Why does the Sun behave like a black body?

[resurgance2001]

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

 

[Orodruin]

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.

 

[resurgance2001]

Thanks - that's a very precise answer. Do you know where I can read more about this? Thanks again.

 

[russ_watters]

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.

 

[uumlau]

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

 

[resurgance2001]

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.

 

[russ_watters]

Flip over the idea of an emission spectrum and you get the concept of an absorption spectrum:
http://en.wikipedia.org/wiki/Absorption_spectroscopy

 

[resurgance2001]

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.

 

[Orodruin]

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.

I don't get that the Sun is a black body just because it is hot.

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.

 

[resurgance2001]

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

 

[resurgance2001]

Yes - thanks Orodruin. It's beginning to make more sense now. Cheers

 

[Orodruin]

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.

 

[D H]

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.

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.

 

[Orodruin]

There is no such thing as an ideal black body;

The CMB is pretty close ...

 

[Ken G]

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

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 continuous 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.

 

[21joanna12]

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)...

 

[Ken G]

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.

 

[BillyT]

... 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.

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.

 

[Ken G]

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.

 

[BillyT]

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, ...

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.

 

[Ken G]

That is the DEFINITON of a BB body, except there is no requirement that there be a "narrow skin depth."

Actually, there is, if you want the object to emit a "blackbody spectrum", as in the original question. Otherwise, you will get light from different temperatures, and you will not get that spectrum.

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.

If you can enforce isothermality, you don't need a narrow skin depth, but note those metal orstars aren't isothermal, so they do need a narrow skin depth in which to absorb light-- or they will not emit blackbody spectra.

I only posted my original comment to correct russ_watter's assertion that the sun was a BB because it was so hot.

Yes, that was wrong, but it's not quite right to say it's a blackbody because it's big, or because it has a large mass. Nebulae in the interstellar medium are hotter and bigger and have more mass than the Sun, yet are terrible blackbodies because they are so low density that they do not absorb light effectively. That's the real reason the solar surface is a good blackbody, it does absorb light quite well, largely because of H minus opacity. That does have something to do with how big it is, but the directly important fact is that it absorbs light very completely over a region that is more or less all at the same temperature, though of course these are all idealizations.

 

[BillyT]

Actually, there is, if you want the object to emit a "blackbody spectrum", as in the original question. Otherwise, you will get light from different temperatures, and you will not get that spectrum.
If you can enforce isothermality, you don't need a narrow skin depth, but note those metal orstars aren't isothermal, so they do need a narrow skin depth in which to absorb light-- or they will not emit blackbody spectra.
Yes, that was wrong, but it's not quite right to say it's a blackbody because it's big, or because it has a large mass. Nebulae in the interstellar medium are hotter and bigger and have more mass than the Sun, yet are terrible blackbodies because they are so low density that they do not absorb light effectively. That's the real reason the solar surface is a good blackbody, it does absorb light quite well, largely because of H minus opacity. That does have something to do with how big it is, but the directly important fact is that it absorbs light very completely over a region that is more or less all at the same temperature, though of course these are all idealizations.

I agree with all you state. I was speaking only of sun mass and density and noting it would still be a BB if just all at room temperature and only Helium. I also noted that a tiny hole in an isothermal chamber is also a BB radiator; However, you are correct when more generally speaking to include objects that have parts at different temperatures radiating to the exterior.

What is really necessary is that ALL the radiation comes volume at same temperature and that the radiation field is thermodynamic equilibrium with that volume. This is even a more general statement, than even yours as it includes the isothermal cavity with small hole, (but no thin skin depth the radiation comes from by necessity, but that is the common case.)

I'm not much of an expert on nebulae and they do have very low particle density, but even so if far from any star, they may be in thermodynamic equilibrium with the cosmic back ground radiation - Have isothermal conditions with temperature of both the particles and their radiation field slightly less than 4K.

I think we agree 100% on the physics; but admit I reserve the "skin depth" term for only electrical conductors and speak of "optical depth" for other cases where skin depth could be used also. This is no doubt because my experimental Ph. D. research was on hot plasma, where optical depth and temperature gradients were important considerations.

PS: Do you know about the "Saha equation"? It tells the relative population of the excited states - they may not be in LTE even if the particles and the radiation field are in Local Thermodynamic Equilibrium. That lack of LTE for an excited state is usually due to very short radative life time - Fast or high transition probability.

 

[Ken G]

What is really necessary is that ALL the radiation comes volume at same temperature and that the radiation field is thermodynamic equilibrium with that volume. This is even a more general statement, than even yours as it includes the isothermal cavity with small hole, (but no thin skin depth the radiation comes from by necessity, but that is the common case.)

Agreed.

