View Full Version : White Dwarf Stars

quanoe

2017-Feb-07, 12:55 PM

I am currently working on a physics project for my International Baccalaureate level physics class. The project I am working on is over the correlation of the luminosity versus the age of white dwarf stars. The reason I chose this is that to my knowledge white dwarf is one of the lesser studied classifications of stars. The bad thing with this is that there are not many resources I have been able to find that are available to high schoolers. I am honestly looking for any level of help whether that be general information about white dwarf stars for my scientific background, or database recommendations so I can find more information.

George

2017-Feb-07, 03:17 PM

There are a variety of catalogs available if you google "white dwarf catalog".

Some general comments from this amateur...

White dwarfs aren't technically stars since their days of fusion are over. They are amazingly hot "ashes" pilled tightly into a small sphere after a star with a range of masses (our Sun is one) blows-off its outer layers at the end of its life. They are simply called white dwarfs.

Their sizes are about that of the Earth. The smaller the size, the less the luminosity. This is an inverse square relationship. But they are, initially, extremely hot (bluish-white hot). Luminosity is a 4th power of temperature. They stay hot for a very long time. From these two factors, their apparent magnitude is not hard to calculate if you assume a temperature and a distance. [This can be tweaked a little given that apparent magnitude is a visual thing and luminosity usually refers to the total energy output. The SED (spectral energy distribution) will shift the proportion of visible light emitted as it cools, thus slightly changing the value of the apparent magnitude.]

Ken G

2017-Feb-07, 03:36 PM

Welcome to the forum! For questions like this, it would help to know where you have already looked. If you google "white dwarf", you will get a lot of basic introductory material, and you will get a rather advanced and complete Wikipedia article with a bibliography for further reading. But you are right that the bibliography will be hard to access and/or understand. Also, beware of false information in the simplified sites-- for example, at

https://imagine.gsfc.nasa.gov/science/objects/dwarfs2.html you will find this oft-repeated falsehood:

"Once a star is degenerate, gravity cannot compress it any more, because quantum mechanics dictates that there is no more available space to be taken up."

The "because" in this statement is wrong, quantum mechanics does not dictate that there is no more available space, because as this very article is about to go on to say, if you add more mass, the white dwarf compresses! So why is there suddenly more "available space" if the squeezing force gets larger? Isn't compression, when you squeeze harder, the very meaning of "available space?" So no, the issue is not that there is no available space, the issue is that in order to get more compression of any gas (that is not undergoing fusion at the time) that is in force balance between gravity and pressure, you must either squeeze harder (and since the squeeze comes from gravity, this means adding mass), or you must disadvantage the pressure by removing heat. The latter is what stars are normally doing, because they shine light, so when that light is not replaced by fusion, the star contracts. What is happening in a white dwarf is that the quantum mechanics of the Pauli exclusion principle dictates that the star cannot lose heat very easily (and ultimately, ,when it becomes a "black dwarf", not at all). If it cannot lose heat, it cannot contract, that's all that is happening with white dwarfs.

Another error you often find is an association between degeneracy (i.e., the Pauli exclusion principle) and super-high densities. White dwarfs only have super-high density because of their spectacular mass and gravity, we find mundane degenerate systems all around us at perfectly normal densities. The electrons in a metal are very much a degenerate system at everyday normal density. It's ironic that many simplified treatments will comment that degenerate stars act a bit like metals do, yet will see no contradiction with saying that degenerate material is super-high density-- even though it isn't in metals! It's pretty difficult to piece together a correct understanding of degenerate gas.

The seminal paper for understanding white dwarfs is by Mestel, and is referenced in the Wiki article, but it is an article written for professional physicists and astronomers so is probably too difficult. So you find yourself in a quandary-- either stay at the superficial (and frankly, sometimes wrong) level of the introductory material, or delve into the advanced physics that explains what is really going on. There should be a middle ground there, explanations that are correct, insightful, and simple, but frankly I don't know where to look for one-- the usual stuff you find falls victim to a host of misconceptions like the ones I've cited. But since your interest is not so much on the detailed physics of a white dwarf, but rather how its luminosity changes as it ages, the situation is quite a bit simpler for you.

Basically, a white dwarf is a star that has lost almost all of the heat that it can lose (that's the Pauli exclusion principle again). So when you talk about the luminosity of a white dwarf, think of it like the last trickle out of a water bottle that is almost dry. It's not that there's no more kinetic energy inside there, there's tons and tons of that, it's that it cannot get out, so you are nearing the end of heat lost in the form of light. So to understand how the luminosity changes, you only need to know how much heat is still in there and how long it is taking to get out, you don't need to worry about any structural changes in the white dwarf any more than you have to worry about structural changes in a nearly empty water bottle. The only thing that will change is the surface temperature, because as was said above, the luminosity is proportional to surface temperature to the fourth power. However, it's not the surface temperature that sets the luminosity, it is the luminosity that sets the surface temperature. So you still need some other way to understand the luminosity. Start with the star having a given internal temperature (not surface temperature, as I said)-- it can be almost anything you want, a white dwarf will have had that internal temperature at some point in its history. That tells you the amount of heat that is in there, because the heat is held by the carbon ions in a completely normal way, like any ideal gas, there's no issues with degeneracy for the ions. (Interesting degeneracy factoid: the heat is in the ions like carbon in a white dwarf, but the kinetic energy is in the electrons, because the ions have much more mass so act like an ideal gas). Also, if you know the temperature, you know the amount of light that is in there (look up the "Stefan-Boltzmann law" for the energy density of light, which is essentially 1/c times that same law for the energy flux density). All you need to know is how long it takes that light to leak out, and that gives you the luminosity by the total amount of light (density times volume) divided by the time to leak out (by radiative diffusion, this is the only hard part). It's not easy to know that, but if you do get that "leakage time" as a function of white dwarf mass, it won't change as the white dwarf cools, so that's good.

Let's call the leak time (the radiative diffusion time) tau, and hope you can find some good way to get at it. then we'll call the heat in the ions Q (which depends only on internal temperature and mass of the star), and the light energy E (which comes from the T and the volume), and we can say the luminosity L obeys L = E/tau. Then L is the rate of dropping of Q, so we have dQ/dt = -E/tau. If you don't like calculus, just define a new time, call it t, which is the time for the amount of heat inside the star to drop by about a factor of 2. Then you will have approximately Q/t ~ E/tau, so you can solve t ~ tau*Q/E, where Q/E is just a simple function of the internal temperature and the mass of the white dwarf. That characterizes the time t for the white dwarf to get noticeably cooler, and its luminosity to get noticeably less (say by about a factor of 2). But you do need tau, and that depends on the physics of radiative diffusion.

However, if you don't want to know tau, there's still a lot you can do because you know that t is the same for all white dwarfs of the same mass. So just focus on a subclass of white dwarfs of the same mass but different age and luminosity. What you can now do, at least, is get the ratio of the cooling time t for two different white dwarfs, call that t1 and t2. Then using t = tau*Q/E, form the ratio t2/t1 = (Q2/Q1) / (E2/E1) , and that result depends entirely on the mass of the white dwarfs and their two internal temperatures. In fact, since Q is proportional to internal temperature, and E is proportional to internal temperature to the 4th power, this means the t ratio is proportional to the internal temperature ratio to the -3 power. That's about the simplest result you will get without a lot of work! Also, we saw above that the ratio L2/L1 equals E2/E1, which is given by T to the 4th power, so simply observing the L ratio tells you the T ratio to the 4th power. Simple algebra thus gives:

t2 / t1 = (L1/L2)3/4.

There's a nifty result you can get quite simply from the basic physics, and don't need to know much about degeneracy except for what was used in the steps above. (And as pointed out in the next answer, the trick in getting the L ratio is knowing the distances, which is the observational challenge there. So I don't know how large of a sample of white dwarfs you can find that have the same mass and known distances, but that's all you need.)

antoniseb

2017-Feb-07, 03:52 PM

One of the hoped for results from the ESA's Gaia mission (currently in operation) is to vastly increase the population of white dwarfs for which we have accurate distances, and from that we should be able to get more accurate numbers about the ranges of their brightnesses, and their initial luminosities. The theory of how rapidly they cool off once they've formed has very little room for variation (a few percent based on composition, more if there is infall from a companion), and so the range of initial conditions is the open detail.

