Thermal Physics Lecture 31

7/30/2019 Thermal Physics Lecture 31

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Physics 301 2-Dec-2002 31-1

Reading

K&K chapter 14.

The Maxwell Velocity Distribution

The beginning of chapter 14 covers some things weve already discussed. Way back inlecture 6, we calculated the pressure for an ideal gas of non-interacting point particles byintegrating over the velocity distribution. In lecture 5, we discovered the Maxwell velocitydistribution for non-interacting point particles. In homework 2, you worked out some ofthe velocity moments for the Maxwell distribution and you also worked out the distributionof velocities of gas molecules emerging from a small hole in an oven. Rather than derivethe earlier results again, well just summarize them here for convenient reference.

First, of all, let f(t,r,v) d3r d3v be the number of molecules with position vectorin the small volume element d3r centered at position r and velocity vector in the smallvelocity volume d3v centered at velocity v at time t. The function f is the distributionfunction (usually, weve assumed it to be a function of the energy, but were working upto bigger and better things!). When we have a Maxwell distribution,

fMaxwell(t,r,v) = n m

2

3/2em(v2x + v2y + v2z)/2 ,

where m is the molecular mass, and n is the concentration. The Maxwell distribution doesnot depend on t nor r. Integrating d3v over all velocities produces n, and integrating d3r

multiplies by the volume of the container, and we get N, the total number of particles. Ifwe want the probability distribution for just the velocity components, we just leave off then (and then we also dont consider it to be a function of r).

Sometimes, we want to know the probability distribution for the magnitude of thevelocity. We change variables from vx, vy , vz to v,,, and dvx dvy dvz = v2 dv sin d d.We integrate over the angles, pick up a factor of 4, and the probability distribution for vis

P(v) dv = 4 m

2

3/2emv2/2v2 dv .

With this distribution, one finds the most probable speed (peak of the probability distri-bution) is

vmp =

2

m= 1.414

m.

The mean speed (denoted by c by K&K) is

c = v =

8

m= 1.596

m.

Copyright c 2002, Princeton University Physics Department, Edward J. Groth


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The root mean square speed (which appears in the energy) is

vrms =

v2 =

3

m= 1.732

m.

Cross Sections

The next topic discussed in K&K is the mean free path. Were going to use this asan excuse to discuss cross sections and develop an expression for reaction rates based oncross sections and the Maxwell velocity distribution. The first thing we need to do is tosee where a cross section comes from and what it means.

Consider a situation in which two particlescan undergo some sort of interaction when theycollide. The interaction may be probabilis-tic; examples might be a nuclear or chemicalreaction, an elastic scattering, or an inelasticscattering (which means one or both particlesmust be in an excited state after the interac-tion). Suppose one has a stationary target par-ticle, and one shoots other particles at it withvelocity v. The concentration of the incidentparticles (number per unit volume) is n. The

flux density of incident particles, or the particle current, is the number crossing a unit areain a unit time and this is just nv. Note that this has the dimensions of number per unitarea per unit time. Now the number of interactions per second must be proportional tonv. If we double the density, we have twice as many particles per second with a chance tointeract. Similarly, if we double the velocity, particles arrive at twice the rate and we havetwice as many particles per second with a chance to interact. Actually, the strength ofthe interaction may depend on the relative velocity, so its not strictly true that the rate isproportional to v, but in a simple process, like the collision of hard spheres, its certainlytrue. We take as a starting point that the interaction rate is proportional to the velocity,and effects having to do with the energy of the collision are included in the proportionalityconstant.

The interaction rate is thenR = (v)nv ,

where is NOT the entropy. Instead its the proportionality constant and is called thecross section. Weve explicitly shown that it might depend on v. There can be morecomplicated dependencies. For example, if the incident and target particles have spin, thecross section might depend on the spins of the particles as well as the relative velocity.



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The proportionality constant is called a cross section because it must have the dimensionsof an area. R has the dimensions of a number (of interactions) per unit time, while nvhas dimensions of number per unit area per unit time. The cross section can be thoughtof as an area that the target particle holds up, perpendicular to the incoming beam. If

an incident particle hits this area presented by the target, an interaction occurs.

Now some jargon. Its often the case that one considers a scattering, in which case thequestion asked is how many particles per second scatter into the solid angle d centered onthe direction (, )? In this case, the number is proportional to d, and the proportionalityconstant is often written as a differential so the rate into d is

R( d) = dd

nv d ,

where d/d is called the differential cross section. If one integrates over all solid angles

to find the rate for all interactions (no matter what the scattering direction), one has

=

d

dd ,

and is called the total cross section in this case. Of course, a reaction (not just ascattering) may result in particles headed into a solid angle d, so reactions may also bedescribed by differential and total cross sections.

If one knows the forces (the interaction) between the incident and target particles,one may calculate the cross sections. Cross sections may also be measured. One uses a

target which has many target particles. One counts the scattered particles or the outgoingreactants using particle detectors. The total number of events is just Nnv(v)t where Nis the number of target particles exposed to the incident beam and t is the duration of theexperiment. This assumes that the chances of a single interaction are so small that thechances of interactions with two target particles are negligible. Of course, sometimes thisisnt true and a multiple scattering correction must be deduced and applied. Sigh. . .



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Example: Cross Section for Smooth Hard Sphere Elastic Scattering

As an example, well work out the differential cross section for smooth hard spherescattering. You might say, what does this have to do with thermal physics? Answer: not

much, but its to motivate the use of the cross section as part of the next topic. Wellconsider a stationary target and an incident particle with velocity v. Both particles aresmooth hard spheres with radius a and mass m. (Think of billiard balls.) We assumethe collision conserves the kinetic energy of motion. This is whats meant by an elasticcollision. The momentum of the incident particle is p0 before the collision and p1 afterthe collision. The momentum of the target particle is p2 after the collision. Conservationof momentum and energy tell us

p0 = p1 + p2 ,1

2mp20 =

1

2mp21 +

1

2mp22 .

If we square the conservation of momentum equation, we have

p20 = p21 + 2p1 p2 + p22 .

If we compare with the conservation of energy equation, we conclude

p1 p2 = 0 ,

which means that either one of the particles is at rest after the collision or their momenta(and velocities) are perpendicular after the collision. Im sure you knew this trick fromPhysics 103/5, right? Suppose the incident particle scatters at an angle relative to its

initial direction. The target particle must recoil at angle /2 from the initial directionof the incident particle. We have

p0 = p1 cos + p2 cos( /2) ,0 = p1 sin + p2 sin( /2) ,

from which we conclude

p1 = p0 cos , p2 = p0 sin .

So far, our analysis has proceeded very much as it would have in Physics 103/5. Weveapplied conservation of momentum and energy, but weve not completely solved thecollision. We have expressed the final momenta in terms of the initial momentum (known)and an angle (unknown). This is typical for collision problems. Conservation of energyand momentum provides four constraints but there are six components of momentum to bedetermined. (In case youre wondering how come we have only one unknown, , the planein which the scattering occurs is also undetermined. That is, we dont know , either!



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To make further progress, we must examine the colli-sion in more detail and specify the initial conditions morecarefully. So we suppose that in the absence of a collision,the centers of the two spheres would pass within distance b

of each other. b is called the impact parameter. The initialmomentum points in the z-direction, as weve been assuming.Finally, the plane containing the initial momentum vector andthe centers of the two spheres makes an angle with respectto the x-axis. Its easy to see that the collision will be con-fined to this plane. Since weve assumed an elastic collisionbetween smooth spheres, the incident particle can exert only a radial, not a tangential,force on the target particle. So we can calculate the angle at which the target particlerecoils just from the geometry of the collision. Inspection of the diagram shows that

sin(/2

) =

b

2a

,

which means

cos =b

2a, sin =

1 b

2

4a2.

At this point, weve completely solved the collision. Given the initial momentum or ve-locity, the impact parameter, and the initial plane of the collision, weve determined thefinal plane (same as initial), the scattering and recoil angles, and all the final momentumcomponents.

However, we still havent figured out the cross

section! We assumed specific values for and b.But we usually dont have such control over the ini-tial conditions. Instead we should assume that theimpact parameter is in the range b b + db and theazimuthal angle is in the range + d. Thenthe cross section is just the area that this implies,

d = b db d .

However, its customary to express the cross sectionin terms of and rather than b and , since b can-not usually be measured, but is relatively easy to

measure (its just the angle describing the locationof the particle counter). From our previous expression relating b and , we have

sin2 = 1 b2

4a2.

Differentiate,

2sin cos d = 2b db 14a2

,



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sob db = |4a2 cos sin d| ,

where we dont care that increasing b means decreasing , and

d = 4a2

cos sin dd = 4a2

cos d ,

ord

d= 4a2 cos .

Note that is confined to the range 0 /2.

The total cross section is found by integrating the differential cross section over allallowed angles,

=

/20

sin d

20

d 4a2 cos = 4a2 .

This makes sense. If the center of the incident sphere passes within 2a of the target sphere,there will be a collision. The area within 2a of a point is 4a2. We didnt really need todo all this work to get the total cross section; in this case, we could have just written itdown by inspection!

Reaction Rates

Consider a gas composed of weakly interacting particles. We want the gas to be essen-tially an ideal gas and the distribution functions to be the Maxwell distribution functions.

Suppose the gas contains two kinds of molecules with masses m1 and m2 and concentra-tions n1 and n2. Suppose these two kinds of molecules, upon colliding, can undergo areaction. The (total) cross section for this reaction is (v) where v is the relative velocityof the molecules. We want to calculate the rate at which the reaction occurs.

Consider a molecule of type 1 moving with velocity v1. The molecules of type 2 arecoming from all directions with different velocities, so how do we calculate anything? Weisolate a particular velocity and direction. Consider molecules with velocity v2 in the ranged3v2. The concentration of such molecules is

f2(v2) d3v2 .

The relative velocity is v = |v2 v1|, so the rate of interactions with our single targetmolecule is

dR = (v)vf2(v2) d3v2 .

We consider target molecules in a small volume d3r1 and with velocities in the range d3v1.

There are f1(v1) d3r1 d3v1 such molecules. The interaction rate for this number of targetparticles is

dR = (v)vf1(v1)f2(v2) d3v1 d

3v2 d

3r1 .



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A little thought shows that dR has the dimensions of inverse time.

To get the overall rate we need to integrate over all differentials. The integral overd3r1 just gives the volume of the container, V. The integrals over the velocity elements

are a little trickier since the integrand includes a dependence on the relative velocity. Wehave

R = V n1n2 m1

2

3/2 m22

3/2 d3v1

d3v2 (v)ve

m1v21 /2em2v22 /2 .

Were going to change variables to the center of mass velocity and the relative velocity.The center of mass velocity is

u =1

M(m1v1 + m2v2) ,

where M = m1 + m2. The relative velocity is

v = v2 v1 .

Expressing v2 and v1 in terms ofu and v, we have

v2 =m1M

v + u , v1 = m2M

v + u .

The kinetic energy can be expressed as

E =

1

2 m1v

2

1 +

1

2 m2v

2

2 =

1

2

m1m2

M v

2

+

1

2 Mu

2

=

1

2 v

2

+

1

2 Mu

2

,

where =

m1m2m1 + m2

,

is called the reduced mass. Weve already used for a number of thingschemical poten-tial, magnetic moment, mobilityoh well. The Jacobian of the transformation is unity,so

R = V n1n2

M

23/2

eMu2/2d3u

23/2

(v)vev2/2d3v

.

We started with an integral written for a pair of particles. Weve now rewritten the integralin terms of the center of mass motion and the motion relative to the center of mass. Thecollision occurs in the relative motion! The integral over the center of mass motion givesunity and finally, the rate per unit volume is

r =R

V= n1n2

2

3/2 (v)vev2/2d3v .



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This expression is generally applicable whenever the gas can be described by a classicalMaxwell Boltzmann distribution and the reaction rate is slow enough that the gas is alwaysessentially in thermal equilibrium and the Maxwell Boltzmann distribution applies. Forexample, expressions like the above are used to calculate the nuclear reaction rates in the

Sun when computing a numerical model of the Sun.

