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 /^2 mH = α^3 EM/^2 M∗. 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 2 πℏ^2
] 3 / 2
where ℏ = h/(2π). From Problem 2, part (c) we found a relation between central temperature and density:
Tc =^1 2
4 π 3
m¯ k
GM 2 /^3 ρ^1 c/^3
Plugging this into our previous expression gives
Ye^ ρc mH
[
mek 2 πℏ^2
4 π 3
m¯ k
GM 2 /^3 ρ^1 c/^3
)] 3 / 2
or
Ye^ ρc mH
4 π
m^3 e/ 2 G^3 /^2 m¯^3 /^2 M ρ^1 c/^2 ℏ^3 or
ρ^1 c /^2 = mH 4 πYe
m^3 e/ 2 G^3 /^2 m¯^3 /^2 ℏ^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 a^3 B
mH α^3 EM m^3 ec^3 ℏ^3
Plugging in we get m^1 H/ 2 α^3 EM/^2 m^3 e /^2 c^3 /^2 ℏ^3 /^2
mH 4 πYe
m^3 e/ 2 G^3 /^2 m¯^3 /^2 ℏ^3 MP
or simplifying
MP = 4πYe
( (^) mH m ¯
) 3 / 2 (^ ℏc Gm^2 H
α^3 EM/^2 mH = 4πYe
( (^) mH m ¯
) 3 / 2 (^ αEM αG
mH
where we have used αG = Gm^2 H /(ℏc) (Phillips Eq. (5.63)). For a pure hydrogen star, ¯m = mH and we get
MP = 4πYe
αEM αG
mH ∼
αEM αG
mH = α^3 EM/^2 M∗
For hydrogen, the ratio of electrons to nucleons Ye = 1 since most hydrogen is one proton and one electron. For heavier elements (such as iron), this ratio is 1:2 or Ye ≈ 0 .5. This is why we see this number used in calculating the Chandrasekhar mass for a white dwarf (which is composed primarily of carbon and oxygen). So for a given central density ρc, using hydrogen instead of heavier elements gives twice the electron density. This would make the mass MP half what we originally calculated. Even more importantly, the average mass per particle for a pure iron star is much larger ( ¯m = 56mH ) and we see that raised to the 3/2 power, it would cut our resulting mass down more than two orders of magnitude! It would make our mass much smaller than what was found (i.e. a pure iron star would be 2(56)^3 /^2 ≈ 840 times smaller than a pure hydrogen star!).
Note: If you used Phillips Eq. (6.4) to get to the answer or worked the reverse of this problem (i.e. started with M = α^3 EM/^2 M∗ = 0. 001 MØ and showed ρatomic was comparable to ρc), that was fine.