George

2017-Feb-07, 03:57 PM

What is happening in a white dwarf is that the quantum mechanics of the Pauli exclusion principle dictates that the star cannot lose heat very easily (and ultimately, ,when it becomes a "black dwarf", not at all). If it cannot lose heat, it cannot contract, that's all that is happening with white dwarfs.That's georgeeze stuff! Thanks. How does this effect its luminosity, since this is the crux of quanoe's interests? Is luminosity (in the context of the Pauli Exclusion principle) affected more so as it cools?

Another error you often find is an association between degeneracy (i.e., the Pauli exclusion principle) and super-high densities. White dwarfs only have super-high density because of their spectacular mass and gravity, we find mundane degenerate systems all around us at perfectly normal densities. The electrons in a metal are very much a degenerate system at everyday normal density. It's ironic that many simplified treatments will comment that degenerate stars act a bit like metals do, yet will see no contradiction with saying that degenerate material is super-high density-- even though it isn't in metals! It's pretty difficult to piece together a correct understanding of degenerate gas. I would guess that this is why when cold-rolling metal, they get thinner, but they also get longer. But metal on Earth has a longer place to go; not so on a WD.

Ken G

2017-Feb-07, 04:18 PM

That's georgeeze stuff! Thanks. How does this effect its luminosity, since this is the crux of quanoe's interests? Is luminosity (in the context of the Pauli Exclusion principle) affected more so as it cools?In my extended answer above, you'll see that luckily, the luminosity can be analyzed independently of the degeneracy, if you control for the mass of the star.

Ken G

2017-Feb-07, 04:25 PM

And actually, to get into it even deeper, one can use the mass-radius relationship that connects radius of a white dwarf to its mass, which says R is proportional to mass M to the -1/3 power, and we can also use radiative diffusion physics, which says the time to leak out is proportional to M/R if I assume constant opacity per gram (I think that might be a good assumption, but I'm not sure). Putting that into my above analysis shows that you don't even need to use white dwarfs of the same mass, you can use any that you know the distance and mass, so you know the luminosity L and the mass M. Then it is fairly straightforward algebra, nothing beyond high school math, to find this for the ratio of the cooling times:

t2/t1 = (M2/M1)7/3 / (L2/L1)3/4.

What that means is, if you know L and M for any two white dwarfs, you can know the ratio of the time it will take them to significantly lower their luminosities, but there's a lot of radiative diffusion and degeneracy physics embedded in there. In particular, in there is the assumption of a constant opacity per gram, so if that isn't valid, or if you don't want to use what you cannot derive, just go back to controlling for the mass and don't put M in there as an explicit variable.

Ken G

2017-Feb-07, 04:35 PM

Oh, and a key point I should have mentioned is, the cooling time t is also pretty much the same thing as the age of the white dwarf, because not only is it the time it will take to look significantly less bright, it is also the time it took to get to the brightness it has. So you can also interpret t, approximately, as the age of the white dwarf, and use the L ratio to estimate the age ratio via

t2 / t1 = (L1 / L2)3/4

or if they have different masses

t2 / t1 = (M2 / M1)7/3 / (L2 / L1)3/4

I just derived these, so check them for possible errors, but they follow pretty simply so I think they are right. The main caveat is that the second one assumes the opacity per gram in a white dwarf is constant, and it assumes the distance the light must diffuse through the crust is proportional to the radius of the dwarf, either of which may be wrong so take that one with a grain of salt.

Ken G

2017-Feb-07, 04:51 PM

Actually, I found a more complete analysis with better assumptions at http://www.astro.sunysb.edu/lattimer/PHY521/degen.pdf. They get

t2/t1 = (M2/M1)5/7 / (L2/L1)5/7

So I had close to the right L dependence but the M dependence was pretty off, because my opacity assumption was too rough. Anyway, this is a nice result because it means the cooling time, and so age of the white dwarf, depends only on M/L, but it's less than proportional to M/L for reasons that are not immediately obvious but will depend on how its crust changes as it ages.

Ken G

2017-Feb-07, 05:18 PM

One final interesting point-- if the cooling time is proportional to L-5/7, it means if you do the calculus, it takes an infinite time to cool completely. So there are no fully degenerate white dwarfs, it's just an approximation we use to understand them. We wouldn't see them anyway, they'd be "black dwarfs", but they all have to have some nonzero L because it takes forever to completely cool.

chornedsnorkack

2017-Feb-07, 06:10 PM

A white dwarf has no internal heat source, unlike a main sequence star.

Is there any reason to think that the ratio of central to surface temperature should remain constant as the dwarf cools?

George

2017-Feb-07, 07:14 PM

However, it's not the surface temperature that sets the luminosity, it is the luminosity that sets the surface temperature. I may shock you.... I think I get this point you have written extensively about. The faster I throw snow balls the hotter I get; I don't increase my throwing rate if I were to get hot first. It's taken me a while to get this... we don't have snow down here. :) [Not a perfect analogy, but seasonally appropriate, perhaps.]

chornedsnorkack

2017-Feb-07, 10:00 PM

Is the total energy of a white dwarf proportional to its interior temperature, or to fourth power of interior temperature?

Ken G

2017-Feb-08, 12:08 AM

A white dwarf has no internal heat source, unlike a main sequence star.

Is there any reason to think that the ratio of central to surface temperature should remain constant as the dwarf cools?There can be changes in that ratio as the crust responds to the cooling interior. The simplest assumption would be that the ratio of internal temperature to surface temperature stays fixed, but it's probably not quite right. That link I gave is the real deal there.

Ken G

2017-Feb-08, 01:53 AM

Is the total energy of a white dwarf proportional to its interior temperature, or to fourth power of interior temperature?Neither-- there is very little connection between the kinetic energy, or total energy, in a white dwarf and its temperature. That's because the temperature only measures the kinetic energy in the ions, but the total kinetic energy is almost exclusively in the degenerate electrons.

chornedsnorkack

2017-Feb-08, 07:30 AM

Neither-- there is very little connection between the kinetic energy, or total energy, in a white dwarf and its temperature. That's because the temperature only measures the kinetic energy in the ions, but the total kinetic energy is almost exclusively in the degenerate electrons.

At absolute zero, both ions and electrons are at their respective ground states, and there is no energy other than zero point energy.

At any nonzero temperature, both excited ions and excited electrons would exist.

The distribution of both excited electrons and excited ions must fit the same temperature, because energy is readily redistributed, e. g. electrons and holes may recombine and emit the energy as phonons.

How is the energy of a white dwarf of nonzero temperature distributed between excited electrons and excited ions? How does the total depend on temperature?

Ken G

2017-Feb-08, 03:05 PM

At absolute zero, both ions and electrons are at their respective ground states, and there is no energy other than zero point energy.

At any nonzero temperature, both excited ions and excited electrons would exist.

The distribution of both excited electrons and excited ions must fit the same temperature, because energy is readily redistributed, e. g. electrons and holes may recombine and emit the energy as phonons.All true. But what I mean by the energy there having "little connection" with temperature is like saying that if you take a billion dollars, and add to it the amount of change you have under your couch cushion, the resulting amount of money you are talking about has "little connection" with what is under your couch-- even though that resulting amount of money is a function only of that amount in your couch.

How is the energy of a white dwarf of nonzero temperature distributed between excited electrons and excited ions? The kinetic energy is virtually all in the electrons, but the exractable heat (which is way less) is virtually all in the ions. This is a situation similar to a warm piece of metal.

How does the total depend on temperature?It looks like the total kinetic energy you have in a fully degenerate gas of electrons at white dwarf density, plus the kinetic energy of a cool ideal gas of ions. That second amount is way less, and is proportional to temperature, while the first amount is huge and has almost no dependence on temperature.

DaCaptain

2017-Feb-08, 03:33 PM

Since a white dwarf isn't burning as a main-sequence star anymore that means its solar wind has probably died as well. Wouldn't the collapse in the solar wind cause more intergalactic material to gravitate towards the white dwarfs center thus cooling it quicker?

Hornblower

2017-Feb-08, 04:37 PM

Since a white dwarf isn't burning as a main-sequence star anymore that means it’s solar wind has probably died as well. Wouldn't the collapse in the solar wind cause more intergalactic material to gravitate towards the white dwarfs center thus cooling it quicker?

