Notes for Class Meeting 8: Reversibility
We saw in the previous set of readings that the direction of time is determined by the
increase of entropy – the tendency of things to go from less likely to more likely states,
or, equivalently, to go from more ordered states to less order states. In this set of
readings we go into this more deeply to understand why in non-closed systems, entropy
can decrease, as happens when a cup of coffee is left on the table to cool, an example
VonKorff discusses in detail.
An Example of Friction
Let’s start with a toy model of a pendulum and see how it loses
energy through friction with the air.1 As shown to the right, a
pendulum is in a box with three molecules of air. The pendulum is
frictionless, except for its interactions with the air. Assume that the
pendulum starts with 145 units of energy and the air molecules each
start with 5 units of energy, 160 units of energy in all. How does the
system evolve in time?
I have prepared a model of how the pendulum could lose energy to
the molecules. It is not physically accurate, but it is adequate to illustrate the point. I
assume that two colliding bodies, A and B, exchange one unit of energy, and that the
probability of body A decreasing its energy is EA / (EA  EB ) . In other words, if an
energetic body collides with a less energetic body, the more energetic body is more likely
to decrease in energy than the less energetic body. That seems reasonable. All four
objects are treated in an equal way, that is, the pendulum is treated as just a big molecule.
The result of one simulation is shown below. The pendulum steadily loses energy until it
has only its “fair share” of the energy, 40 units. It then stays around 40 units, but its
energy will fluctuate due to random exchanges of energy with the air molecules. It will
be bumped around like the styrofoam balls in the demonstration of Brownian motion.
Energy of Pendulum
160
140
Energy
120
100
80
60
40
20
0
1
10
19
28
37
46
55
64
73
82
91
100 109 118 127 136 145 154 163 172 181 190 199
Time
1
This example appears in Alan Lightman’s book Great Ideas in Physics in a slightly modified form.
In thermodynamic terms, the pendulum loses kinetic energy and the air molecules
gain kinetic energy, which is to say that the temperature of the air rises.
To understand these processes in term of entropy, VonKorff introduces a formula for
how many ways n indistinguishable units (usually of energy) can be assigned to m
objects,
 n  m  1  n  m  1 (n  m  1)!
(8.1)

   m  1   n!(m  1)! .
n
There is a simple derivation of this formula, which involves a cute trick. It is given in
Appendix A.
We can use this formula to see how the entropy of the system changes. We can
calculate this entropy by combining the entropy of the pendulum with the entropy of the
air. For all energies, the entropy of the pendulum is zero since we are treating it as one
molecule, and there is only one way to put n units of energy on one object. For the air,
we use the formula in Eq. 8.1 with n = sum of the energy in the air molecules and m = 3.
The result is shown below. The entropy slowly rises from about 5 to about 9 and then
just fluctuates by small amounts.
Entropy of the System
10
Entropy
8
6
4
2
0
1
11
21
31
41
51
61
71
81
91
101
111 121 131
141 151 161
171 181 191 201
Time
Thus, the loss of energy through friction is a process of going from a more ordered
state to a less ordered state, an improbably state to a more probable state. Entropy
increases and the process is asymmetric in time.
Heat Transfer
VonKorff discuss the case of a hot cup of coffee cooling off in detail. In his toy
model there are two molecules of coffee and four molecules of air. If the coffee starts
with 7 units of energy and ends up with 3 units of energy, then using Eq. 8.1, the coffee
will start with an entropy of 2.08 and end with an entropy of 1.39. Its entropy will
spontaneously decrease. However, it is not a closed system. The energy it loses is
transferred to the air molecules. If they start with a total of 2 units of energy, they will
end up with 6 units of energy. The entropy of the air will start at 2.30 and end at 4.43.
Thus, the total energy of the complete system will increase from 2.08 + 2.30 = 4.38 to
1.39 + 4.43 = 5.82. At the end the temperature of the coffee will equal the temperature of
the air, 2 units of energy per molecule.
This is a general phenomenon. When heat flows from hot to cold bodies, the total
number of states, and thus the total entropy of the system increases. It turns out that the
spontaneous flow of heat from a warm body to a cold body is an equivalent statement of
the second law of thermodynamics.
A corollary is that it is impossible to extract mechanical energy from the heat of a
system at a single temperature. The proof is that if it were possible, then the mechanical
energy could be used to heat a hotter body, thus causing heat to flow from a colder
system to a hotter system. It turns out that this corollary is also an equivalent statement
of the second law of thermodynamics.
Perpetual Motion Machines
There are two types of perpetual motion machines. Perpetual motion machines of the
first type create energy, violating energy conservation, or the first law of
thermodynamics. Perpetual motion machines of the second type violate the second law of
thermodynamics, such as extracting mechanical energy from a single temperature. You
might find it interesting to note that the two laws of thermodynamics are so well tested
and understood that the U. S. Patent Office automatically rejects applications for either
kind of perpetual motion machine without further analysis.
