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Eroica
2007-Jul-23, 06:39 PM
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?

Ken G
2007-Jul-23, 07:13 PM
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).

Eroica
2007-Jul-23, 07:41 PM
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.

Jeff Root
2007-Jul-23, 07:58 PM
And I only first heard of their role in generating the Sun's
spectrum here, a few months ago.

-- Jeff, in Minneapolis

Ken G
2007-Jul-23, 08:06 PM
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!

George
2007-Jul-23, 08:32 PM
Is there any significant Doppler component helping the near-Planck distribution?

Ken G
2007-Jul-23, 08:46 PM
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.

George
2007-Jul-23, 09:24 PM
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.

upriver
2007-Jul-23, 10:57 PM
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.

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-answers/60226-opacity-thin-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

Ken G
2007-Jul-24, 02:49 AM
I assume the istotropy and lack of metals are the reason for the near perfect Planck result.

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.

Ken G
2007-Jul-24, 02:51 AM
Which would have been the official explanation(optical thickness) before I introduced the H- explanation

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.

Tim Thompson
2007-Jul-24, 04:43 AM
Yes, I introduced this idea from a different paper as part of my argument against the optical thickness explanation for the solar blackbody spectrum.
Not unless you are a lot older than I thought. See Chandrasekhar & Breen, 1946 (http://adsabs.harvard.edu/abs/1946ApJ...104..430C) (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.

George
2007-Jul-24, 05:13 AM
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.

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

tusenfem
2007-Jul-24, 08:08 AM
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-answers/60226-opacity-thin-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

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.

Ken G
2007-Jul-24, 06:53 PM
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.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.
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.
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!

Tim Thompson
2007-Jul-24, 08:25 PM
Yes, I introduced this idea from a different paper as part of my argument against the optical thickness explanation for the solar blackbody spectrum.

Not unless you are a lot older than I thought. See Chandrasekhar & Breen, 1946 (http://adsabs.harvard.edu/abs/1946ApJ...104..430C) (and the other installements in the 5 part series linked from there).
It appears that the history of this matter dates back to an ApJ paper based on a talk delivered by Rupert Wildt (http://en.wikipedia.org/wiki/Rupert_Wildt), at Yerkes Observatory (http://astro.uchicago.edu/yerkes/), in June 1938 (Wildt, 1939a (http://adsabs.harvard.edu/abs/1939ApJ....89..295W)), and his subsequent, more detailed analysis (Wildt, 1939b (http://adsabs.harvard.edu/abs/1939ApJ....90..611W)). 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 (http://www.amazon.com/Stellar-Atmospheres-books-astronomy-astrophysics/dp/0716703599/ref=sr_1_2/102-2676283-9131318?ie=UTF8&s=books&qid=1185308063&sr=1-2), specifically cites Chandrasekhar (http://en.wikipedia.org/wiki/Subramanyan_Chandrasekhar) & Breen, but only mentions Pannekoek (http://en.wikipedia.org/wiki/Anton_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.

upriver
2007-Jul-25, 03:13 AM
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.

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.

Tim Thompson
2007-Jul-25, 06:20 AM
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.
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 (http://adsabs.harvard.edu/abs/1997PhRvL..79.4770A) 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 (http://adsabs.harvard.edu/abs/1999PhRvA..60..364K)). 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 (http://adsabs.harvard.edu/abs/1988A%26A...193..189J), John, 1989 (http://adsabs.harvard.edu/abs/1989MNRAS.240....1J) and John, 1991 (http://adsabs.harvard.edu/abs/1991A%26A...244..511J)). 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.


There is a difference between density and distance in a plasma, and how it affects a plasma.
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 (http://en.wikipedia.org/wiki/Radiative_transfer), 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 (http://adsabs.harvard.edu/abs/1989MNRAS.240....1J)).


The 255K BB spectrum of the Earth is certainly not from it's surrounding atmosphere.
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 (http://www.amazon.com/Physics-Atmospheres-John-Houghton/dp/0521011221/ref=sr_1_1/105-6815513-5644425?ie=UTF8&s=books&qid=1185344065&sr=1-1) by John Houghton, or Dennis Hartmann's book Global Physical Climatology (http://www.amazon.com/Global-Physical-Climatology-International-Geophysics/dp/0123285305/ref=sr_1_1/105-6815513-5644425?ie=UTF8&s=books&qid=1185344129&sr=1-1).