If particles of types 1 and 2 are in fact the same, then = m/2, n1 = n2 = n, and weneed a double counting correction of 1/2,

r =1

2n2 m

4

3/2 (v)vemv2/4d3v .

Usually, depends only on the magnitude of v, so we can integrate over the angles and

r = 2n2 m4

3/2

(v)v3emv2/4dv .

The Collision Rate and the Mean Free Path

We can use the expression weve just derived to calculate the collision rate amongmolecules in a gas. We assume the molecules are hard spheres of radius d/2. We ignorethe correction to the concentration (as in the van der Waals equation of state). As weworked out earlier, the total cross section for a collision is d2, so the collision rate is

r = 22d2n2

m4

3/2

v3emv2/4dv .

This integral is straightforward and the result is

r = n2d2

4

m.

This is the number of collisions per unit time per unit volume. Now we want to convertthis into the number of collisions per unit time per particle. A unit volume contains nparticles, so we need to divide by n. In addition, each collision involves two molecules, so

we need to multiply by 2. The rate per molecule is

r1 = nd2

16

m.

All the factors in this expression make sense. The bigger the concentration, the higherthe collision rate; the larger the cross sectional area of a molecule, the higher the collisionrate; the faster the molecular speed (

/m), the higher the collision rate. Except for the



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numerical factor of

16, we could have written down this expression just from dimensionalanalysis.

The mean time between collisions is 1/r1. If we assume that between collisions a

particle typically moves with the average speed, we can get an estimate of the typicaldistance a particle moves between collisions. This is called the mean free path,

=c

r1=

8

m

nd2

16

m=

12nd2

.

You will notice that this expression differs by

2 from the expression given in K&K, andderived by a completely different method. The method used by K&K is to imagine aparticle sweeping out a cylinder of base area d2 as it travels. If another particle in thegas has its center within this cylinder, then the first particle will collide with it. As the

height of the cylinder grows, the chance that another particle is in the cylinder grows.When the height of the cylinder is such that theres one particle (on average) within thecylinder, declare that a collision has occurred and the height of the cylinder is the meanfree path. In other words,

nd2K&K = 1 , or K&K =1

nd2.

In actual fact, neither method is completely kosher! Our discussion of the collision rate islegitimate (within the assumptions of hard sphere molecules), but dividing the average rateinto the average velocity is not strictly legal since the average of a ratio is not the same asthe ratio of the averages. The K&K method ignores the relative velocities of the molecules.Why is this important? The simple argument applies to a fast moving molecule. On theaverage, it will run into another molecule after going a distance K&K. But a slow movingmolecule (consider the limit of a molecule that happens to be at rest) will more than likelybe hit by another molecule in the gas before it has a chance to move K&K. Averagingover both slow and fast molecules, the mean free path will be shorter than K&K.

At this point, its useful to plug in some numbers just to get a feel for the collisionrate and the mean free path. We consider a neon atom which has a diameter of about2 A. The mass of neon is about 20.2 atomic mass units. We assume its an ideal gas andevaluate the concentration at 0C and one atmosphere. Under these conditions, one mole

occupies 22.4 liters, and the concentration is

n0 =N0V

= 2.69 1019 molecules cm3 ,

also known as Loschmidts number. The mean free path is

=1

2nd2= 2.10 105 cm ,



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about 1000 times the atomic diameter. The mean speed is

c =

8kT

m= 5.35 104 cm s1 .

The collision rate per atom is

r1 =c

= 2.55 109 collisions s1 .

An interesting thing happens in a vacuum. Since the mean free path and the collisionrate are inversely proportional to and proportional to the concentration (density), loweringthe density increases the mean free path and lowers the collision rate. Just as an example,consider a vacuum of 106 atmospheric pressure (not a particularly good vacuum). The

collision rate drops to about 2550 times per second and the mean free path rises to about21 cm. This is a macroscopic length! Its comparable to the size of laboratory equipment.In good vacuums, the residual gas interacts more with the container than with other gasmolecules.


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Thermal Physics Lecture 31