194 Chapter 6. Entropy, temperature, and free energy[[Student version, January 17, 2003]]
Two-state systems Here’s an immediate example. Suppose the small system has onlytwo
allowed states, and that their energies differ by an amount ∆E=E 2 −E 1 .The probabilities to be
in these states must obey bothP 1 +P 2 =1and
P 1
P 2=
e−E^1 /kBT
e−(E^1 +∆E)/kBT
=e∆E/kBT, simple 2-state system. (6.24)Solving, we find
P 1 =^1
1+e−∆E/kBT,P 2 =^1
1+e∆E/kBT. (6.25)
That makes sense: When the upper state #2 is far away in energy, so that ∆Eis large, the system
is hardly ever excited: It’s almost always in the lower state #1. How far is “far?” It depends on
the temperature. At high enough temperature, ∆Ewill be negligible, andP 1 ≈P 2 ≈ 1 /2. As the
temperature falls below ∆E/kB,however, the distribution (Equation 6.25) quickly changes to favor
state #1.
Here is a more involved example:
Your Turn 6f
Suppose “a” is a small ball tied elastically to some point, and free to move in one dimension
only. The ball’s microstate is described by its positionxand velocityv. Its total energy is
Ea(x, v)=^12 (mv^2 +kx^2 ), wherekis a spring constant.
a. From the Boltzmann distribution, find the average energy〈Ea〉as a function of temperature.
[Hint: Use Equation 3.7 on page 68 and Equation 3.14 on page 70.]
b. Now try it for a ball free to move in three dimensions.
Your result has a name: theequipartition of energy.The name reminds us that energy is being
equally partitioned between all the places it could go. Both kinetic and potential forms of energy
participate.
T 2 Section 6.6.1′on page 208 makes some connections to quantum mechanics.
6.6.2 Kinetic interpretation of the Boltzmann distribution
Imagine yourself having just stampeded a herd of buffalo to a cliff. There they are, grunting,
jostling, crowded. But there’s a small rise before the cliff (Figure 6.8). This barrier corrals the
buffalo, though from time to time one falls over the cliff.
If the cliff is ten meters high, then certainly no buffalo will ever make the trip in the reverse
direction. If it’s only half a meter high, then they’ll occasionally hop back up, though not as easily
as they hop down. In this case we’ll eventually find an equilibrium distribution of buffalo, some up
but most down. Let’s make this precise.
Let ∆E 1 → 2 bethe change in gravitational potential energy between the two stable states;
abbreviate it as ∆E. The key observation is that equilibrium is not a static state, but rather a
situation where the backward flow equals, and socancels,the forward flow.^4 Tocalculate the flow
rates, notice that there’s an activation barrierE‡for falling over the cliff. This is the gravitational
potential energy change needed to pick a buffalo up over the barrier. The corresponding energy
E‡+∆Ein the reverse direction reflects the total height of both the small rise and the cliff itself.
As for buffalo, so for molecules. Here we have the advantage of knowing how rates depend on
barriers, from Section 3.2.4 on page 79. We imagine a population of molecules, each of which can
(^4) Compare our discussion of the Nernst relation (Section 4.6.3 on page 124) or sedimentation equilibrium (Sec-
tion 5.1 on page 142).