Two systems in equilibrium: Entropy, temperature and

Two systems in equilibrium: Entropy, temperature and

chemical potential

PYU33A/P15 Stat Mech 29

PYU33P/A15 Statistical Thermodynamics

Our objective now is to develop statistical expressions for entropy and temperature

(i) We’ll start with entropy...

insulating wall

thermally conducting wall

System1

System2

• Consider two systems of fixed volume in thermal contact such that heat energy is exchanged.

• The systems cannot exchange particles (a thermally conducting wall separates systems).

• Combined system is isolated so the total energy is constant: U U1 U2 constant

Entropy and temperature: K&K Chapter 2


Physics principle: Every accessible state of the combined system is equally probable

Most probable division of energy is that for which the combined system has the maximum number of accessible states

Take the systems 1 and 2 to be our model spin systems in thermal contact in a magnetic field:• The number of spins stays constant on each side• The spin excess is variable on each side• The energy is variable on each side

N1, N2

2m1, 2m2

U1(m1), U2 (m2 )

The number of microstates of the combined system, for given and , ism1 m2

Where is the total number of microstates of the combined system with spin excess 2m.

(N, m)

1 2, 1 1 1 2 2 2( , ) ( , ) ( , )m m N m N m N m

1

1 1

1 2

1 1 1 2 2 1

1 2

( , ) ( , ) ( , )N

m N

N m N m N m m


1 2N N N

1 1 2 2( ) ( ) ( )U m U m U m

For the combined system, we have some constraints:• Total particle count is constant: • Total energy in isolated system is constant:• Total spin excess must be:

1 22 2 2m m m

(think of possible combinations for rolling two dice)

Now and so we can conveniently sum over to count all the states for a given total energy :m2 m m1 m1 ( )U m

Exercise: Work through an example combining two N=4 systems for various m.

Counting microstates for the combined system

Critical observation: For some value of , say , the product will have a maximum value. For

a large system, it can be shown that this maximum is so large that the statistical properties of

the system are completely dominated by this most probable product:

m1 m 1

• Then this is all we will have to determine!

• To ensure that the combined system is large enough, we let one component be a reservoir or heat

bath (Recall from PYU22P10, this is an arbitrarily large system whose energy is effectively constant,

despite any energy exchange with the other component of the total system).

Eg. If N = 100, 2N ~ 1030, the correction will be 10/1030 or 1 part in 1029

(See K&K pp. 33-36 for more in-depth discussion, and pp. 37-39 for estimate of the error due to our assumption above (it’s miniscule!)

1 1 1 2 2 1 max( , ) ( , ) ( , )N m N m N m m


For our model system, with a correcting coefficient term scaling as max ~ 2N ~ N

• Furthermore, the average of a macroscopic physical quantity is accurately determined by the microstates of the most probable product

A A

Multiplicity function

2( ,0) 2NN

N

We can now seek an Equilibrium Condition under the energy exchange allowed through the boundary. Consider the internal energy U=U(m), and differentiate by U to find a maximum for :

For a maximum (For partial differentiation and max/min review, See Woolfson Ch. 4, 11) :

Now, we recall the central equation of thermodynamics (PYU22P10):

Thus we can define the statistical entropy, , and obtain the same functional form!ΩSB ln


We can write where the total volume and particle number are fixed.1 1 1 1 2 2 2 2( , , ) ( , , ) ( , , )U N V U N V U N V

1 1 2 2

1 22 1 1 2

1 2, ,

1 2

0

0

V N V N

d dU dUU U

dU dU dU

1 1 2 2

1 2

1 1 2 2, ,

1 1

V N V NU U

1 1 2 2

1 2

1 2, ,

ln ln

V N V NU U

dU TdS pdV dN

,

1

V N

S

T U

,S V

U

N

Where here is the chemical potential (Gibb’s Free Energy per particle)

1 2T TN.B. at equilibrium also (4)

K&K pp.39-41:ln g

ln 1z z

x z x

Remember in general

(product rule)

(since combined system is isolated)

Hold V, N constant.. Hold S, V constant..

also

Comparison with experiment allows us to set the constant of proportionality as Boltzmann’s constant:

Entropy is thus a quantitative measure of the randomness of a system.

Ludwig Boltzmann1844-1906


ln ( , )B BS S k N U 23 11.381 10Bk JK (5)

The additivity property of entropy

Classical entropy is additive to within experimental error. What about statistical entropy?

