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Anders Starmark
2008-Oct-09, 03:03 PM
According to the latest episode (#109) of Astronomy Cast, small stars (<0,26 solar masses) will turn all their hydrogen into helium-4 and become a white dwarf made entirely from helium - the world's largest helium balloon.

Now, this helium balloon will cool down over the aeons and turn to liquid, superfluid helium (a Bose-Einstein condensate?) as it cools to around 2K. What (if anything) will this mean for the star's future evolution)?

Hornblower
2008-Oct-10, 12:25 PM
The helium will not be in the same state as in a flask here on Planet Earth. It is gravitationally crushed into a super-dense degenerate state and is supported against further collapse by electron degeneracy pressure. I do not know how Bose-Einstein condensate comes into play under these conditions. I will leave further comment on the details to the experts.

Anders Starmark
2008-Oct-10, 01:39 PM
Ah, but here's the rub: IIUC,the degeneracy pressure only applies to fermions (it's a manifestation of the Pauli Exclusion Principle). If the Helium in these very, very cold stars begin behaving according to Bose-Einstein statistics, wouldn't the degeneracy pressure disappear and the star collapse into a Black Hole?

Spaceman Spiff
2008-Oct-10, 02:17 PM
Ah, but here's the rub: IIUC,the degeneracy pressure only applies to fermions (it's a manifestation of the Pauli Exclusion Principle). If the Helium in these very, very cold stars begin behaving according to Bose-Einstein statistics, wouldn't the degeneracy pressure disappear and the star collapse into a Black Hole?

The fermions in question in white dwarf stars are the electrons, not the nucleons. The density within WD stars is nowhere near high enough for quantum degeneracy effects to come into play for them. But as to how the nucleon equation of state evolves for a helium WD is an interesting question. A lot has been written regarding C/O white dwarf stars and the eventual crystallization as the temperature drops over time.

blueshift
2008-Oct-11, 01:34 AM
Ah, but here's the rub: IIUC,the degeneracy pressure only applies to fermions (it's a manifestation of the Pauli Exclusion Principle). If the Helium in these very, very cold stars begin behaving according to Bose-Einstein statistics, wouldn't the degeneracy pressure disappear and the star collapse into a Black Hole?The temperature of a dead star never reaches low enough to become a BEC.

Nature does not make BECs AFAIK. It is made in labs.
http://www.colorado.edu/physics/2000/bec/how_its_made.html

A star that small won't even be able to fuse nuclei into cabon and oxygen. It will simply fade with time. It could accumulate mass from either a very large binary companion or by entering a density wave in the spiral arm of a galaxy. If enough mass was accumulated the star eventually could shed its envelopes and collapse into a neutron star or a BH.

Degeneracy does not simply disappear. Degenerate electrons simply cannot absorb or emit any light as they fail to rise or fall in the required quantized amounts. But they can vibrate faster and faster until special relativity sets the limit. They can't reach speed c. They can only lose their capabilty to both resist gravity and each others' company, causing falling right into the nuclei, causing fusion all over and throughout the star. Neutrino volume skyrockets and heavy elements are produced. Since degeneracy is all that held the star together, the star comes apart.

Eroica
2008-Oct-11, 01:35 PM
According to the latest episode (#109) of Astronomy Cast, small stars (<0,26 solar masses) will turn all their hydrogen into helium-4 and become a white dwarf made entirely from helium - the world's largest helium balloon.

Now, this helium balloon will cool down over the aeons and turn to liquid, superfluid helium (a Bose-Einstein condensate?) as it cools to around 2K. What (if anything) will this mean for the star's future evolution)?
We might not have to wait for aeons to find out. A larger star in a close interacting binary could run through all its core hydrogen fairly quickly, building up a helium-rich core.

If, during its subsequent red-giant phase, it can lose enough mass to its companion, core helium fusion will never begin and the star will become a helium-rich white dwarf. There should be some of these around today, though whether they have had enough time to cool is another matter.

thorkil2
2008-Oct-11, 02:46 PM
Assuming 1) a white dwarf is distinct from a cooler brown dwarf and 2) it is made entirely of helium, then it is called a white dwarf because of its energy output, which can only be the result of fusion reactions: hence it is fusing helium into heavier elements, and will cool only after the material for those fusion reactions is spent. No liquid helium possible because it doesn't cool as a helium object. (and "balloon"? How did we get balloon out of this?)

