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Matthew Schwartz Statistical Mechanics, Spring 2019
Lecture 7: Ensembles
1 Introduction
In statistical mechanics, we study the possible microstates of a system. We never know exactly which microstate the system is in. Nor do we care. We are interested only in the behavior of a system based on the possible microstates it could be, that share some macroscopic proporty (like volume V ; energy E, or number of particles N ). The possible microstates a system could be in are known as the ensemble of states for a system. There are dierent kinds of ensembles. So far, we have been counting microstates with a xed number of particles N and a xed total energy E. We de ned as the total number microstates for a system. That is
(E ; V ; N ) =
X
microstatesk withsameN ;V ;E
Then S = kBln is the entropy, and all other thermodynamic quantities follow from S. For an isolated system with N xed and E xed the ensemble is known as the microcanonical ensemble. In the microcanonical ensemble, the temperature is a derived quantity, with (^) T^1 = (^) @E@S. So far, we have only been using the microcanonical ensemble.
For example, a gas of identical monatomic particles has (E ; V ; N ) (^) N^1! V NE
3 2 N^. From this
we computed the entropy S = kBln which at large N reduces to the Sackur-Tetrode formula.
The temperature is 1 T =^
@S @E =^
3 2
N kB E so that^ E^ =^
3 2 NkBT^. Also in the microcanonical ensemble we observed that the number of states for which the energy of one degree of freedom is xed to "i is
(E ¡ "i). Thus the probability of such a state is Pi = (E(^ ¡E^ )" i) e¡"i/kBT^. This is the Boltzmann distribution. Within the context of the microcanonical ensemble, we also derived the Boltzmann distribution using the principle of maximum entropy. This approach is very general. It uses nothing about the system other than that the total number of degrees of freedom N is large and the total energy is E. To use the maximum entropy principle we counted the number of ways that N particles could be allocated into groups of size ni with energies "i, so that
P
ni = N and
P
ni"i = E. We found that in the most probable allocation of particles to groups, the probability of nding a particle with energy "i was
Pi =
Z
e¡^ "i^ (2)
where Z =
P
i e
¡ "i (^) and = 1 kBT. Sometimes we don't know the total energy, but we know the temperature. This situation is in fact much more common than knowing the energy. For example, in the room you are in, what is the energy of the air molecules? I bet you don't have a clue. But I bet you have a good idea of what the temperature is. When we x temperature instead of energy, we have to allow the energy to uctuate. For example, think of two systems in thermal contact. The thermal contact allows energy to ow in and out of each system, so energy of each system is not xed. We call a set of microstates with N ; V and T xed but variable E the canonical ensemble. In the canonical ensemble, the primary object is not the number of state or the entropy S but rather the partition function
Z( ) =
X
microstatesk withsameN ;V
e¡^ Ek^ (3)
In the partition function, energies of the microstates summed over can vary. Thus the left-hand side, Z( ), cannot depend on energy. Instead, it depends on temperature. Once Z is known, it is straightforward to compute the average energy hE i and other thermodynamic quantities, as we will see.
In both the microcanonical and canonical ensembles, we x the volume. We could instead let the volume vary and sum over possible volumes. Allowing the volume to vary gives the Gibbs ensemble. In the Gibbs ensemble, the partition function depends on pressure rather than volume, just as the canonical ensemble depended on temperature rather than energy. In the microcanonical, canonical, and Gibbs ensembles, the number of particles N in the system is xed. In some situations, we want the number of particles to vary. For example, chemical reactions change the number of each molecule type. So in chemistry we can't x N. Instead we x something called the chemical potential, . Chemical potential is like a pressure for particle number. Chemical potential is a very important concept, but very di cult to grasp, so we will spend a lot of time understanding it in this lecture and beyond. When N can vary we use the grand canonical ensemble. The main object of interest in the grand canonical ensemble is the grand partition function
Z =
X
microstatesk withsameV
e¡^ Eke Nk^ (4)
The grand canonical ensemble is used in chemistry, quantum statistical mechanics, and much of condensed matter physics.