Accretion of interstellar gas and dust, if in sufficient quantity to be significant, will heat it rather than cool it. As for solar wind, I would expect the absence of any to be due to the strong surface gravity, not the lack of internal generation of heat from fusion or anything else. The photosphere temperature of Sirius B is still much greater than that of the Sun, but because of its small radius the surface gravity is about 10,000 times as strong, so atoms at the surface will be less able to escape.

chornedsnorkack

2017-Feb-08, 07:17 PM

The kinetic energy is virtually all in the electrons, but the exractable heat (which is way less) is virtually all in the ions. This is a situation similar to a warm piece of metal.

...while a cold piece of metal is different.

It looks like the total kinetic energy you have in a fully degenerate gas of electrons at white dwarf density, plus the kinetic energy of a cool ideal gas of ions. That second amount is way less, and is proportional to temperature, while the first amount is huge and has almost no dependence on temperature.

Um, are the ions in a white dwarf an ideal gas, or gas at all?

Yes - the Dulong-Petit law is that energy is proportional to temperature, because all translational vibrations are excited.

But that applies only at high temperatures.

At low temperatures, high frequency vibrations are frozen. Only low frequency collective excitations happen - and by Debye law, energy is proportional to fourth power of temperature.

That applies to warm metal - but not to cold metal.

In cold metal, vibrations of nuclei are mostly frozen, as per Debye law - but most remaining energy is in electrons. It is proportional to square of temperature.

Which of these cases applies to white dwarf?

DaCaptain

2017-Feb-08, 07:40 PM

Accretion of interstellar gas and dust, if in sufficient quantity to be significant, will heat it rather than cool it. As for solar wind, I would expect the absence of any to be due to the strong surface gravity, not the lack of internal generation of heat from fusion or anything else. The photosphere temperature of Sirius B is still much greater than that of the Sun, but because of its small radius the surface gravity is about 10,000 times as strong, so atoms at the surface will be less able to escape.

Does that mean the gravitational pull changes as a star becomes a white dwarf?

Noclevername

2017-Feb-08, 08:01 PM

Does that mean the gravitational pull changes as a star becomes a white dwarf?

A star loses some mass when collapsing from Red Giant phase (Helium flash). But the smaller size and higher density of a WD actually increase the escape velocity.

Hornblower

2017-Feb-08, 08:24 PM

Does that mean the gravitational pull changes as a star becomes a white dwarf?

At any given radius outside the photosphere, the gravitational action does not change for whatever remnant is left after the mass loss during the red giant stage. For a given mass, the gravitational attraction at the surface of a spherical body is inversely proportional to the square of its radius.

chornedsnorkack

2017-Feb-08, 09:58 PM

Accretion of interstellar gas and dust, if in sufficient quantity to be significant, will heat it rather than cool it. As for solar wind, I would expect the absence of any to be due to the strong surface gravity, not the lack of internal generation of heat from fusion or anything else. The photosphere temperature of Sirius B is still much greater than that of the Sun, but because of its small radius the surface gravity is about 10,000 times as strong, so atoms at the surface will be less able to escape.

Yet central stars blowing planetary nebulae are obviously white dwarfs - hot ones at that.

As a star evolves from a cold red giant to a hot white dwarf and then into cooler white dwarf, what is the maximum surface temperature reached?

A young white dwarf has a powerful stellar wind, because it is in the process of blowing the planetary nebula. As the flux of the stellar wind decreases, what happens to its speed? Will a planetary nebula whose central star has stopped blowing the nebula keep travelling outwards by inertia and leave behind vacuum around the white dwarf? Or will the speed of stellar wind gradually slow down before it stops completely? Will the planetary nebula fall back in the white dwarf when it has cooled enough to no longer blow the nebula?

Will the wind of ambient interstellar gas blow a planetary nebula away from its central star, exposing the white dwarf to ambient interstellar gas no longer of its own creation? And does a white dwarf leave behind a planetary nebula blown away and separated from its central star?

Hornblower

2017-Feb-08, 10:36 PM

Yet central stars blowing planetary nebulae are obviously white dwarfs - hot ones at that.

As a star evolves from a cold red giant to a hot white dwarf and then into cooler white dwarf, what is the maximum surface temperature reached?

A young white dwarf has a powerful stellar wind, because it is in the process of blowing the planetary nebula. As the flux of the stellar wind decreases, what happens to its speed? Will a planetary nebula whose central star has stopped blowing the nebula keep travelling outwards by inertia and leave behind vacuum around the white dwarf? Or will the speed of stellar wind gradually slow down before it stops completely? Will the planetary nebula fall back in the white dwarf when it has cooled enough to no longer blow the nebula?

Will the wind of ambient interstellar gas blow a planetary nebula away from its central star, exposing the white dwarf to ambient interstellar gas no longer of its own creation? And does a white dwarf leave behind a planetary nebula blown away and separated from its central star?

The material that blows away and forms a nebula was already enormously bloated to a radius where the gravity is very weak. I would not be surprised if some residual envelope material in close, where the gravity is much stronger, falls back onto the newly degenerate core that is becoming a white dwarf. I do not have the mathematical knowhow to do any quantitative calculations.

I have seen temperature as high as 100,000K attributed to newly exposed white dwarfs. The intense ultraviolet radiation makes the nebula shine and keeps it expanding. The nebula typically fades away in a few tens of thousands of years.

If the star is moving rapidly through a dense region of the interstellar medium, the ejecta will be swept out like the tail of a comet instead of remaining centered on the dying star. Mira is doing just that in its far advanced red giant stage.

Hornblower

2017-Feb-08, 10:41 PM

At any given radius outside the photosphere, the gravitational action does not change for whatever remnant is left after the mass loss during the red giant stage. For a given mass, the gravitational attraction at the surface of a spherical body is inversely proportional to the square of its radius.

Let me rephrase this a bit. If we are orbiting a contracting spherical object, the gravity that is keeping the orbit stable will not change during the contraction. If we are sitting on its surface, the gravity we experience will increase as it contracts. If it starts at 1 g and contracts by a factor or 100 we will be squeezed as flat as the proverbial pancake. This of course is a thought exercise in which we assume a solid surface and that we can take the heat.

chornedsnorkack

2017-Feb-09, 05:49 AM

The material that blows away and forms a nebula was already enormously bloated to a radius where the gravity is very weak. I would not be surprised if some residual envelope material in close, where the gravity is much stronger, falls back onto the newly degenerate core that is becoming a white dwarf. I do not have the mathematical knowhow to do any quantitative calculations.

Does the infalling matter cause any observable effects?

If a parcel of gas falls onto a white dwarf, what is the temperature it reaches by decelerating?

I have seen temperature as high as 100,000K attributed to newly exposed white dwarfs. The intense ultraviolet radiation makes the nebula shine and keeps it expanding. The nebula typically fades away in a few tens of thousands of years.

What happens to the nebula in the later stages of planetary nebula? How does a late-stage planetary nebula look like, compared to a young planetary nebula?

Hornblower

2017-Feb-09, 12:29 PM

Does the infalling matter cause any observable effects?

If a parcel of gas falls onto a white dwarf, what is the temperature it reaches by decelerating?To repeat, I do not have the mathematical knowhow or the detailed information on the conditions to give any quantitative answer.

What happens to the nebula in the later stages of planetary nebula? How does a late-stage planetary nebula look like, compared to a young planetary nebula?

I would expect it to become fainter and more rarefied.

Noclevername

2017-Feb-09, 01:50 PM

What happens to the nebula in the later stages of planetary nebula? How does a late-stage planetary nebula look like, compared to a young planetary nebula?

The further it gets from the primary's gravity well, the less gravity it has to hold it there.

Reality Check

2017-Feb-09, 10:55 PM

Properties of Degenerated Fermi-Gas in Astrophysics (PDF) (http://hep.uchicago.edu/~rosner/p342/projs/chen.pdf) shows why astronomers can use the approximations of treating white dwarfs as completely degenerate and neglecting ion pressure.

Although the temperature is finite, we can still consider the electrons in white dwarf or even higher evolution stages of stars as complete degeneracy. Consider a white dwarf with density 109 g=cm3 at T = 107 K, the corresponding alpha = ~3 * 105. However, because its still in nonrelativistic region, E/ kT << alpha and the number density distribution of electrons is therefore much like that of complete degeneracy. If we assume complete degeneracy in the stars, the pressure becomes independent of temperature.

