# Quiz 1 for Physics 176 Solutions Professor Greenside Problems

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Quiz 1 for Physics 176: Solutions
Professor Greenside

Problems That Require Writing

1. A laser consists of a long cylindrical tube of some medium that has mirrors at both ends such that light
bounces back and forth between the mirrors while its intensity is ampliﬁed. (Light not parallel to the
axis of the cylinder is lost rapidly since the outer surface of the cylinder is often transparent.) Assume
that the laser light is monochromatic with frequency f , which means that the laser cavity is ﬁlled with
photons (quantum light particles) that each have momentum p = (h/c)f where h is Planck’s constant
and c is the speed of light. The total energy of the light in the laser is given by E = N hf where N is
the number of photons in the cavity.

(a) (6 points) Use a simple kinetic theory to derive an expression for the pressure P that the light
exerts on a mirror in terms of the photon number N , the volume V of the cylinder, and the light
frequency f .

Answer: This problem is identical to the one discussed in lecture and that you were thinking
expression for the total momentum transfer over a time ∆t and over an area A
N  1
∆ptotal = A(v∆t) ×     × × 2mv,                                  (1)
V  6
and change the height of the prism v∆t → c∆t, change the momentum transfer per particle 2mv →
2[(h/c)f ], and replace the factor 1/6 with 1/2 to get the answer

F   (∆ptotal /∆t)    1             N  1   h                       N
P =     =               =     × A(c∆t) ×   × ×2   f                 =     hf.        (2)
A        A          A∆t            V  2   c                       V

You had to be observant to make the substitution 1/6 → 1/2: unlike the atoms in a gas, the
photons only move along two possible directions with respect to the axis of the laser, so only half
the particles in the laser are moving in a given direction at any given time.
In a quiz, you don’t have time for lengthy descriptions but I do ask that you justify what you are
doing with short phrases or diagrams. Here you needed to say something to the eﬀect that the
pressure on the mirror could be determined by calculating the total momentum change ∆ptotal
over some short arbitrary time ∆t over the area A of a mirror; you needed to say enough that it
was clear that you knew what the symbols meant and what the strategy was for the calculation.
Alternatively, you could have drawn a cylindrical laser with a small prism over one mirror and
sketch in the appropriate details, with short phrases.
In a bit more detail: the total momentum transfer could be calculated by counting the number
of photons that strike the mirror within a short time ∆t and by using the fact that each photon
transfers an amount of momentum 2[(h/c)f ] if the photon reﬂects perfectly from the mirror. (You
lost a point if you didn’t realize that the factor of 2 in the momentum change 2[(h/c)f ] came from
the reversal of direction of the photon by reﬂection.) The number of photons that strike within
time ∆t are all photons that are closer than the distance c∆t from the mirror and therefore lie in
prism of volume A(c∆t). But only half the photons in this prism are moving in the right direction
(toward a given mirror) at any given time because of the wording of the problem. (All photons
moving perpendicular to the laser axis are lost immediately through the transparent surface of
the lasing medium.) These arguments lead to the above algebra and the answer Eq. (2).

(b) (2 points) Discuss brieﬂy how your expression for P is or is not similar to the expression N kT /V
for the pressure of an ideal gas at temperature T .

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Answer: Comparing the expression N hf /V for the photon gas with the expression N kT /V
for a molecular gas, we see that, perhaps to your surprise, that a photon gas has the same
behavior as an ideal molecular gas with respect to the variables P , V , and N . In both
cases, we have P V = N × E where E has units of energy and is kT for the ideal molecular gas
and hf for the photon gas. Thus a photon gas obeys some of the same classic behaviors of an
ideal molecular gas: P ∝ V −1 for ﬁxed N and ﬁxed f , P ∝ N for ﬁxed V and f , V ∝ N for
ﬁxed P and f .
This is not a coincidence but gets to the heart of why a macroscopic description of a substance can
depend weakly on details of what the substance is made of. The ideal gas law is fundamentally a
consequence of momentum conservation and you get the same expression P V = N × E for pretty
much anything such that momentum is conserved as objects collide with a wall and the objects
interact more strongly with the walls than with each other (non-interacting particles).
By the way, you might have been tempted to equate (N/V )kT to (N/V )hf and conclude that

