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Unformatted text preview: ng the expression above: η = 6.3 × 10−4 Tc = 3 × 10 −9 Tc M
M M
M 2/3 M
4/3 R
R0 −2/3 R
R0 2 2 2.8 × 107 M
M 1/3 2 Evaluating this for M = 0.1 M, R = 0.1 R and Tc = 5x106 K, we get R0 = 6.0x107 m and η = 0.4 – the interior of this kind of star is partially degenerate! (d) [20 pts] As a brown dwarf collapses, its core density and temperature increase, so η increases. However, at some point the core temperature stops increasing as the radius gets smaller; i.e., dTc/dR < 0 turns to dTc/dR > 0. Using the equations above for R (part b) and η (part c), find the critical value of η at which this transition occurs, and show that Tc,max = 5×10 M
M 7 4/3 K [NOTE: this involves solving a transcendental equation of η, which is most easily solved with a computer] Rearranging the result from (c), and substituting our expression for R/R0 from (b), we get: Tc = 3 × 108 η M
M 4/3 f (η)−2 The core temperature hits a minimum when dTc
dTc dη
=
=0
dR
dη dR = (f (η)−2 − 2ηf (η)−3 df
2
)( η)
dη R0 η2
2η(1 + η) − η 2
2η
1+η+
− 2η 1 +
=
R0 f (η)3
1+η
(1 + η)2 ignoting the trivial case where η = 0 (zero‐temperature gas) or f(η) = ∞ (requires η = ‐1, which is not possible) then this derivative is equal to 0 when 2η(1 + η) − η 2
η2
= 2η 1 +
1+η+
1+η
(1 + η)2 after a healthy dose of algebraic manipulation: 2η 3 + 4...
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This document was uploaded on 02/26/2014 for the course PHYS 160 at UCSD.
 Fall '08
 Norman,M
 Physics, Work

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