2 Canonical ensemble
In the microcanonical ensemble, we calculated properties of its system by counting the number of
microstates at xed energy. Then, for example, temperature is a derived quantity, 1 kBT =^
@ln @E. In the canonical ensemble, we x the temperature T , and the (average) energy becomes the derived quantity. In order to x the temperature, it is a useful conceptual trick to imagine our system of interest in thermal contact with a heat reservoir. This means the system and heat reservoir can exchange energy through heat, but no work can be done by the system on the reservoir or vice versa. The point of the reservoir is to make concrete the idea of xing the temperature and letting the energy uctuate.
Figure 1. When a system is in thermal contact with a heat reservoir, its temperature is xed. Its energy uctuates around its average value.
We do not allow particles to go from the system to the reservoir, only energy. The number of particles in the system can be small we can have a single atom even it won't matter. This is important because the canonical ensemble will allow us to discuss systems with a limited number of quantum states, in contrast to the microcanonical ensemble where we really did need to expand at large N to make progress. Although the system can be small, the reservoir does need to be large, so that it has much much more energy than the system. But this is not a constraint, just a conceptual trick, since the reservoir does not actually need to exist. We would like to know what is the probability of nding the system in a xed microstate k with energy Ek? To be clear: every momentum and position of every particle in k is xed.
2 Section 2
Thus, knowing the partition function, we can get the expected value for the energy of the system by simply dierentiating. An important point about the canonical ensemble is that we derived a result about the system only. The partition function is a sum over microstates of the system. Pk is the probability of nding the system in microstate k when it is in equilibrium at a temperature T no matter what it is in contact with. We need it to be in contact with something to exchange energy and keep it at nite temperature, but the details of that system are totally irrelevant (except for its temperature). Note that we write hE i for the expected value of energy, rather than E since hE i is calculated rather than xed from the beginning. The thing we compute, hE i is a function of , so hE i is a derived quantity rather than one we x from the start, as in the microcanonical ensemble. In a real system hooked to a thermal bath, the total energy E would uctuate around hE i. If the system is isolated, then we can simply set hE i( ) = E and solve for a relation between E and T. In fact, this is mostly how we will use the canonical ensemble, to compute equilibrium properties of an isolated system. In such cases, we use hE i and E interchangeably. Note that the relation between E and T can be derived from the microcanonical ensemble or from the canonical ensemble. It will be the same relation (as we will check when we can).
3 Example 1: monatomic ideal gas
For an ideal monatomic gas with positions q and momenta p, the energy depends only on momenta E =
P
j
~p^2 2 m. So
Z
Z
d^3 Nqd^3 Np (q)^3 N^ (p)^3 N^ exp
X
j
~ pj^2 2 m
Here p and q are the size of phase space regions that we consider minimal. Classical mechanics gives no indication of what we should take for q and p, and no results that we derive will depend on our choices. As mentioned before, in quantum mechanics, we know to set q p = h (see Lecture 10) so let's take this value. Also, recall that for entropy to be extrinsic, we have to count any state in which the same positions and momenta are occupied as the same state. Thus we need to divide the integration by N! for identical particles. This gives
Z =
N!
Z
d^3 Nqd^3 N h^3 N^
exp
X
j
~ pj^2 2 m
The q integrals trivially give a factor of V N^. The p integrals are the product of 3 N Gaussian integrals. Each one gives Z
¡
1 dpe¡^
p^2 2 m (^) = 2 m
r (16)
So that
Z =
N!
V
h^3
N
2 m
^3
2 N (17)
Mostly we are interested in this at large N , where N!! e¡NN N^ gives
Zmoatomicgas = eN
V
Nh^3
N
2 m
^3
2 N (18)
Once we have Z it is easy to compute the (average) energy:
E = hE i = ¡ @lnZ @
N
ln
2 m
NkBT (19)
4 Section 3
This is in agreement with the result form the equipartition theory (the 3 kinetic degrees of freedom each get 12 kBT of energy per molecule).
Note that this analysis of the ideal gas in the canonical ensemble was a much easier way to compute the average energy than in the microcanonical ensemble, where we had to look at the surface area of a 3 N -dimensional sphere.
