Talk:Gravity darkening

Disputed

This article directly contradicts the Regulus article about which portion of the star is brighter, the equater or the poles. Since this article also confused oblong and oblate shapes, I'm guessing that this is the one in error. --Heywood 16:55, 9 July 2006 (UTC)[reply]

Also, I'm almost positive that 'torusoidal' is not a word and that while a star may appear toroidal, it by no means actually becomes anything like one with anything approximating an actual indention. --Heywood 23:41, 9 July 2006 (UTC)[reply]

Due to fast rotating of such stars the stellar sphere is flattened just as the gas giant planets Jupiter and Saturn. The equatorial radius is larger than the polar one. On stars like Regulus and Fomalhaut it results in higher temperatures at the poles than the equator. So an article like shaped Regulus is OK but the star is not egg shaped but a flattened sphere. <font color=red>Skate</font><font color=blue>biker</font> 21:42, 18 July 2006 (UTC)[reply]

Gravity darkening is a result of the Von Zeipel theorem. The local acceleration due to gravity at the pole of the star is greater than at the equator. This is because the centrifugal force due to the rotation of the star counteracts gravity at the equator but not at the poles. The net result is that the brightness of the star varies across the surface of the star with the poles being brighter than the equator. It is sometimes refered to as gravity brightening because the regions of higher gravity are brigher. The atmosphere of a star in hydrostatic equilibrium requires that the local gravitational force is balanced by the pressure exerted by the net flux of photons flowing outwards. For a radiative atmosphere (one with negligible convection) the luminosity is related directly to the local gravity. In a covective atmoshere gravity darkening is less pronounced. This phenomenon has been observed directly by interferometric observations of Vega and Altair. Pflm 21:10, 11 November 2006 (UTC)[reply]

Color of star

Since the temperature is variable, so too must the color be. But are we talking about a measurable difference? —Soap— 02:07, 26 October 2018 (UTC)[reply]

Not just measurable but it would be visible to the naked eye. Achernar, for example, is something like 20,000 K at the poles and 10,000 K at the equator, that's about the difference between an A0 star and a B2 star. Lithopsian (talk) 13:55, 26 October 2018 (UTC)[reply]

So its a stripey star? OK thanks. That's probably far more powerful than the gravitational redshift phenomenon, which does precisely the opposite thing ... makes the light redder when associated with higher gravity. Im not sure we need the dab I added to the page awhile ago, but Im leaving it there for now. —Soap— 18:26, 27 October 2018 (UTC)[reply]