[User]
Question A2
For this problem, graphical answers should be drawn on the answer sheets graphs
provided. Supporting work is to be written on blank answer sheets. Incorrect graphs
without supporting work will receive no partial credit.
   The current I as a function of voltage V for a certain electrical device is

                                 I = I0 e−qV0 /kB T eqV /kB T − 1

where q is the magnitude of the charge on an electron, kB is Boltzmann’s constant, and T is the
absolute temperature. I0 and V0 are non-zero positive constants. Throughout this problem assume
low temperature values kB T  qV0 .

  a. On the answer sheets, sketch a graph of the current versus voltage for low temperature values
     kB T  qV0 , clearly indicating any asymptotic behavior.

    Shown is a schematic for the device. Positive voltage means that the electric potential of the
left hand side of the device is higher than the right hand side. For this device, I0 = 25 µA and
V0 = 1.0 V.

                                          VL            VR

   Below is a circuit made up of these elements. The voltage supplied the circuit is sinusoidal,
VAB = VA − VB = Vs sin ωt, and is also shown on answer sheets. The resistance is R = 5.0 Ω and
Vs = 5.0 V.

                                     VA

                                            VC               VD
                                                    R

                                    VB

  b. Sketch the potential difference VCD = VC − VD as a function of time on the answer sheet. For
     your convenience, VAB is shown in light gray. Assume that VAB has been running for a long
     time.

   A capacitor is connected to the circuit as shown below. The capacitance is C = 50 mF.

                                     VA
                                                    C

                                            VC               VD
                                                    R

                                    VB

  c. Sketch the new potential difference VCD = VC − VD as a function of time on the answer sheet.
     For your convenience, VAB is shown in light gray. Assume that VAB has been running for a
     long time.

                  Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam                Part A                                               5

Solution:
Question A2
  c. Sketch the potential difference VCD = VC − VD as a function of time. For your convenience,
     VAB is shown in light gray. Assume that VAB has been running for a long time. There is a
     capacitor in this circuit!

                    6

                    5

                    4

                    3

                    2

                    1

     Voltage (V)
                    0

                   −1

                   −2

                   −3

                   −4

                   −5

                   −6
                        0    10     20    30    40    50      60     70   80    90      100   110   120
                                                           Time (ms)

                            Copyright c 2018 American Association of Physics Teachers

USAPhO 2018 Question A2


[User]
3. Consider the Bohr model of the hydrogen atom. Let me and e be the mass and magnitude of the
   charge of the electron respectively. Let a0 be the ground state radius (Bohr radius).
    (a) Obtain an expression for the ionisation energy IH of the ground state of the hydrogen atom     [1]
        in terms of a0 and constants.

          IH =

    (b) Consider a singly ionised helium atom He+ . Obtain the ground state ionisation energy IHe+     [2]
        of He+ in terms of IH .

INPhO 2020 Page 3           Questions & Summary Answers

         IHe+ =

    (c) Now consider a two electron system with arbitrary atomic number Z. Use the Bohr model                [1]
        to obtain the ground state radius (r (Z )) in terms of a0 and Z. Assume the two electrons
        are in the same circular orbit and as far apart as possible.

         r (Z ) =

    (d) Derive an expression for the first ionisation energy IZth for two electron system with arbitrary     [3]
        Z in terms of Z and IH .

         IZ =                                             0
                                       C2
    (e) The table below contains the experimental data of IZexpt (in units of Rydberg where 1 Ryd =
        13.6 eV) versus Z for various two-electron systems.
                                                                                [8]

                 HB                       H−
                                          He
                                          Li+
                                          Be++
                                          B3 +
                                          C4+
                                          N5+
                                                   Z
                                                   1
                                                   2
                                                   3
                                                   4
                                                   5
                                                   6
                                                   7
                                                       IZexpt
                                                       0.055
                                                       1.81
                                                       5.56
                                                       11.32
                                                       19.07
                                                       28.83
                                                       40.60
                                          O6+      8   54.37
                                          F7+      9   70.15

        Experimental values were not found to be equal to the theoretical predictions. This difference
        arises mainly from non-inclusion of Pauli’s principle in the theoretical derivation of part (d).
        It was suggested that if the value of Z was reduced by some fixed amount α (Z ∗ = Z − α)
        in the final expression of IZth obtained in part (d), then IZth∗ ≈ IZexpt . Draw a suitable linear
        plot and from the graph find α. Two graph papers are provided with this booklet in case
        you make a mistake.

