Free Energy in Introductory Physics Reading Reflection need help to Write a reading reflection on paper “Free Energy in Introductory Physics” attached belo

Free Energy in Introductory Physics Reading Reflection need help to Write a reading reflection on paper “Free Energy in Introductory Physics” attached below It should address one or more of the reading reflection prompts. Make sure it is focused and coherent (there is a a unifying theme to your reflection). The reading reflection should be around 250 words long.Reading reflection prompts: what are the author’s purpose in writing this article? What can you take from it that can be applied in your own classroom? How does the article impact your own physics content and pedagogical knowledge? Free Energy in Introductory Physics
Jeffrey J. Prentis and Michael J. Obsniuk
Citation: The Physics Teacher 54, 91 (2016); doi: 10.1119/1.4940172
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Published by the American Association of Physics Teachers
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Free Energy in Introductory Physics
Jeffrey J. Prentis, University of Michigan-Dearborn, Dearborn, MI
Michael J. Obsniuk, Michigan State University, East Lansing, MI
nergy and entropy are two of the most important concepts in science. For all natural processes where a system exchanges energy with its environment, the energy
of the system tends to decrease and the entropy of the system
tends to increase. Free energy is the special concept that specifies how to balance the opposing tendencies to minimize
energy and maximize entropy. There are many pedagogical
articles on energy1 and entropy.2 Here we present a simple
model to illustrate the concept of free energy and the principle
of minimum free energy.
Free energy is not a standard topic in introductory physics.
The recent guidelines established by the National Experiment
in Undergraduate Science Education (NEXUS) call for the
addition of the topic of free energy and its molecular meaning to the introductory physics curriculum for life-science
students.3 Most students encounter free energy for the first
time in a beginning biology or chemistry course, usually as a
computational tool to judge the feasibility of a chemical reaction from tables of standard data. A beginning physics course
is the natural place to provide insight into the mechanical underpinnings of energy, entropy, and free energy.
There are two kinds of free energy commonly used in thermodynamics.4 The Helmholtz free energy F of a system is constructed from the energy E and the entropy S of the system:
F ? E – TS.
The function F stores complete thermodynamic information
on a system in terms of the temperature T and volume V of the
system. The Gibbs free energy, G ? E – TS + pV, is ubiquitous
in chemistry, where the natural thermodynamic variables are
temperature T and pressure p. Here we focus on F, which we
will simply refer to as the “free energy.”
The principle of minimum free energy states that the equilibrium condition for any system in thermal contact with its
environment (energy reservoir) is
F = minimum.
Two short proofs of this principle will be given in the last section of this paper. The condition of “thermal contact” implies
that the system can exchange energy (not volume or matter)
with the environment in the form of heat (not work). Whenever a system in thermal contact with an energy reservoir of
constant temperature undergoes a thermodynamic process
(spontaneous change of state), the free energy of the system
will decrease until a minimum is reached, at which point
further natural change is impossible (improbable), i.e., final
equilibrium is reached.
Just like a ball sits at rest at the bottom of a hill (potentialDOI: 10.1119/1.4940172
energy well), a gas in a container is in a state of stable equilibrium at the bottom of a free energy well. A thermodynamic
system “falls downhill” to minimize its free energy. J. W.
Gibbs, who formulated the general theory of equilibrium in
thermodynamics, regarded the criterion for equilibrium in
Eq. (2) as an “extension of the criterion employed in ordinary
statics” and emphasized that “the transition from the systems
considered in ordinary mechanics to thermodynamic systems
is most naturally made by this formula”.5 Gibbs referred to the
F in Eq. (1) as the “force-function (potential energy) for constant temperature.”5
E = min (falling) vs S = max (spreading)
Entropy S has been interpreted as a “spreading function”
for its role in quantifying the tendency for energy to spread
through space.2 In a similar way, we can view energy E as
a “falling function” because of the tendency for the energy
of a non-isolated system to fall in value.6 The two opposing
“forces” that drive natural processes—the fall and spread of
energy—are automatically built into the free energy function.
The condition E – TS = min represents the exact compromise
of nature to balance the conflicting tendencies to minimize
the amount E of energy within the system and maximize the
spread S of energy throughout the system. In the E versus S
“competition,” the temperature T determines the “winner.”