Nereid
2007-Jul-25, 05:34 PM
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.

Ken G
2007-Jul-25, 07:57 PM
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).

George
2007-Jul-26, 03:36 PM
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. ;)]

Ken G
2007-Jul-26, 03:54 PM
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.
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.

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?Yes, you would then be sampling less of that region of temperature gradient, or "drunkards" who had followed a more similar stumbling history.


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?
Scattering, and opacity variations due to lines etc., always complicate the Planck result, whenever not in strict thermodynamic equilibrium.

Tim Thompson
2007-Jul-26, 05:55 PM
... 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?
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 (http://www.amazon.com/Observation-Analysis-Stellar-Photospheres/dp/0521851866/ref=sr_1_1/105-6815513-5644425?ie=UTF8&s=books&qid=1185471991&sr=1-1), by David F. Gray (http://www.astro.uwo.ca/%7Edfgray/home.html), Cambridge University Press, 2005 (3rd editiion). This book answers the question as well as any book will.

George
2007-Jul-26, 10:19 PM
…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. 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.


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


The same would hold for drunken people stumbling through the doorway of a bar. Yes, it has been said they customarily move opposite that of the brighter lights. :)



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?
Yes, you would then be sampling less of that region of temperature gradient, or "drunkards" who had followed a more similar stumbling history.
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. :)]


Scattering, and opacity variations due to lines etc., always complicate the Planck result, whenever not in strict thermodynamic equilibrium. 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?

Ken G
2007-Jul-26, 10:33 PM
That makes sense. Is it then the radiation flux that determines the CLV so that different mass stars, in equilibrium, will have different CLVs?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.



I have read that the Sun’s surface is not in equilibrium, but I did not understand what they meant.They probably meant it is dynamical-- those darn spicules again.


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


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. Each layer provides a better Planck function than the sum of the layers, because of the temperature gradient.


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?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).

George
2007-Jul-26, 11:16 PM
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. Yes, I had assumed the density variation would be less significant than it apparently is.


Each layer provides a better Planck function than the sum of the layers, because of the temperature gradient. Ok, I like that better than a radial integral producing a better Planck curve.


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). 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.

Ken G
2007-Jul-27, 06:34 AM
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.

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.

George
2007-Jul-27, 01:46 PM
It depends on how one defines the "bottom". If we define the photosphere as what we can directly see, you are right. Indeed, if we take "photos", ain't that as far as we'll see? KISS! ;)


But some people define it by the energy equation: it is the region where the energy flux is carried radiatively rather than convectively. 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?


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). 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.


We need some word for the latter distinction, if not "photosphere" then something else.:hand: First, let's all unite to correct its adjective, then we can work on the noun. :)

Ken G
2007-Jul-27, 02:14 PM
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?
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.


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.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).

Spaceman Spiff
2007-Jul-27, 02:16 PM
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.

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).

Ken G
2007-Jul-27, 02:49 PM
Indeed, there will always be detailed corrections to any overall simplifying description. The real world does not have sharp boundaries, but they are still useful in our minds. (There isn't even a "depth from which we see the Sun"-- an occasional photon can come from very much deeper!)

Nereid
2007-Jul-27, 03:46 PM
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 (http://www.amazon.com/Observation-Analysis-Stellar-Photospheres/dp/0521851866/ref=sr_1_1/105-6815513-5644425?ie=UTF8&s=books&qid=1185471991&sr=1-1), by David F. Gray (http://www.astro.uwo.ca/%7Edfgray/home.html), Cambridge University Press, 2005 (3rd editiion). This book answers the question as well as any book will.Thanks Tim.

Here (http://www.shef.ac.uk/physics/teaching/phy305/) is a set of lecture notes ("Stellar atmospheres (PHY305)") which seems to cover much of what's been discussed so far in this thread.

It's 3rd year undergrad university level, so the math may be somewhat daunting to some readers ...

