Path: utzoo!utgpu!news-server.csri.toronto.edu!helios.physics.utoronto.ca!physics.utoronto.ca!neufeld Newsgroups: sci.space From: neufeld@physics.utoronto.ca (Christopher Neufeld) Subject: Re: Holes,station,vulcan Message-ID: <1990Dec13.193441.27382@helios.physics.utoronto.ca> Organization: University of Toronto Physics/Astronomy/CITA References: Date: 14 Dec 90 00:34:42 GMT In article 18084TM@MSU.BITNET (Tommy Mac) writes: > >Re; Black Holes >Ray, remember your four forces. Star happen because of gravity defeating >the weak force (I think), as do white dwarves and nuetron stars. Black >holes happen when gravity defeats the strong force (this is the strongest >of the four). > Stars form because of gravitational collapse of gas clouds. If there is any countering force, it would be electromagnetic, radiation pressure pushing back on the infalling gas because it is heating up at the centre, even before nuclear ignition. The weak force has much too short a range to be significant in interactions between particles separated even by micrometres. A white dwarf is formed when the star collapses after nuclear burnout, and the Fermi pressure of electrons holds off the gravity. If the star is too massive, the electron Fermi pressure will never equal that of gravity. A proton+electron -> neutron+neutrino reaction converts most of the mass to neutrons, which are able to exert a higher Fermi pressure than electrons, so the collapse stops. If the star is even more massive, it may be heavy enough that neutron pressure can't stop the collapse. There's no known mechanism which can prevent the mass from falling in past its own event horizon in this case, and a black hole is formed. Note that a black hole does not have to have a mass singularity, and it doesn't have to start in the form of degenerate matter. The radius of a black hole (Schwarzschild radius) is proportional to the mass of the body, while the volume goes as the cube of the radius. If all the stars in the galaxy were placed in a sphere whose radius was that of our solar system, the stars wouldn't touch each other, but the resulting construct would be a black hole. A quick explanation of the Fermi pressure I was describing before: from basic physics or chemistry you know that no two fermions can have the same set of quantum numbers. There turns out to be a set of quantum numbers associated with the volume available for the fermions to travel. A white dwarf sets a limit to how far the electrons can travel, they can't leave its surface. The quantum numbers, proportional to momentum, and hence to the square root of kinetic energy, scale up themselves as the reciprocal of the linear dimension of the object, here it would be the radius. So, if the star is compressed, all the momenta of the electrons shift up. This means that the kinetic energy stored in the electrons is higher, so you have to have done some work on the system to get it there. This implies immediately that you compressed it against some pressure. So, a white neutron star isn't stable because some other elementary force is pushing back against gravity, but rather because it has fallen to the bottom of an energy well. Squeezing it farther would reduce the gravitational potential energy of the mass distribution, but it would raise the internal energy of the star by a greater amount, and so it is energetically unfavourable for it to squeeze farther. A more complete discussion can be found in any moderately good book on statistical mechanics, or in chapter 11 of Weinberg's _Gravitation and Cosmology_. -- Christopher Neufeld....Just a graduate student | neufeld@helios.physics.utoronto.ca Ad astra! | S = k log W cneufeld@{pnet91,pro-micol}.cts.com | Boltzmann's epitaph "Don't edit reality for the sake of simplicity" |