h
T =     f,                                       (3)
k
and so discover what the temperature T of a photon gas must be. (This same procedure is how we
discovered how to interpret temperature in terms of the average kinetic energy of the molecules,
(3/2)kt = (1/2)mv 2 .)
But just making a deﬁnition doesn’t meant that Eq. (3) acts like a thermodynamic temperature,
with the ability to indicate the ﬂow of energy between a high-temperature system to a low-
temperature system, and such that thermodynamic equilibrium corresponds to T being constant
throughout the system. For example, this deﬁnition of a T for a photon gas fails when we try
to predict the ﬂow of energy between two lasers that are joined together, say a low-frequency
“cold” red-laser cavity with a higher-frequency “hot” blue-laser cavity. One does not obtain any
equilibration of energy, the red photons continues to bounce back and forth independently of the
blue photons and an intermediate frequency or temperature is not attained.
We will learn later this semester, when we get to Section 7.4 in Schroeder, what it means for
a photon gas to be in thermodynamic equilibrium and how that equilibrium state depends on
temperature T . Suﬃce it say, a monochromatic gas of photons is strongly non-equilibrium so the
concept of temperature is not meaningful to apply.

2. (4 points) A bacterium like E. coli is about d = 2 µm in length (1 µm = one micron = 10−6 m) and
swims with an approximately steady speed v. A bacterium will starve if it depletes resources in the
surrounding water faster than diﬀusion can supply the resources. By comparing the time scale for a
bacterium to swim one body length with the time scale for diﬀusion to supply nutrients over the size
of its body, and given that nearly all small molecules in water at room temperature have a diﬀusion
constant of D ≈ 10−9 m2 /s, deduce an approximate minimum speed v that a bacterium must swim to
“beat” diﬀusion and so survive. Express your answer v in units of microns per second (i.e., approximate
body lengths per second).
Note: by comparing your answer with experimental measurements of v, you can learn whether diﬀusion
plays a role in determining how fast a bacterium has to swim.

Answer:       This problem tested whether you knew that the relaxation time for a diﬀusive process
scales with the size of a system squared. Brieﬂy, you needed to write down

d  d2
≈ ,                                                (4)
v  D
and solve for v to get:

D    10−9 m2 /s            m       µm
v≈     ≈        −6 m
≈ 5 × 10−4   = 500    ≈ 250 body lengths per second.                 (5)
d   2 × 10                 s        s

2
This is really cruising for a little bacterium. Measurements of actual bacterial speeds show that they
move at most about a tenth of this speed, which means something interesting is going on, a bacterium
does not have to beat diﬀusion gradients to survive.
Just writing down the above algebra would not have led to a perfect answer, you had to add some
phrases saying what the symbols mean and why you set up the equality Eq. (4). For example, you
needed to say that the time for a bacterium to travel one of its body lengths was d/v and that the
relaxation time for diﬀusion of some substance over the size of a bacterium had to be d2 /D.

3. (3 points) Give three distinct criteria for a macroscopic system to be in thermodynamic equilibrium.

Answer:     Any three of the following ﬁve criteria would suﬃce:
(a) The system had to be time-independent.
(b) The temperature T had to be the same everywhere inside the system.
(c) The pressure P had to be the same everywhere inside the system unless there was an external
gravitational or electric ﬁeld.
(d) The properties of the system had to be independent of its history (how the system was put
together).
(e) There could be no relative macroscopic motion: an equilibrium system has to translate or rotate
rigidly.
Later in the course, we will add another criterion that replaces the one on pressure, that the chemical
potential µ(T, P ) of a system must be constant throughout the interior of a system.

4. (4 points) Given that
1       1
ex ≈ 1 + x + x2 + x3 ,      for |x|       1,                            (6)
2       6
ﬁnd the Taylor series expansion of the hyperbolic tangent
ex − e−x
tanh(x) =            ,                                      (7)
ex + e−x
up to third-order order in x, i.e., ignore all powers higher than x3 , which provides a 4th-order accurate
approximation.

