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Unformatted text preview: Quiz 3 for Physics 176: Answers Professor Greenside Tuesday, February 23, 2010 The following answers are far more detailed than what was required to get full credit on any of the prob lems. But I hope the details will help you learn the thermal physics and appreciate some of the interesting implications of some of these problems. 1. In a recent homework assignment, you showed (up to a constant factor) that the entropy S of a black hole of mass M is given by S = 8 π 2 k GM 2 hc , (1) where, to the nearest power of ten (zero significant digits): k ≈ 10 23 J / K is Boltzmann’s constant, G ≈ 10 10 N m 2 / kg 2 is the gravitational constant, h ≈ 10 33 J · s is Planck’s constant, and c ≈ 10 8 m / s is the speed of light. (a) (7 points) Given that the energy of a black hole is its relativistic rest mass U = Mc 2 , derive a formula for the temperature T of the black hole in terms of its mass. Answer: The usual procedure is to express the entropy S = S ( U ) as a function of U and then differentiate to get the temperature via 1 /T = dS/dU . Substituting M = U/c 2 into the expression for the black hole entropy, we have: S = 8 π 2 k G hc 5 U 2 . (2) We then differentiate w.r.t. U and replace U by Mc 2 to get the temperature in terms of the mass: T = hc 3 16 π 2 kGM (3) You can check that this answer is reasonable by verifying that the physical units cancel to yield units of kelvin, which they indeed do. (Independently, one could use dimensional analysis to guess this formula with the assumption that T depends on h , c , G , and M , but then you would not know the constant in front of the expression.) Since T ∝ M 1 , the more massive a black hole, the colder it is. You can learn more about Eq. (3), called the Hawking radiation temperature, from the Wikipedia article with title “Hawking Radiation”. (b) (3 points) Estimate to the nearest power of ten the temperature T of a black hole whose mass is one solar mass, M ≈ 10 30 kg. Answer: Formulas rarely mean anything until you have a sense of the magnitude involved, so this simple problem is an important one in terms of gaining some intuition. Substituting the given approximate values of physical parameters and the mass of the Sun, we find: T ≈ 10 33 × (10 8 ) 3 16 × π 2 × 10 23 × 10 10 × 10 30 (4) ≈ 10 33+24+23+10 30 10 × 10 (5) ≈ 10 8 K . (6) Here I approximated 16 by 10 to the nearest power of 10 and I approximated π 2 ≈ 3 2 ≈ 10 to the nearest power of ten. 1 We see that a stellar black hole is a cold object, far colder than the cosmic microwave background radiation of photons left over from the Big Bang, which has a temperature of about 3 K and which you can think of as the temperature of outer space, far from stars. The Milky Way has a ≈ 4 , 000 , 000 solar mass black hole at its center that is much colder still....
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 Spring '08
 Behringer
 Physics, Thermodynamics, Black Holes, Entropy, Black hole

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