Introduction Statistical Thermodynamics · 6 Outline Rewrite History • Atoms ﬁrst! Thermodynamics last! Thermodynamics • First law: conservation of energy • Second law: in

Introduction Statistical Thermodynamics

2

Molecular Simulations◆ Molecular dynamics:

solve equations of motion

◆ Monte Carlo: importance sampling

r1

MD

MC

r2rn

r1r2rn

3

4

5

Questions

5

Questions• How can we prove that this scheme

generates the desired distribution of configurations?

5



• Why make a random selection of the particle to be displaced?

5




• Why do we need to take the old configuration again?

5





• How large should we take: delx?

5





• How large should we take: delx?

What is the desired distribution?

6

Outline

6

OutlineRewrite History

• Atoms first! Thermodynamics last!

6



Thermodynamics• First law: conservation of energy

• Second law: in a closed system entropy increase and takes its maximum value at equilibrium

6





System at constant temperature and volume• Helmholtz free energy decrease and takes its minimum value

at equilibrium

6





System at constant temperature and volume• Helmholtz free energy decrease and takes its minimum value

at equilibrium

Other ensembles:• Constant pressure

• grand-canonical ensemble

Atoms first thermodynamics next

A box of particles

A box of particles

We have given the particles an intermolecular potential

A box of particles


Newton: equations of motion

A box of particles



F(r) = -ru(r)

A box of particles



F(r) = -ru(r)

md2rdt2

= F(r)

A box of particles



F(r) = -ru(r)

Conservation of total energy

md2rdt2

= F(r)

Phase spaceThermodynamics: N,V,E


Molecular:

ΓN = r1,r2 ,…,rN ,p1,p2 ,…,pN{ }


Molecular:

ΓN = r1,r2 ,…,rN ,p1,p2 ,…,pN{ }point in phase space


Molecular:


r1,r2 ,…,rN{ }

p1,p2 ,…,pN{ }


Molecular:


ΓN 0( )

r1,r2 ,…,rN{ }

p1,p2 ,…,pN{ }


Molecular:


ΓN 0( ) trajectory: classical mechanics

r1,r2 ,…,rN{ }

p1,p2 ,…,pN{ }


Molecular:


ΓN 0( )ΓN t( )

trajectory: classical mechanics

r1,r2 ,…,rN{ }

p1,p2 ,…,pN{ }


Molecular:


ΓN 0( )ΓN t( )

trajectory: classical mechanics

r1,r2 ,…,rN{ }

p1,p2 ,…,pN{ }Why this one?

All trajectories with the same initial total energy should describe the same thermodynamic state


r1,r2 ,…,rN{ }

p1,p2 ,…,pN{ }


r1,r2 ,…,rN{ }

p1,p2 ,…,pN{ }


r1,r2 ,…,rN{ }

p1,p2 ,…,pN{ }These trajectories define a probability density in phase space

Intermezzo 1: phase rule


• Question: explain the phase rule?


• Question: explain the phase rule?• Phase rule: F=2-P+C



• F: degrees of freedom




• P: number of phases





• C: number of components





• C: number of components

• Why the 2?

Making a gas

Making a gasWhat do we need to specify to fully define a thermodynamic system?


1. Specify the volume V



2. Specify the number of particles N




3. Give the particles: initial positions initial velocities





More we cannot do: Newton takes over!





More we cannot do: Newton takes over!System will be at constant:





More we cannot do: Newton takes over!System will be at constant: N,V,E





More we cannot do: Newton takes over!System will be at constant: N,V,E

(micro-canonical ensemble)

What is the force I need to apply to prevent the wall from moving?

Pressure

How much work I do?

Collision with a wall

Elastic collisions:

Does the energy change?

Collision with a wall

Elastic collisions:

Does the energy change?

What is the force that we need to apply on the wall?

Pressure

Pressure• one particle: 2 m

vx


vx

• # particles: ρ A vx


vx


• 50% is the positive directions: 0.5


vx


• 50% is the positive directions: 0.5• P A = F = ρ A m vx

2


vx



2

• Kinetic energy: UK = ½ m v2 = ³⁄₂ kB T• (we define temperature)


vx



2

• Kinetic energy: UK = ½ m v2 = ³⁄₂ kB T• (we define temperature)

• Pressure: P V = N kB T

Experiment (1)

NVE1 NVE2

Experiment (1)

NVE1 NVE2 E1 > E2

Experiment (1)

NVE1 NVE2 E1 > E2

What will the moveable wall do?

Experiment (2)

NVE1 NVE2 E1 > E2

Experiment (2)

NVE1 NVE2 E1 > E2

Now the wall are heavy molecules

Experiment (2)

NVE1 NVE2 E1 > E2

What will the moveable wall do?Now the wall are heavy molecules

Newton + atoms

Newton + atoms

• We have a natural formulation of the first law

Newton + atoms


• We have discovered pressure

Newton + atoms


• We have discovered pressure• We have discovered another

equilibrium properties related to the total energy of the system

Thermodynamics (classical)

Experiment

NVE1 NVE2 E1 > E2

Experiment

NVE1 NVE2 E1 > E2

The wall can move and exchange energy: what determines equilibrium ?

