213213 MidtermMidterm comingcoming upup ……

Monday Nov. 11 @ 7 pm (conflict exam @ 5:15pm) Sign up for Conflictconflict by Tuesday night. Covers: Lectures 112 (not including thermal radiation) HW 14 Discussion 14 Labs 12

Review Session Sunday Nov. 10, 35 PM, 141 Loomis

HW 4 is not due until Thursday, Nov. 14 at 8 am, but some of the problems are relevant for the exam.

Lecture 10, p 1 Lecture 10 The Boltzmann Distribution

• Concept of a thermal reservoir

• The Boltzmann distribution

• Paramagnetic Spins – MRI

• Supplement: Proof of equipartition, showing its limits

Reading for this Lecture: Reading for Lecture 11: Elements Ch 8 Elements Ch 9

Lecture 10, p 2 Some Questions We’d Like to Answer

What is the probability What is the range that a DNA molecule will of kinetic energies unfold and replicate? of an O 2 molecule?

Under what conditions

does O 2 break up into What is the vapor pressure two O atoms? of a solid or liquid?

What is the capacity of a When do molecular vibrations myoglobin molecule to carry become important? oxygen to the muscles?

Vitruvian Man, 1490, by Leonardo da Vinci These questions involve the interaction between a small system (atom or molecule) and a much larger system (the environment). This is a basic problem in statistical mechanics. Lecture 10, p 3 Averages from Probabilities

IF: 1 You could list every quantum state of some small system. (this is realistic for small objects, e.g. , oscillators or atoms) 2 And you knew the properties of each state (e.g., energy, magnetic moment, optical density, etc .)

3 And you knew the probability of each state (P 1, P 2, ...Pn)

THEN: You could calculate the average energy, magnetic moment, optical density, etc. for each part . For example, = P 1 E1 + P 2 E2 + ... Pn En

Now if you have a big system, made up of simple little parts, to get , , etc. for the big system, just add all the parts!

We can figure out how things behave, starting from scratch. The key step is 3: The Boltzmann factor tells us the probabilities.

Lecture 10, p 4 Concept of a Thermal Reservoir

We will be considering situations like this:

A small system Note: The systems do not have to be collections of oscillators (with equally spaced energy levels). A large system We only assume that to simplify the math.

The two systems can exchange energy, volume, particles, etc.

If the large system is much larger than the small one, then its temperature will not be significantly affected by the interaction.

We define a thermal reservoir to be a system that is large enough so that its T does not change when interacting with the small system.

The reservoir doesn’t have to be very large, just a lot larger than the small system.

Lecture 10, p 5 Thermal Reservoir

Let’s start with a reservoir that isn’t very large. That makes the problem easier to solve. Consider one oscillator in thermal contact with a system of three oscillators:

En Ek + El + Em = U R = energy of reservoir

4ε

3ε Total energy: U = E n + U R 2ε = # reservoir microstates 1ε R

0 Small Reservoir system

Question : What is the probability Pn that the small system has energy E n = n ε? Answer : The probability is proportional to the corresponding R. Key point: n = 1 for every n, so tot = nR = R. The probability calculation is dominated P ∝ by the behavior of the reservoir. n R

Lecture 10, p 6 One Oscillator Exchanging with Three Oscillators Suppose En UR = U En R(UR ) Pn = R / Σ R U = 4 ε (q+ N − 1)! R = 0 4ε (q = 4) 15 15/35 = 0.43 q!() N − 1 ! ε 3ε 10 10/35 = 0.29 2ε 2ε 6 6/35 = 0.17 N = 3 3ε ε 3 3/35 = 0.09 4ε 0 1 1/35 = 0.03 (q+ N − 1)! 35 = total # states for = tot ∑ R q!( N − 1)! combined system tot P (4+ 4 − 1)! n = = = 35 Why is zero Σ R 4!(4− 1)! 0.5 energy the most probable?

R (n ) Pn= ∝ R ( n ) En (n′ ) ∑n′ R 0 1 2 3 4 ×ε

Probability decreases with E n because # states of the large system decreases as U R = Utot En goes down. Lecture 10, p 7 The Boltzmann Factor

To calculate Pn we only need to know: A small system A large system Temperature = T R • The energy, E n of the small system, • The temperature, T R, of the reservoir.

Let’s do it: σ Pn ∝ = e Because σ = ln( ). 0 Define σ The entropy when E n = 0 ( i.e. , when U R = Utot ). st Calculate σ when E n ≠ 0 using a Taylor expansion (1 derivative only):

∂σ 0 dσ R σ σ σ ()()En=0 + E n =− E n ∂En dU R ∂σ 1 Remember the definition of temperature: R = ∂UR kT R This is the Boltzmann factor. It tells us the probability that E  0 En n a small system is in a state Thereforeσ σ= − and Pn ∝exp  −  kT R kT R  that has energy E n. It is very important !!

Lecture 10, p 8 Normalization of the Probability

−En / kT We now know that Pn ∝ e e−En / kT Let: P = n Z

We can determine the proportionality constant, by requiring that the total probability equal one: e−En / kT e−En / kT ∑ n This is a sum over all ∑Pn = ∑ = = 1 n n Z Z states of the system.

−E/ kT Then Z ≡ ∑ e n n Z is called the “partition function”.

Lecture 10, p 9 Example: Boltzmann Factor

A particular molecule has three states, with energy spacing ε = 10 20 J, as shown; the molecule is in contact with the environment (reservoir), which has a temperature of 1000 K . E 2 ε E E 1 ε Eo

1) What is P 1, the probability that the molecule is in the middle energy state?

2) What is P 2, the probability that it is in the highest energy state?

Lecture 10, p 10 Solution

A particular molecule has three states, with energy spacing ε = 10 20 J, as shown; the molecule is in contact with the environment (reservoir), which has a temperature of 1000 K . E 2 ε E E1 ε You can always pick E=0 Eo at your convenience.

1) What is P 1, the probability that the molecule is in the middle energy state?

e−E1 / kT e −ε / kT P = = 1 −En / kT −0/kT ε −ε / kT − 2/ kT ∑e e+ e + e ε 1020 J == 0.725 e−0.725 0.485 kT 1.38× 10−23 ⋅ 10 3 J = = = 0.282 1+e−0.725 + e − 1.45 1 + 0.485 + 0.235

2) What is P 2, the probability that it is in the highest energy state?

Lecture 10, p 11 Solution

A particular molecule has three states, with energy spacing ε = 10 20 J, as shown; the molecule is in contact with the environment (reservoir), which has a temperature of 1000 K . E 2 ε E E1 ε You can always pick E=0 Eo at your convenience.

1) What is P 1, the probability that the molecule is in the middle energy state?

e−E1 / kT e −ε / kT P = = 1 −En / kT −0/kT ε −ε / kT − 2/ kT ∑e e+ e + e ε 1020 J == 0.725 e−0.725 0.485 kT 1.38× 10−23 ⋅ 10 3 J = = = 0.282 1+e−0.725 + e − 1.45 1 + 0.485 + 0.235

2) What is P 2, the probability that it is in the highest energy state?

e−E 2 / kT .0 235 P2 = = = .0 137 e−E 0 / kT + e−E1 / kT + e−E 2 / kT 1 + 0.485 + 0.235

Lecture 10, p 12 Act 1 E A particular molecule has three states, with E2 energy spacing ε = 10 20 J, as shown. The ε E molecule is in contact with the environment 1 (reservoir) at temperature T. ε Eo

1) What is P 0 when T → 0? a) 0 b) 1/3 c) 1

2) What is P 2 as T → ∞? a) 0 b) 1/3 c) 1

3) What happens to P 2 as we decrease T ? a) decreases b) increases c) decreases, then increases

Lecture 10, p 13 Solution E A particular molecule has three states, with E2 energy spacing ε = 10 20 J, as shown. The ε E molecule is in contact with the environment 1 (reservoir) at temperature T. ε Eo

1) What is P 0 when T → 0? a) 0 b) 1/3 c) 1

P1 and P 2 both → 0, because 1/T → ∞, so P 0 must → 1.

2) What is P 2 as T → ∞? a) 0 b) 1/3 c) 1

3) What happens to P 2 as we decrease T ? a) decreases b) increases c) decreases, then increases

Lecture 10, p 14 Solution E A particular molecule has three states, with E2 energy spacing ε = 10 20 J, as shown. The ε E molecule is in contact with the environment 1 (reservoir) at temperature T. ε Eo

1) What is P 0 when T → 0? a) 0 b) 1/3 c) 1

P1 and P 2 both → 0, because 1/T → ∞, so P 0 must → 1.

2) What is P 2 as T → ∞? a) 0 b) 1/3 c) 1 Now,1/T → 0, so all the probabilities become equal.

3) What happens to P 2 as we decrease T ? a) decreases b) increases c) decreases, then increases

Lecture 10, p 15 Solution E A particular molecule has three states, with E2 energy spacing ε = 10 20 J, as shown. The ε E molecule is in contact with the environment 1 (reservoir) at temperature T. ε Eo

1) What is P 0 when T → 0? a) 0 b) 1/3 c) 1

P1 and P 2 both → 0, because 1/T → ∞, so P 0 must → 1.

2) What is P 2 as T → ∞? a) 0 b) 1/3 c) 1 Now,1/T → 0, so all the probabilities become equal.

3) What happens to P 2 as we decrease T ? a) decreases b) increases c) decreases, then increases

As T decreases, there is less chance to find the molecule with energy

E2, because that’s the highest E. The ratio of its probability to that of every other state always decreases as T increases.

Lecture 10, p 16 How to apply the Boltzmann Factor if there are degenerate states

You often come across a system that is “degenerate”, which means that more than one state has the same energy . The Boltzmann calculation still applies – just make sure to sum over all states. In this case, you simply need to count the number of degenerate states. For example, three states have energy E:

3 states E

1 state 0 The Boltzmann factor tells us the probability per state!! The probability that the system has a particular energy E depends on the number of states at that energy: d e −En / kT −0 / kT n P(0)=e / Z = 1/ Z Pn = d e −En / kT −E/ kT ∑ n P(E )= 3e / Z n −E/ kT Z=1 + 3 e dn = degeneracy of state n Three states have the same This is the probability Boltzmann factor. of a particular energy . Lecture 10, p 17 Example: Paramagnetism A System of Independent Magnetic Spins

In a magnetic field , spins can only point parallel or antiparallel to the field (a result of quantum mechanics) .

B → N spins, each with magnetic moment , in contact with a thermal bath at temperature T.

→→ Each spin has potential energy En = •B = ±B (a result from P212) . The probability Pn that a spin will have energy E n is given by the Boltzmann distribution (which gives the probability that the spin is up or down): −En / kT Pn = e/ Z

Lecture 10, p 18 Application: Magnetic Resonance Imaging (MRI)

MRI exploits the paramagnetic behavior of the protons in your body. What happens when you place your body in a magnetic field? The protons (hydrogen nuclei) align their magnetic moments (spins) with the magnetic field. This is the basis of Magnetic Resonance Imaging

Consider the small ‘system’ to be a single proton spin; the ‘reservoir’ is your body. Here are the energy levels of the proton: Edown = +B B Eup = −B

We are interested in the net magnetic moment, M , of the N protons in the magnetic field: M = m, where m ≡ Nup Ndown , the “spin excess”.

2003 Nobel Prize in Medicine UIUC’s Paul Lauterbur

Lecture 10, p 19 Magnetic Resonance Imaging (2)

Solve this problem using Boltzmann factors.

exp(−B / kT ) Edown= + B P down = B Z exp(+B / kT ) E= − B P = up up Z

The partition function, Z, is the sum of the Boltzmann factors:

Z= e B/ kT + e − B / kT

The total magnetic moment is:

M= (Nup − N down) = NPP ( up − down ) e BkT/− e − BkT / = N e BkT/+ e − BkT / = N tanh() B / kT Let’s plot this function...

Lecture 10, p 20 Magnetic Resonance Imaging (3)

HighB or lowT: Most spins lined up LowB or highT: M “Saturation ” Few spins lined up N Linear response

1 +1 B/kT M= N ⋅ tanh( B / kT )

−N

At sufficiently high temperatures , the ratio x ≡ B / kT << 1 . Using tanh(x) ~ x for small x, the total magnetic moment of the spin system is: M N 2 B B B/T M ≈ ∝ Curie’s Law kT T T

Lecture 10, p 21 ACT 2

Consider a collection of N spins in magnetic field.

1) What is the entropy of these N spins as T→→→ ∞?

a) 0 b) Nln(2) c) 2ln(N)

2) What is the entropy as T →→→ 0 ?

a) 0 b) Nln(2) c) 2ln(N)

Lecture 10, p 22 Solution

Consider a collection of N spins in magnetic field.

1) What is the entropy of these N spins as T→→→ ∞?

a) 0 b) Nln(2) c) 2ln(N)

At high temperature, each spin is as likely to point up as down. That is, each spin as two possible equally likely microstates.

2) What is the entropy as T →→→ 0 ?

a) 0 b) Nln(2) c) 2ln(N)

Lecture 10, p 23 Solution

Consider a collection of N spins in magnetic field.

1) What is the entropy of these N spins as T→→→ ∞?

a) 0 b) Nln(2) c) 2ln(N)

At high temperature, each spin is as likely to point up as down. That is, each spin as two possible equally likely microstates.

2) What is the entropy as T →→→ 0 ?

a) 0 b) Nln(2) c) 2ln(N)

Every spin is stuck in the lowestenergy state, aligned with the field. This is a general result: For any realistic system (even a big one) there are only one or two ground states. Therefore:

As T → 0: S → 0 “The third law” of thermodynamics

Lecture 10, p 24 Summary: Collection of Spins

We used the Boltzmann factor (and remembering that the sum of the probabilities is always 1) to tell us the probabilities of each of the two energy states of a single magnetic moment in a magnetic field. eB/ kT e −B / kT P=; P = up ee BkT//+− BkTdown ee BkT // + − BkT

In a collection, the average number pointing up and down is just N times the probabilities:

Nup = NP up , and Ndown = NP down

Using these averages, we can calculate macroscopic properties (see Appendix): • total magnetic moment, M • internal energy, U

• heat capacity, CB • entropy, S

Lecture 10, p 25 Two -state Systems in General

Consider a twostate system with an energy E2 difference E between the two states. E

E1 How do the occupation probabilities of the states vary with T?

− E 1.0 1 e kT P = 0.8 1 − E P2 = − E 1+ e kT 1+ e kT P1 The low energy state is preferentially 0.6 occupied at low T, but the states 0.4

Probabilities P approach equal occupancy at high T. 0.2 2

0.0 0 1 2 3 4 5 kT/E

This behavior will be exactly the same for every “twostate system” with the same E.

Lecture 10, p 26 Next Time Applying Boltzmann Statistics

• Polymers • Simple Harmonic Oscillators:

CV of molecules – for real! When equipartition fails • Planck Distribution of Electromagnetic Radiation

Lecture 10, p 27 Supplement: Internal Energy of a Collection of Spins

E Recall how to calculate the internal energy U:
down B

U = NuEu +NdEd = (N uNd)B Eup = NB tanh( B/kT)

Eup = Edown = B What does this look like as a function of T?

U/ BN 2 4 6 8kT/ B 0

Low T (kT << B): Boltzmann factor ~ 0. All spins are stuck in low energy state. 0.5

U = NE up = BN, independent of T

1.0 High T (kT >> B): Boltzmann factor approaches 1. Almost equal numbers in the up and down states.

U ≈ (N/2)(E up +E down ) = 0, independent of T Lecture 10, p 28 Supplement: Heat Capacity of a Collection of Spins

We now have U(T), for fixed B, so we can get the heat capacity, CB (at constant B), by taking ∂ U/ ∂T.

2 ∂U  B 2 CB =  = Nk  sech() B / kT ∂TB= const  kT

For kT << B, C B vanishes, because all are stuck in “ground state”.

C /kN B For kT >> B, C vanishes, because the probabilities of the two states each approach 0.5, and cease to depend on T. 1 2 3 kT/ B

A collection of 2state spins does not behave anything like an ideal gas.

Lecture 10, p 29 Supplement: Entropy of a Collection of Spins

Given C B, We can calculate S at any T. At T = ∞, each spin has two equally likely microstates. Therefore, the system has = 2 N ⇒ σ = Nln2, and S = Nkln(2).

At fixed B, dS = CBdT/T . This is just like fixed V, where dS=CvdT/T. If B is not kept constant, some of the energy goes into other forms (work is done).

T Integrals of hyperbolic functions CB   B  B  B  ST()=∞+ S () dTNk = ln2cosh  − tanh   are similar to integrals of trig ∫ functions. ∞ T kTkT   kT 

This is a bit messy. Here’s the graph: S/Nk It has the behavior we expect: ln(2) As T → 0, S(T) → 0. At T = 0, there is only one available microstate (all spins up). 1 2 3 kT/ B

Lecture 10, p 30