...

In this case, the degeneracy pressure is ten times larger than the pressure from ions. In the high evolution stages of stars such as white dwarf, the degeneracy is much higher (alpha is much more negative), and the ratio between degeneracy pressure and ion pressure is much larger. In conclusion, degeneracy pressure dominates the gas pressure in the high evolution stages of stars.

chornedsnorkack

2017-Feb-10, 05:20 AM

To repeat, I do not have the mathematical knowhow or the detailed information on the conditions to give any quantitative answer.

Sun notoriously has a corona.

The energy mysteriously winding up in corona is alleged to be 1/40 000 the total energy emitted by Sun - 10 times more than moonlight.

Some of it is emitted by corona, some spent to lift solar wind out of Sunīs gravity, and some spent to give kinetic energy to solar wind escaping.

If that whole energy were emitted by a black body surface of Sun, its temperature would be mere +130 degrees.

Yet the temperature of corona is above a million degrees.

Sunīs corona emits x-rays which are not just the blueward Wien law tail of blackbody radiation.

Concerning a white dwarf: if any gas fell onto a white dwarf, whether dropping back from planetary nebula or from ambient interstellar gas, by coming to halt the gas should from simple physical considerations heat up to hundreds of millions of degrees!

While the black body surface of the white dwarf does not seem to reach 200 000 degrees even in the young central stars of planetary nebulae.

So: do white dwarfs possess visible coronae emitting x-rays other than Wien tail of black body surface?

I would expect it to become fainter and more rarefied.

Fainter, by definition. But more rarified?

Do old planetary nebulae fade because they expand and get less dense? Or do they fade because, as the white dwarf cools and emits less ultraviolet, the gas is still there around the white dwarf, but no longer lit as brightly?

Also, while the nebula is active, itīs expanding. Suggesting that its pressure is higher than that of ambient interstellar gas (otherwise the interstellar gas would stop the nebula from expanding).

But are planetary nebulae denser than the ambient interstellar gas? They are hotter. They might have higher pressure but lower density.

If a planetary nebula is a heated and rarefied hole in interstellar gas, rather than a gas cloud, then fading of the white dwarf might cause the pressure to fall as the gas cools - fall below the ambient interstellar gas pressures, causing the ambient interstellar gas to refill the hole, and make the planetary nebula denser as it fades...

Ken G

2017-Feb-12, 05:09 PM

Properties of Degenerated Fermi-Gas in Astrophysics (PDF) (http://hep.uchicago.edu/~rosner/p342/projs/chen.pdf) shows why astronomers can use the approximations of treating white dwarfs as completely degenerate and neglecting ion pressure.

Yes, it's a common simplification to use the zero temperature limit, and neglect ion pressure, though of course that will get the answer wrong if you ask what is the change in pressure when heat is added. Also, note the common misconception being promoted by the clearly contradictory language used. You quoted his statement "degeneracy pressure dominates gas pressure", but if you look at the derivation he uses for degeneracy pressure, right there in the paragraph before section 1.1 he says "In the assumption of perfect gas, the pressure is calculated by..." So right there, we have a pressure being calculated under the assumption that the pressure is the pressure of a perfect gas. In short, it's gas pressure. No one can ever understand the first thing about degeneracy pressure until they understand that it comes from the sum over the momenta of the particles, just as he gets it-- it's gas pressure.

It is insightful to think of any equation of state of a gas as a mapping from the kinetic energy per particle, and the volume per particle (or the density), into the pressure and temperature of the gas. Hence, it has two parts, one that gives us pressure and one that gives us temperature. Do you know which of those parts is exactly the same for degenerate and ideal gases, and which part is different? Answer: the pressure part is exactly the same-- it's gas pressure. Degeneracy only affects the temperature part of that mapping. Ergo, the oft-mistaken meaning of the term "degeneracy pressure" is simply the gas pressure you get from that mapping when the temperature part of that mapping gives zero temperature. This fact justifies the following statement: degeneracy pressure is not a type of pressure, it is a normal pressure when you have a degenerate type of temperature.

Reality Check

2017-Feb-14, 09:54 PM

Ye Also, note the common misconception being promoted by the clearly contradictory language used. ....

The language used does not matter - he calculates that "degeneracy pressure dominates the gas pressure in the high evolution stages of stars". There is no "common misconception" there.

ETA: Maybe you are just complaining that someone whose English seems to be a second language does not write clear enough English? The author is Hsin-Yu Chen and their English reminds me of the English of my Chinese friends. A probable Hsin-Yu Chen (https://astro.uchicago.edu/people/hsin-yu-chen.php) is Taiwanese.

Part 1 Degeneracy of Elections starts with the case of compete degeneracy of an electron gas which of course has a "gas pressure" and mentions the perfect gas assumption before section 1.1.

The paragraph you refer to has a "sum over the momenta of the particles" so the author does "understand the first thing about degeneracy pressure". The pressure equation P is an integration over the momenta p (look at the start of Part I - small p = momentum), velocity v and total number density of electrons n(p).

Part 1 does use a perfect gas approximation which is removed in Part II: Pressure ionization

In previous part, we consider all particle as perfect gas, which means that there is not interaction between particles. However, in the interior of white dwarf, the density is so high that the particle seperation are compressed. The electron energy levels of nearby atoms would overlapped. Because electron is fermi gas, the mutual wave function of overlapping electrons must be antisymmetric. Due to Pauli exclusion principle, the many degenerated energy levels of discrete energy reform a continuous band of energy shared by all atoms, similar to that in a metal. The electrons are then said to be free or ionized, and indeed the wave function of electron can be expressed by that of free electron. This phenomenon is what we called pressure ionization.

Ken G

2017-Feb-15, 06:03 AM

The language used does not matter - he calculates that "degeneracy pressure dominates the gas pressure in the high evolution stages of stars".What you just said is nonsense-- you just said he calculated a sentence. No, people don't calculate sentences, they calculate quantities which they then may choose to interpret via sentences of English. In so doing, they sometimes insert obvious contradictions, as here. Probably it was just a typo-- he meant "dominates the ion pressure" or "dominates the ideal gas pressure", but what he actually said was clearly wrong, for the reason I pointed out already. And it doesn't make the slightest difference to me why he inserted that misconception, if he is not adept at the use of English it is still a misconception. I am not running down his character, I am pointing out his error-- period.

Reality Check

2017-Feb-15, 08:17 PM

What you just said is nonsense-- you just said he calculated a sentence. ...

What

The language used does not matter - he calculates that "degeneracy pressure dominates the gas pressure in the high evolution stages of stars".

means is that calculations were done that support the text. More explicitly:

calculations then

"In this case, the degeneracy pressure is ten times larger than the pressure from ions. In the high evolution stages of stars such as white dwarf, the degeneracy is much higher (alpha is much more negative), and the ratio between degeneracy pressure and ion pressure is much larger. In conclusion, degeneracy pressure dominates the gas pressure in the high evolution stages of stars."

and

"Although the temperature is finite, we can still consider the electrons in white dwarf or even higher evolution stages of stars as complete degeneracy."

then calculations

I said that her calculations are what matter, not her language. I said that her English is probably a second language which may explain her imprecise language in some parts. The language and numbers make it clear that for white dwarfs:

Complete degeneracy is a good approximation.

Ion pressure can be neglected because election pressure dominates.

What is the "common misconception", Ken G? Describe it coherently, cite and quote other sources for it to make the misconception "common".

Ken G

2017-Feb-15, 11:06 PM

What

means is that calculations were done that support the text. And that's what I said is wrong-- her calculation is contradicted by her text. I said exactly why.

I said that her calculations are what matter, not her language.And of course that's just complete nonsense. If only calculations mattered, there would be no need for the text. But there is a need for the text, to interpret the calculations, and hence there is a need for the text to reflect a correct interpretation. It did not. The calculation is elementary and can be found in a host of places, what is interesting and important is figuring out what it means.

I said that her English is probably a second language which may explain her imprecise language in some parts. And I said I don't care why the text is wrong, only that it is wrong. That's how text works-- people read it.

Ion pressure can be neglected because election pressure dominates. Only that's not what was said in the text that I said was wrong, now is it? I mean, it's right there in black and white.

