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Non-Spherical Stars

An anonymous reader writes "Now that the large interferometers are coming on line, the stars are no longer dots. Achernar (Alpha Eridani), is a huge ellipsoid whose polar radius (due to fast spinning) is 50% smaller than the equatorial one!"

70 comments

  1. Husky stars by SandSpider · · Score: 2, Funny

    I blame large plates. And a lack of exercise.

    =Brian

    --
    There is nothing so good that someone, somewhere, will not hate it.
    1. Re:Husky stars by Randolpho · · Score: 2, Funny

      No, clearly the problem is sunspots. I suggest an aluminium foil helmet.

      --
      "Times have not become more violent. They have just become more televised."
      -Marilyn Manson
    2. Re:Husky stars by mhesseltine · · Score: 1

      Maybe we should implement a tax on them? We could use this as a model.

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      Overrated / Underrated : Moderation :: Anonymous Coward : Posting
  2. Press release here: by molo · · Score: 5, Informative

    More details at the press release:

    http://www.eso.org/outreach/press-rel/pr-2003/pr-1 4-03.html

    Including more technical drawings.

    -molo

    --
    Using your sig line to advertise for friends is lame.
  3. Energy output by VendingMenace · · Score: 4, Interesting

    50% smaller? Wow, this must be spinning incredibly fast. With so must mass being displaced from where it would be in a sphere, it must effect the pressure inside the star. As such, i wonder how much this effects the fusion within the star. Since fusion is driven by the compressional forces of the suns mass, the effective reduced mass must reduce the energy output of this star. RIght?

    Perhps i don't really kow what i am talking about.

    1. Re:Energy output by BigBir3d · · Score: 2, Informative

      Quick and dirty calculations:

      Earth spins about .25 km/s

      Said star spins 220-300 km/s

      Obviously the star isn't a body of a uniform density. Possibly not conforming to known ideas regarding rotating solid masses in general.

      I wish I had a better physics comprehension in times like this...

    2. Re:Energy output by Fulcrum+of+Evil · · Score: 2, Insightful

      ,i>Possibly not conforming to known ideas regarding rotating solid masses in general.

      Well that makes sense, seeing as how this isn't a solid mass in the first place.

      --
      "We returned the General to El Salvador, or maybe Guatemala, it's difficult to tell from 10,000 feet"
    3. Re:Energy output by L7_ · · Score: 2, Insightful

      The earth is not a 'solid mass' either.

      IIRC, it is composed of a 'liquid' core that is rotating as well (and faster than the rotation of the earth about the sun).

      No planet, to anything other than a zeroth order calculation, follows the I=MR^2 rule of solid spheres for inertial mass.

  4. Cosmic Rugby by follower_of_christ · · Score: 3, Funny

    Hey God!

    I found your rubgy ball!

    1. Re:Cosmic Rugby by Lars+T. · · Score: 1

      Now, a star shaped like a rugby ball (or American Football ball), that would be something to "pose an unprecedented challenge for theoretical astrophysics" ;-)

      --

      Lars T.

      To the guy who modded me down from perfect to terrible Karma - Apple haters still suck

    2. Re:Cosmic Rugby by kamukwam · · Score: 2, Funny

      So what will be the next discovery? The intergalactic frisbee?

    3. Re:Cosmic Rugby by edwinkite · · Score: 1

      A rugby ball a surprise? Not at all. It's been predicted. Check out Chandresekhar's "Ellipsoidal Figures of Equilibrium". The *degree* of rugby-ball-isation might be unexpected, but not the fact.

  5. Amateur Astronomy by Anonymous Coward · · Score: 3, Funny

    Nicole Kidman or Gwyneth Paltrow are the flattest stars that can be seen with the naked eye or possibly binoculars.

  6. I crush you... by FroMan · · Score: 5, Funny

    You know when you take your index finger and thumb and look at something pretty far away. Then you squish them till they touch.

    I think someone was doing that at the end of the telescope.