3.1 Heat capacity
Recall that the heat capacity CV is the amount of heat required to change the temperature at
constant volume: CV =
Q T
V
@E @T
V
. Recalling that = (^) k^1 BT^ we have
CV =
@ hE i @T
@ hE i @
@T
kBT 2
@lnZ @
kBT 2
@^2 lnZ @ 2
This equation lets us compute the heat capacity directly from the partition function. Let's check for the monatomic ideal gas. Using Eq. (18) we nd that
CV =
kBT 2
@^2
ln
¡ 32 N
N
(^2) kBT 2 =
NkB (21)
in agreement with our previous results.^2
3.2 Entropy
How do we extract the entropy from the partition function. The easiest way is using the Gibbs entropy:
S = ¡kB
X
k
Pkln Pk = ¡kB
X
k
e¡^ Ek Z
ln e¡^ Ek Z
= kB
X
k
e¡^ Ek Z
( Ek + ln Z) (24)
The rst term on the right is the average energy timesP kB = (^1) T. For the second term we just use 1 Z e
¡ Ek (^) = P^ Pn = 1. Therefore
S = hE i T
For example, using the partition function for the monatomic ideal gas, Eq. (18) we get
S =
NkB + kBln
eN
V
Nh^3
N
2 m
^3
2 N^
which reduces to:
S = NkB
ln
V
Nh^3
ln[2mkBT ] +
- It's interesting to write the calculation another way. Note that @ @
¡ @ @lnZ
= (^) @@
1 Z
X Ee¡^ E
= ¡ (^) Z^12
@Z @
X Ee¡^ E^ ¡ (^1) Z
X E^2 e¡^ E^ (22)
using that ¡ (^) Z^1 @Z@ = hE i we see that this is ¡hE^2 i. Thus,
CV = ¡ (^) k^1 BT^2
@ @
¡ @ @lnZ
= hE
(^2) i ¡ hE i 2 kBT 2 (23) In other words, the heat capacity is given by the RMS energy uctuations. This tells us that how a system changes when heated can be determined from properties of the system in equilibrium (the RMS energy uctuations). In other words, to measure the heat capacity, we do not ever have to actually heat up the system. Instead, we can let the system heat up itself through thermal uctuations away from the mean. This is a special case of a very general and powerful result in statistical physics known as the uctuation dissipation theorem. Another example was our computation of how the drag coe cient in a viscous uid related to the uctuations determined by random walks (Brownian motion): if you drag something, the energy dissipates in the same way that statistical uctuations dissipate.
Example 1: monatomic ideal gas 5
(Feel free to use mathematica to take derivatives or do it yourself by writing the expression in terms of exponentials and using the chain rule.) Comparing to Eq. (31) we see that the average excitation number is
hni =
e
~! kBT (^) ¡ 1
for kBT. ~!, hni 0 and only the ground state is occupieda and from (35), the energy atlines at its
zero point: E 0 = 12 ~!. At higher temperatures, the hE i and hni grow linearly with the temperature. The heat capacity is
CV =
@E
@T
= kB
kBT
e
¡ (^) k~B!T 1 ¡ e
¡ (^) k~B!T
Note that heat capacity is very small until the rst energy state can be excited, then it grows linearly. For H 2 , the vibrational mode has ~vib = 4342cm¡^1 corresponding to Tvib = ch~ kB^ =^
~! kB^ =^6300 K. So at low energies, the vibrational mode cannot be excited which is why the heat capacity for
hydrogen is CV = 5 2 NkBT^ rather than^
7 2 NkBT^. We discussed this in Lecture 4, but now we have explained it and can make more precise quantitative predictions of how the heat capacity changes with temperature. Including the kinetic contribution, and a factor of N for the N molecules that can be excited in the vibration mode we see
CV =
NkB + NkB
@ Tvib T
e¡
Tvib 2 T
1 ¡ e¡^
Tvib T
A
2 = (38)
This shows how the heat capacity goes up as the vibrational mode starts to be excitable. Note that although the temperature for the vibrational mode is 6300 K, the vibrational mode starts to be excited well below that temperature. The dots are data. We see good agreement! Can you gure out why the heat capacity dies o at low temperature? What do you think explains the small oset of the data from the theory prediction in the plot? We'll eventually produce a calculation in even better agreement with the data, but we need to incorporate quantum indistinguishability to get it right, as we will learn starting in Lecture 10.
