Thermal & Kinetic Lecture 7

Maxwell-Boltzmann distribution, Equipartition of energy

(….and some problems with classical physics)

Deriving the Maxwell-Boltzmann distribution function

LECTURE 7 OVERVIEW

Equipartition and degrees of freedom

Last time….

Boltzmann factors and probabilities

Distribution of velocities in an ideal gas

Velocity distribution – TAKE NOTES

Consider velocities of molecules first:

)2

exp()(

)/exp()(2

kT

mvAvf

kTEAEf

xx

So, our velocity distribution in one dimension is: )2

exp(2

)(2

kT

mv

kT

mvf xx

How might we determine what the constant, A, should be?[See Q7(b) of the ’04-’05 Thermal and Kinetic paper for a similar question].

Given:

??

dxx )exp( 2

Gaussian function, <vx> = 0

)21

exp(2

1)(

2

x

xg

Distribution of molecular speeds

We need to do a few more steps to get a formula for the distribution of molecular speeds.

First, we can combine the expressions for molecular velocities to get:

zyxzyx

zyxzyx dvdvdvkT

vvvm

kT

mdvdvdvvvvf )

2

)(exp(

2),,(

2222/3

This expression is written in Cartesian co-ordinates (x,y,z).

Switch to spherical polar coordinates.

Sphericalpolar

coords

vz

vy

vx

222zyx vvvv

Surface element, dS, shown – need to consider volume element,

dV

vz

vy

vx

Considering all directions, the tip of the velocity vector ‘sweeps’ out a spherical volume (only one ‘octant’ shown above). How many velocity states within v and v + dv ?

Molecular speeds: Polar coordinates

Consider thin shell of sphere whose radius changes from v to v + dv. ?? What is the volume of this

thin shell?

What is the volume of the thin shell?

4p v

3/3

2p v

dv

4p v

2 d

v

None

of thes

e

15%

28%

57%

0%

a) 4v 3/3b) 2v dvc) 4v 2 dvd) None of these

Maxwell-Boltzmann distribution

zyxzyx

zyxzyx dvdvdvkT

vvvm

kT

mdvdvdvvvvf )

2

)(exp(

2),,(

2222/3

We had:

Taking into account discussion of spherical polar coordinates:

dvkT

mv

kT

mvdvvf )

2exp(

24)(

22/32

Maxwell-Boltzmann distribution of speeds of molecules in a gas at thermal equilibrium.

Maxwell-Boltzmann distribution

To convert from velocity to speed we have carried out two steps:

1. Convert from 1D to 3D probability.2. Consider all directions.

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/maxspe.html#c1

)2

exp(2

4)(22/3

2

kT

mv

kT

mvvf

Function no longer a Gaussian!

293 K

600 K

Maxwell-Boltzmann distribution

Maxwell-Boltzmann distribution for N2 molecules

293 K

600 K

!!• Maximum not at v=0;• most probable speed is less than mean speed; (CW 2)• curve broadens as T increases

Maxwell-Boltzmann distribution

293 K

600 K

Shaded part of graph gives fraction of molecules with speeds between 500 and 1000 ms-1. (Integrate under curve with appropriate limits).

Coursework Set 2 includes a number of questions on this distribution function.

½ m<v2> = (3/2)kT

In our derivation of the ideal gas law, we set the constant of proportionality between mean kinetic energy and temperature as 3k/2

From the Maxwell-Boltzmann distribution function, we can now show why the mean kinetic energy is given by 3k/2

0

24

2/32

0

22/3222

)2

exp(2

4

)2

exp(2

4

dvkT

mvv

kT

mv

dvkT

mv

kT

mvvv

The integral can be found in standard integral tables (see Lecture Notes Set 2b) and leads to the result:

kTvm

m

kTv

2

3

2

1

3

2

2

Typographical errors in Section 2.6: (i) Boltzmann’s (not Planck’s) constant!; (ii) Factor of 4missing from Eqn. 2.32. Apologies.

Equipartition of Energy and Degrees of Freedom

kTvm2

3

2

1 2 This statement has very important implications for both classical and quantum theory.

We’ve found that the mean kinetic energy for the gas molecules is3kT/2. However, this expression was derived by considering motionsof the molecules in the x, y and z directions.

We say that each molecule has three degrees of freedom

Theorem of equipartition of energy

“Each quadratic term in the expression for the average total energy of a particle in thermal equilibrium with its surroundings contributes on average ½ kT to the total energy”

or “Each degree of freedom contributes an average energy of ½ kT”