Answer: Over half the class had trouble with this problem, from which I learned a hard lesson: that
a single homework problem on a new topic is not suﬃcient time and practice for you to gain mastery.
The goal of this problem was to determine a Taylor series approximation with up to third-order terms
for the hyperbolic tanh function (pronounced in English as “tanch”), i.e., to ﬁnd a third-order polyno-
mial c0 + c1 x + c2 x2 + c3 x3 that best approximates this function in the vicinity of the location x = 0.
From the ﬁrst homework assignment, the wrong way to do this would have been to start diﬀerentiat-
ing tanh(x) up to third order, the algebra is excessive and unnecessary.
Instead, the idea was for you to use the given Taylor series of the exponential function to generate a
ratio of polynomials, which you could then simplify using the identity 1/(1 − x) ≈ 1 + x to convert a
reciprocal to a product.
The steps go like this. First, the numerator of Eq. (7) can be written as:
1    1                  1    1                  1
ex − e−x =    1 + x + x2 + x3       − 1 − x + x2 − x3         = 2 x + x3       + O x5 .       (8)
2    6                  2    6                  6
while the denominator can be written as
1    1                  1    1                  1
ex + e−x =    1 + x + x2 + x3       + 1 − x + x2 − x3         = 2 1 + x2       + O x4 .       (9)
2    6                  2    6                  2

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The “big-oh” notation indicates the power of the ﬁrst term being ignored in the Taylor series and is a
convenient way to indicate the magnitude of terms that are being ignored. Substituting into Eq. (7)
gives

ex − e−x
tanh(x) =                                                       (10)
ex + e−x
x + 1 x3
6
≈                                                    (11)
1 + 1 x2
2
1            1
≈   x + x3    × 1 − x2                               (12)
6            2
1      1 3
= x − x3 +     x + h.o.t.s                           (13)
2      6
1
= x − x3 .                                           (14)
3
This last line is the answer. Here is a line-by-line justiﬁcation of the above algebra:
(a) Eq. (11) was obtained by substituting the Taylor series approximations up to third-order terms
from Eq. (8) and Eq. (9) into Eq. (10).
(b) Eq. (12) was obtained by using the two lowest-order terms of the geometric series 1/(1 − x) ≈
1 + x + · · · with x replaced by −(1/2)x2 to replace division by the polynomial 1 + (1/2)x2 with
multiplication by the polynomial 1 − (1/2)x2 :
1          1 2      4
1 2 ≈ 1 − 2x + O x   .                               (15)
1 + 2x

I ignored all higher-order terms in the geometric series because the next term in the geometric
series involved squaring (1/2)x2 to give a 4th-order power of x, and that can be thrown out since
we are only retaining terms to 3rd-order.
(c) Eq. (13) was obtained by multiplying out the two polynomials in Eq. (12), and by immediately
setting to zero any power of x higher than 3. The abbreviation “h.o.t.s” means “higher order
terms” and is a common way to refer to inﬁnitely many terms that can be ignored because they
are suﬃciently small in the limit x → 0.

The following Mathematica plot shows that the third-order Taylor series Eq. (14) in purple accurately
approximates the tanh function in blue (relative error smaller than 10%) over the range [-1,1] and
then deviates substantially, if anything because tanh(x) is bounded between -1 and 1 while the cubic
polynomial is unbounded.

2

1

4           2                        2           4

1

2

We will use the answer Eq. (14) later in the course, it shows up when calculating properties of a
paramagnet and some polymers.

4
True or False Questions (2 points each)
For each of the following statements, please circle T or F to indicate respectively whether a given statement
is true or false. No justiﬁcation is needed, but small drawings or comments in the margins can be taken into
account.

1.    T/ F       There can be macroscopic motion at absolute zero, T = 0,

Answer: T As discussed in lecture, and as you worked through in one of the homework problems, a
gas of He4 atoms at temperatures below about 2 K and for pressures below about 25 atmospheres will
condense into a quantum superﬂuid state that has the property of ﬂowing forever because the ﬂuid
viscosity vanishes in the limit of approaching absolute zero. So bulk motion is possible at arbitrarily
low temperatures although only for ﬂuids made up of helium atoms.

2.     T / F      If you are blindfolded and then touch a ﬂat polished piece of metal with your left hand
and simultaneously touch a ﬂat polished piece of wood with your right hand, and if both the wood and
metal have the same temperature as your body (about 37◦ C), then the piece of metal will feel colder
than the wood.