NB. We will use the maximum product from now on.


1

1 1 1 2 2 1ln ( , ) ln ( , ) ( , )B B

U

S k N U k N U N U U

1

1 1 1 2 2 1 max( , ) ( , )U

N U N U U

max 1 1 1 2 2 1( , ) ( , )N U N U U

max 1 1 1 2 2 1ln ln ( , ) ln ( , )B B BS k k N U k N U U

1 2S S S

But

where

Thus

I.e.

Ie. Log of a sum of a product becomes a simple log of a product

Sum becomes simple product

For most likely state 1U

Irreversible heat flow example K&K p. 44-45Entropy increase on heat flow: 10 g specimen of copper at 350 K is placed in contact with an identical specimen

at 290 K. Note for copper:


1 1 1 1

( ) ( )

0.389 10 ( 290 ) 0.389 10 (350 )

Cu Cu

P cold final cold P hot hot final

final final

U c m T T c m T T

J g K g T K J g K g K T

1 10.389Cu

Pc J g K

290 350 / 2 320finalT K K K 11.7U J

Consider now change in entropy after small (~0.1 J) energy exchange (hence temperatures remain almost same):

4 111.72.86 10

350hot

JS JK

K

4 111.73.45 10

290cold

JS JK

K

4 10.59 10combinedS JK A net increase in entropy has occurred:

4 118

23 1

exp

0.59 10exp exp(4.3 10 )

1.381 10

combinedcombined

B

S

k

JK

JK

Equivalently, a net increase in multiplicity (number of available microstates) has occurred which is rather large:

A rather large number

Consider an isolated system of 2 parts that are brought together (U, N, V constant). Initially

• If we prepare a system in a non-equilibrium state, will increase to a maximum with time, giving us a statistical understanding of a system coming to equilibrium.

SG

Internalenergy

time t

U

0

U

U1

U2

U1

U2

time t

SG S(U)

When two systems are brought into thermal contact, entropy will tend to increase.

Generalized entropy SG

• Statistical thermodynamics allows us easily to generalise entropy (call it SG) to non-equilibrium situations.


1 1 1 1 2 1( , ) ln ( ) ( )G BS U U U k U U U

1 1 max( , ) ln ( )G BS U U U k S U

1U U 2 0U 0t E.g. and at

1 2U U U

1 2U U U

1 2 maxln( )finalG BS S k

The system will evolve to be in its most probable state at equilibrium 1 1U U

max max1 2ln( )initB Gk S

See K&K pp. 45-46

Ways to increase the entropy


Add particles

Add energy(Eg. Temperature)

Expand volume

Decompose molecules

Allow polymer to curl up

Temperature is defined such that two systems in thermal equilibrium with each other will have the same value of this quantity.

To check this, consider removing a positive quantity of energy, , from system 1 to system 2U

For a spontaneous process, entropy must increase:

i.e. the transfer of energy is from the hotter body to the colder body, as we require.


1 1 2 2

1 2

1 2, ,

ln ln

V N V NU U

1 1 2 2

1 2

1 2, ,V N V N

S S

U U

1 1 2 2

1 2

1 2 1 2, ,

1 1

V N V N

S SS U U U

U U T T

1 2

1 10S U

T T

1 2T T

At equilibrium from (4)

So, via (5), we have a statistical definition of temperature:

, ,

ln1

i i i i

i iB

i iV N V N

Sk

T U U

(6)

Temperature

or

Consider our model spin system in a magnetic field:

• We will be making an adiabatic (thermally insulated) and reversible change – which is isentropic (constant entropy) We will thus first express S in terms of T and B:

• Adiabatic cooling is a technique using a trade-off between the dependence of entropy S on temperature Tand other parameters such as volume V or here, for our model spin system, the magnetic field B


Example: Adiabatic cooling by magnetic field (Mandl Chapter 5)

2

0

2( , ) exp

mN m

N

0 2

! 2( ,0) 2

2 !

