Talk:White dwarf

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[edit] Are there atoms in a white dwarf or not?

This is a subject of great confusion to me as I can not get the clarification I need from the article.

We know there are different "kinds" of neutron stars: carbon, oxygen, helium, oxygen-neon-magnesium, etc.

The article says that it is supported by electron degeneracy pressure. Okay all makes sense so far.

Then, lower, it talks about that White dwarfs are not composed of atoms, but degenerate plasma. That's how its possible for their density to be so great and diameter is so small.

Huh?

If they are ALREADY degenerate then: 1.) HOW can you say that they are being held up by electron degeneracy pressure? 2.) HOW can you have different kinds of white dwarfs (based on their chemical composition) ??

Please help. Thanks

Latrosicarius 15:44, 20 July 2007 (UTC)

The interiors of white dwarfs contain nuclei, but not atoms. So, a carbon-oxygen white dwarf will contain carbon and oxygen nuclei, a helium white dwarf will contain helium nuclei, etc. However, the nuclei are so close together that normal electron orbitals cannot exist, so no atoms exist. Rather, the electron orbitals merge into a so-called Fermi sea. The existence of electron degeneracy pressure supports the material against collapsing to greater degeneracy, but does not stop it from being degenerate.

Spacepotato 23:05, 1 August 2007 (UTC)

Hello and thank you very much for responding Spacepotato. This is starting to make more sense to me. However, it is confusing me on one point. As I understand (which might not be correct) electron degeneracy pressure is due to the Pauli Ex Principal which means that WITHIN ATOMIC ORBITALS, electrons may be compressed until there is an electron filling every atomic state. It cannot be compressed anymore without stripping all electrons off of the nucleus. I did not think that the Pauli Ex Principal applies to non-atomic matter where the electrons are ALREADY stripped off... there aren't even any orbitals to compress. So how can one say that it is being held up by electron degeneracy pressure when the electrons already degenerate?

Latrosicarius 20:08, 2 August 2007 (UTC)

The Pauli exclusion principle, which says that no two electrons can be in the same quantum state, is always applicable, whether atomic orbitals exist or not. One way of seeing where electron degeneracy pressure comes from is to look at a small cube of electron-degenerate matter. Each electron in the cube can be thought of as having a quantum state which is delocalized over the cube. By the exclusion principle, the quantum states of different electrons must be distinguished somehow; this is done by placing different electrons in states with different momenta. Now, if we compress the cube, the positions of the electrons become less uncertain, so by the Heisenberg uncertainty principle, the momenta of the electrons must become more uncertain. The average electron momentum therefore increases, so the kinetic energy of the electrons also increases. This means that we must have done work on the degenerate matter when we compressed it. This work came from opposing the electron degeneracy pressure.

Spacepotato 00:00, 3 August 2007 (UTC)

Okay, I think I understand better. So the electrons will continue to support the matter from completely degenerating until electron capture occurs? I wish there were different terminologies for the different levels of degenerate matter. It should be distinguished:

(1) degenerate matter implying all electrons are in the lowest possible orbitals within their atom,

(2) degenerate matter implying non-atomic matter in which electrons still exist, but along-side the nuclei, not in orbitals, and

(3) degenerate matter implying electrons have been integrated into the protons, releasing neutrinos and yielding only neutrons via the inverse beta decay process (neutron star)

Thank you for your clarification, and I hope to hear your impression about distinguishing the term "degenerate matter" based on it's several possible contextual meanings.

Latrosicarius 18:14, 6 August 2007 (UTC)

(1) is the state of e.g. NaCl at 1 atm and 0K. Even though all electrons are in states with the lowest energy possible, this would not be called degenerate matter. (2) is electron-degenerate matter and (3) is neutron-degenerate matter.

Spacepotato 20:42, 6 August 2007 (UTC)

Spacepotato, you are a great person for having this continued discussion. Thank you very much for the proper terminology of (2) and (3). However, I'd have to disagree with (1) because I have heard this definition multiple times before. For instance:

<http://abyss.uoregon.edu/~js/cosmo/lectures/lec27.html>

...degenerate is a physical word to describe the state of matter that has cooled to densities where all the electron shell orbits are filled and in their lowest states.