Answer: F The two blocks should feel the same temperature since your ability to feel how hot or
cold some object is depends on your body having a diﬀerent temperature than the object, which causes
heat to transfer from or into your ﬁngertips. A metal object has a higher thermal diﬀusivity κ than
wood and so your ﬁngers will lose more energy in a short period of time with the piece of metal (if the
metal is at room temperature rather than body temperature) which your brain interprets as the metal
being colder.

3.     T / F      If k is the Boltzmann constant and T is the temperature in kelvin, then kT has units of
joules/second.

Answer: F From the ideal gas equation P V = N kT , one can see immediately that kT has units
of P V which in turn has units of energy: P V = (F/A) × V = F × x which is units of force times a
distance or energy.
At this point in the course, however, you should realize that kT not only has units of energy but, as
suggested by the equipartition theorem, is an important fundamental unit of energy associated with
microscopic objects that are inside a macroscopic equilibrium object at temperature T .

4.     T / F As a non-equilibrium system approaches thermodynamic equilibrium, it typically reaches
thermal equilibrium before it reaches diﬀusive equilibrium.

Answer: F The relaxation time for mechanical equilibrium, which involves responses to imbalanced
forces and so net accelerations, generally is much shorter than the the relaxation times for thermal
or diﬀusive equilibrium, which involves diﬀusion of information by random collisions. But there is
no general rule about whether thermal relaxation times are slower or faster than diﬀusive relaxation,
i.e. whether thermal diﬀusivities κ must be bigger or smaller than diﬀusion constants D. Many small
molecules at room temperature in water have diﬀusion constants of order 10−9 m2 /s which is on the
smaller side of many thermal diﬀusivities. But there is enough variation in material properties and
molecular properties that one can ﬁnd diﬀusion constants as large as 10−5 m2 /s and thermal diﬀusivities
smaller than this value (for good insulators).

5.    T / F Rooms A and B are the same size and room A is kept warmer than room B by a heating
unit. Then after the rooms are connected by an open door and allowed to reach a steady state, room B
will have a greater mass of air than room A.

5
Answer: T Air close to STP is ideal so is described by the equation P V = N kT . Here we are
told that two rooms have the same volume V and they also must have the same pressure P since
they are connected by an open door. (If the pressure were not equal between the rooms, there would
be a macroscopic ﬂow of air back and forth arising from the pressure diﬀerence that would cause the
pressure to become equal everywhere.) Thus the idea gas law takes the form N T = P V /k = constant.
Room B with the lower temperature T must therefore have more particles N than room A, which also
means that it contains a greater mass of air.

6.    T/ F       If x = 2 + y 1/2 then x ∝ y 1/2 .

Answer: F This question was intended to test whether you understand the notation and meaning
of “something being proportional to something else”. I had used this notation several times in one
lecture, e.g., when we estimated that the time to defrost an 8 tonne Siberian mammoth would be of
order 9 months (because τ ∝ L2 ∝ V 2/3 ∝ M 2/3 where M is the mass of the volume).
A quantity A is proportional to some quantity B, written A ∝ B, if there is a nonzero constant c such
that A = cB. The above statement is false, although x − 2 ∝ y 1/2 would be a true statement.
Closely related to the idea of proportionality is that a quantity A scales with a quantity B. This means
that A ∝ B α for some exponent α which is called the scaling exponent. You should be able to see
that scaling is an equivalence relation, if A scales a certain way with B and B scales in a certain way
with C then A scales with C.

7.    T/ F
1        x
1
dx       dy x2 y =        .
0        0                  6

Answer: F      You should have been able to evaluate this double integral directly as follows:
1        x                         1               x
dx       dy x2 y =                 dx x2 ×         dy y       (16)
0        0                         0               0
1            2 x
y
=            dx x2 ×                    (17)
0               2       0
1
x2
=            dx x2 ×                    (18)
0             2
1
=          .                            (19)
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The only slight subtlety was for you to realize that the x2 factor acts like a constant when inside an
integral over y—even if one of the y-integration bounds depends on x—and so the x2 factor can be
moved to the left, outside the y-integral.

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