Lowering T tilts the balance toward E = min while raising T
favors S = max.
To illustrate the interplay of E and S in determining the
stable equilibrium situation, consider the gas-piston system in
Fig. 1. The temperature is kept constant by placing the system
in thermal contact with an energy reservoir. In a state of equilibrium, how does the
height of the piston depend on the temperature
of the gas?
The condition that
this system (gas, piston,
Earth) is in equilibrium
can be viewed as dynamLow T
High T
ic (net force on piston is
zero) or thermodynamic
(free energy of system
is minimum). For T = 0, Fig. 1. (a) For low T, E–TS = min
is approximately E = min (falling
E – TS = min reduces to
dominates), which is satisfied if the
E = min, which means
piston height is low. (b) For high
the piston will be low to
T, E–TS = min is approximately –TS
= min or S = max (spreading domiminimize the gravitational potential energy of nates), which is satisfied if the gas
molecules have access to a large
the piston-Earth. Therspace in which to wander around
modynamically speaking, and explore many microstates.
The Physics Teacher ? Vol. 54, February 2016
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the piston tends to fall to increase the entropy of the reservoir.6 For example, a fall of E in the system from 5 J to 3 J is
accompanied by a spread of 2 J in the reservoir since the total
energy must be conserved. For T ? 0, the mechanical condition E = min is not the whole story. Molecular impacts (thermal agitation at finite T) push the piston up. In statistical
mechanics, this molecular force is related to the probability
to acquire volume. Gaining volume increases the number of
spatial microstates accessible to the molecules of the gas. In
short, the piston tends to fall (decrease E) and the gas tends
to spread (increase S). When the free-energy function E –TS
reaches its lowest value, the mechanical force of gravity is in
perfect balance with the “thermodynamic force” of entropy.
The bottom of the free-energy well uniquely determines the
equilibrium height of the piston.
Gas-piston model
We introduce a simple model that contains the essential
energetic and entropic elements of a real gas-piston system.
Consider a ball (molecule) bouncing up and down at high
speed between the ends of a vertical cylinder as shown
in Fig. 2. The top end is a movable weight (piston). The ball
Fig. 2. A “gas” consists of a ball of
mass m moving up and down at high
speed v in a cylinder fitted with a
movable piston of mass M. The rapid
series of impacts of the light ball
against the heavy piston hold the piston up at a height H. Adding energy by
heating causes the ball to move faster
and the piston to float higher.
translates up and down, but does not rotate. There is no
friction in the system. The ball and piston are rigid bodies
whose internal state does not change. The mass of the piston
is much greater than the mass of the ball. All collisions are
elastic. During each upward motion of the ball, the ball collides with the piston. The piston is free to fall but the collision
frequency is so high that the piston remains in a suspended
state of equilibrium.
A one-particle “gas” can be assigned a constant temperature T if the particle is allowed to exchange energy with its
surroundings of temperature T. For this purpose, we can
imagine placing the container in Fig. 2 in thermal contact
with an energy reservoir.7 This reservoir allows us to treat the
speed v of the ball as a parameter whose constant (root mean
square) value is simply related to the temperature of the reservoir.
The basic question is: How does the equilibrium height H
of the piston depend on the speed v of the ball? The principle
of minimum free energy provides a simple solution to the
general problem of finding H as a function of v, m, and M.
Calculating the free energy
The goal is to find how F depends on H and then minimize
the function F(H) with respect to H. The energy and entropy
functions of the system in Fig. (2) have simple forms:
E = MgH + E0
S = k ln H + S0 .
The additive terms E0 and S0 are independent of H. These
H-independent terms are not important in the free energy
method because when the derivative with respect to H is
taken to minimize F, these constant terms will vanish.
To understand the origin of Eqs. (3) and (4), imagine the
piston to be at the height H. Let the system consist of the ball,
the piston, and the Earth. The energy of the system is E =
MgH + mgH/2 + K, where MgH is the potential energy of the
piston-Earth, mgH/2 is the average potential energy of the
ball-Earth, and K is the average kinetic energy of the ball and
piston. For M>>m, the total energy reduces to E = MgH + K.
Since K only depends on speed (temperature), the H-dependent energy function has the form given by Eq. (3).