<http://www.britannica.com/eb/article-9029753/degenerate-gas>

Degenerate gas is a particular configuration, usually reached at high densities, of a gas composed of subatomic particles with half-integral intrinsic angular momentum (spin). Such particles are called fermions, because their microscopic behaviour is regulated by a set of quantum mechanical rules--Fermi-Dirac statistics. These rules state, in particular, that there can be only one fermion occupying each quantum-mechanical state of a system. As particle density is increased, the additional fermions are forced to occupy states of higher and higher energy, because the lower-energy states have all been progressively filled. This process of gradually filling in the higher-energy states increases the pressure of the fermion gas, termed degeneracy pressure. A fermion gas in which all the energy states below a critical value (designated Fermi energy) are filled is called a fully degenerate, or zero-temperature, fermion gas. Such particles as electrons, protons, neutrons, and neutrinos are all fermions and obey Fermi-Dirac statistics. The electron gas in ordinary metals and in the interior of white dwarf stars constitute two examples of a degenerate electron gas.

Latrosicarius 22:26, 6 August 2007 (UTC)

In (1), each electron is bound to its respective nucleus, so the electrons do not form a gas. Since they are not a gas, they are not a degenerate gas. (The first definition you give is so abbreviated that it does not mention electron gas, but if you click on the word "degenerate", you will be taken to <http://abyss.uoregon.edu/~js/glossary/degenerate.html>, which is another copy of the Britannica definition you have given.)

Spacepotato 23:06, 6 August 2007 (UTC)

Spacepotato, yes I know that the Britanica definition was through that page... that's how I found it. I suppose that the Professor at U Oregon (Dr. James Schombert) who wrote that website, wrote seemingly misleading or unclear information for some reason. Viewing his profile <http://abyss.uoregon.edu/~js/>, he seems quite eccentric. I will try to take what he says with a grain of salt.

Latrosicarius 15:39, 9 August 2007 (UTC)

[edit] Confusing. Seemingly contradictory. Need clarification.

[edit] Temp - Cooling over time vs increasing temperature

The article says that White Dwarfs will cool down because they are not fusing anything. Okay. Then it says that if more matter is added (say, from a companion red giant), that it will heat up to the point where it may fuse carbon.

If more mass = more pressure = more heat, then WHY would it cool down over time if it's just sitting there??? It's not losing mass, is it? Or maybe it is losing mass in the form of radiating heat energy? Please clarify this point. I know it's covered lower in the article, but it's confusing until that point is read.

Also, it may be good to go over exactly why it will slow its rotation with age. It's just stated with no rational.

[edit] Mass - Degenerating into neutron star vs carbon detonation

I am trying to figure out what will happen when a White Dwarf surpasses the Chandrasekhar Limit (assuming its not rotating).

The CL means that electrons will no longer be able to support the density and it will collapse into degenerate matter (like a neutron star). Degenerate matter doesn't "fuse".

Yet it also says that once it passes the CL, it will be hot enough to fuse carbon and explode. But it can't fuse carbon, since the carbon will not be carbon any more once it passes the CL--it will be a neutron soup. There is no fusion process to fuse anything at this point.

So please help me clarify this. Thank you. —The preceding unsigned comment was added by 129.139.144.14 (talk • contribs) 21:07, 3 July 2007 (UTC)

A white dwarf cools because it radiates away its heat. It is like an ember left after a fire, which at first glows red but eventually cools to black.
The article doesn't discuss whether or how white dwarf rotation slows with time.
If a white dwarf explodes via carbon fusion, it will do so slightly before it reaches the Chandrasekhar limit. Spacepotato 21:56, 3 July 2007 (UTC)

Spacepotato, thank you for replying. In terms of carbon detonation immediately before degeneration--is this just coincidence, that this pressure just happens to be sufficient to fuse carbon? Or is there a hard linkage between the two, where carbon fusing MUST occur before degeneration? Because, just a reminder--degeneration pressure will vary with different rotational rates (e.g. faster rotating white dwarfs will have a higher threshold Chandrasekhar Limit).

So specifically I'm asking, are there any circumstances where the white dwarf will degenerate rather than undergo carbon fusing --> supernova?

You are correct about it not talking about rotation slowing over time. This is the paragraph I was referencing. I'm sorry I misread non-radiating for "non-rotating". Apologies. A white dwarf will eventually cool and become a non-radiating black dwarf in approximate thermal equilibrium with its surroundings and with the cosmic background radiation. However, no black dwarfs are thought to exist yet.