    --
    Norris/Palin 2012
    Fact: We deserve leaders who can kick your ass and field dress your carcass.
    1. Re:I crush you... by DustMagnet · · Score: 2, Funny
      --
      'SBEMAIL!' is better than a goat!!
  7. Black holes must be flat dishes by Frans+Faase · · Score: 2, Interesting

    If you extend this idea to very fast spinning black holes, you end up with the idea of a spinning disk which "radius" in one direction is
    maybe only a few percent of the radius in the other directions.

    1. Re:Black holes must be flat dishes by PhuCknuT · · Score: 2, Informative

      The arguement fails for the same reason black holes are called black. Once the anything is inside the event horizon, it's impossible for it to stop falling inwards, no matter how fast it is moving. So no matter how fast it spins, it will collapse to a point (in theory).

    2. Re:Black holes must be flat dishes by Atario · · Score: 1

      More to the point, the event horizon of a black hole (which is all you can ever sense about its shape) is a perfect sphere because your overall gravitational attraction to a thing is determined by your distance from its center, not from its surface. Therefore, the escape velocity will always be the same at any spherical shell around the gravitational center. And, of course, all the event horizon is is the spherical shell where the escape velocity equals the speed of light.

      --
      "A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
    3. Re:Black holes must be flat dishes by uberdave · · Score: 1

      I don't know about that one. Imagine two people with ropes, pulling a load. Now, imagine those two people pull at ninety degrees to each other. The load will experience a certain force. Now, imagine the two people pulling with the same force, but this time being only thirty degrees apart. The force on the load is much greater, because the component along the direction of travel is greater, and the force perpendicular is less.

      Now, imagine two points equidistant from the black hole, one above the pole, and one in the equatorial plane. The black hole is going to subtend more of an "arc" from the polar point's point of view than the equatorial point of view. The polar point is going to be pulled from a wide angle. The equatorial point is going to be pulled from a narrower angle. In other words, the equatorial point will have a larger centerward component and a smaller radial component to the gravitational pull than the polar point.

      So, the shape of the isogravitational shell is not going to be spherical, but will be larger about the equator than from pole to pole.

      Thus ends the handwaving lesson for today.

    4. Re:Black holes must be flat dishes by CheshireCatCO · · Score: 2, Informative

      Er, I don't beleive so, no. You're restricting yourself to the Schwartzschild solution, there. Schwartschild assumed that the black hole wasn't spinning and was uncharged. So of course it's spherically symmetrical, there's nothing to break the symmetry.

      Real black holes are likely to be spinning. And then they aren't spherical, as I recall. Also, their horizons start to seperate. Things get a *leeetle* bit weird from there on out.

    5. Re:Black holes must be flat dishes by Atario · · Score: 1
      Now, imagine those two people pull at ninety degrees to each other. The load will experience a certain force. Now, imagine the two people pulling with the same force, but this time being only thirty degrees apart.
      Key phrase: "with the same force." More in a sec.
      Now, imagine two points equidistant from the black hole, one above the pole, and one in the equatorial plane. The black hole is going to subtend more of an "arc" from the polar point's point of view than the equatorial point of view. The polar point is going to be pulled from a wide angle. The equatorial point is going to be pulled from a narrower angle.

      Right. BUT. The sum of the pulls from all the points in 3-space (an integral) is going to be the same either way. This is a basic result of physics and calculus. The points from the view of the polar point will be more spread out laterally, but will be closer together in distance, whereas the points from the view of the equatorial point will be closer together laterally, but more spread out in distance. Remember, the gravitational attraction falls off with the square of the distance, so closer points have a much higher attraction than far ones.

      I wish I could remember the integral and show you, but I simply remember this result from physics class in college: the attraction is purely dependent on the distance to center and mass within the sphere defined by that center and radius. (Which means that the mass above your sphere cancels itself out -- if the star were oblate enough (almost a disk), and you sat on the surface at the center of that disk, you would hardly feel any gravity at all: only the mass contained in the sphere defined by the star's center and your radius from that center (only a small portion of the star) would count toward your gravitational acceleration.)