NNN

NN

where (3)

22

0 0

22( , ) ln ln B

B B

k mkmS N m k k S

N N

2

0 2( , ) ( )

2

Bk US N U S U S

B N

( ) 2U m m B Switching our representation of the macrostate from m to U using :

We can then easily express S in terms of T and B

2 2

1 B

N

k US

T U B N

2 2

( )B

B NU T

k T

2 2

0 2( , , )

2 B

B NS T B N S

k T

Ie. A function of directly measurable

and controllable quantities

Step 1:• Magnetic field is switched on to isothermally magnetise the sample at 1 K. This pushes spins parallel into lower energy states with some initial entropy

• During this process step (AB in lower right diagram), a heat of magnetisationis transferred from the sample to liquid He coolant through He gas G

• Sample is then thermally isolated by pumping out the He gas around it

• B field switched off slowly, in practice over a few seconds which is sufficient for effectively reversible field reduction. The spins tend to re-randomize. This process step (BC) is adiabatic and reversible and thus has constant entropy.


Magnetic adiabatic cooling experiment

2 2

0 22

initinit

B init

B NS S

k T

2 2

0 22

final

final init

B final

B NS S S

k T

final

final init

init

BT T

B

Ie. In order to compensate for moving back to state with no magnetic field and higher disorder, the temperature must drop in an isentropic process

initTfinalT

init finalS S

0S

Binit>Bfinal

Step 2:

• Helium (He) gas present in a space G around sample P of paramagnetic salt

• is the residual field due to partial ordering of the magnetic spins, typically ~0.01 T

for a paramagnetic salt e.g. cerium magnesium nitrate.

• To get lower, can use nuclear spins with smaller residual magnetic fields, e.g. Cu, with spin 3/2.

(Bohr magneton )eB mmme if 2/

• Current record is 0.28 nK using rhodium metal.

• These coolers are used in astronomical satellites to cool detectors.


Magnetic cooling experiment

finalB

~ 0.01finalT K ~ 1initB Tfor

1initT K1initB T• Process BC is carried where S(B) dependence is strong:

Can repeat process:

initTfinalT

init finalS S

0S

We now allow particle numbers to change by moving between the two systems (diffusion) and also by combining together or breaking up (which is chemistry: Bond forming/breaking).

As before, we need the maximum in subject to these constraints:

Chemical potential: Kittel & Kroemer Ch. 5

• We consider two systems in both thermal and diffusive contact (wall between systems is porous)

• The constraints are now , , each fixed

• In particular, particle numbers can vary1 2U U U 1 2N N N 1 2V V V

1 2,N N

1 1 1 1 2 2 2 2( , , ) ( , , )U N V U N V

1 1 2 2 1 1 2 2

1 2 1 22 1 1 2 2 1 1 2

1 2 1 2, , , ,

1 2 1 2

0

0 ; 0

V N V N V U V U

d dU dU dN dNU U N N

dU dU dU dN dN dN

1 1 2 2

1 2

1 2, ,V N V N

S S

U U

1 1 2 2

1 2

1 2, ,V U V U

S S

N N

(6) When V, U held constant (as before), and

1 2T T

This is just as before, except with the extra terms in N. Reorganising and substituting :lnBS k


Using the central equation (4), we can identify the chemical potential of the system:

• As the particle number varies, the number of states varies and measures this dependence.

• In general, a system with high particle concentration has high . In approaching equilibrium, particles

diffuse from regions of high concentration to low concentration, increasing the entropy.

• If particles break up or combine, thus changing particle number, we are dealing with chemistry, and

chemical equilibrium is controlled by the chemical potential:

dU TdS pdV dN

,V U

ST

N

Where is the chemical potential, so

And, from (6), 1 = 2 at equilibrium(7)

, , s

r

r V U N

ST

N

0r r

r

For each chemical species r, holding all others s constant, with r being the stoichiometric coefficient or number of moles

and


Reservoir

System

These contain all the information necessary to calculate the thermodynamic properties of any system and are more convenient than using the multiplicity or statistical weight, i.e. equation (2).

• We consider two systems in both thermal and diffusive (particle exchange) contact, but one

system is a very large reservoir, while the other will be a single multi-particle microstate.

• Consider members of an ensemble comprising identical replicas of the system + reservoir,

one copy for each quantum state of the combination.