—The preceding unsigned comment was added by 68.196.79.244 (talk • contribs) 05:53, 4 July 2007 (UTC)

The reason rotating white dwarfs have a higher limiting mass than non-rotating white dwarfs is not because the pressures at which electron-degenerate matter collapses change, but because the global equilibrium of the star changes.
The collapse ("accretion-induced collapse") of an accreting white dwarf to a neutron star may be possible for ONeMg or massive CO white dwarfs. See e.g. Nomoto and Kondo, ApJ 367, pp. L19–L22, Timmes and Woosley, ApJ 396, pp. 649–667, Dessart et al., ApJ 644, pp. 1063–1084. Spacepotato 23:28, 4 July 2007 (UTC)

[edit] Headers

Hmm, this article could use some headers and better structure. Fredrik 16:59, 23 Feb 2004 (UTC)

You think so? I am currently doing a paper of the ending of stars lives, and found this to be actually one of the clearer, more structured articles. As I am just being introduced to these things (first course this semester), clarity and order is key, and I found it quite easy to follow, and appreciate, this article.

[edit] Nova

In the article it states that a white dwarf of a certain mass undergoes a supernova, it is said the mass is usually gained from a companion star. When this occurs to a white dwarf, is it not a nova instead of a supernova? Harley peters 01:10, 4 Feb 2005 (UTC)

No, the article is correct. A core-collapse supernova results (in theory) when the mass of the core exceeds the limit that electron degeneracy can support. It destroys the star. A nova is a lesser event where hydrogen from the companion star ignites (nuclearly) near the surface and blows away some fraction of its mass, but it lives to explode another day. Novae can reccur; supernovae cannot.

[edit] Unanswered Question

I read thru the article and couldn't figure out whether white dwarfs more often contained helium, carbon, or whether other elements were found in them. If there is any information then perhaps the article could be clearer which elements could be found in these starts and how that would depend on their ealier mass.

As an addition, I looked at the Russian version of the article (translated by www.translate.ru) and it seems to have a lot more information, but the translator couldn't get all the words in the article so I didn't totally get what it was talking about. Anyway, if anybody knows russian they could probably bring a lot more information just from reading the russian version.

HI there.

Here are your answers.

White dwarfs are observed in greater number to be what we call "Hydrogen white dwarfs", and relatively fewer are "helium white Dwarfs". The difference in composition between the two is only in the outer atmosphere of the star where the spectral lines are formed. Small changes here result in big changes in what we observe. All white dwarfs contain Carbon in large amounts in their cores (So theory dictates) and also in small amounts in their atmospheres (which we detect by observing carbon absorption lines in the far ultraviolet). Other elements are seen in the white dwarfs atmospheres and the amounts of these roughly follow universal element abundances. Again we see these elements as absorption lines in the FUV.

In fact nearly all elements have been spotted in the spectra from FUSE up to Iron in atomic mass. We havn't seen much (though with some exceptions) heavier than this as nobody has calculated the lab wavelengths for these spectral lines. As heavier and heavier atoms are considered you have to consider more and more electrons (whos change in energy creates the lines in the spectra) and so much beyond Iron and it starts getting messy with 20,000 possible lines in the FUSE wavelength reagion.

Any more questions would be helpful to me, as I am doing a phd on white dwarfs and it is helpful to experience questions

cheers Soloist 20:48, 9 November 2006 (UTC)

I keep seeing references in other articles to different types of white dwarves, namely oxygen/carbon and neon/oxygen types ; but the wiki article doesn't seem to go into these - maybe someone more knowledgable can add these details and how they differ, etc. ? The Yeti 13:10, 28 January 2007 (UTC)

[edit] White dwarfs don't fuse anything

The Formation section claims that white dwarfs avoid collaps by having fuel... They generally need not have any fuel, nor burn any fuel to avoid collaps – they simply are not massive enough to "reach the Chandrasekhar limit", kind of... Rursus 22:52, 8 January 2007 (UTC)

You are wrong when you say White dwarfs don't fuse anything. Our Sun is a white dwarf and it certainly does fuse hydrogen into helium. Otherwise we would not exist. Maybe you confuse white dwarfs with brown dwarfs (sub-stellar objects too small to support hydrogen fusion)? --Friendly Neighbour 06:36, 9 January 2007 (UTC)

Friendly Neighbour! Sol (our sun) is a "yellow dwarf", which is an alias for main sequence star of class G. "White dwarfs" aren't stars in the ordinary sense, especially they aren't class A main sequence stars. "White dwarfs" and "dwarfs" are separated by some 7-8 magnitudes, where the "white dwarfs" are the fainter ones. Astronomers are known to be sloppy with terms and definitions, so "white dwarf" are real dwarfs in comparison to "dwarfs". White dwarfs are upholding inner pressure by electron degeneracy, not fusion - this approximatelly means that the matter is so compressed that it's impossible to compress more electrons into it, while "paradoxically" the more heavy atom nuclei move freely within this electron degenerate matter. I've never seen it by myself, only heard it spoken of. Rursus 18:51, 30 March 2007 (UTC)

He is not wrong (*). White dwarf is the endpoint of a low-mass star, supported by electron degeneracy pressure. Our Sun is a yellow dwarf, or a yellow main sequence star. Sirius B, Procyon B, and Van Maanen's star are white dwarfs. One could call Sirius A also a "white dwarf" as it is a a class A main sequence star, but that is misleading.