5 Gibbs ensemble
In the microcanonical ensemble, we computed the number of states at a given energy (V ; N ; E) and used it to derive the entropy S(V ; N ; E) = kBln (V ; N ; T ). In the canonical ensemble, we computed the partition function by summing over Boltzmann factors, Z(N ; V ; ) =
P
k e
¡ Ek. In
both cases we have been holding V and N xed. Now we want to try varying V.
Gibbs ensemble 7
First, let's quickly recall why the temperature is the same in any two systems in thermal equilibrium. The quickest way to see this is to recall that entropy is extensive, so a system with energy E 1 = E and another with energy E 2 = Etot ¡ E has total entropy
S 12 (Etot; E) = S 1 (E) + S 2 (Etot ¡ E) (39)
Then the state with maximum entropy is the one where
@S 12 (Etot; E) @E
@S 1 (E)
@E E =E 1
@S 2 (E)
@E E =E 2
where the @S @E^1 (E ) E =E 1
means evaluate the partial derivative at E = E 1. In the second term we set
E = E 2 = Etot ¡ E. Thus we see that 1 T =^
@S @E is the same in the two systems. Now let's consider an ensemble that lets V vary. This is sometimes called the Gibbs ensemble. In the Gibbs ensemble you have two systems in equilibrium that can exchange energy and volume. Exchanging volume just means we have a moveable partition in between them. So the total volume is conserved
Figure 2. An ensemble where volume is allowed to vary
Now we just apply the same formal argument as in Eqs. (39) and (40): the entropy is the sum of the entropy of the two sides, and the total volume is xed: V 1 + V 2 = Vtot. This implies that
S 12 (Etot; Vtot; E ; V ) = S 1 (E ; V ) + S 2 (Etot ¡ E ; Vtot ¡ V ) (41)
And so maximizing entropy, by demanding both the partial derivative with respect to E and the
one with respect to V vanish give that the temperature (^1) T = (^) @E@S is the same on both sides (from the E derivative) and that
@S 12 (Etot; Vtot; E ; V ) @V
@S 1 (E ; V )
@V V =V 1
@S 2 (E ; V )
@V V =V 2
Thus @S@V is another quantity that is the same for any system in equilibrium. It is related to pressure,
but is @S @V =^ P^ or^
1 P or^2 ^
T P? We can gure out the^ T^ factor by dimensional analysis, and adjust the constant we can also check by matching on to the ideal gas law. To gure out what @S @V is, all we have to do is compute it in some sample system, such as a monatomic ideal gas. Using the entropy of a monatomic ideal gas in the canonical ensemble, Eq. (27), we nd (^) @S @V
E
@V
NkB
ln
V
N
ln
4 mE 3 Nh^2
NkB V
P
T
We have established that the left-hand side of this equation @S@V is the same for any two systems
in equilibrium. We also already know that (^) T^1 =
@S @E
N ;V
is the same for any two systems in
equilibrium. We conclude that the quantity T @S@V is the same for any two systems in equilibrium, and give this quantity the name pressure :
P T
@S
@V
E
8 Section 5
A useful way to think about chemical potential is as a pressure for number density. For example, suppose you have an atom that has two states, a ground state 0 and an excited state 1. In equilibrium, there will be some concentrations hn 0 i and hn 1 i of the two states, and the two chemical potentials 1 and 2 will be equal. Since the excited states have more energy, we expect fewer of them, so hn 1 i < hn 0 i. Say we add to the system some more atoms in the ground state. This would push more atoms into the excited state to restore equilibrium. This pushing is due to the number density pressure of the chemical potential. Adding to n 0 pushes up 0 , so 0 =/ 1 anymore; the number densities then change until equilibrium is restored. While there is only one kind of temperature and pressure there are lots of chemical potentials: one for every type of conserved N. If there are 4 dierent types of particles involved, there are 4 chemical potentials. The more general formula is
@S 1 (E ; V ; N 1 ; N 2 ; ) @N 1
T
@S 1 (E ; V ; N 1 ; N 2 ; )
@N 2
T
You should not think of the chemical potential as being connected to the grand canonical ensemble in any essential way. The chemical potential is property of the system, like pressure or temperature, relevant no matter what statistical system we use to perform the calculation. To see how chemical potential is embedded in the microcanonical ensemble, recall our microcanonical maximum entropy calculation, where we imposed ni = N and
P
ni"i = E as constraints. Then we maximized entropy by maximizing
S kB = ln = ¡N
X
i=
m fi ln fi ¡
¡X^
ni ¡ N
¡X^
ni"i ¡ E
Since @ln @E =^ , we identi ed this Lagrange multiplier^ with the usual^ =^
1 kBT. Since^
@ln @N = we can now identify = ¡ kBT as the chemical potential. Thus given in the microcanonical ensemble, we compute the chemical potential as
= ¡ kBT = ¡kBT
@ln (E ; V ; N ) @N
= ¡T
@S
@N
E ;V
in agreement with Eq. (50). As in Eq. (46) we can now consider the total derivative of energy, letting E ; V and N all vary:
dS =
@S
@E
dE +
@S
@V
dV +
@S
@N
dN =
T
dE +
P
T
dV ¡
T
dN (54)
That is,
dE = TdS ¡ PdV + dN (55)
This implies that (^) @E @N
S ;V
So the chemical potential represents the change in energy when a particle is added at constant V and S. This is almost intuitive. Unfortunately for the constant S constraint makes Eq. (56) hard to interpret. Don't worry though, we'll come up with better ways to understand in Section 7.
6.1 Grand partition function
As in Section 2 let us now hook a small system up to a reservoir to derive the Boltzmann factor. This time the reservoir should have large energy and large particle number, and both energy and particle number can ow between the system and reservoir. As before, think about picking one microstate k of the system with energy Ek and Nk particles. Once Ek and N are xed, the total number of microstates is determined only by the states in the reservoir. Eq. (7) becomes
ln (^) res(Etot ¡ Ek; Ntot ¡ Nk) = ln (^) res(Etot; Ntot) ¡ Ek + Nk (57)
where Eq. (53) was used. This leads to a Boltzmann factor
Pk =
Z
e¡^ Ek+^ Nk^ (58)
10 Section 6
where
Z(V ; ; ) =
X
k
Pk =
X
k
e¡^ Ek+^ Nk^ (59)
is called the grand partition function. The grand partition function lets us calculate the expected number of particles
hN i =
X
k
NkPk =
Z
X
k
Nke¡^ Ek+^ Nk^ =
Z
@Z
1 @lnZ @
We can also calculate the usual things the partition function lets us calculate, such as the average energy.
hE i =
X
k
EkPk =
Z
X
k
Eke¡^ Ek+^ Nk^ = ¡
Z
@Z
Z
X
k
Nke¡^ "k+^ Nk^ (61)
@lnZ @ ¡ hN i (62)
Particle number and chemical potential are conjugate, like pressure and volume. If you know N
for a system then you can calculate by @E@N. This is like how if you know the energy for a system,
you can calculate temperature from (^1) T = (^) @E@S. If you know the chemical potential instead of N , then
you can compute average number by hN i = 1 @ @lnZ. This is like how if you know temperature and
not the energy, you can compute the average energy form hE i = ¡@ln@ Z.
Finally, let's compute the entropy, in analogy to Eq. (25). We start with Eq. (24), which goes through to the grand canonical ensemble with Z! Z and E! (E ¡ N ):
S = kB
X (^) e¡ Ek+ Nk Z [ (Ek ¡ Nk) + ln Z] (63)
hE i T
hN i T
Thus,
¡kBT lnZ = hE i ¡ TS ¡ hN i (65)
This will be a useful relation.
7 Chemical potential
To get a feel for , let's do some examples. We'll work in the microcanonical ensemble for now since is independent of the way we calculate it and trying to understand Z and at the same time is unnecessarily challenging. We'll come back to Z in the next lecture and use it a lot in quantum statistical mechanics.