Classical Thermodynamics


• 1st law of Thermodynamics• Energy is conserved


• 1st law of Thermodynamics• Energy is conserved

• 2nd law of Thermodynamics• Heat spontaneously flows from hot

to cold


Carnot: Entropy difference between two states:



�S = SB � SA =

�B

A

dQrev

T



�S = SB � SA =

�B

A

dQrev

T

Using the first law we have:



�S = SB � SA =

�B

A

dQrev

T




�S = SB � SA =

�B

A

dQrev

T

Using the first law we have:�U = Q + W



�S = SB � SA =

�B

A

dQrev

T


If we carry out a reversible process, we have for any point along the path

�U = Q + W



�S = SB � SA =

�B

A

dQrev

T



�U = Q + W

dU = TdS + dW



�S = SB � SA =

�B

A

dQrev

T



�U = Q + W

If we have work by a expansion of a fluid

dU = TdS + dW



�S = SB � SA =

�B

A

dQrev

T



�U = Q + W

If we have work by a expansion of a fluid

dU = TdS + dW

dU = TdS − pdV


H L

Let us look at the very initial stage

H L


dq is so small that the temperatures of the two systems do not change

H L



For system HH L



dSH = �dq

TH

For system HH L



dSH = �dq

TH

For system H

For system L

H L



dSH = �dq

TH

For system H

For system LdSL =

dq

TL

H L



Hence, for the total system

dSH = �dq

TH

For system H

For system LdSL =

dq

TL

H L




dSH = �dq

TH

For system H

For system LdSL =

dq

TL

dS = dSL + dSH = dq

�1

TL⇥

1

TH

⇥

H L




dSH = �dq

TH

For system H

For system LdSL =

dq

TL

dS = dSL + dSH = dq

�1

TL⇥

1

TH

⇥

Heat goes from warm to cold: or if dq > 0 then TH > TL

H L




dSH = �dq

TH

For system H

For system LdSL =

dq

TL

dS = dSL + dSH = dq

�1

TL⇥

1

TH

⇥


dS > 0This gives for the entropy change:

H L




dSH = �dq

TH

For system H

For system LdSL =

dq

TL

dS = dSL + dSH = dq

�1

TL⇥

1

TH

⇥


Hence, the entropy increases until the two temperatures are equal

dS > 0This gives for the entropy change:

H L

Question

Question

• Thermodynamics has a sense of time, but not Newton’s dynamics

• Look at a water atoms in reverse

• Look at a movie in reverse

Question

• Thermodynamics has a sense of time, but not Newton’s dynamics

• Look at a water atoms in reverse

• Look at a movie in reverse

• When do molecules know about the arrow of time?

Thermodynamics (statistical)

Statistical Thermodynamics


Basic assumption


For an isolated system any microscopic configuration is equally likely

Basic assumption



Basic assumption

Consequence



Basic assumption

Consequence

All of statistical thermodynamics and equilibrium thermodynamics



Basic assumption

Consequence

All of statistical thermodynamics and equilibrium thermodynamics

... b

ut cl

assic

al

ther

modyn

amics

is bas

ed o

n law

s

Ideal gas

Let us again make an ideal gas

Ideal gas


We select:

Ideal gas


We select: (1) N particles,

Ideal gas


We select: (1) N particles, (2) Volume V,

Ideal gas


We select: (1) N particles, (2) Volume V, (3) initial velocities

Ideal gas


We select: (1) N particles, (2) Volume V, (3) initial velocities + positions

Ideal gas



This fixes; V/n, U/n

Ideal gas

Basic assumption



This fixes; V/n, U/n


What is the probability to find this configuration?


The system has the same kinetic energy!!



Our basic assumption must be seriously wrong!



Our basic assumption must be seriously wrong!

... but are we doing the statistics correctly?

Question

• Is it safe to be in this room?

... lets look at our statistics correctly









Basic assumption:



Basic assumption:


P =1

total # of configurations


Basic assumption:

number 1 can be put in M positions, number 2 at M positions, etc


P =1



Basic assumption:



P =1


Total number of configurations:


Basic assumption:



P =1


Total number of configurations: MN


Basic assumption:



P =1


Total number of configurations: with MN M =V

dr


Basic assumption:



P =1


Total number of configurations: with MN

the larger the volume of the gas the more configurations

M =V

dr

1 2

3

4

5

6

7

8

9

1 2

3

4

5

What is the probability to find the 9 molecules exactly at these 9 positions?

6

7

8

9

1 2

3

4

5


6

7

8

9

✓�V

V

◆N

1

2

3

45


6

7

8

9

1

2

3

45


6

7

8

9

✓�V

V

◆N

1

2

3

45


6

7

8

9

1

2

3

45


6

7

8

9

✓�V

V

◆N

1

2

3

45


6

7

8

9

1

2

3

45


6

7

8

9

✓�V

V

◆N



exactly equal as to any other configuration!!!!!!



This is reflecting the microscopic reversibility of Newton’s equations of motion. A microscopic system has no “sense” of the direction of time


Are we asking the right question?


This is reflecting the microscopic reversibility of Newton’s equations of motion. A microscopic system has no “sense” of the direction of time


Are we asking the right question?These are microscopic properties; no irreversibility

Are we asking the right question?These are microscopic properties; no irreversibility

Thermodynamic is about macroscopic properties:


Measure densities: what is the probability that we have all our N gas particle in the upper half?

These are microscopic properties; no irreversibilityThermodynamic is about macroscopic properties:






N P(empty)




N P(empty)

1 0.5




N P(empty)

1 0.5

2 0.5 x 0.5




N P(empty)

1 0.5

2 0.5 x 0.5

3 0.5 x 0.5 x 0.5




N P(empty)

1 0.5

2 0.5 x 0.5

3 0.5 x 0.5 x 0.5

1000 10 -301


Summary

Summary

• On a microscopic level all configurations are equally likely

Summary


• On a macroscopic level; as the number of particles is extremely large, the probability that we have a fluctuation from the average value is extremely low

Summary


• On a macroscopic level; as the number of particles is extremely large, the probability that we have a fluctuation from the average value is extremely low

• Let us quantify these statements

Basic assumptionE1 > E2

NVE1 NVE2

Basic assumptionE1 > E2

Let us look at one of our examples; let us assume that the total system is isolate but heat can flow between 1 and 2.NVE1 NVE2

Basic assumption

All micro states will be equally likely!