      --
      "A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
    6. Re:Black holes must be flat dishes by Atario · · Score: 1
      Real black holes are likely to be spinning. And then they aren't spherical, as I recall.
      The black hole itself (that is, the matter that constitutes it) can be whatever shape it wants to be. What we're talking about is the shape of the event horizon, which is purely defined by the escape velocity, which is purely defined by the gravitational attraction at a certain distance, which is purely defined by the mass and distance from center point. All points at a certain distance from a point a sphere; that's the mathematical definition of a sphere.
      --
      "A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
    7. Re:Black holes must be flat dishes by CheshireCatCO · · Score: 1

      "Black Hole" isn't a terrifically well-defined term when you come right down to it. If you mean the thing in the middle (the "matter"), that's the singularity. It can have no real shape, as it is, as far as we can figure, a point. (The math goes all wonky, so it's better not to ask too many questions of theorists, as they tend to get kinda touchy about this.) If you mean the edge of where all hell breaks lose, then you're referring to the event horizon. (Formally defined as the point of no return, where spacetime is falling into the black hole faster than the speed of light.) Usually, that's what you care about with a black hole. (Particularly as theory has little to say about the singularity and observations have even less to tell us. And, for that matter, it's the event horizon you need to avoid. If you're dodging the singularity, you're already dead.)

      Now, as for the shape of the event horizon, I'm sorry, you're flat out wrong. A spinning black hole will be sqashed at the poles and bulged at the equator. The case of a spinning BH is the Kerr solution to Einstein's equations and it clearly shows this. See http://astronomy.colorado.edu/astr2030/Kerr_files/ v3_document.htm I'd further suggest Kip Thorne's book ("Black Holes and Time Warps"), an excellent read which covers an amazing about of black hole theory while remaining thorougly readable.

    8. Re:Black holes must be flat dishes by BungoMan85 · · Score: 1

      meh... black holes arent black =P in order to not violate the laws of thermodynamics they must have a non zero temperature. and ALL matter with a non zero temperature emits electromagnetic radiation of some kind. so black holes arent as black as we thought... about them bein spherical or not... well i dunno... just thought id interject that lil tidbit (sorry if its off topic). if you wanna know where i got that info, its in Stephen Hawking's book "The Theory of Everything". great book. goes into WAY more detail that i could ever hope to.

      --
      Bungo!
    9. Re:Black holes must be flat dishes by Atario · · Score: 1

      Maybe I'm dense or something (Get it? Dense? Black holes? Ehh.), but nothing on that site tells me that the event horizon is anything but a sphere. It seems to be saying that the "static limit" (region within which nothing can remain stationary) can be an oblate spheroid.

      --
      "A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
    10. Re:Black holes must be flat dishes by CheshireCatCO · · Score: 1

      Hm, Mitch must have said it in the actual lecture, then.

      Then read the book. It's a good read, anyway.

      In any event, the event horizon need not be a sphere for the simple reason that you're not worried about how far from the singularity you are, you cannot know that. (Information cannot propogate outward from inside the horizon to any point further out.) What you care about is local spacetime. Since spacetime gets dragged by spinning objects, there is no real reason to expect spinning black holes to look like non-spinning ones.

  8. This has been done before. by mph · · Score: 4, Informative

    The oblateness of Altair was measured using the Palomar Testbed Interferometer (PTI) in 1999-2000.

    1. Re:This has been done before. by Angry+Toad · · Score: 1

      I'm reasonably sure that it was done decades before that, even. Burnham's Celestial Handbook mentions some figures for the oblateness of Altair, and the handbook was last updated in, IIRC, the late 1970s.

      That being said, I'm sure the latest work is much more advanced.

  9. our sun, the planets by kamukwam · · Score: 4, Insightful

    I really don't think the fact that the star isn't a perfect sphere is surprising. The fact that we can measure it is a breakthrough. If we look at the sun, we can see that isn't a perfect sphere. It's not very much an ellipsoid either, but you could imagine a star (much younger) that spins very fast and is more like an ellipsoid. Even Jupiter (and also the earth!) are somewhat flattened.

    1. Re:our sun, the planets by Atario · · Score: 1

      The fact that we can measure it is a breakthrough.

      You got that right. Who would've thought we could start to resolve the diameters of other stars within our lifetimes?? Astromony never ceases to amaze.

