[User]
This long problem illustrates some important features of free energy. A rubber
    molecule can be modeled in one dimension as a chain consisting of a series of N = N+ + N− ≫ 1
    links of length a, where N+ point in the +z direction and N− point in the −z direction. The
    orientations of each of the links are independent, and it costs no energy to rotate a link. Thus all
    configurations are equally likely at any temperature.

 (a) Find the length L of the chain in terms of N+ and N− .

 (b) Using the approximation mentioned in problem 14 and assuming L ≪ N a, show that

                                                       L2

                                   S = N kB log 2 −
                                                     2N 2 a2

     and hence show that the entropy decreases as L increases. This model is very basic; for
     instance, it tells us nothing about the tension in the chain, but the sign here is correct. (Hint:
     notice that the changes in entropy are of order L2 /N , and hence are very small. You will have
     to be careful with your approximations to avoid dropping an important contribution.)

 (c) Note that the entropy doesn’t depend on temperature, because all configurations have the
     same energy. In a more realistic model, the total entropy of the molecule would be the sum
     of the entropy computed above, plus the entropy due to additional thermal motions, such as
     vibrations of the links. Given this, if a rubber band is quickly stretched, does it feel warm or
     cold to the touch? (You can check your answer experimentally!)

 (d) Now suppose a mass m is hung from the bottom of the chain. Find the average length of the
     chain at temperature T , using the Boltzmann distribution. Do not assume L ≪ N a. (The
     force that raises this mass is another example of an entropic force.)

 (e) It turns out that you can also arrive at this answer by maximizing the entropy of the entire
     universe Suniv . Argue in general that Suniv is maximized when the Helmholtz free energy

                                               F = U − TS

     is minimized, where U is the total energy of the rubber band and mass, and T is the tem-
     perature of the system and its surroundings. (Hint: use the expression for dSuniv , along with
     conservation of energy.)

 (f) Show that in the limit L ≪ N a, F is indeed minimized at the equilibrium length.

  Remark: Heat vs. Work
  What is the difference between work and heat? If we have a gas in a cylinder with a piston,
  moving the piston counts as work. Clearly, if we split the piston into two halves and moved
  them independently, this would still be work. But we can keep splitting the piston until
  we are down to individual atoms of the piston. These can transfer energy to gas particles
  because they can slam into them as the gas particles are bouncing off, like a batter hitting a
  baseball. But as mentioned in T1, this is precisely how a hot wall transfers energy to a gas!
  So when did we transition from doing work to supplying heat?

  The point is that there is no sharp distinction. There’s just energy transfer; work is the subset
  of energy transfers that we keep track of detail, and heat is the subset we don’t. Entropy
  quantifies our ignorance of the precise state a system is in. That’s why supplying heat changes
  the entropy and doing work doesn’t.

  Remark: Is Entropy Subjective?

  Our definition of entropy involved the number of states the system “could” be in, but of
  course in reality the system is actually in just one state. We really mean the number of
  states the system could be in given the information you know. For example, for an ideal gas
  we usually know the temperature, pressure, and volume.

  You might find this disturbing because it implies the entropy is subjective, depending on
  what each person personally knows. But the entropy determines things like F , and hence
  the amount of work that can be extracted from the system, so how could it differ?

  In fact, there is no contradiction between these two things. For example, suppose that
  besides the usual properties, particles in an ideal gas can be slightly bigger or smaller. If
  you happen to somehow know that all of the big particles are on the left side of a box and
  all the small particles are on the right side, then the entropy you assign to the box is lower
  than somebody who doesn’t know about this, and accordingly you can extract more work
  from the box. For instance, you can put a filter that only allows small particles through in
  the middle, then collect energy as small particles move from right to left.

  On the other hand, if you are hit on the head and forget that the particles are sorted by size,
  your entropy for the gas goes up. And accordingly, you can now extract less work, because
  you wouldn’t know it made sense to put in such a filter in the first place!

  While this example may seem contrived, it’s actually exactly the same as our earlier mixing
  entropy example! If you don’t have any equipment that can distinguish between oxygen and
  nitrogen molecules, then the “unmixed” state already has just as much entropy as the mixed
  state. In this case, the entropy doesn’t change at all when you allow the gases to mix.

  What this example shows is that you don’t have to know everything about a system to use
  thermodynamics. Even if you are wrong about how the system behaves microscopically, you
  can still use the theory to describe what you can do with the system, using your knowledge.
  That’s the core reason for the longevity of thermodynamics. As a recent paper put it:

       If physical theories were people, thermodynamics would be the village witch. Over
       the course of three centuries, she smiled quietly as other theories rose and withered,
       surviving major revolutions in physics, like the advent of general relativity and
       quantum mechanics. The other theories find her somewhat odd, somehow different
       in nature from the rest, yet everyone comes to her for advice, and no one dares
       to contradict her.

  For more about this perspective, see the insightful articles by E.T. Jaynes, The Gibbs Paradox
  and Information Theory and Statistical Mechanics. (But don’t read too much of his later
  work, since he got pretty cranky with age.)

   4    Blackbody Radiation
       Idea 9: Planck’s Law
       As we will show in X1, the radiation power per unit area from an object at temperature T
       at frequency f , per unit frequency, is given by Planck’s law,

                                                2πhf 3         1
                                      I(f ) =      2     hf /k   T −1
                                                                      e(f )
                                                  c    e       B

       where e(f ) is called the emissivity. For an ideal blackbody, e(f ) = 1. By integrating over all
       frequencies, one finds that the total power per unit area from an ideal blackbody is
                                        Z ∞
                                                                   2π 5 kB

                                   J=        I(f ) df = σT 4 , σ =         .
                                          0                        15c2 h3

       This is the Stefan–Boltzmann law, which can also be derived with dimensional analysis. It’s
       a compact and useful result, which means it comes up quite often in Olympiads.

Solution:
(a) We simply have L = (N+ − N− )a. For convenience, we’ll set kB = a = 1 for now.

 (b) We have
                                                        N!
                                            S = log            .
                                                      N+ !N− !
    Applying the approximation log N ! ≈ N log N − N , the linear terms cancel out, giving

                S = N log N − N+ log N+ − N− log N−
                               1                        1
                  = N log N − (N + L) log((N + L)/2) − (N − L) log((N − L)/2)
                               2                        2
                                       1                    1
                  = N log 2 + N log N − (N + L) log(N + L) − (N − L) log(N − L).
                                       2                    2
    To simplify further, we write the logarithms as

                                                               L
                                  log(N ± L) = log N + log 1 ±     .
                                                               N

     Plugging this in and defining x = L/N gives
                                     N
                      S = N log 2 −    ((1 + x) log (1 + x) + (1 − x) log (1 − x)) .

     Expanding the quantity in parentheses is a little tricky, because it vanishes at first order in
     x. So since the answer is second order, we need to expand the logarithms to second order to
     make sure we have everything. We have
                                                               x2
                                          log(1 ± x) = ±x −       .

     and plugging this in gives
                          ((1 + x) log (1 + x) + (1 − x) log (1 − x)) = x2 + . . . .
     If we hadn’t included the second-order term in the logarithm, we would have gotten 2x2
     instead, leading to an erroneous result.
     Putting everything together and restoring the units, we have
                                          N x2                    L2

                            S = N log 2 −       = N kB log 2 −
                                            2                   2N 2 a2
     as desired.

 (c) Since the process is fast, it is adiabatic, so the entropy of the entire rubber band stays the
     same. But we have shown the entropy of the chain configuration degrees of freedom (i.e. ones
     which determine the length L) goes down, so the entropy of the other degrees of freedom must
     go up. Hence these other degrees of freedom receive heat, and the rubber band feels hot.
     By the way, this fact provides a new perspective on a classic puzzle. If you take an ordinary
     metal spring, stretch it with a clamp, and put the whole thing in a bath of acid, then when
     the acid dissolves the spring, where does the elastic potential energy go? For a metal spring,
     the answer is that when you stretch it, you’re pulling the chemical bonds inside of it longer,
     so that they’re displaced from the minimum of potential energy. The spring’s constituent
     atoms thus start out in a less deep potential well, so when they are dissolved they emerge
     with higher energy, which ultimately heats up the bath.
     Howeer, for a rubber band the answer is different, because the elastic force is entropic. Specif-
     ically, though you can describe the force from a rubber band macroscopically with a potential
     energy, at the microscopic level it doesn’t correspond to any actual potential energy; we have
     been assuming throughout this problem that it costs no energy to rotate any of the links!
     Accordingly, when you stretch a rubber band, the work you put in is not actually stored
     within it as potential energy. It instead warms up the rubber band, and therefore quickly
     turns into thermal energy in the room. When the rubber band’s temperature returns to its
     original value, its total energy is the same as before you stretched it. If you put this rubber
     band in an acid bath, the acid’s final temperature will be the same regardless of how the band
     was stretched, because the energy you put in has already left.

 (d) It’s easiest to look at each link individually. The energy of the +z configuration now is 2mga
     greater than the −z configuration, so the ratio of their probabilities is α = e−2mga/kB T . Then
     the probability of ±z is
                                                 1              α
                                         p− =        , p+ =
                                               1+α            1+α

    so the expected contribution to the length is
                                   1   α
                                     −    = tanh(mga/kB T ).
                                  1+α 1+α
    By summing over all links, the expected length is
                                                          mga
                                         ⟨L⟩ = N a tanh        .
                                                          kB T
                                                     √
    Since N is large, thermal fluctuations fall as 1/ N by the logic of P2, so this is also approxi-
    mately the observed length at all times.
    As mentioned in the problem, this raising of the mass is due to an entropic force. Here’s the
    fascinating thing: we haven’t specified anything about the dynamics of the molecule in this
    problem. For example, it could be made of rigid rods connected with pivots, designed to snap
    to the upward or downward position, and immersed inside an ideal gas of temperature T which
    occasionally flip them in a collision. Or maybe the rods could be made of iron, and immersed
    inside a box of permanent magnets flying around with temperature T , which occasionally flip
    them with the magnetic force. The point is that it doesn’t matter! As long as the whole
    system is at equilibrium at temperature T , and the expressions above correctly describe the
    rod’s entropy, we’ll get the same result for the length ⟨L⟩. We don’t need to talk about where
    the entropic force comes from, because the result doesn’t depend on that.

 (e) The entropy of the universe is, up to a constant, −F/T , because

                                                F    U
                                            −     =S− .
                                                T    T
    The first term is the rubber band’s entropy, while −U/T accounts for the change of entropy
    of the environment by ∆Senv = ∆Uenv /Tenv = −∆U/T . For several steps in this argument,
    it’s essential that everything is at temperature T .