E1 > E2


Basic assumption


... but the number of micro states that give an particular energy distribution (E1,E-E1) not ...

E1 > E2


Basic assumption


... but the number of micro states that give an particular energy distribution (E1,E-E1) not ...

E1 > E2


... so, we observe the most likely one ...

In a macroscopic system we will observe the most likely one


P(E1, E2) =N1(E1)�N2(E ⇤ E1)

�E1=EE1=0 N1(E1)�N2(E ⇤ E1)


P(E1, E2) =N1(E1)�N2(E ⇤ E1)

�E1=EE1=0 N1(E1)�N2(E ⇤ E1)

The summation only depends on the total energy:


P(E1, E2) =N1(E1)�N2(E ⇤ E1)

�E1=EE1=0 N1(E1)�N2(E ⇤ E1)

The summation only depends on the total energy:P(E1, E2) = C�N1(E1)�N2(E ⇤ E1)


P(E1, E2) =N1(E1)�N2(E ⇤ E1)

�E1=EE1=0 N1(E1)�N2(E ⇤ E1)

lnP(E1, E2) = lnC + lnN1(E1) + lnN2(E ⌅ E1)

The summation only depends on the total energy:P(E1, E2) = C�N1(E1)�N2(E ⇤ E1)


P(E1, E2) =N1(E1)�N2(E ⇤ E1)

�E1=EE1=0 N1(E1)�N2(E ⇤ E1)



We need to find the maximum

P(E1, E2) = C�N1(E1)�N2(E ⇤ E1)


P(E1, E2) =N1(E1)�N2(E ⇤ E1)

�E1=EE1=0 N1(E1)�N2(E ⇤ E1)




P(E1, E2) = C�N1(E1)�N2(E ⇤ E1)

d lnP(E1, E2)

dE1=

d lnN (E1, E2)

dE1= 0


P(E1, E2) =N1(E1)�N2(E ⇤ E1)

�E1=EE1=0 N1(E1)�N2(E ⇤ E1)




P(E1, E2) = C�N1(E1)�N2(E ⇤ E1)

d lnP(E1, E2)

dE1=

d lnN (E1, E2)

dE1= 0

d [lnN1(E1) + lnN2(E ⌅ E1)]

dE1= 0

We need to find the maximumd [lnN1(E1) + lnN2(E ⌅ E1)]

dE1= 0


d lnN1(E1)

dE1= ⇤

d lnN2(E ⇤ E1)

dE1

d [lnN1(E1) + lnN2(E ⌅ E1)]

dE1= 0


As the total energy is constant

d lnN1(E1)

dE1= ⇤

d lnN2(E ⇤ E1)

dE1

d [lnN1(E1) + lnN2(E ⌅ E1)]

dE1= 0



E2 = E � E1

d lnN1(E1)

dE1= ⇤

d lnN2(E ⇤ E1)

dE1

d [lnN1(E1) + lnN2(E ⌅ E1)]

dE1= 0



dE1 = ⇤d(E ⇤ E1) = ⇤dE2

E2 = E � E1

d lnN1(E1)

dE1= ⇤

d lnN2(E ⇤ E1)

dE1

d [lnN1(E1) + lnN2(E ⌅ E1)]

dE1= 0



dE1 = ⇤d(E ⇤ E1) = ⇤dE2

Which gives as equilibrium condition:

E2 = E � E1

d lnN1(E1)

dE1= ⇤

d lnN2(E ⇤ E1)

dE1

d [lnN1(E1) + lnN2(E ⌅ E1)]

dE1= 0



dE1 = ⇤d(E ⇤ E1) = ⇤dE2

Which gives as equilibrium condition:

E2 = E � E1

d lnN1(E1)

dE1=

d lnN2(E2)

dE2

d lnN1(E1)

dE1= ⇤

d lnN2(E ⇤ E1)

dE1

d [lnN1(E1) + lnN2(E ⌅ E1)]

dE1= 0

Let us define a property (almost S, but not quite) : S

* = lnN E( )

Let us define a property (almost S, but not quite) :

Equilibrium if:

S* = lnN E( )

d lnN1 E1( )dE1

=d lnN2 E2( )

dE2


Equilibrium if:

S* = lnN E( )

d lnN1 E1( )dE1

=d lnN2 E2( )

dE2

∂S1*

∂E1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N1 ,V1

=∂S2

*

∂E2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N2 ,V2

or


Equilibrium if:

And for the total system:

S* = lnN E( )

d lnN1 E1( )dE1

=d lnN2 E2( )

dE2

∂S1*

∂E1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N1 ,V1

=∂S2

*

∂E2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N2 ,V2

S* = S1

* + S2*

or


Equilibrium if:


For a system at constant energy, volume and number of particles the S* increases until it has reached its maximum value at equilibrium

S* = lnN E( )

d lnN1 E1( )dE1

=d lnN2 E2( )

dE2

∂S1*

∂E1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N1 ,V1

=∂S2

*

∂E2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N2 ,V2

S* = S1

* + S2*

or


Equilibrium if:


For a system at constant energy, volume and number of particles the S* increases until it has reached its maximum value at equilibrium

S* = lnN E( )

d lnN1 E1( )dE1

=d lnN2 E2( )

dE2

∂S1*

∂E1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N1 ,V1

=∂S2

*

∂E2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N2 ,V2

S* = S1

* + S2*

or

What is this magic property S*?