      The truth of observation that astronomy has advanced more in the last 30 years than the over entire previous history of man must surely give the power of Moore's Law a run for its money.

      --
      "A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
    2. Re:our sun, the planets by Kotetsu · · Score: 2, Informative

      Who would've thought we could start to resolve the diameters of other stars within our lifetimes??

      Considering that Albert Michelson (yes, that Michelson) made the first measurement of another star (not the Sun) in 1920 (about a third or the way down the page for that detail), the question is probably more like how old are you? My parents weren't born yet when that happened.

      --

      "Bite me, it's fun!" - Crowe T. Robot
    3. Re:our sun, the planets by Atario · · Score: 1

      Not measure the diameters, resolve the diameters. As in, take a picture of, with sufficient resolution to see the dang thing.

      --
      "A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
    4. Re:our sun, the planets by hubie · · Score: 2, Informative
      You are correct, but the results are quite different. Though the technique used now is fundamentally the same as what Michelson used, Michelson would have been very hard pressed to measure oblatness because he (and Pease) were very limited in how they could change their baselines. In effect, Michelson and Pease could only measure the diameter across one direction of the star, so they could not have made an oblateness measurement.

      The modern interferometers, besides having very long observing baselines, also make such a large number of baseline observations that they can actually do an inverse transform and get an image.

      If you are interested, some nice info is found here, and the best collection of stellar interferometry links is found here.

    5. Re:our sun, the planets by hubie · · Score: 1
      Michelson measured the diameters by resolving them. When you look at a point source with two apertures (which is the method he used), what you see are interference fringes. As you move your apertures apart, their resolving power increases. When you reach the point where you start resolving your source, your interference fringes go away.

      The large interferometers use this same method where they make a whole bunch of fringe observations (for N apertures, you can measure N(N-1)/2 fringes), but the basic method is the same. If you make enough measurements, you can then do a transform on your data to turn it into an image.

    6. Re:our sun, the planets by Atario · · Score: 1

      Wait a sec. I thought they only came up with usably accurate and precise equipment for doing interferometry with visible wavelength recently (say, in the last 10 years or so). We could do this since the 20s?? Where the heck are our pictures of the moons of the fifth planet of Regulus V?

      --
      "A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
    7. Re:our sun, the planets by hubie · · Score: 1

      Michelson pushed the limits of what could be accomplished at the time. He used apertures that were spaced 20 feet apart (I believe). After he died his collaborator Pease tried a 50 ft baseline but it wasn't stable enough. It took another 40-50 years until the technology caught up to stabilize longer baselines.

  10. The Very Large Telescope Interferometer by jwachter · · Score: 5, Informative

    This site describes the telescopes that comprise the interferometer used to make the observations:
    http://www.eso.org/projects/vlti/

    Quote:
    The Very Large Telescope Interferometer (VLTI) consists in the coherent combination of the four VLT Unit Telescopes and of several moveable 1.8m Auxiliary Telescopes. Once fully operational, the VLTI will provide both a high sensitivity as well as milli-arcsec angular resolution provided by baselines of up to 200m length.

  11. Shape of the earth by xyrw · · Score: 2, Insightful

    The earth is not a sphere either. Any celestial body with a reasonable angular velocity will be slightly elliptical.

    1. Re:Shape of the earth by Anonymous Coward · · Score: 0

      RTFA and you won't be redundant.

      (translation: the article already mentioned it)

    2. Re:Shape of the earth by hplasm · · Score: 1

      Oblate. That's the word.

      --
      ...and he grinned, like a fox eating shit out of a wire brush.
  12. Who writes these articles? Or am I iggernint? by Atario · · Score: 3, Interesting

    Due to its daily rotation, the solid Earth is slightly flattened...

    Solid Earth? Only the surface (and part of the core) is solid, right? The rest is [Dr. Evil] liquid hot magma.

    The observed flattening cannot be reproduced by the "Roche-model" that implies solid-body rotation and mass concentration at the center of the star.

    I thought stars were pretty much all plasma, which is to say, a fluid. Why, therefore, should stars obey any "solid-body" rule at all?