 (f) In this case the argument of the hyperbolic tangent is small, so

                                                  mgN a2
                                            L≈           .
                                                   kB T
    On the other hand, we have

                                                                    L2

                                F = −mgL − N kB T log 2 −
                                                                   2N 2 a2

    and we have
                                         ∂F         kB T L
                                            = −mg +
                                         ∂L          N a2
    which implies that
                                                     mgN a2
                                             L=
                                                      kB T
    as expected.

  Remark: Heat vs. Work
  What is the difference between work and heat? If we have a gas in a cylinder with a piston,
  moving the piston counts as work. Clearly, if we split the piston into two halves and moved
  them independently, this would still be work. But we can keep splitting the piston until
  we are down to individual atoms of the piston. These can transfer energy to gas particles
  because they can slam into them as the gas particles are bouncing off, like a batter hitting a
  baseball. But as mentioned in T1, this is precisely how a hot wall transfers energy to a gas!
  So when did we transition from doing work to supplying heat?

  The point is that there is no sharp distinction. There’s just energy transfer; work is the subset
  of energy transfers that we keep track of detail, and heat is the subset we don’t. Entropy
  quantifies our ignorance of the precise state a system is in. That’s why supplying heat changes
  the entropy and doing work doesn’t.

  Remark: Is Entropy Subjective?

  Our definition of entropy involved the number of states the system “could” be in, but of
  course in reality the system is actually in just one state. We really mean the number of
  states the system could be in given the information you know. For example, for an ideal gas
  we usually know the temperature, pressure, and volume.

  You might find this disturbing because it implies the entropy is subjective, depending on
  what each person personally knows. But the entropy determines things like F , and hence
  the amount of work that can be extracted from the system, so how could it differ?

  In fact, there is no contradiction between these two things. For example, suppose that
  besides the usual properties, particles in an ideal gas can be slightly bigger or smaller. If
  you happen to somehow know that all of the big particles are on the left side of a box and
  all the small particles are on the right side, then the entropy you assign to the box is lower
  than somebody who doesn’t know about this, and accordingly you can extract more work
  from the box. For instance, you can put a filter that only allows small particles through in
  the middle, then collect energy as small particles move from right to left.

  On the other hand, if you are hit on the head and forget that the particles are sorted by size,
  your entropy for the gas goes up. And accordingly, you can now extract less work, because
  you wouldn’t know it made sense to put in such a filter in the first place!

  While this example may seem contrived, it’s actually exactly the same as our earlier mixing
  entropy example! If you don’t have any equipment that can distinguish between oxygen and
  nitrogen molecules, then the “unmixed” state already has just as much entropy as the mixed
  state. In this case, the entropy doesn’t change at all when you allow the gases to mix.

  What this example shows is that you don’t have to know everything about a system to use
  thermodynamics. Even if you are wrong about how the system behaves microscopically, you
  can still use the theory to describe what you can do with the system, using your knowledge.

       That’s the core reason for the longevity of thermodynamics. As a recent paper put it:

             If physical theories were people, thermodynamics would be the village witch. Over
             the course of three centuries, she smiled quietly as other theories rose and withered,
             surviving major revolutions in physics, like the advent of general relativity and
             quantum mechanics. The other theories find her somewhat odd, somehow different
             in nature from the rest, yet everyone comes to her for advice, and no one dares
             to contradict her.

       For more about this perspective, see the insightful articles by E.T. Jaynes, The Gibbs Paradox
       and Information Theory and Statistical Mechanics. (But don’t read too much of his later
       work, since he got pretty cranky with age.)

   4    Blackbody Radiation
       Idea 9: Planck’s Law
       As we will show in X1, the radiation power per unit area from an object at temperature T
       at frequency f , per unit frequency, is given by Planck’s law,

                                                 2πhf 3     1
                                       I(f ) =                      e(f )
                                                   c2 ehf /kB T − 1
       where e(f ) is called the emissivity. For an ideal blackbody, e(f ) = 1. By integrating over all
       frequencies, one finds that the total power per unit area from an ideal blackbody is
                                        Z ∞
                                                                   2π 5 kB

                                   J=        I(f ) df = σT 4 , σ =         .
                                          0                        15c2 h3

       This is the Stefan–Boltzmann law, which can also be derived with dimensional analysis. It’s
       a compact and useful result, which means it comes up quite often in Olympiads.


[User]
A point charge q of mass m is released from rest a distance d from a grounded
    conducting plane. Find the time until the point charge hits the plane. (Hint: use Kepler’s laws.)

Solution:
This is an incredibly classic problem, which has been appearing in various forms on
    competitions for decades. By using image charges, we see that the particle always experiences a
    force
                                                       kq 2
                                                  F = 2
                                                       4z
    directly towards the plane, where z is its separation from the plane. Let the particle impact the
    plane at point O.
       This force has the form of an inverse-square law. In particular, we would get the exact same
    result if the force were always directed towards O (rather than always directed towards the plane),

                                                        kq 2
                                                  F=−        r̂.
                                                        4r2
    But in this case, the problem is perfectly analogous to the central force of gravity, where O serves
    as the location of the Sun, and one of the foci of the charge’s orbit. In particular, releasing the
    charge from near rest and waiting for it to hit the plane corresponds to performing the first half of
    an extremely eccentric elliptic orbit.
       The trick is now to use Kepler’s third law. If the charge had performed a circular orbit of radius
    d about O, then
                                            kq 2   mv 2

                                                 =       = mω 2 d
                                            4d       d
    which gives a period of                               s
                                                  2π           md3
                                            T =      = 4π           .
                                                  ω            kq 2

    We can use Kepler’s third law to find the period of the eccentric elliptic orbit the charge actually
    follows. This orbit has semimajor axis d/2, so it has period

                                                           T
                                                      T′ = √ .
                                                          2 2
    The actual path of the charge is only the first half of this orbit, so the answer is
                                                            s
                                         T′      T       π     md3
                                            = √ =√                  .
                                          2     4 2       2 kq 2

    Of course, the problem can also be solved by directly integrating the differential equation. If you
    do it that way, you’ll get the same integral as in a similar example, given in P1.

Kevin Zhou Physics Olympiad Handouts — educational use, attribute to Kevin Zhou (knzhou.github.io). LICENSE FLAG: must reconfirm permissions before public release.


[User]
A simple polarizer contains many very thin, closely spaced wires. If the wires are
    vertical, they block vertical electric fields, allowing only horizontally polarized light to go through;
    this is a horizontal polarizer. One can similarly make diagonal and vertical polarizers.

      (a) Suppose that perfectly monochromatic, but unpolarized light is incident on a double slit. (In
          this case, assume “unpolarized” means that at each instant in time, the polarization of the
          light passing through each slit is the same, but over longer timescales that polarization can
          vary.) What does the intensity pattern on the screen look like?

      (b) Next, suppose a vertical polarizer is placed in front of one slit, and a horizontal polarizer is
          placed in front of the other slit. Now what does the intensity pattern look like?

        (c) Finally, we further modify the setup of part (b) by placing many diagonal polarizers, at 45◦
            to the vertical and horizontal, right in front of the screen. What does the intensity pattern
            on the screen look like now?

      (d) On an unrelated note, suppose we wish to rotate the polarization of linearly polarized light by
          using N ≫ 1 intermediate polarizers. What’s the best way to do this, and what’s the fraction
          of light that passes through the stack?

         Idea 3
         For a plane wave propagating along the z-axis with general polarization, it’s useful to write

                                        E(z, t) = Re E0 ei(kz−ωt)

       where E0 is a complex two-component vector, describing both its amplitude and polarization.
       For example, if E0 = E0 x̂, the light wave is horizontally polarized, if E0 = iE0 x̂, it’s
       horizontally polarized with a phase shifted by π/2, if E0 = E0 ŷ it’s vertically polarized, and
       if E0 = E0 (x̂ + ŷ)/2 it’s diagonally polarized.

       When linear polarizations are combined with a relative phase, the result√is circular (or more
       generally, elliptical) polarization. For example, when E0 = E0 (x̂ + iŷ)/ 2, we have

                                     E0                                    E0
                         Ex (z, t) = √ cos(kz − ωt),           Ey (z, t) = √ sin(kz − ωt)
                                      2                                     2
       which is a circularly polarized light wave; the electric field at a fixed point rotates in a circle
       over time, and if one draws the electric field vectors in a line along k̂, they trace out a spiral.
       Birefringent materials, which have different indices of refraction in different directions, cause
       such phase shifts, and thus can convert linear polarizations into other polarizations.

       Example 3

       A plane wave with amplitude E0 = E0 x̂ enters a linear optical device, which does not absorb
       or reflect any energy.
                          √ When the plane wave exits the device, it has circular polarization,
       E0 = E0 (x̂ + iŷ)/ 2. What does the device do to light with vertical polarization?

       Solution
       Vertical polarized √
                          light has to exit with circular polarization of the other handedness, i.e. with
       E0 = E0 (x̂ − iŷ)/ 2, because this is the only possibility consistent with energy conservation.

       To see this, note that the energy of a light wave is proportional to the time-averaged value
       of |E|2 , which is turn proportional to |E0 |2 . Since horizontal and vertical polarizations are
       orthogonal, they don’t interfere, so sending in both a horizontal and vertical light wave of
       amplitude E0 at the same time just doubles the input energy. This must also double the
       output energy, and indeed, under the above ansatz we have
                                                  x̂ + iŷ x̂ − iŷ √
                                      x̂ + ŷ →      √    + √      = 2 x̂
                                                       2       2
       which indeed has double the energy of one wave by itself.

       The more general principle here is that, since x̂ and ŷ were orthogonal to each other, they
       must be mapped to two other unit vectors which are still orthogonal, as complex vectors.
       That is indeed true, because

                                   (x̂ + iŷ)† (x̂ − iŷ) = x̂ · x̂ + i2 ŷ · ŷ = 0.

       By the way, now that we know what the device does to both horizontal and vertically polarized
       light, we also know what it does to any polarization of light, by superposition.

[2] Problem 8 (MPPP 127). A birefringent material is placed between two orthogonal polarizers. The

    material has thickness d, and has an index of refraction of n1 for light linearly polarized along the
    axis e, and n2 for light polarized about an orthogonal axis.

    If the system is illuminated with light of wavelength λ, give a value for d and orientation of e that
    maximizes the transmitted light.
[2] Problem 9 (MPPP 128). In the first 3D movies, spectators would wear glasses with one eye tinted
    blue and the other tinted red. This was quickly abandoned in favor of a system that used the
    polarization of light.
      (a) If you wear an old pair of 3D movie glasses, close one eye, and look in the mirror, then
          you can only see the open eye. Explain how these glasses employ light polarization. What
          disadvantages might this system have?