What is the "common misconception", Ken G? That degeneracy pressure is not gas pressure. I'm just repeating now, but she derived the degeneracy pressure by starting with the common expression for gas pressure. Then she said the degeneracy pressure she derived, using the expression for gas pressure, "dominates gas pressure", and I quote. So that's the obvious contradiction, which plays quite clearly into the common misconception I am telling you about. The misconception is repeated in so many places, that it has become extremely important to point it out as often as necessary to defeat it. Because text does matter.

Reality Check

2017-Feb-16, 02:02 AM

And that's what I said is wrong-- her calculation is contradicted by her text.

Read page 4.

How does a calculation that degeneracy pressure is ten times larger than the pressure from ions contradict the text "degeneracy pressure is ten times larger than the pressure from ions"?

How does a calculation that the number density distribution of electrons is close to complete degeneracy contradict the text "the number density distribution of electrons is therefore much like that of complete degeneracy"?

N.B. "From numerical calculation, the gas becomes degenerated after alpha -< 2" and she calculates alpha about 3 * 105.

You wrote (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2391313#post2391313)

No one can ever understand the first thing about degeneracy pressure until they understand that it comes from the sum over the momenta of the particles, just as he gets it-- it's gas pressure.

My emphasis added. My reply seems to have been ignored:

The paragraph you refer to has a "sum over the momenta of the particles" so the author does "understand the first thing about degeneracy pressure". The pressure equation P is an integration over the momenta p (look at the start of Part I - small p = momentum), velocity v and total number density of electrons n(p).

You know that integration is a summation and thus she derived the degeneracy pressure using your criteria for the degeneracy pressure :eek:!

It seems all you are complaining about is the word "gas" used in the context of an completely degenerate electron gas.

Ken G

2017-Feb-16, 03:15 PM

Read page 4.

[LIST]

How does a calculation that degeneracy pressure is ten times larger than the pressure from ions contradict the text "degeneracy pressure is ten times larger than the pressure from ions"?Hmm, now, is that the text I said was contradicted, or did I specifically quote a different text in my very last post? It is silly logic to think that a claim that an argument is internally inconsistent requires that every statement it makes must be inconsistent with every other statement, such that it would make any sense for you to counter that by citing random statements that are not contradictory. Surely this is obvious! The article makes contradictory statements. I told you what they were. I did not say all the statements are contradictory, including the random few you chose to list above. However, the contradictory statements do promote a common misconception, and one that you held yourself before I explained why it is wrong (remember when you claimed the degeneracy pressure was not a sum over the momentum of the particles? That was the misconception I was able to correct for you, and which this author does know correctly). I said what the misconception was, quite clearly: it is the misconception that degeneracy pressure is not gas pressure, and it is perhaps the single most widespread of any misconception I have ever found among people who know enough to see why it is a misconception. So if you have any comments on anything I actually did say, I'd be happy to repeat the logic, but you will need to make relevant points.

Reality Check

2017-Feb-19, 09:39 PM

Hmm, now, is that the text I said was contradicted, ...

Still no indication what the actual "common misconception" is or any evidence that it is common, Ken G.

What you stated was "she derived the degeneracy pressure by starting with the common expression for gas pressure" but that "common expression" is the sum over momenta that you, Ken G, state is needed for degeneracy pressure.

The pressure equation P is an integration over the momenta p (look at the start of Part I - small p = momentum), velocity v and total number density of electrons n(p).

So let us rewind:

Ken G, quote the text from the PDF that contains that "common misconception" so we can be sure where that misconception is.

Ken G, if that is the paragraph starting "In the assumption of perfect gas, the pressure is calculated by" in the Complete Degeneracy section, why is her "sum over momenta" equation not the pressure of a completely degenerate gas of electrons?

You know that a perfect gas (https://en.wikipedia.org/wiki/Perfect_gas) is not an ideal gas and she does not use the ideal gas law so that should not be an issue.

Cite me claiming that "degeneracy pressure was not a sum over the momentum of the particles", Ken G.

I have cited Properties of Degenerated Fermi-Gas in Astrophysics (PDF) (http://hep.uchicago.edu/~rosner/p342/projs/chen.pdf) in other threads. In this thread, I have quoted and implicitly agreed with the PDF:

2017-Feb-14: The paragraph you refer to has a "sum over the momenta of the particles" so the author does "understand the first thing about degeneracy pressure".? (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2391804#post2391804)

2017-Feb-16: You know that integration is a summation and thus she derived the degeneracy pressure using your criteria for the degeneracy pressure! (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2392015#post2392015)

Ken G

2017-Feb-21, 02:43 AM

Still no indication what the actual "common misconception" is or any evidence that it is common, Ken G.You are quite wrong on both counts, of course.

1) The common misconception is that degeneracy pressure is not gas pressure, as I've repeated ad nauseum.

2) It is indeed very common, you thought it, the author you just cited said it (even though she used an equation that refutes it, so we can't tell how deep the misconception is for her or if it is just her misuse of the language), and many other people on this forum have demonstrated that they did not understand degeneracy pressure is gas pressure. It is just not ideal gas pressure, because it does not obey the ideal gas law, but that's more about temperature anyway.

What you stated was "she derived the degeneracy pressure by starting with the common expression for gas pressure" but that "common expression" is the sum over momenta that you, Ken G, state is needed for degeneracy pressure.Your argument is getting even sillier. I did not say that expression is needed, she did: because she used it. Obviously.

So let us rewind:

Ken G, quote the text from the PDF that contains that "common misconception" so we can be sure where that misconception is.You want me to cite if for the third time? OK, here's the misconception from the text: "degeneracy pressure dominates gas pressure." That's the misconception that degeneracy pressure is not gas pressure.

Ken G, if that is the paragraph starting "In the assumption of perfect gas, the pressure is calculated by" in the Complete Degeneracy section, why is her "sum over momenta" equation not the pressure of a completely degenerate gas of electrons?Oh goodness, you have not understood anything I've said this whole thread! Her sum over momenta is the pressure of a completely degenerate gas, that's the whole point-- it's also the common expression for gas pressure. I see that this is hopeless, you just don't understand.

Reality Check

2017-Feb-21, 03:44 AM

1) The common misconception is that degeneracy pressure is not gas pressure, as I've repeated ad nauseum.

2) It is indeed very common, ...

That is what you imagine is a common misconception without any evidence that it is a misconception or common :eek:!

The context you missed out is:

For example, I consider gas composed of C12 and density of 2.5*105 g/cm3 at T = 108 K. The corresponding alpha ~ -12 and Pe~ 1022dyne/cm2. On the other hand, the pressure from ions is about 1021 dyne/cm2. In this case, the degeneracy pressure is ten times larger than the pressure from ions. In the high evolution stages of stars such as white dwarf, the degeneracy is much higher (alpha is much more negative), and the ratio between degeneracy pressure and ion pressure is much larger. In conclusion, degeneracy pressure dominates the gas pressure in the high evolution stages of stars.

Where does she state that "degeneracy pressure is not gas pressure", Ken G?

That is a calculation that electron degeneracy pressure dominates the gas pressure of a gas of C12. Followed by the observation that for white dwarfs the degeneracy pressure (electrons) domination is even higher over the gas (electrons + ions) pressure.

Cite me claiming that "degeneracy pressure was not a sum over the momentum of the particles", Ken G (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2392622#post2392622).

Ken G

2017-Feb-21, 04:36 PM

That is what you imagine is a common misconception without any evidence that it is a misconception or common :eek:!

My evidence is that the author falls under it, by their quote, and so do you, by your inability to recognize it, even now after I have explained it. But I also happen to know from personal experience that very few people understand that degeneracy pressure is gas pressure, and one good way to calculate gas pressure is by summing over the momenta of the gas particles.

The context you missed out is:Nah, I didn't miss it, it is merely irrelevant. There's a difference, you know. I am well aware that the author has calculated the ion pressure, and I am well aware that it is much less than the degeneracy pressure, for fairly obvious reasons for anyone who understands degeneracy. My point is that they said the degeneracy pressure dominates the gas pressure, not the other statement you keep harping on, that it dominates the ion pressure. You can see the difference in those two statements, can you not? (One of the words is different, I italicized the difference just in case.)

Where does she state that "degeneracy pressure is not gas pressure", Ken G?

What I have quoted so many times now is the last line of the second-to-last paragraph in Part I: " In conlusion, degeneracy pressure dominates the gas pressure in the high evolution satges of stars."

That's wrong, and the very fact that you still don't know it's wrong shows how common the misconception is. What's wrong about it is that degeneracy pressure is gas pressure, so it cannot dominate gas pressure. I also pointed out that the degeneracy pressure in that very article was derived by starting from the expression for gas pressure, as a sum over the momenta of the gas particles (which is also how ideal gas pressure is calculated, by the way, but the temperature dependence is very different there because the thermodynamics is so different). But you haven't understood any of this yet, so I'm not sure this is going to make it click.

While we are on the topic of common misconceptions that you don't realize are even wrong, the last sentence in Part I is another prime example:

" If we assume complete degeneracy in the stars, the pressure becomes independent of temperature."

Of course that statement is complete nonsense, but the author is to be forgiven because it can be found in so many places. It is obviously nonsense, however, because if you assume complete degeneracy, it means you have zero temperature. Yes, that's what it means. So the statement is equivalent to saying that at zero temperature, the pressure is independent of temperature. Which makes no sense, of course. It's not even true that the derivative of the pressure with respect to temperature is zero at T=0, because of the ions. But these are just the common misconceptions that you neither think are common, even though you hold them yourself, nor misconceptions, even though I just proved with the most basic logic that the claim makes no sense. But that's just the state of things there, there appears to be no changing it.

That is a calculation that electron degeneracy pressure dominates the gas pressure of a gas of C12. Yes quite, it is a calculation that the gas pressure of the electrons dominates the gas pressure of the ions. Now, is that what I'm objecting to here? (No, that's the irrelevant issue you keep returning to.)

Cite me claiming that "degeneracy pressure was not a sum over the momentum of the particles", Ken G.Your misconception seems to be that the sum over momenta doesn't give you gas pressure. That's precisely the "common misconception" I have been talking about this whole thread. It's so common, you still don't even see it.

Reality Check

2017-Feb-22, 12:15 AM

My evidence is ...

not listed in a mostly irrelevant post. No evidence that the "misconception" exists. No evidence that it is common. Only your repeated opinion that the author committed it and that 1 person did it somehow makes it common.

The paragraph is easy enough to understand which you do up to the last sentence.

For example, I consider gas composed of C12 and density of 2.5*105 g/cm3 at T = 108 K. The corresponding alpha ~ -12 and Pe~ 1022dyne/cm2. On the other hand, the pressure from ions is about 1021 dyne/cm2. In this case, the degeneracy pressure is ten times larger than the pressure from ions. In the high evolution stages of stars such as white dwarf, the degeneracy is much higher (alpha is much more negative), and the ratio between degeneracy pressure and ion pressure is much larger. In conclusion, degeneracy pressure dominates the gas pressure in the high evolution stages of stars.

Her "gas pressure" in the last sentence is obviously the combination of degenerate pressure from electrons and ion pressure from ions.

2017-Feb-19 Ken G: Cite me claiming that "degeneracy pressure was not a sum over the momentum of the particles", Ken G (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2392622#post2392622).

T tends to zero in the simple case of complete degeneracy. That is a limit. It is not setting T = 0. We do not have "zero temperature". That is obvious from the equations above "1. Compete Degeneracy" in the PDF. Set T = 0 and we get division by zero :eek:!

The assumption of complete degeneracy by looking at the limit as T -> 0 is literally textbook physics: Electron degeneracy pressure (https://en.wikipedia.org/wiki/Electron_degeneracy_pressure)

Griffiths (1994). Introduction to Quantum Mechanics. London, UK: Prentice Hall. ISBN 0131244051.Equation 5.46

To make this more explicit:

Ken G: Cite the derivation of electron degeneracy pressure using your suggestion of T = 0 (zero temperature).

Textbook answer - there is none and T is not zero for complete degeneracy.

But you have raised a valid point - "complete" degeneracy is misleading because it makes people think T = 0 rather than T -> 0. On the other hand "very nearly complete" degeneracy is a bit cumbersome.

Ken G

2017-Feb-22, 12:36 AM

No evidence that it is common. Only your repeated opinion that the author committed it and that 1 person did it somehow makes it common.

It is extremely common, I can find you any number of examples on the web. I'll give you a typical quote, from a college lecture at http://teacher.pas.rochester.edu/Ast102/LectureNotes/Lecture11/Lecture11.pdf:

"Stars supported by degeneracy pressure instead of gas pressure would have sizes close to that determined from astronomical observations of Sirius B."

Still, not all places make this mistake, it's not so important to say what fraction do, only that it is a misconception to be on the alert for.

But what is more important is that I now realize I was misinterpreting the author you cited-- that particular one does not actually have the misconception!

Her "gas pressure" in the last sentence is obviously the combination of degenerate pressure from electrons and ion pressure from ions.Yes, you're right, I misinterpreted what she meant by "dominated the gas pressure." I interpreted that as "dominated over the gas pressure", but that's not what she said, nor apparently what she meant. So I now agree with you that I was wrong about her falling under the misconception-- I've just seen it so many places that I was too quick to see it again here.

T tends to zero in the simple case of complete degeneracy. That is a limit. It is not setting T = 0. Sure it's the same, since nothing singular happens in that limit. And this really is a misconception that she falls under too. Here are two facts about degeneracy pressure:

1) Everyone knows that T=0 is always a limit, there's obviously no such thing as true T=0. Complete degeneracy still means T=0, it is an idealized way of treating a gas. Idealizations in science are completely routine. The limit of extreme degeneracy is the limit of extremely low temperature. Hence, it makes no sense to say that a completely degenerate gas has a pressure that does not depend on temperature, as this is precisely the same thing as saying that a gas at zero temperature has a pressure that does not depend on its temperature. That's just as true in a limit as it is if T=0 was actually possible, the assertion simply makes no sense. The correct assertion is that the pressure is not zero even when the temperature is zero.

2) Even if one means dP/dT when one talks about the way the pressure "depends on temperature", that derivative is not zero at T=0, as I said, because of the ions. So it is also just plain wrong to claim that a gas that combines highly degenerate electrons with ions that are all being treated in the low temperature limit have dP/dT = 0. What is correct is that, due to the ions, a gas of carbon 12 mixed with highly degenerate electrons has a dP/dT that is precisely the same as the ion ideal gas dP/dT, and that's not zero, it's nik, where ni is the number density of ions. Since the number density of ions is 1/7 the total number density, we easily see that dP/dT is 1/7 what it would be if the whole business was an ideal gas. That's not at all zero, nor can it be said that P "does not depend on" T there, it's just plain wrong, either way you interpret the statement.

We do not have "zero temperature". That is obvious from the equations above "1. Compete Degeneracy" in the PDF. Set T = 0 and we get division by zero :eek:!If you think that, then you are looking at the wrong equations. If you think "complete degeneracy" does not imply "zero temperature", you are wrong. If you look at the correct equations, you will find no divisions by zero anywhere, T=0 is a well behaved limit. Of course it has to be, because stars really do approach that limit if you wait long enough. What, do you think the pressure in stars just blows up if you wait long enough? Of course not, you are just not looking at the right equations.

To make this more explicit:

Ken G: Cite the derivation of electron degeneracy pressure using your suggestion of T = 0 (zero temperature).

Textbook answer - there is none and T is not zero for complete degeneracy.

Totally wrong. I'm sure it's not the textbooks that have messed this up, it must be how you are reading them.

But you have raised a valid point - "complete" degeneracy is misleading because it makes people think T = 0 rather than T -> 0. On the other hand "very nearly complete" degeneracy is a bit cumbersome.No, there is no difference at all between T=0 and T-->0, the former always means the latter as there is never really any such thing as the former (that's called' "Nernst's theorem.") So, every single time you ever saw T=0 in your whole life, in absolutely any context whatsoever, it always meant T-->0.

Reality Check

2017-Feb-22, 01:16 AM

But what is more important is that I realize I was misinterpreting this particular author-- this particular one does not actually have the misconception!

Good :D!

Using a limit such as "T -> 0" can never be a "misconception".

Here are two facts about degeneracy pressure: ...

