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Why does the Sun behave like a black body?

  1. Mar 24, 2015 #1
    My question is, given that the Sun is composed almost entirely of just hydrogen and helium, why is the spectrum from it continuous and not an emission line spectrum?

    This applies to other stars as well.

    There are other elements of course present in the Sun, but they are present in very small quantities relatively, so I don't really see how the presence of the other elements would account for the continuos nature of the Sun's spectrum. Is the Sun's spectrum really continuous? If one is able to look at it at fine enough detail would one find hundreds of fine emission lines?

    Thanks
     
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  3. Mar 24, 2015 #2

    Orodruin

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    The light from the Sun is not due to transitions between atomic energy levels (in fact, the Sun is mostly a plasma). Instead, it is a thermodynamic system, including a photon gas of the same temperature as the surroundings. As some of these photons near the solar surface escape, you obtain a thermal blackbody spectrum.
     
    Last edited: Mar 24, 2015
  4. Mar 24, 2015 #3
    Thanks - that's a very precise answer. Do you know where I can read more about this? Thanks again.
     
  5. Mar 24, 2015 #4

    russ_watters

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    http://en.wikipedia.org/wiki/Black-body_radiation
    [edit] I post that, but I'm not sure that's what you really need. You are already aware of the term "black body", so perhaps you already understand how they work and just didn't realize the sun is one because it is hot. There really isn't anything more to explain in the answer than that.
     
  6. Mar 24, 2015 #5
    An interesting aspect of black body radiation (mentioned in passing in the wiki link above) is that it was explained by Planck as being the result of light being emitted in discrete particles, now called photons. The black body radiation equation has Planck's constant in it. It is the first instance of quantum mechanics discovered in physics, and as such is also the first instance of using quantum statistics as a new theory underlying classical thermodynamics.

    So quantum mechanics plays a very strong role explaining black body radiation - it just doesn't imply any emission lines as you would have in a cooler system such as a gas-discharge lamp: http://en.wikipedia.org/wiki/Gas-discharge_lamp
     
  7. Mar 24, 2015 #6
    But then how do we know what the composition of the Sun is? Every where you read that the composition of the Sun is known because of spectroscopic studies of the light coming from the photosphere and chromosphere - that is by studying the emission spectra. So now I am at a loss.

    I don't get that the Sun is a black body just because it is hot. After all, when people previously use to study emission spectra of elements, I thought that this was done by heating the element in a flame, no? Also the spectrum tube which do produce emission line spectra are not completely cold. I know that the outside of the tube does not get very hot but how hot is the actual gas in the tube?

    Any way the big confusion or contradiction here seems to be that people are supposedly able to analyze the composition of the Sun and other stars by looking at emission line spectra and yet we are saying that the spectrum of light coming from the Sun and other stars are continuous and very nearly approximately like that of a black body.
     
  8. Mar 24, 2015 #7

    russ_watters

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  9. Mar 24, 2015 #8
    Yes if you place a cool gas between a source of a broad continuous sprectrum you get an ansorbtion spectrum. But when I looked up the question of how we know about the compilation of the Sun I found a thread on this forum and links that say that we know the composition of the Sun and other stars by analyzing line emission spectra from the photosphere and chromosphere. So now I just don't understand how that spectrum from the Sun can be both conitnous and also produce line emission spectra at the same time . I looked up another thread on physics forums but there seemed to be so much disagreement between the various members that it did not really shed any light - excuse the pun. Orodruin's answer seemed to be leading somewhere which is why I asked if he knew of any other reading material I could study about this.
     
  10. Mar 24, 2015 #9

    Orodruin

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    Generally, the very outer layers of a star are cooler and give you absorption lines. These lines can be identified in the continuous spectrum and give you some information on the stellar composition. However, this is never going to give you information on the inner parts of the star.

    Everything (essentially) gives you a blackbody spectrum because it is hot. There is a reason heated metal first looks red and then tends to whiten as the temperature increases. It is different when you oxidise different elements. The light then does not come from the temperature per se, but from the emission lines.
     
  11. Mar 24, 2015 #10
    Well I just found this answer which seems to be fairly satisfying: "

    Black body radiation is thermal radiation. The sun initially generates non-thermal radiation, of course, in the course of nuclear fusion. Photons of wavelengths/energies appropriate to the nuclear reaction, not to the temperature of the hydrogen being fused, are generated. However, the core of the sun, where nucleosynthesis takes places, is far from transparent. It is estimated that the energy generated in the core takes anything from hundreds of thousands to tens of millions of years before it is finally radiated into space. That gives the photons all the time needed to come into thermal equilibrium with the material of the sun.

    Of course, if you look at a solar spectrum you will see any number of Fraunhofer lines, absorption lines corresponding to the material in the outer layers of the sun. To that extent, the solar radiation is not exactly black-body."