(*) Except that white dwarfs occasionally do fuse hydrogen. The event is explosive and is called a nova.--JyriL ^talk 11:26, 9 January 2007 (UTC)

[edit] Weight of a white dwarf

I remember being told back in high school that "one teaspoonful" of a white dwarf would way about the same as our Earth. Is this anywhere where near true or was it just my science teacher exaggerating again?? --Ukdan999 17:55, 10 January 2007 (UTC)

That is a huge exaggeration. But still would weight many tons.--JyriL ^talk 17:59, 10 January 2007 (UTC)

Speaking of odd fantasies, I used to read that if Saturn would be dropped in a huge ocean of water, then it would float. I could never properly imagine such an ocean of water. Rursus 18:59, 30 March 2007 (UTC)

[edit] Helium white dwarf

This article doesn't cover helium white dwarfs. That is, a white dwarf with a helium core, formed when the outer envelope of a red giant is stripped away,[1][2][3] or from a precursor star with less than 0.3 solar masses.[4] Thanks. — RJH (talk) 18:48, 2 March 2007 (UTC)

Fixed, in the lead section. — RJH (talk) 16:21, 5 March 2007 (UTC)

[edit] The maximum mass of ideal white dwarfs

The text reads:

S. Chandrasekhar discovered in 1930 (Astroph. J. 1931, vol. 74, p. 81-82 [5]) in an article called "The maximum mass of ideal white dwarfs" that no white dwarf can be more massive than about 1.4 solar masses.

If you read the paper, however, it says 0.91 solar masses. — RJH (talk) 19:38, 3 March 2007 (UTC)

As written in the text, "Chandrasekhar solved the hydrostatic equation together with the nonrelativistic Fermi gas equation of state,[8] and also treated the case of a relativistic Fermi gas, giving rise to the value of the limit shown above [0.91]." --Gspinoza 16:00, 20 March 2007 (UTC)

Yes, but that would be my point. The paper says 0.91 solar masses. So there must be a later paper with the currently accepted value of 1.44, making the current reference outdated. — RJH (talk) 16:18, 20 March 2007 (UTC)

The limit is a function of the average atomic weight per electron, μ, of the star. In the 1931 Astrophysical Journal paper, μ is taken to be 2.5. In other papers by Chandrasekhar (e.g., Monthly Notices of the Royal Astronomical Society, vol. 95, pp. 207–225) the limit is computed as a function of μ. When this is done the limit is approximately 5.7/μ² solar masses. Taking μ=2 gives the modern value.

The discrepancy in μ arises from changing beliefs about the composition of the stars. In the past it was believed that stars were composed of heavy atoms, for which μ=2.5 would be approximately correct. For a carbon-oxygen white dwarf however one should clearly take μ=2. Spacepotato 20:06, 20 March 2007 (UTC)

Thanks. So it might make sense to clarify this in the history section. — RJH (talk) 20:43, 20 March 2007 (UTC)

The article has been revised. Spacepotato 21:38, 4 May 2007 (UTC)

Intro reads: typical mass of that of the sun contained in a volume about equal to that of the Earth,

section 1 reads: typical white dwarf has around half the mass of the Sun yet is only slightly bigger than the Earth

which is correct? amend correspondingly?

Ben

The article has been revised. Spacepotato 21:38, 4 May 2007 (UTC)

[edit] Diamond star? Really?

The text says: "the carbon interior ...... may have solidified to form an enormous diamond." Really? I think there's an opportunity of serious misunderstanding. Isn't the interior of a white dwarf a degenerate material, completely different from a real diamond? If anything, it must be much(x10) more different from diamond than graphite is.

I think the term "diamond star" is just an insider joke for astronomers who of course know better, but the term would confuse most people. Someone should clarify this point. (I'd like to, but I'm not an astronomy major, so I might screw up.) - Jick 210.216.114.50 06:25, 30 April 2007 (UTC)

The article has been revised. Spacepotato 23:27, 30 April 2007 (UTC)

[edit] Grow Smaller?!

