Theoretical Statistical Physics

Transcription

1 Janosh Riebesell, Adrian van Kan Lecturer: Manfred Salmhofer December nd, 06 Theoretical Statistical Physics Solution to Exercise Sheet 5 Ideal gas work (3 oints Within the kinetic model of an ideal gas, show that the work done to the gas when changing the volume is dv. Kinetic theory traces the macroscoic henomenon of ressure on a surface back to a constant bombardment by microscoic articles, each of which obeys Newton s laws of motion. Uon imact, a tiny amount of momentum is transferred onto the surface. The resulting average force can be calculated exlicitly by considering a simle toy model, a cubic box of length L containing N articles, each of mass m. We assume that a article travelling with momentum v x in the x-direction bounces off a wall erfectly elastically so that it returns with velocity v x. The resulting momentum transfer is x = mv x. Since the article is traed in a box, it will again hit the same wall after t = L/v x. The force due to this single article is thus F = x t = mv x L. ( Summing u the contributions from all N articles in the container, the total average force is F = N m v x. ( L v x is the square of the velocity in x-direction averaged over all articles. The x-direction is in no way distinguished from y or z, meaning v x = v /3. Thus the differential work required to imress one of the container s walls by a distance dx is δw = F dx = N m v 3L dx = N E kin 3L 3 L dx = N E kin 3V dv. (3 The sign above stems from the fact that if dv < 0, we need to exert a force to squeeze the box, thereby increasing its energy, whereas for dv > 0, the system itself is doing the work, thus decreasing its energy. Inserting E kin = 3 k BT and the ideal gas law V = N k B T, we get δw = N k BT V dv = dv. (4 Density of states ( oints Consider a system of N identical, uncouled quantum mechanical oscillators. Comute the number of states at a given total energy of the system. A quantum harmonic oscillator features the well-known ladder of equidistant energy states E n = ( n+ ω, with n N0. (5 For N identical oscillators, we can thus immediately write down the ground state energy as E min = N ω. Since this energy is attained only by a single state n i = 0 i {,...,N}, the number of microstates with energy E min is Ω(E min =.

2 At the first excited level E min + ω, we have one energy quantum to allocate. We could use it to excite any of the N oscillators, so the number of states increases to Ω(E min + ω = N. (6 At E min + ω, we have quanta to distribute. Either we give both quanta to one oscillator for which there are again N ossibilities, or to two different oscillators, resulting in N(N ossibilities. However, order doesn t matter since first giving a quantum to oscillator i followed by exciting oscillator j results in the same state as doing it the other way round. We therefore have to halve the number of states resulting from the second configuration. In total, this gives Ω(E min + ω = N + N (N = N (N +. (7 The counting roblem we are dealing with is simly that of how many ways we can distribute m identical quanta amongst N oscillators? The answer is rovided by the binomial coefficient, ( N +m (N +m! Ω m = Ω(E m = = m m!(n!, (8 where E m = E min + m ω = ( N + m ω. For m {0,,,3,4}, we thus get the following numbers of states. m Ω m N N (N + N 6 (N +(N + N 4 (N +(N +(N +3 Now that we have the number of states at a given energy, it is a trivial matter to derive the entroy S m of N oscillators with total energy E m. Using Stirlings aroximation for large factorials, ln(n! = n ln(n n+o(lnn, we get S m = k B ln(ω m = k B (ln[(n +m!] ln(m! ln[(n!] k B ((N +m ln(n +m mln(m (N ln(n (9 k B ((N +mln(n +m mln(m N ln(n = k B (N ln ( N+m N +mln ( N+m m. 3 Stationary distribution ( oints Consider the Boltzmann equation with external force F(x = x V(x. Find the stationary distribution f 0 (x,. The Boltzmann equation describes the dynamical evolution of hase sace densities for systems with a large number of constituents such as a gas. It is an integro-differential equation whose significance derives from its ability to describe out-of-equilibrium rocesses. It reads ( t + m x +F f(x,,t = d 3 kd 3 d 3 k,k T,k [ ] f f k f f k. (0 The above formulation already incororates the Stosszahlansatz, also known as molecular chaos, which assumes that the collision term results solely from two-body collisions between articles that are uncorrelated rior to the collision. This was the key assumtion by Boltzmann, as it Boltzmann assumed that the influence of the external force F on the collision rate is negligible to derive (0. Molecular chaos can also intuitively be interreted as the assumtion that velocity and osition of a constituent article are uncorrelated.