      (b) If you wear a new pair of 3D movie glasses and do the same, then you can only see the closed
          eye. Explain why.
[2] Problem 10 (HRK). A quarter-wave plate is a birefringent plate that causes a π/2 phase shift
    between light polarized along e and perpendicular to e. Similarly, a half-wave plate causes a π
    phase shift. Suppose you are given an object, which may be a quarter-wave plate, a half-wave plate,
    a linear polarizer, or just a semi-opaque disk of glass. How can you identify the object? You can use
    an unpolarized light source, and any number of polarizers and quarter-wave and half-wave plates.
[2] Problem 11 (HRK). A polarizer and a quarter-wave plate are glued together so that, if the
    combination is placed with face A against a shiny coin, the face of the coin can be seen when
    illuminated by light of appropriate wavelength. When the combination is placed with face A away
    from the coin, the coin cannot be seen. Which component is on face A and what is the relative
    orientation of the components?

Solution:
(a) It’s just an ordinary double slit interference pattern. Since the light goes through
         the slits with the same polarization at each moment, it shows up at the screen with the
         same polarization at each moment. Then the amplitudes just add, so the intensities add in
         quadrature, yielding interference effects.

      (b) Let’s let the horizontal and vertical directions be x and y. Then at a given point on the
          screen, the light from one slit gives Ex and the light from the other slit gives Ey . The intensity
          is proportional to Ex2 + Ey2 , so the intensities just add, with no interference term. So the
          interference pattern is completely destroyed. The general point is that stable interference
          patterns can only exist when the things you’re interfering have all the same properties.
                                                             √          √
      (c) The amplitude of the diagonal light is now (Ex / 2) + (Ey / 2), so the intensity at the screen
          is now proportional to (Ex + Ey )2 , so it’s now possible to see some interference effects again.

     However, it’s not perfect. For example, if the light entering the slits was horizontal at some
     moment, then it would get totally blocked at one slit, so the result shows no interference
     pattern. The same is true if the light entering the slits is vertical. However, if the light entered
     the slits polarized
                  √      at 45◦ to the vertical, then its amplitude would get penalized by the same
     factor of 1/ 2 at each slit, and at the screen we would get a perfect interference pattern. In
     practice, the incoming polarization of “unpolarized” light rapidly changes over time, so on
     average we get a partially visible interference pattern, i.e. there are still minima and maxima
     but the minima don’t have zero intensity.
     This is an interesting result because you might think, based on the result of part (b), that
     polarizers can only destroy interference patterns; however, adding more of them at the screen
     can actually bring them back!
     Exactly the same logic can be used to explain the famous “delayed choice quantum eraser”. In
     popular science articles, people often say this experiment proves that quantum mechanics can
    “rewrite the past”. But it’s really just a tricky interference effect that also shows up classically.

 (d) For a small misalignment, the fraction of intensity lost goes up quadratically with the mismatch
     in angle, so we should have the polarizers uniformly spaced by angle θ/N . Then each filter
     multiplies the intensity by cos2 (θ/N ) ≈ 1 − θ2 /(2N 2 ), so the overall intensity is multiplied by
                                               N
                                           θ2              θ2

                                       1−         ≈ exp  −      .
                                          2N 2             2N

     In the limit N → ∞, no intensity is lost. So in principle you can rotate polarization this way,
     though it’s better to just use a half-wave plate, which will be described below.

  Idea 3
  For a plane wave propagating along the z-axis with general polarization, it’s useful to write

                                 E(z, t) = Re E0 ei(kz−ωt)

  where E0 is a complex two-component vector, describing both its amplitude and polarization.
  For example, if E0 = E0 x̂, the light wave is horizontally polarized, if E0 = iE0 x̂, it’s
  horizontally polarized with a phase shifted by π/2, if E0 = E0 ŷ it’s vertically polarized, and
  if E0 = E0 (x̂ + ŷ)/2 it’s diagonally polarized.

  When linear polarizations are combined with a relative phase, the result√is circular (or more
  generally, elliptical) polarization. For example, when E0 = E0 (x̂ + iŷ)/ 2, we have

                                 E0                                E0
                     Ex (z, t) = √ cos(kz − ωt),       Ey (z, t) = √ sin(kz − ωt)
                                  2                                 2
  which is a circularly polarized light wave; the electric field at a fixed point rotates in a circle
  over time, and if one draws the electric field vectors in a line along k̂, they trace out a spiral.
  Birefringent materials, which have different indices of refraction in different directions, cause
  such phase shifts, and thus can convert linear polarizations into other polarizations.

       Example 3

       A plane wave with amplitude E0 = E0 x̂ enters a linear optical device, which does not absorb
       or reflect any energy.
                          √ When the plane wave exits the device, it has circular polarization,
       E0 = E0 (x̂ + iŷ)/ 2. What does the device do to light with vertical polarization?

       Solution
       Vertical polarized √
                          light has to exit with circular polarization of the other handedness, i.e. with
       E0 = E0 (x̂ − iŷ)/ 2, because this is the only possibility consistent with energy conservation.

       To see this, note that the energy of a light wave is proportional to the time-averaged value
       of |E|2 , which is turn proportional to |E0 |2 . Since horizontal and vertical polarizations are
       orthogonal, they don’t interfere, so sending in both a horizontal and vertical light wave of
       amplitude E0 at the same time just doubles the input energy. This must also double the
       output energy, and indeed, under the above ansatz we have
                                                  x̂ + iŷ x̂ − iŷ √
                                      x̂ + ŷ →      √    + √      = 2 x̂
                                                       2       2
       which indeed has double the energy of one wave by itself.

       The more general principle here is that, since x̂ and ŷ were orthogonal to each other, they
       must be mapped to two other unit vectors which are still orthogonal, as complex vectors.
       That is indeed true, because

                                   (x̂ + iŷ)† (x̂ − iŷ) = x̂ · x̂ + i2 ŷ · ŷ = 0.

       By the way, now that we know what the device does to both horizontal and vertically polarized
       light, we also know what it does to any polarization of light, by superposition.

[2] Problem 8 (MPPP 127). A birefringent material is placed between two orthogonal polarizers. The
    material has thickness d, and has an index of refraction of n1 for light linearly polarized along the
    axis e, and n2 for light polarized about an orthogonal axis.

    If the system is illuminated with light of wavelength λ, give a value for d and orientation of e that
    maximizes the transmitted light.

    Solution. We want to turn vertical polarization into horizontal polarization. Note that the hori-
    zontal polarization x̂ and vertical polarization ŷ can both be viewed as equal superpositions of the

                                      √
    diagonal polarizations (x̂ ± iŷ)/ 2, but with opposite relative sign, i.e. a relative phase shift of π.
       Thus, we want ê to point diagonally, φ = 45◦ , and set d so that it flips the relative sign between
    these diagonal polarizations. This occurs if d = (k + 1/2)λ/(|n1 − n2 |) for a whole number k. This
    system is called a half-wave plate.

[2] Problem 9 (MPPP 128). In the first 3D movies, spectators would wear glasses with one eye tinted
    blue and the other tinted red. This was quickly abandoned in favor of a system that used the
    polarization of light.

      (a) If you wear an old pair of 3D movie glasses, close one eye, and look in the mirror, then
          you can only see the open eye. Explain how these glasses employ light polarization. What
          disadvantages might this system have?

      (b) If you wear a new pair of 3D movie glasses and do the same, then you can only see the closed
          eye. Explain why.

    Solution. (a) One eye only lets vertically polarized light through, and the other only lets hor-
         izontally polarized light through, so that you can see different images with each eye. The
         disadvantage if that if you tilt your head, the images for each eye will get mixed together.

      (b) One eye only lets clockwise polarized light through, and the other only lets counterclockwise
          polarized light through; the mirror flips the direction of circular polarization.

[2] Problem 10 (HRK). A quarter-wave plate is a birefringent plate that causes a π/2 phase shift
    between light polarized along e and perpendicular to e. Similarly, a half-wave plate causes a π
    phase shift. Suppose you are given an object, which may be a quarter-wave plate, a half-wave plate,
    a linear polarizer, or just a semi-opaque disk of glass. How can you identify the object? You can use
    an unpolarized light source, and any number of polarizers and quarter-wave and half-wave plates.

    Solution. There are many ways to approach this, but here’s a way that just uses an unpolarized
    light source and up to two polarizers. First, a quarter-wave plate and a half-wave plate don’t change
    the intensity of light, so if the intensity is reduced, it’s either a polarizer or a semi-opaque disk. To
    distinguish between the latter two, you can check if the output is linearly polarized, by applying a
    polarizer. (For example, there should be an orientation of the polarizer where the output light is
    blocked entirely.)
        To distinguish between a quarter-wave and half-wave plate, one could use a polarizer to send in
    linearly polarized light. A half-wave plate transforms linearly polarized light into linearly polarized
    light, with possibly a different polarization axis. That means that the output must be linearly
    polarized, which you can check by seeing if it can be totally blocked by a polarizer.
        On the other hand, a quarter-wave plate can transform linearly polarized light into an arbitrary
    elliptical polarization, and if you orient it correctly, the output will be circularly polarized. When
    circular polarized light enters a polarizer of any orientation, exactly half of the intensity is blocked.
    So if the object is a quarter-wave plate, there will exist an orientation of the object so that when you
    when you put a polarizer after it, the output intensity is independent of that polarizer’s orientation.

[2] Problem 11 (HRK). A polarizer and a quarter-wave plate are glued together so that, if the
    combination is placed with face A against a shiny coin, the face of the coin can be seen when
    illuminated by light of appropriate wavelength. When the combination is placed with face A away
    from the coin, the coin cannot be seen. Which component is on face A and what is the relative
    orientation of the components?

    Solution. Let the components of the object be A and B. One simple way to think about it is to
   “unfold” the reflection, i.e. to replace the coin with a inverted version of this system behind it. Then
    the problem tells us that some light can pass through BAAB, but none can pass through ABBA.
        This is only possible if A is the polarizer and B is the quarter wave plate. Some fraction of the
    light will always pass through BAAB. As for ABBA, suppose we let A be a vertical polarizer, and
    let the optical axis of B be at 45◦ to the vertical. Then two copies of B is effectively a half-wave
    plate, which converts vertically polarized light to horizontally polarized light, which gets blocked
    by the second copy of A. So no light goes through ABBA.