Text says: white dwarfs grow smaller! Extremely funny, but I couldn't continue my reading for this. Isn't there any better way to formulate this? Said: Rursus 18:15, 30 April 2007 (UTC)

[edit] PG 1159 and other Pre-White dwarfs

Any mention on Pre-White dwarfs, white dwarfs with an exposed core. Theses are recent created white dwarf. I am not familar with it. Thanks, CarpD 5/21/07.

[edit] Super-Chandrasekhar-Mass White Dwarfs

Should this article cover the topic of super-Chandrasekhar-mass white dwarfs that are theoretically allowed by rapid rotation? This paper, for example, appears to allow stable white dwarfs with up to 2-4 solar masses. — RJH (talk) 16:15, 24 May 2007 (UTC)

[edit] Extended class of white dwarfs, based on the White dwarf section

Not sure where this get added in.

DAB: a hydrogen- and neutral helium-rich white dwarf
DAO: a hydrogen- and ionized helium-rich white dwarf
DAZ: a hydrogen-rich cool metallic white dwarf
DBZ: a helium-rich cool metallic white dwarf

Thanks, CarpD 6/12/07.

These are examples of the use of the use of letters to indicate secondary features of the spectrum after the indication of the first, primary, feature of the spectrum. For DAB, for example, the primary feature is A and the secondary feature is B. This is already mentioned in White dwarf#Atmosphere and spectra. See Sion, Greenstein, et al., 1983, § III(a). Spacepotato 19:24, 13 June 2007 (UTC)

I've added another example to White dwarf#Atmosphere and spectra. Spacepotato 19:45, 13 June 2007 (UTC)

Cool, thanks, CarpD, 6/15/07.

[edit] GA review process

I'll leave fixes needed to pass GA here - give me a few hours. cheers, Casliber (talk · contribs) 03:50, 21 June 2007 (UTC)

~~In lead - As a class, white dwarfs are fairly common; they .... - first 3 words are redundant.~~

In lead - ...dense, with a typical mass on the order of the Sun's contained in a volume on the order of the Earth's. ungainly segment - "a mass comparable to the Sun in a volume comparable to the Earth" -is a little better but have a play.

The first 2 paras of the lead could do with some massaging. There are alot of intertwining concepts so think shorter sentences. The last two sentences shoudl be incorporated higher into one of the first couple of paras.

The first white dwarf to be discovered was.. - remove the "to be"

type A, i.e., white - i.e. looks funny in text like this, try "namely" ?

The spectral type of 40 Eridani B entered the literature in 1914 when it was given by... -"The spectral type of 40 Eridani B was officially described in 1914 by.."

identified with the theoretically predicted companion - "identified as the predicted companion"

he had determined the spectrum of Sirius B and found that it was similar to that of Sirius - "had found the spectrum of Sirius B to be similar to Sirius."

The high density of white dwarfs was discovered early on. - odd sounding sentence. How about "White dwarves were noted to be extremely dense very soon after their discovery" or something simialr

However, since white dwarfs have an extremely small surface area from which to radiate this heat, they remain hot for a long period of time - choppy - try "White dwarfs will remain hot for a very long period of time as they have an extremely small surface area to radiate heat from" (I don't mind prepositions at the end of clauses -it is an old english construction)

Although white dwarf material is initially plasma, i.e., a fluid composed of nuclei and electrons, in the 1960s, it was theoretically predicted that at a late stage of cooling, it should crystallize, starting at the center of the star - remove teh ie, link plasma and move 1960s cluase to after "predicted" to make this clunky section flow better

However, since a white dwarf spends more of its lifetime at cooler temperatures than at hotter temperatures, we should expect that there are more cool white dwarfs than hot white dwarfs - flip clauses - However, since is an ugly construction

This trend, however, stops - remove however

Eventually, a white dwarf will cool into.. - "A white dwarf will eventually cool and become..."

For example: (1) a white dwarf with.. - put the 2 examples as bulleted text.

The composition of the white dwarf produced, however, will differ.. remove "however"

I'd put the quotes in quote box templates to further delineate them from text

I'm not a fan of See also sections -all should be linked somewhere in the text and removed from here.

