2 Problems 1. The Minkowski metric is written as nur, and is a real, symmetric metric. How many independent components does it have? A general metric for a curved spacetime is written as guy; how many independent components does this metric have? 2. Show that the interval with the Minkowski metric ds? = -dt2 + dx2 + dy?+ dz?, can be transformed into the interval with the Rindler metric ds? = -a?X?dT" + dx2+ dy2+ dz2 with the following coordinate transformation, t = X cosh (aT) I = X sinh (aT) y = Y Z = Z. The Rindler metric describes a constantly accelerating observer. 3. The acceleration g of a freely falling body near a Schwarzschild black hole is given by the formula GNM/72 9 = (1 - 2GNM )1/2' when at a radius where 2ar 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 = Rs, 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 M and angular momentum J). (c) Let At 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 compar- ison 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) 1/2 At = Rs Rs 2to fall from a "radius" r to the event horizon. Calculate At numerically. if Rs = 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). 4. In the previous problem, you showed that the "surface gravity" gs of a nonrotating black hole is inversely proportional to the mass M. Given that the surface area is proportional to M', show that the infinitesimal change dA is proportional to MdM. In this manner, show that gadA o dM. On the other hand, the first law of thermo- dynamics states TdS = 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, and note the results of the next problem. What is the entropy of the black hole in terms of the surface area? 5. Stephen Hawking found that black holes emit radiation as though they were a hot object (a blackbody) at a certain temperature, Ty, called the "Hawking Tempera- ture." 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 AE for a time At such that AEAt ~ h/2. If the maximum distance that a virtual pair can separate is comparable to half the cir- cumference of the event horizon, cAt/2 ~ #2GM/c', then one of the pair has a reasonable chance of escaping to infinity while its partner falls into the black hole. If the energy AE that materializes at infinity is characteristic of a thermal distribution, AE ~ kBT, determine the temperature of the black hole. (Notice, of course, that we have deliberately "fine-tuned" our argument unjustifiably to obtain the correct numer- ical coefficient.) Use your expression to compute the temperature, T, when M is one solar mass, and when M = 10% grams, the mass of a mini-black hole. What is the size of the Schwarzschild radius for M = 10 grams? Observationally, there seem to be few, if any, such mini-black holes in the Universe. 6. 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 = AROT where Rs is the Schwarzschild radius, T is the temperature of the black hole which you found in a previous problem, and o is the Stefan-Boltzmann constant, 0 = 60 h3c2'(a) Show now that the energy-loss rate can be written as L = ; 3072072GRM2 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 M = 10 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, Mc', of an isolated black hole must equal the above expression for L. Argue thereby that Mc?/L gives roughly the time t required for a black hole of mass M to evaporate completely. Because the evaporation process accelerates as M decreases, calculus makes the actual time shorter by a factor of 3. Show now (using calculus) that t = 256012 ( 2GNMI ) (M) Calculate t for M = 2 x 10 grams. Convert your answer to billions of years. Draw appropriate astrophysical conclusions. 7. Suppose that an evaporating black hole reaches a mass m such that its Compton wave- length h/me, the scale on which it is "fuzzy" becomes comparable to its Schwarzschild radius, 2Gym/c'. Show that the argument of the previous problem must break down, because any particle or antiparticle emitted would have E = Ame', comparable to the mass-energy 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, he MPI = IGN Compute Mp numerically, and compare it with the mass of the proton, my- To discuss black holes of mass Mp 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.8. Show that the current proper distance to our particle horizon, defined as the most distant place we can see (in principle), for a matter-dominated k = 0 Universe with no cosmological constant, is rado = 3cto, where ry is the comoving radial coordinate of the particle horizon, an is the scale factor today, and to is the present age of the Universe. Thus, more and more distant regions of the Universe "enter the horizon" and become visible as time progresses. Why is the answer different from the naively expected result, cto? Hint: Light moves along null geodesics, defined as paths for which ds = 0, and there- fore, in the FRW metric, light reaching us from a comoving coordinate r will obey dr2 0 = o'dt2 - a" (t) _ kr2 Replace a(t) with do () , appropriate for a matter-dominated cosmology, separate the variables, and integrate from r = 0 at t = 0 (the Big Bang), and ry at t = to (today). 9. (a) For a k = 0 Universe with ?A = 1, that at t = 0 already has a scale do, show that the exact solution to the Friedmann equations for these cosmological parameters is a(t) = do exp (Hot) . (b) Find the comoving radial coordinate, "#, of galaxies that will be on the particle horizon (see the previous problem) at a time t in the future. Show that in this case TH approaches a constant, c/co Ho, and therefore galaxies beyond this ry will never become visible. Hint: Proceed as in the previous problem, but now with a(t) = do exp (Hot). 10. (a) For the same cosmology as in the previous problem (k = 0 and 2A = 1), find the comoving radius TEH of galaxies that will be on our event horizon at a time t in the future; i.e., galaxies with which we will be unable to communicate. In other words, light signals sent by us at time t will never reach those galaxies, light signals sent out by those galaxies at time t will never reach us, and therefore we will never see those galaxies as they appeared at time t and thereafter. Show that, in this case, TEH shrinks exponentially, and we thus lose the possibility of communication with more and more of our neighbors. (b) Assume that Ho = 70 km/s/Mpc, and find, for such a Universe (which approx- imates the actual world we live in), within how many years will the galaxies in the nearby Virgo cluster (distance ~ 15 Mpc) reach the event horizon. Hint: Proceed as in the previous problem, but integrate from r = 0 at future emission time t to ren at t = co (the photons never reach us). Then equate TEH to the comoving radius of Virgo, 15 Mpc/ag