[User]
A lens with a focal length $f_1=\SI{4}{cm}$ is positioned so that a parallel light beam of diameter $d_0=\SI{1}{cm}$ is focused at one point on the screen. Sometimes a bigger area is needed to be lighted on the screen but moving the lens or changing the source of the light is impossible. How big has to be the focal length $f_2$ of an additional lens positioned to the right of the initial lens, so that an evenly lighted area with a diameter $d=\SI{2}{cm}$ would appear on the screen? The distance between the lenses is $L$.
\begin{center}
\includegraphics[scale=1.5]{2015-v3g-02-valgustamine-yles_ing}
\end{center}

Solution:
Vaatleme kõige äärmiste valguskiirte liikumist läbi süsteemi.
\begin{center}
\includegraphics[scale=1.5]{2015-v3g-02-valgustamine-lah}
\end{center}
Pärast lisaläätse läbimist koonduvad valguskiired punktiks lisaläätsest kaugusel $f_2$ ehk kaugusel $L-f_2$ algsest läätsest. Läätse valemi
\begin{equation}
\frac{1}{a}+\frac{1}{L-f_2}=\frac{1}{f_1}.
\end{equation}
põhjal tekib sellest punktist omakorda punktkujutis kaugusele $a$ esimesest läätsest. Sarnastest kolmnurkadest saame veel kaks võrrandit:
\begin{equation}
\frac{d_1}{d_0}=\frac{L-f_2}{f_2},
\end{equation}
\begin{equation}
\frac{d}{a-f_1}=\frac{d_1}{a}.
\end{equation}
Kahest esimesest võrrandist saame suuruse $\frac{1}{L-f_2}$ avaldamisel
\begin{equation}
\frac{1}{a}+\frac{d_0}{d_1f_2}=\frac{1}{f_1}\implies \frac{d_0}{d_1f_2}=\frac{a-f_1}{af_1}.
\end{equation}
Viimasest kahest võrrandist saame avaldada suuruse $d_1(a-f_1)/a$, mille põhjal saame $d=d_0\frac{f_1}{f_2}.$ Seega valguslaigu suurus ekraanil sõltub ainult lisatud läätse fookuskaugusest, aga mitte läätsedevahelisest kaugusest $L$. Valguslaik diameetriga \SI{2}{cm} tekib kui $f_2=f_1d_0/d=\SI{2}{cm}$.

Valgustamine

Lääts fookuskaugusega $f_1=\SI{4}{cm}$ on paigutatud nii, et läätsele suunatud paralleelsete valguskiirte kimp diameetriga $d_0=\SI{1}{cm}$ koondub ekraanil ühte punkti. Mõnikord on tarvis valgustada ekraanil suuremat ala, kuid läätse nihutamine või valgusallika vahetamine pole võimalik. Kui suur peab olema olemasolevast läätsest paremale paigutatava lisaläätse fookuskaugus $f_2$, et ekraanil tekiks ühtlaselt valgustatud laik diameetriga $d=\SI{2}{cm}$, kui läätsede vahekaugus on $L$?
\begin{center}
\includegraphics[scale=1.5]{2015-v3g-02-valgustamine-yles}
\end{center}

Antud juhul tasub vaadelda kõige äärmisi süsteemi sisenevaid kiiri.


[User]
01T
                        IZhO 2021, problem 3. A problem on the propagation of light through a
    waveguide, unifying material from E7 and W1.

    3      Water Waves
    Water waves are the most familiar examples of waves in everyday life, but you won’t find them
    mentioned often in introductory textbooks, because they’re far more complicated than any other
    kind of wave we’ll consider. In all the problems below, we will completely neglect viscosity, surface
    tension, and compressibility of the water. Despite this, our results will still only be approximate.

Solution:
See the official solutions here.

    3    Water Waves
    Water waves are the most familiar examples of waves in everyday life, but you won’t find them
    mentioned often in introductory textbooks, because they’re far more complicated than any other
    kind of wave we’ll consider. In all the problems below, we will completely neglect viscosity, surface
    tension, and compressibility of the water. Despite this, our results will still only be approximate.


[User]
At time t = 0, the position and transverse velocity of a string obeying the wave
    equation are given by y(x) and vy (x). Find an explicit expression for y(x, t) in terms of these
    functions; this is called d’Alembert’s solution. (Hint: construct solutions with initial position y(x)
    and zero initial velocity, and vice versa, and add them together. Your final answer should contain
    an integral involving vy .)

Solution:
We write y(x, t) = g(x, t) + h(x, t), where g has the correct initial position y(x) but no
    initial velocity, and h has the correct initial velocity vy (x) but zero initial position. The lesson of
    problem 2 was precisely that
                                                  y(x − vt) + y(x + vt)
                                       g(x, t) =                         .

    Now let’s construct h(x, t). We know it must vanish at t = 0, and that when it’s differentiated with
    respect to time at t = 0, we get vy (x). So an initial guess might be
                                                      Z t
                                          h(x, t) =         vy (x − vt) dt.

    That’s because, by the fundamental theorem of calculus, the only contribution to ∂h/∂t at t = 0 is
    from the change of the upper bound of the integral, so

                                       ∂h
                                              = vy (x − vt)     = vy (x).
                                       ∂t t=0               t=0

    However, this solution doesn’t work, because it doesn’t quite satisfy the wave equation. In particular,
    at t = 0 we have h = 0, which implies ∂ 2 h/∂x2 = 0, which implies ∂ 2 h/∂t2 = 0 by the wave equation.
    In other words, the solution for h should have no initial acceleration because the string begins flat.
    But taking the second time derivative of the solution above gives something that’s certainly nonzero.
        To fix this, we use exactly the same trick that was used to derive g. We superpose a solution
    with dependence on x + vt rather than x − vt,

                                           1 t
                                              Z
                                 h(x, t) =      (vy (x − vt) + vy (x + vt)) dt.
                                           2 0

    This still has the right initial conditions, and it does solve the wave equation.
      We therefore conclude that the general solution is
                                                         Z t

                    y(x, t) =     y(x − vt) + y(x + vt) +     (vy (x − vt) + vy (x + vt)) dt .
                              2                             0

    Using this solution you can show, for example, that if you poke the string and thereby change either
    y or vy at one point in space, the influence of that change spreads out at speed v. In other words,
    the physics is “local”: there is no way to send a signal to a distant place instantaneously. This
    solution gives the basic prototype for many arguments involving locality in more advanced physics.


[User]
This problem is just for fun; the techniques used here are too advanced to appear on
    Olympiads. We will prove Rayleigh’s monotonicity law, which states that increasing the resistance
    of any part of a resistor network increases the equivalent resistance between any two points. This
    may seem obvious, but it’s actually tricky to prove. The following is the slickest way.
      (a) Consider a graph of resistors, where a battery is attached across two of the vertices, fixing
          their voltages. Write an expression for the total power dissipated, assuming the voltages at
          each vertex are Vi and the resistances are Rij .
     (b) The voltages Vi at all the other vertices are determined by Kirchoff’s rules. But suppose you
         didn’t know that, or didn’t want to set up those equations. Remarkably, it turns out that

     you can derive the exact same results by simply treating the voltages Vi as free to vary, and
     setting them to minimize the total power dissipated! Show this result. (This is an example of
     a variational principle, like the principle of least action in mechanics.)

 (c) For any network of resistors, show that P = V 2 /R when V is the battery voltage applied
     across two vertices, R is the equivalent resistance between them, and P is the total power
     dissipated in the resistors. (This is intuitive, but it’s worth showing in detail to assist with
     the next part.)

 (d) By combining all of these results, prove Rayleigh’s monotonicity law.

 (e) We can use Rayleigh’s monotonicity law to prove some mathematical results. Consider the
     resistor network shown below, where the variables label the resistances.

     By considering the resistances before and after closing the switch P Q, show that the arithmetic
     mean of two numbers is at least the geometric mean.

 (f) Consider the resistor network shown below.

     By closing all the switches, show that the arithmetic mean of n numbers is at least the
     harmonic mean.

  Remark
  You might think that Rayleigh’s monotonicity law is too obvious to require a proof; if you
  decrease a resistance, how could the net resistance possibly go up? In fact, this kind of
  non-monotonicity occurs very often! For example, Braess’s paradox is that fact that adding
  more roads can slow down traffic, even when the total number of cars stays the same. A U.S.
  Physics Team coach has argued that allowing more team strategies can make a basketball
  team score less. For more on this subject, see the paper Paradoxical behaviour of mechanical
  and electrical networks or this video.

    Remark
    Circuit questions can get absurdly hard, but at some point they start being more about
    mathematical tricks than physics. As a result, I haven’t included any such problems here;
    they tend not to appear on the USAPhO or IPhO, or in college physics, or in real life, or really
    anywhere besides a few competitions. On the other hand, you might find such questions fun!
    For some examples, see the Physics Cup problems 2013.6, 2017.2, 2018.1, and 2019.4.

2    RC Circuits
Next we’ll briefly cover RC circuits, our first exposure to a situation genuinely changing in time.

    Example 4: CPhO

    The capacitors in the circuit shown below were initially neutral. Then, the circuit is allowed
    to reach the steady state.

    After a long time, what is the charge stored on the 10 mF capacitor?

    Solution
    After a long time, no current flows through the capacitors, so there is effectively a single loop
    in the circuit. It has a total resistance 60 Ω and a total emf 6 V, so the current is I = 0.1 A.
    Using this, we can straightforwardly label the voltages everywhere on the outer loop.

       To finish the problem, we need to know the voltage V0 of the central node, so we need
       one more equation. That equation is charge conservation. We note that the central part
       of the circuit, containing the inner plates of the three capacitors, begins uncharged and is
       not directly connected to anything else, so it must remain uncharged. (Charge can’t move
       through a capacitor!) Suppressing units, this means

                            20(26 − V0 ) + 20(7 − V0 ) + 10(0 − V0 ) = 0,   V0 =      V

       from which we read off the answer,

                                               Q = CV = 0.132 C.

Solution:
(a) The power is
                                                X (Vi − Vj )2
                                          P =                   .
                                                       Rij
                                                i<j

     Here the sum over i < j counts all pairs of vertices once. If there is no direct connection
     between i and j, the resistance Rij is infinite.

 (b) The power is minimized when its derivative is zero, and we are free to vary all voltages except
     for the two points where the battery is connected. Let Vi be one of these voltages. Then
                                      ∂P      X 2(Vi − Vj )
                                           =                = 0.
                                      ∂Vi            Rij
                                              j̸=i

     Now compare this to how we would solve the problem using Kirchoff’s laws. The fact that
     the sum of the voltage drops along a loop is zero is already accounted for, because we already
     have specified the voltages at each vertex. The only new equations we would write down
     would be charge conservation at each vertex,
                                              X
                                                  Iij = 0.
                                                j̸=i

     However, applying Ohm’s law, we see this is precisely the equation that power minimization
     has given us!

 (c) By the definition of the equivalent resistance, V = IR where I is the total current going
     through the circuit. By the definition of power, the power put in by the battery is P = IV ,
     since any current going through the circuit must go through the battery. By conservation of
     energy, the power dissipated in the circuit is equal to the power put in by the battery. So the
     power dissipated is P = IV = V 2 /R.

 (d) Put a battery of voltage V across the points we are considering. By part (c) Rayleigh’s
     monotonicity law is equivalent to the statement that, if we increase any of the Rij , the total
     power P dissipated in the resistor network goes down.
     We can account for the effect of increasing one of the Rij in two steps. First, suppose we
     do so while artificially keeping all the voltages Vi constant. Then by part (a), P decreases.
     Second, in reality the voltages quickly rearrange themselves to satisfy Kirchoff’s laws, which
     we saw in part (b) is equivalent to minimizing the power. So this further rearrangement can
     only further decrease P . This shows the desired result.

 (e) Before closing the switch, the resistance is

                                                        a+b
                                               Ri =         .

     After closing the switch, the resistance is
                                                         −1
                                                   1 1                     2ab
                                      Rf = 2        +                  =       .
                                                   a b                     a+b

     Closing the switch is equivalent to decreasing RP Q from infinity to zero, so Rf ≤ Ri by
     Rayleigh’s monotonicity law. This gives
                                               √          a+b
                                                   ab ≤

     which is the AM-GM inequality.

 (f) Before closing the switches,
                                                        1X
                                               Ri =        ai
                                                        n
                                                              i

     which is the arithmetic mean. After closing the switches,
                                                                       !−1
                                                        X 1
                                           Rf = n
                                                                  ai
                                                          i

     which is the harmonic mean. Thus, the arithmetic mean is at least the harmonic mean.

  Remark
  You might think that Rayleigh’s monotonicity law is too obvious to require a proof; if you
  decrease a resistance, how could the net resistance possibly go up? In fact, this kind of
  non-monotonicity occurs very often! For example, Braess’s paradox is that fact that adding
  more roads can slow down traffic, even when the total number of cars stays the same. A U.S.
  Physics Team coach has argued that allowing more team strategies can make a basketball
  team score less. For more on this subject, see the paper Paradoxical behaviour of mechanical
  and electrical networks or this video.

    Remark
    Circuit questions can get absurdly hard, but at some point they start being more about
    mathematical tricks than physics. As a result, I haven’t included any such problems here;
    they tend not to appear on the USAPhO or IPhO, or in college physics, or in real life, or really
    anywhere besides a few competitions. On the other hand, you might find such questions fun!
    For some examples, see the Physics Cup problems 2013.6, 2017.2, 2018.1, and 2019.4.

2    RC Circuits
Next we’ll briefly cover RC circuits, our first exposure to a situation genuinely changing in time.

    Example 4: CPhO

    The capacitors in the circuit shown below were initially neutral. Then, the circuit is allowed
    to reach the steady state.

    After a long time, what is the charge stored on the 10 mF capacitor?

    Solution
    After a long time, no current flows through the capacitors, so there is effectively a single loop
    in the circuit. It has a total resistance 60 Ω and a total emf 6 V, so the current is I = 0.1 A.
    Using this, we can straightforwardly label the voltages everywhere on the outer loop.

       To finish the problem, we need to know the voltage V0 of the central node, so we need
       one more equation. That equation is charge conservation. We note that the central part
       of the circuit, containing the inner plates of the three capacitors, begins uncharged and is
       not directly connected to anything else, so it must remain uncharged. (Charge can’t move
       through a capacitor!) Suppressing units, this means

                           20(26 − V0 ) + 20(7 − V0 ) + 10(0 − V0 ) = 0,   V0 =      V

       from which we read off the answer,

                                              Q = CV = 0.132 C.


[User] Let there be an absolutely black cube made of a material with very good heat conductivity. The cube is placed in front of a parallel light beam with an intensity (power per cross section area) $I$. What is the maximal and minimal stable temperature $T_\text{max}$ and $T_\text{min}$ that the cube attains depending on its position with respect to the direction of the radiation’s spreading?

Solution:
Ülesandes esitatud tingimuste kohaselt on see kuup musta keha kiirgur ja see neelab kõik sellele langenud kiirguse. Olenemata asendist on kuubi poolt kiiratud koguvõimsus sõltuv ainult kuubi temperatuurist ja selle küljetahu pindalast $A$. Kuubil on teatavasti 6 tahku. Seega on kuubi kiiratav koguvõimsus Stefani-Boltzmanni seaduse järgi
$$
P=6A\sigma T^4 .
$$
Tasakaaluolukorras on kuubi poolt neelatav võimsus ja kiiratav võimsus võrdsed. Kuubi poolt neelatav võimsus on võrdeline kuubi projektsiooniga valguskiirtega risti olevale tasandile. Selle projektsiooni suurus sõltub kuubi asendist valguskiirte suhtes. Olgu $\alpha$ tegur, mis näitab kui palju on kuubi projektsioon suurem selle küljepindalast. Sellel juhul saame tasakaaluolukorra
$$
6A\sigma T^4=\alpha AI,
$$
millest saame tasakaalulise temperatuuri sõltuvuse kuubi asendist.
$$
T(\alpha)=\sqrt[4]{\frac{\alpha I}{6\sigma}}.
$$
Nüüd tuleb leida kõrgeim ja madalaim temperatuur. Selleks on vaja leida suurim ja vähim $\alpha$. Vaadates kuubi geomeetriat, on üsnagi lihtne järeldada, et minimaalne võimalik $\alpha$ on 1. See vastab olukorrale, kus kuubi üks külg on valgusvooga risti.
$$
T\idx{min}=\sqrt[4]{\frac{ I}{6\sigma}}.
$$
Maksimaalse juhu leidmine on aga keerukam. Selleks leidub geomeetrilisi meetodeid, kuid üks lihtsaim meetod on kasutada teadmist, et pinnaühiku projektsioon pinnale on võrdeline selle pinnanormaali $\vec{n}$ ja valgusvoo suuna $\vec{i}=\vec{I}/I$ vahelise nurga koosinusega. Ühikvektorite korral
$$
A\idx{projektsioon}=A\vec{n}\cdot\vec{i} \Rightarrow \alpha=\Sigma\vec{n}\cdot\vec{i}.
$$
Kuubil saavad olla valgusvoo suunas kõige rohkem kolm tahku korraga. Tähistame need kui $x$, $y$ ja $z$ ning nende pinnanormaalid kui $\vec{n}_x$, $\vec{n}_y$ ja $\vec{n}_z$. Seega
$$
\alpha= (\vec{n}_x + \vec{n}_y + \vec{n}_z) \cdot \vec{i}.
$$
Kui me defineerime taustsüsteemi, kus kuubi küljed on risti vastavate telgedega, siis lihtsustub antud valem $\vec{i}$ komponentide summaks
$$
\alpha = i_x + i_y + i_z.
$$
Kuna $\vec{i}$ on ühikvektor, siis
$$
i_x^2+i_y^2+i_z^2=1.
$$
On näha, et $\alpha$ on maksimaalne kui kuubi diagonaal on suunatud valgusvoo suunas ehk kõikide külgede komponendid on võrdsed. Seega
$$
\begin{array}{c}
\alpha\idx{max} = 3i_x \\ 3i_x^2=1
\end{array}
\Rightarrow \alpha\idx{max}=\sqrt{3}.
$$
Järelikult
$$
T\idx{max}=\sqrt[4]{\frac{\sqrt{3} I}{6\sigma}}.
$$
ehk temperatuur varieerub $\sqrt[8]{3} \approx \num{1,15}$ korda.

Must kuup

Olgu väga hea soojusjuhtivusega materjalist absoluutselt must kuup paralleelses valgusvihus, mille intensiivsus (võimsus ristlõikepindala kohta) on $I$. Milline on maksimaalne ja minimaalne stabiilne temperatuur $T_\text{max}$ ja $T_\text{min}$, mille kuup saavutab sõltuvalt selle asendist kiirguse leviku suuna suhtes?

Soojustasakaalu korral kiirgab kuup sama palju kui see neelab. Kiiratav võimsus on leitav Stefan-Boltzmanni seadusest ning neelatav võimsus kuubi projektsiooni pindalast kiirguse leviku suunaga risti oleval tasandil. On selge, et maksimaalse temperatuuriga peab kuubi projektsiooni pindala olema maksimaalne ja minimaalse temperatuuriga minimaalne.


[User]
APhO 2015, problem 2. A nice but somewhat hard to read problem on the
    aurora and the solar wind.

    3    Compact Objects
    Compact objects such as white dwarfs and neutron stars must be handled with quantum statistical

                      01^
    mechanics, as introduced in X1.

Solution:
See the official solutions here. But also note that there are some minor typos in it, as
    pointed out here.

    3    Compact Objects
    Compact objects such as white dwarfs and neutron stars must be handled with quantum statistical
    mechanics, as introduced in X1.


[User]
Question B1
Suppose a domino stands upright on a table. It has height h, thickness t, width w (as shown below),
and mass m. The domino is free to rotate about its edges, but will not slide across the table.

                                                                          p

                                                            h
                                            h

                                                                 t
                                                          table surface
                             l         w
                                  t

  a. Suppose we give the domino a sharp, horizontal impulsive push with total momentum p.
        i. At what height H above the table is the impulse p required to topple the domino smallest?
       ii. What is the minimum value of p to topple the domino?
  b. Next, imagine a long row of dominoes with equal spacing l between the nearest sides of any
     pair of adjacent dominos, as shown above. When a domino topples, it collides with the next
     domino in the row. Imagine this collision to be completely inelastic. What fraction of the
     total kinetic energy is lost in the collision of the first domino with the second domino?
  c. After the collision, the dominoes rotate in such a way so that they always remain in contact.
     Assume that there is no friction between the dominoes and the first domino was given the
     smallest possible push such that it toppled. What is the minimum l such that the second
     domino will topple?
     You may work to lowest nontrivial order in the angles through which the dominoes have
     rotated. Equivalently, you may approximate t, l  h.
  d. A row of toppling dominoes can be considered to have a propagation speed of the length l + t
     divided by the time between successive collisions. When the first domino is given a minimal
     push just large enough to topple and start a chain reaction of toppling dominoes, the speed
     increases with each domino, but approaches an asymptotic speed v.

     Suppose there is a row of dominoes on another planet. These dominoes have the same density
     as the dominoes previously considered, but are twice as tall, wide, and thick, and placed with
     a spacing of 2l between them. If this row of dominoes topples with the same asymptotic
     speed v previously found, what is the local gravitational acceleration on this planet?

                   Copyright c 2017 American Association of Physics Teachers

2017 USA Physics Olympiad Exam                       Part B                                                       9

Solution:
Question B1
Suppose a domino stands upright on a table. It has height h, thickness t, width w (as shown below),
and mass m. The domino is free to rotate about its edges, but will not slide across the table.

                                                                          p

                                                            h
                                            h

                                                                 t
                                                          table surface
                             l          w
                                  t

  a. Suppose we give the domino a sharp, horizontal impulsive push with total momentum p.

        i. At what height H above the table is the impulse p required to topple the domino smallest?
       ii. What is the minimum value of p to topple the domino?

                                                Solution

     First we’ll look for the height H at which we should push to topple the domino with the least
     momentum. A convenient method is to look at the angular momentum in the domino because
     this is easy to calculate and describes rotational motion. Because the push is horizontal, the
     moment arm of the push (about the domino’s rotational axis) is purely vertical. That means
     the angular momentum of the push is pH, where p is the momentum imparted and H is the
     height of the push. There is some minimum angular momentum Lmin to topple the domino,
     so we set Lmin = pH. The bigger H, the smaller p, so we should choose the largest possible
     H. In other words, we should push at the very top of the domino, H = h. While we’re on this
     part, note that if the push weren’t constrained to be horizontal, the push could be a little bit
     smaller since the moment arm could be the entire diagonal of the thin edge of the domino.
     Next we calculate the minimum p using energy. As the domino rotates, it converts kinetic
     energy to potential energy, so we’ll calculate both. The domino’s potential energy is greatest
     when its center of mass is directly over the contact point of the domino and the table. That
     height is half the diagonal
                         √        of the domino, so the distance from the contact point to the center
                       1   2     2
     of the domino is 2 t + h . From the push until it reaches this point, the domino’s potential
     energy increases by
                                              1    p
                                        ∆U = mg        t2 + h2 − h .
                                              2
     Then the domino topples. By conservation of energy, ∆U is how much rotational kinetic
     energy the domino must have begun with.
     Next we find the kinetic energy using the rotational kinetic energy formula, T = L2 /2I. The
     moment of inertia of the domino about its contact point with the table is I = 13 m(h2 + t2 ).

                   Copyright c 2017 American Association of Physics Teachers

2017 USA Physics Olympiad Exam                Part B                                               16

    You can find this with an integral, or if you know the moment of inertia about the center
      1
    ( 12 m(h2 + t2 )), you can use the parallel axis theorem. We know the momentum L = ph, so
    the kinetic energy is

                                  1 L2   1      (ph)2         3   p2 h2
                            T =        =                    =               .
                                  2 I    2 (1/3)m(h2 + t2 )   2 m(h2 + t2 )

    Setting the initial kinetic energy equal to the gain in potential energy and solving for p,
                                           r
                                       1 m       p
                               pmin = √       g    t2 + h2 − h (h2 + t2 ).
                                        3h

  b. Next, imagine a long row of dominoes with equal spacing l between the nearest sides of any
     pair of adjacent dominos, as shown above. When a domino topples, it collides with the next
     domino in the row. Imagine this collision to be completely inelastic. What fraction of the
     total kinetic energy is lost in the collision of the first domino with the second domino?

                                              Solution
                                                             √
    Right after the collision, the dominoes touch at a height h2 − l2 above the table. The second
    domino is vertical, while the first is rotated so that the angle between its leading edge and
    the table is θ = arccos(l/h). Let the dominoes’ angular velocities be ω1 and ω2 respectively.
    After the dominoes collide, they stick together. If we take the two parts of the dominoes
    that are in contact, they must have the same horizontal velocity component in order  √ for the
                                                                                2    2
    dominoes to stay in contact. For the first domino, this velocity is ω1 h sin
                                                                               √  θ = ω 1 h −l .
    For the second domino it is ω2 times the height of the impact point, so ω2 h2 − l2 . Because
    these velocity components must be equal, ω1 = ω2 .
    This means the dominoes have the same angular momentum as each other, measured relative
    to their respective rotation axes. If we look at the collision, the forces between the dominoes
    are purely horizontal because the dominoes’ faces are frictionless, so they only exert normal
    forces. The second domino is vertical, so all its normal forces are purely horizontal. These
    horizontal normal forces exchange angular momentum between the two dominoes. They have
    the same moment arm (i.e. the height of the collision), so the amount of angular momentum
    transferred out of the first domino is equal to the amount gained by the second domino,
    both measured relative to the dominoes’ respective rotation axes. Because the dominoes
    are identical and have the same angular velocity, they have the same angular momentum.
    This means each domino has half as much angular momentum about its rotation axis as the
    first domino had just before the collision. (Note that forces from the table cannot change
    the angular momentum of the dominoes about their respective rotational axes because such
    forces have zero moment arm, so only the inter-domino forces need to be examined here.)
    Kinetic energy scales with the square of angular momentum, so each domino has a quarter
    as much kinetic energy after the collision as the first domino had before it. That means the
    total kinetic energy after the collision is half what it was before the collision. The fraction of
    kinetic energy lost is one half.

                 Copyright c 2017 American Association of Physics Teachers

2017 USA Physics Olympiad Exam               Part B                                               17

  c. After the collision, the dominoes rotate in such a way so that they always remain in contact.
     Assume that there is no friction between the dominoes and the first domino was given the
     smallest possible push such that it toppled. What is the minimum l such that the second
     domino will topple?
    You may work to lowest nontrivial order in the angles through which the dominoes have
    rotated. Equivalently, you may approximate t, l  h.

                                              Solution

    As in part (a), we will find the highest potential energy of the system and make sure the
    initial kinetic energy is high enough to get the system to that point.
    At any time after the collision, let us call the angle that the first domino has rotated past its
    point of highest potential energy α, and the angle that the second domino has rotated past
    its point of highest potential energy β. Also, let’s call the angle a domino rotates from its
    standing position up to its point of highest potential energy φ, so the dominoes have rotated
    φ + α and φ + β respectively.
    The dominoes need to be in contact. The top right corner of the first domino has moved
    horizontally a distance h sin(φ + α) which we will approximate as h(φ + α) using the small
    angle approximation. The top left corner of the second domino moves horizontally forward
    by h sin(φ + β) + t(1 − cos(φ + β)). We ignore the cosine term as second order in the rotation
    angle and approximate this as h(φ + β).
    The y-coordinate of the upper right corner of the first domino is h(1 − cos(φ + α)) ≈ h. The
    y-coordinate of the upper left corner of the second domino is h(1 − cos(φ + β)) + t sin(φ + β) ≈
    h + t(φ + β) There is a first-order difference in y-coordinates of the two corners, but this
    means the difference in x coordinate between the top left corner of the second domino and
    the top right corner of the first domino is second-order in the rotation angles. We conclude
    that to first order
                                                                   l
                              h(φ + α) = l + h(φ + β) ⇒ α = + β
                                                                   h
    because this condition puts the top right corner of the first domino at the same position as
    the top left of the second domino.
    The potential energy of the first domino, setting zero potential energy to be when the domino
    is upright, is U1 = ∆U (1 − (α/φ)2 ) to second order in α, and for the second domino, U2 =
    ∆U (1 − (β/φ)2 ). These figures come from fitting a quadratic whose peak is when the center
    of mass is above the rotation point and which is zero when the domino is upright. The total
    potential energy is U1 + U2 , and using the relation between α and β, it is minimized for
                                                         l
                                             βmax = −      .
                                                        2h
    In other words, the second domino is as far away from rotating to the top of arc as the first
    domino has rotated past the top of its arc. This gives a maximum potential energy

                                                           l2

                                     Umax = 2∆U 1 − 2
                                                          4t

    where we have used the approximation φ ≈ t/h.

                  Copyright c 2017 American Association of Physics Teachers

2017 USA Physics Olympiad Exam                Part B                                             18

     At impact, Uimpact ≈ ∆U (−l2 /t2 + 2l/t). Before impact, the kinetic energy is ∆U − Uimpact
     because the maximum potential energy before impact was ∆U , and the kinetic energy was
     zero there. The kinetic energy just after the collision is then (∆U − Uimpact )/2. Setting this
     equal to the potential energy gain as the two dominoes rotate to their highest potential energy
     Umax ,
                                   1
                                     (∆U − Uimpact ) = Umax − Uimpact
                                   2
     or
                                        1                1
                                          ∆U = Umax − Uimpact .
                                        2                2
     Plugging in the earlier expressions for all these gives

                                                     l2            l2

                               1                               l
                                 ∆U = 2∆U 1 − 2 − ∆U             − 2
                               2                    4t         t 2t

     and solving this yields
                                                   3
                                                 l= t
                                                   2
     to first order in t.

  d. A row of toppling dominoes can be considered to have a propagation speed of the length l + t
     divided by the time between successive collisions. When the first domino is given a minimal
     push just large enough to topple and start a chain reaction of toppling dominoes, the speed
     increases with each domino, but approaches an asymptotic speed v.

     Suppose there is a row of dominoes on another planet. These dominoes have the same density
     as the dominoes previously considered, but are twice as tall, wide, and thick, and placed with
     a spacing of 2l between them. If this row of dominoes topples with the same asymptotic
     speed v previously found, what is the local gravitational acceleration on this planet?

                                              Solution

     This part is independent of the others and requires only dimensional analysis. The speed v
     can depend on g, h, w, l. To get a quantity with dimensions [LT −1 ], we must take
                                                   p
                                             v = c gL

     where L is some length made from h, w, and l. On the new planet, v is the same and L is
     twice as much, so g must be half as great on the new planet.

                   Copyright c 2017 American Association of Physics Teachers

2017 USA Physics Olympiad Exam                       Part B                                                     19

USAPhO 2017 Question B1


[User]
\begin{wrapfigure}{r}{0.35\textwidth}
	\vspace{-1.8ex}
	\includegraphics[width=\linewidth]{2010-lahg-09-Red_eyed_tree_frog}
	\vspace{-4ex}
\end{wrapfigure}
Väike puukonn suudab ronida mööda seinu ja lagesid, luues enda ja seina vahele
seinaga risti oleva tõmbejõu (nt iminappade tekitatud vaakumiga) ning vältides libisemist
selle tagajärjel tekkiva hõõrdejõu abil.
Millise nurga all maapinna suhtes peab olema sein, et tal oleks end kõige raskem paigal hoida
(mil libisemise vältimiseks vajalik seinaga risti olev jõud on maksimaalne)?
Hõõrdetegur seina ja konna vahel on $\mu$.

