Chapter 3: Interactions and Implications.Start with Thermodynamic Identities

, , ln , ,BS U V N k U V N

, , ,N V N U U V

S S SdS dU dV dN

U V N

TU

S

NV

1

,

,U N

S P

V T

,

?U V

S

N

Diffusive Equilibrium and Chemical Potential

0,

AA NVA

AB

U

S0

,

AA VUA

AB

N

S

B

B

A

A

N

S

N

S

VUN

ST

,

BA

Sign “-”: out of equilibrium, the system with the larger S/N will get more particles. In other words, particles will flow from from a high /T to a low /T.

Let’s fix VA and VB (the membrane’s position is fixed), but assume that the membrane becomes permeable for gas molecules (exchange of both U and N between the sub-systems, the molecules in A and B are the same ).

UA, VA, NA UB, VB, NB

TN

S

VU

,

For sub-systems in diffusive equilibrium:

In equilibrium, BA TT

- the chemicalpotential

0.2

0.4

0.6

0.8

0.2

0.4

0.6

0.8

-6

-5

-4

0.2

0.4

0.6

0.8

S

AN

AU

Chemical Potential: examples

Einstein solid: consider a small one, with N = 3 and q = 3.

10

!)1(!

!1)3,3(

Nq

NqqN 10ln)3,3( BkqNS

let’s add one more oscillator: 20ln)3,4( BkqNS

NdT

UdT

dS

1

SVN

U

,

SN

U

To keep dS = 0, we need to decrease the energy, by subtracting one energy quantum.

Thus, for this system

S

N

U

2

5ln

3

4ln),,( 2/5

2/3

2NU

h

mVkNUVNS B

N

VTmgTkTk

h

m

N

VTk

N

ST BBB

VU

2/32/3

2,

ln2

ln

Monatomic ideal gas:

At normal T and P, ln(...) > 0, and < 0 (e.g., for He, ~ - 5·10-20 J ~ - 0.3 eV.

Sign “-”: usually, by adding particles to the system, we increase its entropy. To keep dS = 0, we need to subtract some energy, thus U is negative.

The Quantum Concentration

At T=300K, P=105 Pa , n << nQ. When n nQ, the quantum statistics comes into play.

2/3

2

2

Tkh

mn BQ

- the so-called quantum concentration (one particle per cube of side equal to the thermal de Broglie wavelength). When nQ >> n, the gas is in the classical regime.

2/52/3

32/322/3

2 2ln

2ln

2ln

Tk

P

m

hTk

Tmk

hnTkTk

h

m

N

VTk

B

BB

BBB

n=N/V – the concentration of molecules

The chemical potential increases with the density of the gas or with its pressure. Thus, the molecules will flow from regions of high density to regions of lower density or from regions of high pressure to those of low pressure .

0

QB

BB n

nTk

Tmk

hnTk ln

2ln

2/32

when n nQ, 0

2/3

23

1

h

Tmkn

Tmk

h

p

h B

dB

Q

B

dB

when nincreases

Entropy Change for Different Processes

Type of interaction

Exchanged quantity

Governing variable

Formula

thermal energy temperature

mechanical volume pressure

diffusive particles chemical potential

NVU

S

T ,

1

NUV

S

T

P

,

VUN

S

T ,

The last column provides the connection between statistical physics and thermodynamics.

The partial derivatives of S play very important roles because they determine how much the entropy is affected when U, V and N change:

Thermodynamic Identity for dU(S,V,N)

NVUSS ,, if monotonic as a function of U (“quadratic” degrees of freedom!), may be inverted to give NVSUU ,,

dNN

UdV

V

UdS

S

UdU

VSNSNV ,,,

compare withTU

S

VN

1

,

SVSNVNN

UP

V

UT

S

U

,,,

pressure chemical potential

This holds for quasi-static processes (T, P, are well-define throughout the system).

NddVPSdTUd

shows how much the system’s energy changes when one particle is added to the system at fixed S and V. The chemical potential units – J.

- the so-called thermodynamic identity for U

Thermodynamic Identities

is an intensive variable, independent of the size of the system (like P, T, density). Extensive variables (U, N, S, V ...) have a magnitude proportional to the size of the system. If two identical systems are combined into one, each extensive variable is doubled in value.

With these abbreviations:

shows how much the system’s energy changes when one particle is added to the system at fixed S and V. The chemical potential units – J.