Defined a property S* (that is almost S):

S*(E1,E−E1)= lnℵ(E1,E−E1)= lnℵ1(E1)+ lnℵ2 (E−E1)= S1

*(E1)+ S2*(E−E1)



*(E1)+ S2*(E−E1)

Why is maximizing S* the same as maximizing N?



*(E1)+ S2*(E−E1)

Why is maximizing S* the same as maximizing N?The logarithm is a monotonically increasing function.



*(E1)+ S2*(E−E1)


Why else is the logarithm a convenient function?



*(E1)+ S2*(E−E1)


Why else is the logarithm a convenient function?Makes S* additive! Leads to extensivity.



*(E1)+ S2*(E−E1)



Why is S* not quite entropy?



*(E1)+ S2*(E−E1)



Why is S* not quite entropy?

Units! The logarithm is just a unitless quantity.

Thermal Equilibrium (Review)

E1 > E2

Isolated system that allows heat flow between 1 and 2.NVE1 NVE2


E1 > E2


ℵ(E1,E−E1)=ℵ1(E1)iℵ2 (E−E1)


Number of micro states that give an particular energy distribution (E1,E-E1) is maximized with respect to E1.

E1 > E2


ℵ(E1,E−E1)=ℵ1(E1)iℵ2 (E−E1)

For a partitioning of E between 1 and 2, the number of accessible states is maximized when:

∂S1*

∂E1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N1 ,V1

=∂S2

*

∂E2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N2 ,V2


∂S1*

∂E1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N1 ,V1

=∂S2

*

∂E2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N2 ,V2

What do these partial derivatives relate to?


∂S1*

∂E1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N1 ,V1

=∂S2

*

∂E2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N2 ,V2


Thermal equilibrium --> Temperature!


∂S1*

∂E1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N1 ,V1

=∂S2

*

∂E2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N2 ,V2



dE = TdS-pdV + µidNii=1

M

∑


∂S1*

∂E1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N1 ,V1

=∂S2

*

∂E2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N2 ,V2




M

∑

T= ∂E∂S

⎛⎝⎜

⎞⎠⎟V ,Ni

Temperature


∂S1*

∂E1

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N1 ,V1

=∂S2

*

∂E2

⎛

⎝⎜⎜⎜⎜

⎞

⎠⎟⎟⎟⎟N2 ,V2




M

∑

T= ∂E∂S

⎛⎝⎜

⎞⎠⎟V ,Ni

Temperature or1T= ∂S

∂E⎛⎝⎜

⎞⎠⎟V ,Ni

Summary

• Statistical Mechanics:• basic assumption:

• all microstates are equally likely

• Applied to NVE• Definition of Entropy: S = kB ln Ω

• Equilibrium: equal temperatures

Question

How large is Ω for a glass of water?

Question

How large is Ω for a glass of water?• For macroscopic systems, super-astronomically

large.

Question


large.

• For instance, for a glass of water at room temperature

Question


large.


Question


large.


• Macroscopic deviations from the second law of thermodynamics are not forbidden, but they are extremely unlikely.

Question

Systems at Constant Temperature (different ensembles)

The 2nd law

The 2nd law

Entropy of an isolated system can only increase; until equilibrium were it takes its maximum value

The 2nd law


Most systems are at constant temperature and volume or pressure?

The 2nd law


Most systems are at constant temperature and volume or pressure?

What is the formulation for these systems?

Constant T and V

1

Constant T and V

We have our box 1 and a bath1

Constant T and V

We have our box 1 and a bath

fixed volume but can exchange energy

1

Constant T and V

We have our box 1 and a bathTotal system is isolated and the volume is constant


1

Constant T and V


First law


1

Constant T and V


First law


dU = dq � pdV = 0

1

Constant T and V


First law


Second lawdU = dq � pdV = 0

1

Constant T and V


First law


Second law dS � 0

dU = dq � pdV = 0

1

Constant T and V


First law

Box 1: constant volume and temperature


Second law dS � 0

dU = dq � pdV = 0

1

Constant T and V


First law



Second law dS � 0

1st law:

dU = dq � pdV = 0

1

Constant T and V


First law



Second law dS � 0

1st law:

dU = dq � pdV = 0

dU1 + dUb = 0

1

Constant T and V


First law



Second law dS � 0

1st law:

dU = dq � pdV = 0

dU1 + dUb = 0 or

1

Constant T and V


First law



Second law dS � 0

1st law:

dU = dq � pdV = 0

dU1 + dUb = 0 or dU1 = �dUb

1

Constant T and V


First law



Second law dS � 0

1st law:The bath is so large that the heat flow does not influence the temperature of the bath + the process is reversible

dU = dq � pdV = 0


1

Constant T and V


First law



Second law dS � 0


dU = dq � pdV = 0


2nd law:

1

Constant T and V


First law



Second law dS � 0


dU = dq � pdV = 0


2nd law: dS1 + dSb = dS1 +dUb

T� 0

1

Constant T and V


First law



Second law dS � 0


dU = dq � pdV = 0


2nd law: dS1 + dSb = dS1 +dUb

T� 0

TdS1 � dU1 � 0

1

Constant T and V

Total system is isolated and the volume is constantBox 1: constant volume and temperature 2nd law: TdS1 � dU1 � 0


1

Constant T and V


d(U1 ⇤ TS1) � 0


1

Constant T and V


d(U1 ⇤ TS1) � 0

Let us define the Helmholtz free energy: A


1

Constant T and V


d(U1 ⇤ TS1) � 0

Let us define the Helmholtz free energy: AA � U � TS


1

Constant T and V


d(U1 ⇤ TS1) � 0


For box 1 we can write


1

Constant T and V


d(U1 ⇤ TS1) � 0


For box 1 we can write dA1 � 0


1

Constant T and V


d(U1 ⇤ TS1) � 0


For box 1 we can write dA1 � 0

Hence, for a system at constant temperature and volume the Helmholtz free energy decreases and takes its minimum value at equilibrium


1

51

Canonical ensembleConsider a small system that can exchange heat with a big reservoir

51


51


1/kBT

51


51


51


Hence, the probability to find Ei:

51



51



51



Boltzmann distribution

51



Boltzmann distribution

52

ThermodynamicsWhat is the average energy of the system?

