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

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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

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(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,

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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 `:

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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

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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)

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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

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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

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