The coefficients may be identified as: TN

S

T

P

V

S

TU

S

UVUNVN

,,,

1

This identity holds for small changes S provided T and P are well defined.

- the so-called thermodynamic identity

The 1st Law for quasi-static processes (N = const):

The thermodynamic identity holds for the quasi-static processes (T, P, are well-define throughout the system)

dU TdS PdV dN

dU TdS PdV

1 PdS dU dV dN

T T T

, , ,N V N U U V

S S SdS dU dV dN

U V N

The Equation(s) of State for an Ideal Gas

2/,, NfNUVNgNVU Ideal gas:(fN degrees of freedom) / 2, , ln f

BS U V N Nk g N VU

V

kNT

V

STP B

NU

,

TkNPV B

UkN

f

U

S

T BNV

1

2

1

,

The “energy” equation of state (U T):

TkNf

U B2

The “pressure” equation of state (P T):

- we have finally derived the equation of state of an ideal gas from first principles! In other words, we can calculate the thermodynamic information for an isolated system by counting all the accessible microstates as a function of N, V, and U.

Ideal Gas in a Gravitational Field

Pr. 3.37. Consider a monatomic ideal gas at a height z above sea level, so each molecule has potential energy mgz in addition to its kinetic energy. Assuming that the atmosphere is isothermal (not quite right), find and re-derive the barometric equation.

NmgzUU kin

NddVPSdTUd

3/ 2

2,

2( ) lnB B

S V

U V mz mgz k T k T mgz

N N z h

In equilibrium, the chemical potentials between any two heights must be equal:

2/3

2

2/3

2

2

)0(ln

2

)(ln Tk

h

m

N

VTkmgzTk

h

m

zN

VTk BBBB

)0(ln)(ln NTkmgzzNTk BB Tk

mgz

BeNzN

)0()(

note that the U that appears in the Sackur-Tetrode equation represents only the kinetic energy

Pr. 3.32. A non-quasistatic compression. A cylinder with air (V = 10-3 m3, T = 300K, P =105 Pa) is compressed (very fast, non-quasistatic) by a piston (A = 0.01 m2, F = 2000N, x = 10-3m). Calculate W, Q, U, and S.

VPSTU

WQU holds for all processes, energy conservation

quasistatic, T and P are well-defined for any intermediate state

quasistatic adiabatic isentropic non-quasistatic adiabatic

JmPa 11010101

111

11

11

1

1)(

2335

1

1

11

x

xVP

x

xVP

V

VVP

VV

VP

dVV

VPconstPVdVVPW

ii

ii

f

iii

if

ii

V

V

ii

V

V

f

i

f

i

fi

V

V

VVPdVPWf

i

J2

m10m10Pa102 23225

The non-quasistatic process results in a higher T and a greater entropy of the final state.

S = const along the isentropic

line

P

V Vi Vf

1

2

2*

Q = 0 for both

Caution: for non-quasistatic adiabatic processes, S might be non-zero!!!

An example of a non-quasistatic adiabatic process

NfkUkNf

VkNNVUS BBB lnln2

ln,, Direct approach:

adiabatic quasistatic isentropic 0Q WU VPTkNf

B 2

VV

TkNTkN

f BB

2constTV f 2/

f

f

i

i

f

V

V

T

T/2

0ln2

2ln

i

fB

i

fB V

V

fkN

f

V

VkNS

adiabatic non-quasistatic

5 5 5

ln ln2 2

1

2 10 10 10 1J/K

300 300

f f f i f iB B B B

i i i i

i i i

i i i i

i

i

V U V V U Uf fS Nk Nk Nk Nk

V U V U

P U P V P V P VV U

T T T T

P P V

T

J 250m 10Pa 102

5

2

5

233-5 iiiBi VPTkN

fUJWU 2210

V

V

K/J300

2

K300

2J

T

Q

T

US

The entropy is created because it is an irreversible, non-quasistatic compression.

P

V Vi Vf

To calculate S, we can consider any quasistatic process that would bring the gas into the final state (S is a state function). For example, along the red line that coincides with the adiabata and then shoots straight up. Let’s neglect small variations of T along this path ( U << U, so it won’t be a big mistake to assume T const):

K/J300

1

K300

1J2J(adiabata) 0

T

QS

U = Q = 1J

For any quasi-static path from 1 to 2, we must have the same S. Let’s take another path – along the isotherm and then straight up:

1

2

P

V Vi Vf

1

2

U = Q = 2J

K/J300

1

K300

10m 10Pa 10

ln )(1

2-33-5

x

x

T

VP

V

V

T

VP

V

dV

T

VPdVVP

TS

ii

i

fii

V

V

ii

V

V

f

i

f

i

isotherm:

“straight up”:

Total gain of entropy: K/J300

1K/J

300

2K/J

300

1S

The inverse process, sudden expansion of an ideal gas (2 – 3) also generates entropy (adiabatic but not quasistatic). Neither heat nor work is transferred: W = Q = 0 (we assume the whole process occurs rapidly enough so that no heat flows in through the walls).

i

fBrev V

VTkNW ln

The work done on the gas is negative, the gas does positive work on the piston in an amount equal to the heat transfer into the system

J/K300

1ln0

ii

ii

f

iB

revrevrevrev V

V

T

VP

V

VNk

T

W

T

QSWQ

P

V Vi Vf

1

2

3 Because U is unchanged, T of the ideal gas is unchanged. The final state is identical with the state that results from a reversible isothermal expansion with the gas in thermal equilibrium with a reservoir. The work done on the gas in the reversible expansion from volume Vf to Vi:

Thus, by going 1 2 3 , we will increase the gas entropy by J/K300

2321 S

Systems with a “Limited” Energy Spectrum

1

,

NV

U

ST

The definition of T in statistical mechanics is consistent with our intuitive idea of the temperature  (the more energy we deliver to a system, the higher its temperature) for many, but not all systems.

“Unlimited” Energy Spectrum

T > 0

the multiplicity increase monotonically with U : U f N/2

U

S

C

T

U

U

NVU

S

T ,

1

VV T

UC

S

U

T

U C

U

Pr. 1.55

0

0

,

2

2

,

NV

NV

U

S

U

S

ideal gas in thermal equilibriumself-gravitating ideal

gas(not in thermal

equilibrium)

T > 0

0VC0VC

Pr. 3.29. Sketch a graph of the entropy of H20 as a function of T at P = const, assuming that CP is almost const at high T.

dTT

CS PP

T

C

T

S P

P

At T 0, the graph goes to 0 with zero slope. At high T, the rate of the S increase slows down (CP const). When solid melts, there is a large S at T = const, another jump – at liquid–gas phase transformation.

S

T

ice water vapor

“Limited” Energy Spectrum: two-level systems

e.g., a system of non-interacting spin-1/2 particles in external magnetic field. No “quadratic” degrees of freedom (unlike in an ideal gas, where the kinetic energies of molecules are unlimited), the energy spectrum of the particles is confined within a finite interval of E (just two allowed energy levels).

-1.0 -0.5 0.0 0.5 1.00.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

T < 0T > 0

S / N

k B

U / NB

S

U

E

the multiplicity and entropy decrease for some range of U

in this regime, the system isdescribed by a negative T

S

U

T U

2NB

01

,

VNU

S

T

Systems with T < 0 are “hotter” than the systems at any positive temperature - when such systems interact, the energy flows from a system with T < 0 to the one with T > 0 .

½ Spins in Magnetic Field

The total energy of the system:

NNBNNBMBU 2

NNN

B

N - the number of “up” spins

N - the number of “down” spins

Our plan: to arrive at the equation of state for a two-state paramagnet U=U (N,T,B) using the multiplicity as our starting point.

- the magnetic moment of an individual spin

E

E1 = - B

E2 = + B

0an arbitrary choice

of zero energy

The magnetization:

NNNNM 2

(N,N) S (N,N) = kB ln (N,N) U =U (N,T,B)

1),(

U

UNST

BN

UNN

1

2

BN

UNN

1

2Express and with and , N N N U

From Multiplicity – to S(N) and S(U)

!!

!),(

NN

NNNThe multiplicity of any macrostate

with a specified N:

NNNNNNNNNNNN

NNNNNNN

N

k

NNS

B

lnlnlnln!ln

!ln!ln!ln!!

!ln

,

Max. S at N = N (N= N/2): S=NkBln2

BN

U

BN

UN

BN

U

BN

UNNk

BN

UN

BN

UN

BN

UN

BN

UNNNkUNS

B

B

1ln12

1ln12

2ln

12

ln12

12

ln12

ln,

-1.0 -0.5 0.0 0.5 1.00.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

T < 0T > 0

S / N

k B

U / NB

ln2 0.693

1 , 12 2 2

N N N N U N UN N

U B N N N B N B

From S(U,N) – to T(U,N)

2exp exp

B B

N E EB

N k T k T

expB

N E E

N k T

Energy

E

E

1

/

/ln

2

BUN

BUN

k

BT

B

BNUBNU

B

k

BBN

U

BBBN

U

Bk

BN

U

BN

UN

BN

U

BN

UNN

Uk

U

S

T

BB

BBN

1

1ln

22

11ln

2

1

2

11ln

2

1

1ln12

1ln12

2ln1

,

The same in terms of N and N : NNNNNNNNk

S

B

lnlnln

1 12 1

ln ln2 211

2

B B

N U UN N NN B k kN B

UN T B N B NN UN BN B

Boltzmann factor!

The Temperature of a Two-State Paramagnet

0

E1

E2

E1

E2

Tk

EE

N

N

B

12exp

Energy

E1

E2

T

Energy

E1

E2

Tk

EE

N

N

B

12exp

- N B

N B

U

11

/

/ln

2ln

2

BUN

BUN

k

B

N

N

k

BT

BB

E1

E2

T = + and T = - are (physically) identical – they correspond to the same values of thermodynamic parameters.

Systems with neg. T are “warmer” than the systems with pos. T: in a thermal contact, the energy will flow from the system with neg. T to the systems with pos. T.

The Temperature of a Spin GasThe system of spins in an external magnetic field. The internal energy in this model is entirely potential, in contrast to the ideal gas model, where the energy is entirely kinetic.

Tk

E

iB

i

en

B

E6

E5

E4

E3

E2

E1

B

At fixed T, the number of spins ni of energy Ei decreases exponentially as energy increases.

spin 5/2(six levels)

- lnni

Ei

- lnni

Ei

- lnni

Ei

T =

- lnni

Ei

T = 0

For a two-state system, one can always introduce T - one can always fit an exponential to two points. For a multi-state system with random population, the system is out of equilibrium, and we cannot introduce T.

Boltzmann distribution

no T

Tk

En

B

ii ln

the slope T

The Energy of a Two-State Paramagnet

Tk

BBN

e

eBNU

BTkB

TkB

B

B

tanh1

1/2

/2

BUN

BUN

B

k

TB

/

/ln

2

1

U approaches the lower limit (-NB) as T decreases or, alternatively, B increases (the effective “gap” gets bigger).

U

B/kBT

N B

- N B

U

T

- N B

1

The equation of state of a two-state paramagnet:

(N,N) S (N,N) = kB ln (N,N) U =U (N,T,B)1

),(

U

UNST

T < 0T > 0

S(B/T) for a Two-State Paramagnet

NNNNNNNN

NNN

N

k

NNS

B

lnlnln!!

!ln

,

S

U

0 N B - N B Problem 3.23 Express the entropy of a two-state

paramagnet as a function of B/T .

NNBU 2

Tk

BBNU

B

tanh

xTk

B

B

2

tanh1tanh2

xNNxBNNNB

2

tanh1ln

2

tanh1

2

tanh1ln

2

tanh1

2

tanh1ln

2

tanh1

2

tanh1ln

2

tanh1ln

,

xxxx

xN

xxN

xN

kN

xNS

B

x

e

x

xxx

x

e

x

xxx

xx

coshcosh

sinhcoshtanh1

coshcosh

sinhcoshtanh1

?)(TNN

NkBln2

xxxxx

ee

x

eex

xxx

exx

x

e

x

e

x

e

x

e

x

e

kN

xNS

xxxx

xx

xxxx

B

tanhcosh2lncosh2lncosh2cosh2

cosh2lncosh2

cosh2lncosh2

cosh2ln

cosh2cosh2ln

cosh2

,

Tk

B

Tk

B

Tk

BkN

Tk

BNS

BBBB

B

tanhcosh2ln,

B/T 0, S = NkB ln2

B/T , S = 0

S(B/T) for a Two-State Paramagnet (cont.)

0 2 40

0.5

10.692

0

fi

50.02 1

xi

S/NkB

ln2 0.693

kBT/B = x-1

high-T (low-B) limit

low-T(high-B)

limit0 2 4 6

0.0

0.2

0.4

0.6

0.8

1.0S / NkB

x = B / kBT

ln2 0.693

0 2 4 60.0

0.2

0.4

0.6

0.8

1.0S / NkB

x = B / kBT

Low-T limit

1221

1

11ln

22ln

tanhcosh2ln,

222

2

22

xekNexexkN

e

exexkN

ee

eex

eekN

Tk

B

Tk

B

Tk

BkN

Tk

BNS

xB

xxB

x

xx

Bxx

xxxx

B

BBBB

B

Which x can be considered large (small)?

1Tk

Bx

B

ln2 0.693

2

2

. ., 2, 0.018 1.8%

2 1 0.1

x

x

e g x e

e x

0 1 2 3 4 50.0

0.1

0.2

0.3

0.4

0.5CB / Nk

B

kBT / B

The Heat Capacity of a Paramagnet

TkB

TkBkN

T

UC

B

BB

BNB /cosh

/2

2

,

The low -T behavior: the heat capacity is small because when kBT << 2 B, the thermal fluctuations which flip spins are rare and it is hard for the system to absorb thermal energy (the degrees of freedom are frozen out). This behavior is universal for systems with energy quantization.

The high -T behavior: N ~ N and again, it is hard for the system to absorb thermal energy. This behavior is not universal, it occurs if the system’s energy spectrum occupies a finite interval of energies.

kBT

2 B

kBT2 B

compare withEinstein solid

E per particle

C

equipartition theorem(works for quadratic degrees

of freedom only!)

NkB/2

2cosh

xx eex

The Magnetization, Curie’s Law

B/kBT

N

M

1

xxx tanh1

Tk

BNM

B

2

The high-T behavior for all paramagnets (Curie’s Law)

Tk

BN

B

UNNM

B

tanh B/kBT

N

- N

The magnetization:

Negative T in a nuclear spin system NMR MRI

Fist observation – E. Purcell and R. Pound (1951)

Pacific Northwest National Laboratory

By doing some tricks, sometimes it is possible to create a metastable non-equilibrium state with the population of the top (excited) level greater than that for the bottom (ground) level - population inversion. Note that one cannot produce a population inversion by just increasing the system’s temperature. The state of population inversion in a two-level system can be characterized with negative temperatures - as more energy is added to the system, and S actually decrease.

An animated gif of MRI images of a human head.- Dwayne Reed

Metastable Systems without Temperature (Lasers)

For a system with more than two energy levels, for an arbitrary population of the levels we cannot introduce T at all - that's because you can't curve-fit an exponential to three arbitrary values of #, e.g. if occ. # = f (E) is not monotonic (population inversion). The latter case – an optically active medium in lasers.

E1

E2

E3

E4

Population inversionbetween E2 and E1

Tk

E

B

exp

Sometimes, different temperatures can be introduced for different parts of the spectrum.

Problem

A two-state paramagnet consists of 1x1022 spin-1/2 electrons. The component of the electron’s magnetic moment along B is B = 9.3x10-24 J/T.  The paramagnet is placed in an external magnetic field B = 1T which points up.

• Using Boltzmann distribution, calculate the temperature at which N= N/e.• Calculate the entropy of the paramagnet at this temperature.• What is the maximum entropy possible for the paramagnet? Explain your reasoning.

(a)

K35.1J/K 101.38

T 1J/T 103.92223

24

B

B

k

BT

BEETk

EE

N

NB

B

2exp 1212

kBT

E1= - BB

E2 = + BB

B

spin 1/2(two levels)

B

- B

J/T 103.92

24e

B m

e1212

Tk

B

Tk

EE

B

B

B

N

N

Problem (cont.)

J/K012.009.0J/K1038.1101

tanhcosh2ln,

2322

Tk

B

Tk

B

Tk

BkN

Tk

BNS

B

B

B

B

B

BB

B

B

xx

xxxxxx

ee

eex

eex

eex

tanh2

sinh2

cosh

If your calculator cannot handle cosh’s and sinh’s:

0 2 40

0.5

1

fi

1

xi

kBT/ B

S/NkB

0.09

J/K1.0693.0J/K1038.1101, 2322

Tk

BNS

B

B

(b)

323

24

105.4K 300J/K 101.38

T 2J/T 103.9

Tk

B

B

B

the maximum entropy corresponds to the limit of T (N=N): S/NkB ln2

For example, at T=300K:

E1

E2

kBT

0 2 40

0.5

10.692

0

fi

50.02 1

xi

kBT/ B

S/NkB

T S/NkB ln2

T 0S/NkB 0

ln2

Problem (cont.)