Ignoring the absorption lines, the Sun's photosphere gives rise to a continuous spectrum that approximates the thermal spectrum of a perfect blackbody.

What is the origin of this spectrum? Specifically, what processes produce the photons that make up the spectrum?

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The surface temperature of the Sun has to be about 6000 K to make the interior structure work. Given that, the Sun would have to be a blackbody at nearly that effective temperature, no matter what the thermalizing opacity was. All that can change is whether the actual temperature at the surface is pretty much the same as the "effective" temperature, and that requires some kind of significant source of thermalizing opacity-- but what the source is does not matter terribly much ("thermalizing" opacity is opacity that is capable of destroying pre-existing photons and creating new ones based entirely on the local temperature).

Having said that, it seems your question remains, what is the thermalizing opacity that serves this purpose? The answer is, "H minus opacity". What this means is, there are a few neutral H atoms near the Sun's surface that have acquired a kind of "hitch-hiker"-- an additional electron in one very weakly bound state that makes the H a negative ion. That electron is easily knocked off the atom, but when a photon does it, that destroys the photon. The energy goes into free electrons, which can later make new H minus atoms-- in turn emitting new photons in a way that depends on the temperature of that bath of free electrons. So you have just what you need-- thermalizing opacity that is capable of acting over a wide range of frequencies (because photons of almost any energy can knock the electrons off, and can be created when electrons that have various amounts of thermal energy are captured in that bound state).

Thanks, Ken G. I did a bit of Googling on this and came across a paper by Rupert Wildt from 1939 attributing the continuous spectrum to H^- ions and the free electrons they give rise to. I had never heard of such ions before.

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And I only first heard of their role in generating the Sun's
spectrum here, a few months ago.

-- Jeff, in Minneapolis

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Yes, it is a remarkably obscure ion, given that the vast majority of all the light that enters our eyes in our lifetime comes from the generation of that ion!

Is there any significant Doppler component helping the near-Planck distribution?

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Doppler thermal broadening, you mean? No, that only shows up in sharp lines. The thermal broadening is only about 1 part in 10,000 of the spectrum-- the solar continuum must rely on the fact that the photon can be created at many different energies even in the rest frame of the emitter, because it is a bound-free process involving H minus.

Ok, thanks.

Another application of what you have stated is the even better example found in the CMBR which is a more true Planck distribution caused during Recombination. I assume the istotropy and lack of metals are the reason for the near perfect Planck result.

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Hi, Eroica,

My name is upriver.

Yes, I introduced this idea from a different paper as part of my argument against the optical thickness explanation for the solar blackbody spectrum.

Here;
http://www.bautforum.com/questions-a...in-plasma.html

Notice how quickly H^- was seized as the official explanation. There is no word of optical depth or 1/tau or 2/3 extinction depth in the explanation to Eroica.

Which would have been the official explanation(optical thickness) before I introduced the H^- explanation for the solar blackbody spectrum to BAUT forum on 10-June-2007, 11:38 PM.

I think I proved my point. Say thank you to the ATMer.

Upriver

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I'm not sure the presence of metals would have changed things that much-- it would have kept free electrons around longer, so would have delayed the "decoupling" somewhat, but you put your finger on it-- the isotropy was highly conducive to generating thermodynamic equilibrium, and would have sufficed even in the presence of metals.

Not at all, I've known that very explanation for over 20 years, and my answer above would have been exactly the same then as now.

Not unless you are a lot older than I thought. See Chandrasekhar & Breen, 1946 (and the other installements in the 5 part series linked from there).

In any case, I think that the radiation is already thermal once it reaches the bottom of the photosphere from the deep interior. In cool stars like the sun, the free-free & bound-free transitions involving the H^- ion are the principle sources of opacity in the stellar atmosphere. But they don't create the thermal spectrum, rather they allow the spectrum to remain thermal, since it was already thermal at the base of the photosphere. Besides, the solar spectrum is only approximately thermal, since we are seeing emission simultaneousy from all parts of the photosphere, which are at different temperatures. That's why the sun us said to have an effective temperature of about 5700 Kelvins.

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Yes, the Sun's outer layers are not in thermodynamic equilibrium. Bhatnagar and Livingston state, "...different methods of observation yield different values, especially when observations refer to different atmospheric layers."

