763620SS STATISTICAL PHYSICS Solutions 8 Autumn 2015

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Warren Harrison
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1 76360SS STATISTICAL PHYSICS Solutions 8 Autumn 05. Thermodynamics of a Black Hole The energy and entropy for a non-rotating black hole of mass M can be defined as E = Mc S = k BA, 8πl P where A = 4πR is the surface area of the black hole, with R = GM/c the Schwarzschild radius, and l P = G/c 3 is the Planck length, and the gravitation constant G = Nm /(kg). Calculate the temperature of the black hole as a function of mass using thermodynamical identities. Black hole radiates like an (almost) perfect black body (why?). Calculate the radiation power P and the timescale associated with the loss of mass, τ = E/P. Calculate the temperature T and lifetime τ in years for a black hole of mass 0 kg. Currently the Universe is filled with radiation of T.7 K, as a remnant of the Big Bang. How does this affect the development of the above black hole (qualitatively). Solution: We can find the temperature of the black hole by calculating ( ) S(E, V, N) T =. E We can easily express S in terms of E by plugging in the area of black hole, then its radius in terms of its mass, and finally setting M = E/c : V N S = k B 4πR = 8πl P k ( ) B GM 4π = k ( ) B GE 4π = k BG E. 8πl P c 8πl P c 4 c 8 l P Temperature turns out to be inversely proportional to E and consequently to M as well: T = 4k BG c 8 l P E = 4k BG M T = c6 l P c 6 l P 4k B G M = c3 4Gk B M.

2 Obviously, a black hole is a perfect absorber (how it can emit particles is a longer story!). The radiation power can be found using the formula, derived elsewhere (Problem 7.5): P = 4πR σt 4. One merely plugs in R, T and σ = π kb 4 as a function of M to get the power and lifetime 60 3 c τ: ( ) ( ) GM c 3 4 P = 4π σ = c6 π 3 c 4Gk B M 960G M, τ = 960G π 3 c 4 M 3. More massive the black hole is, the longer it will be around. Fixing the mass, M = 0 kg, we get some definite values for T, P, and τ: T = K, P = W, τ = a. Although the cosmic microwave background radiation will slow down the evaporation of this black hole, it will only get hotter as its mass radiates away. Should it have been more (much more) massive, it could have become cool enough to reach thermal equilibrium with the CMB and thus survive much longer.. Gregarious Bosons Consider a Bose-Einstein condensate with all N bosons in the single-particle ground state ψ 0 Ψ(r, r,..., r N ) = ψ 0 (r ) ψ(r N ). Suppose a new particle is gently injected into the system, into an equal superposition of the M lowest single-particle states φ(r ) = M ( ψ0 (r ) + ψ (r ) + + ψ (r ) ). The state Φ(r,..., r ) of the N + -boson system after the insertion is obtained by symmetrizing the product function Ψ(r, r,..., r N )φ(r ). Calculate the symmetrized initial state of the system with the injected particle. Show that the probability that the new boson enters the ground state is enhanced by a factor N + of its probability to enter the ground state in an empty space. (Hint: First do it for N =.)

3 Solution: To symmetrize the given wavefunction, we simply sum over all the possible permutations of the parameters {r i }. Denoting by P (s) all the permutations of a sequence s, the symmetrized Φ reads: Φ(r,..., r ) Ψ(r p, r p,..., r pn )φ(r p ), p P (,...,) where N is a normalization factor (which will magically absorb any constant coefficients), to be determined later. For N = the above becomes: Φ(r, r ) Ψ(r )φ(r ) + Ψ(r )φ(r ) ψ 0 (r ) ψ m (r ) + ψ 0 (r ) ψ m (r ) M M m=0 ψ 0 (r )ψ 0 (r ) + ψ 0 (r ) m=0 ψ m (r ) + ψ 0 (r )ψ 0 (r ) + ψ 0 (r ) ψ 0 (r )ψ 0 (r ) + ψ 0 (r ) ψ m (r ) + ψ 0 (r ) ψ m (r ). ψ m (r ) Note now that the terms appearing in the last line above are all mutually orthogonal, and that only the first term (appearing with twice the amplitude of the other ones) has all the particles in the ground state. Normalization constant is then N = + (M ), and the probability of observing the new particle in the ground state is P (All at ground state) = N. The asked ratio of ground state to excited state probabilities is P (All at ground state) P (Not all at ground state) = P (All at ground state) P (All at ground state) = (M ) 4 = M. Since the probability of finding φ(r) = m ψ m(r)/ M in the ground state is clearly /M, the above ratio is simply /(M ). We see that in the symmetrized case, that probability is + times higher. So does this hold for general N? One can pretty much now guess the form of the symmetrized Φ from what we got above: adding one more particle into the ground state will increase the amplitude of the ground state by one, and introduce one more M term 3