    --
    "A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
  13. Re:Who writes these articles? Or am I iggernint? by mwtown · · Score: 3, Informative

    Actually, I think they mean "solid-body" as "cohesive object" in this case.

    While I'm getting technical, Plasma can't be considered a fluid either, as it's not a liquid, it's a different state of matter altogether.

  14. Theoretical maximum for common stellar materials? by Raindance · · Score: 3, Interesting

    Any physics buffs know what the largest theoretical ratio would be between a star's polar radius and equatorial radius, for the stuff that stars are made out of? Is the ratio for this star anywhere close to that?

    I'd imagine one can only attain this through centrifugal force, which necessarily puts structural stress on the star, and past a certain amount of structural stress stars should disintegrate.

  15. Re:Who writes these articles? Or am I iggernint? by Atario · · Score: 1
    Actually, I think they mean "solid-body" as "cohesive object" in this case.
    Ermmmm...and how is that different from solid in the usual sense?
    While I'm getting technical, Plasma can't be considered a fluid either, as it's not a liquid, it's a different state of matter altogether.
    Unless I'm mistaken, a fluid is any state of matter in which the molecules flow freely and assume the shape of the container (i.e., liquid, gas, or plasma). How is a plasma not a fluid?
    --
    "A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
  16. Re:Theoretical maximum for common stellar material by CheshireCatCO · · Score: 4, Interesting

    I believe a star has zero tensile strength* (it's just a fluid), so once you're spinning too fast for gravity to hold you together, it's bye-bye time.

    The better question is this: how did that star form? If it was spinning too fast to hold together, how did it accrete matter with that much angular momentum at all?

    * Barring magnetic fields, mind you. But you'd need an ass-kicking field to hold a star together very long, I would think.

  17. Re:Theoretical maximum for common stellar material by arvindn · · Score: 2, Informative
    The article has some info on grandparent's question:

    The indicated ratio between the equatorial and polar radii of Achernar constitutes an unprecedented challenge for theoretical astrophysics, in particular concerning mass loss from the surface enhanced by the rapid rotation (the centrifugal effect) and also the distribution of internal angular momentum (the rotation velocity at different depths).

    The astronomers conclude that Achernar must either rotate faster (and hence, closer to the "critical" (break-up) velocity of about 300 km/sec) than what the spectral observations show (about 225 km/sec from the widening of the spectral lines) or it must violate the rigid-body rotation.

    From this I think we can conclude that the star is very close to the theoretical limit for polar/eq radius for stable stars, but that this theoretical model might be inaccurate.

    Your question about how the star formed at all is interesting. IANAP, but it could be that when the star formed it wasn't spinning fast enough to break apart, but as it loses mass due to fusion, it becomes more elongated until the weakened gravity isn't able to hold it together any longer.

  18. Re:Who writes these articles? Or am I iggernint? by mwtown · · Score: 1
    Ermmmm...and how is that different from solid in the usual sense?
    A cohesive object doesn't have to be made of matter in a solid state. You can't move through three feet of ice (by yourself anyways), but you can jump into a lake or walk through a fog.
    Unless I'm mistaken, a fluid is any state of matter in which the molecules flow freely and assume the shape of the container (i.e., liquid, gas, or plasma). How is a plasma not a fluid?
    Plasma is a collection of charged particles that have some of the properties of a gas, but is different in that it's a good conductor of electricity and can be affected by magnetic fields. It won't try to fill the space it's in, and you can't pour it. Do you remember seeing a Jacob's Ladder? That electric arc is actually plasma.

    On the other hand, you could be right as well. It's been years since my last Physics class :)

  19. Re:Who writes these articles? Or am I iggernint? by Christopher+Thomas · · Score: 1

    Plasma is a collection of charged particles that have some of the properties of a gas, but is different in that it's a good conductor of electricity and can be affected by magnetic fields. It won't try to fill the space it's in, and you can't pour it.

    Actually, it will try to fill the space it's in, just like any other gas. It can be impeded from doing so by magnetic fields (including fields generated by its own motion), but this is not a permanent state of affairs.

  20. The Earth is largely solid by Anonymous Coward · · Score: 0

    I just have to remind everyone: the Earth is mostly solid.