52


52


52


52


Compare:

53

ThermodynamicsFirst law of thermodynamics

53


Helmholtz Free energy:

53



53



53



53



54

What is the average energy of the system?

54


54


54


54


Compare:

54


Compare: Hence:

We have assumed that we can count states

Atoms?


Quantum Mechanics: energy discreet

Atoms?



What to do for classical model such as an ideal gas, hard spheres, Lennard-Jones?

Atoms?




Energy is continue: • potential energy • kinetic energy

Atoms?





Particle in a box:

Atoms?





Particle in a box:εn =

nh( )28mL2

Atoms?






nh( )28mL2

Atoms?






nh( )28mL2

What are the energy levels for Argon in a 1-dimensional box of 1 cm?

Atoms?

εn =nh( )28mL2


εn =nh( )28mL2


Argon: m=40 g/mol=6.63×10-26 kg h=6.63×10-34 J s

εn =nh( )28mL2



εn = 5 ×10−39n2 J( )

Kinetic energy of Ar at room temperature ≈4.14 × 10-21 J

εn =nh( )28mL2



εn = 5 ×10−39n2 J( )


εn =nh( )28mL2



εn = 5 ×10−39n2 J( )

Many levels are occupied: only at very low temperatures or very small volumes one can see quantum effect!


εn =nh( )28mL2



εn = 5 ×10−39n2 J( )


qtranslational = e−

nh( )28mL2kBT

n=1

∞

∑


εn =nh( )28mL2



εn = 5 ×10−39n2 J( )



nh( )28mL2kBT

n=1

∞

∑


nh( )28mL2kBT dn

0

∞

∫


εn =nh( )28mL2



εn = 5 ×10−39n2 J( )



nh( )28mL2kBT

n=1

∞

∑


nh( )28mL2kBT dn

0

∞

∫ qtranslational =2πmkBT

h2⎛⎝⎜

⎞⎠⎟

12L


εn =nh( )28mL2



εn = 5 ×10−39n2 J( )



nh( )28mL2kBT

n=1

∞

∑


nh( )28mL2kBT dn

0

∞


h2⎛⎝⎜

⎞⎠⎟

12L

3D: qtranslational =2πmkBT

h2⎛⎝⎜

⎞⎠⎟

32V = V

Λ3


εn =nh( )28mL2



εn = 5 ×10−39n2 J( )



nh( )28mL2kBT

n=1

∞

∑


nh( )28mL2kBT dn

0

∞


h2⎛⎝⎜

⎞⎠⎟

12L

3D: qtranslational =2πmkBT

h2⎛⎝⎜

⎞⎠⎟

32V = V

Λ3De Broglie wavelength

Partition function; q = e−Enn=1

∞

∑ = e−En dn∫


∞

∑ = e−En dn∫

Hamiltonian: H =Ukin +Upot =pi2

2mi

+Upot rN( )i∑


∞

∑ = e−En dn∫


2mi

+Upot rN( )i∑

Z1,V ,T = C e− p2

2m kBT dp3N∫ e−Up r( )kBT dr3N∫


∞

∑ = e−En dn∫


2mi

+Upot rN( )i∑

Z1,V ,T = C e− p2


One ideal gas molecule: Up(r)=0


∞

∑ = e−En dn∫


2mi

+Upot rN( )i∑

Z1,V ,T = C e− p2



Z1,V ,TIG = C e

− p2

2m kBT dp∫ dr∫ = CV 2πmkBT( )32

Partition function;

qtranslational =2πmkBT

h2⎛⎝⎜

⎞⎠⎟

32V = V

Λ3

q = e−Enn=1

∞

∑ = e−En dn∫


2mi

+Upot rN( )i∑

Z1,V ,T = C e− p2



Z1,V ,TIG = C e

− p2


Partition function;

qtranslational =2πmkBT

h2⎛⎝⎜

⎞⎠⎟

32V = V

Λ3

q = e−Enn=1

∞

∑ = e−En dn∫


2mi

+Upot rN( )i∑

Z1,V ,T = C e− p2



Z1,V ,TIG = C e

− p2


Z1,V ,TIG = CV 2πmkBT( )

32 = V

Λ3

N gas molecules:

ZN ,V ,T =1h3N

e− p2

2m kBT dpN∫ e−U r( )kBT drN∫

WRONG!Particles are

indistinguishable

N gas molecules:

ZN ,V ,T =1h3N

e− p2


WRONG!Particles are

indistinguishable

N gas molecules:

ZN ,V ,T =1

h3NN!e− p2


ZN ,V ,T =1h3N

e− p2


WRONG!Particles are

indistinguishable

N gas molecules:

ZN ,V ,T =1

h3NN!e− p2


ZN ,V ,T =1h3N

e− p2


Configurational part of the partition function:

WRONG!Particles are

indistinguishable

N gas molecules:

ZN ,V ,T =1

h3NN!e− p2


QN ,V ,T =1

Λ3NN!e−U r( )kBT drN∫

ZN ,V ,T =1h3N

e− p2


Configurational part of the partition function:

Question

• For an ideal gas, calculate:• the partition function

• the pressure

• the energy

• the chemical potential

ideal gas molecules: QN ,V ,TIG = V N

Λ3NN!