The statistical-mechanical meaning of entropy S for any
system is given by Boltzmann’s principle:2,8
S = k ln W.
In this universal formula, k is Boltzmann’s constant and
W is the multiplicity function that measures the number of
microstates with a certain energy that are accessible to the
system. To fix the notions of “microstate” and “multiplicity”
for the system in Fig. 2, we subdivide the space inside the cylinder into cells, each of length l, as shown in Fig. 3.
Fig. 3. The ball occupies locations in
the container (cells in state space)
like a checker occupies squares
on a checkerboard. Entropy is a
logarithmic measure of the number
of cells (H/l ) accessible to the ball.
A microstate of the ball is defined by specifying the position
(cell location) and the velocity of the ball. The number of
possible cells that the ball can occupy is H/l. For each of these
H/l position states, there are two velocity states—the ball can
move up or down in a cell.9 Thus the total number of microstates is W = H/l 3 2. The entropy in Eq. (5) becomes S = k ln
H + k ln (2/l), which has the form stated in Eq. (4).10
Given Eqs. (1) and (5), all free energy functions have the
form F = E – kT ln W. The factor kT is the natural unit of
The Physics Teacher ? Vol. 54, February 2016
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“thermal energy.” For a one-dimensional ideal gas of molecules at temperature T, the average translational kinetic energy of one molecule is kT/2.11 Therefore, the thermal parameter kT of the gas-piston model is associated with the kinetic
energy of the ball:
kT = mv2.
Given E, S, and T in Eqs. (3), (4), and (6), we can assemble
the free energy F ? E ? TS:
F = MgH – mv2 ln H + F0 .
The H-independent constant is F0 ? E0 – TS0.
The form of F in Eq. (7) reveals the competition between
lowering E and raising S. To make F in Eq. (7) as small as possible, we need to make E ~ H small or S ~ ln H large. A low
H will minimize the amount E of energy (piston falls down).
A high H will maximize the spread S of matter (ball explores
larger space). For large v, the entropy term (mv2 ln H) in Eq.
(7) “wins” and H will be large, i.e., a fast ball implies a high
piston. For small v, the energy term (MgH) in Eq. (7) dominates F, i.e., a slow ball implies a low piston. Since T ~ v2, this
energy-entropy analysis explains why the piston is low (high)
if the temperature is low (high).
Fig. 4. The free energy well for the gas-piston model. The
equilibrium height He of the piston is the special value of H
determined by the condition F(He) = min.
Minimizing the free energy
The free energy function F(H) in Eq. (7) is sketched in Fig.
4. This function exhibits a minimum at H = He. To find this
equilibrium value of H, we take the derivative of F(H) in Eq.
(7) and set the result equal to zero:
Solving Eq. (8) for He gives
Note that He is independent of the “length quantum” l, which
was introduced solely to digitize space and count states. In
terms of the temperature T of the surroundings, the equilibrium height is He = kT/Mg.
Suppose the mass ratio is M/m = 1000, i.e., a Ping-Pong
ball “molecule” (5 g) colliding with a bowling-ball “piston”
(5 kg). For this mass ratio, Eq. (9) gives He = v2/1000g. If
v = 50 m/s, then He = 0.25 m. If energy is added until the ball
reaches a new speed v = 100 m/s, then the piston will rise to
He = 1.0 m. If v = 50 m/s and we assume this speed is constant,
then the ball hits the piston at the rate of v/2He = 100 collisions per second. During the time 0.01 s between collisions,
the piston falls from rest a distance of 0.00049 m, which is
0.2% of the equilibrium height. For v = 100 m/s, the value
of temperature in energy units, kT = mv2 = 50 J, provides a
meaningful measure of the average kinetic energy of the ball
or the vibration energy of the wall.7 Note that the corresponding value of temperature in kelvin units, T = 3.631024 K, has
no practical meaning as a thermometer reading when dealing
with a Ping-Pong ball “gas.”
