We’ve spent quite a bit of time learning about how the individual fundamental particles that compose the universe behave.

Can we start with that “microscopic” knowledge and learn something about

bulk properties of matter such as temperature?

Will the peculiar rules governing the behavior of an individual particle or small

group of particles(such as the Pauli principle) influence

the properties of the matter they compose?

Naturally, I am sure you are wondering:

Particles have identical physical properties…but can be

distinguished by following their(well defined) classical paths.

In equilibrium, the energy distribution of the particles will converge

to the most probable allowed.

In principle, there is no limit on the number of particles

occupying each state.

ASSUMPTIONSASSUMPTIONS

Imagine 6 particles with 9 indivisible quanta of energy divided among them.

•If the particles are indistinguishable, we only care about how many particles are in each state, and there are 26 unique ways to distribute the energy among them—26 unique combinations.

•If the particles are distinguishable (we make a distinction as to which particle is in which energy state), there are 2002 unique permutations.

Remember that for now we have assumed that these particles are distinguishable.

If you are making choices from n objects, then on your first pick you have n choices. On your second pick, you have n-1 choices, n-2 for your third choice

and so forth. As illustrated before for 5 objects, the number of ways to pick from 5 objects is 5! .

Suppose you are going to pick a subset r out of the total number of objects n, like drawing 5 cards from a deck of 52. For the first pick, you have n choices,

then n-1 and so on down to n-r+1 for the last pick. The number of ways you can do it is:

rnPrn

nrnnnn

)!(

!)1)...(2)(1(

If we care about which particle is in which

‘state’, there are six different states where

one particle one particle has all of the

energy.

If each of these “microstates” is equally likely (and we assumed that we will converge on the most probable), it seems that nature doesn’t

favor the situation where one particle has all of the energy.

Here we have to choose more than one particle for each state,

and we can distinguish between

different combinations of particles in each

state, so the multiplicity gets

bigger.

We assume that each “microstate” (unique permutation) is equally probable.

In other words, if there are 180 permutations that will produce a particular energy distribution, then that distribution is more probable than a distribution that can only be produced by six permutations.

To find the average number of particles in each state

2211 pnpnn jjjaverage number

of particles in the jth energy level count the

number of particles in

each state for this

distribution

multiply by the number of permutations that can produce this distribution divided by the total number of permutation for all distributions

Energylevel

Averagenumber

0 2.143

1 1.484

2 0.989

3 0.629

4 0.378

5 0.210

6 0.105

7 0.045

8 0.015

9 0.003

There are only 6 (out of 2002) permutations that can produce a situation where one particle has all 9E. Apparently it’s not very probable.

x

y

z

x

y

z

vd

constantv

The Maxwell-Boltzmann distribution can be shown graphically as the plot of the number of molecules traveling at a given speed versus the speed. As the temperature increases, this curve broadens and extends to higher speeds.

m

TkBrms

3

TkKm B2

3

2

1 2

m

TkBrms

3Using:

The equipartition theoremThe equipartition theorem

follows

A classical molecule in thermal equilibrium has an average energy of kT/2 per degree of freedom.

But generally, there are more than three degrees of freedom (more than just the translational motion in each of x, y, and z):

22

2

1

2

1kxmE x

2

2

1 IErot For molecules that can rotate, you can have a rotational degree of freedom.

A one dimensional harmonic oscillator has two degrees of freedom, one corresponding to its potential energy, the other to its kinetic energy.

There is an energy kT/2 associated with each of these degrees of freedom.

??

a

a

a

aab

b

b

The first two pictures give the same outcome. Even though a and b are identical, you can tell them apart by following them along their unique paths.

Quantum mechanically, each particle has some probability of being somewhere at a particular time, which overlaps greatly at the collision point.

Which particle emerges where? In wave terms, they interfered.

bb

The Maxwell-Boltzmann distribution assumes that the particles they describe are distinguishable. Two particles can be considered distinguishable if the distance separating the particles is large compared to their DeBroglie wavelength.

Since we know that particles are really “wavicles” and Maxwell Boltzmann statistics is only good for distinguishable particles, what good is it?