         α=

                           Detailed answers can be found on page numbers:

                           Detailed answers can be found on page numbers:

INPhO 2020 Page 4             Questions & Summary Answers               Last four digits of Roll No.:

Solution:
3. Consider the Bohr model of the hydrogen atom. Let me and e be the mass and magnitude of the
   charge of the electron respectively. Let a0 be the ground state radius (Bohr radius).
    (a) Obtain an expression for the ionisation energy IH of the ground state of the hydrogen atom       [1]
        in terms of a0 and constants.

         IH =

INPhO 2020 Page 5          Questions & Summary Answers

          Solution: Centripetal acceleration is given by Coulomb force.

                                     me v 2  Ke2         1             –
                                            = 2 ; K=         ; me vr = h                     (3.1)
                                      a0      a0        4π0
                                               –2
                                               h
                                        a0 =                                                 (3.2)
                                             Kme e2
                                             me v 2 Ke2    Ke2
                              Total energy =       −    =−                                   (3.3)
                                               2     a0     2a0
                                             Ke  2
                                        IH =                                                 (3.4)
                                             2a0

    (b) Consider a singly ionised helium atom He+ . Obtain the ground state ionisation energy IHe+   [2]
        of He+ in terms of IH .

         IHe+ =

          Solution:
                                                                0
                                     C2
                                    me v 2
                                      r
                                         r=
                                           =
                                             2Ke2

                                             1
                                               r2
                                                –
                                                h2
                                             2Kme
                                                   and me vr = h

                                                   2

                      Total Energy (T.E.) = me v 2 −
                                                     =
                                                       a0
                                                       2
                                                      2Ke 2
                                                            =−
                                                               –

                                                                Ke2
                                                                    =
                                                                      2Ke2
                                                                           = −4IH
                                                                                (3.5)

                                                                                (3.6)

                                                                                (3.7)

                HB                           2
                                     IHe+ = 4IH
                                                       r         r     a0

    (c) Now consider a two electron system with arbitrary atomic number Z. Use the Bohr model
                                                                                (3.8)

        to obtain the ground state radius (r (Z )) in terms of a0 and Z. Assume the two electrons
        are in the same circular orbit and as far apart as possible.
                                                                                [1]

         r (Z ) =

          Solution:
                                 me v 2   KZe2    Ke2
                                        =      −        and me vr = –
                                                                    h                        (3.9)
                                  r        r2    (2r )2

                                                        1
                                                                                –
                                                                                h2
                                me v 2 r = Ke2 Z −              and me v 2 r =              (3.10)
                                                        4                            me r
                                                 –2
                                                 h                       a0
                                     r=                         =                       (3.11)
                                          Kme e2       Z − 14           Z − 14

INPhO 2020 Page 6           Questions & Summary Answers                     Last four digits of Roll No.:
    (d) Derive an expression for the first ionisation energy IZth for two electron system with arbitrary      [3]
        Z in terms of Z and IH .

         IZ =

          Solution:

                                                  Ke2 Z − 41
             Kinetic Energy (K.E.) = me v 2 =                                                        (3.12)
                                                           r
                                                               2
                                                  Ke2 Z − 14
                                              =                                                      (3.13)
                                                       a0
                                                         1 2

                                             =2 Z−            IH                                     (3.14)
                                                         4
                                                 2KZe2 Ke2
                     Potential Energy (P.E.) = −           +                                         (3.15)
                                                     r          2r
                                             =    0
                                               −2Ke2
                                                  r

                                                           Z−
                                                                 1
                                                                 4

                                                                                (3.16)

                               C2            =
                                               −2Ke
                                                  a0

                                             = −4 Z −

                                     (T.E.)i = −2 Z −
                                                       2
                                                           Z−

                                                           1
                                                           4
                                                           1 2
                                                           4
                                                             2

                                                                 1 2
                                                                 4

                                                                IH

                                                                IH

                                                                                (3.17)

                                                                                (3.18)

                                                                                (3.19)

         HB                          (T.E.)f = −Z 2 IH

                                          IZth = (T.E)f − (T.E)i = 2 Z −

                                          IZth = Z 2 − Z +
                                                                1
                                                                8
                                                                  IH

                                                                              1
                                                                              4
                                                                                2
                                                                                IH − Z 2 IH
                                                                                (3.20)

                                                                                (3.21)

                                                                                (3.22)

    (e) The table below contains the experimental data of IZexpt (in units of Rydberg where 1 Ryd =           [8]
        13.6 eV) versus Z for various two-electron systems.