All thermodynamics follows from the partition function!


Λ3NN!

Free energy:



Λ3NN!

Free energy: FIG = −kBT lnQN ,V ,TIG = kBTN lnΛ3 − ln V N( )⎡⎣ ⎤⎦



Λ3NN!


FIG = F0 + kBTN lnρ



Λ3NN!




Pressure:


Λ3NN!




Pressure: p = − ∂F∂V

⎛⎝⎜

⎞⎠⎟ T ,N

= kBTN1V


Λ3NN!





⎛⎝⎜

⎞⎠⎟ T ,N

= kBTN1V

Energy:


Λ3NN!





⎛⎝⎜

⎞⎠⎟ T ,N

= kBTN1V

Energy: E = ∂F T∂1 T

⎛⎝⎜

⎞⎠⎟V ,N

= 3kBN∂lnΛ∂1 T

⎛⎝⎜

⎞⎠⎟V ,N


Λ3NN!





⎛⎝⎜

⎞⎠⎟ T ,N

= kBTN1V


⎛⎝⎜

⎞⎠⎟V ,N

= 3kBN∂lnΛ∂1 T

⎛⎝⎜

⎞⎠⎟V ,N

Λ = h2

2πmkBT⎛⎝⎜

⎞⎠⎟

12


Λ3NN!





⎛⎝⎜

⎞⎠⎟ T ,N

= kBTN1V


⎛⎝⎜

⎞⎠⎟V ,N

= 3kBN∂lnΛ∂1 T

⎛⎝⎜

⎞⎠⎟V ,N

Λ = h2

2πmkBT⎛⎝⎜

⎞⎠⎟

12

E = 32NkBT

61

Chemical potential: µi =∂F∂Ni

⎛⎝⎜

⎞⎠⎟ T ,V ,N j

61


⎛⎝⎜

⎞⎠⎟ T ,V ,N j

βF = N lnΛ3 + N ln N

V⎛⎝⎜

⎞⎠⎟

61


⎛⎝⎜

⎞⎠⎟ T ,V ,N j


V⎛⎝⎜

⎞⎠⎟

βµ = lnΛ3 + lnρ +1

61


⎛⎝⎜

⎞⎠⎟ T ,V ,N j


V⎛⎝⎜

⎞⎠⎟

βµ = lnΛ3 + lnρ +1

βµIG = βµ0 + lnρ

62

Summary:Canonical ensemble (N,V,T)

62


Partition function:

62


Partition function:

Probability to find a particular configuration

62


Partition function:


Free energy

63

Summary:micro-canonical ensemble (N,V,E)

63


Partition function:

63


Partition function:


63


Partition function:


Free energy

Other Ensemble

65

Other ensembles?

65

Other ensembles?In the thermodynamic limit the thermodynamic properties are independent of the ensemble: so buy a bigger computer …

65


However, it is most of the times much better to think and to carefully select an appropriate ensemble.

65



For this it is important to know how to simulate in the various ensembles.

65




But for doing this wee need to know the Statistical Thermodynamics of the various ensembles.

65





COURSE: MD and MC different

ensembles

65





66

Example (1): vapour-liquid equilibrium mixture

composition

T

L

VL+V

66


Measure the composition of the coexisting vapour and liquid phases if we start with a homogeneous liquid of two different compositions:

composition

T

L

VL+V

66



• How to mimic this with the N,V,T ensemble?

composition

T

L

VL+V

66



• How to mimic this with the N,V,T ensemble?

• What is a better ensemble?composition

T

L

VL+V

67

Example (2): swelling of clays

67

Example (2): swelling of clays

Deep in the earth clay layers can swell upon adsorption of water:

• How to mimic this in the N,V,T ensemble?

• What is a better ensemble to use?

68

Ensembles

68

Ensembles

• Micro-canonical ensemble: E,V,N

• Canonical ensemble: T,V,N

• Constant pressure ensemble: T,P,N

• Grand-canonical ensemble: T,V,µ

Constant pressure

fixed N but can exchange energy + volume

1

We have our box 1 and a bath


1



1


First law


1


First law dU = dq � pdV = 0


1


First lawSecond law

dU = dq � pdV = 0


1


First lawSecond law dS � 0

dU = dq � pdV = 0


1


First law

Box 1: constant pressure and temperature Second law dS � 0

dU = dq � pdV = 0


1


First law


1st law:

dU = dq � pdV = 0


1


First law


1st law:

dU = dq � pdV = 0

dU1 + dUb = 0


1


First law


1st law:

dU = dq � pdV = 0

or dU1 + dUb = 0


1


First law


1st law:

dU = dq � pdV = 0

or dU1 = �dUbdU1 + dUb = 0


1


First law


1st law:

dU = dq � pdV = 0

or dU1 = �dUbdU1 + dUb = 0dV1 + dVb = 0


1


First law


1st law:

dU = dq � pdV = 0

or dU1 = �dUbdU1 + dUb = 0dV1 + dVb = 0 or


1


First law


1st law:

dU = dq � pdV = 0

or dU1 = �dUbdU1 + dUb = 0dV1 + dVb = 0 or dV1 = �dVb


1


First law


1st law:

The bath is very large and the small changes do not change P or T; in addition the process is reversible

dU = dq � pdV = 0

or dU1 = �dUbdU1 + dUb = 0dV1 + dVb = 0 or dV1 = �dVb


1


First law


1st law:


dU = dq � pdV = 0

or dU1 = �dUb

2nd law:

dU1 + dUb = 0dV1 + dVb = 0 or dV1 = �dVb


1


First law


1st law:


dU = dq � pdV = 0

or dU1 = �dUb

2nd law:


dS1 + dSb = dS1 +dUb

T+

p

TdVb � 0


1


First law


1st law:


dU = dq � pdV = 0

or dU1 = �dUb

2nd law:


TdS1 � dU1 � pdV1 � 0

dS1 + dSb = dS1 +dUb

T+

p

TdVb � 0


1

Total system is isolated and the volume is constantBox 1: constant pressure and temperature 2nd law:TdS1 � dU1 � pdV1 � 0


1

Total system is isolated and the volume is constantBox 1: constant pressure and temperature 2nd law:

d(U1 ⌅ TS1 + pV1) � 0

TdS1 � dU1 � pdV1 � 0


1


Let us define the Gibbs free energy: G

d(U1 ⌅ TS1 + pV1) � 0

TdS1 � dU1 � pdV1 � 0


1


Let us define the Gibbs free energy: G

d(U1 ⌅ TS1 + pV1) � 0

G � U ⇥ TS + pV

TdS1 � dU1 � pdV1 � 0


1


Let us define the Gibbs free energy: GFor box 1 we can write

d(U1 ⌅ TS1 + pV1) � 0

G � U ⇥ TS + pV

TdS1 � dU1 � pdV1 � 0


1



d(U1 ⌅ TS1 + pV1) � 0

G � U ⇥ TS + pV

dG1 � 0

TdS1 � dU1 � pdV1 � 0


1



Hence, for a system at constant temperature and pressure the Gibbs free energy decreases and takes its minimum value at equilibrium

d(U1 ⌅ TS1 + pV1) � 0

G � U ⇥ TS + pV

dG1 � 0

TdS1 � dU1 � pdV1 � 0


1

N,P,T ensemble

Consider a small system that can exchange volume and energy with a big reservoir

N,P,T ensemble


N,P,T ensemble


The terms in the expansion follow from the connection with Thermodynamics:

N,P,T ensemble


S = kB lnΩThe terms in the expansion follow from the connection with Thermodynamics:

N,P,T ensemble



We have:

N,P,T ensemble



dS = 1TdU + p

TdV − µi

TdNi∑We have:

N,P,T ensemble



dS = 1TdU + p

TdV − µi

TdNi∑We have:

∂S∂U

⎛⎝⎜

⎞⎠⎟V ,Ni

= 1T

N,P,T ensemble



dS = 1TdU + p

TdV − µi

TdNi∑We have:

∂S∂U

⎛⎝⎜

⎞⎠⎟V ,Ni

= 1T

and

N,P,T ensemble



dS = 1TdU + p

TdV − µi

TdNi∑We have:

∂S∂U

⎛⎝⎜

⎞⎠⎟V ,Ni

= 1T

∂S∂V

⎛⎝⎜

⎞⎠⎟ E ,Ni

= pT

and

lnΩ V −Vi ,E − Ei( ) = lnΩ V ,E( )− ∂ lnΩ

∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂V

⎛⎝⎜

⎞⎠⎟ E ,N

Vi +!

lnΩ V −Vi ,E − Ei( ) = lnΩ V ,E( )− Ei

kBT− pkBT

Vi


∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂V

⎛⎝⎜

⎞⎠⎟ E ,N

Vi +!


kBT− pkBT

Vi


∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂V

⎛⎝⎜

⎞⎠⎟ E ,N

Vi +!

Hence, the probability to find Ei,Vi:


kBT− pkBT

Vi


∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂V

⎛⎝⎜

⎞⎠⎟ E ,N

Vi +!

Hence, the probability to find Ei,Vi:


kBT− pkBT

Vi


∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂V

⎛⎝⎜

⎞⎠⎟ E ,N

Vi +!

Partition function:

Partition function: Δ N ,P,T( ) = expi, j∑ − Ei

kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥


kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥

Ensemble average:


kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥

V =Vj expi, j∑ − Ei

kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥

Δ N , p,T( ) = −kBT∂ lnΔ∂p

⎛⎝⎜

⎞⎠⎟ T ,N

Ensemble average:


kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥


kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥


⎛⎝⎜

⎞⎠⎟ T ,N

Ensemble average:

Thermodynamics


kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥


kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥


⎛⎝⎜

⎞⎠⎟ T ,N

Ensemble average:

dG = −SdT +Vdp + µidNi∑Thermodynamics


kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥


kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥


⎛⎝⎜

⎞⎠⎟ T ,N

Ensemble average:


V = ∂G∂p

⎛⎝⎜

⎞⎠⎟ T ,N


kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥


kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥


⎛⎝⎜

⎞⎠⎟ T ,N

Ensemble average:


V = ∂G∂p

⎛⎝⎜

⎞⎠⎟ T ,NHence:


kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥


kBT−pVj

kBT⎡

⎣⎢

⎤

⎦⎥


⎛⎝⎜

⎞⎠⎟ T ,N

Ensemble average:


V = ∂G∂p

⎛⎝⎜

⎞⎠⎟ T ,NHence:

GkBT

= − lnΔ N , p,T( )

75

SummaryIn the classical limit, the partition function becomes

75


75


The probability to find a particular configuration:

grand-canonical ensemble

Grand-canonical ensemble

Classical• A small system that can exchange heat and

particles with a large bath

Statistical• Taylor expansion of a small reservoir

Constant T and μ

1

Constant T and μ

1

exchange energy and particles

Constant T and μ

Total system is isolated and the volume is constant1


Constant T and μ

Total system is isolated and the volume is constantFirst lawdU = TdS − pdV + µdN = 0