    The crust and upper mantle are solid, 99% of the partially melted asthenospheric mantle is solid rock, the mantle beneath that is solid rock, and the inner core is solid. That really just leaves the outer core as liquid.

    It's easy to go on and on with observations demonstrating this is true, to pull out stacks of earthquake wave data, etc., but I think the simplest and most convincing evidence is in the shape of the Earth itself. It has been realized since the 1800s that for the Earth to retain its nearly spherical shape as it spins, the average strength of the Earth must be greater than that of a similar sized ball of glass or iron.

    So again, with the exception of the outer core, the Earth is made of solid rock -- all apologies to Dr. Evil. The next time the structure of the Earth comes up on Slashdot, as it does from time-to-time, don't forget: the mantle is also made of solid rock, not liquid magma.

    Thank you for your time and attention to this sniveling detail.

    - Anonymous Coward

    1. Re:The Earth is largely solid by Atario · · Score: 1
      Wait a second.
      The crust and upper mantle are solid, 99% of the partially melted asthenospheric mantle is solid rock, the mantle beneath that is solid rock, and the inner core is solid. That really just leaves the outer core as liquid.
      According to this nice summary, the mantle (asthenosphere part of it, anyway) is "semi-solid". Other sources describe it as akin to toothpase or silly putty.
      It has been realized since the 1800s that for the Earth to retain its nearly spherical shape as it spins, the average strength of the Earth must be greater than that of a similar sized ball of glass or iron.
      That's a pretty good trick, seeing as how iron is stronger than stone, which you're saying most of the planet is made of. Anyway, Jupiter spins over twice as fast, is immensely larger, and made almost entirely of fluids (i.e., non-solids), and even so is only just visibly squished! Earth, being so much cooler, smaller, slower, and more solid, would have no reason to be any other shape than "almost spherical", would it? Where are you getting this?
      --
      "A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
    2. Re:The Earth is largely solid by Anonymous Coward · · Score: 0

      Ok, lets be real quick here.

      You're basically wrong, although I'll grant you a *tiny* bit on the asthenosphere.

      The asthenosphere has no effective yield strength given reasonable strains and strain rates, but it is not molten liquid. It composed of crystals -- primarily a kind of mineral called olivine -- that are lubricrated with a tiny amount of melt. This melt fraction is around 1% of the total material. So this miniscule amount of liquid lets the grains slowly slide past eachother as they interact with the tectonic plates above them. This portion of the mantle does not, however, behave like a liquid because it does transmit seismic shear waves, although they are attenuated a little bit (I wanted to keep earthquake waves out of this). If you were to dig a deep hole in to the asthenosphere, then hit a piece of it with a hammer, you could break a chunk off. The asthenosphere is probably about 1/100 the viscosity of the overlying lithosphere, but keep in mind that's still pretty damn stiff. If you lived off a diet of rock, and someone gave you a bowl of asthenosphere, you wouldn't choose to eat it with a spoon.

      Are you certain that iron is less viscous than high pressure mantle rock? Irrespective of the viscosity of iron (I didn't find a convenient reference to this value), I think we can agree that a) glass is solid, and b) rock is more viscous than glass. So I think my point stands. (Never mind that the Earth has to be solid given the observed propogation of earthquake shear waves.)

      The Earth is not spherical. It is slightly flattened at the poles, as it must be given the physics and mechanical properties involved. Jupiter is also flattened, but considerably more so than the Earth. Both are almost spherical depending on what you call "almost," but either one would be subtly more or less spherical if it had different physical properties. I'm afraid I don't know where you were going with this.

      Finally, the Earth absolutely must be solid for shear waves to propogate through it. They don't go through the liquid outer core, but can be traced with quite reasonable precision through the rest of the planet.

      Again, thank you for your time and attention to this matter.

      - Anonymous Coward

  21. Re:Who writes these articles? Or am I iggernint? by i+chose+quality · · Score: 1
    How is a plasma not a fluid?
    because it's plasma!