Bearing in mind its stated scope, does anyone feel there's anything in this set of material that may be misleading (or downright wrong)*, wrt the scope of this thread?

*Of course, much of it is well beyond the scope of this thread!

Ken G
2007-Jul-27, 04:04 PM
A quick look doesn't find any problems, but it's very "mathy"-- sort of, here's the equations, here's what you get when you solve them. Most people won't find that terribly illuminating, but when instead "explanations" are given, they are sometimes wrong-- so this is kind of the "safe" approach to the subject.

Eroica
2007-Jul-27, 04:14 PM
Here (http://www.shef.ac.uk/physics/teaching/phy305/) is a set of lecture notes ("Stellar atmospheres (PHY305)") which seems to cover much of what's been discussed so far in this thread.
Thanks, Nereid. The sixth lecture was very relevant and interesting: Stellar Opacity (hydrogen).

The binding energy of the extra electron on the H- ion is 0.754 eV. Photons with wavelengths of 1640 nm (Infrared) or less can ionize the ion back to neutral H. The extra electrons needed to form H- ions come from ionized matals such as Ca+. But only 2 out of every 108 hydrogen atoms is a H- ion.


For Solar-like stars, it turns out that H- is the dominant continuum opacity source at optical wavelengths. In early-type stars H- is too highly ionized to play a role, whilst in late-type stars there are too few free electrons (since no ionized metals).

Ken G
2007-Jul-27, 06:15 PM
I hadn't seen that part-- those do sound like quite helpful explanations! It does look generally like a pretty complete and well organized course.

George
2007-Jul-27, 08:32 PM
Doesn't fusion aggravate the free electron problem by eating two electrons to convert to one He? This raises an interesting question, Eroica.

Ken G
2007-Jul-27, 09:34 PM
Yes, but all that is in the core, not the atmosphere. Unless you are talking about very highly evolved stars!

George
2007-Jul-27, 09:40 PM
So you think a 400,000 km non-convective zone might insulate it enough? You have my vote! [Does the core develop a large postive net charge?]

Ken G
2007-Jul-27, 09:44 PM
No-- the helium eats protons too!

George
2007-Jul-27, 10:50 PM
No-- the helium eats protons too! Ah, that makes sense; those are my favorite. :)

captain swoop
2007-Jul-28, 12:39 AM
Protons taste like Chicken.

Hmmm Protons!

George
2007-Jul-28, 12:55 AM
:) Have you tried the croutons? They come in a lattice.

Ken G
2007-Jul-28, 01:25 AM
I recommend the quark-gluon soup.

George
2007-Jul-28, 04:16 AM
Which of the six flavors are best? :)

Meanwhile, back to asking about black-body radiation from a white-body radiator :doh: ....Neried's link to another (305-2) states that opacity is inversely proportional to the mean free path of a photon, so I'm still a bit confussed.

Summarizing only the basics of what is happening given a view into a cylindrical region penetrating directly toward the solar center....

1) The vast majority of the light is from the photons generated as the weakly bound electrons become attached to the atom.

2) The near blackbody radiation from the Sun is due to the extensive range of kinetic energies of the free electrons allowing such a wide distribution of photon energies. These photons form when free electrons combine with hydrogen atoms to form the hydrogen ion (H-).

3) The visible depth into the atmosphere is approximately the depth of the photosphere. The scale height more accurately sets the actual depth.

4) The rising adiabatic convective cells from the lower convective zone become turbulent and non-adiabatic in the photosphere.

5) A slight majority of the light comes from the lower-half region of the scale height due to the higher density in this lower region.

6) The total energy flux of the Sun from its surface is the same as the energy flux from the core.

7) The photon flux of the Sun from its surface is far greater than that of the core.

8) Normal absorption lines will be found with some blurring due to minor Doppler, Zeeman, and pressure broadening effects.

9) Hydrogen and helium constitute over 99% of the surface atoms.

10) The other elements produce the majority of the absorption lines in the spectrum.

11) The number of absorption lines exceed 25,000.

12) Emission lines are also found in the spectrum.

13) The peak energy intensity is near 455nm (blue, almost violet).

