Click here to load reader
Upload
richard-zhu
View
221
Download
5
Embed Size (px)
DESCRIPTION
Filippone
Citation preview
Phys. 12c 2015, Lecture Notes (Week 10)
Kinetic Theory
Equilibrium and non-equilbrium processes due to particle motion. Consider velocity distributions andtransport phenomena non-equilibrium driven by gradient (density, temperature, voltage,. . . velocity)
Velocity and Speed DistributionsConsider classical, ideal, non-relativistic gas with constant , V,NI. What is probability that particle of mass m has velocity componentbetween vx and vx + dvx?
Recall: Classical probability for particle to occupy point in phase space and to havemomentum between px and px + dpx and py and py + dpy, . . . in volume betweenx and x+ dx, etc. is
Volume phase space element dpxdpydpzdx dy dz
occupation probability Boltzmann factor eE/
Therefore
P (px)dpx =
[V olume
py
pzeE/dxdydzdpydpz
]dpx
V olume
Mom.
dpxdpydpzdxdydz eE/
Normalization factor
Volume integration givesV ol dxdydz = V and divides out.
Get P (vx)dvx via change of variables vx =pxm and use
E = 12m|~v|2, |~v|2 = v2x + v2y + v2z
P (vx)dvx =
[ dvydvze
m2 (v2y + v2z)
dvxdvydvzem2 (v2x + v
2y + v
2z)
]e
mv2x2 dvx
Now let ui =
m2 vi, dvi =
2mui
Giving Gaussian integrals:
P (vx)dvx =e
mv2x2 dvx(
2m )
duyduze
u2yeu2z(
2m
) 32 duxduyduze
u2xeu2yeu2z
with e
x2dx =pi and finally:
P (vx)dvx =( m
2pi
) 12
emv2x2 dvx
Simple Gaussian
1
Note:1.
vx = vx = P (vx)vxdvx
= 0 [since P (vx) is symmetric]
2. v2x =m [=(standard deviation of Gaussian)
2=2] then energy per particleE = 12mv2x + v2y + v2z = 32 (Ideal Gas Energy) since v2x = v2y = v2z = m .Gives Equipartition of Energy each translational degree of freedom givesan energy of 12 per particle (Classical Result!) also true for other quadratic degrees of freedom in energy (e.g. x2 or p2).(eg. harmonic oscillator, rotation ...) since more Gaussian integrals
II. Distribution of speeds (Maxwell-Boltzmann velocity distribution)
What is probability that particle has speed (v = |~v|) betweenv and v + dv? Use spherical symmetry (all directions of v equally likely) then withdvxdvydvz = v
2 sin vdvdvdv
P (v)dv = pi0
2pi0e
mv2
2 v2 sin vdvdvdv pi0
2pi0
0e
mv22 v2 sin vdvdvdv
= emv22 v2dv
( 2m )320eu2u2du
where u =
m2 v
Note:0 u
2eu2
du =pi4 . Also
0 ueu2du = 12
0 u3eu
2
du = 120 u
4eu2
du = 3pi
80 u
5eu2
du = 1
Thus
P (v)dv = 4pi( m
2pi
) 32 v2e
mv2
2
MaxwellBoltzmann Distributiondv
v
P( )vv
vv
mp
rms
< >
Note:1.
v = 0 vP (v)dv= 4pi
(m2pi
) 32(2m
)2 0 u
3eu2
du; u =
m2 v
=(8pim
) 12 = 1.60
(m
) 12
2
2.
v2 = 0 v2P (v)dv = 4pi ( m2pi ) 32 (2m ) 52 0
u4eu2
du 3pi
8
= 3 mthen
vrms =
(3
m
) 12
= 1.73( m
) 12
3. vmp vmost probable v that maximizes P (v).Set Pv = 0 = ()v
2(m2 )(2v)emv22 + ()2ve
mv2
2 v2mp = 2mvmp =
2m = 1.41
(m
) 12
Application of velocity distribution:
Effusion: consider a box of atoms at with small hole in box (hole doesntperturb equilibrium inside of box).Outside of box is vacuum.
