the dynamic range of the eye alone. – user22620 Dec 15 '14 at 02:16
  • My mistake - I'll correct. I do mean photospheric density. – ProfRob Dec 15 '14 at 07:11
  • Have you calculated the flux density at visible wavelengths? Perhaps the mass is too low to distinguish the emitted radiation from background. – user22620 Dec 15 '14 at 15:46
  • @user22620 Indeed, my question could be hypothetical. Perhaps my next question would be can it be done? However, if you had say a 10cm cube, you could increase the density by a factor of $\sim 10^{4}$ and it would still be optically thin. That would help. – ProfRob Dec 15 '14 at 15:55
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    We define the sun's surface as the point where it becomes optically thick. However, we know the entire thing is a giant ball of ionized gas and has no solid surface. We only see the corona due to Thomsen scattering of photons off the electrons. So if the gas is optically thin, how would you "see" it? Have I missed something? – honeste_vivere Jan 01 '15 at 15:51
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    @honeste_vivere Optically thin simply means that the optical depth is much less than 1 so that emitted photons leave the gas without further interaction. e.g. The corona is not only seen by Thomson scattering (and very few escaping white light photons interact in this way) it is seen via optically thin (X-ray) emission from a $>10^6$K plasma. – ProfRob Jan 01 '15 at 15:59
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    @RobJeffries - Ah, I understand now. I was inappropriately assuming by "look like" you meant what one could see with the naked eye. If you meant what is measurable then that is different, as you correctly state. Haven't astronomers gone through all sorts of emissivity calculations for optically thin gases? – honeste_vivere Jan 01 '15 at 17:05
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    @honeste_vivere I do mean what you could see with the naked eye. What is the optical appearance of an optically thin gas at a range of temperatures. For example I know there are some highly excited iron ion optical transitions that could be seen in 1 million degree gas, but I am interested in more modest temperatures. – ProfRob Jan 01 '15 at 17:39
  • @RobJeffries - So would you consider the plasma produced in the combustion of magnesium to be an optically thin gas? I think fire, in the conventional sense, is a kind of optically thin gas (though ionized). I cannot think of how to produce light without nuclear reactions (e.g., electron transitions) or thermal emission in a tenuous gas, but both of those require high "temperatures" for visible wavelengths, right? – honeste_vivere Jan 01 '15 at 17:45
  • @honeste_vivere Yes, I would think at temperatures of a few thousand degrees we are talking about "doing a flame test", but on material that has the solar photospheric composition. So what does that look like? – ProfRob Jan 01 '15 at 18:43
  • @RobJeffries - I would assume that a blackbody spectrum is independent of composition, right? So it would "look like" whatever your eyes responded to depending on temperature, due the response function of your eye. I had a similar discussion here. – honeste_vivere Jan 01 '15 at 19:23
  • @honeste_vivere This has got nothing to do with blackbody radiation (other than that the source function is the Planck function). If you have an answer - write it down. – ProfRob Jan 01 '15 at 19:27
  • Is this really a question as to what the boundary radiation source flux distribution needs to be to maintain equilibrium? – user22620 Feb 21 '15 at 03:17

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You would expect it to radiate roughly a black body spectrum, but with lower intensity due to the thinness. You would also be able to see through it to things beyond. So if you had a container like that in front of you, you would see the background with a slight overlay of black body spectrum.

Ross Millikan
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  • An optically thin gas sample cannot radiate like a blackbody, which by definition is optically thick. – ProfRob Apr 17 '16 at 19:00