They show different ways to obtain temperature values, including:
Effective temperature (calculate from total Solar irradiance using Stefan-Boltzmann's law). [They get 5760K, but I have seen 5777K, too.]
Brightness temperature (a Planck temperature. My work shows ~5850K)
Color temperature (UBV filter system)
Kinetic tempeature (derived from average speed of the ions, from P=nkT)
Excitation temperature (from the distribution of energy level comparisons)
Ionization temperature (compares the relative no. of atoms in ionization states)

Then there is the variation in temperature found due to the center-to-limb variation (CLV). This ranges from about 5000K at the limb to about 6400K at the center.

PS_h: Or, in heliochromological terms, from a white limb to a bluish-white center.

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Well, that's taking liberty with reality, or at least context. That whole discussion was started up because you don't believe that a plasma can create a BB spectrum and try to do anything to prove you are right, including misinterpreting papers.

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But what it was below this layer is quite irrelevant-- thermalizing opacity will thermalize that which is thermal and that which isn't just as easily, within limits of course. It still has to cool its temperature, after all. Yes, every concept in thermodynamics, as in all physics, is really just an approximation. It just depends on how deeply one chooses to delve and what level of accuracy is desired!

It appears that the history of this matter dates back to an ApJ paper based on a talk delivered by Rupert Wildt, at Yerkes Observatory, in June 1938 (Wildt, 1939a), and his subsequent, more detailed analysis (Wildt, 1939b). There are several more papers authored or co-authored by Wildt on the topic of the negative hydrogen ion in the following years. The Chandrasekhar & Breen paper appears to be the one that gets credit for being the first really detailed analysis of the problem. Mihalas, in his book on stellar atmospheres, specifically cites Chandrasekhar & Breen, but only mentions Pannekoek & Wildt without specific references. Wildt makes use of data from Pannekoek, but I can't find any earlier papers than Wildt's, and I can't find any by Pannekoek.

So we can trace our understanding, that the continuum SED of the sun is due mainly to photospheric negative hydrogen ions back to 69 years ago.

P.S. After typing all that, I see that Wheeler & Wildt, 1942, credit Pannekoek (MNRAS, vol 96, p. 162, 1931) as the original proponent, which pushes the idea back to 77 years ago. The Pannekoek paper is not online, at least it is not in the NASA ADS server.

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Nobody ever said a word in their arguments here on BAUT about H^- being the cause of the solar blackbody spectrum. At least not in their arguments with me.

I dont believe that distance creates a blackbody spectrum in a thin plasma.
And you cant prove that it does in a lab on earth.

I know that a dense plasma will create a blackbody spectrum. Remember the bubble spectrum that was halfway between line and blackbody emission?
There is a difference between density and distance in a plasma, and how it affects a plasma.

I did not misinterpret papers. That paper says that H^- are responsible for the BB spectrum of the sun, and that density is normally responsible for a blackbody spectrum, except in the case of the sun.
I brought that paper up as evidence that density is the cause of blackbody spectrum, and that a thin plasma is not capable of generating a blackbody spectrum.

The 255K BB spectrum of the Earth is certainly not from it's surrounding atmosphere.

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Well, you know physics trumps what you, or anyone else believes, every time. And yes you can "prove it", in a lab. After all, the absorption coefficients of the H^- ion are actually measured in laboratories (i.e., Andersen, et al., 1997 and references thereto). Those absorption coefficents are consistent with those which are calculated from first principles, also known as physics (i.e., Kuan, Jiang & Chun, 1999). And all of the observed & calculated absorption coefficients, applied to radiative transfer models of the solar photosphere, are consistent with observations of the sun (i.e., John, 1988, John, 1989 and John, 1991). This is a direct application of laboratory data to stellar atmospheres, which clearly shows that H^- opacity is responsible for the solar continuum emission. The fact that we cannot build a 400 kilometer replica of the solar photosphere is not something to hide behind.

Indeed so, but neither density nor distance is the key to understanding absorption in a plsama, or in a neutral gas. Absorption is the key, and a strong absorption coefficient will trump both density and distance. All you have to do is look at the radiative transfer equation, where dI/ds (I is the spectral intensity and s is path length or distance) is set equal to emission minus absorption minus scattering. A strong absorption coefficient will be effective over a short distance. You don't find density explicitly in the equation anywhere, beause it is included in the absorption coefficient. So a small absorption can produce dramatic results over a long distance (which increases the path length) or in a dense environment (which increases the mass absorption coefficient). And a large absorption coefficient does not need much of either path length or density to get the job done.

This means that you can't make absolute statements like "I dont believe that distance creates a blackbody spectrum in a thin plasma" and expect to get away with it. In all cases, plasma or not, you need to know the specifics of the situation first. Distance without any question or doubt at all, certainly will produce a blackbody (or near blackbody) spectrum, in a thin plasma, or a thin neutral gas, if the absorption coefficient is strong enough. The depth of the solar photosphere is roughly 400 km, and decades of both laboratory data & physical theory clearly show that in the solar photosphere, the H^- absorption coefficient is strong enough to do exactly that (i.e, John, 1989).