3 allows to write the collision term as a momentum-sace integral in which the two-article correlator F(x,,k,t factorizes into two one-article distribution functions f(x,,tf(x,k,t. The term [ f f k f f k ] in(0isashorthandnotationfor [ f(x,,tf(x,k,t f(x,,tf(x,k,t ]. For a stationary system, the Boltzmann equation greatly simlifies in two ways. On the one hand, the article distribution loses its exlicit time-deendence, f(x,, t. On the other hand, stationarity imlies that the Boltzmann H-function must be time-indeendent, since its timedeendence derives exclusively from f(x,, t, H(t = d 3 x d 3 f(x,,tln[f(x,,t]. ( A stationary H results in a condition known as detailed balance(see lecture notes from November, in which the number of articles leaving a certain mode due to a given scattering rocess is exactly equal to the number of articles entering that mode by the reverse rocess. Concetually: k k = k k Under these circumstances, the loss and gain terms f f k and f f k in (0 exactly cancel, meaning the r.h.s. of the Boltzmann equation vanishes. We are left with m xf 0 (x, = F(x f 0 (x, = x V(x f 0 (x,. This artial differential equation is solved by the ansatz ( β f 0 (x, = α ex m ( 0 +γv(x +δ. (3 Reinserting (3 into ( gives ( m γ xv(x = x V(x β m ( 0, (4 from which we infer β = γ and 0 = ( 0. Moreover, normalizability of the hase sace density requires δ = 0. Thus, f 0 (x, = αe β m. +V(x For α = ( d ( m d d d xe βv(x, β = πk B T k B T, (5 this is recisely the Maxwell-Boltzmann distribution in d dimensions. 4 Pressure on a wall (3 oints Comute the ressure of an ideal gas in three dimensions uon a wall at x = 0 that attracts molecules at large distance and reels them at smaller distance. Let the force be given by the otential U(x = Ae αx +Be αx, (6 with A,B > 0. Consider searately the cases where the range of the force is a small comared to the mean free ath l, and b comarable to it. 3

4 U(x α l α l x The energy of a article in the vicinity of the wall where U(x 0 is E(x,ẋ = m ẋ +U(x. (7 Energy must be conserved during collisions with the wall. Since the otential deends only on x (rather than x, the transverse energy E t = m (ẏ +ż is searately conserved from the normal contribution E n = E E t = m ẋ +U(x. (8 We can solve the latter for the velocity in x-direction, ẋ(x = ± m[ En U(x ]. (9 The ressure on the wall is determined by the total momentum transfer from all article collisions. If a single article encounters the wall at time t 0, its change in momentum is x = x (t 0 +τ x (t 0 τ = m [ ẋ(t 0 +τ ẋ(t 0 τ ], (0 where τ = l/ v x is the characteristic scattering time inversely roortional to the average velocity in x-direction v x = E n /m. a In the weak scattering case where the range of the force /α is much smaller than the mean free ath l, the velocity ẋ(t 0 ±τ ẋ(l in (0 will be evaluated at a distance l from the wall. This is because x(t 0 = 0 and the article moves towards/away from the wall with v x carrying it to a distance of aroximately v x τ = l within the scattering time τ. At x l, the otential becomes negligible. Inserting (9 into (0 for U(l 0 gives x,a = me n ( b In the strong scattering case, the scattering time τ = l/ v x is much shorter and the mean free ath decreases, becoming of the order of the range of the force α l. To comute the momentum transfer, the velocity will now be evaluated at a shorter distance l from the wall where the otential still exerts a significant attraction on the article, F x = x U(l < 0. This increases the momentum transfer onto the wall and thus the ressure, assuming B A x,b = m [ E n U(x ] m ( E n +Ae αx > me n = x,a. ( To get a more quantitative result, rather than this rough aroximation, we can searate variables in (9 to get dx [ ] = dt. (3 ± m En U(x 4

5 The solution to this differential equation is x(t = [ α ln ξcosh[α v x (t t 0 ] A ] ( B, where ξ = + A. (4 E n E n 4En Differentiating (4 w.r.t. time results in the velocity ẋ(t = sinh[α v x (t t 0 ] cosh[α v x (t t 0 ] A E nξ v x, (5 and the momentum transfer x,b = m [ ẋ(t 0 +τ ẋ(t 0 τ ] [ ] sinh[α v x τ] sinh[ α v x τ] = m v x cosh[α v x τ] A E nξ cosh[ α v x τ] A E nξ sinh[α v x τ] = x,a cosh[α v x τ] A, E nξ (6 where we used sinh( x = sinh(x and cosh( x = cosh(x. Since α v x τ = αl, we can aroximate this as ( x,b = x,a + A e α vxτ. (7 E n ξ Again, this is larger than the momentum transfer we obtained in the weak scattering case, resulting in an increased ressure on the wall. 5