VacuumWhat is flux of atoms leaving box?The number of atoms/unit area/unit time=TotCrude estimate: All atoms heading in correct directionwithin cylinder will exit hole in the time dtif ` = vdt (v = v).
l
z
Hole area dA
Then
TotdAdt =n`dA
6=nvdtdA
6
(TotdAdt = the number of exiting atoms, n = density of atoms The factor of 6 results since 1/6 of atoms heading in +z direction.Therefore
Tot =nv
6CrudeEstimate
Exact calculation:
Include velocity distribution (but not geometric effects).Then Tot =
0 (vz)dvz for vz > 0 only.
(vz) is the flux through hole for particular vz,
3
Zero lower limit since only particles moving towards hole will exit. Then
(vz)dAdt # exiting atoms with vz
= n`dAP (vz) = nvzdtdAP (vz)
Therefore using P (vz) from above
Tot = n0 vz
(m2pi
) 12 e
mv2z2 dvz
= n(m2pi
) 12(2m
) 0
ueu2
du 12
, u =
m2 vz
= n
2pim
Recall from earlier v = v =
8pim .
Therefore
Tot =1
4nv
Exact result has modest difference ( 50%) with respect to crude estimate.Now with E.O.S. n = P , we have
Tot =P
2pim
Thus flux effusing from hole is larger if m is smaller. Used in 40s to separate U 238F6
nonfissionableand U 235F6
0.7% abundancefissionableKinetic Theory of Transport
Characterize transport of heat, velocity, e, particles in non-equilibrium situations.Approximate treatment good to factor of 2(Exact treatmentorders of magnitude harder)Assumptions:1. Time between collisions collision time (impulse only)2. Two particle colllisions dominate (ignore 3-body interactions)3. Particles behave classically (de Broglie distance between particles)To model transport, first need to understand interaction of atoms (non-ideal)
Collisions:
Model particles as hard spheres with radius RParticles collide if they are within 2R of each other. R
RCross sectional area for collisions 0 = cross section = pi(2R)2Probability of colliison with distance `:
4
For dilute gas and small d`Probability of Collision=Area of particlesTotal Area 2R 2R
2R2R
2R
2R
2R
2R
2R
L
L
dl
Area of particles = the number of particles pi(2R)2 = ndensity
(L2d`)0 V olume
Then
probabiliy of collision =nL2d`0L2
= n0d`
To find Probability of collision for a large distance ` [PC(`)]define probability of no collision in ` as PNC(`) [PC = 1 PNC ]. Then
PNC(`+ d`) No collision in `+d`
= PNC(`) No collision in `
(1 n0d`) No collision in d`
ordPNCd`
= PNC(`)n0solving
PNC(`) = en0` (since PNC(0) = 1)
thenPc(`) = 1 en0`
1 e`
`mfp
with `mfp = mean free path length =1
n00= mean distance between collisions.
