emc wrote:Makes sense to me... the rotation speed is greater at the equator and less at the poles so I believe ejected matter would tend to travel further from the equator.
I have my doubts, so lets do a "back of an envelope" calculation. According to
the solar wiki the equatorial rotation is 14.18º/day. From º to radians multiply by π and divide by 180, from days to seconds multiply by 86400. The conversion factor is about 2E-8, so the circle frequency ω=3E-6 radians/second. The centrifugal acceleration is ω²R, with R=7E8 m. That leads to an equatorial centrifugal acceleration of about 6E-3 m/s²
Now assume the sun is nearly spherical, not as irregularly shaped as the Kuiperbelt object named Santa. In a reasonable approximation the acceleration of gravity at the solar equator is GM/R² where G is the gravitational constant (6.67E-11 Nm²/kg²), M is the solar mass (2E30 kg) and R = 7E8 m. The equatorial acceleration of gravity is then approximately 260 m/s². Approximately and about mean 1 digit, so 260 might be 300 or 200 as well.
Compare the equatorial centrifugal force 6E-3 with the acceleration of gravity 260 m/s² Their quotient is of the order of 2E-5 or 20 ppm. In my opinion that is minute. Effects of turbulence in the ejected mass are larger than the difference as caused by centrifugal acceleration.
Doum wrote:I was seeing superconductivity in a star strarting to collapse before explosion thus creating higher magnetic field.
Superconductivity is explained by the formation of Cooper pairs. Bardeen, Cooper and Schrieffer received a Nobel prize for their explication of superconductivity in 1972, which Heike Kamerlingh Onnes has discovered in 1911. A Cooper pair consists of two electrons with opposite spin. They are attracted by each other, in spite of the repulsive forces due to their equal signed charge. It is a kind of Bose-Einstein condensate, the latter two predicted in the nineteen twenties. The bond is kept togehther as long as the thermal exitations (phonons) are sufficiently low. That is the reason why superconductivity takes place at low temperatures. At least you will need liquid nitrogen (77K) to cool the matter down, so that Cooper pairs may form. Temperatures in the photosphere of a red giant, of about 2000 K, are a bit too high, thermal exitations are 1.5 to 2 orders of magnitude too high to allow for the formation of Cooper pairs.
Another argument why superconductivity is an unlikely phenomenon to play a significant role in a stellar explosion: the Meisner effect. This effect prevents magnetic fields to penetrate into a superconducting material. So superconducting stellar material moving in the stellar magnetic field is rather unlikely.
Finally Doum wrote:So it is not possible that the magnetic field of a star about to explode, will be estimate of what it was looking just before the explosion by looking at the gas expansion it create.
IMHO that is rather unlikely. Vincent Icke et al. have developed a mathematical model that describes the interaction between the originally ejected material and the stellar wind. The popular version of the article i would like to quote now, is at the desk of the editor of the magazine of our local astronomy club. The less popular (easy to read) version can be found on the
website of Vincent Icke. The images and the Flash movies on his site bare similarity with the shape of the "Red rectangle" nebula. It might be possible that i did not fully understood your final remark. So, alternatively, if by any means it would be possible to estimate the current magnetic field of the imploded star, an estimation (of a few orders of magnitude) can be given of the strength of the magnetic field of the star before it exploded. The shape of a stellar field, well, i fully agree with Art Neuendorffer, that is dipolar, just like a bar magnet.
[quote="emc"]Makes sense to me... the rotation speed is greater at the equator and less at the poles so I believe ejected matter would tend to travel further from the equator. [/quote]
I have my doubts, so lets do a "back of an envelope" calculation. According to [url=http://en.wikipedia.org/wiki/Solar_rotation]the solar wiki[/url] the equatorial rotation is 14.18º/day. From º to radians multiply by π and divide by 180, from days to seconds multiply by 86400. The conversion factor is about 2E-8, so the circle frequency ω=3E-6 radians/second. The centrifugal acceleration is ω²R, with R=7E8 m. That leads to an equatorial centrifugal acceleration of about 6E-3 m/s²
Now assume the sun is nearly spherical, not as irregularly shaped as the Kuiperbelt object named Santa. In a reasonable approximation the acceleration of gravity at the solar equator is GM/R² where G is the gravitational constant (6.67E-11 Nm²/kg²), M is the solar mass (2E30 kg) and R = 7E8 m. The equatorial acceleration of gravity is then approximately 260 m/s². Approximately and about mean 1 digit, so 260 might be 300 or 200 as well.
Compare the equatorial centrifugal force 6E-3 with the acceleration of gravity 260 m/s² Their quotient is of the order of 2E-5 or 20 ppm. In my opinion that is minute. Effects of turbulence in the ejected mass are larger than the difference as caused by centrifugal acceleration.
[quote="Doum"]I was seeing superconductivity in a star strarting to collapse before explosion thus creating higher magnetic field. [/quote]
Superconductivity is explained by the formation of Cooper pairs. Bardeen, Cooper and Schrieffer received a Nobel prize for their explication of superconductivity in 1972, which Heike Kamerlingh Onnes has discovered in 1911. A Cooper pair consists of two electrons with opposite spin. They are attracted by each other, in spite of the repulsive forces due to their equal signed charge. It is a kind of Bose-Einstein condensate, the latter two predicted in the nineteen twenties. The bond is kept togehther as long as the thermal exitations (phonons) are sufficiently low. That is the reason why superconductivity takes place at low temperatures. At least you will need liquid nitrogen (77K) to cool the matter down, so that Cooper pairs may form. Temperatures in the photosphere of a red giant, of about 2000 K, are a bit too high, thermal exitations are 1.5 to 2 orders of magnitude too high to allow for the formation of Cooper pairs.
Another argument why superconductivity is an unlikely phenomenon to play a significant role in a stellar explosion: the Meisner effect. This effect prevents magnetic fields to penetrate into a superconducting material. So superconducting stellar material moving in the stellar magnetic field is rather unlikely.
[quote="Finally Doum"]So it is not possible that the magnetic field of a star about to explode, will be estimate of what it was looking just before the explosion by looking at the gas expansion it create.[/quote]
IMHO that is rather unlikely. Vincent Icke et al. have developed a mathematical model that describes the interaction between the originally ejected material and the stellar wind. The popular version of the article i would like to quote now, is at the desk of the editor of the magazine of our local astronomy club. The less popular (easy to read) version can be found on the [url=http://www.strw.leidenuniv.nl/~icke/html/VincentRR.html]website of Vincent Icke[/url]. The images and the Flash movies on his site bare similarity with the shape of the "Red rectangle" nebula. It might be possible that i did not fully understood your final remark. So, alternatively, if by any means it would be possible to estimate the current magnetic field of the imploded star, an estimation (of a few orders of magnitude) can be given of the strength of the magnetic field of the star before it exploded. The shape of a stellar field, well, i fully agree with Art Neuendorffer, that is dipolar, just like a bar magnet.