    So it would seem that people are able to study the composition of the Sun and other stars by looking at the absorption spectra due to the gases in the outer layers of the Sun. I kind of half buy that and it seems to agree with other things in have read. Hmmm
     
  12. Mar 24, 2015 #11
    Yes - thanks Orodruin. It's beginning to make more sense now. Cheers
     
  13. Mar 24, 2015 #12

    Orodruin

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    Yes, but as I said, this only a priori gives information about the outer layers of the star. In order to infer things about the inner parts, you need to use other methods and model the star.
     
  14. Mar 24, 2015 #13

    D H

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    The key word in your question is "approximately". There is no such thing as an ideal black body; they're idealizations. In reality, you'll always see some deviation from this ideal behavior. In the case of the Sun, there are absorption lines and emission lines on top of the thermal radiation. Those absorption and emission lines are what less scientists "see" the composition of stars. After smoothing over those absorption and emission lines, the radiation from the Sun is very close to that of an ideal black body (but it's still not a perfect match).

    The reason for that near-black body radiation is simple: The Sun is a plasma. When charged particles interact electromagnetically, they emit photons; that's the mechanics by which charged particles interact electromagnetically. The frequency of the emitted photon depends on how close the particles came to one another and on the relativity velocities, both of which are random and continuous rather than discrete and well-defined). Since the particles in that plasma are more or less in thermal equilibrium, the relative velocities come very close to following the Maxwell–Boltzmann distribution, and thus the emitted photons come very close to the body distribution.
     
  15. Mar 25, 2015 #14

    Orodruin

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    The CMB is pretty close ...
     
  16. Mar 25, 2015 #15

    Ken G

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    There are a few important things that have not been mentioned yet. One very important consideration for the difference between Fraunhofer absorption lines, and the Sun's continous spectrum, is density. At very low density, lines are rather pristine, and you don't expect blackbody emission (consider low density nebulae, for example). But at higher density, found deeper in the Sun, there are more in the way of continuum sources-- plasma emission having been already mentioned.

    So the Sun would be a decent blackbody even if it were plasma emission that caused its continuum spectrum. However, it isn't-- sunlight comes from a very surprising source not a lot of people have heard about. The photospheric layers of the Sun are cool enough that there is a lot of neutral hydrogen there, even though the rest of the Sun is indeed plasma. What's more, neutral hydrogen actually has one weakly bound state available for capturing a second electron. That creates what is called the "H minus ion", because it has an overall negative charge. The ionization energy of H minus is only a little over an electron volt, which is well into the infrared, so you might think the creation of this ion would not generate visible light. However, the average energy of the free electrons in the photosphere is in the visible regime, so it is not the release of the ionization energy that makes the photons visible, it is the excess kinetic energy the electrons had prior to getting captured by the H. Since such electrons have a continuous amount of energy, they make a continuum of light as they are captured by H, and that's largely what sunlight is.
     
    Last edited: Mar 25, 2015
  17. Mar 27, 2015 #16
    Sorry to tag another question onto this thread, but I was reading a wikipedia article and it said that "
    Whilst the photosphere has an absorption line spectrum, the chromosphere's spectrum is dominated by emission lines. In particular, one of its strongest lines is the Hα at a wavelength of 656.3 nm; this line is emitted by a hydrogen atom whenever its electron makes a transition from the n=3 to the n=2 energy level. A wavelength of 656.3 nm is in the red part of the spectrum, which causes the chromosphere to have its characteristic reddish colour.

    By analysing the spectrum of the chromosphere, it was found that the temperature of this layer of the solar atmosphere increases with increasing height in the chromosphere itself..." So it seems that the photosphere does produce an absorption spectrum? I thought that a continuous (black body) spectrum is emitted from the photosphere and the chromosphere and corona produce absorption lines (not emission)...
     
  18. Mar 27, 2015 #17

    Ken G

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    Yes, the photosphere does produce an absorption line spectrum, including a continuum spectrum. That's because in the continuum, you mostly see to a single depth, the depth over which visible light can get through the H minus ions without being absorbed. This depth varies a bit depending on where on the surface of the Sun you look, because of the different angle you are seeing the surface from, but you can still associate a single average temperature because you have a single average depth. That's the "blackbody continuum" you are talking about. But at wavelengths within an absorption line, like the sodium D lines, there is a different opacity source that is much stronger, so you don't see nearly as deep into the Sun. Since the temperature in the photosphere is dropping as you go up, if you don't see as deep, you see something cooler, and that makes the absorption line.