I'm not much of an expert on nebulae and they do have very low particle density, but even so if far from any star, they may be in thermodynamic equilibrium with the cosmic back ground radiation - Have isothermal conditions with temperature of both the particles and their radiation field slightly less than 4K.

Nebulae are often optically thin, so they don't absorb all the light that impinges on them. Their temperature is often around 10,000 K, due to internal heating mechanisms often due to UV light from stars, so they are not in thermodynamic equilibrium with their surroundings.

I think we agree 100% on the physics; but admit I reserve the "skin depth" term for only electrical conductors and speak of "optical depth" for other cases where skin depth could be used also. This is no doubt because my experimental Ph. D. research was on hot plasma, where optical depth and temperature gradients were important considerations.

Yes, I didn't mean "skin depth" in any technical way, just a means of associating a single temperature to the "surface" of the Sun. It's always going to be an idealization, because the Sun is a broiling mass of convecting gas at various different temperatures, and we see to different depths in the photosphere if we look at different places on the disk.

 

[resurgance2001]

There are a lot of very interesting replies to this discussion. Thanks to all contributors, especially Ken.

I am still trying to find a way of distilling all of this information and then summarizing it in as simple as possible terms for my students to understand. They are only GCSE students but the chap who asked the question is quite bright and I want to give him a solid answer, but one which is not too lengthy.

I just want to check my understanding again. There are a lot of very high energy photons emitted in the core. These are scattered continually and presumably many (most?) are aborbed and re-emitted trillions of times before they eventually reach the photosphere. So is this light that we are seeing mostly coming from the core and being effectively filtered by the photosphere? I suspect my understanding is off at this point. Or is it that the photosphere itself is basically emitting a continuous spectrum plus some absorbtion Franhoffer lines? And if that is the case we saying
that the photosphere itself emits a mostly continuous spectrum because of H- ions? I get the bit about emission lines from the chromosphere.

Also can it then be said that a fairly rare low density gas will produce emission lines when heated, where as a massive amount of high density gas will tend to emit a continuous spectrum ?

 

[Ken G]

That's basically all correct. There are a few interesting details to add-- the photons emitted in the core should probably be regarded as heating the gas, which then emits new photons to replace the old ones, which could loosely be regarded as the "same photons" that are being "re-emitted," but that's not quite true for several reasons. First of all, the photons keep getting re-emitted at lower and lower energies, because the gas is at lower and lower temperatures as you go out-- so to conserve energy, you keep getting more photons as their energy drops (so we can't really say they are the "same photons" that get "re-emitted" if there are more of them!). Also, most of the energy in the Sun at any given moment is in the thermal motions of the gas particles, not in the light, so if we are imagining that what is happening is a bunch of photons are being absorbed and re-emitted, that energy must spend more time waiting between re-emissions than it spends in the moving photons themselves. And finally, there is an outer region of the Sun called the convection zone, and in that zone it is actually convective gas motions (like in a convection oven) that is carrying the energy outward, the role of the moving photons is negligible there. In that region, you have rising hot gas, and falling cool gas, so that transports the energy flux in the Sun, and would do so even if there was no light emission going on in there (though it is still plenty bright in there, of course).

Anyway, the upshot of all that is eventually the hot gas gets to the surface, and light emitted by it escapes into deep space, so at that point (the photosphere), the energy flux is back in its original form of being carried by light, though by then you have a much larger number of much lower energy photons. Because of H minus opacity, that emission is over a continuous range of energies, and because of low density layers above that, at special wavelengths the light does not get out so well, and that makes the Frauhoefer absorption lines. Then the emission lines come from even higher up, where the temperature rises due to mechanical heating of that very low density gas, related to the convection (most likely involving interaction with magnetic fields).

 

[resurgance2001]

D

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.

DH thanks for this very helpful reply. I feel like I am getting somewhere now. So from what you are saying we have three things going on: 1/ Continuous (nearly black body) spectrum due to plasma ion interactions. 2/ Absorbtion lines mostly due to relatively cooler gases in thephotosphere 3/ Emission lines particularly from the hotter less dense chromosphere. Is that correct?

 

[resurgance2001]

Ken, I have just seen your post also. It is once again very helpful. My next task then will be to try to get all if this down in a single page or so of A4. I want to have something that I can basically give the more inquisitive advanced students which is manageable. I had forgotten about the convective zone. I am still a little bit confused about the absorption lines. It seems that you are saying they are really due to the H minus ions in the photosphere. Is that correct? Previously people had been saying that it was because the photosphere is relatively cooler. But surely 5800 k is not cool? Or are we saying that the photosphere is also mainly consisting of ions that really this hot surface is actually emitting the radiation that we 'see' and that this is mainly continuos due to ion / ion interactions in the photosphere. Sorry I am being really picky here but I know that these are the sort of questions my own (one or two) students will ask me.