Actually, yes it is. The global average surface temperature of Earth is 288 K (about 59 F), whereas its effective radiative temperature is 255 K (about -1 F). The radiative temperature is cooler than the surface kinetic temperature because an outside observer is looking at the cooler outer atmosphere, and therefore sees a cooler temperature. You will find this explained in any book on atmospheric physics, i.e., The Physics of Atmospheres by John Houghton, or Dennis Hartmann's book Global Physical Climatology.

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If we consider only the steady state spectrum, and ignore (for now) stars which are quite variable (e.g. those in the instability strip); if we focus only on the SED around the blackbody peak - say an OOM or two either side* - what is the answer to the question in the OP ... for all the kinds of stars we can see?

Factors to consider include: temperature, pressure, composition.

And it might be interesting to take a deeper look at this part of the OP: "Ignoring the absorption lines": under what circumstances do absorption (or emission!) lines become important re the "continuous spectrum"? For example, can absorption lines be so broad and/or so close together that there's no continuous spectrum to be found across a large part of spectrum? What about stellar atmospheres with lots of molecules (which have much richer absorption spectra than atoms)?

*So we don't look at the x-ray or higher energy part (except if the star in question is extremely hot!), nor the microwave or radio.

The molecular absorption lines dominate in very cool stars, and do cover the spectrum. Even cool regions of the Sun apparently do that, in CO lines. For much hotter stars, you don't have molecules or H minus, but you have bound-free edges of various species. The degree of scattering gets quite large-- from free electron opacity-- but we need to understand the thermalizing opacity to understand where the blackbody spectrum comes from. Free electrons in the vicinity of protons have some photon-thermalizing abilities, called "brehmstrahlung" (braking radiation) or "free-free emission" (it's the same thing), but I don't know if that ever exceeds the bound-free edges (generally I think not).

I would be highly interested in learning how the scale height affects the spectrum from the observers point of view here on Earth. The center-to-limb variation (CLV) reveals a significant difference in temperature of about 1400K between limb and the central zone of the disk.

Normally, the solar spectral irradiance is the integral of all emissions from the disk. Due to the CLV, this will not result in a nice Planck distribution, albeit, it is pretty close -- the difference in effective temperature and Planck temperature is only about 70K, or so.

If, however, a spectral irradiance observation is obtained of only a small region of the Sun, then a much nicer Planck distribution should be found, right?

What is unclear to me is whether or not the central disk temperature of 6400K will produce a very nice Planck distribution for that temperature (ignoring the emission & absorption bands)? In other words, would scattering or other effects [observable to the bottom of the photosphere] complicate the Planck result?

The advancement of heliochromology appreciates all scientific comments. [Reasonable fluff is acceptable as the textbook draft is still rather thin. ]

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That's not really an effect of the scale height per se, in that you could increase or reduce the scale height without altering the limb darkening. Yes, it does relate to the temperature gradient over the scale height, but the logic goes the other way-- the physics of diffusive radiation causes both the limb darkening and the temperature gradient; neither of those cause the other. You can use either to establish the other, but the logical cause is simply the physics of radiative diffusion in radiative equilibrium. That's why changing the scale height will change the temperature gradient so as to end up with the same limb darkening.

The limb darkening is caused by the fact that diffusive radiation is more likely to emerge more or less directly out of the boundary than into a similar-sized angular bin that is at a steep angle from the boundary. The same would hold for drunken people stumbling through the doorway of a bar.
Yes, you would then be sampling less of that region of temperature gradient, or "drunkards" who had followed a more similar stumbling history.
Scattering, and opacity variations due to lines etc., always complicate the Planck result, whenever not in strict thermodynamic equilibrium.

I think this can & will quickly become an answer that is too long & complex to delve into here in detail. For the sun, most of the continuum comes from the H^- ion. In cooler stars we see a lot more absorption lines, including molecular lines, and the continuum now becomes a combination of line wings and true continuum sources. And in hotter stars, the H^- absorption goes down.

With that in mind, I would like to suggest a book: The Observation and Analysis of Stellar Photospheres, by David F. Gray, Cambridge University Press, 2005 (3rd editiion). This book answers the question as well as any book will.

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That makes sense. Is it then the radiation flux that determines the CLV so that different mass stars, in equilibrium, will have different CLVs?