(Above ignored other particles motion small correction)Checking:
` =0`PNC(`)d`
0PNC(`)d`
=`mfp
0e ``mfpd`
`mfp
= `mfp
Diffusion and Effusion DemosTransport Theory:
Elementary Treatment:Simple linear law good for many physical system (Phenomenological).Flux of something (particles, heat, momentum) Driving force gradient (something)Ohms-like Laws:
~Jelectric current = (1R
)~Q (Q = (electric potential)
Self diffusion ~Jparticles = D~n (D = diffusivity, n = density)~Jheat = K~T (K = thermal conductivity)Jz momentum = vxz ( = viscosity) see figure below
5
zx
Fluid
Moving plate
Stationary plate
vfluid velocityx
v0
Example: DiffusionUse simple kinetic theory to calculate D.Consider density gradient along zaxis, calc. flux of particlescrossing fixed z = z0 plane:
z=z
z=z
z=zn(z )0
n
n
l
l
n(z )
For distances < `mfp we will ignore collisions (particles move freely). Then
Flux up = [# particles going up/volume at z = z0 `mfp] [Average Velocity]Jup =
n(z0`mfp)6 (Vrms) Note : Vrms independent of density
Jdown =n(z0+`mfp)
6 (Vrms)
Then net flux:
Jparticles = Jup Jdown= Vrms6 [(n(z0 `mfp) n(z0 + `mfp]= Vrms6 {[(n(z0) n(z0 `mfp)]
nz `mfp
+ [n(z0 + `mfp) n(z0) nz `mfp with dz`mfp
}
= (Vrms`mfp
3
)nz
Therefore D = 13Vrms`mfp =13
3( m)
120p
For N2 at room and 1 atm., Vrms 5 104 cm/s,`mfp 3 105 cm gives D ' 0.5 cm2/s(Exp. gives D = 0.2 cm2/s) (within factor 2, OK for order of magnitude)
6
Note:
1. Can also model diffusion as random walk.Random Walk
Eg. How long for photon made at center of sun to reach surface?Without collisions Rsunc , Rsun 7 1010 cmSpeed of photons = c = 3 1010 cm/sTime to surface without collisions = Rsunc = 2 sectc = time to surface with collisions: random walkRecall D2 = N`2 (N is the number of steps, ` is the average step size= `mfp)tc = Ntime per step= N(`mfpC ) but N = D
2`2 =
Rsun`2mfp
Therefore, tc =R2sun`mfpC
, ne 8 1023/cm3, `mfp = 1n0n is density of electrons, 0 is the photon-electron cross section, 0 0.67 barns(1 barn= 1024 cm2)Therefore `mfp 1 cm and tc 2 2 (pi 107 s)
1 year
' 4000 years
2. Particle flow must obey a continuity equation(the number of particles is conserved)
t(n Adz
volume
) = AJz(z) AJz(z + dz)
nt
=Jzz
Therefore since Jz = Dnz , n obeys differential equation nt = D2nz2
Advanced Treatment (recall effusion example-yesterday)Want to correctly include velocity distribution in calculations:Consider f(~r,~v, t)d3~rd3~v where f is the particle distribution function,= mean number of particles with ~r between ~r and ~r + d~r and ~v between ~v + d~v
at time tKnowledge of f solves transport problem, therefore need Eq. for fConsider first in absence of collisions, particles under influence of force. Then
~r = ~r + ~vdt~v = ~v + ~vdt = ~v + ~FM dt
and we havef(~r, ~v, t)d3~rd3~v = f(~r,~v, t)d3~rd3~v
since total number of particles is unchanged.Now from Liouvilles Theorem of Classical Mechanics (Ph106, Ph1a?):(Volume of phase space is conserved) d3~rd3~v = d3~rd3~vTherefore f(~r, ~v, t) = f(~r,~v, t) or f(~r + ~rdt, ~v + ~vdt, t+ dt) f(~r, ~N, t) = 0
7
or we can rewrite above as Df = 0 Boltzmann Transport Eq. without collisionsor since
f(x+ dx, y + dy) f(x, y) = fxdx+
f
ydy = df
we have
Df =
(f
xx+
f
yy +
f
zz
)dt+
(f
vxvx +
f
vyvy +
f
vzvz
)dt+
f
tdt = 0
or
~v ~f +~F
m ~vf + f
t= 0
Boltzmann Transport Equation (without collisions)
Then with collisions
f(~r, ~v, t)d3~vd3~r = f(~r,~v, t)d3~rd3~v +Dcollf(~r,~v, t)dtd3~vd3~r
(Dcollf(~r,~v, t)dt is net increase in number of particles in d3~rd3~v due to collisions)
GivingDf = Dcollf
Full Boltzmann equation
8