Physics 212: Statistical mechanics II Lecture IV

Transcription

1 Physics 22: Statistical mechanics II Lecture IV Our program for kinetic theory in the last lecture and this lecture can be expressed in the following series of steps, from most exact and general to most specific and approximate (but also useful!): Liouville s theorem BBGKY hierarchy Boltzmann eqn. Navier-Stokes eqn. () The first step, from Liouville s theorem to the BBGKY hierarchy, will be completed in this lecture. The only assumption is that the system being dealt with is a gas with binary collisions, and the BBGKY hierarchy is still exact for this specific system. We will say a bit about the approximations required to get from the BBGKY hierarchy to the closed Boltzmann equation for f toward the end of this lecture, and start deriving the Navier-Stokes equation in the next lecture. Let s review what we learned last time. Suppose we have a problem of N particles moving in the Hamiltonian (switching back to the usual variables (r, p) rather than the generalized conjugate variables (q, p)) p 2 H = 2m + N U i + v ij i<j U i = U(r i ), v ij = v ji = v( r i r j ). (2) One of the main results from last lecture is Liouville s equation for the time evolution of g, the full N-particle distribution normalized to unity: Nd g t + ( ) g g ṗ i + ṙ i = 0. (3) p i r i For the specific case of the dilute gas with binary collisions, Liouville s equation becomes (where F = U, K ij = ri v ij ) g N t = F i K ij pi g p i g. (4) j=,j i Here g is the distribution function in N-particle phase space. For compactness, in most of this lecture we will suppress the time and sometimes other variables appearing in g, when the meaning is clear. Now rewrite Liouville s equation in the form [ t + h N(r, p,..., r N, p N )] g = 0, (5) where the differential operator h N is defined as h N (r, p,..., r N, p N ) = [ ] pi + F i pi

2 + 2 i,j= K ij ( pi pj ). (6) Our goal for most of this lecture will be to find an exact equation for the evolution of the n-body distribution functions f n by doing a partial integration of Liouville s equation. The definition of the -body distribution function f (z ), with z = (p, r), is f(z ) = δ(z z i ) = N g(z,..., z n )δ(z z ) dz... dz N = N g(z,..., z n ) dz 2... dz N. (7) Here the normalization factor N arose because the sum over i in the second form contains N identical terms. That is, the one-body distribution function f is a particle density (normalized to N), while the function g is normalized to over all of phase space. g differs from f N by a normalization factor N!. This choice of normalization may seem strange, but it will make the BBGKY equation derived below simpler. is Now the general definition of the n-body distribution function f n (z,..., z n ), with z = (p, r), f n (z,..., z n) = = j=,j i N! (N n)!... δ(z z i )δ(z 2 z j )... g(z,..., z n, z n+,..., z N ) dz n+... dz N, (8) where, as before for f, the combinatorial factor N!/(N n)! arises because each term in the sums over i, j,... contributes an equal result, given by the integral in the final form. A simple way to remember this normalization is that, just as f is normalized to give a particle density, f 2 is normalized to give a density of ordered pairs, f 3 of ordered triples, etc. as Now we show a surprising property of the differential operator h n, which is defined for n N Recall that h n (r, p,..., r n, p n ) = [ ] pi + F i pi + K ij ( pi pj ). (9) 2 i,j= ( ) t + h N g = 0. (0) Since the normalization of g is constant, integrating g t over all the space and momentum variables should give zero. Then, from (??), we also have h N (z,..., z N ) dz... dz N = 0. () This actually holds for all the h n for n < N as well: h n (z,..., z n ) dz... dz n = 0. (2) 2

3 To see this, go back to the definition of h n. It contains one set of terms which are spatial gradients of g multipled by space-independent terms, and another set of terms which are momentum gradients of g multipled by momentum-independent terms. To see that the first set vanish, consider holding the momentum variables constant and integrating the spatial variables. This integral over the gradient of g evaluates to g on the boundary of spatial volume, which for a bounded system we can take to be zero. Similarly the momentum-gradient terms can be argued to integrate to zero, under the assumption of a bounded system so boundary terms vanish. The operator h N can be broken down into terms containing only the first n particles, terms containing only the later N n particles, and terms which contain one particle from the first group and one particle from the second group: h N (z,..., z N ) = h n (z,..., z n ) + h N n (z n+,..., z N ) + j=n+ K ij ( pi pj ). (3) The factor of 2 in (??) disappears in the sum here because we no longer have both possibilities i < j and i > j. Now integrate out the last N n variables in the evolution of g: then the h N n part integrates to zero, by the lemma (??) shown above. Finally, t + h n(z,..., z n ) + j=n+ K ij ( pi pj ) g dz n+... dz N = 0. (4) The p j -gradient terms in the above integrate to zero for a bounded system as their coefficient is momentum-independent (same argument as above). Multiplying by N!/(N n)! then gives t + h n(z,..., z n )) f n (z,..., z n ) = N! (N n)! n j=n+ n K ij pi g dz n+... dz N N! = K i,n+ pi g dz n+... dz N (N n + )! n = K i,n+ pi f n+ (z,..., z n+ ) dz n+. (5) Here in passing from the first line to the second we used the fact that the sum over j contains N n equal terms, because of the integration. (We could not similarly get rid of the sum over i because the z i variables aren t integrated.) It is essential that the Hamiltonian contains only binary collisions (no three-body or higher interactions) so that the evolution of f at time t is determined by f and f 2 at that t, not by any higher f i. The expression (??) for the time evolution of f n in terms of f n+ is known as the BBGKY hierarchy. The key assumption that allowed us to write the hierarchy in this simple way was that collisions involve only two particles. Now we can say a bit about the physical meaning of the hierarchy. The first equation of the hierarchy is + F p f (t, z ) = K 2 p f 2 (t, z, z 2 ) dz 2. (6) 3