A mistake:

1) "Everyone knows that T=0 is always a limit, there's obviously no such thing as true T=0." Compete degeneracy defined is the limit as T goes to 0. Thus complete degeneracy means T is never 0.

2) Relies on the mistake that T = 0.

The equations are right there in the PDF so I am looking at the correct equations. Or you claiming that a working astrophysicist is wrong about textbook quantum mechanics? In that case:

Ken G: Please cite the textbook equation for "density of electrons is described by Fermi-Dirac statistics" (the n(p)dp equation).

You imply that the answer to:

Ken G: Cite the derivation of electron degeneracy pressure using your suggestion of T = 0 (zero temperature).

is that the textbooks plug in T = 0. Please confirm or deny that.

I know that that the word "complete" in the phrase "complete degeneracy" implies T = 0. I also know the definition of complete degeneracy is the limit as T tends to zero and T never reaches zero.

Nernst's theorem (https://en.wikipedia.org/wiki/Third_law_of_thermodynamics) is a statement of the third law of thermodynamics

In 1912 Nernst stated the law thus: "It is impossible for any procedure to lead to the isotherm T = 0 in a finite number of steps."[4]

So not only is T never zero for complete degeneracy, it is impossible to reduce T to zero in a finite number of steps.

Ken G

2017-Feb-22, 01:34 AM

Good :D!

Yes, we have established that the author was not wrong about that.

Compete degeneracy defined is the limit as T goes to 0. Thus complete degeneracy means T is never 0.Nonsense. Every time you have ever seen T=0 referred to in any context at any time in your life, it always meant T goes to 0. There is no difference in those two things, if you think otherwise, you are mistaken.

The equations are right there in the PDF so I am looking at the correct equations.The equations refer to a function that is smooth at T=0. That means they can be written in a form that is not singular if T is set to zero. Your problem is that you think there is a difference between the value of a smooth function at T=0, and its limit as T goes to zero. Your misconception violates basic mathematics.

I know that that the word "complete" in the phrase "complete degeneracy" implies T = 0. I also know the definition of complete degeneracy is the limit as T tends to zero and T never reaches zero.Excellent, then let us not hear any more nonsense about complete degeneracy not meaning T=0. It does. This also means that Nernst's theorem tells you that complete degeneracy is just as impossible as T=0. Do you understand that?

So not only is T never zero for complete degeneracy, it is impossible to reduce T to zero in a finite number of steps.You have drawn the wrong conclusion. You think this means complete degeneracy must not be T=0 because you think complete degeneracy can be reached in a finite number of steps. You are wrong, it cannot. The two are exactly the same in every way.

Reality Check

2017-Feb-22, 04:10 AM

Nonsense. Every time you have ever seen T=0 referred ...

Followed by irrelevant stuff because this is not about setting T = 0.

Is there any point in trying to teach you about the definition of completes degeneracy where T tends to zero and never gets there? I will try one last time:

Read Properties of Degenerated Fermi-Gas in Astrophysics (PDF) (http://hep.uchicago.edu/~rosner/p342/projs/chen.pdf) again. Put T = 0 in the first equations. See the division by zero :eek:!

When we apply the limit as T tends to 0, there is no division by zero because we never get to T = 0.

We have another unsupported assertion so just for curiosity:

Ken G: Show that the two equations are smooth (at least their first derivative is continuous) at T = 0.

One answer is that you do not know where they are smooth or not at T = 0. Another would be citing the literature on these functions. Another would be plugging them into a math program. Or actually do the differentiation.

A quick glance suggests that differentiation will not get rid of the "exp(alpha + E/(kT))" part and division by 0 will occur but I am prepared to be wrong.

But you also stated these are not the correct equations so:

Ken G: Please cite the textbook equation for "density of electrons is described by Fermi-Dirac statistics" (the n(p)dp equation). (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2393004#post2393004)

We seem to be back to an assertion of putting T = 0 so:

Ken G: Cite the derivation of electron degeneracy pressure using your suggestion of T = 0 (zero temperature). (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2392988#post2392988)

2017-Feb-19 Ken G: Cite me claiming that "degeneracy pressure was not a sum over the momentum of the particles" (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2392622#post2392622).

Ken G

2017-Feb-22, 03:27 PM

Followed by irrelevant stuff because this is not about setting T = 0.You are the one who thinks there is something important about setting T=0, not me. I see no distinction between setting T[ to zero and taking the limit as T approaches zero, they always mean the same thing to a physicist.

Is there any point in trying to teach you about the definition of completes degeneracy where T tends to zero and never gets there?Only if you want to teach me a bunch of false distinctions. It's nonsense. You have zero evidence that "complete degneracy" means anything different from "zero temperature", because for fermions, they are the precise same thing. I don't suppose you are going to understand that, because you asking me silly things like:

: Show that the two equations are smooth (at least their first derivative is continuous) at T = 0.What on Earth are you talking about? There is only one equation, here, not two, it is the pressure as a function of temperature for a gas of ionized carbon. And if you think it is not completely smooth as T goes to zero, I have to wonder if you understand anything about what you are talking about here. That function is smooth as silk as T goes to zero, just plot it. Of course it would be ridiculous to think anything else, as the star's T really does approach zero with time, so how on Earth could its pressure do something unsmooth?

Reality Check

2017-Feb-22, 10:45 PM

You are the one who thinks there is something important about setting T=0, not me

The fact is that we cannot put T = 0 in the equations because that is division by zero. That is extremely important :eek:! What is becoming more obvious is that you seem not not know what the equations are, despite claiming that they are smooth.

I wrote That is obvious from the equations above "1. Compete Degeneracy" in the PDF. Set T = 0 and we get division by zero! (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2392988#post2392988)

"1 Complete degeneracy We first consider a simple case which is T -> 0." is on page 1 of the PDF.

The 2 equation are above that section.

The 2 equation are the second and third on that page.

The 2 equations are indented.

The first equation starts with "ne(p)dp". The second equation starts with "P(p)".

The 2 equations are the ones that are undefined when setting T = 0. As any physicist knows that means that there is a difference between setting T = 0 and letting T tend to 0. In the first case the result is undefined. In the second case we can do valid math and get actual physics out of the equations. This is done on the next page where the first of the 2 equations is used to derive the total number density of electrons for complete degeneracy.

The total number density of electrons for complete degeneracy is used in section "1.1. Nonrelativistic complete degeneracy" to derive nonrelativistic pressure.

The total number density of electrons for complete degeneracy is used in section "1.2. Relativistic complete degeneracy" to derive relativistic pressure.

2017-Feb-19 Ken G: Cite me claiming that "degeneracy pressure was not a sum over the momentum of the particles" (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2392622#post2392622).

These questions may be moot but just in case:

Ken G: Show that the two equations are smooth (at least their first derivative is continuous) at T = 0. (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2393036#post2393036)

Ken G: Please cite the textbook equation for "density of electrons is described by Fermi-Dirac statistics" (the n(p)dp equation). (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2393004#post2393004)

Ken G: Cite the derivation of electron degeneracy pressure using your suggestion of T = 0 (zero temperature). (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2392988#post2392988)

Ken G

2017-Feb-23, 12:02 AM

The fact is that we cannot put T = 0 in the equations because that is division by zero.We deal with this all the time in physics, we have functions like e-E/kT. No physicist has any trouble at all telling that that function is zero at T=0, it's trivial. All the functions you are talking about are similar, it's a complete non-issue. But I notice you never answered my question, so I will ask it again:

You seem to realize, since I quoted Nernst's theorem, that it is impossible to actually reach T=0. Do you understand, or not, that it is equally impossible to reach complete degeneracy?

Second question: If you do realize that, then why do you think complete degeneracy is any different from T=0? I see no basis at all to support your claim, to me they are quite obviously the same thing.

Reality Check

2017-Feb-23, 12:37 AM

We deal with this all the time in physics, we have functions like e-E/kT. No physicist has any trouble at all telling that that function is zero at T=0, it's trivial.

All physicists know basic mathematics, e.g. that dividing by zero is not valid for ordinary mathematics systems: Division by zero (https://en.wikipedia.org/wiki/Division_by_zero). Thus "at T=0" is wrong.

e-E/kT at T = 0 is undefined.

e-E/kT tends to 0 as T tends to zero which is the condition used in Properties of Degenerated Fermi-Gas in Astrophysics (PDF) (http://hep.uchicago.edu/~rosner/p342/projs/chen.pdf).