Put in mathematical terms, if their average separation is larger than the the uncertainty in their momentum. xd

Using the uncertainty principle: 2

xp

At thermal equilibrium: 0xpthe particles are moving randomly and the

directions will cancel each other

02 xp but their magnitudes aren’t zero

22

2 Tk

m

pKE Bx Tmkppp Bxxx 22

for each degree of freedom

Tmkx

B2

31

using

N

Vd

18 2

3

TmkV

N

B

Maxwell Boltzmann

statistics are valid for low density and high

temperature and particle mass

Before, we had distinguishable particles….

)()( 21 rr ba

2

2

2

1212*

1** )()()()()()( rrrrrr aaaaaa

)]()()()([2

11221 rrrr baba )]()()()([

2

11221 rrrr baba

)()( 12 rr ba

)()(2)]()()()([2

1211221 rrrrrr aaaaaa

0)]()()()([2

11221 rrrr aaaa

*2

2

2

1* 2)()(2 rr aa

assuming they are in two different states, you get two distinct wavefunctions, where r1 and r2 are the positions of the toe particles…

the probabililty of both particles being in the same state is then…

…but indistinguishable particles require a more complicated wavefunction.

or

If both particles are in the same state these become…

“bosons” as they are called, are more likely to be found in

the same energy state than apart…

“fermions”, however, can never be found in the same state! The

uncertainty principle…again.

We are now considering indistinguishable particles we are now only counting how many particles are in each energy state, rather than which particles are in each energy state.

Now each of the 26 energy combinations show occur with equal probability.

However, our assumption that there is no theoretical limit on the number of particles occupying each state still holds.

Energylevel

AveragenumberMaxwell-Boltzmann

AveragenumberBose-Einstein

0 2.143 2.269

1 1.484 1.538

2 0.989 0.885

3 0.629 0.538

4 0.378 0.269

5 0.210 0.192

6 0.105 0.115

7 0.045 0.077

8 0.015 0.038

9 0.003 0.038

2211 pnpnn jjj

This probability is now the same for each

energy distribution, just 1 divided by the

total number of distributions.

Not surprisingly, the distribution for bosons show a higher probability at the extremes, where there were fewer permutations producing the observed energy spectrum.

The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

dT

dUC

TkTk BB 3322/

energy per degree of freedom (from the equipartition theorem)

number of degrees of freedom per dimension (kinetic and potential)

in real life, there are three dimensions

here, U is the thermal energy

Thermal energy per atom:

To get the energy per mole, multiply by Avagadro’s number:

RTTkNU BA 33

Kmolcal97.533 RRT

dT

d

dT

dU Note that the specific heat predicted by the classical equipartition theorem is

constant with temperature!

This prediction disagrees with the

data at low temperature.

Einstein’s solution:

Treat each atom as an independent quantum simple harmonic oscillator, and quantize its energy.

1

TkBeE

The energy is then:

this approaches the classical result, kT, at high

temperatures as we know it should (the correspondence

principle)

The specific heat is then:

2/

2

13

Tk

Tk

BB

B

e

e

TkR

dT

dUC

Einstein’s approach worked (mostly). Some fine tuning was required, however. Debye pointed out that the atoms in a solid do not move independently but interacts with its neighbors. The result - continuous vibrational waves-sound.

•Once again we assume that the particles are distinguishable.

•Thanks to the Pauli exclusion principle, we must, however, remove assumption that there is no theoretical limit on the number of particles in each state!

As in the case for bosons, each allowed energy distribution has an equal probability.

Unlike the boson case, only five energy distributions are allowed! All others have more than the permitted two particles in each state in violation of the exclusion principle.

E

AveragenumberMaxwell-Boltzmann

AveragenumberBose-Einstein

AveragenumberFermi-Dirac

0 2.143 2.269 1.8

1 1.484 1.538 1.6

2 0.989 0.885 1.2

3 0.629 0.538 0.8

4 0.378 0.269 0.4

5 0.210 0.192 0.2

6 0.105 0.115 0

7 0.045 0.077 0

8 0.015 0.038 0

9 0.003 0.038 0

2211 pnpnn jjj

Now there are few allowed states. Since the particles are indistinguishable, each

state is again equally likely.

Note that the energy distribution will flatten out at higher energies. If one particle has all of the energy state, that would require all of the other particle to be in the same energy state – zero - in violation of the Pauli principle.