1


Constant T and μ

Total system is isolated and the volume is constantFirst law

Second lawdU = TdS − pdV + µdN = 0

1


Constant T and μ


Second law dS � 0

dU = TdS − pdV + µdN = 0

1


Constant T and μ


Box 1: constant chemical potential and temperature

Second law dS � 0


1


Constant T and μ



Second law dS � 0

1st law:


1


Constant T and μ



Second law dS � 0

1st law: or dU1 = �dUbdU1 + dUb = 0


1


Constant T and μ



Second law dS � 0

1st law: or dU1 = �dUbdU1 + dUb = 0or


dN1 + dNb = 0 dNb = −dN1

1


Constant T and μ



Second law dS � 0

1st law:

The bath is very large and the small changes do not change μ or T; in addition the process is reversible

or dU1 = �dUbdU1 + dUb = 0or



1


Constant T and μ



Second law dS � 0

1st law:


or dU1 = �dUb

2nd law:

dU1 + dUb = 0or



1


Constant T and μ



Second law dS � 0

1st law:


or dU1 = �dUb

2nd law:

dU1 + dUb = 0or



dS1 + dSb = dS1 +1TbdUb −

µb

TbdNb ≥ 0

1



µb

TbdNb ≥ 0


µb

TbdNb ≥ 0

We can express the changes of the bath in terms of properties of the system


µb

TbdNb ≥ 0


dS1 −1TdU1 +

µTdN1 ≥ 0


µb

TbdNb ≥ 0


dS1 −1TdU1 +

µTdN1 ≥ 0 d TS1 −U1 + µN1( ) ≥ 0


µb

TbdNb ≥ 0


dS1 −1TdU1 +

µTdN1 ≥ 0 d TS1 −U1 + µN1( ) ≥ 0d U −TS − µN( ) ≤ 0


µb

TbdNb ≥ 0


dS1 −1TdU1 +


For the Gibbs free energy we can write:


µb

TbdNb ≥ 0


dS1 −1TdU1 +


G ≡U −TS + pVG = µN



µb

TbdNb ≥ 0


dS1 −1TdU1 +




− pV =U −TS − µNor


µb

TbdNb ≥ 0


dS1 −1TdU1 +




− pV =U −TS − µNor

Giving:


µb

TbdNb ≥ 0


dS1 −1TdU1 +




− pV =U −TS − µN

d − pV( ) ≤ 0

or

Giving:


µb

TbdNb ≥ 0


dS1 −1TdU1 +





d − pV( ) ≤ 0

or

or d pV( ) ≥ 0Giving:


µb

TbdNb ≥ 0


dS1 −1TdU1 +





d − pV( ) ≤ 0

or

or

Hence, for a system at constant temperature and chemical potential pV increases and takes its maximum value at equilibrium

d pV( ) ≥ 0Giving:

80

µ,V,T ensemble

Consider a small system that can exchange particles and energy with a big reservoir

80

µ,V,T ensemble


lnΩ E − Ei ,N − N j ,( ) = lnΩ E,N( )− ∂ lnΩ

∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂N

⎛⎝⎜

⎞⎠⎟ E ,V

N j +!

80

µ,V,T ensemble



∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂N

⎛⎝⎜

⎞⎠⎟ E ,V

N j +!

The terms in the expansion follow from the connection with Thermodynamics:

80

µ,V,T ensemble



∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂N

⎛⎝⎜

⎞⎠⎟ E ,V

N j +!


80

µ,V,T ensemble



∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂N

⎛⎝⎜

⎞⎠⎟ E ,V

N j +!


dS = 1TdU + p

TdV − µ

TdN

80

µ,V,T ensemble



∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂N

⎛⎝⎜

⎞⎠⎟ E ,V

N j +!


dS = 1TdU + p

TdV − µ

TdN

∂S∂U

⎛⎝⎜

⎞⎠⎟V ,N

= 1T

80

µ,V,T ensemble



∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂N

⎛⎝⎜

⎞⎠⎟ E ,V

N j +!


dS = 1TdU + p

TdV − µ

TdN

∂S∂U

⎛⎝⎜

⎞⎠⎟V ,N

= 1T

∂S∂N

⎛⎝⎜

⎞⎠⎟ E ,V

= − µT


∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂N

⎛⎝⎜

⎞⎠⎟ E ,V

N j +!

lnΩ E − Ei ,N − N j( ) = lnΩ E,N( )− Ei

kBT+ µkBT

N j


∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂N

⎛⎝⎜

⎞⎠⎟ E ,V

N j +!


kBT+ µkBT

N j


∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂N

⎛⎝⎜

⎞⎠⎟ E ,V

N j +!

lnΩ E − Ei ,N − N j( )

Ω E,N( ) = − Ei

kBT+ µkBT

N j

Hence, the probability to find Ei,Nj:


kBT+ µkBT

N j


∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂N

⎛⎝⎜

⎞⎠⎟ E ,V

N j +!


Ω E,N( ) = − Ei

kBT+ µkBT

N j

Hence, the probability to find Ei,Nj:


kBT+ µkBT

N j


∂E⎛⎝⎜

⎞⎠⎟V ,N

Ei −∂ lnΩ∂N

⎛⎝⎜

⎞⎠⎟ E ,V

N j +!


Ω E,N( ) = − Ei

kBT+ µkBT

N j

P Ei ,N j( ) = Ω E − Ei ,N − N j( )Ω E − Ek ,N − Nl( )k ,l∑ ∝ exp − Ei

kBT+ µNi

kBT⎡

⎣⎢

⎤

⎦⎥

82

μ,V,T ensemble (2)In the classical limit, the partition function becomes

82


82


The probability to find a particular configuration:

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