phys212ln3

phys212ln3 - Physics 212 Statistical mechanics II Fall 2005...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 212: Statistical mechanics II, Fall 2005 Lecture III : Equilibrium conditions, Liouville theorem, BBGKY Let us begin by finding what one-particle distributions f are preserved by the Boltzmann equation. The collision term derived in Lecture II for a monatomic gas at one spatial point r is C ( f ) = Z w ( f 1 f 2- f 1 f 2 ) d p 2 d p 1 d p 2 . (1) This collision term will vanish identically at a given point in space if for all incoming and outgoing momenta with p 1 + p 2 = p 1 + p 2 , (2) we have f 1 f 2 = f 1 f 2 , or log f ( p 1 ) + log f ( p 2 ) = log f ( p 1 ) + log f ( p 2 ) . (3) The algebra we went through in proving the H theorem shows that this is actually a necessary condition to obtain dS/dt = 0, assuming that w is nonzero. Now (3) will hold as long as log f is a sum of quantities which are conserved in the collision process. The conservation of momentum and energy in collisions forces some constraints on the collision term C ( f ): we have Z C ( f ) d p 1 = Z C ( f ) p 1 i d p 1 = Z C ( f ) p 1 2 2 m d p 1 = 0 . (4) Assuming that the only conserved quantities are energy, momentum, and particle number, then we take log f ( p ) = c + c 1 · p + c 2 p 2 = c 2 ( p- p ) 2 + C. (5) Here p =- c 1 / 2 c 2 and C = c- c 1 2 / 4 c 2 . Now we can start to recognize the familiar Maxwellian distribution of velocities in a gas. The integral over all momenta should give the local number density, so π c 2 d/ 2 e C = n ( r ) , (6) which determines e C = n ( r )( c 2 / 2 π ) d/ 2 in d dimensions. The local momentum density is n ( r ) p , and the local energy density is n ( r ) Z p 2 2 m f ( p ) d p = n ( r ) p 2 2 m + d 4 c 2 m ! . (7) It remains to derive c 2 , which is best done by constructing the equation of state (cf. Huang). Taking a shortcut and invoking the answer from an undergraduate course or equipartition, we know the mean energy for a gas at rest is dkT/ 2, so c 2 = (2 mKT )- 1 . Finally, the equilibrium distribution given by the collision term is f ( p , r ) = n ( r ) e- ( p- p ) 2 / 2 mkT (2 πmkT ) d/ 2 . (8) 1 There are still at this point multiple possibilities for equilibrium: it could be that at every point the collision integral vanishes because the distribution is Maxwellian, but the local mean energy and velocity vary. It remains to show that any gradient in local density, momentum, or energy will lead to a current directed so as to eliminate the gradient. Such gradients can survive for quite a long time relative to the collisional time scale, and are best studied using the Navier-Stokes equation to be derived below. We now change gears and detour into classical dynamics to understand the physical content of the Boltzmann equation and how it can be reconciled with the microscopic equations of motion....
View Full Document

{[ snackBarMessage ]}

Page1 / 5

phys212ln3 - Physics 212 Statistical mechanics II Fall 2005...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online