Thus we "deal with this all the time in physics" by not setting T = 0 but by looking at the limit as T goes toward 0.

2017-Feb-19 Ken G: Cite me claiming that "degeneracy pressure was not a sum over the momentum of the particles" (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2392622#post2392622).

These questions may be moot but just in case:

Ken G: Show that the two equations are smooth (at least their first derivative is continuous) at T = 0. (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2393036#post2393036)

Ken G: Please cite the textbook equation for "density of electrons is described by Fermi-Dirac statistics" (the n(p)dp equation). (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2393004#post2393004)

I understand that your questions are a strawman argument because complete degeneracy is not degeneracy at T = 0: "1 Complete degeneracy We first consider a simple case which is T -> 0." is on page 1 of the PDF. At no, point during that derivation of compete degeneracy pressure is T =0. If the author and I am wrong then show us by answering:

2017-Feb-22 Ken G: Cite the derivation of electron degeneracy pressure using your suggestion of T = 0 (zero temperature). (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2392988#post2392988)

Ken G

2017-Feb-23, 02:22 AM

e-E/kT at T = 0 is undefined.I'm afraid you are the only person who thinks that, because the function, as used in physics, has a very obvious value at T=0, even though if you don't know how that function is used in physics it might just seem like a blind function that you plug T=0 into and don't know whether it corresponds to the positive limit or negative limit. Since in physics, this is the Boltzmann factor, and it's obvious that the Boltzmann factor should be zero at T=0, we have no problem at all with defining that value. But all this is utterly irrelevant, and has nothing to do with degeneracy. Complete degeneracy is just as easy a situation to understand as T=0, as they are the same. But have you answered my questions yet? No. I wonder why not?

These questions may be moot but just in case:

Ken G: Show that the two equations are smooth (at least their first derivative is continuous) at T = 0. (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2393036#post2393036)It's perfectly obvious, do you need me to show you elementary mathematics? Just take the derivative with respect to T, and see what happens as T approaches zero from the physically relevant side, the positive side. That's what is meant by complete degeneracy, it is precisely what is meant by complete degeneracy, since these are all smooth functions there. Do the calculus, I already know the answer. Complete degeneracy is what you get when you take the limit as T goes to 0 from the positive side, which is precisely what is meant by the T=0 value of the pressure function. Why you even want to make that distinction is a complete red herring, I have no idea what you think it gains you. I know exactly what the pressure is for complete degeneracy, and I know exactly why Nernst's theorem tells us it can never be achieved. You seem to think that complete degeneracy can be achieved, but T=0 cannot, so they are somehow different. You are just completely wrong, I don't know what else to tell you.

Unfortunately, your error there is also completely irrelevant to what I'm saying. This what I'm saying:

The author is wrong that the pressure of a carbon/electron gas is independent of temperature when the electrons are completely degenerate. This claim is wrong for two reasons:

1) when the electrons are completely degenerate, this is specifying that T=0, so it makes no sense to say a function is independent of its argument when its argument is zero.

2) it is not even correct that dP/dT = 0 at T=0, for the reasons I said. So the claim is just plain wrong, and it's just as wrong if T=0 means you have to take a limit as T gets small from the positive side.

I understand that your questions are a strawman argument because complete degeneracy is not degeneracy at T = 0:Yes, it is. But there's a more concrete question here: do you think complete degeneracy can be achieved in a finite number of steps, or don't you? Simple question, asked three times now. But ducking this over and over doesn't even matter, because your entire objection there has no merit in the first place.

Roger E. Moore

2017-Feb-23, 02:05 PM

Astronomy discussed with vigor and passion. Much approved.

Reality Check

2017-Feb-23, 08:28 PM

A rather incoherent post seemingly repeating that division by zero has a value does not add anything.

I can show that division by zero has no meaning: Division by zero (https://en.wikipedia.org/wiki/Division_by_zero). Any function of a meaningless value has no meaning. Thus:

e-E/kT at T = 0 is meaningless (undefined).

e-E/kT tends to 0 as T tends to zero.

From you we have unsupported assertions. I will make this even more clear for you in simple steps. Let f(T) = -E/(k*T). Set T = 0 as you demand.

2017-Feb-19 Ken G:

What is the value of k*T when T = 0?

What is the value of f(T) when T = 0?

What is the value of exp(f(T) when T = 0?

I am citing a physicist who does not divide by zero to get degeneracy pressure ( Properties of Degenerated Fermi-Gas in Astrophysics (PDF) (http://hep.uchicago.edu/~rosner/p342/projs/chen.pdf).). Can you back up your "physicists" assertion with evidence?

2017-Feb-19 Ken G: Cite physicists or even physics textbooks where division by zero is done to get a numeric value, e.g. f(T) = -E/(k*T) at T = 0. N.B. This is not T tends to zero.

2017-Feb-19 Ken G: Cite me claiming that "degeneracy pressure was not a sum over the momentum of the particles". (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2392622#post2392622)2017-Feb-22 Ken G: Cite the derivation of electron degeneracy pressure using your suggestion of T = 0 (zero temperature). (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2392988#post2392988)

2017-Feb-23 Ken G: Show that the two equations are smooth (at least their first derivative is continuous) at T = 0. (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2393036#post2393036)

2017-Feb-23 Ken G: Please cite the textbook equation for "density of electrons is described by Fermi-Dirac statistics" (the n(p)dp equation). (https://forum.cosmoquest.org/showthread.php?163945-White-Dwarf-Stars&p=2393004#post2393004)

2017-Feb-19 Ken G: Your questions are a strawman argumentbecause complete degeneracy is not calculated as degeneracy at T = 0.

I will do what you do not and cite a physicist stating that complete degeneracy is degeneracy as T tends to zero: Properties of Degenerated Fermi-Gas in Astrophysics (PDF) (http://hep.uchicago.edu/~rosner/p342/projs/chen.pdf).

2017-Feb-19 Ken G: Another strawman argument - there is no finite number of steps in the limit of T -> 0.

Limit (https://en.wikipedia.org/wiki/Limit_(mathematics))

In mathematics, a limit is the value that a function or sequence "approaches" as the input or index approaches some value.[1] Limits are essential to calculus (and mathematical analysis in general) and are used to define continuity, derivatives, and integrals.

It is an infinite "number" of steps. T tends to zero is that we get ever and ever closer to zero without reaching zero. An analogy is a converging infinite series (shades of my thesis :eek:!) which can have a defined sum (e.g. zero) but has infinite "steps".

Ken G

2017-Feb-23, 09:53 PM

A rather incoherent post seemingly repeating that division by zero has a value does not add anything.Only if you fail to read essentially any of it. Again: division by zero has nothing whatever to do with what I'm talking about, you are the only having an issue with it. I have no problem with it at all, I understand completely what zero temperature means, and I never have to divide anything by zero to achieve that understanding. I do it by taking the limit as T gets small from the positive side, that's also how Nernst theorem does it, you can look that up.

Now, complete degeneracy is exactly the same thing as zero temperature for fermions. If you think it isn't, give one single difference. If you cannot, you are just saying nonsense.

Worse, your nonsense is irrelevant to everything I'm saying. To repeat, this is what I'm saying:

1) It makes no sense to say that the pressure of a gas as temperature goes to zero does not depend on temperature. Do you understand that, or don't you?

2) It is even wrong if you think that phrase means that dP/dT = 0 at T = 0 (or even, goes to zero as T goes to zero, which is the exact same thing but you might somehow prefer it), because dP/dT doesn't go to zero as T goes to zero, for a gas of carbon and electrons.

Now, points #1 and #2 are just plain facts, and have nothing to do with any divisions by zero. Your problem with that is a complete red herring, as I have already pointed out. If you have no evidence to refute any of those points, I suggest you stop going on about it. And, I note you still haven't answered this question:

Can you, or can you not, achieve complete degeneracy in a finite number of steps? If you don't know, just admit it.

Swift

2017-Feb-25, 04:04 PM

A rather incoherent post seemingly repeating that division by zero has a value does not add anything.

Only if you fail to read essentially any of it.

You two are welcome to debate vigorously, but you both need to learn to do it without the snide comments. Infractions for both of you.

As the OP left a long time ago and you two can't play nice, this thread is closed.

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