385lec19 - PHYS 390 Lecture 19 - Degenerate stars Lecture...

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PHYS 390 Lecture 19 - Degenerate stars 19 - 1 © 2003 by David Boal, Simon Fraser University. All rights reserved; further resale or copying is strictly prohibited. Lecture 19 - Degenerate stars What's Important: pressure of a degenerate Fermi gas equilibrium of degenerate stars neutron stars Text : Gasirowicz, Chap. 9; Carroll and Ostlie, Chap. 15 Pressure of a degenerate Fermi gas From the previous lecture, the total energy of a degenerate Fermi gas of spin 1/2 particles is E TOT = π 3 h 2 10 m 3 N π 5/3 V - 2/3 . (1) The pressure of the gas can be obtained from this by differentiation P deg = - E TOT / V at fixed N , (2) so P deg = - - 2 3 π 3 h 2 10 m 3 N π 5/3 V - 5/3 = π 3 h 2 15 m 3 N π V 5/3 (3) The density can replace N/V in the last expression. Equilibrium of degenerate stars In normal stars, the nuclear fires generate energy which saves the star from gravitational collapse. But even in the absence of an energy source, the pressure from degenerate electrons (or neutrons, for dense stars) may balance the gravitational pressure and maintain mechanical equilibrium. To find this condition, we first must determine the pressure from gravity. First, recall that the total gravitational potential energy is given by V grav = -(3/5) GM 2 / R . (4) Students may have seen this derived in earlier courses; the calculation involves the following steps: 1. the mass of a thin shell of material at radius r is 4 π r 2 dr 2. the gravitational potential of the shell arising from its attraction to the mass within
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385lec19 - PHYS 390 Lecture 19 - Degenerate stars Lecture...

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