Sections 3,4 and 5 of contents could all be subsections of 2, which makes for a nice hierarchy of headings

Alot of the writing is choppy and could do with a copyedit. I think correcting the above examples is fine for GA but would need to be much more throughly massaged fro FA. A great read and certainly comprehensive cheers, Casliber (talk · contribs) 04:24, 21 June 2007 (UTC)

Thanks for your review. I've edited the article in an attempt to address these points, except that:

I haven't flipped the clauses in two of the sentences in the section on cooling, because I believe the logical flow of the text is clearer if the sentences are left in the form α implies β rather than changed to β because α.
I haven't snipped the See also section, as I think it contains a useful short list of links for the browser who wishes to get somewhere else fast. All the links in this section, except for Timeline of white dwarfs, neutron stars, and supernovae, are present elsewhere in the article.

Spacepotato 03:15, 22 June 2007 (UTC)

No probs - they're not deal breakers. Yer through. :) cheers, Casliber (talk · contribs)

[edit] disputed: "Mass-radius relationship and mass limit" is bogus

Zero radius at equilibrium?! which is out of mind and maths. The E=pc approximation only works as a substituand if the elèctròns' velocity is c, not even if it's near celerity. As pp/2m and pc share the same units and dimensions, the former more accurate for lower speeds, the formula should interpolare between both expressions and the radius should never drop out both equations. They can onely find a singularity if they presuppos a singularity, which is why the calculations are bogus; there should still be a relativistic asumptotè, not a dumbassed root.

Also, if the radius drops out both equations, then the body is at equilibrium for all radii and its mass is constant for all radii. It doesn't yield a greattest mass. -lysdexia 07:08, 27 September 2007 (UTC)

In the analysis in question, if you substitute for p²/2m the exact relativistic formula, $\sqrt{m^2c^4+p^2c^2}-mc^2$ , for the kinetic energy, you obtain the equation

$N\left(\sqrt{\frac{\hbar^2 N^{2/3} M^{2/3} c^2}{R^2}+m^2 c^4}-mc^2\right)=\frac{GM}{R},$

which, after some algebra, reduces to

$R=\frac{N^{5/3} \hbar^2}{2 m G M^{1/3}} - \frac{GM}{2 Nc^2 m}.$

For small M the second term will be negligible compared to the first, and we recover the non-relativistic formula. However, as M increases the second term becomes larger and larger until at

$M= N^2 \left(\frac{\hbar c}{G}\right)^{3/2},$

the terms are equal in magnitude and the equation gives R=0. For larger M, we find R<0, which is unphysical, so this value of M is the limiting mass produced by this analysis. As you can see, it is identical with the mass, M_limit, found when we use the E=pc approximation. As explained in the article, this approximation is used not because it gives a correct picture of the curve of R against M (it does not), but, rather, because as the electron momentum becomes large, we expect the E=pc approximation to become more and more accurate, and, therefore, M to approach the value, M_limit, of the mass that results from the approximation. Spacepotato 20:33, 27 September 2007 (UTC)

[edit] still bogus

First of all, a relativistic solution over a classic should not bear a mass limit or a singularity unless a stray (unnatural, wrong) factor comes in. You took out my disputed tag even thouh my objections were true, and the proof didn't work. From the text:

"Δ x will be on the order of the average distance between electrons, which will be approximately n^-1/3, i.e., the reciprocal of the cube root of the number density, n, of electrons per unit volume."

", i.e." is a runon, BTW. In the proof the dimensionality of momentum is 3 yet that of gravity is 1, the former a locally-free manibody state and the latter a radial onebody potential. The radius for each is nonparallel; the latter should be the potential between individual charges and should use something with quantum gravity. Otherwise, momentum in the former should not be equiparted in three dimensions. Elèctric interactions are always stronger than gravital interactions at all sizes and masses; you should see that the proof uses wrong conditions. There cannot be a upper limit for the mass of any elèctric or coloral body due to gravity alone. Black holes (relativistic) also do not exist and are mathematicly impossibil. -lysdexia 68.123.4.253 04:42, 28 September 2007 (UTC)

White dwarfs are not sufficiently dense to require the use of quantum gravity, nor are electric interactions always stronger than gravitational ones. It's not electromagnetism that holds us to the floor. The rest of your objections here are likewise without merit. If you do not believe that there is a limiting mass for white dwarfs, I must point out that the physics literature differs with you. You might wish to read one of Chandrasekhar's papers on the subject or his Nobel Prize lecture. Spacepotato 06:35, 28 September 2007 (UTC)

[edit] wrong method

Quantum gravity has nothing to do with scale unification; the point was that the gravital potential was treated between one body—the star—and a test mass in one dimension, whereas the momentum or Planckian energy was between a bunch of distributed bodies. If this were a two-body problem with variabil mass and charge, one could see that no amount of relativistic correction could make its gravity stronger than its elèctricity. Why is its momentum subject to special relativity but the other side of the equation, its gravital potential, not subject to general relativity and its attendent proper length contraction-expansion? Thas both sides of the equation are counted in unlike dimensions means that mathematic artefacts can creep out of poor conjecture.