 

[resurgance2001]

When charged particle collide they are accelerated so they emit broad spectrum thermal radiation?

 

[resurgance2001]

The creation of those photons (light particles) is explained by quantum field theory QED?

 

[resurgance2001]

Well I have now got into this topic so much that I've thrown the towel in and just ordered another textbook from Amazon. It is called 'An Introduction to the Sun and Stars' and is the first course book for a series produced by the Open Uni. I first read it about twenty years ago when I was studying their introductory course in astronomy and planetary science. I remember then being impressed with it and the clarity of the explanations. I think that what every one has said here pretty much covers it, but there is some more useful detail and nice diagrams also in the book as well as, if I remember correctly, a nice intro to HR diagrams as well as the PP cycle and CNO cycles. Let's see. :) I do love science and it does impress me how much humans have been able to find out about it all, especially nucleosynthesis. Cheers

 

[Ken G]

I am still a little bit confused about the absorption lines. It seems that you are saying they are really due to the H minus ions in the photosphere. Is that correct?

No, you had it right before-- the continuum comes from H minus opacity, but H minus opacity is over all energies throughout the visible range, because all visible photons are capable of being absorbed by the H minus ion (which ionizes it back to neutral H), and all neutral H is capable of emitting any visible photon (if it can in the process grab an electron and make H minus). So that will not give any lines at all. The absorption lines come from the lower density gas above that, which is largely at cooler temperatures, and that tends to remove photons at the specific energies of the line transitions in that low density gas (a line transition involves electrons jumping between bound states in the atom).

Previously people had been saying that it was because the photosphere is relatively cooler.

There are several ways to get an absorption line, but the simplest and most important comes from the fact that the photosphere is not just a single sphere, it is more like a shell with finite width, and the gas in that shell gets cooler the farther out you go. So that means there is cooler gas overlying warmer gas, but this cooler gas only has an effect on the line frequencies, not the rest of the continuum because the overlying cool layer is transparent except at line frequencies. Perhaps what is confusing you is that the atoms responsible for the absorption lines are mixed in with the H minus atoms responsible for the continuum, but the lines are much more opaque, so are seen at higher and cooler altitudes.

Or are we saying that the photosphere is also mainly consisting of ions that really this hot surface is actually emitting the radiation that we 'see' and that this is mainly continuos due to ion / ion interactions in the photosphere.

The key process that generates the continuum that we call "sunlight" is when a free electron is grabbed into the single bound state that neutral H has for adding an electron and becoming H minus. Any process that can emit light has an inverse process that can absorb it, so when light hits the H minus and ionizes it, that is what makes the photosphere opaque and stops us from seeing even deeper and hotter layers.

 

[resurgance2001]

Thanks Ken - I think I finally got it! H minus ions in the hotter part of the photosphere produce the broad continuous spectrum, and then in upper, slightly cooler (are we able to say what temperature?) parts of the photosphere ordinary H atoms (and presumably some other elements) produce absorption lines. Yes? So obviously within the photosphere there is a fairly continuous temperature gradient and a mixture of H minus ions together with ordinary H atoms. Do we know why the corona is so hot? I seem to remember that the temperature of that soars again but when I first studied this twenty years ago I learned then that scientists were not sure of the reason for this. That of course could have changed since I first read about it. Cheers

 

[Ken G]

Thanks Ken - I think I finally got it! H minus ions in the hotter part of the photosphere produce the broad continuous spectrum, and then in upper, slightly cooler (are we able to say what temperature?) parts of the photosphere ordinary H atoms (and presumably some other elements) produce absorption lines. Yes? So obviously within the photosphere there is a fairly continuous temperature gradient and a mixture of H minus ions together with ordinary H atoms. Do we know why the corona is so hot? I seem to remember that the temperature of that soars again but when I first studied this twenty years ago I learned then that scientists were not sure of the reason for this. That of course could have changed since I first read about it. Cheers

Yes that's right. The continuum comes from a range of temperatures somewhere around 6000 K, and the temperature drops to around 5000 K at the minimum, though there's not much point in trying to be too precise because the mottled surface of the Sun shows you that the temperature varies a lot from place to place. You are right that the chromosphere gets hotter and the corona gets much much hotter (it gets to temperatures similar to those found in the center of the Sun where fusion occurs, but the corona is too low of a density to have much fusion). The reason the corona gets so hot is that it is so low density that any appreciable heating can easily make it very hot, though the exact nature of that heating is still not known in detail. It must be some form of "mechanical" heating, meaning it is not heated by sunlight. The players are the convection, which excites waves and eddies, and magnetic fields, which can twist up and store energy that can then be released in the coronal plasma. You can amaze your students with online pictures from things like the TRACE satellite.