Physics 161 Homework 3 - Solutions Wednesday September 21, 2011

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1 Physics 161 Homework 3 - Solutions Wednesday September 21, 2011 ake sure your name is on every page, and please box your final answer. Because we will be giving partial credit, be sure to attempt all the problems, even if you don t finish them. The homework is due at the beginning of class on Wednesday, September 28th. Because the solutions will be posted immediately after class, no late homeworks can be accepted! You are welcome to ask questions during the discussion session or during office hours. 1. The acceleration g of a freely falling body near a Schwarzschild black hole is given by the formula G N /r 2 g = 1 2G N 1/2, when at a radius where 2πr is the circumference. a Show that far from the event horizon, the above formula reduces to the usual Newtonian expression. Show also that when r = R S, the gravitational force becomes infinite, and therefore irresistible. b An infinite quantity is not very convenient to work with; hence, it is convenient to define the surface gravity which is the numerator of the proceeding formula, evaluated at the event horizon. Determine this value. A similar quantity can be defined for rotating black holes, which do not possess complete angular symmetry, but the surface gravity also works out to a constant depending only on the mass and angular momentum J. c Let t be the time interval elapsed for an observer at infinity. Using the Schwarzschild metric, find the time elapsed for a fixed local observer at radial distance r. Show that a local observer far from the black hole ages at the same rate as the observer at infinity, and that the former s aging rate stops in comparison with the latter s when the event horizon is approached. d A particle falling with a parabolic velocity zero binding energy takes the time observed by the falling particle [ t = 2 ] 1/2 r r R S 3 c R S c to fall from a radius r to the event horizon. Calculate t numerically. if R S = 3 km, and r = 10 km. Note that your results from part c say that an observer at infinity thinks that it takes an infinite time for the particle to reach the event horizon.

2 Solution a When the radius r, then g = G N r 1 2G N 1/2 2 c 4 r G N r 1 + G N 2 c 4 r = G N + G r 2 N 2 c 4 r 3 = G N, r 2 where we have expanded the second term in a binomial series, and neglected the term r 3, since we are taking the r limit. b The surface gravity is g S = G N /r 2 where r is evaluated at the event horizon, c 2 2 g S = G N = c4 2G N 4G N. c From the Schwarzschild metric, 1 2G N c 2 dt 2 + ds 2 dr 2 = 1 2G N + r 2 dω, where t represents the time that passes for an observer at infinity. Now, for a fixed position, when dr = dω = 0, then ds 2 = 1 2G N c 2 dt 2, from which the proper time is dτ 2 = ds 2 /c 2, and so dτ 2 = 1 2G N dt 2. Taking the square root and integrating noting that r =constant we find t = 1 2G 1/2 N t c 2. r Far from the black hole i.e., when r, then t t, which means that the local observer will measure the same amount of time as an observer at infinity. When r = R S, then t = 0, meaning that the observer at the event horizon appears to be frozen in time to an observer at infinity. d Just plugging in for the numbers gives [ ] 1/2 t = 2 r r 3 c R S R S c [ ] 1/2 = 2 r r 3 c R S R S r [ = ] 1/ = seconds, or 34 µs, which is a very short time!

3 2. You showed that the surface gravity g S of a nonrotating black hole is inversely proportional to the mass. Given that the surface area is proportional to 2, show that the infinitesimal change da is proportional to d. In this manner, show that g S da d. On the other hand, the first law of thermodynamics states T ds = dq, where T is the temperature, S is the entropy, and Q the heat. Summarize in words how these two statements are related. Hint: recall that E = mc 2, and note the results of the next problem. What is the entropy of the black hole in terms of the surface area? The surface gravity is Solution g S = c4 4G N. The surface area of the black hole at the event horizon is A = 4πR 2 S, or Thus, such that A = 4πRS 2 2GN = 4π c 2 da = 32πG2 N c 4 2 = 16πG2 N c 4 2. d, g S da = 8πG N d d = gs 8πG N da. Now, because E = c 2, then de = dc 2, and from the first law of thermodynamics then dq = de = T ds, we see that d = T c 2 ds = g S 8πG N da ds = g Sc 2 8πG N T da. Now, plugging in for the temperature from problem 3 and the surface gravity, we find ds = g Sc 6 8πG N T da = c4 4G N k B c da = k3 B 3 4G N da. The expression in front of the da is a constant, and so we find that the entropy is proportional to the surface area of the black hole, kb c 3 S = A. 4G N

4 3. Stephen Hawking found that black holes emit radiation as though they were a hot object a blackbody at a certain temperature, T H, called the Hawking Temperature. The exact derivation of this temperature is complicated, needing quantum field theory and the full apparatus of General Relativity, but a heuristic derivation can be found as follows. The Uncertainty Principle allows vacuum fluctuations to produce particles and antiparticles of energy E for a time t such that E t /2. If the maximum distance that a virtual pair can separate is comparable to half the circumference of the event horizon, c t/2 π2g N /c 2, then one of the pair has a reasonable chance of escaping to infinity while its partner falls into the black hole. If the energy E that materializes at infinity is characteristic of a thermal distribution, E k B T, determine the temperature of the black hole. Notice, of course, that we have deliberately fine-tuned our argument unjustifiably to obtain the correct numerical coefficient. Use your expression to compute the temperature, T, when is one solar mass, and when = grams, the mass of a mini-black hole. What is the size of the Schwarzschild radius for = grams? Observationally, there seem to be few, if any, such mini-black holes in the Universe. Solution The Uncertainty Principle says that E t /2, while t 4πG N. Thus, E c 3 = c3, and since E k 2 t 8πG N BT, then T = c 3 8πG N k B. This is the Hawking temperature. It can be rewritten in a form useful for calculations by writing c 3 T = 8πG N k B = c 3 S, 8πG N k B S where S is the solar mass. Now, the factor in front of the parenthesis is a constant, and so c 3 S T = = S Kelvins, 8πG N k B S after plugging in for all the constants. Thus, the temperature of a one solar-mass black hole = S is Kelvins, while for a mass of = grams, then S / = /10 15 = 10 18, and so T = = Kelvins, which is very much hotter! The temperature depends inversely on mass, and so the smaller the black hole gets, the hotter it gets!