7.1 Ideal gas
For a monatomic gas, using the Sackur-Tetrode equation, we nd
= ¡T
@S
@N E ;V
= ¡T
@N
NkB
ln
V
N
ln
4 mE 3 Nh^2
=¡kBT
ln
V
N
ln
4 mE 3 Nh^2
Note that the 5 2 has dropped out. Using^ E^ =^
3 2 NkBT^ for this gas, we can write this relation in an abbreviated form
= kBT ln
n
h^2 2 mkBT
= kBT ln n^3 (68)
Chemical potential 11
One can also derive this from the canonical ensemble. Say the partition function without the energy shift is Z 0. Then with the energy oset Z = Z 0 E¡^ N". This gives hE i = ¡@ln@ Z= hE 0 i + N". Then
S =
hE i T
N"
T
Thus using the canonical ensemble we nd that the entropy S with the shift has the functional form as S 0 without the shift, it is only the energy where we evaluate S that changes. This is in agreement with Eq. (71). From the entropy, we can compute the chemical potential
= ¡T
@S
@N
E ;V
= kBT ln n^3 + " (73)
with in Eq. (69), so that for an ideal gas with ground-state energy "
n =
^3
exp
kBT
Thus when the energies shift, the chemical potential can shift to compensate. Dierences in chem- ical potential are independent of an overall energy shift. This is consistent with our interpretation of chemical potential as a potential. It is like a potential energy, and shifts with energy as potential energy does. With the energy oset, we can re ne our observation about the chemical potential being neg- ative for an ideal gas. Now we see that a more precise statement is that the chemical potential is less than the ground state energy " for a classical gas. In general, the chemical potential gets two contributions: one from the density and one from the energy. The density contribution is of entropic origin and depends on how many molecules are in the system. The energetic contribution is due to the internal structure of the molecule and independent of whatever else is going on. Equilibrium, where chemical potentials are equal, comes from a balance between these two contributions. This will be clearer with some examples.
7.3 Chemical reactions
Chemical potentials are useful in situations where particles turn into other types of particles. When there are more than one type of particle in the system (as there typically are when we consider problems involving chemical potential), we need a dierent for each particle. So Eq. (55) becomes
dE = TdS ¡ PdV +
X
jNj (75)
As a concrete example, consider the Haber process for the production of ammonia
3 H 2 + N 2 2 NH 3 (76)
Note that the number of each individual molecule is not conserved, but because the number of hydrogen atoms and nitrogen atoms is conserved, the relative coe cients (3,1 and 2) in Eq. (76) are xed. In chemistry, the concentrations or molar number densities of molecule j are denoted as [j] = njNA, with nj = N Vj and NA = 6 (^1023) mol^1 Avogadro's number. In equilibrium, there will
be some relationship among the concentrations [H 2 ] of hydrogen, [N 2 ] for nitrogen and [NH 3 ] for ammonia that we can compute using chemical potentials. As the concentrations change, at xed volume and xed total energy, the entropy changes as
dS =
@S
@[H 2 ]
d[H 2 ] +
@S
@[N 2 ]
d[N 2 ] +
@S
@[NH 3 ]
d[ NH 3 ] (77)
The changes in concentrations are related by Eq. (76): for every mole of N 2 consumed, exactly three moles of H 2 are consumed and 2 moles of NH 3 are produced. Thus, d[N 2 ] = 3d[H 2 ] = ¡ 2 d[NH 3 ]. Thus, using Eq. (75) with dV = dE = 0, or equivalently equation (50), we have
0 = 3H 2 + N 2 ¡ 2 NH 3 (78)
Chemical potential 13
This constraint among the chemical potentials is a generalization of 1 = 2 in equilibrium for two systems that can exchange particles. Here there are 3 systems that can exchange particles. Now, from Eq. (74) we know how to relate the number of particles to the chemical potential for a monatomic ideal gas:
[X] =
^3
exp
"X ¡ X
kBT
where "X is the ground state energy for molecule X. To get the 's to drop out, we can take the ratio of concentrations to appropriate powers:
[H 2 ]^3 [N 2 ] [NH 3 ]^2
NH^63
H^9 2 N^3
exp
3 "H 2 + "N 2 ¡ 2 "NH 3
kBT
exp
3 H 2 + N 2 ¡ 2 NH 3
kBT
|||||||||||||||||||||||| |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||{z}}}}}}}}}}}}}}}}}}}}}} }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}
The second exponential is just 1 because of Eq. (78), which is why we chose the powers of [H 2 ] and [NH 3 ] that we did on the left hand side. The means that we are approximating everything as monatomic ideal gases (not a great approximation but it's a start) The sum of energies is just the net energy change in the reaction, ". For the Haber process,
which is exothermic, " = 92.4 (^) molkJ. So
[H 2 ]^3 [N 2 ] [NH 3 ]^2
NH^63
H^9 2 N^3
exp
kBT
(assuming monatomic gases) (81)
This is special case (for monatomic ideal gases) of the law of mass action. It says that the rela- tive concentrations of reacting molecules in equilibrium are determined by the Boltzmann factor dependent on the change in energy associated with the reaction. This formula arises from a balance between entropic contributions to the chemical potentials on both sides (though their number densities) and energetic contributions (in the exponential factor). We have written explicitly the reminder that this formula assumes that the reactants and products are monatomic gases. This is not a bad assumption in some cases. More generally though, for chemicals reacting, we will need to add corrections to the right hand side. These corrections will be included in the next lecture, where the law of mass action is derived in full.