• System + reservoir have total particles and internal energy .0N

Partition Functions: Kittel & Kroemer Ch. 3,5

0 0( , ) ( , )i i i iP N E N N U E 0 0

0 0

( , ) ( , )

( , ) ( , )

i i i i

j j j j

P N E N N U E

P N E N N U E

and


0U

• We use the properties of the reservoir to simplify our derivation

• Probability of the system being in a single microstate i, with particle number and energy , is proportional

to the number of microstates for both system and reservoir: ie. 1 for system x number of reservoir microstates

• But microstate of the system requires macrostate of the reservoir:

iNiE

0 0( , )i iN N U E ( , )i iN E

Ni, Ei

UPDATE: K&K Uses e for N-particle microstate energy! We will use E here, reserving e for single particle (orbital) microstates like Mandl for developing Quantum statistics

lnBS k ( , ) exp BN U S k Using we have

( , )

exp( , )

i iB

j j

P N ES k

P N E i jS S S So where

But the reservoir is very large:0

1iN

N

0

1iE

Uand

( ) ( ) ( ) ( )f x f a x a f a

0 0

0 0 0 0( , ) ( , )i i i i

i iU N

S SS N N U E S N U N E

N E

0 0

1( ) ( ) ( ) ( )i j i j i j i j

U N

S SS N N U U N N E E

N U T T

exp( , )

( , ) exp

i i Bi i

j j j j B

N E k TP N E

P N E N E k T

Using a Taylor series expansion (Ch. 4, Woolfson):

Thus: (8)

For the reservoir we have


(higher order terms)

(two variable expansion to first order)

is chemical potential

• Fixing the number of particles, we obtain , the Boltzmann Factor (Canonical Distribution).

Josiah Willard Gibbs 1839-1903Yale University

(probably the greatest American physicist of the 19th century)

• The term is called the Gibbs Factor (Grand Canonical Distribution) exp i i BN E k T

exp i BE k T


0

3

2

1

The microstate occupancies n within N-particle microstate 3 (E3,T3)

Microstate j Occupancy nj

1

2

3

4

These factors describe the relative population of (or distribution between) occupied microstates (i.e. the relative number of particles in each microstate):

• Can visualize using energy states of single particle system (“orbital”) from quantum theory (with energy eigenvalues ei )

• Lower energy states are more highly populated

• As T increases, the population of higher energy states increases

E.g. N=10 particle boson system


E1, T1

e3

e2

e1

e0

< <

0

3 2 1 01 2 3 4

r j j

j

E n e

e e e e

Single particle microstate energy

level

N-particle states r

for r=3

E2, T2 E3, T3

Grand Partition Function (K&K p. 138-140)

• Is the normalising factor which converts relative probabilities to absolute probabilities.• Is convenient for calculating the thermodynamic properties of the system • We take sums over Gibbs Factors for all states of our system with a size N, over all N from 0 to ∞

[NB: degenerate states must also all be counted.]

We can take the ensemble (thermal) average of any thermodynamic quantity A such as N or E :

0

( , , ) expN

N

i B

N i

T V N E k T

(9)

(10)

(11)

0

( , ) 1i

N i

P N E

1

( , ) expi i BP N E N E k T

0 0

1( , ) ( , ) ( , ) expi i B

N i N i

A A N i P N E A N i N E k T

Note that


The absolute probability of finding the system in a microstate (dropping subscript on iN) is:, iN E

For both variable, we have the Grand Partition Function,N E

Count over microstates iN for each N

(hence we can find the average number of particles in an open system)

0

1exp

N

i i i B

N i

U E E N E k T

0

1 lnexp

1N

i i B

N i B

N U N E N E k Tk T

Example: Look for an expression for internal energy U in terms of .

(12)

(13)

0

1 lnexp

N

i B B

N i

N N N E k T k T

and

But N U N U U U N N U

ln ln

1B

B

U k Tk T

Thus we have

1 lnz z

z x x

Since in general


by (11)

Suppose that a solid hydrogen (H) atom has four possible electronic microstates

Q: Show that, to have an average of one electron per atom (ie. <N>=1), we must have chemical potential /2

• System: A single H atom. Reservoir: Rest of H-atom crystal.

• We are looking at electrons and electron states, so particle number N refers to number of electrons.

• N varies, so we must use the Grand Partition Function


Single H atom microstate Number of electrons N Energy Ei

Ground 1 -/2

Positive Ion 0 -/2

Negative Ion 2 +/2

Excited 1 +/2

Example: Electronic states of solid hydrogen

(In reality, these are just the most likely ones!)


