Central Density and Degenerate Electrons: Comparison with Normal Atomic Densities - Prof. , Assignments of Physics

Download Central Density and Degenerate Electrons: Comparison with Normal Atomic Densities - Prof. and more Assignments Physics in PDF only on Docsity! Phys 132 Homework 21 Solutions Professor Crystal Martin TA: Ellie Hadjiyska Problem 21: (Phillips Problem 6.1) We now want to show that the central density of a body supported by degenerate electrons becomes comparable with normal atomic densities when the mass of the body is comparable with MP = (αEM/αG)3/2mH = α 3/2 EMM∗. We begin by looking at what relationship degeneracy gives us between central temperature and density. The electron density is related to the central mass density by (Phillips Eq. (6.1)) ne = Ye ρc mH where Ye is the ratio of electrons to nucleons and m is the mass of hydrogen. If we assume the star is supported by degeneracy pressure, we can set this equal to the quantum concentration (Phillips Eq. (2.22)) ne = Ye ρc mH = nQ = [ mekTc 2π~2 ]3/2 where ~ = h/(2π). From Problem 2, part (c) we found a relation between central temperature and density: Tc = 1 2 ( 4π 3 )1/3 m̄ k GM2/3ρ1/3c Plugging this into our previous expression gives Ye ρc mH = [ mek 2π~2 ( 1 2 ( 4π 3 )1/3 m̄ k GM2/3ρ1/3c )]3/2 or Ye ρc mH = 1 4π √ 3 m 3/2 e G3/2m̄3/2Mρ 1/2 c ~3 or ρ1/2c = mH 4πYe √ 3 m 3/2 e G3/2m̄3/2 ~3 MP We now look at what happens when this central density from degeneracy is comparable to the density we’d expect from ordinary, uncompressed matter. i.e. ρc ∼ ρatomic = mH a3B = mHα 3 EMm 3 ec 3 ~3 Plugging in we get m 1/2 H α 3/2 EMm 3/2 e c3/2 ~3/2 = mH 4πYe √ 3 m 3/2 e G3/2m̄3/2 ~3 MP