大学物理(热学) 1) Mean Free Path 2)Diffusion, Thermal conductivity, and viscosity 2006-10-12
大学物理（热学） 1) Mean Free Path 2) Diffusion, Thermal conductivity, and Viscosity 2006-10-12
Equilibrium distribution v2n1n2+n22+ ∑(舞)一 n -the number of molecules of dn is the number of molecules in vx>vx+dvx near v V,→V,+dv ν→)1+dv
Equilibrium distribution N v v v v N 2 2 2 2 2 1 + + + = dn is the number of molecules x x x v →v + dv vy vy + dv → y z z z v → v + dv in → = N n v N n v i i i 2 2 d N n v + n v + = 2 2 2 2 1 1 ni ⎯ the number of molecules of i v near v
Meaning of dn/n among N molecules, how many are near probability that one molecule is near v proportional to avx fly dv dn(vr,vy,v =/()如yv,y,fvk N
among N molecules, how many are near v probability that one molecule is near v proportional to dvx ... ( ) x x f v dv Meaning of dn/N ( )d ( )d ( )d d ( , , ) x x y y z z x y z f v v f v v f v v N n v v v =
Isotropic assumption (v,/(v,)(2)=m2)=V2+y+n2) The solution is (v,)=Cexp(2/a2) The probability should be normalized dn
Isotropic assumption ( ) ( ) ( ) ( ) ( ) 2 2 2 2 x y z x y z f v f v f v = v = v + v + v ( ) ( ) 2 2 f vx =Cexp − vx The solution is The probability should be normalized = N dn 1
∫=c=/) 2/a2 1=Cl explvx ly each normalized √兀 a exp C C
( ) 3 3 2 2 exp d d 1 = − = x x C v v N n − = − x x v v C exp d 1 2 2 ( ) = − x x 1 C exp v dv 2 2 12 = 1 each normalized