Solution:
\begin{wrapfigure}{r}{0.5\textwidth}
	\vspace{-4ex}
	\includegraphics[width=65mm]{2010-lahg-09-lah}
	\vspace{-6ex}
\end{wrapfigure}
Konnale mõjub kolm jõudu: raskusjõud $m\vec g = \vec {BC}$, mis on suunatud vertikaalselt alla, jõud $\vec F =\vec {CA^\prime}$,
mis on suunatud kaldpinna pinnanormaali sihis (pinna sisse),
ning rõhumisjõu ja hõõrdejõu resultant $\vec {A^\prime B}$, mille nurk pinnanormaaliga $\alpha=\angle CA^\prime B$ ei ületa väärtust
$\arctan \mu$. Need kolm vektorit moodustavad tasakaalu korral
joonisel toodud kolmnurga $A^\prime BC$. Jooniselt on ilmne, et antud kaldenurga
($A^\prime C$ sihi) puhul saab konn minimeerida vajalikku jõudu (st lõigu $A^\prime C$ pikkust)
suurendades nurga $\alpha$ maksimaalse võimaliku väärtuseni $\alpha = \arctan \mu$ (mil $A^\prime = A$). Kui nüüd muuta pinna kaldenurka,
siis joonistab punkt $A$ ringjoone kaare (sest punktid $B$ ja $C$ on fikseeritud ning $\angle BAC$ on konstantne ($\arctan \mu$).
Vajalik jõud (lõik $AC$) on maksimaalne, kui lõik $AC$ on ringi diameetriks $OC$, st pinnanormaal moodustab horisondiga
nurga $\arctan\mu$ (sest $\angle OBC=90^\circ$). Seega, sein on vertikaali suhtes kaldus $\arctan \mu$ võrra, moodustades põrandaga teravnurga.

\vspace{0.5\baselineskip}

{\em Alternatiivne lahendus}\\
Olgu iminapa tekitatud tõmbejõud $F$, hõõrdejõud $F_h$ ja normaaljõud $N$. Jõudude tasakaalu tingimusest saame
$F_{h}=mg\sin\theta$ ja $F=N+mg\cos\theta$. Kui konn rakendab minimaalset tarvilikku jõudu, siis

$$F_{h}=\mu N=mg\sin\theta,$$

ehk

$$N=\frac{mg\sin\theta}{\mu}.$$

Niisiis,

$$F=N+mg\cos\theta=\frac{mg\sin\theta}{\mu}+mg\cos\theta=mg\left( \frac{\sin\theta}{\mu}+\cos\theta\right).$$

Nüüd on vaja leida, millise $\theta$ korral on $F$ maksimaalne. Selleks katsume siinuse ja koosinuse summa avaldada ühe siinusena.

Otsime A ja B nii, et kehtiks võrdus
\[
A\sin (\theta+B)=\frac{\sin\theta}{\mu}+\cos\theta.
\]
Kuna
\[
A\sin (\theta+B)=A\sin\theta\cos B + A\sin B \cos\theta,
\]
siis
$A\sin B=1$ ja $A\cos B=\frac{1}{\mu}$. Siit saame $\tan B=\mu$ ja
\[
A=\sqrt{\frac{1}{{\mu}^2} + 1}=\frac{1}{\mu}\sqrt{1+{\mu}^2}.
\]
Seega
\[
F=\frac{mg}{\mu}\sqrt{1+{\mu}^2}\sin(\theta + \arctan\mu).
\]
Siinuse suurim võimalik väärtus on $\sin(\ang{90})=1$, ehk $\theta + \arctan\mu=\ang{90}$. Seega on otsitav nurk $\theta = \ang{90}-\arctan\mu$.

Konn

\begin{wrapfigure}{r}{0.35\textwidth}
	\vspace{-1.8ex}
	\includegraphics[width=\linewidth]{2010-lahg-09-Red_eyed_tree_frog}
	\vspace{-4ex}
\end{wrapfigure}
Väike puukonn suudab ronida mööda seinu ja lagesid, luues enda ja seina vahele
seinaga risti oleva tõmbejõu (nt iminappade tekitatud vaakumiga) ning vältides libisemist
selle tagajärjel tekkiva hõõrdejõu abil.
Millise nurga all maapinna suhtes peab olema sein, et tal oleks end kõige raskem paigal hoida
(mil libisemise vältimiseks vajalik seinaga risti olev jõud on maksimaalne)?
Hõõrdetegur seina ja konna vahel on $\mu$.

Konnale mõjub kolm jõudu: raskusjõud, rõhumisjõu ja hõõrdejõu resultant ning iminapa tekitatud tõmbejõud. Rõhumisjõu ja hõõrdejõu resultandi nurk pinnanormaali suhtes on kriitilisel juhul $\arctan \mu$. Tasakaalu korral peavad antud jõud üksteist tasakaalustama, ehk moodustuma kolmnurga.


[User]
The left face of a thin lens is flat while the right face is concave and silvered. How big is the focal length $f$ of such an optical element for a light falling from the left? The radius of the lens’s concave part is $R$, the refractive index of the lens with respect to the surrounding environment is $n$.
\begin{center}
\begin{tikzpicture}[scale=0.7, transform shape]
\def\tmp{65.92}
\fill [black] (0.7,2) arc[start angle=90+\tmp, end angle=270-\tmp,radius=4.9] -- (0.8,-2) -- (0.8,-2) arc[start angle=270-\tmp, end angle=90+\tmp,radius=4.9] -- (0.7,2);
\draw (0.8,-2) -- (0,-2) -- (0,2) -- (0.8,2);
\end{tikzpicture}
\end{center}

Solution:
Langegu läätsele vasakult valguskiir, mille suund on paralleelne optilise teljega ning mille kaugus sellest on $d$. Läätse sisenedes valgus oma levimissuunda ei muuda, kuna langeb läätsele selle pinnaga risti. Hõbetatud pind toimib kumerpeeglina, mille fookuskaugus on $\frac{R}{2}$. Kui peegeldunud valguskiire nurk optilise peatelje suhtes on $\alpha$, siis läätsest väljunud kiire ning optilise peatelje vaheline nurk on murdumisseaduse põhjal $\sin\gamma=n\sin\alpha$. Kuna lääts on õhuke, asetsevad punktid, kus valguskiir peegeldus ning kus see murdus, üksteisele väga lähedal. Seetõttu võime kirjutada
\[ d=\frac{R}{2}\tan\alpha=f\tan\gamma. \]
Õhukese läätse korral on selle nõgusa osa kõverusraadius oluliselt suurem fookuskaugusest, mis lubab kasutada väikeste nurkade lähendust ehk $\alpha\approx\sin\alpha\approx\tan\alpha$ ning $\gamma\approx\sin\gamma\approx\tan\gamma$. Tulemuseks saame
\[ f=\frac{R}{2n}. \]
Tasub teada, et avaldis sfäärilise peegli fookuskauguse jaoks kehtibki ainult väikeste peegeldumisnurkade korral. Suurte peegeldumisnurkade korral ilmneb sfääriline aberratsioon ning ei ole võimalik rääkida ühest fookusest.

\begin{center}
\begin{tikzpicture}[scale=1]
\def\tmp{65.92}
\fill [gray] (0.7,2) arc (90+\tmp:270-\tmp:4.9) -- (0.8,-2) -- (0.8,-2) arc (270-\tmp:90+\tmp:4.9) -- (0.7,2);
\draw[thick] (0.8,-2) -- (0,-2) -- (0,2) -- (0.8,2);
\draw[dashed] (-1.5,0) -- (3,0);
\draw[->,thick] (-1.5,1) -- (-0.6,1);
\draw[thick] (-0.6,1) -- (0.38,1);
\draw[dashed] (0.38,1) -- (2.73,0);
\draw[->,thick] (0.38,1) -- (0.1,1.12);
\draw[thick] (0.1,1.12) -- (0,1.16);
\draw[->,thick] (0,1.16) -- (-1,2.16);
\draw[thick] (-1,2.16) -- (-1.5,2.66);
\draw[dashed] (0,1.16) -- (1.16,0);
\draw[<->] (-1,0) -- (-1,1);
\node [left] at (-1,0.5) {d};
\draw[dashed] (-1.5,1.16) -- (3,1.16);
\node [left] at (-0.25,1.3) {$\gamma$};
\draw (-0.7,1.16) arc (180:135:0.7);
\node [right] at (0.55,1) {$\alpha$};
\draw (1,1.16) arc (0:-24:1);
\draw[dashed] (1.16,0) -- (1.16,-0.2);
\draw[<->] (0,-0.2) -- (1.16,-0.2);
\node [below] at (0.58,-0.2) {f};
\draw[dashed] (2.73,0) -- (2.73,-0.7);
\draw[<->] (0,-0.7) -- (2.73,-0.7);
\node[below] at (1.36,-0.7) {$\frac{\text{R}}{2}$};
\node[left] at (2.4,0.1) {$\alpha$};
\draw (2,0) arc (180:156:0.7);
\node[left] at (1,0.1) {$\gamma$};
\draw (0.65,0) arc (180:135:0.5);
\end{tikzpicture}
\end{center}

Fookuskaugus

Õhukese tasanõgusa läätse parem külg on hõbetatud. Kui suur on sellise optilise elemendi fookuskaugus $f$ vasakult langeva valguse jaoks? Läätse nõgusa osa kõverusraadius on $R$, läätse materjali murdumisnäitaja ümbritseva keskkonna suhtes on $n$.

\begin{center}
\begin{tikzpicture}[scale=0.7, transform shape]
\def\tmp{65.92}
\fill [black] (0.7,2) arc[start angle=90+\tmp, end angle=270-\tmp,radius=4.9] -- (0.8,-2) -- (0.8,-2) arc[start angle=270-\tmp, end angle=90+\tmp,radius=4.9] -- (0.7,2);
\draw (0.8,-2) -- (0,-2) -- (0,2) -- (0.8,2);
\end{tikzpicture}
\end{center}

Läätse fookuskauguse leidmiseks tasub vaadelda valguskimpu, mille suund on paralleelne optilise peateljega. Sellisel juhul tekitavad kiired näilise kujutise läätse fookusesse.


[User]
A particle in a uniform vertical gravitational field is constrained to move on a curve
    y(x). If y(x) is a circular arc, then this system is just a simple pendulum, and we know that its
    period is not perfectly independent of its amplitude. Find a differential equation relating dy/dx
    and y, so that the period of oscillation is exactly 2π/ω0 for a fixed parameter ω0 , independent of
    amplitude. (Hint: work in terms of the coordinate s, the arc length along the curve.)

[3] Problem 8 (Cahn). A particle of mass M is constrained to move on a frictionless horizontal plane.
    A second particle of mass m is constrained to a vertical line. The two particles are connected by a
    massless string which passes through a hole in the plane.

    The system is set up so that the mass M moves in a circle of radius r, while the mass m remains still.
    Show that this motion is stable with respect to small changes in r, and find the angular frequency

                   01@
    of small oscillations.

Solution:
The reason s is useful is because

                                                K = mṡ2

    exactly. So, we would get motion with fixed period 2π/ω0 if the potential energy had the form

                                               V = ω02 ms2 .