Force and motion interpretation of F = min
A pedagogical feature of the gas-piston model is that it can
also be analyzed using force and motion without mention of
temperature or entropy. This force-motion analysis complements the energy-entropy analysis. The downward force of
gravity on the piston is Mg. The average upward force on the
piston due to the hail of impacts (elastic collisions) of the ball
against the piston is mv2/H. This force formula, familiar from
the kinetic theory of gases,12 derives from the product of two
terms: the momentum gained (2mv) by the piston per elastic
collision times the number of collisions (v/2H) per second:
2mv 3 v/2H = mv2/H. The balance-of-force condition, Mg =
mv2/H, is identical to the minimum-of-free-energy condition
in Eq. (8). For this gas-piston model, the mechanical (Newtonian) principle “net force = zero” is equivalent to the thermal
(Gibbsian) principle “free energy = minimum.”
We can gain additional dynamical insight into the nature
of thermodynamic equilibrium by analyzing the rise-and-fall
motion of the piston after colliding with the ball. If the collisions are timed “just right,” then the piston will “hover” at a
fixed height, simply undergoing small oscillations. To hover
at H, the piston must complete one rise-and-fall motion (after being hit by the ball at H) and return to H at exactly the
same time that the ball arrives back at H after completing
one down-and-up motion in the cylinder. Thus, the hovering condition is Vt – ½ gt2 = 0 (net displacement of piston is
zero), where t = 2H/v is the round-trip time of the ball, and
V = mv/M (momentum conservation) is the speed of the
piston just after colliding with the ball. The solution to the
hovering condition, mv/M – ½ g(2H/v) = 0, is H = mv2/Mg,
which is identical to the value of H that solves F = min.13 In
other words, thermodynamic equilibrium for this special case
is established if the ball-piston dynamics satisfy a “resonance
condition”: piston period (2V/g) = ball period (2H/v).
The Physics Teacher ? Vol. 54, February 2016
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Ideal gas of N molecules
We can generalize the free-energy analysis of the one-ball
gas to an analysis of a gas of N balls at temperature T in a
three-dimensional container with a piston. The H-dependent
energy function, MgH, in Eq. (3) is unchanged, while the Hdependent entropy function, k ln H, in Eq. (4) generalizes to
Nk ln H.14 The free energy principle, E – TS = min, is
Using the ideal gas law, pV = NkT, where p is the pressure of
the gas on the piston of area A and V = AH is the volume of
the gas, Eq. (10) can be written as
Mg ? pA = 0.
Equation (11) says that the downward force of gravity on
the piston is equal to the upward force of the molecules on
the piston. Once again, we see how balancing opposite forces
(net force = 0) is equivalent to balancing opposing tendencies
(E –TS = min).
Two proofs of F = min
• Thermodynamic proof: For any system (sys) in thermal
contact with its environment (env), the second law of thermodynamics states
DSsys + DSenv ? 0.
The conservation of energy states DEenv + DEsys = 0. The
entropy change of the environment (reservoir) is equal to the
energy it absorbs from the system divided by the temperature
of the environment: DSenv = ?DEsys/T. The second law in Eq.
(12) can therefore be written in the form
DEsys – TD Ssys ? 0,
or simply DFsys ? 0.
The remarkable thing about Eq. (13) is that it expresses
the total entropy change of the universe (sys + env) in terms
of a property of the system alone: T(DSsys+DSenv) = ? DFsys.
The increase in the entropy of the universe is equivalent to a
decrease in the free energy of the system. The following equilibrium conditions are therefore equivalent:
Fsys = min ó Ssys + Senv = max.
P(H) = a exp(–MgH/kT) 3 H.
Note that P(H) is a product of two factors: [Boltzmann factor] 3 [multiplicity factor]. The constant a is independent of
H. The Boltzmann factor exp(?E/kT) for the energy level E =
MgH determines the probability to find the piston at H when
the ball is in one particular cell. Multiplying this single-cell
probability by the total number of possible cells, W ~ H, gives
the total probability P(H) for level H. Equation (15) can be
written in the form
P(H) = a exp[–F(H)/kT],
• Statistical mechanical proof. Consider the ball-piston
model. In statistical mechanics,4 the probability P(H) to find
the piston at height H is
where F(H) ? MgH ? kT ln H. The free energy function
F(H), which first appeared in Eq. (7), now emerges as the “energy exponent” in a Boltzmann factor.
The concept of free energy is usually applied to a macroscopic system. The statistical analysis presented here shows
how F = min applies to a one-particle system in thermal
contact with a reservoir. According to Eq. (16), the condition F(H) = min is equivalent to P…
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