14) The photon flux distribution is almost flat across the visible spectrum.

15) The net color of this region, with proper attenuation for the observing eye, is.....white or bluish-white.


Original attempt:
1) The vast majority of the light is from the photons generated as the weakly bound electrons become attached to the atom.
[I assume it is too cold for Helium ions?]
2) The near bb radiation is due to the extensive range of photon energies allowable because of the same extensive range of electron states of the outer electron of the hydrogen ion.
3) The visible depth into the atmosphere is the region known as the photosphere.
4) The depth of the photosphere is the scale height for this region, ignoring some variations such as covection currents.
5) A slight majority of the light comes from the lower-half region of the scale height due to the higher density in this lower region.
6) The total energy flux of the Sun from its surface is the same as the energy flux from the core.
7) The photon flux of the Sun from its surface is far greater than that of the core.
8) Normal absorption lines will be found with some blurring due to minor Doppler effects.
9) Additional blurring of the absorption bands may occur due to magnetic activity (Zeeman effect).
10) Hydrogen and helium constitute about 98%(?) of the surface atoms.
11) The other elements produce the majority of the absorption bands in the spectrum.
12) The number of absorption bands exceed 25,000. [I didn't count them!]
13) Emission bands are also found in the spectrum.
14) The peak energy intensity is near 455nm (blue, almost violet).
15) The photon flux distribution is almost flat across the visible spectrum.
16) The net color of this region, with proper attenuation for the observing eye, is.....white or bluish-white.

Eroica
2007-Jul-28, 11:54 AM
16) The net color of this region, with proper attenuation for the observing eye, is.....white or bluish-white.
Is that your final answer? :)

George
2007-Jul-28, 02:47 PM
Is that your final answer? :)

:) It is a subtle shout for a call-to-arms!!! Which is it: white or bluish-white? [Sad that it isn't green. :(]

I hope the answer comes after the heliochromology textbooks are printed -- I should have enough for 6 or 7 pages of text (50 or 60 with sunset pictures). Offering them for sale after the fact would be of little use except as collector items. Perhaps we could add Poor Pluto's story in a chapter, too. I could use some sympathetic padding. ;)

[Added. BTW, I recall it was you, Eroica, that presented us with the 25000 lines in the solar spectrum (Item 12). Do I have that correct?]

Ken G
2007-Jul-30, 02:32 PM
2) The near bb radiation is due to the extensive range of photon energies allowable because of the same extensive range of electron states of the outer electron of the hydrogen ion.The extensive range is in the free electron energies, prior to making an ion of much more definite energy.


3) The visible depth into the atmosphere is the region known as the photosphere.
Unless defined in terms of the energy equation. If we use your definition, we still need a name for what lies between the convection zone and the photosphere.


4) The depth of the photosphere is the scale height for this region, ignoring some variations such as covection currents.The currents at the surface aren't really convection, they are mechanically driven by dynamical energy releases, i.e., they are not adiabatic like standard convection.

8) Normal absorption lines will be found with some blurring due to minor Doppler effects.There is also some blurring due to "pressure broadening" by the perturbing effect of high density, especially important in dwarfs. Indeed, this effect is pretty much the way that dwarfs are identified, though I expect we'd know the Sun was one anyway!

10) Hydrogen and helium constitute about 98%(?) of the surface atoms.Even more.

13) Emission bands are also found in the spectrum.
Note the word "line" is better than "band", unless dealing with molecules.

George
2007-Jul-30, 06:09 PM
The extensive range is in the free electron energies, prior to making an ion of much more definite energy.
Ah, that's a little different than I thought. It is those energies posessed by the free electrons that correlate to the spectral distrubtion. Is this correct?


Unless defined in terms of the energy equation. If we use your definition, we still need a name for what lies between the convection zone and the photosphere. Maybe that is why I have not seen a very specific depth stated for the photosphere. I suppose "the phovectosphere" would not eschew obfuscation. How 'bout "the blueocline" for its distincitive color (once established, that is)? ;) [Perhaps whoever came up with "the tachocline", for the intermediary zone between the convective and the radiative zones, could help, as their term does make some sense (it is a rotational speed interface boundary). :)]


The currents at the surface aren't really convection, they are mechanically driven by dynamical energy releases, i.e., they are not adiabatic like standard convection. Interesting, though I don't get it. Are the convective currents stable, similar to a laminar flow, until they reach the photosphere where they, apparently, become turbulent?