    "Although plasma includes electrons and ions and conducts electricity, it is macroscopically neutral: in measurable quantities, the number of electrons and ions are equal. The charged particles are affected by electric and magnetic fields applied to the plasma, and the motions of the particles in the plasma generate fields and electric currents from within. This complex set of interactions makes plasma a unique, fascinating, and complex state of matter."
    --
    the computer is online
    i am not at it
    what a waste of ressources
  22. Re:Who writes these articles? Or am I iggernint? by X-rated+Ouroboros · · Score: 4, Informative

    How is a plasma not a fluid?
    because it's plasma!

    This exchange is about on par with "How is a liquid not a fluid?" "Because it's a liquid."
    "Fluid" is not a state of matter, no one's claiming it's a state of matter, saying plasma can't be a fluid because plasma is the 4th state of matter is a category error. Liquid is the second state of matter. Gas is the third state of matter. Both are fluids.

    A fluid is any substance which undergoes continuous deformation when subjected to a shear stress. The problem we're probably having is that the obvious sources for the shear stresses in the couse of, say, water being poured from a cup (normal force of the side of the cup vs gravity) are paralleled for the case of plasma by electromagnetic feilds. It just don't grok intuitively but, plasma behaves like a fluid... ergo, it is a fluid.

    --
    Simple Machines in Higher Dimensions
  23. Re:Who writes these articles? Or am I iggernint? by spakka · · Score: 1

    This complex set of interactions makes plasma a unique, fascinating, and complex state of matter.

    Gases and liquids are different states of matter, both fluids.
  24. Hal Clement by IPFreely · · Score: 2, Interesting
    As soon as I read this, I thought of Mission of Gravity, a book about a flattened planet, Jupiter size, 18 minute rotation, Surface gravity at the equator was about 3 Earth G, while at the pole was more like 600+ Earth G, flat just like this. It was written in 1953 I believe, and included some detailed physics in the back of the book covering how the planet maintains it shape.

    The story is about natives on the planet, but the physics alone is worth the read. It's quite a strange place.

    --
    There is nothing so silly as other peoples traditions, and nothing so sacred as our own.
  25. Re:Theoretical maximum for common stellar material by CheshireCatCO · · Score: 1

    A star like the Sun will only convert about 0.01% of it's mass into energy over its entire lifetime. (According to my quick, back of the envelope calculation.) Which means you still have to form very close to the breakup spin rate. Which is still quite difficult to pull off, saith the dynamicists. (The same problem comes up in fusion-formation models of moons.)

  26. 50% smaller? Ah, no... by Anonymous Coward · · Score: 1, Informative

    The equatorial radius is 50% larger than the polar radius. This does not make the polar radius 50% smaller than the equatorial radius.

  27. yes, it does affect luminosity of the star by Doctor+Fishboy · · Score: 3, Informative

    In the more recent surveys of bright stars in a cluster, they've seen that faster rotating stars (seen indirectly by the rotational broadening of spectral lines of the star) of the same spectral type have a wider scatter of observed brightnesses. The explanation for this is that:

    (i) Faster rotating stars are brighter at their poles than their equators (because of centripetal force slightly expanding the distance of the equator from the core of the star), and:

    (ii) The spin axes of stars are randomly oriented with respect to telescopes on Earth, so:

    (iii) For a large sample of fast rotating stars, you sample all the brightnesses from the equator to the poles, hence a large scatter in measured brightness. You can assume that all stars are effectively at the same distance if they are in a distant cluster.

    Hope that's reasonably clear,

    Dr Fish

    1. Re:yes, it does affect luminosity of the star by VendingMenace · · Score: 1

      OK, so are you saying that the line broadening is cuased by the fact that you see several diffferent orientations or stars at the same time? So this is an inhomogenious line broadening. Much like that observed for phenomenon like doppler broadening? Where you would not see line braodening if you observed only one star, but you would see line broadening when you observe the entire cluster.

      Or are you saying that there is line broadening do the the fast rotation in the individual stars. that is, you would still see line broadeneing if you were only able to observe a single star.

    2. Re:yes, it does affect luminosity of the star by Alsee · · Score: 1

      Or are you saying that there is line broadening do the the fast rotation in the individual stars. that is, you would still see line broadeneing if you were only able to observe a single star.