Hornblower
2008-Oct-11, 05:19 PM
Assuming 1) a white dwarf is distinct from a cooler brown dwarf and 2) it is made entirely of helium, then it is called a white dwarf because of its energy output, which can only be the result of fusion reactions: hence it is fusing helium into heavier elements, and will cool only after the material for those fusion reactions is spent. No liquid helium possible because it doesn't cool as a helium object. (and "balloon"? How did we get balloon out of this?)
You are mistaken. A white dwarf is in the final stage in which the fusion has ceased. It still is glowing because it takes a very long time, perhaps billions of years, to cool down. Remember, a large hot object cools off more slowly than a smaller one.

cjameshuff
2008-Oct-11, 06:09 PM
You are mistaken. A white dwarf is in the final stage in which the fusion has ceased. It still is glowing because it takes a very long time, perhaps billions of years, to cool down. Remember, a large hot object cools off more slowly than a smaller one.

They are physically small, but extraordinarily dense, composed of degenerate matter, so there's far more thermal mass than there is in the largest brown dwarf. And their small size gives them little surface area for radiation...a red dwarf would radiate more than a "white dwarf" of equivalent mass and temperature, if there are any red dwarfs still around when the universe gets old enough for white dwarfs to cool to that degree.

Also, objects cool more slowly as their temperature drops. After a white dwarf cools to red heat, it will take much longer for it to cool enough that it mainly emits in infrared. IIRC, it would take several hundred thousand times as much time as the universe has been around for it to cool a temperature close to the current background.

JustAFriend
2008-Oct-12, 03:29 PM
To lighten things up:

Lionel Ritchie demonstrates why there are no helium stars HERE. (http://www.youtube.com/watch?v=549P2XRImUc)

Anders Starmark
2008-Oct-14, 06:33 PM
We might not have to wait for aeons to find out. A larger star in a close interacting binary could run through all its core hydrogen fairly quickly, building up a helium-rich core.

If, during its subsequent red-giant phase, it can lose enough mass to its companion, core helium fusion will never begin and the star will become a helium-rich white dwarf. There should be some of these around today, though whether they have had enough time to cool is another matter.

Yes, but it will still take some time for it to reach a temperature of about 2K... (that's when the superfluid phase begins to exist.)

thorkil2
2008-Oct-14, 09:13 PM
You are mistaken. A white dwarf is in the final stage in which the fusion has ceased. It still is glowing because it takes a very long time, perhaps billions of years, to cool down. Remember, a large hot object cools off more slowly than a smaller one.

Sorry, I took a wrong tack on that. My point was that fusion doesn't cease with full conversion to helium.

Hornblower
2008-Oct-14, 10:06 PM
Sorry, I took a wrong tack on that. My point was that fusion doesn't cease with full conversion to helium.

That depends on the mass of the star. Lightweight K and M main-sequence stars may well stop at helium. More massive ones get hot enough in the core to start fusing helium into carbon. I do not know just where the cutoff is.

Spaceman Spiff
2008-Oct-14, 10:41 PM
That depends on the mass of the star. Lightweight K and M main-sequence stars may well stop at helium. More massive ones get hot enough in the core to start fusing helium into carbon. I do not know just where the cutoff is.

It's about 0.5 solar masses. Stars less massive than this are not expected to fuse helium into anything else, and are expected to ultimately end up as a helium white dwarf. Those with initial masses between about 0.25 and near 0.5 solar masses (will eventually) move up the red giant branch before swooping over the H-R diagram in {L, T_eff} onto the white dwarf branch, while those less massive than about 0.25 solar masses never evolve up the RGB and evolve directly onto the white dwarf (http://en.wikipedia.org/wiki/White_dwarf) branch. See, for example, this paper (http://adsabs.harvard.edu/abs/1997ApJ...482..420L).

The question is and remains - if I understood it correctly - what happens to the equation of state (http://en.wikipedia.org/wiki/Equation_of_state) of the helium nuclei in the helium white dwarf as the temperature approaches values near absolute zero (assuming we don't have to wait so long that protons decay first)? The nucleons in carbon/oxygen white dwarf stars crystallize into a sodium metal type lattice long before they become cold. And the electrons continue to support the structure under highly (electron) degenerate (http://en.wikipedia.org/wiki/Degenerate_matter) conditions. But I honestly have never heard of the fate of the helium nuclei of helium white dwarf stars. I do know that you'll have to wait a long, long time. :whistle: