11/1/21

1

Lecture 20 PHYS 416 Tuesday November 2 Fall 2021

1. Comment on rubber bands (Sethna 5.12 T/F questions)2. Sethna 6.6 Chemical equilibrium & reaction rates3. HW 6.12 Michaelis-Menten & Hill (and Mb & Hb)4. HW 6.26 Nucleosynthesis as a chemical reaction

1

Copyright Oxford University Press 2006 v2.0 --

122 Entropy

plex water-solvent interactions give for osmoticpressure (Section 5.2.2).(d) If we increase the temperature of our rubberband while it is under tension, will it expand orcontract? Why?In a more realistic model of a rubber band, theentropy consists primarily of our configurationalrandom-walk entropy plus a vibrational entropyof the molecules. If we stretch the rubber bandwithout allowing heat to flow in or out of therubber, the total entropy should stay approxi-mately constant. (Rubber is designed to bouncewell; little irreversible entropy is generated in acycle of stretching and compression, so long asthe deformation is not too abrupt.)(e) True or false?(T) (F) When we stretch the rubber band, it willcool; the configurational entropy of the randomwalk will decrease, causing the entropy in the vi-brations to decrease, causing the temperature todecrease.(T) (F) When we stretch the rubber band, it willcool; the configurational entropy of the randomwalk will decrease, causing the entropy in the vi-brations to increase, causing the temperature todecrease.(T) (F) When we let the rubber band relax, itwill cool; the configurational entropy of the ran-dom walk will increase, causing the entropy inthe vibrations to decrease, causing the tempera-ture to decrease.(T) (F) When we let the rubber band relax, theremust be no temperature change, since the entropyis constant.This more realistic model is much like the idealgas, which also had no configurational energy.(T) (F) Like the ideal gas, the temperaturechanges because of the net work done on the sys-tem.(T) (F) Unlike the ideal gas, the work done onthe rubber band is positive when the rubber bandexpands.You should check your conclusions experimen-tally; find a rubber band (thick and stretchy isbest), touch it to your lips (which are very sen-sitive to temperature), and stretch and relax it.

(5.13) How many shuffles? (Mathematics) ©iFor this exercise, you will need a deck of cards,either a new box or sorted into a known order(conventionally with an ace at the top and a kingat the bottom).56

How many shuffles does it take to randomize adeck of 52 cards?The answer is a bit controversial; it dependson how one measures the information left inthe cards. Some suggest that seven shuffles areneeded; others say that six are enough.57 We willfollow reference [186], and measure the growingrandomness using the information entropy.We imagine the deck starts out in a known order(say, A♠, 2♠, . . . , K♣).(a) What is the information entropy of the deckbefore it is shuffled? After it is completely ran-domized?(b) Take a sorted deck of cards. Pay attentionto the order; (in particular, note the top and bot-tom cards). Riffle it exactly once, separating itinto two roughly equal portions and interleavingthe cards in the two portions.. Examine the cardsequence, paying particular attention to the topfew and bottom few cards. Can you tell whichcards came from the top portion and which camefrom the bottom?The mathematical definition of a riffle shuffle iseasiest to express if we look at it backward.58

Consider the deck after a riffle; each card inthe deck either came from the top portion orthe bottom portion of the original deck. A riffleshuffle makes each of the 252 patterns tbbtbttb . . .(denoting which card came from which portion)equally likely.It is clear that the pattern tbbtbttb . . . deter-mines the final card order: the number of t’stells us how many cards were in the top portion,and then the cards are deposited into the finalpile according to the pattern in order bottom totop. Let us first pretend the reverse is also true:that every pattern corresponds one-to-one witha unique final card ordering.(c) Ignoring the possibility that two different rif-fles could yield the same final sequence of cards,what is the information entropy after one riffle?You can convince yourself that the only way two

56Experience shows that those who do not do the experiment in part (b) find it challenging to solve part (c) by pure thought.57More substantively, as the number of cards N → ∞, some measures of information show an abrupt transition near 3/2 log2 N ,while by other measures the information vanishes smoothly and most of it is gone by log2 N shuffles.58The forward definition mimics an expert riffler by splitting the deck into two roughly equal portions by choosing n cards ontop with a binomial probability 2−52

(52n

). Then the expert drops cards with probability proportional to the number of cards

remaining in its portion. This makes each of the 252 choices we found in the backward definition equally likely.

Copyright Oxford University Press 2006 v2.0 --

122 Entropy

plex water-solvent interactions give for osmoticpressure (Section 5.2.2).(d) If we increase the temperature of our rubberband while it is under tension, will it expand orcontract? Why?In a more realistic model of a rubber band, theentropy consists primarily of our configurationalrandom-walk entropy plus a vibrational entropyof the molecules. If we stretch the rubber bandwithout allowing heat to flow in or out of therubber, the total entropy should stay approxi-mately constant. (Rubber is designed to bouncewell; little irreversible entropy is generated in acycle of stretching and compression, so long asthe deformation is not too abrupt.)(e) True or false?(T) (F) When we stretch the rubber band, it willcool; the configurational entropy of the randomwalk will decrease, causing the entropy in the vi-brations to decrease, causing the temperature todecrease.(T) (F) When we stretch the rubber band, it willcool; the configurational entropy of the randomwalk will decrease, causing the entropy in the vi-brations to increase, causing the temperature todecrease.(T) (F) When we let the rubber band relax, itwill cool; the configurational entropy of the ran-dom walk will increase, causing the entropy inthe vibrations to decrease, causing the tempera-ture to decrease.(T) (F) When we let the rubber band relax, theremust be no temperature change, since the entropyis constant.This more realistic model is much like the idealgas, which also had no configurational energy.(T) (F) Like the ideal gas, the temperaturechanges because of the net work done on the sys-tem.(T) (F) Unlike the ideal gas, the work done onthe rubber band is positive when the rubber bandexpands.You should check your conclusions experimen-tally; find a rubber band (thick and stretchy isbest), touch it to your lips (which are very sen-sitive to temperature), and stretch and relax it.

(5.13) How many shuffles? (Mathematics) ©iFor this exercise, you will need a deck of cards,either a new box or sorted into a known order(conventionally with an ace at the top and a kingat the bottom).56

How many shuffles does it take to randomize adeck of 52 cards?The answer is a bit controversial; it dependson how one measures the information left inthe cards. Some suggest that seven shuffles areneeded; others say that six are enough.57 We willfollow reference [186], and measure the growingrandomness using the information entropy.We imagine the deck starts out in a known order(say, A♠, 2♠, . . . , K♣).(a) What is the information entropy of the deckbefore it is shuffled? After it is completely ran-domized?(b) Take a sorted deck of cards. Pay attentionto the order; (in particular, note the top and bot-tom cards). Riffle it exactly once, separating itinto two roughly equal portions and interleavingthe cards in the two portions.. Examine the cardsequence, paying particular attention to the topfew and bottom few cards. Can you tell whichcards came from the top portion and which camefrom the bottom?The mathematical definition of a riffle shuffle iseasiest to express if we look at it backward.58

Consider the deck after a riffle; each card inthe deck either came from the top portion orthe bottom portion of the original deck. A riffleshuffle makes each of the 252 patterns tbbtbttb . . .(denoting which card came from which portion)equally likely.It is clear that the pattern tbbtbttb . . . deter-mines the final card order: the number of t’stells us how many cards were in the top portion,and then the cards are deposited into the finalpile according to the pattern in order bottom totop. Let us first pretend the reverse is also true:that every pattern corresponds one-to-one witha unique final card ordering.(c) Ignoring the possibility that two different rif-fles could yield the same final sequence of cards,what is the information entropy after one riffle?You can convince yourself that the only way two

56Experience shows that those who do not do the experiment in part (b) find it challenging to solve part (c) by pure thought.57More substantively, as the number of cards N → ∞, some measures of information show an abrupt transition near 3/2 log2 N ,while by other measures the information vanishes smoothly and most of it is gone by log2 N shuffles.58The forward definition mimics an expert riffler by splitting the deck into two roughly equal portions by choosing n cards ontop with a binomial probability 2−52

(52n

). Then the expert drops cards with probability proportional to the number of cards

remaining in its portion. This makes each of the 252 choices we found in the backward definition equally likely.

I claimed: STRETCH: ∆𝑆!"##$! #%&' < 0 → ∆𝑆()#!%*)+&, > 0 → ∆𝑇 > 0

RELAX: ∆𝑆!"##$! #%&' > 0 → ∆𝑆()#!%*)+&, < 0 → ∆𝑇 < 0

Why does the temperature rise when the rubber band stretches?

2

3H2 + N2! 2NH3

6.6 Chemical equilibrium and reaction rates

First, let’s consider a simplistic interpretation of chemical reactions.

[ ][ ][ ]

23

eq32 2

NH( )

N HK T=This leads to the law of mass action:

Originally, “mass action” meant that rates depended on the amount of the material (the mass) within the active region. Today we would emphasize the role of concentration. (See Wikipedia.) The key is that it is a product of the concentrations. Where have we seen this before?

3

3H2 + N2! 2NH3

[ ][ ][ ]

23

eq32 2

NH( )

N HK T=

Think of this as a probability argument. Within a certain active region V, what is the probability that one N2 and three H2 molecules are present? It is the product of the probabilities of each individually being present:

P(N2 ,H2 ,H2 ,H2 ) = P(N2 )P(H2 )P(H2 )P(H2 )

= P(N2 )P(H2 )3

= cN2× cH2

3

= N2⎡⎣ ⎤⎦ H2⎡⎣ ⎤⎦3

4

Forward reaction rate per unit volume:

Backward reaction rate per unit volume: KB NH3⎡⎣ ⎤⎦2

KF N2⎡⎣ ⎤⎦ H2⎡⎣ ⎤⎦3

Then in steady state these are equal, soNH3⎡⎣ ⎤⎦

2

N2⎡⎣ ⎤⎦ H2⎡⎣ ⎤⎦3 =

KFKB

= Keq (T )

This derivation of the law of mass action assumes a lot of simultaneous collisions are taking place at the same spot, which is physically unrealistic. We now use statistical mechanics to derive it. Also note, this simple theory did NOT give us the value of the rate constant.

5

Forward reaction rate per unit volume:

Backward reaction rate per unit volume:

Then in steady state these are equal, so

What would the law of mass action look like if instead of 2 product molecules, we required 2n product molecules?

KF N2⎡⎣ ⎤⎦nH2⎡⎣ ⎤⎦

3n

KB NH3⎡⎣ ⎤⎦3n

NH3⎡⎣ ⎤⎦2n

N2⎡⎣ ⎤⎦nH2⎡⎣ ⎤⎦

3n =KFKB

⎛

⎝⎜⎞

⎠⎟

n

= (Keq (T ))n

6

11/1/21

2

NH3⎡⎣ ⎤⎦2n

N2⎡⎣ ⎤⎦nH2⎡⎣ ⎤⎦

3n =KFKB

⎛

⎝⎜⎞

⎠⎟

n

= (Keq (T ))n

NH3⎡⎣ ⎤⎦2

N2⎡⎣ ⎤⎦ H2⎡⎣ ⎤⎦3 =

KFKB

= Keq (T )

The difference between these two functions completely explains the difference between Michaelis-Menton reaction theory and the Hill theory for cooperative reactions.

This is in every biochemistry book ever written.

It is essential for understanding how hemoglobin delivers oxygen from the lungs to the places that need it.

Copyright Oxford University Press 2006 v2.0 --

Exercises 165

formulæ from parts (a) and (b), give your esti-mate of the decay rate for our system. Relatedformula:

∫∞0

x exp(−x2/2σ2) dx = σ2.How could we go beyond this one-dimensionalcalculation? In the olden days, Kramers stud-ied other one-dimensional models, changing theways in which the system was coupled to the ex-ternal bath. On the computer, one can avoid aseparate heat bath and directly work with thefull multidimensional configuration space, lead-ing to transition-state theory. The transition-state theory formula is very similar to the oneyou derived in part (c), except that the prefac-tor involves the product of all the frequenciesat the bottom of the well and all the positivefrequencies at the saddlepoint at the top of thebarrier (see [72]). Other generalizations arisewhen crossing multiple barriers [81] or in non-equilibrium systems [109].

(6.12) Michaelis–Menten and Hill. (Biology, Com-putation) ©3Biological reaction rates are often saturable; thecell needs to respond sensitively to the introduc-tion of a new chemical S, but the response shouldnot keep growing indefinitely as the new chem-ical concentration [S] grows.61 Other biologi-cal reactions act as switches; a switch not onlysaturates, but its rate or state changes sharplyfrom one value to another as the concentrationof a chemical S is varied. These reactions givetangible examples of how one develops effectivedynamical theories by removing degrees of free-dom; here, instead of coarse-graining some largestatistical mechanics system, we remove a singleenzyme E from the equations to get an effectivereaction rate.The rate of a chemical reaction,

NS +B → C, (6.80)

where N substrate molecules S combine with a Bmolecule to make a C molecule, will occur witha reaction rate given by the law of mass-action:

d[C]dt

= k[S]N [B]. (6.81)

Saturation and the Michaelis–Menten equation.Saturation is not seen in ordinary chemical reac-tion kinetics. Notice that the reaction rate goesas the Nth power of the concentration [S]; far

from saturating, the reaction rate grows linearlyor faster with concentration.The archetypal example of saturation in biolog-ical systems is the Michaelis–Menten reactionform. A reaction of this form converting a chem-ical S (the substrate) into P (the product) has arate given by the formula

d[P ]dt

=Vmax[S]KM + [S]

, (6.82)

where KM is called the Michaelis constant(Fig. 6.17). This reaction at small concentra-tions acts like an ordinary chemical reaction withN = 1 and k = Vmax/KM , but the rate saturatesat Vmax as [S]→∞. The Michaelis constantKM

is the concentration [S] at which the rate is equalto half of its saturation rate (Fig. 6.17).

0 KM, KHSubstrate concentration [S]

0

Vmax

Rat

e d[

P]/d

t

Michaelis-MentenHill, n = 4

Fig. 6.17 Michaelis–Menten and Hill equationforms.

We can derive the Michaelis–Menten form by hy-pothesizing the existence of a catalyst or enzymeE, which is in short supply. The enzyme is pre-sumed to be partly free and available for bind-ing (concentration [E]) and partly bound to thesubstrate (concentration62 [E : S]), helping it toturn into the product. The total concentration[E] + [E : S] = Etot is fixed. The reactions areas follows:

E + S !k−1k1

E : Skcat→ E + P. (6.83)

We must then assume that the supply of sub-strate is large, so its concentration changesslowly with time. We can then assume that theconcentration [E : S] is in steady state, and re-move it as a degree of freedom.

61[S] is the concentration of S (number per unit volume). S stands for substrate.62The colon denotes the bound state of the two, called a dimer.

7

But first, let’s do the HW problem 6.12 Michaelis-Menten & Hill, because we only need the simple form of the law of mass action to solve it.

Copyright Oxford University Press 2006 v2.0 --

Exercises 165

formulæ from parts (a) and (b), give your esti-mate of the decay rate for our system. Relatedformula:

∫∞0

x exp(−x2/2σ2) dx = σ2.How could we go beyond this one-dimensionalcalculation? In the olden days, Kramers stud-ied other one-dimensional models, changing theways in which the system was coupled to the ex-ternal bath. On the computer, one can avoid aseparate heat bath and directly work with thefull multidimensional configuration space, lead-ing to transition-state theory. The transition-state theory formula is very similar to the oneyou derived in part (c), except that the prefac-tor involves the product of all the frequenciesat the bottom of the well and all the positivefrequencies at the saddlepoint at the top of thebarrier (see [72]). Other generalizations arisewhen crossing multiple barriers [81] or in non-equilibrium systems [109].

(6.12) Michaelis–Menten and Hill. (Biology, Com-putation) ©3Biological reaction rates are often saturable; thecell needs to respond sensitively to the introduc-tion of a new chemical S, but the response shouldnot keep growing indefinitely as the new chem-ical concentration [S] grows.61 Other biologi-cal reactions act as switches; a switch not onlysaturates, but its rate or state changes sharplyfrom one value to another as the concentrationof a chemical S is varied. These reactions givetangible examples of how one develops effectivedynamical theories by removing degrees of free-dom; here, instead of coarse-graining some largestatistical mechanics system, we remove a singleenzyme E from the equations to get an effectivereaction rate.The rate of a chemical reaction,

NS +B → C, (6.80)

where N substrate molecules S combine with a Bmolecule to make a C molecule, will occur witha reaction rate given by the law of mass-action:

d[C]dt

= k[S]N [B]. (6.81)

Saturation and the Michaelis–Menten equation.Saturation is not seen in ordinary chemical reac-tion kinetics. Notice that the reaction rate goesas the Nth power of the concentration [S]; far

from saturating, the reaction rate grows linearlyor faster with concentration.The archetypal example of saturation in biolog-ical systems is the Michaelis–Menten reactionform. A reaction of this form converting a chem-ical S (the substrate) into P (the product) has arate given by the formula

d[P ]dt

=Vmax[S]KM + [S]

, (6.82)

where KM is called the Michaelis constant(Fig. 6.17). This reaction at small concentra-tions acts like an ordinary chemical reaction withN = 1 and k = Vmax/KM , but the rate saturatesat Vmax as [S]→∞. The Michaelis constantKM

is the concentration [S] at which the rate is equalto half of its saturation rate (Fig. 6.17).

0 KM, KHSubstrate concentration [S]

0

Vmax

Rat

e d[

P]/d

t

Michaelis-MentenHill, n = 4

Fig. 6.17 Michaelis–Menten and Hill equationforms.

We can derive the Michaelis–Menten form by hy-pothesizing the existence of a catalyst or enzymeE, which is in short supply. The enzyme is pre-sumed to be partly free and available for bind-ing (concentration [E]) and partly bound to thesubstrate (concentration62 [E : S]), helping it toturn into the product. The total concentration[E] + [E : S] = Etot is fixed. The reactions areas follows:

E + S !k−1k1

E : Skcat→ E + P. (6.83)

We must then assume that the supply of sub-strate is large, so its concentration changesslowly with time. We can then assume that theconcentration [E : S] is in steady state, and re-move it as a degree of freedom.

61[S] is the concentration of S (number per unit volume). S stands for substrate.62The colon denotes the bound state of the two, called a dimer.

Copyright Oxford University Press 2006 v2.0 --

Exercises 165

formulæ from parts (a) and (b), give your esti-mate of the decay rate for our system. Relatedformula:

∫∞0

x exp(−x2/2σ2) dx = σ2.How could we go beyond this one-dimensionalcalculation? In the olden days, Kramers stud-ied other one-dimensional models, changing theways in which the system was coupled to the ex-ternal bath. On the computer, one can avoid aseparate heat bath and directly work with thefull multidimensional configuration space, lead-ing to transition-state theory. The transition-state theory formula is very similar to the oneyou derived in part (c), except that the prefac-tor involves the product of all the frequenciesat the bottom of the well and all the positivefrequencies at the saddlepoint at the top of thebarrier (see [72]). Other generalizations arisewhen crossing multiple barriers [81] or in non-equilibrium systems [109].

(6.12) Michaelis–Menten and Hill. (Biology, Com-putation) ©3Biological reaction rates are often saturable; thecell needs to respond sensitively to the introduc-tion of a new chemical S, but the response shouldnot keep growing indefinitely as the new chem-ical concentration [S] grows.61 Other biologi-cal reactions act as switches; a switch not onlysaturates, but its rate or state changes sharplyfrom one value to another as the concentrationof a chemical S is varied. These reactions givetangible examples of how one develops effectivedynamical theories by removing degrees of free-dom; here, instead of coarse-graining some largestatistical mechanics system, we remove a singleenzyme E from the equations to get an effectivereaction rate.The rate of a chemical reaction,

NS +B → C, (6.80)

where N substrate molecules S combine with a Bmolecule to make a C molecule, will occur witha reaction rate given by the law of mass-action:

d[C]dt

= k[S]N [B]. (6.81)

Saturation and the Michaelis–Menten equation.Saturation is not seen in ordinary chemical reac-tion kinetics. Notice that the reaction rate goesas the Nth power of the concentration [S]; far

from saturating, the reaction rate grows linearlyor faster with concentration.The archetypal example of saturation in biolog-ical systems is the Michaelis–Menten reactionform. A reaction of this form converting a chem-ical S (the substrate) into P (the product) has arate given by the formula

d[P ]dt

=Vmax[S]KM + [S]

, (6.82)

where KM is called the Michaelis constant(Fig. 6.17). This reaction at small concentra-tions acts like an ordinary chemical reaction withN = 1 and k = Vmax/KM , but the rate saturatesat Vmax as [S]→∞. The Michaelis constantKM

is the concentration [S] at which the rate is equalto half of its saturation rate (Fig. 6.17).

0 KM, KHSubstrate concentration [S]

0

Vmax

Rat

e d[

P]/d

t

Michaelis-MentenHill, n = 4

Fig. 6.17 Michaelis–Menten and Hill equationforms.

We can derive the Michaelis–Menten form by hy-pothesizing the existence of a catalyst or enzymeE, which is in short supply. The enzyme is pre-sumed to be partly free and available for bind-ing (concentration [E]) and partly bound to thesubstrate (concentration62 [E : S]), helping it toturn into the product. The total concentration[E] + [E : S] = Etot is fixed. The reactions areas follows:

E + S !k−1k1

E : Skcat→ E + P. (6.83)

We must then assume that the supply of sub-strate is large, so its concentration changesslowly with time. We can then assume that theconcentration [E : S] is in steady state, and re-move it as a degree of freedom.

61[S] is the concentration of S (number per unit volume). S stands for substrate.62The colon denotes the bound state of the two, called a dimer.

Copyright Oxford University Press 2006 v2.0 --

Exercises 165

formulæ from parts (a) and (b), give your esti-mate of the decay rate for our system. Relatedformula:

∫∞0

x exp(−x2/2σ2) dx = σ2.How could we go beyond this one-dimensionalcalculation? In the olden days, Kramers stud-ied other one-dimensional models, changing theways in which the system was coupled to the ex-ternal bath. On the computer, one can avoid aseparate heat bath and directly work with thefull multidimensional configuration space, lead-ing to transition-state theory. The transition-state theory formula is very similar to the oneyou derived in part (c), except that the prefac-tor involves the product of all the frequenciesat the bottom of the well and all the positivefrequencies at the saddlepoint at the top of thebarrier (see [72]). Other generalizations arisewhen crossing multiple barriers [81] or in non-equilibrium systems [109].

(6.12) Michaelis–Menten and Hill. (Biology, Com-putation) ©3Biological reaction rates are often saturable; thecell needs to respond sensitively to the introduc-tion of a new chemical S, but the response shouldnot keep growing indefinitely as the new chem-ical concentration [S] grows.61 Other biologi-cal reactions act as switches; a switch not onlysaturates, but its rate or state changes sharplyfrom one value to another as the concentrationof a chemical S is varied. These reactions givetangible examples of how one develops effectivedynamical theories by removing degrees of free-dom; here, instead of coarse-graining some largestatistical mechanics system, we remove a singleenzyme E from the equations to get an effectivereaction rate.The rate of a chemical reaction,

NS +B → C, (6.80)

where N substrate molecules S combine with a Bmolecule to make a C molecule, will occur witha reaction rate given by the law of mass-action:

d[C]dt

= k[S]N [B]. (6.81)

Saturation and the Michaelis–Menten equation.Saturation is not seen in ordinary chemical reac-tion kinetics. Notice that the reaction rate goesas the Nth power of the concentration [S]; far

from saturating, the reaction rate grows linearlyor faster with concentration.The archetypal example of saturation in biolog-ical systems is the Michaelis–Menten reactionform. A reaction of this form converting a chem-ical S (the substrate) into P (the product) has arate given by the formula

d[P ]dt

=Vmax[S]KM + [S]

, (6.82)

where KM is called the Michaelis constant(Fig. 6.17). This reaction at small concentra-tions acts like an ordinary chemical reaction withN = 1 and k = Vmax/KM , but the rate saturatesat Vmax as [S]→∞. The Michaelis constantKM

is the concentration [S] at which the rate is equalto half of its saturation rate (Fig. 6.17).

0 KM, KHSubstrate concentration [S]

0

Vmax

Rat

e d[

P]/d

t

Michaelis-MentenHill, n = 4

Fig. 6.17 Michaelis–Menten and Hill equationforms.

We can derive the Michaelis–Menten form by hy-pothesizing the existence of a catalyst or enzymeE, which is in short supply. The enzyme is pre-sumed to be partly free and available for bind-ing (concentration [E]) and partly bound to thesubstrate (concentration62 [E : S]), helping it toturn into the product. The total concentration[E] + [E : S] = Etot is fixed. The reactions areas follows:

E + S !k−1k1

E : Skcat→ E + P. (6.83)

We must then assume that the supply of sub-strate is large, so its concentration changesslowly with time. We can then assume that theconcentration [E : S] is in steady state, and re-move it as a degree of freedom.

61[S] is the concentration of S (number per unit volume). S stands for substrate.62The colon denotes the bound state of the two, called a dimer.

Copyright Oxford University Press 2006 v2.0 --

Exercises 165

formulæ from parts (a) and (b), give your esti-mate of the decay rate for our system. Relatedformula:

∫∞0

x exp(−x2/2σ2) dx = σ2.How could we go beyond this one-dimensionalcalculation? In the olden days, Kramers stud-ied other one-dimensional models, changing theways in which the system was coupled to the ex-ternal bath. On the computer, one can avoid aseparate heat bath and directly work with thefull multidimensional configuration space, lead-ing to transition-state theory. The transition-state theory formula is very similar to the oneyou derived in part (c), except that the prefac-tor involves the product of all the frequenciesat the bottom of the well and all the positivefrequencies at the saddlepoint at the top of thebarrier (see [72]). Other generalizations arisewhen crossing multiple barriers [81] or in non-equilibrium systems [109].

(6.12) Michaelis–Menten and Hill. (Biology, Com-putation) ©3Biological reaction rates are often saturable; thecell needs to respond sensitively to the introduc-tion of a new chemical S, but the response shouldnot keep growing indefinitely as the new chem-ical concentration [S] grows.61 Other biologi-cal reactions act as switches; a switch not onlysaturates, but its rate or state changes sharplyfrom one value to another as the concentrationof a chemical S is varied. These reactions givetangible examples of how one develops effectivedynamical theories by removing degrees of free-dom; here, instead of coarse-graining some largestatistical mechanics system, we remove a singleenzyme E from the equations to get an effectivereaction rate.The rate of a chemical reaction,

NS +B → C, (6.80)

where N substrate molecules S combine with a Bmolecule to make a C molecule, will occur witha reaction rate given by the law of mass-action:

d[C]dt

= k[S]N [B]. (6.81)

Saturation and the Michaelis–Menten equation.Saturation is not seen in ordinary chemical reac-tion kinetics. Notice that the reaction rate goesas the Nth power of the concentration [S]; far

from saturating, the reaction rate grows linearlyor faster with concentration.The archetypal example of saturation in biolog-ical systems is the Michaelis–Menten reactionform. A reaction of this form converting a chem-ical S (the substrate) into P (the product) has arate given by the formula

d[P ]dt

=Vmax[S]KM + [S]

, (6.82)

where KM is called the Michaelis constant(Fig. 6.17). This reaction at small concentra-tions acts like an ordinary chemical reaction withN = 1 and k = Vmax/KM , but the rate saturatesat Vmax as [S]→∞. The Michaelis constantKM

is the concentration [S] at which the rate is equalto half of its saturation rate (Fig. 6.17).

0 KM, KHSubstrate concentration [S]

0

Vmax

Rat

e d[

P]/d

t

Michaelis-MentenHill, n = 4

Fig. 6.17 Michaelis–Menten and Hill equationforms.

We can derive the Michaelis–Menten form by hy-pothesizing the existence of a catalyst or enzymeE, which is in short supply. The enzyme is pre-sumed to be partly free and available for bind-ing (concentration [E]) and partly bound to thesubstrate (concentration62 [E : S]), helping it toturn into the product. The total concentration[E] + [E : S] = Etot is fixed. The reactions areas follows:

E + S !k−1k1

E : Skcat→ E + P. (6.83)

We must then assume that the supply of sub-strate is large, so its concentration changesslowly with time. We can then assume that theconcentration [E : S] is in steady state, and re-move it as a degree of freedom.

61[S] is the concentration of S (number per unit volume). S stands for substrate.62The colon denotes the bound state of the two, called a dimer.

Copyright Oxford University Press 2006 v2.0 --

Exercises 165

formulæ from parts (a) and (b), give your esti-mate of the decay rate for our system. Relatedformula:

∫∞0

x exp(−x2/2σ2) dx = σ2.How could we go beyond this one-dimensionalcalculation? In the olden days, Kramers stud-ied other one-dimensional models, changing theways in which the system was coupled to the ex-ternal bath. On the computer, one can avoid aseparate heat bath and directly work with thefull multidimensional configuration space, lead-ing to transition-state theory. The transition-state theory formula is very similar to the oneyou derived in part (c), except that the prefac-tor involves the product of all the frequenciesat the bottom of the well and all the positivefrequencies at the saddlepoint at the top of thebarrier (see [72]). Other generalizations arisewhen crossing multiple barriers [81] or in non-equilibrium systems [109].

(6.12) Michaelis–Menten and Hill. (Biology, Com-putation) ©3Biological reaction rates are often saturable; thecell needs to respond sensitively to the introduc-tion of a new chemical S, but the response shouldnot keep growing indefinitely as the new chem-ical concentration [S] grows.61 Other biologi-cal reactions act as switches; a switch not onlysaturates, but its rate or state changes sharplyfrom one value to another as the concentrationof a chemical S is varied. These reactions givetangible examples of how one develops effectivedynamical theories by removing degrees of free-dom; here, instead of coarse-graining some largestatistical mechanics system, we remove a singleenzyme E from the equations to get an effectivereaction rate.The rate of a chemical reaction,

NS +B → C, (6.80)

where N substrate molecules S combine with a Bmolecule to make a C molecule, will occur witha reaction rate given by the law of mass-action:

d[C]dt

= k[S]N [B]. (6.81)

Saturation and the Michaelis–Menten equation.Saturation is not seen in ordinary chemical reac-tion kinetics. Notice that the reaction rate goesas the Nth power of the concentration [S]; far

from saturating, the reaction rate grows linearlyor faster with concentration.The archetypal example of saturation in biolog-ical systems is the Michaelis–Menten reactionform. A reaction of this form converting a chem-ical S (the substrate) into P (the product) has arate given by the formula

d[P ]dt

=Vmax[S]KM + [S]

, (6.82)

where KM is called the Michaelis constant(Fig. 6.17). This reaction at small concentra-tions acts like an ordinary chemical reaction withN = 1 and k = Vmax/KM , but the rate saturatesat Vmax as [S]→∞. The Michaelis constantKM

is the concentration [S] at which the rate is equalto half of its saturation rate (Fig. 6.17).

0 KM, KHSubstrate concentration [S]

0

Vmax

Rat

e d[

P]/d

t

Michaelis-MentenHill, n = 4

Fig. 6.17 Michaelis–Menten and Hill equationforms.

We can derive the Michaelis–Menten form by hy-pothesizing the existence of a catalyst or enzymeE, which is in short supply. The enzyme is pre-sumed to be partly free and available for bind-ing (concentration [E]) and partly bound to thesubstrate (concentration62 [E : S]), helping it toturn into the product. The total concentration[E] + [E : S] = Etot is fixed. The reactions areas follows:

E + S !k−1k1

E : Skcat→ E + P. (6.83)

We must then assume that the supply of sub-strate is large, so its concentration changesslowly with time. We can then assume that theconcentration [E : S] is in steady state, and re-move it as a degree of freedom.

61[S] is the concentration of S (number per unit volume). S stands for substrate.62The colon denotes the bound state of the two, called a dimer.

Copyright Oxford University Press 2006 v2.0 --

Exercises 165

formulæ from parts (a) and (b), give your esti-mate of the decay rate for our system. Relatedformula:

∫∞0

x exp(−x2/2σ2) dx = σ2.How could we go beyond this one-dimensionalcalculation? In the olden days, Kramers stud-ied other one-dimensional models, changing theways in which the system was coupled to the ex-ternal bath. On the computer, one can avoid aseparate heat bath and directly work with thefull multidimensional configuration space, lead-ing to transition-state theory. The transition-state theory formula is very similar to the oneyou derived in part (c), except that the prefac-tor involves the product of all the frequenciesat the bottom of the well and all the positivefrequencies at the saddlepoint at the top of thebarrier (see [72]). Other generalizations arisewhen crossing multiple barriers [81] or in non-equilibrium systems [109].

(6.12) Michaelis–Menten and Hill. (Biology, Com-putation) ©3Biological reaction rates are often saturable; thecell needs to respond sensitively to the introduc-tion of a new chemical S, but the response shouldnot keep growing indefinitely as the new chem-ical concentration [S] grows.61 Other biologi-cal reactions act as switches; a switch not onlysaturates, but its rate or state changes sharplyfrom one value to another as the concentrationof a chemical S is varied. These reactions givetangible examples of how one develops effectivedynamical theories by removing degrees of free-dom; here, instead of coarse-graining some largestatistical mechanics system, we remove a singleenzyme E from the equations to get an effectivereaction rate.The rate of a chemical reaction,

NS +B → C, (6.80)

where N substrate molecules S combine with a Bmolecule to make a C molecule, will occur witha reaction rate given by the law of mass-action:

d[C]dt

= k[S]N [B]. (6.81)

Saturation and the Michaelis–Menten equation.Saturation is not seen in ordinary chemical reac-tion kinetics. Notice that the reaction rate goesas the Nth power of the concentration [S]; far

from saturating, the reaction rate grows linearlyor faster with concentration.The archetypal example of saturation in biolog-ical systems is the Michaelis–Menten reactionform. A reaction of this form converting a chem-ical S (the substrate) into P (the product) has arate given by the formula

d[P ]dt

=Vmax[S]KM + [S]

, (6.82)

where KM is called the Michaelis constant(Fig. 6.17). This reaction at small concentra-tions acts like an ordinary chemical reaction withN = 1 and k = Vmax/KM , but the rate saturatesat Vmax as [S]→∞. The Michaelis constantKM

is the concentration [S] at which the rate is equalto half of its saturation rate (Fig. 6.17).

0 KM, KHSubstrate concentration [S]

0

Vmax

Rat

e d[

P]/d

t

Michaelis-MentenHill, n = 4

Fig. 6.17 Michaelis–Menten and Hill equationforms.

We can derive the Michaelis–Menten form by hy-pothesizing the existence of a catalyst or enzymeE, which is in short supply. The enzyme is pre-sumed to be partly free and available for bind-ing (concentration [E]) and partly bound to thesubstrate (concentration62 [E : S]), helping it toturn into the product. The total concentration[E] + [E : S] = Etot is fixed. The reactions areas follows:

E + S !k−1k1

E : Skcat→ E + P. (6.83)

We must then assume that the supply of sub-strate is large, so its concentration changesslowly with time. We can then assume that theconcentration [E : S] is in steady state, and re-move it as a degree of freedom.

61[S] is the concentration of S (number per unit volume). S stands for substrate.62The colon denotes the bound state of the two, called a dimer.

Copyright Oxford University Press 2006 v2.0 --

Exercises 165

formulæ from parts (a) and (b), give your esti-mate of the decay rate for our system. Relatedformula:

∫∞0

x exp(−x2/2σ2) dx = σ2.How could we go beyond this one-dimensionalcalculation? In the olden days, Kramers stud-ied other one-dimensional models, changing theways in which the system was coupled to the ex-ternal bath. On the computer, one can avoid aseparate heat bath and directly work with thefull multidimensional configuration space, lead-ing to transition-state theory. The transition-state theory formula is very similar to the oneyou derived in part (c), except that the prefac-tor involves the product of all the frequenciesat the bottom of the well and all the positivefrequencies at the saddlepoint at the top of thebarrier (see [72]). Other generalizations arisewhen crossing multiple barriers [81] or in non-equilibrium systems [109].

(6.12) Michaelis–Menten and Hill. (Biology, Com-putation) ©3Biological reaction rates are often saturable; thecell needs to respond sensitively to the introduc-tion of a new chemical S, but the response shouldnot keep growing indefinitely as the new chem-ical concentration [S] grows.61 Other biologi-cal reactions act as switches; a switch not onlysaturates, but its rate or state changes sharplyfrom one value to another as the concentrationof a chemical S is varied. These reactions givetangible examples of how one develops effectivedynamical theories by removing degrees of free-dom; here, instead of coarse-graining some largestatistical mechanics system, we remove a singleenzyme E from the equations to get an effectivereaction rate.The rate of a chemical reaction,

NS +B → C, (6.80)

where N substrate molecules S combine with a Bmolecule to make a C molecule, will occur witha reaction rate given by the law of mass-action:

d[C]dt

= k[S]N [B]. (6.81)

Saturation and the Michaelis–Menten equation.Saturation is not seen in ordinary chemical reac-tion kinetics. Notice that the reaction rate goesas the Nth power of the concentration [S]; far

from saturating, the reaction rate grows linearlyor faster with concentration.The archetypal example of saturation in biolog-ical systems is the Michaelis–Menten reactionform. A reaction of this form converting a chem-ical S (the substrate) into P (the product) has arate given by the formula

d[P ]dt

=Vmax[S]KM + [S]

, (6.82)

where KM is called the Michaelis constant(Fig. 6.17). This reaction at small concentra-tions acts like an ordinary chemical reaction withN = 1 and k = Vmax/KM , but the rate saturatesat Vmax as [S]→∞. The Michaelis constantKM

is the concentration [S] at which the rate is equalto half of its saturation rate (Fig. 6.17).

0 KM, KHSubstrate concentration [S]

0

Vmax

Rat

e d[

P]/d

t

Michaelis-MentenHill, n = 4

Fig. 6.17 Michaelis–Menten and Hill equationforms.

We can derive the Michaelis–Menten form by hy-pothesizing the existence of a catalyst or enzymeE, which is in short supply. The enzyme is pre-sumed to be partly free and available for bind-ing (concentration [E]) and partly bound to thesubstrate (concentration62 [E : S]), helping it toturn into the product. The total concentration[E] + [E : S] = Etot is fixed. The reactions areas follows:

E + S !k−1k1

E : Skcat→ E + P. (6.83)

We must then assume that the supply of sub-strate is large, so its concentration changesslowly with time. We can then assume that theconcentration [E : S] is in steady state, and re-move it as a degree of freedom.

61[S] is the concentration of S (number per unit volume). S stands for substrate.62The colon denotes the bound state of the two, called a dimer.

Copyright Oxford University Press 2006 v2.0 --

Exercises 165

formulæ from parts (a) and (b), give your esti-mate of the decay rate for our system. Relatedformula:

∫∞0

x exp(−x2/2σ2) dx = σ2.How could we go beyond this one-dimensionalcalculation? In the olden days, Kramers stud-ied other one-dimensional models, changing theways in which the system was coupled to the ex-ternal bath. On the computer, one can avoid aseparate heat bath and directly work with thefull multidimensional configuration space, lead-ing to transition-state theory. The transition-state theory formula is very similar to the oneyou derived in part (c), except that the prefac-tor involves the product of all the frequenciesat the bottom of the well and all the positivefrequencies at the saddlepoint at the top of thebarrier (see [72]). Other generalizations arisewhen crossing multiple barriers [81] or in non-equilibrium systems [109].

(6.12) Michaelis–Menten and Hill. (Biology, Com-putation) ©3Biological reaction rates are often saturable; thecell needs to respond sensitively to the introduc-tion of a new chemical S, but the response shouldnot keep growing indefinitely as the new chem-ical concentration [S] grows.61 Other biologi-cal reactions act as switches; a switch not onlysaturates, but its rate or state changes sharplyfrom one value to another as the concentrationof a chemical S is varied. These reactions givetangible examples of how one develops effectivedynamical theories by removing degrees of free-dom; here, instead of coarse-graining some largestatistical mechanics system, we remove a singleenzyme E from the equations to get an effectivereaction rate.The rate of a chemical reaction,

NS +B → C, (6.80)

where N substrate molecules S combine with a Bmolecule to make a C molecule, will occur witha reaction rate given by the law of mass-action:

d[C]dt

= k[S]N [B]. (6.81)

Saturation and the Michaelis–Menten equation.Saturation is not seen in ordinary chemical reac-tion kinetics. Notice that the reaction rate goesas the Nth power of the concentration [S]; far

from saturating, the reaction rate grows linearlyor faster with concentration.The archetypal example of saturation in biolog-ical systems is the Michaelis–Menten reactionform. A reaction of this form converting a chem-ical S (the substrate) into P (the product) has arate given by the formula

d[P ]dt

=Vmax[S]KM + [S]

, (6.82)

where KM is called the Michaelis constant(Fig. 6.17). This reaction at small concentra-tions acts like an ordinary chemical reaction withN = 1 and k = Vmax/KM , but the rate saturatesat Vmax as [S]→∞. The Michaelis constantKM

is the concentration [S] at which the rate is equalto half of its saturation rate (Fig. 6.17).

0 KM, KHSubstrate concentration [S]

0

Vmax

Rate

d[P

]/dt

Michaelis-MentenHill, n = 4

Fig. 6.17 Michaelis–Menten and Hill equationforms.

We can derive the Michaelis–Menten form by hy-pothesizing the existence of a catalyst or enzymeE, which is in short supply. The enzyme is pre-sumed to be partly free and available for bind-ing (concentration [E]) and partly bound to thesubstrate (concentration62 [E : S]), helping it toturn into the product. The total concentration[E] + [E : S] = Etot is fixed. The reactions areas follows:

E + S !k−1k1

E : Skcat→ E + P. (6.83)

We must then assume that the supply of sub-strate is large, so its concentration changesslowly with time. We can then assume that theconcentration [E : S] is in steady state, and re-move it as a degree of freedom.

61[S] is the concentration of S (number per unit volume). S stands for substrate.62The colon denotes the bound state of the two, called a dimer.

Copyright Oxford University Press 2006 v2.0 --

166 Free energies

(a) Assume the binding reaction rates ineqn 6.83 are of traditional chemical kinetics form(eqn 6.81), with constants k1, k−1, and kcat, andwith N = 1 or N = 0 as appropriate. Write theequation for d[E : S]/dt, set it to zero, and use itto eliminate [E] in the equation for dP/dt. Whatare Vmax and KM in the Michaelis–Menten form(eqn 6.82) in terms of the ks and Etot?We can understand this saturation intuitively:when all the enzyme is busy and bound to thesubstrate, adding more substrate cannot speedup the reaction.Cooperativity and sharp switching: the Hill equa-tion. Hemoglobin is what makes blood red;this iron-containing protein can bind up to fourmolecules of oxygen in the lungs, and carriesthem to the tissues of the body where it releasesthem. If the binding of all four oxygens wereindependent, the [O2] concentration dependenceof the bound oxygen concentration would havethe Michaelis–Menten form; to completely de-oxygenate the Hemoglobin (Hb) would demanda very low oxygen concentration in the tissue.What happens instead is that the Hb bindingof oxygen looks much more sigmoidal—a fairlysharp transition between nearly four oxygensbound at high [O2] (lungs) to nearly none boundat low oxygen concentrations. This arises be-cause the binding of the oxygens is enhanced byhaving other oxygens bound. This is not be-cause the oxygens somehow stick to one another;instead, each oxygen deforms the Hb in a non-local allosteric63 fashion, changing the configura-tions and affinity of the other binding sites. TheHill equation was introduced for hemoglobin todescribe this kind of cooperative binding. Likethe Michaelis–Menten form, it is also used to de-scribe reaction rates, where instead of the carrierHb we have an enzyme, or perhaps a series oftranscription binding sites (see Exercise 8.11).We will derive the Hill equation not in terms ofallosteric binding, but in the context of a sat-urable reaction involving n molecules binding si-multaneously. As a reaction rate, the Hill equa-tion is

d[P ]dt

=Vmax[S]

n

KnH + [S]n

(6.84)

(see Fig. 6.17). For Hb, the concentration ofthe n-fold oxygenated form is given by the right-

hand side of eqn 6.84. In both cases, the transi-tion becomes much more of a switch, with the re-action turning on (or the Hb accepting or releas-ing its oxygen) sharply at a particular concen-tration (Fig. 6.17). The transition can be mademore or less sharp by increasing or decreasing n.The Hill equation can be derived using a sim-plifying assumption that n molecules bind in asingle reaction:

E + nS !kukb

E : (nS), (6.85)

where E might stand for hemoglobin and S forthe O2 oxygen molecules. Again, there is a fixedtotal amount Etot = [E] + [E : nS].(b) Assume that the two reactions in eqn 6.85have the chemical kinetics form (eqn 6.81) withN = 0 or N = n as appropriate. Write the equi-librium equation for E : (nS), and eliminate [E]using the fixed total Etot. What are Vmax andKH in terms of kb, ku, and Etot?Usually, and in particular for hemoglobin, thiscooperativity is not so rigid; the states with one,two, and three O2 molecules bound also competewith the unbound and fully bound states. Thisis treated in an approximate way by using theHill equation, but allowing n to vary as a fittingparameter; for Hb, n ≈ 2.8.Both Hill and Michaelis–Menten equations areoften used in biological reaction models evenwhen there are no explicit mechanisms (enzymes,cooperative binding) known to generate them.

(6.13) Pollen and hard squares.64 ©i

Q B b

63Allosteric comes from Allo (other) and steric (structure or space). Allosteric interactions can be cooperative, as in hemoglobin,or inhibitory.64This problem was inspired by conversations with Ard Louis. Copyright Oxford University Press 2006 v2.0 --

166 Free energies

(a) Assume the binding reaction rates ineqn 6.83 are of traditional chemical kinetics form(eqn 6.81), with constants k1, k−1, and kcat, andwith N = 1 or N = 0 as appropriate. Write theequation for d[E : S]/dt, set it to zero, and use itto eliminate [E] in the equation for dP/dt. Whatare Vmax and KM in the Michaelis–Menten form(eqn 6.82) in terms of the ks and Etot?We can understand this saturation intuitively:when all the enzyme is busy and bound to thesubstrate, adding more substrate cannot speedup the reaction.Cooperativity and sharp switching: the Hill equa-tion. Hemoglobin is what makes blood red;this iron-containing protein can bind up to fourmolecules of oxygen in the lungs, and carriesthem to the tissues of the body where it releasesthem. If the binding of all four oxygens wereindependent, the [O2] concentration dependenceof the bound oxygen concentration would havethe Michaelis–Menten form; to completely de-oxygenate the Hemoglobin (Hb) would demanda very low oxygen concentration in the tissue.What happens instead is that the Hb bindingof oxygen looks much more sigmoidal—a fairlysharp transition between nearly four oxygensbound at high [O2] (lungs) to nearly none boundat low oxygen concentrations. This arises be-cause the binding of the oxygens is enhanced byhaving other oxygens bound. This is not be-cause the oxygens somehow stick to one another;instead, each oxygen deforms the Hb in a non-local allosteric63 fashion, changing the configura-tions and affinity of the other binding sites. TheHill equation was introduced for hemoglobin todescribe this kind of cooperative binding. Likethe Michaelis–Menten form, it is also used to de-scribe reaction rates, where instead of the carrierHb we have an enzyme, or perhaps a series oftranscription binding sites (see Exercise 8.11).We will derive the Hill equation not in terms ofallosteric binding, but in the context of a sat-urable reaction involving n molecules binding si-multaneously. As a reaction rate, the Hill equa-tion is

d[P ]dt

=Vmax[S]

n

KnH + [S]n

(6.84)

(see Fig. 6.17). For Hb, the concentration ofthe n-fold oxygenated form is given by the right-

hand side of eqn 6.84. In both cases, the transi-tion becomes much more of a switch, with the re-action turning on (or the Hb accepting or releas-ing its oxygen) sharply at a particular concen-tration (Fig. 6.17). The transition can be mademore or less sharp by increasing or decreasing n.The Hill equation can be derived using a sim-plifying assumption that n molecules bind in asingle reaction:

E + nS !kukb

E : (nS), (6.85)

where E might stand for hemoglobin and S forthe O2 oxygen molecules. Again, there is a fixedtotal amount Etot = [E] + [E : nS].(b) Assume that the two reactions in eqn 6.85have the chemical kinetics form (eqn 6.81) withN = 0 or N = n as appropriate. Write the equi-librium equation for E : (nS), and eliminate [E]using the fixed total Etot. What are Vmax andKH in terms of kb, ku, and Etot?Usually, and in particular for hemoglobin, thiscooperativity is not so rigid; the states with one,two, and three O2 molecules bound also competewith the unbound and fully bound states. Thisis treated in an approximate way by using theHill equation, but allowing n to vary as a fittingparameter; for Hb, n ≈ 2.8.Both Hill and Michaelis–Menten equations areoften used in biological reaction models evenwhen there are no explicit mechanisms (enzymes,cooperative binding) known to generate them.

(6.13) Pollen and hard squares.64 ©i

Q B b

63Allosteric comes from Allo (other) and steric (structure or space). Allosteric interactions can be cooperative, as in hemoglobin,or inhibitory.64This problem was inspired by conversations with Ard Louis.

Copyright Oxford University Press 2006 v2.0 --

166 Free energies

(a) Assume the binding reaction rates ineqn 6.83 are of traditional chemical kinetics form(eqn 6.81), with constants k1, k−1, and kcat, andwith N = 1 or N = 0 as appropriate. Write theequation for d[E : S]/dt, set it to zero, and use itto eliminate [E] in the equation for dP/dt. Whatare Vmax and KM in the Michaelis–Menten form(eqn 6.82) in terms of the ks and Etot?We can understand this saturation intuitively:when all the enzyme is busy and bound to thesubstrate, adding more substrate cannot speedup the reaction.Cooperativity and sharp switching: the Hill equa-tion. Hemoglobin is what makes blood red;this iron-containing protein can bind up to fourmolecules of oxygen in the lungs, and carriesthem to the tissues of the body where it releasesthem. If the binding of all four oxygens wereindependent, the [O2] concentration dependenceof the bound oxygen concentration would havethe Michaelis–Menten form; to completely de-oxygenate the Hemoglobin (Hb) would demanda very low oxygen concentration in the tissue.What happens instead is that the Hb bindingof oxygen looks much more sigmoidal—a fairlysharp transition between nearly four oxygensbound at high [O2] (lungs) to nearly none boundat low oxygen concentrations. This arises be-cause the binding of the oxygens is enhanced byhaving other oxygens bound. This is not be-cause the oxygens somehow stick to one another;instead, each oxygen deforms the Hb in a non-local allosteric63 fashion, changing the configura-tions and affinity of the other binding sites. TheHill equation was introduced for hemoglobin todescribe this kind of cooperative binding. Likethe Michaelis–Menten form, it is also used to de-scribe reaction rates, where instead of the carrierHb we have an enzyme, or perhaps a series oftranscription binding sites (see Exercise 8.11).We will derive the Hill equation not in terms ofallosteric binding, but in the context of a sat-urable reaction involving n molecules binding si-multaneously. As a reaction rate, the Hill equa-tion is

d[P ]dt

=Vmax[S]

n

KnH + [S]n

(6.84)

(see Fig. 6.17). For Hb, the concentration ofthe n-fold oxygenated form is given by the right-

hand side of eqn 6.84. In both cases, the transi-tion becomes much more of a switch, with the re-action turning on (or the Hb accepting or releas-ing its oxygen) sharply at a particular concen-tration (Fig. 6.17). The transition can be mademore or less sharp by increasing or decreasing n.The Hill equation can be derived using a sim-plifying assumption that n molecules bind in asingle reaction:

E + nS !kukb

E : (nS), (6.85)

where E might stand for hemoglobin and S forthe O2 oxygen molecules. Again, there is a fixedtotal amount Etot = [E] + [E : nS].(b) Assume that the two reactions in eqn 6.85have the chemical kinetics form (eqn 6.81) withN = 0 or N = n as appropriate. Write the equi-librium equation for E : (nS), and eliminate [E]using the fixed total Etot. What are Vmax andKH in terms of kb, ku, and Etot?Usually, and in particular for hemoglobin, thiscooperativity is not so rigid; the states with one,two, and three O2 molecules bound also competewith the unbound and fully bound states. Thisis treated in an approximate way by using theHill equation, but allowing n to vary as a fittingparameter; for Hb, n ≈ 2.8.Both Hill and Michaelis–Menten equations areoften used in biological reaction models evenwhen there are no explicit mechanisms (enzymes,cooperative binding) known to generate them.

(6.13) Pollen and hard squares.64 ©i

Q B b

63Allosteric comes from Allo (other) and steric (structure or space). Allosteric interactions can be cooperative, as in hemoglobin,or inhibitory.64This problem was inspired by conversations with Ard Louis. Copyright Oxford University Press 2006 v2.0 --

166 Free energies

(a) Assume the binding reaction rates ineqn 6.83 are of traditional chemical kinetics form(eqn 6.81), with constants k1, k−1, and kcat, andwith N = 1 or N = 0 as appropriate. Write theequation for d[E : S]/dt, set it to zero, and use itto eliminate [E] in the equation for dP/dt. Whatare Vmax and KM in the Michaelis–Menten form(eqn 6.82) in terms of the ks and Etot?We can understand this saturation intuitively:when all the enzyme is busy and bound to thesubstrate, adding more substrate cannot speedup the reaction.Cooperativity and sharp switching: the Hill equa-tion. Hemoglobin is what makes blood red;this iron-containing protein can bind up to fourmolecules of oxygen in the lungs, and carriesthem to the tissues of the body where it releasesthem. If the binding of all four oxygens wereindependent, the [O2] concentration dependenceof the bound oxygen concentration would havethe Michaelis–Menten form; to completely de-oxygenate the Hemoglobin (Hb) would demanda very low oxygen concentration in the tissue.What happens instead is that the Hb bindingof oxygen looks much more sigmoidal—a fairlysharp transition between nearly four oxygensbound at high [O2] (lungs) to nearly none boundat low oxygen concentrations. This arises be-cause the binding of the oxygens is enhanced byhaving other oxygens bound. This is not be-cause the oxygens somehow stick to one another;instead, each oxygen deforms the Hb in a non-local allosteric63 fashion, changing the configura-tions and affinity of the other binding sites. TheHill equation was introduced for hemoglobin todescribe this kind of cooperative binding. Likethe Michaelis–Menten form, it is also used to de-scribe reaction rates, where instead of the carrierHb we have an enzyme, or perhaps a series oftranscription binding sites (see Exercise 8.11).We will derive the Hill equation not in terms ofallosteric binding, but in the context of a sat-urable reaction involving n molecules binding si-multaneously. As a reaction rate, the Hill equa-tion is

d[P ]dt

=Vmax[S]

n

KnH + [S]n

(6.84)

(see Fig. 6.17). For Hb, the concentration ofthe n-fold oxygenated form is given by the right-

hand side of eqn 6.84. In both cases, the transi-tion becomes much more of a switch, with the re-action turning on (or the Hb accepting or releas-ing its oxygen) sharply at a particular concen-tration (Fig. 6.17). The transition can be mademore or less sharp by increasing or decreasing n.The Hill equation can be derived using a sim-plifying assumption that n molecules bind in asingle reaction:

E + nS !kukb

E : (nS), (6.85)

where E might stand for hemoglobin and S forthe O2 oxygen molecules. Again, there is a fixedtotal amount Etot = [E] + [E : nS].(b) Assume that the two reactions in eqn 6.85have the chemical kinetics form (eqn 6.81) withN = 0 or N = n as appropriate. Write the equi-librium equation for E : (nS), and eliminate [E]using the fixed total Etot. What are Vmax andKH in terms of kb, ku, and Etot?Usually, and in particular for hemoglobin, thiscooperativity is not so rigid; the states with one,two, and three O2 molecules bound also competewith the unbound and fully bound states. Thisis treated in an approximate way by using theHill equation, but allowing n to vary as a fittingparameter; for Hb, n ≈ 2.8.Both Hill and Michaelis–Menten equations areoften used in biological reaction models evenwhen there are no explicit mechanisms (enzymes,cooperative binding) known to generate them.

(6.13) Pollen and hard squares.64 ©i

Q B b

63Allosteric comes from Allo (other) and steric (structure or space). Allosteric interactions can be cooperative, as in hemoglobin,or inhibitory.64This problem was inspired by conversations with Ard Louis.

8

9 10

11 12

11/1/21

3

2 2 3

2 2 3

2 2 3

H N NHH N NH

H N NH 3 2

A A AA N N NN N N

µ µ µ

¶ ¶ ¶D = D + D + D

¶ ¶ ¶

= - - +

2 2 3H N NH3 2 0µ µ µ- - + =

For a system at fixed volume and temperature, consider the Helmholtz free energy:

A(T ,V ,NH2,NN2

,NNH3)

When the reaction takes place, the free energy changes:

As we argued before, the free energy goes to a minimum as the system moves to equilibrium, so at equilibrium we will have:

13

A(N ,V ,T ) = NkBT log((N /V )λ3)−1⎡⎣ ⎤⎦ + NF0

µ(N ,V ,T ) = ∂A∂N

= kBT log((N /V )λ3)−1⎡⎣ ⎤⎦ + NkBT (1/ N )+ F0

= kBT log((N /V )λ3)+ F0

= kBT log(N /V )+ c + F0

Now we assume that each of the three species is behaving like uncorrelated ideal gases (not too bad of an assumption). Each Helmholtz free energy looks like

(λ = h / 2πmkBT thermal deBroglie wavelength)

The last term F0 is the internal free energy of the molecule. We can write the chemical potential as:

14

−3log H2⎡⎣ ⎤⎦ − log N2⎡⎣ ⎤⎦ + 2log NH3⎡⎣ ⎤⎦ = log(Keq ),

⇒ NH3⎡⎣ ⎤⎦

2

H2⎡⎣ ⎤⎦3

N2⎡⎣ ⎤⎦= Keq .

Insert that result into2 2 3H N NH3 2 0µ µ µ- - + =

Also X⎡⎣ ⎤⎦ = NX V

This leads to:

15

Keq ∝ exp(−ΔEnet / kBT )

eq 0 netexp( / )BK K F k T= -D

ΔFnet = −3F0H2 − F0

N2 + 2F0NH3

Let’s solve for the equilibrium constant. First we define the difference in internal free energies as:

This gives us:

Then going back to all of the other neglected constants, we get: K0 =

λH29 λN2

3

λNH36 =

h6mNH33

8π 3kB3T 3mH2

9/2mN23/2 ∝T

−3

Usually, the temperature dependence of the reaction constant is dominated by the Boltzmann factor:

16

Also consider Sethna 6.22 Nucleosynthesis as a chemical reaction

Copyright Oxford University Press 2006 v2.0 --

176 Free energies

In this exercise, we explore a hypothetical uni-verse with faster nucleosynthesis reactions, orslower expansion, so that the reactions remainedin equilibrium. Clearly the nucleons for the equi-librated Universe will all fuse into their most sta-ble form.77 The binding energy ∆E released byfusing nucleons into 56Fe is about 5 × 108 eV(about half a proton mass). In this exercise, weshall ignore the difference between protons andneutrons,78 ignore nuclear excitations (assumingno internal entropy for the nuclei, so ∆E is thefree energy difference), and ignore the electrons(so protons are the same as hydrogen atoms,etc.)The creation of 56Fe from nucleons involves acomplex cascade of reactions. We argued in Sec-tion (6.6) that however complicated the reac-tions, they must in net be described by the over-all reaction, here

56p→ 56Fe, (6.98)

releasing an energy of about ∆E = 5×108 eV orabout ∆E/56 = 1.4 × 10−12 joules/baryon. Weused the fact that most chemical species are di-lute (and hence can be described by an ideal gas)to derive the corresponding law of mass–action,here

[Fe]/[p]56 = Keq(T ). (6.99)

(Note that this equality, true in equilibrium nomatter how messy the necessary reactions path-ways, can be rationalized using the oversimpli-fied picture that 56 protons must simultaneouslycollect in a small region to form an iron nucleus.)We then used the Helmholtz free energy for anideal gas to calculate the reaction constant ex-plicitly Keq(T ) for a similar reaction, in the idealgas limit.(a) Give symbolic formulas for Keq(T ) ineqn 6.99, assuming the nucleons form an ap-proximate ideal gas. Reduce it to expressionsin terms of ∆E, kBT , h, mp, and the atomic

number A = 56. (For convenience, assumemFe = Amp, ignoring the half-proton-mass en-ergy release. Check the sign of ∆E: should itbe positive or negative for this reaction?) Eval-uate Keq(T ) in MKS units, as a function of T .(Note: Do not be alarmed by the large numbers.)Wikipedia tells us that nucleosynthesis hap-pened “A few minutes into the expansion, whenthe temperature was about a billion . . . Kelvinand the density was about that of air”. To figureout when half of the protons would have fused([Fe] = [p]/A), we need to know the temperatureand the net baryon density [p] + A[Fe]. We canuse eqn 6.99 to give one equation relating thetemperature to the baryon density.(b) Give the symbolic formula for the net baryondensity79 at which half of the protons would havefused, in terms of Keq and A.What can we use for the other equation re-lating temperature to baryon density? One ofthe fundamental constants in the Universe as itevolves is the baryon to photon number ratio.The total number of baryons has been conservedin accelerator experiments of much higher en-ergy than those during nucleosynthesis. The cos-mic microwave background (CMB) photons havemostly been traveling in straight lines80 since thedecoupling time, a few hundred thousand yearsafter the Big Bang when the electrons and nu-cleons combined into atoms and became trans-parent (decoupled from photons). The ratio ofbaryons to CMB photons η ∼ 5×10−10 has beenconstant since that point: two billion photonsper nucleon.81 Between the nucleosynthesis anddecoupling eras, photons were created and scat-tered by matter, but there were so many morephotons than baryons that the number of pho-tons (and hence the baryon to photon ratio η)was still approximately constant.We know the density of blackbody photons at atemperature T ; we integrate eqn (7.66) for the

77It is usually said that 56Fe is the most stable nucleus, but actually 62Ni has a higher nuclear binding energyIron-56 is favored by nucleosynthesis pathways and conditions inside stars, and we shall go with traditioncalculation.78The entropy cost for our reaction in reality depends upon the ratio of neutrons to protons. Calculations implyfalls out of equilibrium a bit earlier, and is about 1/7 during this period. By ignoring the difference betwprotons, we avoid considering reactions of the form Mp + (56 −M)n + (M − 26)e →56Fe.

79Here we ask you to assume, unphysically, that all baryons are either free protons or in iron nuclei – ignoring the baryonsforming helium and other elements. These other elements will be rare at early (hot) times, and rare again at late (cold) times,but will spread out the energy release from nucleosynthesis over a larger range than our one-reaction estimate would suggest.80More precisely, they have traveled on geodesics in space-time.81It is amusing to note that this ratio is estimated using the models of nucleosynthesis that we are mimicking in this exercise.

Copyright Oxford University Press 2006 v2.0 --

176 Free energies

In this exercise, we explore a hypothetical uni-verse with faster nucleosynthesis reactions, orslower expansion, so that the reactions remainedin equilibrium. Clearly the nucleons for the equi-librated Universe will all fuse into their most sta-ble form.77 The binding energy ∆E released byfusing nucleons into 56Fe is about 5 × 108 eV(about half a proton mass). In this exercise, weshall ignore the difference between protons andneutrons,78 ignore nuclear excitations (assumingno internal entropy for the nuclei, so ∆E is thefree energy difference), and ignore the electrons(so protons are the same as hydrogen atoms,etc.)The creation of 56Fe from nucleons involves acomplex cascade of reactions. We argued in Sec-tion (6.6) that however complicated the reac-tions, they must in net be described by the over-all reaction, here

56p→ 56Fe, (6.98)

releasing an energy of about ∆E = 5×108 eV orabout ∆E/56 = 1.4 × 10−12 joules/baryon. Weused the fact that most chemical species are di-lute (and hence can be described by an ideal gas)to derive the corresponding law of mass–action,here

[Fe]/[p]56 = Keq(T ). (6.99)

(Note that this equality, true in equilibrium nomatter how messy the necessary reactions path-ways, can be rationalized using the oversimpli-fied picture that 56 protons must simultaneouslycollect in a small region to form an iron nucleus.)We then used the Helmholtz free energy for anideal gas to calculate the reaction constant ex-plicitly Keq(T ) for a similar reaction, in the idealgas limit.(a) Give symbolic formulas for Keq(T ) ineqn 6.99, assuming the nucleons form an ap-proximate ideal gas. Reduce it to expressionsin terms of ∆E, kBT , h, mp, and the atomic

number A = 56. (For convenience, assumemFe = Amp, ignoring the half-proton-mass en-ergy release. Check the sign of ∆E: should itbe positive or negative for this reaction?) Eval-uate Keq(T ) in MKS units, as a function of T .(Note: Do not be alarmed by the large numbers.)Wikipedia tells us that nucleosynthesis hap-pened “A few minutes into the expansion, whenthe temperature was about a billion . . . Kelvinand the density was about that of air”. To figureout when half of the protons would have fused([Fe] = [p]/A), we need to know the temperatureand the net baryon density [p] + A[Fe]. We canuse eqn 6.99 to give one equation relating thetemperature to the baryon density.(b) Give the symbolic formula for the net baryondensity79 at which half of the protons would havefused, in terms of Keq and A.What can we use for the other equation re-lating temperature to baryon density? One ofthe fundamental constants in the Universe as itevolves is the baryon to photon number ratio.The total number of baryons has been conservedin accelerator experiments of much higher en-ergy than those during nucleosynthesis. The cos-mic microwave background (CMB) photons havemostly been traveling in straight lines80 since thedecoupling time, a few hundred thousand yearsafter the Big Bang when the electrons and nu-cleons combined into atoms and became trans-parent (decoupled from photons). The ratio ofbaryons to CMB photons η ∼ 5×10−10 has beenconstant since that point: two billion photonsper nucleon.81 Between the nucleosynthesis anddecoupling eras, photons were created and scat-tered by matter, but there were so many morephotons than baryons that the number of pho-tons (and hence the baryon to photon ratio η)was still approximately constant.We know the density of blackbody photons at atemperature T ; we integrate eqn (7.66) for the

77It is usually said that 56Fe is the most stable nucleus, but actually 62Ni has a higher nuclear binding energyIron-56 is favored by nucleosynthesis pathways and conditions inside stars, and we shall go with traditioncalculation.78The entropy cost for our reaction in reality depends upon the ratio of neutrons to protons. Calculations implyfalls out of equilibrium a bit earlier, and is about 1/7 during this period. By ignoring the difference betwprotons, we avoid considering reactions of the form Mp + (56 −M)n + (M − 26)e →56Fe.

79Here we ask you to assume, unphysically, that all baryons are either free protons or in iron nuclei – ignoring the baryonsforming helium and other elements. These other elements will be rare at early (hot) times, and rare again at late (cold) times,but will spread out the energy release from nucleosynthesis over a larger range than our one-reaction estimate would suggest.80More precisely, they have traveled on geodesics in space-time.81It is amusing to note that this ratio is estimated using the models of nucleosynthesis that we are mimicking in this exercise.

eq 0 netexp( / )BK K F k T= -D

K0 =λH29 λN2

3

λNH36 =

h6mNH33

8π 3kB3T 3mH2

9/2mN23/2 ∝T

−3

17

Copyright Oxford University Press 2006 v2.0 --

176 Free energies

In this exercise, we explore a hypothetical uni-verse with faster nucleosynthesis reactions, orslower expansion, so that the reactions remainedin equilibrium. Clearly the nucleons for the equi-librated Universe will all fuse into their most sta-ble form.77 The binding energy ∆E released byfusing nucleons into 56Fe is about 5 × 108 eV(about half a proton mass). In this exercise, weshall ignore the difference between protons andneutrons,78 ignore nuclear excitations (assumingno internal entropy for the nuclei, so ∆E is thefree energy difference), and ignore the electrons(so protons are the same as hydrogen atoms,etc.)The creation of 56Fe from nucleons involves acomplex cascade of reactions. We argued in Sec-tion (6.6) that however complicated the reac-tions, they must in net be described by the over-all reaction, here

56p→ 56Fe, (6.98)

releasing an energy of about ∆E = 5×108 eV orabout ∆E/56 = 1.4 × 10−12 joules/baryon. Weused the fact that most chemical species are di-lute (and hence can be described by an ideal gas)to derive the corresponding law of mass–action,here

[Fe]/[p]56 = Keq(T ). (6.99)

(Note that this equality, true in equilibrium nomatter how messy the necessary reactions path-ways, can be rationalized using the oversimpli-fied picture that 56 protons must simultaneouslycollect in a small region to form an iron nucleus.)We then used the Helmholtz free energy for anideal gas to calculate the reaction constant ex-plicitly Keq(T ) for a similar reaction, in the idealgas limit.(a) Give symbolic formulas for Keq(T ) ineqn 6.99, assuming the nucleons form an ap-proximate ideal gas. Reduce it to expressionsin terms of ∆E, kBT , h, mp, and the atomic

number A = 56. (For convenience, assumemFe = Amp, ignoring the half-proton-mass en-ergy release. Check the sign of ∆E: should itbe positive or negative for this reaction?) Eval-uate Keq(T ) in MKS units, as a function of T .(Note: Do not be alarmed by the large numbers.)Wikipedia tells us that nucleosynthesis hap-pened “A few minutes into the expansion, whenthe temperature was about a billion . . . Kelvinand the density was about that of air”. To figureout when half of the protons would have fused([Fe] = [p]/A), we need to know the temperatureand the net baryon density [p] + A[Fe]. We canuse eqn 6.99 to give one equation relating thetemperature to the baryon density.(b) Give the symbolic formula for the net baryondensity79 at which half of the protons would havefused, in terms of Keq and A.What can we use for the other equation re-lating temperature to baryon density? One ofthe fundamental constants in the Universe as itevolves is the baryon to photon number ratio.The total number of baryons has been conservedin accelerator experiments of much higher en-ergy than those during nucleosynthesis. The cos-mic microwave background (CMB) photons havemostly been traveling in straight lines80 since thedecoupling time, a few hundred thousand yearsafter the Big Bang when the electrons and nu-cleons combined into atoms and became trans-parent (decoupled from photons). The ratio ofbaryons to CMB photons η ∼ 5×10−10 has beenconstant since that point: two billion photonsper nucleon.81 Between the nucleosynthesis anddecoupling eras, photons were created and scat-tered by matter, but there were so many morephotons than baryons that the number of pho-tons (and hence the baryon to photon ratio η)was still approximately constant.We know the density of blackbody photons at atemperature T ; we integrate eqn (7.66) for the

77It is usually said that 56Fe is the most stable nucleus, but actually 62Ni has a higher nuclear binding energyIron-56 is favored by nucleosynthesis pathways and conditions inside stars, and we shall go with traditioncalculation.78The entropy cost for our reaction in reality depends upon the ratio of neutrons to protons. Calculations implyfalls out of equilibrium a bit earlier, and is about 1/7 during this period. By ignoring the difference betwprotons, we avoid considering reactions of the form Mp + (56 −M)n + (M − 26)e →56Fe.

79Here we ask you to assume, unphysically, that all baryons are either free protons or in iron nuclei – ignoring the baryonsforming helium and other elements. These other elements will be rare at early (hot) times, and rare again at late (cold) times,but will spread out the energy release from nucleosynthesis over a larger range than our one-reaction estimate would suggest.80More precisely, they have traveled on geodesics in space-time.81It is amusing to note that this ratio is estimated using the models of nucleosynthesis that we are mimicking in this exercise.

Copyright Oxford University Press 2006 v2.0 --

Exercises 177

number of photons per unit frequency to get

ρphotons(T ) = (2ζ(3)/π2)(kBT/!c)3; (6.100)

the current density ρphotons(TCMB) of microwavebackground photons at temperature TCMB =2.725K is thus a bit over 400 per cubic centime-ter, and hence the current density of a fractionof a baryon per cubic meter.(c) Numerically matching your formula for thenet baryon density at the halfway point inpart (b) to ηρphotons(Treaction), derive the tem-perature Treaction at which our reaction wouldhave occurred. (You can do this graphically, ifneeded.) Is it roughly a billion degrees Kelvin?Is the nucleon density roughly equal to that ofair? (Hint: Air has a density of about 1 kg/m3.)

(6.23) Pistons in probability space.82 (Mathemat-ics, Information Geometry) ©4Fig. 5.3 shows the Carnot cycle as a path in theP–V space of presure and volume – parametersvaried from the outside. Once could draw a simi-lar diagram in the space of pressure and temper-ature, or volume and temperature. Here we shallexplore how to describe the path in the spaceof probability distributions. In the process, weshall compute the model manifold of the idealgas (Exercise 1.14), and show that it is a two-dimensional plane.As discussed in Exercise 1.15, there is a naturaldistance, or metric, in the space of probabilitydistributions:

gµν = −⟨∂2 log(ρ)∂θµ∂θν

⟩, (6.101)

the Fisher information metric. So, a system inthe Gibbs ensemble is described in terms of twoparameters, usually P and T . We shall insteaduse θ1 = p = βP and θ2 = β, where β = 1/kBT .The squared distance in probability space be-tween two systems with tiny changes in pressureand temperature is then

d2(ρ(X|θ), ρ(X|θ + dθ)) = gµνdθµdθν . (6.102)

(a) Compute g(p,β)µν = −〈∂2 log(ρ)/∂θµ∂θν〉 using

the results of part (a) in Exercise 6.19.

The metric tensor g(p,β) for the Gibbs ensem-ble of the piston tells us the distance in proba-bility space between neighboring pressures andtemperatures. What kind of surface (the modelmanifold) is formed by this two-parameter fam-ily of probability distributions? Does it have anintrinsic curvature?(b) Show that one can turn the metric tensor intothe identity g(x,y)µν = δµν by a coordinate trans-formation (p,β)→ (x = A log(p), y = B log(β)).What are the necessary scale factors A and B?Hence the model manifold of the piston in theGibbs ensemble is a plane! We can draw ourcontrol paths in the (x, y) plane. We label thefour steps of the Carnot cycle as in Fig. 5.3.(c) Draw the Carnot cycle path in as a param-eterized curve in (x, y), with Pa = 1, Pb = 0.5,T1 = 1 and T2 = 0.8, for N = 1. (Hint: eqn 5.8will be helpful in finding the adiabatic parts ofthe path p(β).) Is the length of the expansionat fixed pressure the same as you calculated inExercise 6.19?

(6.24) FIM for Gibbs.83 (Mathematics, Thermody-namics, Information Geometry) ©4In this exercise, we study the geometry in thespace of probability distributions defined by theGibbs ensemble84 of a general equilibrium sys-tem. We compute the Fisher Information Metric(FIM, Exercises 1.15 and 6.23)

gµν = −⟨∂2 log(ρ)∂θµ∂θν

⟩, (6.103)

of the Gibbs phase space ensemble ρ(P,Q) interms of thermodynamic properties of the sys-tem.In Exercise 6.23 we calculated gµν for the idealgas, using the variables θ1 = p = βP and θ2 = β,rather than P and T . Why are these coordinatesspecial? The log of the Gibbs probability distri-bution for an arbitrary interacting collection ofparticles with Hamiltonian H (eqn 6.91) is

log(ρ) = −βH(P,Q)− βPV − log Γ

= −βH(P,Q)− pV − log Γ.(6.104)

This is the logarithm of the partition function Γplus terms linear in p and β.85 So the second

82This exercise was developed in collaboration with Ben Machta, Archishman Raju, Colin Clement, and Katherine Quinn83This exercise was developed in collaboration with Ben Machta, Archishman Raju, Colin Clement, and Katherine Quinn84The Fisher information distance is badly defined except for changes in intensive quantities. In a microcanonical ensemble,for example, the energy E is constant and so the derivative ∂ρ/∂E would be the derivative of a δ function. So we study pistonsvarying P and β = 1/kBT , rather than at fixed volume or energy.85In statistics, log probability distributions which depend on parameters in this linear fashion are called exponential families.Many common distributions, including lots of statistical mechanical models like ours, are exponential families.

Copyright Oxford University Press 2006 v2.0 --

176 Free energies

In this exercise, we explore a hypothetical uni-verse with faster nucleosynthesis reactions, orslower expansion, so that the reactions remainedin equilibrium. Clearly the nucleons for the equi-librated Universe will all fuse into their most sta-ble form.77 The binding energy ∆E released byfusing nucleons into 56Fe is about 5 × 108 eV(about half a proton mass). In this exercise, weshall ignore the difference between protons andneutrons,78 ignore nuclear excitations (assumingno internal entropy for the nuclei, so ∆E is thefree energy difference), and ignore the electrons(so protons are the same as hydrogen atoms,etc.)The creation of 56Fe from nucleons involves acomplex cascade of reactions. We argued in Sec-tion (6.6) that however complicated the reac-tions, they must in net be described by the over-all reaction, here

56p→ 56Fe, (6.98)

releasing an energy of about ∆E = 5×108 eV orabout ∆E/56 = 1.4 × 10−12 joules/baryon. Weused the fact that most chemical species are di-lute (and hence can be described by an ideal gas)to derive the corresponding law of mass–action,here

[Fe]/[p]56 = Keq(T ). (6.99)

(Note that this equality, true in equilibrium nomatter how messy the necessary reactions path-ways, can be rationalized using the oversimpli-fied picture that 56 protons must simultaneouslycollect in a small region to form an iron nucleus.)We then used the Helmholtz free energy for anideal gas to calculate the reaction constant ex-plicitly Keq(T ) for a similar reaction, in the idealgas limit.(a) Give symbolic formulas for Keq(T ) ineqn 6.99, assuming the nucleons form an ap-proximate ideal gas. Reduce it to expressionsin terms of ∆E, kBT , h, mp, and the atomic

number A = 56. (For convenience, assumemFe = Amp, ignoring the half-proton-mass en-ergy release. Check the sign of ∆E: should itbe positive or negative for this reaction?) Eval-uate Keq(T ) in MKS units, as a function of T .(Note: Do not be alarmed by the large numbers.)Wikipedia tells us that nucleosynthesis hap-pened “A few minutes into the expansion, whenthe temperature was about a billion . . . Kelvinand the density was about that of air”. To figureout when half of the protons would have fused([Fe] = [p]/A), we need to know the temperatureand the net baryon density [p] + A[Fe]. We canuse eqn 6.99 to give one equation relating thetemperature to the baryon density.(b) Give the symbolic formula for the net baryondensity79 at which half of the protons would havefused, in terms of Keq and A.What can we use for the other equation re-lating temperature to baryon density? One ofthe fundamental constants in the Universe as itevolves is the baryon to photon number ratio.The total number of baryons has been conservedin accelerator experiments of much higher en-ergy than those during nucleosynthesis. The cos-mic microwave background (CMB) photons havemostly been traveling in straight lines80 since thedecoupling time, a few hundred thousand yearsafter the Big Bang when the electrons and nu-cleons combined into atoms and became trans-parent (decoupled from photons). The ratio ofbaryons to CMB photons η ∼ 5×10−10 has beenconstant since that point: two billion photonsper nucleon.81 Between the nucleosynthesis anddecoupling eras, photons were created and scat-tered by matter, but there were so many morephotons than baryons that the number of pho-tons (and hence the baryon to photon ratio η)was still approximately constant.We know the density of blackbody photons at atemperature T ; we integrate eqn (7.66) for the

77It is usually said that 56Fe is the most stable nucleus, but actually 62Ni has a higher nuclear binding energyIron-56 is favored by nucleosynthesis pathways and conditions inside stars, and we shall go with traditioncalculation.78The entropy cost for our reaction in reality depends upon the ratio of neutrons to protons. Calculations implyfalls out of equilibrium a bit earlier, and is about 1/7 during this period. By ignoring the difference betwprotons, we avoid considering reactions of the form Mp + (56 −M)n + (M − 26)e →56Fe.

79Here we ask you to assume, unphysically, that all baryons are either free protons or in iron nuclei – ignoring the baryonsforming helium and other elements. These other elements will be rare at early (hot) times, and rare again at late (cold) times,but will spread out the energy release from nucleosynthesis over a larger range than our one-reaction estimate would suggest.80More precisely, they have traveled on geodesics in space-time.81It is amusing to note that this ratio is estimated using the models of nucleosynthesis that we are mimicking in this exercise.

Copyright Oxford University Press 2006 v2.0 --

Exercises 177

number of photons per unit frequency to get

ρphotons(T ) = (2ζ(3)/π2)(kBT/!c)3; (6.100)

the current density ρphotons(TCMB) of microwavebackground photons at temperature TCMB =2.725K is thus a bit over 400 per cubic centime-ter, and hence the current density of a fractionof a baryon per cubic meter.(c) Numerically matching your formula for thenet baryon density at the halfway point inpart (b) to ηρphotons(Treaction), derive the tem-perature Treaction at which our reaction wouldhave occurred. (You can do this graphically, ifneeded.) Is it roughly a billion degrees Kelvin?Is the nucleon density roughly equal to that ofair? (Hint: Air has a density of about 1 kg/m3.)

(6.23) Pistons in probability space.82 (Mathemat-ics, Information Geometry) ©4Fig. 5.3 shows the Carnot cycle as a path in theP–V space of presure and volume – parametersvaried from the outside. Once could draw a simi-lar diagram in the space of pressure and temper-ature, or volume and temperature. Here we shallexplore how to describe the path in the spaceof probability distributions. In the process, weshall compute the model manifold of the idealgas (Exercise 1.14), and show that it is a two-dimensional plane.As discussed in Exercise 1.15, there is a naturaldistance, or metric, in the space of probabilitydistributions:

gµν = −⟨∂2 log(ρ)∂θµ∂θν

⟩, (6.101)

the Fisher information metric. So, a system inthe Gibbs ensemble is described in terms of twoparameters, usually P and T . We shall insteaduse θ1 = p = βP and θ2 = β, where β = 1/kBT .The squared distance in probability space be-tween two systems with tiny changes in pressureand temperature is then

d2(ρ(X|θ), ρ(X|θ + dθ)) = gµνdθµdθν . (6.102)

(a) Compute g(p,β)µν = −〈∂2 log(ρ)/∂θµ∂θν〉 using

the results of part (a) in Exercise 6.19.

The metric tensor g(p,β) for the Gibbs ensem-ble of the piston tells us the distance in proba-bility space between neighboring pressures andtemperatures. What kind of surface (the modelmanifold) is formed by this two-parameter fam-ily of probability distributions? Does it have anintrinsic curvature?(b) Show that one can turn the metric tensor intothe identity g(x,y)µν = δµν by a coordinate trans-formation (p,β)→ (x = A log(p), y = B log(β)).What are the necessary scale factors A and B?Hence the model manifold of the piston in theGibbs ensemble is a plane! We can draw ourcontrol paths in the (x, y) plane. We label thefour steps of the Carnot cycle as in Fig. 5.3.(c) Draw the Carnot cycle path in as a param-eterized curve in (x, y), with Pa = 1, Pb = 0.5,T1 = 1 and T2 = 0.8, for N = 1. (Hint: eqn 5.8will be helpful in finding the adiabatic parts ofthe path p(β).) Is the length of the expansionat fixed pressure the same as you calculated inExercise 6.19?

(6.24) FIM for Gibbs.83 (Mathematics, Thermody-namics, Information Geometry) ©4In this exercise, we study the geometry in thespace of probability distributions defined by theGibbs ensemble84 of a general equilibrium sys-tem. We compute the Fisher Information Metric(FIM, Exercises 1.15 and 6.23)

gµν = −⟨∂2 log(ρ)∂θµ∂θν

⟩, (6.103)

of the Gibbs phase space ensemble ρ(P,Q) interms of thermodynamic properties of the sys-tem.In Exercise 6.23 we calculated gµν for the idealgas, using the variables θ1 = p = βP and θ2 = β,rather than P and T . Why are these coordinatesspecial? The log of the Gibbs probability distri-bution for an arbitrary interacting collection ofparticles with Hamiltonian H (eqn 6.91) is

log(ρ) = −βH(P,Q)− βPV − log Γ

= −βH(P,Q)− pV − log Γ.(6.104)

This is the logarithm of the partition function Γplus terms linear in p and β.85 So the second

82This exercise was developed in collaboration with Ben Machta, Archishman Raju, Colin Clement, and Katherine Quinn83This exercise was developed in collaboration with Ben Machta, Archishman Raju, Colin Clement, and Katherine Quinn84The Fisher information distance is badly defined except for changes in intensive quantities. In a microcanonical ensemble,for example, the energy E is constant and so the derivative ∂ρ/∂E would be the derivative of a δ function. So we study pistonsvarying P and β = 1/kBT , rather than at fixed volume or energy.85In statistics, log probability distributions which depend on parameters in this linear fashion are called exponential families.Many common distributions, including lots of statistical mechanical models like ours, are exponential families.

Copyright Oxford University Press 2006 v2.0 --

176 Free energies

In this exercise, we explore a hypothetical uni-verse with faster nucleosynthesis reactions, orslower expansion, so that the reactions remainedin equilibrium. Clearly the nucleons for the equi-librated Universe will all fuse into their most sta-ble form.77 The binding energy ∆E released byfusing nucleons into 56Fe is about 5 × 108 eV(about half a proton mass). In this exercise, weshall ignore the difference between protons andneutrons,78 ignore nuclear excitations (assumingno internal entropy for the nuclei, so ∆E is thefree energy difference), and ignore the electrons(so protons are the same as hydrogen atoms,etc.)The creation of 56Fe from nucleons involves acomplex cascade of reactions. We argued in Sec-tion (6.6) that however complicated the reac-tions, they must in net be described by the over-all reaction, here

56p→ 56Fe, (6.98)

releasing an energy of about ∆E = 5×108 eV orabout ∆E/56 = 1.4 × 10−12 joules/baryon. Weused the fact that most chemical species are di-lute (and hence can be described by an ideal gas)to derive the corresponding law of mass–action,here

[Fe]/[p]56 = Keq(T ). (6.99)

(Note that this equality, true in equilibrium nomatter how messy the necessary reactions path-ways, can be rationalized using the oversimpli-fied picture that 56 protons must simultaneouslycollect in a small region to form an iron nucleus.)We then used the Helmholtz free energy for anideal gas to calculate the reaction constant ex-plicitly Keq(T ) for a similar reaction, in the idealgas limit.(a) Give symbolic formulas for Keq(T ) ineqn 6.99, assuming the nucleons form an ap-proximate ideal gas. Reduce it to expressionsin terms of ∆E, kBT , h, mp, and the atomic

number A = 56. (For convenience, assumemFe = Amp, ignoring the half-proton-mass en-ergy release. Check the sign of ∆E: should itbe positive or negative for this reaction?) Eval-uate Keq(T ) in MKS units, as a function of T .(Note: Do not be alarmed by the large numbers.)Wikipedia tells us that nucleosynthesis hap-pened “A few minutes into the expansion, whenthe temperature was about a billion . . . Kelvinand the density was about that of air”. To figureout when half of the protons would have fused([Fe] = [p]/A), we need to know the temperatureand the net baryon density [p] + A[Fe]. We canuse eqn 6.99 to give one equation relating thetemperature to the baryon density.(b) Give the symbolic formula for the net baryondensity79 at which half of the protons would havefused, in terms of Keq and A.What can we use for the other equation re-lating temperature to baryon density? One ofthe fundamental constants in the Universe as itevolves is the baryon to photon number ratio.The total number of baryons has been conservedin accelerator experiments of much higher en-ergy than those during nucleosynthesis. The cos-mic microwave background (CMB) photons havemostly been traveling in straight lines80 since thedecoupling time, a few hundred thousand yearsafter the Big Bang when the electrons and nu-cleons combined into atoms and became trans-parent (decoupled from photons). The ratio ofbaryons to CMB photons η ∼ 5×10−10 has beenconstant since that point: two billion photonsper nucleon.81 Between the nucleosynthesis anddecoupling eras, photons were created and scat-tered by matter, but there were so many morephotons than baryons that the number of pho-tons (and hence the baryon to photon ratio η)was still approximately constant.We know the density of blackbody photons at atemperature T ; we integrate eqn (7.66) for the

77It is usually said that 56Fe is the most stable nucleus, but actually 62Ni has a higher nuclear binding energyIron-56 is favored by nucleosynthesis pathways and conditions inside stars, and we shall go with traditioncalculation.78The entropy cost for our reaction in reality depends upon the ratio of neutrons to protons. Calculations implyfalls out of equilibrium a bit earlier, and is about 1/7 during this period. By ignoring the difference betwprotons, we avoid considering reactions of the form Mp + (56 −M)n + (M − 26)e →56Fe.

79Here we ask you to assume, unphysically, that all baryons are either free protons or in iron nuclei – ignoring the baryonsforming helium and other elements. These other elements will be rare at early (hot) times, and rare again at late (cold) times,but will spread out the energy release from nucleosynthesis over a larger range than our one-reaction estimate would suggest.80More precisely, they have traveled on geodesics in space-time.81It is amusing to note that this ratio is estimated using the models of nucleosynthesis that we are mimicking in this exercise.

Copyright Oxford University Press 2006 v2.0 --

Exercises 177

number of photons per unit frequency to get

ρphotons(T ) = (2ζ(3)/π2)(kBT/!c)3; (6.100)

the current density ρphotons(TCMB) of microwavebackground photons at temperature TCMB =2.725K is thus a bit over 400 per cubic centime-ter, and hence the current density of a fractionof a baryon per cubic meter.(c) Numerically matching your formula for thenet baryon density at the halfway point inpart (b) to ηρphotons(Treaction), derive the tem-perature Treaction at which our reaction wouldhave occurred. (You can do this graphically, ifneeded.) Is it roughly a billion degrees Kelvin?Is the nucleon density roughly equal to that ofair? (Hint: Air has a density of about 1 kg/m3.)

(6.23) Pistons in probability space.82 (Mathemat-ics, Information Geometry) ©4Fig. 5.3 shows the Carnot cycle as a path in theP–V space of presure and volume – parametersvaried from the outside. Once could draw a simi-lar diagram in the space of pressure and temper-ature, or volume and temperature. Here we shallexplore how to describe the path in the spaceof probability distributions. In the process, weshall compute the model manifold of the idealgas (Exercise 1.14), and show that it is a two-dimensional plane.As discussed in Exercise 1.15, there is a naturaldistance, or metric, in the space of probabilitydistributions:

gµν = −⟨∂2 log(ρ)∂θµ∂θν

⟩, (6.101)

the Fisher information metric. So, a system inthe Gibbs ensemble is described in terms of twoparameters, usually P and T . We shall insteaduse θ1 = p = βP and θ2 = β, where β = 1/kBT .The squared distance in probability space be-tween two systems with tiny changes in pressureand temperature is then

d2(ρ(X|θ), ρ(X|θ + dθ)) = gµνdθµdθν . (6.102)

(a) Compute g(p,β)µν = −〈∂2 log(ρ)/∂θµ∂θν〉 using

the results of part (a) in Exercise 6.19.

The metric tensor g(p,β) for the Gibbs ensem-ble of the piston tells us the distance in proba-bility space between neighboring pressures andtemperatures. What kind of surface (the modelmanifold) is formed by this two-parameter fam-ily of probability distributions? Does it have anintrinsic curvature?(b) Show that one can turn the metric tensor intothe identity g(x,y)µν = δµν by a coordinate trans-formation (p,β)→ (x = A log(p), y = B log(β)).What are the necessary scale factors A and B?Hence the model manifold of the piston in theGibbs ensemble is a plane! We can draw ourcontrol paths in the (x, y) plane. We label thefour steps of the Carnot cycle as in Fig. 5.3.(c) Draw the Carnot cycle path in as a param-eterized curve in (x, y), with Pa = 1, Pb = 0.5,T1 = 1 and T2 = 0.8, for N = 1. (Hint: eqn 5.8will be helpful in finding the adiabatic parts ofthe path p(β).) Is the length of the expansionat fixed pressure the same as you calculated inExercise 6.19?

(6.24) FIM for Gibbs.83 (Mathematics, Thermody-namics, Information Geometry) ©4In this exercise, we study the geometry in thespace of probability distributions defined by theGibbs ensemble84 of a general equilibrium sys-tem. We compute the Fisher Information Metric(FIM, Exercises 1.15 and 6.23)

gµν = −⟨∂2 log(ρ)∂θµ∂θν

⟩, (6.103)

of the Gibbs phase space ensemble ρ(P,Q) interms of thermodynamic properties of the sys-tem.In Exercise 6.23 we calculated gµν for the idealgas, using the variables θ1 = p = βP and θ2 = β,rather than P and T . Why are these coordinatesspecial? The log of the Gibbs probability distri-bution for an arbitrary interacting collection ofparticles with Hamiltonian H (eqn 6.91) is

log(ρ) = −βH(P,Q)− βPV − log Γ

= −βH(P,Q)− pV − log Γ.(6.104)

This is the logarithm of the partition function Γplus terms linear in p and β.85 So the second

82This exercise was developed in collaboration with Ben Machta, Archishman Raju, Colin Clement, and Katherine Quinn83This exercise was developed in collaboration with Ben Machta, Archishman Raju, Colin Clement, and Katherine Quinn84The Fisher information distance is badly defined except for changes in intensive quantities. In a microcanonical ensemble,for example, the energy E is constant and so the derivative ∂ρ/∂E would be the derivative of a δ function. So we study pistonsvarying P and β = 1/kBT , rather than at fixed volume or energy.85In statistics, log probability distributions which depend on parameters in this linear fashion are called exponential families.Many common distributions, including lots of statistical mechanical models like ours, are exponential families.

Copyright Oxford University Press 2006 v2.0 --

176 Free energies

In this exercise, we explore a hypothetical uni-verse with faster nucleosynthesis reactions, orslower expansion, so that the reactions remainedin equilibrium. Clearly the nucleons for the equi-librated Universe will all fuse into their most sta-ble form.77 The binding energy ∆E released byfusing nucleons into 56Fe is about 5 × 108 eV(about half a proton mass). In this exercise, weshall ignore the difference between protons andneutrons,78 ignore nuclear excitations (assumingno internal entropy for the nuclei, so ∆E is thefree energy difference), and ignore the electrons(so protons are the same as hydrogen atoms,etc.)The creation of 56Fe from nucleons involves acomplex cascade of reactions. We argued in Sec-tion (6.6) that however complicated the reac-tions, they must in net be described by the over-all reaction, here

56p→ 56Fe, (6.98)

releasing an energy of about ∆E = 5×108 eV orabout ∆E/56 = 1.4 × 10−12 joules/baryon. Weused the fact that most chemical species are di-lute (and hence can be described by an ideal gas)to derive the corresponding law of mass–action,here

[Fe]/[p]56 = Keq(T ). (6.99)

(Note that this equality, true in equilibrium nomatter how messy the necessary reactions path-ways, can be rationalized using the oversimpli-fied picture that 56 protons must simultaneouslycollect in a small region to form an iron nucleus.)We then used the Helmholtz free energy for anideal gas to calculate the reaction constant ex-plicitly Keq(T ) for a similar reaction, in the idealgas limit.(a) Give symbolic formulas for Keq(T ) ineqn 6.99, assuming the nucleons form an ap-proximate ideal gas. Reduce it to expressionsin terms of ∆E, kBT , h, mp, and the atomic

number A = 56. (For convenience, assumemFe = Amp, ignoring the half-proton-mass en-ergy release. Check the sign of ∆E: should itbe positive or negative for this reaction?) Eval-uate Keq(T ) in MKS units, as a function of T .(Note: Do not be alarmed by the large numbers.)Wikipedia tells us that nucleosynthesis hap-pened “A few minutes into the expansion, whenthe temperature was about a billion . . . Kelvinand the density was about that of air”. To figureout when half of the protons would have fused([Fe] = [p]/A), we need to know the temperatureand the net baryon density [p] + A[Fe]. We canuse eqn 6.99 to give one equation relating thetemperature to the baryon density.(b) Give the symbolic formula for the net baryondensity79 at which half of the protons would havefused, in terms of Keq and A.What can we use for the other equation re-lating temperature to baryon density? One ofthe fundamental constants in the Universe as itevolves is the baryon to photon number ratio.The total number of baryons has been conservedin accelerator experiments of much higher en-ergy than those during nucleosynthesis. The cos-mic microwave background (CMB) photons havemostly been traveling in straight lines80 since thedecoupling time, a few hundred thousand yearsafter the Big Bang when the electrons and nu-cleons combined into atoms and became trans-parent (decoupled from photons). The ratio ofbaryons to CMB photons η ∼ 5×10−10 has beenconstant since that point: two billion photonsper nucleon.81 Between the nucleosynthesis anddecoupling eras, photons were created and scat-tered by matter, but there were so many morephotons than baryons that the number of pho-tons (and hence the baryon to photon ratio η)was still approximately constant.We know the density of blackbody photons at atemperature T ; we integrate eqn (7.66) for the

77It is usually said that 56Fe is the most stable nucleus, but actually 62Ni has a higher nuclear binding energyIron-56 is favored by nucleosynthesis pathways and conditions inside stars, and we shall go with traditioncalculation.78The entropy cost for our reaction in reality depends upon the ratio of neutrons to protons. Calculations implyfalls out of equilibrium a bit earlier, and is about 1/7 during this period. By ignoring the difference betwprotons, we avoid considering reactions of the form Mp + (56 −M)n + (M − 26)e →56Fe.

79Here we ask you to assume, unphysically, that all baryons are either free protons or in iron nuclei – ignoring the baryonsforming helium and other elements. These other elements will be rare at early (hot) times, and rare again at late (cold) times,but will spread out the energy release from nucleosynthesis over a larger range than our one-reaction estimate would suggest.80More precisely, they have traveled on geodesics in space-time.81It is amusing to note that this ratio is estimated using the models of nucleosynthesis that we are mimicking in this exercise.

Copyright Oxford University Press 2006 v2.0 --

Exercises 177

number of photons per unit frequency to get

ρphotons(T ) = (2ζ(3)/π2)(kBT/!c)3; (6.100)

the current density ρphotons(TCMB) of microwavebackground photons at temperature TCMB =2.725K is thus a bit over 400 per cubic centime-ter, and hence the current density of a fractionof a baryon per cubic meter.(c) Numerically matching your formula for thenet baryon density at the halfway point inpart (b) to ηρphotons(Treaction), derive the tem-perature Treaction at which our reaction wouldhave occurred. (You can do this graphically, ifneeded.) Is it roughly a billion degrees Kelvin?Is the nucleon density roughly equal to that ofair? (Hint: Air has a density of about 1 kg/m3.)

(6.23) Pistons in probability space.82 (Mathemat-ics, Information Geometry) ©4Fig. 5.3 shows the Carnot cycle as a path in theP–V space of presure and volume – parametersvaried from the outside. Once could draw a simi-lar diagram in the space of pressure and temper-ature, or volume and temperature. Here we shallexplore how to describe the path in the spaceof probability distributions. In the process, weshall compute the model manifold of the idealgas (Exercise 1.14), and show that it is a two-dimensional plane.As discussed in Exercise 1.15, there is a naturaldistance, or metric, in the space of probabilitydistributions:

gµν = −⟨∂2 log(ρ)∂θµ∂θν

⟩, (6.101)

the Fisher information metric. So, a system inthe Gibbs ensemble is described in terms of twoparameters, usually P and T . We shall insteaduse θ1 = p = βP and θ2 = β, where β = 1/kBT .The squared distance in probability space be-tween two systems with tiny changes in pressureand temperature is then

d2(ρ(X|θ), ρ(X|θ + dθ)) = gµνdθµdθν . (6.102)

(a) Compute g(p,β)µν = −〈∂2 log(ρ)/∂θµ∂θν〉 using

the results of part (a) in Exercise 6.19.

The metric tensor g(p,β) for the Gibbs ensem-ble of the piston tells us the distance in proba-bility space between neighboring pressures andtemperatures. What kind of surface (the modelmanifold) is formed by this two-parameter fam-ily of probability distributions? Does it have anintrinsic curvature?(b) Show that one can turn the metric tensor intothe identity g(x,y)µν = δµν by a coordinate trans-formation (p,β)→ (x = A log(p), y = B log(β)).What are the necessary scale factors A and B?Hence the model manifold of the piston in theGibbs ensemble is a plane! We can draw ourcontrol paths in the (x, y) plane. We label thefour steps of the Carnot cycle as in Fig. 5.3.(c) Draw the Carnot cycle path in as a param-eterized curve in (x, y), with Pa = 1, Pb = 0.5,T1 = 1 and T2 = 0.8, for N = 1. (Hint: eqn 5.8will be helpful in finding the adiabatic parts ofthe path p(β).) Is the length of the expansionat fixed pressure the same as you calculated inExercise 6.19?

(6.24) FIM for Gibbs.83 (Mathematics, Thermody-namics, Information Geometry) ©4In this exercise, we study the geometry in thespace of probability distributions defined by theGibbs ensemble84 of a general equilibrium sys-tem. We compute the Fisher Information Metric(FIM, Exercises 1.15 and 6.23)

gµν = −⟨∂2 log(ρ)∂θµ∂θν

⟩, (6.103)

of the Gibbs phase space ensemble ρ(P,Q) interms of thermodynamic properties of the sys-tem.In Exercise 6.23 we calculated gµν for the idealgas, using the variables θ1 = p = βP and θ2 = β,rather than P and T . Why are these coordinatesspecial? The log of the Gibbs probability distri-bution for an arbitrary interacting collection ofparticles with Hamiltonian H (eqn 6.91) is

log(ρ) = −βH(P,Q)− βPV − log Γ

= −βH(P,Q)− pV − log Γ.(6.104)

This is the logarithm of the partition function Γplus terms linear in p and β.85 So the second

82This exercise was developed in collaboration with Ben Machta, Archishman Raju, Colin Clement, and Katherine Quinn83This exercise was developed in collaboration with Ben Machta, Archishman Raju, Colin Clement, and Katherine Quinn84The Fisher information distance is badly defined except for changes in intensive quantities. In a microcanonical ensemble,for example, the energy E is constant and so the derivative ∂ρ/∂E would be the derivative of a δ function. So we study pistonsvarying P and β = 1/kBT , rather than at fixed volume or energy.85In statistics, log probability distributions which depend on parameters in this linear fashion are called exponential families.Many common distributions, including lots of statistical mechanical models like ours, are exponential families.

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