      Right. The left side of the star may be coming towards you dopplershifting it to the blue, and the right side may be moving away dopplershifting it to the red.

      As you look from one side of the the star to the other to smoothly range between the two effects. Every line gets "smeared" across a range.

      -

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  28. Re:Who writes these articles? Or am I iggernint? by i+chose+quality · · Score: 1

    thanks for not flaming. i reread the whole thread and recognised my error. (liquid fluid)

    that's what you get for posting before the coffee break... ;)

    --
    the computer is online
    i am not at it
    what a waste of ressources
  29. Re:Theoretical maximum for common stellar material by Anonymous Coward · · Score: 0

    If it's a gas plasma it's quite likely that it doesn't conform to rigid-body models. We believe that the interior and exterior of Jupiter rotate at somewhat different angular velocity, which creates the red spot. Similar spots have been created in (practical) models in the laboratory where fluids are spun at differential rates.

    There's no reason to believe that the outside of the lenticular star spins at the same rate as the pole. It would be interesting to get a good enough image to determine if it has a "spot

  30. More Please by IPFreely · · Score: 1
    I've heard of strange theories where a rapidly spinning Black Hole might separate enough to form a torus, and the event horizon would follow it. Thus, if you flew through the hole but still outside the horizon, you go to strange places.

    Any more insite on that, or is it complete wash?

    --
    There is nothing so silly as other peoples traditions, and nothing so sacred as our own.
  31. Re:Who writes these articles? Or am I iggernint? by Atario · · Score: 1

    Thank you!

    --
    "A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
  32. Re:Who writes these articles? Or am I iggernint? by Atario · · Score: 1
    A cohesive object doesn't have to be made of matter in a solid state. You can't move through three feet of ice (by yourself anyways), but you can jump into a lake or walk through a fog.
    Fog is a cohesive object? Then what the heck is non-cohesive? A ghost?
    --
    "A great democracy must be progressive or it will soon cease to be a great democracy." --Theodore Roosevelt
  33. The difference ... by hubie · · Score: 1
    From van Belle, et al., ApJ, 559, 2, p. 1155, 2001:
    Altair is the first main-sequence star for which direct observations of an oblate photosphere have been reported and the first star for which vsini has been established from observations of the star's photospheric geometry.
  34. Peace! by fm6 · · Score: 1
    A fluid is any substance which undergoes continuous deformation when subjected to a shear stress.
    OK, thanks for educating us as to the physicist's definition of the word fluid. And in that context it does make sense to call gas or plasma (or sand or granulated sugar) a fluid. It also makes sense with the original Latin word fluidus, "flowing".

    But most people aren't physicists or classical scholars. So they use "fluid" as a synonym for "liquid". That's not a sign of rampant stupidity, it's just the way imprecise usages creep into any language.

  35. Implications on stellar death by MeowmiXXX · · Score: 2, Interesting

    I wonder how this will affect it's distruction, the decreased pressure would decrease the rate of fusion while the spinning would make it easier to fly apart, and how would it die off? Since, it wouldn't have much growth before the centrifugal forces rip it apart it should be hotter and more compact.

  36. Re:Theoretical maximum for common stellar material by CheshireCatCO · · Score: 1

    Uh, it isn't a different rotation rate in the interior that creates the Great Red Spot. It's solar and internal heat, causing convection, and the spin, causing belts and zones. Admittedly, the GRS isn't totally understood, but a differentially rotating interior doesn't seem to be required. (As for whether the interior does rotate differentially, it's kind of hard to say. Since you have the belts and zones moving at all different speeds, it turns out to be difficult to define a rotation period for the cloud level. At least 2 attempts were made to define a rotation period based on the clouds, but the rotation period cited is usually the interior rotation perdio. The interior is easy to clock, thanks to the magnetic field and the radio emissions due to it.)

    That said, of course you don't expect a star to be rotating as a solid body. As you suggest, there's no reason it has to. But you can make an easier, and far better, case if you just point to the Sun. The Sun's rotation period varies from equator to poles (the equator is fastest, which doesn't help a star stay together any). The interior also rotates differentially, as determined by some of the GONG results.