There is also some blurring due to "pressure broadening" by the perturbing effect of high density, especially important in dwarfs.
Ok. Is there a specific term for it?


Indeed, this effect is pretty much the way that dwarfs are identified, though I expect we'd know the Sun was one anyway!
You mean...... a yellow dwarf! :wall: :)


Note the word "line" is better than "band", unless dealing with molecules. Yes.

Ken G
2007-Jul-30, 07:07 PM
It is those energies posessed by the free electrons that correlate to the spectral distrubtion. Yes, the electrons can be characterized by a single local electron temperature, and that shows up in the blackbody radiation they are responsible for, for all energies above the binding energy of H minus.


Are the convective currents stable, similar to a laminar flow, until they reach the photosphere where they, apparently, become turbulent?Convection is pretty turbulent everywhere, but the need to carry a heat flux is just a tiny "hitchhiking" effect that maintains the instability, it doesn't alter the buoyancy-controlled structure that drives the turbulence. In contrast, the flows at the surface are highly non-adiabatic-- they are getting heated and cooled will nilly and following ballistic trajectories due to "squirting" and/or "explosive" types of forces that are highly out of force balance (unlike buoyancy). So the gas in the convection zone is all at the same entropy, pretty much, but above that the entropy becomes very dynamical.


Ok. Is there a specific term for it?Yeah-- pressure broadening!


You mean...... a yellow dwarf! :wall: :)Of course, just ask kindergardners.

George
2007-Jul-30, 10:16 PM
Yes, the electrons can be characterized by a single local electron temperature, and that shows up in the blackbody radiation they are responsible for, for all energies above the binding energy of H minus. Is the binding energy for H- a set value as found in the other transition levels? If so, then is the photon energy simply the difference in this energy state and the free electron's kinetic energy?


Convection is pretty turbulent everywhere, but the need to carry a heat flux is just a tiny "hitchhiking" effect that maintains the instability, it doesn't alter the buoyancy-controlled structure that drives the turbulence. In contrast, the flows at the surface are highly non-adiabatic-- they are getting heated and cooled will nilly and following ballistic trajectories due to "squirting" and/or "explosive" types of forces that are highly out of force balance (unlike buoyancy). So the gas in the convection zone is all at the same entropy, pretty much, but above that the entropy becomes very dynamical.So does this mean that somewhere below the photosphere that the convective cells are essentially adiabatic?


Yeah-- pressure broadening!
Wow, deja vu, I am sure I have recently heard this very term. ;) I think, however, it deserves a scientific name. Even the acronym looks wimpy.


Of course, just ask kindergartners. Alas, we arrive finally at a first cause issue in astronomy! ;)

Ken G
2007-Jul-31, 02:55 AM
Is the binding energy for H- a set value as found in the other transition levels? If so, then is the photon energy simply the difference in this energy state and the free electron's kinetic energy?Yes and yes. The energy was stated earlier in the thead, a bit under 1 eV as I recall.


So does this mean that somewhere below the photosphere that the convective cells are essentially adiabatic?Yes, pretty much throughout the convective region, except at the top and bottom where you convert back to and from radiative energy flux. Note this implies that the convection zone is a lot like a very poor efficiency Carnot engine. Where does the work go? A lot of it, apparently, into magnetic activity (solar flares and so forth).

George
2007-Jul-31, 03:15 AM
Note this implies that the convection zone is a lot like a very poor efficiency Carnot engine. :) Nice oxymoron. I never thought I'd see a Carnot engine and now you've shown me another; the first in the Cepheid ionization zones, possibly.

Ken G
2007-Jul-31, 03:22 AM
Yes, they really have two uses, one to find the maximal efficiency, but another to just see a general mechanism that is approximately followed, to some good or poor level, by a host of periodic and quasi-periodic phenomena that are driven by heat flow.