    The chromosphere makes emission lines for a different reason-- you have to look "off the edge" of the photosphere, where you never get optically thick in the H minus continuum. You are looking along a line of sight that "misses" the photosphere, and so there is not much of a background continuum against which to see the continuum. In that case, any line you see must be an emission line instead, and that's the only place the chromosphere will look red. If you get something like a solar flare, or if you look at stars that are constantly flaring, then the chromospheric lines can be emission lines even if you look at the center of the star and there is a continuum in the background, because then you have a thick hot chromosphere and you can actually see this reversal in the temperature gradient. In our Sun, when there is not a flare, you see this as a mini emission line sitting at the bottom of a photospheric absorption line like H alpha.
     
    Last edited: Mar 27, 2015
  19. Mar 28, 2015 #18
    That is at best mis-leading, but really completely wrong. The sun (or any other object that absorbs all wave lengths) is a BB because for any wave length, a = e are nearly unity, especially for the huge sun, where a is the absorption coefficient and e is the emissivity coefficient.

    To show how wrong the idea that sun is a BB because it is very hot:
    Consider a small sphere of molten metal "floating" inside an orbiting satellite. It will be very visible as for metals the coefficient of reflectivity, r, is high - say 0.8 and it is always true that a + e + r = 1. Thus a & e being equal for same wave lengths would have:
    a = 0.1 = e = 0.1 for r = 0.8 case. No one would even dream of calling this very hot sphere a BB as it is highly reflective and a poor absorber.

    The sun, except for minor mainly absorption effects in the outer layers, is a BB because it is big. That solar mass* would be a BB even if it were at room temperature as a very high percentage of the photons headed towards its center from outside space will not reflect, but get "lost inside." A room temperature gas mass* as big as the sun would have a > 0.85 so e is also high, like at least a "grey body."

    * Especially if it is helium whose first excited state can be reached only by a very harsh UV photon absorption from the ground state; where at room temperature all would be. All most all others would just "Compton scatter" off the two bound atoms, with tiny change in angular direction and little energy loss in almost all cases. I.e. they would "random walk" ever deeper into the mass, almost never getting back out of it.
     
    Last edited: Mar 28, 2015
  20. Mar 28, 2015 #19

    Ken G

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    Blackbodies have little to do with either temperature or size. What you need for a blackbody is an object that absorbs all light (or at least all frequencies of interest) within a narrow skin depth, such that when the light is re-emitted it will all come out at a similar temperature. It doesn't matter if the temperature is high or low, and it doesn't matter if the object is large or small, it matters that it absorbs light well at its surface. The Sun absorbs light well at its surface because of all that H minus opacity, though hotter stars that don't have H minus opacity are decent blackbodies too because their density is not as low as things like nebulae in the interstellar medium. Nebulae, on the other hand, are terrible blackbodies, despite being much larger, significantly hotter, and often more massive, than the Sun.

    Also, scattering is not a good way to make a blackbody, it will lead to lots of reflection at the surface. You want absorption, not elastic scattering, to get a good blackbody, but a little elastic scattering won't cause too much of a problem because enough of the photons do get lost inside long enough to get absorbed. We should note that treating stars as blackbodies is just a convenient approximation, not to be trusted in too much detail.
     
    Last edited: Mar 28, 2015
  21. Mar 28, 2015 #20
    That is the DEFINITON of a BB body, except there is no requirement that there be a "narrow skin depth."

    In fact there is no "narrow skin depth" for the case of the best physically realizable BB, which is any isotheral chamber with walls made of any high temperature capable material such as high melting temperature metal or even a white ceramic like Al2O3 that has a tiny hole in the wall. ("Tiny" meaning the area of the hole is less than 0.01% of the interior surface area of the chamber.) The photons falling on the hole have less than 0.0001 chance of getting back out. I. e. a, the hole's absorption coefficient is greater than 0.9999 and no flat black paint or physically flat surface has: a = .999 so that is why this isothermal chamber is the "gold standard" that the NBS uses to calibrate secondary standards. Very few, none I can think of quickly, have a = 0.99 !

    {I could design such a "surface" but its "micro" (or actual) surface would be thousands of times greater than its "gross surface" - I. e. it would be made of many, very tall, (extreme aspect ratios) tiny, four-sided, pyramids, close packed, and made of reasonable high "a" material. Even then it would have a > 0.999 only for near normally incident light.}

    There is a radiation field coming out of that "tiny hole" that is almost exactly given by Planck's BB equation with T being the temperature (in Kelvin) of the isothermal walls.

    I only posted my original comment to correct russ_watter's assertion that the sun was a BB because it was so hot. I'm glad you agree that temperature of BB need not be hot. I illustrated that his assertion was false by noting that the same mass of room temperature Helium would be a BB also* as its radiation field also would be described by Planck's BB equation with T being room temperature (in Kelvin, about 300K)
    -
    * In this case (room temp solar mass of Helium) the "optical thickness" or "depth" (both terms are in common use) would be at least a few Kilo meters - not a "thin skin depth" - that is best achieved with an electrical conductor and they are rarely if ever BBs.
     
    Last edited: Mar 28, 2015
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