Why does it even matter how the star's energy quanta behave? The proof uses the Planckian relation between momentum-energy and length as the elèctric state rather than the elèctric potential. A charge's de Broglie wavelength is immaterial; the equality should use the charges' separation. That proof you ripd off that site and wrote in wrongly equals two different lengths as one radius, and cancels that radius. The problem was whether any charge had the energy at a given lengthscale to overcome the gravital potential at the same lengthscale; however, your proof uses the energy of a charge at the scale of the star's size rather than the size between any two charges. As the latter size shrinks with the cubic root of the former, any charge will be more than springy to meet any weiht. Either both expressions must use respective potentials to get a proper equation of state, or they must use respective quantal wavelengths in the same dimension.

[edit] bad maths

Yes, I should read Chandrasekar's work, but many of his conclusions are without the work. Where does he show how the Emden function is empiricly supported by measurements of stars? Nowhere did he even explicitly use the gravital potential to derive the mass limit; we are supposed to trust thas the Emden function has that work—is it even relativistic? The fun stuff starts at page 147, where a bunch of meaningles mistakes follow. For the classic states he has the same units and dimensions for pressure and energy. He also introduces these, the Fermi conditions, and the mass limit without any explanation. How is anyone supposed to work with these? Later on he labels a plot's pressure axis in dynes.

Here are some of the nonsensical units in his equations:

(27,28) P,E ~ 1/ms⁵ = m(v²a⁵)^/3
(30) P ~ v⁴/m^7/3
(40) P ~ m/st³
(42) P ~ m/s⁴

(34) What in the hell. Shouldn't there be a minus in the argument of the relativistic relations? Get rid of the m and it should recover the familiar contraction formula.
In his other paper, he omits radiation pressure and says β=1, but β is radiation pressure over whole pressure! And for (67'), his "length" isn't even a length, but m²s²/t³.

If you understand his papers, you should know this subject and fill in these gaps and defend your position. However, you don't show any of these but handwave with speciose claims.

[edit] get a clue

Elèctricity holds me to the floor, not gravity; gravity pulls me throuh the floor but elèctricity pushes me over the floor. If one were to switch off gravity, the Earth would blow up and I would be shot into spase because of Its finite Young's modulus. Review the strengths of each fundamental interaction. These come from ratios of specific potentials with gravity; that is, each potential over mass for a fundamental carrier. They tell gravity will never overtake the other interactions at any norm, relativistic or not. Any renorm applied to one interaction must also be applied to gravity. A heavy white dwarf doesn't collapse because gravity wins out as the scientists claim, as gravity is only a enabler, but because at that energy it's more favorabil to bear neutròns. This is because the elèctrocoloral (weak) neutrino carries a stronger interaction than gravity; without the neutrino and the neutròn, gravity could not collapse a white dwarf.

It's likely thas the masses of neutronic stars are near the Chandrasekar limit is coincidental. His expression yields a limit of 1.44 M_sun. However, because there are neutronic stars under this limit is proof thas the limit is born out of wrong maths; on page 156 he says pulsars are consistently near 1.4 M_sun, and neutron star says there are some of 1.35 M_sun. There can be leihter neutronic stars because the other interactions are stronger than gravity and can accomodare potentials at any scale, and cannot collapse into a singularity. There is also a finite probability that the star will skip the transformation and stay elèctric; and by the author's argument there should be a few dwarf holes; that is, there should be singularities near the sun's weiht.

