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
         about in a homework assignment. To obtain the answer, all you had to do was start with the
         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 .


                                                     1
        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

                                                   3
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)
                                                                        10
     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.




                                                                6