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Space Science

Non-Spherical Stars 70

Posted by michael from the first-you-assume-a-spherical-cow dept.
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!"
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Non-Spherical Stars More

Non-Spherical Stars

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  • Husky stars (Score:3, Funny)

    by SandSpider ( 60727 ) on Thursday June 12, 2003 @03:00PM (#6184869) Homepage Journal
    I blame large plates. And a lack of exercise.

    =Brian

  • Press release here: (Score:5, Informative)

    by molo ( 94384 ) on Thursday June 12, 2003 @03:06PM (#6184912) Journal
    More details at the press release:

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

    Including more technical drawings.

    -molo

  • Energy output (Score:4, Interesting)

    by VendingMenace ( 613279 ) on Thursday June 12, 2003 @03:07PM (#6184920)
    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.

    • Re:Energy output (Score:3, Informative)

      by BigBir3d ( 454486 )
      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...

      • ,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.

      • Re:Energy output (Score:3, Insightful)

        by L7_ ( 645377 )
        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.

    • yes, it does affect luminosity of the star (Score:4, Informative)

      by Doctor Fishboy ( 120462 ) on Friday June 13, 2003 @09:59AM (#6190835)
      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
      • 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
        • 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.

          -

  • Cosmic Rugby (Score:3, Funny)

    by follower_of_christ ( 626504 ) <phatcoder@yahoo.com> on Thursday June 12, 2003 @03:09PM (#6184935)
    Hey God!

    I found your rubgy ball!

  • Amateur Astronomy (Score:3, Funny)

    by Anonymous Coward on Thursday June 12, 2003 @03:23PM (#6185099)
    Nicole Kidman or Gwyneth Paltrow are the flattest stars that can be seen with the naked eye or possibly binoculars.

  • I crush you... (Score:5, Funny)

    by FroMan ( 111520 ) on Thursday June 12, 2003 @03:42PM (#6185275) Homepage Journal
    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.

  • 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.
    • 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).
      • 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.
        • 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

          • 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 pol

        • 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.
          • 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?

          • 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'

            • "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, wher
              • 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.
                • 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 loo
        • 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 de

  • This has been done before. (Score:5, Informative)

    by mph ( 7675 ) <mph@freebsd.org> on Thursday June 12, 2003 @04:51PM (#6185990)
    The oblateness of Altair was measured [caltech.edu] using the Palomar Testbed Interferometer (PTI) in 1999-2000.

    • I'm reasonably sure that it was done decades before that, even. Burnham's Celestial Handbook [amazon.com] 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.

      • 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.

  • our sun, the planets (Score:4, Insightful)

    by kamukwam ( 652361 ) on Thursday June 12, 2003 @05:27PM (#6186298) Journal
    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.

    • 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.

      • 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 [aavso.org] (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.
        • Not measure the diameters, resolve the diameters. As in, take a picture of, with sufficient resolution to see the dang thing.
          • 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 me

            • 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?
              • 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.

        • Re:our sun, the planets (Score:3, Informative)

          by hubie ( 108345 )
          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,

  • The Very Large Telescope Interferometer (Score:5, Informative)

    by jwachter ( 319790 ) <`moc.liamg' `ta' `rethcaw'> on Thursday June 12, 2003 @06:58PM (#6186937) Homepage
    This site describes the telescopes that comprise the interferometer used to make the observations: [eso.org]
    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.

  • Shape of the earth (Score:2, Insightful)

    by xyrw ( 609810 )
    The earth is not a sphere either. Any celestial body with a reasonable angular velocity will be slightly elliptical.

  • Who writes these articles? Or am I iggernint? (Score:3, Interesting)

    by Atario ( 673917 ) on Thursday June 12, 2003 @08:38PM (#6187399) Homepage

    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?

    • 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.

      • 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?

        • 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

          • 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.
          • 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?

        • How is a plasma not a fluid?

          because [spacescience.org] it's [pppl.gov] plasma [nasa.gov]!

          "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."

          • Re:Who writes these articles? Or am I iggernint? (Score:5, Informative)

            by X-rated Ouroboros ( 526150 ) on Friday June 13, 2003 @07:25AM (#6189678) Homepage

            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.

            • 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... ;)

            • Peace! (Score:2)

              by fm6 ( 162816 )

              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 i

          • 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.
  • 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.

    • Re:Theoretical maximum for common stellar material (Score:5, Interesting)

      by CheshireCatCO ( 185193 ) on Thursday June 12, 2003 @10:19PM (#6187920) Homepage
      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.
      • 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

        • 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.)

  • Hal Clement (Score:3, Interesting)

    by IPFreely ( 47576 ) <mark@mwiley.org> on Friday June 13, 2003 @07:45AM (#6189809) Homepage Journal
    As soon as I read this, I thought of Mission of Gravity [amazon.com], 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.

  • 50% smaller? Ah, no... (Score:1, Informative)

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

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