7.4 Example: matter antimatter asymmetry
For another example, consider the process of a proton-antiproton annihilation. Antiprotons p¡ are anti-particles of protons. They have the same mass as protons but opposite electric charge. Protons and anti-protons can annihilate into photons
p+^ + p¡^ + (82)
The reverse reaction is photons converting into proton-antiproton pairs. These annihilations and conversions happen constantly when the temperature is well above the threshold energy for pair production
kBT " = 2mpc^2 = 2 GeV (83)
We don't care so much about the details of why or how this process occurs, just that it does occur. This threshold temperature is around 2 1013 K. So in most systems of physical interest (stars, your body, etc.) this doesn't happen. It did happen however, in the early universe, until 0. seconds after the big bang. Note that while the above reaction conserves the number of protons minus the number of antiprotons, it does not conserve the number of photons. Indeed, other reactions can easily change photon number, such as
A more down to earth example is the light in your room photons are constantly being produced, not absorbed. Eq. (84) implies that
+ e¡ = e¡ + 2 (85)
14 Section 7
As the universe continues to expand from Tf down to 3 K its size scales with temperature so
[p+] = [p¡] 1023
m^3
3 K
Tf
= 1.68 10 ¡^10
m^3
This is the honest-to-goodness prediction of cosmology for the density of protons left over from the big bang (i.e. close to the best estimate physicists can make). Eq. (95) is much more reasonable than the equilibrium prediction in Eq. (90), but still in stark disagreement with data: the average number density of protons in the universe is [p+] = 0.26 (^) m^13. This is a problem. In fact, this is one of the great unsolved problems in physics, called the mystery of baryogenesis or the matter-antimatter asymmetry. One possible solution is to set the initial conditions so that [p+] =/ [p¡] to start with. Once these are set, if all the processes are symmetric in p+^ and p¡^ then [p+] =/ [p¡] will persist. Note however, that the universe is currently 10^26 m wide, and growing. There are 10^80 more protons than antiprotons in the observable universe today. So it would be a little strange to set this enormous asymmetry at the big bang. When the universe is only 10¡^35 m across, this would correspond to a shocking number density of 10^185 m^13. Moreover, the current cosmological model involves in ation, which produces exponential growth at early times, so whatever initial asymmetry we set would be completely washed away when in ation ends. In other words, it's possible, but would be very unsettling, to solve the baryogenesis problem by tuning the initial conditions. Another option is to start o symmetric but have processes that are not symmetric between particles and antiparticles. In turns out in the Standard Model of particle physics, there are none: for every way of producing an electron or proton, there is also a way of producing a positron or antiproton with exactly the same rate. In fact, this equality is guaranteed by symmetries (lepton number and baryon number). Moreover, if you made a modi cation so that the symmetries were violated, then eectively protons could turn into antiprotons. Thus, since protons and antiprotons have the same mass (and value of ") their chemical potentials would push them towards the same concentrations, which by Eq. (90) is zero. The story is again a little more complicated, since there is in ation, and reheating and the expansion of the universe is not quite quasi-static, and there is actually a super-tiny violation of the symmetry between protons and antiprotons within the Standard Model. Even when you include all these things, it doesn't work, you still get no matter out once the universe cools. So we are stuck. Why is there so much matter in the universe? Why is there more matter than antimatter? Nobody knows.