I have read that the Sun’s surface is not in equilibrium, but I did not understand what they meant.

Ok, and this is how the photons got there in the first place, right?

Yes, it has been said they customarily move opposite that of the brighter lights.

So, using this colorful analogy, the observer of a smaller, central region would be seeing more of those who are less drunk –- those who at least are sober enough to see the door and, thus, will tend to get through in greater numbers. As for the more wobbly characters, they will be bouncing off the walls in random and staggered walk. This poor behavior will tend to make them more embarrassed and causing them to be more red-faced once they finally exit. [I hope I have your analogy correct.]

If we increase the size of the bar, we will increase the number of drunks. Further, if we keep the same-sized a/c unit, more will be hitting the walls and more will be exiting, and they will be hotter as the exit, too. [I know analogies aren’t suppose to stretch this far, but yours was worth the try. ]

Yes, but what I keep thinking about is how there must be only a small number of photons that can make it “out the door” from the bottom of the photosphere (the scale height). These will tend to be of shorter wavelength since they come from a hotter region. The photons from higher and higher regions, along a radial line, will have fewer atoms to encounter above and will have a greater flux coming "out the door".

So I see the final spectrum as the result of all these distributions from the different radial layers. For all I know, the emissions throughout the scale height are needed to get a Planck distribution, but I am curious how to take these dynamics into account.

I also wonder what kind of Planck temperature result we would discover here if a spectrometer (adiabatic ) were placed at the bottom of the photosphere?

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Larger flux just scales up the CLV proportionately. What changes it are details like bumpy spicules, departures from radiative equilibrium, nonspherical geometry due to rotation, all kinds of nasty things we'd rather idealize away. Then if you want to talk about the apparent temperature as part of the CLV, you also have to look at how much scattering vs. thermalizing opacity you have, and how it varies with frequency. Basically, you need a very powerful computer, but you can squint your eyes and get the basic gist much more easily, in terms of what emerges naturally from radiative diffusion.

They probably meant it is dynamical-- those darn spicules again.
Right, they diffused up from the top of the convection zone, where they were released by rising parcels of hot gas. And got destroyed and re-created so many times along the way, often at lower energies, that they are not really even the "same photons". Yet the flux of energy was maintained.
Yes, but the density is lower, and so is the temperature-- so fewer photons come from there, not more. It doesn't "turn over" until you look too deep, and they can't get out easily enough.
Each layer provides a better Planck function than the sum of the layers, because of the temperature gradient.
You'd get pretty much the local temperature at that point, if you put it down there deep enough-- that's the meaning of "local thermodynamic equilibrium" despite the presence of a flux (so it's not full-blown thermodynamic equilibrium).

Yes, I had assumed the density variation would be less significant than it apparently is.

Ok, I like that better than a radial integral producing a better Planck curve.

I wonder then if the observed temp. of ~6400K for the central disk would be close to the temperature found at the bottom of the photosphere? It is probably a little hotter down at that point, I suppose.

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It depends on how one defines the "bottom". If we define the photosphere as what we can directly see, you are right. But some people define it by the energy equation: it is the region where the energy flux is carried radiatively rather than convectively. That means it is the surface region that is in radiative equilibrium, in contrast to the region where the entropy is the same at all depths (adiabatic convection does that). We need some word for the latter distinction, if not "photosphere" then something else.

Indeed, if we take "photos", ain't that as far as we'll see? KISS!

Actually, I assumed this is really the better view. Perhaps I've already asked, but, is there any reason not to simply consider this just like the mean free path (mfp) for a photon?

I would guess that when convective zones are measured in thousands of kilomters, that adiabatic is a fair description. However, I don't understand the contrast you refer to.

First, let's all unite to correct its adjective, then we can work on the noun.

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The depth of such a layer would extend to many mean-free-paths, it's fixed by where the star is convectively stable, not by the photon mfp.
In radiative equilibrium, the radiative energy flux has to be constant. In energy balance, it's just the total energy flux, including convective energy of hot gases, that has to be constant. In adiabatic convection, the entropy is also constant, because you are looking at more or less the same gas everywhere, just in different stages of adiabatic expansion (which in turn makes it buoyant or heavy in rapid succession).

Except that the convection zones in the outer envelopes of stars are super-adiabatic (i.e., large departures above the adiabatic temperature gradient) in the outermost regions (in our Sun r/R > 0.95). These super-adiabatic zones have lower efficiencies (I believe this is primarily results of too little mass in these outermost layers). In addition, the role of convective energy transport cannot be ignored even in the 'photosphere' (small, but not zero).