    On the other hand, since the system is in a uniform gravitational field, V = mgy, so we need

                                                       ω02 2
                                                  y=      s .
                                                       2g
    Taking the derivative with respect to x of both sides, we have
                                                  s
                                 dy    ω02 ds       2ω02 y p
                                    =     s     =           1 + (dy/dx)2 .
                                 dx     g dx           g

    Solving this for dy/dx gives
                                           dy              y
                                                  r
                                              =                     .
                                           dx         (g/2ω02 ) − y

    As a check, when y is small, this equation is approximately solved by y(x) = ω02 x2 /(2g), which
    in turn is approximately an arc of a circle of radius L, where ω02 = g/L. Solving the differential
    equation exactly is a bit nasty, but it turns out to be a cycloid. This fact was first discovered by
    Huygens, who invented the cycloidal pendulum for accurate timekeeping.

[3] Problem 8 (Cahn). A particle of mass M is constrained to move on a frictionless horizontal plane.
    A second particle of mass m is constrained to a vertical line. The two particles are connected by a
    massless string which passes through a hole in the plane.

    The system is set up so that the mass M moves in a circle of radius r, while the mass m remains still.
    Show that this motion is stable with respect to small changes in r, and find the angular frequency
    of small oscillations.

    Solution. In equilibrium we have
                                                    M v2     L2
                                             mg =         =
                                                      r     M r3
    where r is the radius of the circle, and L is the conserved angular momentum. If the hanging mass
    goes downward, then r decreases, so the tension in the string increases providing a restoring force;
    hence the orbit is stable.
       To find the angular frequency of small oscillations, we’ll use the energy method, with generalized
    coordinate r. The “kinetic energy”, which is the part of the energy dependent on ṙ, is
                                                1       1
                                             T = M ṙ2 + mṙ2 .
                                                2       2
    The “potential energy”, which is the part of the energy dependent on r, is

                                                               L2
                                             V = mgr +              .
                                                              2M r2
    Note that L2 /2M r2 is treated as potential energy here even though it is associated with the motion
    of the large mass. From the kinetic energy, we see the “effective mass” is meff = M + m, as one
    might expect. From the potential energy, we see the “effective spring constant” is

                                                          3L2    3mg
                                          keff = V ′′ =      4
                                                               =     .
                                                          Mr      R
    Since keff /meff is positive, the motion is stable, and the angular frequency is
                                              s          r r
                                                 keff      3g     m
                                         ω=           =                .
                                                 meff      R m+M


[User]
\begin{wrapfigure}{r}{0.4\linewidth}
\includegraphics[width=\linewidth]{2012-lahg-09-n88r_ipe}
\end{wrapfigure}
Two plates make up a V-shaped horizontal groove. Both of the plates are at an angle $\theta$ with respect to the ground. In the groove there is a section of rope of length $L$ and of even mass distribution. All of the rope is located on the plane perpendicular to the groove so that both of the plates are touching the same amount of the rope. Above the bottom of the groove a part of the rope of length $fL$ does not anymore rest on the plates. Find $f$ if the rope is on the verge of slipping. The coefficient of friction between the rope and the plates is $\mu = 1$.

Solution:
Olgu kogu nööri mass $m$. Vaatleme nööri paremat poolt. Nööri pinge vertikaalkomponent nööri ja plaadi kokkupuutepunktis, $T \sin \theta$, peab tasakaalustama õhus rippuva osa kaalu $\frac{f}{2}mg$. Nüüd vaatleme plaadil lebavat osa, selle mass on $\frac{1-f}{2}m$. Seega nii toereaktsioon kui ka maksimaalne hõõrdejõud (kuna $\mu = 1$) on $\frac{1-f}{2}mg \cos \theta$. Hõõrdejõud peab olema tasakaalus raskusjõu komponendiga piki nööri, $\frac{1-f}{2}mg \sin \theta$, ning pingega $T=\frac{f m g}{2 \sin \theta}$: $$\frac{f m g}{2 \sin \theta} + \frac{1-f}{2}mg \sin \theta = \frac{1-f}{2}mg \cos \theta.$$
Siit saame avaldada vastuse $$f = \frac{1-\tan \theta}{1+\tan \theta} \tan \theta. $$

Nöör rennis

\begin{wrapfigure}{r}{0.4\linewidth}
\includegraphics[width=\linewidth]{2012-lahg-09-n88r_ipe}
\end{wrapfigure}
Kaks plaati moodustavad V-kujulise horisontaalse renni. Mõlemad plaadid on
horisontaaltasapinna suhtes nurga $\theta$ all.
Rennis on jupp ühtlase
massijaotusega nööri pikkusega $L$, mis asub tervikuna renniga ristuvas tasandis
nii, et mõlema plaadiga puutub kokku sama palju nööri.
Renni põhja kohal ei toetu nöör enam pikkuse $fL$
ulatuses plaatidele. Leidke $f$, kui nöör on libisemise piiril. Hõõrdetegur nööri ja plaatide
vahel on $\mu = 1$.

Nöörile mõjuvad kolm jõudu: raskusjõud, nööri toereaktsioon ning hõõrdejõud. Selleks, et tasakaalutingimusi kirja panna on mugav vaadelda õhus rippuvat osa ja plaadil lebavat osa eraldi. Seejuures peab arvestama ka nööri ja plaadi kokkupuutepunktis mõjuva nööri pingega.


[User]
The energy of a photon is E = |p|c.

     (a) Treating the photon classically, compute the average energy of a photon moving in one
         dimension, assuming it obeys the Boltzmann distribution with temperature T . Check that
         the result obeys the equipartition theorem.

     (b) Generalize this result to three dimensions to compute γ for a 3D photon gas.

Solution:
Since we’re talking about both momenta p and probabilities p here, let’s temporarily
   rename the probability distribution function to f .

     (a) Using the Boltzmann distribution and E = pc,

                                                  f (p) ∝ e−pc/kB T .

         Then the expected energy is
                                          R∞       −pc/kB T dp
                                                                     R ∞ −x
                                           0R (pc)e                       xe dx
                             ⟨E⟩ = ⟨pc⟩ =     ∞ −pc/kB T       = kB T R0 ∞ −x    .
                                             0 e          dp            0 e   dx

         Note that the numerator can be related to the denominator using integration by parts,
                                        Z ∞             Z ∞
                                               −x
                                             xe dx =        e−x dx.
                                              0                0

         Then the fraction is just 1, so
                                                    ⟨E⟩ = kB T.
         Evidently, every degree of freedom where the energy is linear in the momentum contributes
         energy kB T . This is just as we expect by the equipartition theorem, since there’s one degree
         of freedom with n = 1.

     (b) The reasoning is very similar, but now

                                                 f (p) ∝ e−|p|c/kB T .

         The expected energy is then, using spherical coordinates,
                                       R∞        −pc/kB T (4πp2 ) dp
                                                                            R ∞ 3 −x
                                           (pc)e                               x e dx
                       ⟨E⟩ = ⟨|p|c⟩ = 0R ∞ −pc/k T           2
                                                                     = kB T R0∞ 2 −x .
                                          0 e          (4πp ) dp             0 x e   dx
                                                    B

         Again the numerator can be related to the denominator using integration by parts,
                                     Z ∞                Z ∞
                                          x3 e−x dx = 3     x2 e−x dx.
                                         0                      0

         Then the fraction is 3, so ⟨pc⟩ = 3kB T , which is just what we expected since there are three
         linear degrees of freedom. Hence we have γ = 4/3.


[User] Kolm ühesuguse raadiusega kuuli $A$, $B$ ja $C$ on ühendatud kergete varraste abil võrdkülgseks kolmnurgaks $ABC$, mis lebab siledal (kuid nullist erineva hõõrdeteguriga) horisontaalpinnal. Kuuli $C$ lükatakse hästi aeglaselt nii, et selle kiirusvektor on kogu aeg risti sirgega $AC$. Kui kuul $A$ on piisavalt raske (st masside suhe $M_A/M_B$ on piisavalt suur), siis jääb kuul $A$ paigale. Millise suhte $M_A/M_B$ puhul hakkab kuul $A$ libisema?

Solution:
Et kuuli $C$ kiirusvektor on risti sirgega $AC$, siis hetkeline pöörlemiskese $O$ asub sellel sirgel. Seega, kui kuul $A$ hakkab libisema, siis on selle kiirusvektor samuti (ning järelikult ka hõõrdejõu vektor) risti sirgega $AC$. Piirjuhtumil, kui kuulide masside suhe on selline, et kuul $A$ hakkab vaevu liikuma, on punkt $O$ väga lähedal punktile $A$ ning seega on punkti $B$ kiirusvektor (ja hõõrdejõu suund) risti sirgega $AB$. Hõõrdejõudude jõumomentide summa punkti $C$ suhtes peab olema null; et punkti $A$ rakendatud hõõrdejõu õlg on $|AC|$ ning punkti $B$ puhul on õlg $|AC|/2$, saame kriitiliseks masside suhteks $1/2$, st kuul $A$ jääb paigale, kui $M_A/M_B > 1/2$.

\emph{Märkus}. Uurides olukorda edasi ja vaadeldes üha vähenevaid $M_A$ väärtusi (alustades $M_B/2$-st ja lõpetades tühiselt väikeste massidega) paneme tähele, et punkt $O$ nihkub piki sirget $AC$, alustades punkti $A$ juurest, üha kaugemale punktidest $A$ ja $C$ ning läheneb piiril $M_A \rightarrow 0$ punktile $D$, mis asub punktist $A$ kaugusel $|AC|$ --- nõnda, et kolmnurk $BCD$ on täisnurkne ning seetõttu punkti $B$ rakendatud hõõrdejõu õlg läheneb nullile.

Kuulid

Kolm ühesuguse raadiusega kuuli $A$, $B$ ja $C$ on ühendatud kergete varraste abil võrdkülgseks kolmnurgaks $ABC$, mis lebab siledal (kuid nullist erineva hõõrdeteguriga) horisontaalpinnal. Kuuli $C$ lükatakse hästi aeglaselt nii, et selle kiirusvektor on kogu aeg risti sirgega $AC$. Kui kuul $A$ on piisavalt raske (st masside suhe $M_A/M_B$ on piisavalt suur), siis jääb kuul $A$ paigale. Millise suhte $M_A/M_B$ puhul hakkab kuul $A$ libisema?

Et kuuli $C$ kiirusvektor on risti sirgega $AC$, siis hetkeline pöörlemiskese asub sellel sirgel. Seega, kui kuul $A$ hakkab libisema, siis on selle kiirusvektor samuti (ning järelikult ka hõõrdejõu vektor) risti sirgega $AC$. Piirjuhtumil, kui kuulide masside suhe on selline, et kuul $A$ hakkab vaevu liikuma, on hetkeline pöörlemiskese väga lähedal punktile $A$.