2 2 2 2 2exp exp exp exp

B B B Bk T k T k T k T

0

1exp

N

i B B

N i

N N N E k T k T

0

( , , ) expN

i B

N i

T V N E k T

(9)(12) and

2

0

11 exp

N

i B

N i

N N E k T

We require so1N

2 2

0 0

exp expN N

i B i B

N i N i

N E k T N N E k T

Now we just do the required summations:

2 2 2 2 20 exp 1 exp exp 2 exp

B B B Bk T k T k T k T

Recall

N=0 N=1 N=2

(left)

(right)

2 2 2exp exp

B Bk T k T

2exp exp

B Bk T k T

2

N ranges 0 to 2 in this system

The Partition Function

If we fix the number of particles we obtain the Partition Function:


( , , ) exp i

i B

EZ N T V

k T

1( ) exp i

i

B

EP E

Z k T

(14)

(15)

2

1exp

ln ln

(1 )

ii

i B

B

B

EU U E

Z k T

Z Zk T

k T T

Note that Z and are extensive quantities

Absolute probability:

(1 )Bk T

To evaluate :

Consider1

2(1 ) 1 ( ) 1B

B B

k T TT

T k T k

2

1(1 )B

B

k T Tk T

and thus2

(1 )B

B

k Tk T T

(16)

Ie. A Boltzmann Factor normalized by the Partition Function

We use the central equation of thermodynamics (4) and the partition function (14) (Ie. with fixed particle number):

Entropy expressed as an absolute probability

Boltzmann expression in absolute probabilities – Gibbs expression is similar, but with double sum


1( ) exp i

i i

B

EP P E

Z k T

ln lni B iE k T P Z From (15):

i i

i

U E P i i i i

i i

dU E dP dE P

dU TdS pdV

Now and

But so we equate the first terms (variation in probability)

ln lnB i i B i

i i

dS k PdP k ZdP ln lni i B i i

i i

TdS E dP k T P Z dP

lnB i i

i

S k P P

There is no additive constant in the integration because, by the Third Law, P0=1 implies S0=0.

lnB i i

i

dS k d P P

and so

1i

i

P 0i

i

dP and thus and further,

Then by integrating:

ln ZAs is independent of i

(17)

The (Helmholtz) Free energy F

F can be expressed particularly simply in terms of Z (fixed particle number), and so is used as the main link to other thermodynamic quantities:

Reminder: F is a minimum for a system of constant N and V, in contact with a reservoir (constant T ), when the system is at equilibrium.


1( ) exp i

i i

B

EP P

Z k Te

ln lnii

B

EP Z

k T

ln lni iB i i B i

i i B

PES k P P k P Z

k T

1ln lni i B i B

i i

US PE k Z P k Z

T T

and so

F U TS lnBF k T Z Now and so we have

(from last slide)

1i

i

P Remembering

(18)

By (15) again:

Partition Function Summary

The macroscopic properties of a (closed) system are determined from statistical thermodynamics by:

• Using Z to find F from (18)

• Using to find the Partition Function, Z, from (14)

• Using F to find other macroscopic thermodynamic quantities:


• Using quantum mechanics to find the energy of N-particle microstates of the system iE

iE

, ,

lnB

N T N T

F Zp k T

V V

, ,

lnB

N V N V

k T ZFS

T T

2

,

lnB

N V

ZU F TS k T

T

• Expressions for are more complex, but particle number is useful: ln

BN k T

The Third Law of Thermodynamics (Mandl, ch. 4)

An example would be a perfect crystal where, in its ground state, each atom is on its lattice site, so the system is completely determined (no uncertainty). This form of the Third Law is used in the calculation of entropies of pure materials.

Both the Second and Third Laws of Thermodynamics emerge naturally from Statistical Thermodynamics.

• If we assume the ground state is 0–fold ( > 1) degenerate, then:

An example would be a solid solution of an alloy of two elements, A and B, where the atoms are randomly distributed on lattice sites, so the system has a residual entropy of mixing.


The entropy of a system has the property that as , where is a constant independent of the external parameters which act on the system (3rd Law).

0S S 0T 0S

• If we assume the ground state of the system is non-degenerate ( ), then as the system enters its lowest energy state of statistical weight (and ):

0T

0 1 1 0U U

0 1P U

0 0ln ln1 0B BS k k 0 0 0ln 0BS k P P or

0 0lnBS k

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Two systems in equilibrium: Entropy, temperature and