Types of Energy Increases
Four common ways of increasing entropy in simple systems
are illustrated to the right. In each case it is clear that there are
more states after the change, and thus increased entropy.
(a) Add particles.
(b) Add energy.
(c) Add volume.
(d) Mix two types of particles.
The Ideal Gas Law
(a)
(b)
(c)
(d)
Mechanical energy can be extracted from thermal energy if
there are two different temperatures in the system, increasing the entropy of the system.
The system can also be run in reverse, inputting energy and decreasing entropy. To see
how this works, it is convenient to consider an ideal gas, which was discussed in detail by
VonKorff. The formula for ideal gasses is
PV  kNT ,
(8.2)
where
P = pressure = force per unit area,
V = volume,
N = number of molecules,
T = temperature in the kelvin or absolute scale = degrees Celsius + 273.15,2
k = Boltzmann’s constant = 1.38  10-23 m2 kg s-2 K-1.
I will supplement just a few of VonKorff’s justifications for this law. Pressure is
caused by molecules hitting the wall of the vessel, scattering elastically, and transferring
2 mv of momentum to the wall. By Newton’s 2nd law, this creates a force on the wall.
It is easy to see where the ideal gas law comes from by holding two of the variables
fixed and seeing why a change in one causes the corresponding change in the other.
Examples:
(1) Hold V and T constant. If N doubles, then P doubles because twice as many
molecules hit the wall.
(2) Hold V and N constant. If T doubles, then 12 mv 2 doubles, and v increases by 2 .
This causes P to double because the momentum transfer increases by 2 and it happens
2 times as often because of the increase in the molecules’ velocity.
(3) Hold T and N constant. If V halves, then P doubles because the collisions happen
twice as fast. This is independent of the vessel shape, but is easiest to visualize if you
consider a vessel shaped like a thin disk.
These three observations are sufficient to establish the ideal gas law.
Reversibility
A closed system is reversible if it can come back to its original state with no change.
Since there is no change in any of the variables, then there cannot be any change in the
entropy either.3 In practice, no process is completely reversible, but systems can
approach it arbitrarily closely. An irreversible process is one that cannot return to its
initial state, or technically, one that is extraordinarily unlikely to. We will discover later
in the course that irreversible processes will play an important role in quantum
mechanical measurements.
We can use a cylinder of gas with a piston to illustrate both irreversible and reversible
processes. Consider the diagram shown on the next page. Suppose we pull back the
piston very fast, faster than the molecules can move. Then we have just effectively
increased the volume, but left the temperature the same, since no work has been done on
the molecules. To return to the initial position, we need to use mechanical energy to
overcome the pressure created by the gas. The piston will do work on the molecules,
2
In 1967 the 13th General Conference on Weights and Measures (CGPM) declared that temperatures on
this scale are called kelvins, not degrees kelvin, kelvin is not capitalized, and the symbol (capital K) stands
alone with no degree symbol. (I have been trying to think of a social situation in which you could use this
tantalizing factoid, but I have not come up with one yet.)
3
We are normally only interested the thermodynamic variables from which we calculate the entropy.
Thus, if we want to know whether a system is reversible, it would be normally sufficient for it to have the
same volume, temperature, pressure, etc. at the beginning and end, but we would not require that every
molecule be in the same place.
which will increase their temperature. Thus, the final state will have a higher temperature
than the initial state, and thus more entropy. This is clearly an irreversible process.
Consider the process again, but this time we allow the piston to be pushed back by the
pressure of the gas, moving very slowly. Mechanical energy is generated since the
molecules are doing work on the piston, and the molecules cool off, since they are losing
energy to the piston. In the second step, the mechanical energy (assuming that we were
able to store it in some way) can be reapplied to push the molecules back to their original
volume very slowly. The piston is now doing work on the molecules, heating them back
up to their original temperature. Thus the system has returned to its original state and
thus the process has been a reversible one.
The Entropy of the Earth
In a reversible process, the increase in entropy, S , due to a quantity of heat, Q,
being added to a system at a temperature T (on the kelvin scale) is given by4
Q
S  .
(8.3)
T
If there is an additional entropy increase due to irreversible process, it just adds additively
to right side of Eq. 8.3. We can use this equation to understand the entropy of the earth.
Leaving global warming aside, the earth is in a steady state. Its energy does not
change as evidenced by its temperature not steadily increasing or decreasing over time.
There is no obvious evidence of its entropy changing markedly over time either. Things
are not noticeably more or less orderly this year than they were last year. How is this
possible, since irreversible processes, which always increase entropy, are occurring
constantly?
The answer is found in the energy balance of the earth. We are receiving energy from
radiation from the sun. This is from a high temperature, so Eq. 8.3 tells us that that it is
adding a low amount of entropy to the earth. For the earth not to increase its temperature
over time, it needs to radiate an equal amount of radiation back into space. But the earth
is much cooler than the sun. Therefore, Eq. 8.3 tells us that the earth is radiating away a
large amount of entropy. Thus, all of the entropy we are producing is being radiated
away into space!
Of course, this situation cannot go on forever, since someday the sun will run out of
fuel. However, we do not need to worry about that now, since that day is still five billion
years in the future.
4
The derivation of this formula is beyond the scope of this course.
Extracting Mechanical Energy from Thermal Energy and Refrigeration
We can use Eq. 8.3 to understand the limits imposed by thermodynamics on how
much thermal energy can be converted to mechanical energy and how little energy is
need for refrigeration, moving thermal energy from a cold source to a warmer source.
We have already shown that one cannot extract mechanical energy from a source at a
single temperature. Therefore, we must have two heat reservoirs at different
temperatures, dumping some of the energy from the hot source into the cold sink, with
the remaining energy from the hot source going into mechanical energy. The situation is
shown in the diagram below on the left.
TH
TH
Heat
Heat
Mechanical
Energy
Mechanical
Energy
TC
TC
The highest efficiency allowed by thermodynamics for converting thermal energy to
mechanical energy would be for a reversible process. Thus, using Eq. 8.3 will give us an
upper limit on the efficiency, , which is defined as the fraction of the heat from the hot
source, QH , that is converted into mechanical energy, E,
E TH  TC


.
(8.4)
QH
TH
The derivation of this equation can be found in Appendix B. The temperatures in Eq. 8.4
are in the kelvin scale. So, for example, the maximum efficiency for getting energy from
boiling water (100 oC) to room temperature water (20 oC) is 80/373.15 = 21.4%.
Refrigeration is exactly the same process, but with the direction of energy flow
reversed, as shown by the diagram above on the right. We want to extract thermal energy
from the cold source and dump it into the hot sink. To do this we have to input
mechanical energy into the system. We can define the coefficient of performance, C, as
the ratio of extracted heat from the cold source, QC , to the mechanical energy, E, needed
to be input,
QC
TC

.
(8.5)
E TH  TC
The derivation of this formula is also found in Appendix B. As an example, C for your
refrigerator is 277.15/16 = 17.3, assuming that the refrigerator temperature is 4 oC and
that room temperature is 20 oC. Thus, it does not really take very much energy to cool
down a beer.5
C
Other Topics
There are several other interesting topics concerning the second law of
thermodynamics. If we have time, we can discuss some of them in our class meeting.
One topic you should think about is key to our investigation of time: We have seen that
the direction of time is completely correlated with the increase in entropy. However,
correlation does not imply causation. So, does the increase in entropy determine the
direction of time, or is the direction of time independent of the increase in entropy, or is
this a meaningless question?
Appendix A: Derivation of Eq. 8.1
Let a string of o’s and lines such as
oo|ooo||ooooo
(8.6)
represent how n indistinguishable units (of energy, perhaps) might be apportioned among
m objects. The o’s represent the indistinguishable units and the lines represent the
boundaries of the m objects. For example, Eq. 8.2 represents 2, 3, 0, and 5 units being
apportioned to the 1st, 2nd, 3rd, and 4th objects, respectively. Note that there is no need to
put lines at the ends of the string because we can always assume that the start and end of
the string are boundaries of the starting and ending object. Thus, we only need (m –1)
lines to delineate m objects.
Thus, the permutations of the n o’s and the (m - 1) lines represent all of the possible
ways that n units can be distributed among m objects. There are clearly (n + m – 1)!
permutations. However, the n units are indistinguishable, so we do not care what order
they are in, so we can divide by the n! possible orders. Further, the lines are just markers
of the boundaries of the objects, so we do not care what order they are in, so we can also
divide by their (m – 1)! possible orders. (The objects are distinguishable, but they are
distinguished by the order in which they come in the string.) This gives us the expression
in Eq. 8.1:
 n  m  1  n  m  1 (n  m  1)!

   m  1   n!(m  1)!
n
5
We have assumed reversible process to get an upper limit on the coefficient of performance. Real
refrigerators will have irreversible aspects (friction, motion at non-zero speeds, etc.) and thus will probably
have only about half as large a coefficient of performance.
Appendix B: Derivation of Eqs. 8.4 and 8.5
We are assuming reversible processes, so, in both cases, the total entropy must be
zero, thus using Eq. 8.3,
Q
Q
 H  C  0.
(8.7)
TH TC
Conservation of energy requires that
QH  QC  E .
(8.8)
To obtain Eq. 8.4, solve Eq. 8.8 for QC and substitute in Eq. 8.7. To obtain Eq. 8.5,
solve Eq. 8.8 for QH and substitute in Eq. 8.7.