4 Here we have restored time as a variable in order to connect to the Boltzmann equation derived previously. The left side of the first hierarchy equation is just that of the Boltzmann equation, so somehow the right side must contain information about collisional processes that increase the entropy S of f. The second equation of the hierarchy is t + p + F p + p 2 m x 2 + F 2 p2 + ) 2 K 2 ( p p2 ) f 2 (t, z, z 2 ) = (K 3 p + K 23 p2 )f 3 (t, z, z 2, z 3 ) dz 3. (7) There is an important difference between the left sides of the first and second hierarchy equations. In the first hierarchy equation, there are no terms involving collisions, only streaming terms resulting from single-particle motion. These streaming terms, as seen before, cannot change the entropy, so important physics is lost if we neglect the collisional term on the right side of the first equation. In the second equation, however, there is already a collisional term on the left side. In fact, a simple estimate shows that for the dilute gas with λ = na 3, the left side collisional term K 2 dominates the right side collisional terms. Hence it is a good approximation for the dilute gas to truncate the hierarchy at the second equation, t + p + F p + p 2 m x 2 + F 2 p2 + ) 2 K 2 ( p p2 ) f 2 (t, z, z 2 ) = 0. (8) This is the first of several approximations involved in deriving the Boltzmann equation from the BBGKY hierarchy: it just amounts to neglect of any correlations beyond f 2. Some of the correlations in f 2 will also have to be neglected in order to obtain a closed equation for f. The second assumption will be that, on a relatively rapid time scale, f 2 equilibrates to be nearly a product distribution f 2 (t, z, z 2 ) f (t, z )f (t, z 2 ) except for rare collision events. Then f 2 continues to evolve in time, but its evolution is due to changes in f, rather than to changes in the pair correlations. Under this assumption f 2 t 0 and the second hierarchy equation can be used to simplify the first to + F p f (t, z ) = (p x + p 2 x2 )f 2 (t, z, z 2 ) dz 2. (9) m The remaining part of obtaining the Boltzmann equation requires working through the kinematics of collisions in some detail, and can be found in Huang and other textbooks (part of this may be an exercise on problem set 2). Instead we will focus in this lecture and the next on understanding more physical consequences of the Boltzmann equation. By asking the question of what distributions make zero the collisional term of the Boltzmann equation, we were led to the equation f f(t, r, p ) = exp(c 0 + c p + c 2 p 2 ). (20) Here and for the rest of this lecture we revert to the Boltzmann equation notation, where f, f 2, f, f 2 are shorthand for the one-particle distribution function evaluated at four different points in momentum space and one position. The above form for the distribution function came about from the 4

5 requirement that log f be a sum of quantities conserved in collisions: the collision term will vanish if for all incoming and outgoing momenta with we have f f 2 = f f 2, or p + p 2 = p + p 2, (2) log f(p ) + log f(p 2 ) = log f(p ) + log f(p 2). (22) Understanding conservation laws more generally will lead to the Navier-Stokes equation and other fundamental equations of hydrodynamics. We will then study a few examples of hydrodynamics such as Stokes law (used later for Brownian motion) and turbulence. Let be a conserved quantity like energy or momentum that moves with a particle, and define χ = χ(r, p ) as the value of χ for particle, etc. Then for any collision, χ + χ 2 = χ + χ 2. (23) We now show a local conservation law for such quantities: the collisional term in the Boltzmann equation does not change them, so ( ) f dp = 0. (24) t coll To show this, we make use of the same variable-switching tricks used in the proof of the H theorem in Lecture II to write the desired expression in a more symmetric form: ( ) f dp = t coll 4 w(f f 2 f f 2 )(χ + χ 2 χ χ 2) dp dp 2 dp dp 2 = 0. (25) Note that we did not need to require that the distribution f satisfied the Boltzmann transport equation. (In the derivation of the H theorem, we had log f in place of χ and so on.) Now integrating both sides of the Boltzmann equation over momentum (but not position) and using the above gives m x + F p f(t, r, p) dp = 0. (26) In the next lecture we will discuss physical interpretation of this law for conserved quantities such as energy and momentum. 5