5 4. It might be thought that if a black hole has temperature and radiates, losing surface area and therefore entropy, this would violate the second law of thermodynamics. In fact, as Bekenstein has explicitly shown, the entropy lost by the black hole is more than made up by the entropy produced in the thermal radiation. The second law of thermodynamics, stated in its most general form, is thus still satisfied, since the total entropy of the Universe a black hole plus the outside, has increased in the evaporation process. Ignore particulate emission, and suppose for simplicity that a black hole radiates only electromagnetic energy at a rate L = 4R 2 SσT 4, where R S is the Schwarzschild radius, T is the temperature of the black hole which you found in a previous problem, and σ is the Stefan-Boltzmann constant, σ = π c. 2 a Show now that the energy-loss rate can be written as L = k 4 B c π 2 G 2 N 2, which demonstrates that a black hole would not radiate if h were zero, and that low-mass black holes emit more quantum radiation than high-mass black holes. Compute L numerically for = grams. In what part of the spectrum is this radiation mostly emitted? Hint: Look up Wein s displacement law. b The time-rate of decrease of the mass-energy content, c 2, of an isolated black hole must equal the above expression for L. Argue thereby that c 2 /L gives roughly the time t required for a black hole of mass to evaporate completely. Because the evaporation process accelerates as decreases, calculus makes the actual time shorter by a factor of 3. Show now using calculus that t = 2560π 2 2GN c 2 2 Calculate t for = grams. Convert your answer to billions of years. Draw appropriate astrophysical conclusions. Solution a We have that L = 4RS 2 σt 4, where R S problem, T = c 3 /8πG N k B. Thus, L = 4RS 2 σt 4 = 4 2G N 2 π 2 k 4 c 2 B 60 3 c 2 = 16G2 N 2 π 2 k 4 B 4 c c 6 8π 4 G 4 N k4 B 4 = c π 2 G N 2,. = 2G N /c 2, while, from the previous 4 c 3 8πG N k B

6 as expected. Once again, we can rewrite this in terms of a solar mass, c 6 L = 30720π 2 G N = c 6 2 S π 2 G N S 2 Now, the first term, not involving is a pure constant, and evaluating it gives c π 2 G N 2 S = Watts, which means that the luminosity of a solar-mass black hole is precisely this value. So, the luminosity of a black hole with mass is 2 L = S Watts. Now, for a black hole of mass = grams, then 2 S = = , and so L = = 57 megawatts. So, the luminosity of a kg black hole is 57 W. The temperature of this black hole is c 3 T H = 8πG N k B = c 3 S = S Kelvins. 8πG N k B S Thus, plugging in for the mass = we find that the temperature is 2 10 T H = = Kelvins Now, Wein s displacement law says that the peak wavelength of a blackbody at temperature T is given by λ max = b T, where b = nm K, which gives, finally λ max = which is in the gamma ray region = nm, b The luminosity is L = d dt c2, where the minus sign is because the mass is decreasing. Thus, d c 4 = dt 30720π 2 G 2 N. 2

7 Now, integrating this expression from a mass at time t = 0 to a final mass of 0 at time t gives 3 3 = c 4 t π 2 G 2 N Solving for the lifetime gives 2 t = 2560π 2 2GN c 2. Finally, plugging in for = grams we find t = 2560π 2 2G N 2 c 2 2 = 2560π = seconds, which is the same as 13.4 billion years. Black holes of this small mass are expected to possibly be made during the Big Bang. If this is the case, then we should expect these primordial black holes to be decaying today.

8 5. Suppose that an evaporating black hole reaches a mass m such that its Compton wavelength h/mc, the scale on which it is fuzzy becomes comparable to its Schwarzschild radius, 2G N m/c 2. Show that the argument of Problem 3 must break down, because any particle or antiparticle emitted would have E = mc 2, comparable to the massenergy contained by the black hole as a whole no pun intended. Except for factors of order unity, show that the value of m when this happens is given by the Planck mass, c m Pl =. G N Compute m Pl numerically, and compare it with the mass of the proton, m p. To discuss black holes of mass m Pl or less requires a quantum theory of gravity. Such a theory does not yet exist; general relativity is a classical theory of the gravitational field the structure of spacetime. Hawking s discussion of evaporating black holes is therefore semiclassical, in that it treats particles quantum-mechanically, but the gravitational field spacetime classically. Solution Once again, we rely on the Uncertainty Principle, which says that E t /2, such that E /2 t. Now, we have c t/2 π2g N m/c 2, but at this stage of the evaporation, we have assumed that 2G N m/c 2 h/mc, and so t h/mc 2 = 2π /mc 2. Thus, the energy is t 2 t = 2 2π /mc 2 mc2. Now, the change in energy is E = mc 2, and so E = mc 2 mc 2, meaning that the change in mass of the black hole is of the same order as the mass, itself! This means that these fluctuations of energy are not small, and the idea of a long-term evaporation breaks down and the final stages are essentially an explosion! The black hole mass at which this occurs is h/mc 2G N m/c 2, or m = hc/2g N c/gn, which is the Planck mass. Plugging in the values gives m Pl = c G N = g. The mass of a proton is m p = grams, and so m Pl /m p , and since the mass of a proton is about 1 GeV, then m Pl GeV.