So where are all the data for mass and radius to prove his tabula? -lysdexia 16:44, 29 September 2007 (UTC)

[edit] Fate

Shouldn't there be something about black dwarf in the "Fate" section? It's mentioned elsewhere in the article but I think it should also appear in this section. Also, some content of this section is quite similar to that of black dwarf. Took 02:23, 29 September 2007 (UTC)

I concur with my colleague, Took, about modifying the "Fate" section. Also, isn't the following sentence in the opening incorrect for similar reasons: "White dwarfs are thought to be the final evolutionary state of all stars whose mass is not too high..." Isn't it pretty inevitable that all white dwarfs will change into black dwarfs given enough time? If so, the white dwarf stage is not the star's final evolutionary state, is it? Jacob1207 (talk) 20:02, 1 March 2008 (UTC)

A black dwarf is just a black white dwarf, so the sentence in the opening is correct. I've edited the "Fate" section to address your other point. Spacepotato (talk) 21:02, 1 March 2008 (UTC)

[edit] This article is missing a rather crucial bit of info

There's no mention of who first determined that white dwarfs were in fact dead/dying stars. Serendi ^pod ous 09:01, 24 April 2008 (UTC)

[edit] Graph: A vs. B

I believe A is better for the following reasons:

Higher resolution.
Labels on curves are clearer.
Axes cut off at 0, so omits unphysical region where M < 0 or R < 0.
Grid is conventional (grid in B is very odd.)
Radius is in solar radii to match article text and astronomical literature.

Spacepotato (talk) 19:29, 28 May 2008 (UTC)

I can't really agree to that.

A is too big. One has to scroll around to see something and the graphics are less smooth than B's.
I don't think so. Blue and Black are polytropes, only B tells this. Green is not relativistic, but from non-relativistic to ultra-relativistic, because it uses the general EoS.
Doesn't seem to apply anymore.
Grid in A is useless, grid in B seems to be intended to read off some values - good idea.
Using solar radii for such small values is not useful, because one has no feeling of how much it is. Luckily, Wikipedia is no journal, but free for everybody. --79.220.199.21 (talk) 21:16, 28 May 2008 (UTC)

A is a bad copy of B anyway.—Preceding unsigned comment added by 79.220.199.21 (talk • contribs) 21:16, 28 May 2008

Re your comments:

A looks OK on my browser (you can scroll or not, as you please.)
Yes, the blue and black curves are for polytropes, but this fact is merely incidental. We are interested in these curves because they are the non-relativistic and ultra-relativistic limits of the general solution, and would still be interested in them even if they were not polytropes. So, the labels in A are better. The statement that the green curve is not relativistic is a misunderstanding of the word "relativistic". A correct fully relativistic solution will also apply in the non-relativistic limit.
The grid in B is helpful for some questions (such as finding the radius of a 0.5 solar mass white dwarf in the general case), but not in other ways. For example, if one wishes to find the radius of a 1.0 solar mass white dwarf in the non-relativistic model, the grid line will direct one to the right-hand side of the graph, where there is no scale. Also, the gray horizontal grid line near the bottom is superfluous.
The scale in the figure must match the explanation in the article, which uses solar radii. Otherwise, the article will not be intelligible. Stellar radii are generally given in units of solar radii here on Wikipedia (see e.g. 40 Eridani.) By the way, as you may know, much of the astronomical literature is freely available on line, courtesy of the Astrophysics Data System.

Spacepotato (talk) 23:43, 28 May 2008 (UTC)

There is no use in scrolling, I don't see much in the big version.
Yet they are polytropes and A doesn't mention it.
Obsolete.
Obsolete.

I agree that A is just a copy of B. --Mira7 (talk) 13:03, 29 May 2008 (UTC)

Although the revisions to B are a big improvement (especially the dual scale for radius), in my view the labels on A are still superior, for the reasons I've explained above. Spacepotato (talk) 22:06, 29 May 2008 (UTC)

You are the only person with this opinion here, therefore you must not revert picture B to A when someone changes it. B has better scaling, smoother graphics, better positioned grid lines that are also not too prominent, a much better radius scale that allows the viewer to decide which values to look at. And you challenge this with just one thing that you personally like better (because of bias, not reason). --Mira7 (talk) 23:53, 29 May 2008 (UTC)

I agree with Spacepotato. And Wikipedia isn't a democracy, FYI... :) What goes into an article isn't a matter of "vote". --Etacar11 02:10, 30 May 2008 (UTC)

You can't just say "I agree" and change it without any arguments, overruling other people's opinions. To me, this is lunacy, there is no reasonable argument to use a worse and stolen(!!) picture because of some labels, which are not even false. Picture B has many more positive points to offer, while A has just 1 at the most. I'll get the author to change those labels and then we'll see. --Mira7 (talk) 02:32, 30 May 2008 (UTC)

Fine: I agree with Spacepotato's points and reasoning for using graphic A. Is that enough of an argument for you? Not that I'm saying your edits are in bad faith, cause I don't think they are. I just think more input from other editors is needed for a consensus. Just my opinion. I'll leave it at that. --Etacar11 02:50, 30 May 2008 (UTC)