8 Partition function and the spectrum (optional)
Some of you may nd it illuminating to think about the partition function in a big-picture, more abstract sense (as if it's not abstract enough already!). The following discussion is just included because some students may nd it illuminating. It is not required reading for the course. The partition function is computed by summing over energies. As you probably know, the energies of a system contain a tremendous amount of information. Indeed, in classical mechanics, the energy at a point in phase space is given by the the Hamiltonian function H(~qi; p~i; t). If you know H, every possible behavior of the system can be determined by solving Hamilton's equations of motion. In quantum mechanics, the same is true: if you know the Hamiltonian operator H^
~q^i; p~^i
you can determine the time-evolution of the system completely through the Schrödinger equation. Eigenvalues of the Hamiltonian are the energies of the system. For example, consider the Hamiltonian for interactions among water molecules. We can approx- imate the Hamiltonian as depending on the distances Rij = j~q 1 ¡ ~q 2 j between the centers of mass of the molecules. We should nd that if two molecules are far away, the energy becomes independent of Rij. If we try to put the molecules on top of each other, it should be impossible, so the energy should blow up. Because of hydrogen bonding, we expect a weak attractive force at intermediate distances with a shallow potential minimum. That is, we expect something roughly like:
16 Section 8
Figure 4. (left) The potential energy between two water molecules is close to a Lennard-Jones potential. The corresponding density of states has singularities at the bound state energy and at zero energy the when the molecules are far apart.
This is called the Lennard-Jones potential. A pairwise potential model like this can explain many the properties of water surface tension, boiling, freezing, heat capacity, etc. For example, the
force from water molecule i is given by F~^ = ¡ (^) @q@~ i H(~qi; p~i). More sophisticated classical models,
including 3-body interactions and so on, can explain even more emergent behavior of water. If you know the quantum Hamiltonian H^^ exactly, you would be able to determine everything about water exactly (in principle). If you really want to determine the complete time evolution of a system, you need the full functional form of the classical Hamiltonian, or in a quantum system, the energy eigenvalues "i and eigenvectors (^) i(x) of the quantum Hamiltonian. However, you get a long way to understanding the behavior of a system just knowing the spectrum. The density of states for the Lennard Jones potential is shown on the right of Fig. 4. Its singularities indicate the bound state at E = ¡1. and the continuum at E = 0. The only thing you can't get from the density of states is the bound state distance r 0 , since there is no information about position in the density of states (of course, in this case, you could reconstruct the distance from the energy by dimensional analysis). So the spectrum itself has almost all of the information we care about. In quantum mechanics, the spectrum is the set of eigenvalues. Knowing only the spectrum, we don't have the information contained in the eigenvectors. Most of the time we don't actually care about the eigenvectors. One way to see this is that the eigenvectors are just projections from changing basis, (^) i(x) = h"ijxi. In an energy basis, the Hamiltonian is diagonal. Thus if we are interested in basis-independent properties of a system (as we almost always are), the spectrum is su cient. The point of the above discussion is to motivate why the spectrum of a system is extremely powerful, and contains almost all the physical information we would care to extract about a system. Now observe that the partition function carries the same information as the spectrum, just represented dierently. In fact, the partition function is just the Laplace transform of the spectrum. A Laplace transform is a way of constructing a function F ( ) from a function f(E) by integrating over E:
Z( ) = L[f (E)] =
Z
0
1 dE e¡^ Ef(E) (96)
So the partition function is the Laplace transform of the spectrum Z = L[E(qi; pi)]. Note that a Laplace transform is just a real version of a Fourier transform (take! i ). Thus working with the partion function instead of the spectrum is like working in Fourier space, with representing the frequency. The average over con gurations is like the average over time taken to produce a frequency spectrum. Equilibrium corresponds to a particular value of which is analogous to a particular frequency component dominating, like the resonance on a ute. The shape in Fourier space of the frequency spectrum around a resonance gives the timbre of a note, explaining why a C on a ute sounds dierent from an C on a trumpet. Thus, in a way, the partition function, through its derivatives at a xed , give the timbre of a physical system.
Partition function and the spectrum (optional) 17