4 into Φ, just with different permutation of r,,3. But let us for completeness go through the above steps for general N: Φ(r,..., r ) p P (,...,) Ψ(r p, r p,..., r pn )φ(r p ), p P (,...,k,k+,...,) Ψ(r p, r p,..., r pn ) φ(r k ), N!Ψ(r,..., r k, r k+,..., r )φ(r k ) Ψ(r,..., r k, r k+,..., r ) ψ m (r k ) M m=0 Single out the p index; rename it to say k and sum it separately, taking care to remove the k term from the permutation p summation. Since Ψ(r,..., r N ) is already symmetric, each term in the permutation summation is just the same; replace the sum by the factor N!Ψ(r,..., r k, r k+,..., r ). Next just remember that Ψ is simply a product of ground state wave functions: Φ(r,..., r ) Ψ(r,..., r k, r k+,..., r )ψ 0 (r k ) }{{} Ψ(r,...,r ) (N + )Ψ(r,..., r ) + + Ψ(r,..., r k, r k+,..., r ) ψ 0 (r ) ψ 0 (r k )ψ m (r k )ψ 0 (r k+ ) ψ 0 (r ) ψ m (r k ) Again, we have the ground state term appearing with the amplitude N +, and the rest (N + )(M ) terms with amplitudes. In each term a single-particle state with a different r k is raised to a different excited state m. Therefore, all terms are mutually orthogonal and so the normalization is N = (N + ) + (N + )(M ) = (N + )(N + M). The ratio of ground state to excited state probabilities is then P (All at ground state) P (All at ground state) = (N + ) (N + )(N + M) (N + ) + M. Indeed. This is again N + times larger than the value for φ. The preferrability of the ground state increases by a factor of N + after the new particle is injected into the condensate, compared to it being in empty space. 4

5 3. Bohr-van Leeuwen Theorem Consider a system of charged particles (not dipoles), obeying classical mechanics and classical statistics. Show that the magnetic susceptibility of this system is identically zero. Solution: Suppose B = Bẑ, and use the Landau gauge to express the vector potential A, A = Bxŷ. The Hamiltonian (expressed in canonical momenta that include the vector potential) reads H = P x + (P y qbx) + Pz + eφ(x, y, z) m where φ is the scalar potential. Magnetization can again be evaluated using the formula M = ln Z, β B Z = Z N, Z = e βh d 3 x d 3 P. It is quite easy to now see that Z is independent of B: Z = e βh d 3 x d 3 P = = { P exp β x + (P y qbx) + Pz m { exp β P x + P y + P z m } + eφ(x, y, z) } + eφ(x, y, z) d 3 x d 3 P d 3 x d 3 P Change of variable P y P y + qbx; integration bounds still... Notice that B has now disappeard from Z! Since Z / B = 0, must then the magnetization M, and consequently, the susceptibility be also zero. As a quick proof of the applicability of the our equation for the magnetization, note that we can write the Hamiltonian also in terms of the kinetic momenta p, and a potential associated with the external magnetic field: H = p m m B + V (r) where m = qr p/m is the magnetic moment of a point charge on a circular path. Upon inspection of ln Z / B, the formula for the magnetization M quickly follows. 4. Cluster Expansion Derive the following expressions for the virial coefficients in terms of cluster integrals b l : B = b = d 3 r e βu(r) B 3 = 4b b 3 B 4 = 0b 3 + 8b b 3 3b 4. 5

6 Calculate the nd virial coefficient for a gas of hard spheres (u(r < d) =, u(r d) = 0), and a gas with a hard core and attractive box -potential u(r < d) =, u(d r < R) = a, u(r R) = 0. Solution: We have the following two expansions: p = k B T b k λ 3k T e bkµ, ρ = k b k e bkµ, λ 3k T and we are to find the values of the coefficients B, B 3, and B 4 in Let us denote x = e bµ λ 3 T, and p = k B T (ρ + B ρ + B 3 ρ 3 + B 4 ρ 4 + ). p = k B T r, r = k b k x k, ρ = k kb k x k. We can simply set r ρ B ρ B 3 ρ 3 B 4 ρ 4 + = 0 and expand everything in powers of x, equate the coefficients of like powers, and from the resulting set of three equations solve the asked quantities. One only needs to expand up to the fourth power: b x + b x + b 3 x 3 + b 4 x 4 + = 0 b x b x 3b 3 x 3 4b 4 x 4 B b x + 4b b x 3 + (6b b 3 + 4b )x 4 + B 3 b 3 x 3 + 6b b x 4 + B 4 b 4 x 4 + = b B b x = 0. + b 3 4B b b B 3 b 3 x 3 + 3b 4 B (6b b 3 + 4b ) 6B 3 b b B 4 b 4 x 4 + 6

7 It is worthwhile to notice that b =. So, set the coefficients of x,3,4 all separately to zero and solve: = = b B = 0 b 3 4B b B 3 = 0 3b 4 B (6b 3 + 4b ) 6B 3 b B 4 = 0 B = b B 3 = 4b b 3 B 4 = 3b 4 + 8b b 3 0b 3 Let us calculate b next: b = U (r, r ) d 3 r d 3 r!v = e βu( r r ) d 3 r d 3 r V = V d 3 r } {{ } V e βu( r ) d 3 r = e βu(r) d 3 r. Set r r = r We also should take that a bit further: Since the potential only depends on r = r, we can do the integral over the angle part, giving for B : B = π e βu(r) r dr. Next, for the hard sphere potential, we find: And likewise for the box potential: B = π d 0 B = π d 0 = π 3 d3. r dr + π r dr R d e βa r dr = π 3 d3 + π e βa R 3 d 3 { = πd3 e βa ( ) } 3 R 3 d 7

ψ s a ˆn a s b ˆn b ψ Hint: Because the state is spherically symmetric the answer can depend only on the angle between the two directions.

ψ s a ˆn a s b ˆn b ψ Hint: Because the state is spherically symmetric the answer can depend only on the angle between the two directions. 1. Quantum Mechanics (Fall 2004) Two spin-half particles are in a state with total spin zero. Let ˆn a and ˆn b be unit vectors in two arbitrary directions. Calculate the expectation value of the product