                                                   Z   IZexpt
                                          H−       1   0.055
                                          He       2   1.81
                                          Li+      3   5.56
                                          Be++     4   11.32
                                          B3 +     5   19.07
                                          C4 +     6   28.83
                                          N5 +     7   40.60
                                          O6+      8   54.37
                                          F7 +     9   70.15

        Experimental values were not found to be equal to the theoretical predictions. This difference
        arises mainly from non-inclusion of Pauli’s principle in the theoretical derivation of part (d).
        It was suggested that if the value of Z was reduced by some fixed amount α (Z ∗ = Z − α)

INPhO 2020 Page 7          Questions & Summary Answers
       in the final expression of IZth obtained in part (d), then IZth∗ ≈ IZexpt . Draw a suitable linear
       plot and from the graph find α. Two graph papers are provided with this booklet in case
       you make a mistake.

        α=

         Solution:
                                                                   1                          1

            ∆IZ = IZth − IZexpt = IZth − IZth∗ =       Z2 − Z +      − (Z − α )2 − (Z − α ) +          IH
                                                                   8                          8
                                                                                (3.23)
                     h                 i
            ∆IZ = −α2 + 2Zα − α IH                                                                 (3.24)

                                            Z      (IZexpt )    IZth       (IZth − IZexpt )
                                   H −
                                            1       0.055      0.125            0.07
                                   He       2        1.81       2.13            0.32
                                   Li+      3        5.56      6.13             0.57
                                  Be++      4       11.32      12.13            0.81
                                  B3 +      5       19.07      20.13            1.06
                                  C4 +
                                  N5+
                                  O6+
                                            6
                                            7
                                            8
                                                    28.83
                                                    40.60
                                                    54.37
                                                                0
                                                               30.13
                                                               42.13
                                                               56.13
                                                                                1.30
                                                                                1.53
                                                                                1.76
                                   F7+

                                         C2 9
         A plot of ∆IZ vs Z is a linear graph.
         Slope of the graph =2α = 0.24
         α = 0.12
                                                    70.15

         One can also linearize in the following way:
                                                               72.13            1.98

               HBq
                                   r

                                            IZexpt = IZth∗ = Z − α −

                                       IZexpt +
                                                  1
                                                  8
                                                    = Z −α−
                                                            1
                                                            2
                                                                                1
                                                                                2
                                                                                2
                                                                                −

         Plot of IZexpt + 18 vs Z is a linear graph. This method gives α = 0.10. Both methods
         are acceptable.
                                                                                1
                                                                                8
                                                                                (3.25)

                                                                                (3.26)

         Accepted range of α is : 0.10 ≤ α ≤ 0.14.
         Graph is on the last page of this booklet.

                          Detailed answers can be found on page numbers:

INPhO 2020 Page 8                  Questions & Summary Answers                Last four digits of Roll No.:

INPhO 2020 Problem 3


[User]
Question B3
Radiation pressure from the sun is responsible for cleaning out the inner solar system of small
particles.

  a. The force of radiation on a spherical particle of radius r is given by

                                                F = P Qπr2

     where P is the radiation pressure and Q is a dimensionless quality factor that depends on
     the relative size of the particle r and the wavelength of light λ. Throughout this problem
     assume that the sun emits a single wavelength λmax ; unless told otherwise, leave your answers
     in terms of symbolic variables.

        i. Given that the total power radiated from the sun is given by L , find an expression for
           the radiation pressure a distance R from the sun.
                                                                                Fradiation
       ii. Assuming that the particle has a density ρ, derive an expression for the ratio
                                                                                Fgravity
           in terms of L , mass of sun M , ρ, particle radius r, and quality factor Q.
      iii. The quality factor is given by one of the following
            • If r  λ, Q ∼ (r/λ)2
            • If r ∼ λ, Q ∼ 1.
            • If r  λ, Q = 1
          Considering the three possible particle sizes, which is most likely to be blown away by
          the solar radiation pressure?

  b. The Poynting-Robertson effect acts as another mechanism for cleaning out the solar sys-
     tem.

        i. Assume that a particle is in a circular orbit around the sun. Find the speed of the
           particle v in terms of M , distance from sun R, and any other fundamental constants.
       ii. Because the particle is moving, the radiation force is not directed directly away from the
           sun. Find the torque τ on the particle because of radiation pressure. You may assume
           that v  c.
      iii. Since τ = dL/dt, the angular momentum L of the particle changes with time. As such,
           develop a differential equation to find dR/dt, the rate of change of the radial location of
           the particle. You may assume the orbit is always quasi circular.
      iv. Develop an expression for the time required to remove particles of size r ≈ 1 cm and
          density ρ ≈ 1000 kg/m3 originally in circular orbits at a distance R = Rearth , and use
          the numbers below to simplify your expression.

   Some useful constants include
                                   M        =     1.989 × 1030 kg
                                   L        =     3.828 × 1026 W
                                   Rearth   =        1.5 × 1011 m
                                   λmax     =             500 nm

                  Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam      Answer Sheets                          10

                         Answer Sheets

 Following are answer sheets for some of the graphical portions of the test.

               Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam                       Answer Sheets                                 11

                                                                     Student AAPT ID #:

                                                                     Proctor AAPT ID #:

Solution:
Question B3
Radiation pressure from the sun is responsible for cleaning out the inner solar system of small
particles.

  a. The force of radiation on a spherical particle of radius r is given by

                                                F = P Qπr2

     where P is the radiation pressure and Q is a dimensionless quality factor that depends on
     the relative size of the particle r and the wavelength of light λ. Throughout this problem
     assume that the sun emits a single wavelength λmax ; unless told otherwise, leave your answers
     in terms of symbolic variables.

        i. Given that the total power radiated from the sun is given by L , find an expression for
           the radiation pressure a distance R from the sun.

                                                    Solution
          We will assume that light from the sun is completely absorbed, and then re-radiated as
          blackbody isotropically. In that case,
                                                      I    L
                                                P =     =        .
                                                      c   4πR2 c
          The relationship between the pressure P and the energy density I can be derived from
          the equation p = E/c for photons, or simply postulated by dimensional analysis, since
          there is no other relevant speed.
          Alternatively, this relation can be derived using classical electromagnetism, though this
          was not required. For a plane wave, the momentum density (i.e. pressure) of the electro-
          magnetic field is P = 0 |E×B|, while the energy density is I = |S|/c where S = E×B/µ0 .
          Then P = I/c since c2 = 1/0 µ0 .

                                                                                Fradiation
       ii. Assuming that the particle has a density ρ, derive an expression for the ratio
                                                                                Fgravity
          in terms of L , mass of sun M , ρ, particle radius r, and quality factor Q.

                                                    Solution
          We have
                                     GM 4                           L             L
                        Fgravity =         πρr3 ,   Fradiation =          Qπr2 =       Qr2
                                      R2 3                         4πR2 c        4R2 c
          which gives
                                           Fradiation      3L    Q
                                                      =            .
                                            Fgravity    16πGcM ρ r

      iii. The quality factor is given by one of the following
            • If r  λ, Q ∼ (r/λ)2
                  Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam                Part B                                               17

           • If r ∼ λ, Q ∼ 1.
           • If r  λ, Q = 1
         Considering the three possible particle sizes, which is most likely to be blown away by
         the solar radiation pressure?

                                                Solution
         In order to be blown away, the ratio should be greater than one. Since it is independent
         of distance from the sun, if it is blown away, it will be blown away at any distance. For
         r  λ, the ratio is proportional to 1/r, so smaller particles are more likely to be blown
         away. For r  λ, the ratio is proportional to r, so larger particles are more likely to be
         blown away. Thus particles of size near λ are most likely to be blown away, and even
         then, only if the density is small enough.

  b. The Poynting-Robertson effect acts as another mechanism for cleaning out the solar sys-
     tem.

       i. Assume that a particle is in a circular orbit around the sun. Find the speed of the
          particle v in terms of M , distance from sun R, and any other fundamental constants.

                                                Solution
         Using the circular motion equation

                                                 GM          v2
                                                         =
                                                  R2         R
         we have                                     r
                                                         GM
                                                v=          .
                                                          R

      ii. Because the particle is moving, the radiation force is not directed directly away from the
          sun. Find the torque τ on the particle because of radiation pressure. You may assume
          that v  c.

                                                Solution
         Work in the reference frame of the particle. In this frame, the radiation hits the particle
         at an angle θ = v/c from the radial direction. The particle then re-emits the radiation
         isotropically, contributing no additional radiation pressure. (We ignore relativistic effects
         because they occur at second order in v/c, while the effect we care about is first order.)
         The tangential component of the force is
                                            v L            v L
                                      F =        2
                                                    Qπr2 =         Qr2
                                            c 4πR c        c 4R2 c
         so
                                                       v L
                                               τ =−          Qr2
                                                       c 4Rc

                   Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam                   Part B                                             18

         where the negative sign is because this tends to decrease the angular momentum.
         The problem can also be solved in the reference frame of the Sun. In this frame, the
         radiation hits the particle radially, but the particle does not re-emit the radiation isotrop-
         ically since it is moving; instead the radiation is Doppler shifted. This eventually leads
         to the same result, after a much more complicated calculation.

      iii. Since τ = dL/dt, the angular momentum L of the particle changes with time. As such,
           develop a differential equation to find dR/dt, the rate of change of the radial location of
           the particle. You may assume the orbit is always quasi circular.

                                                   Solution
         The angular momentum is
                                                    r
                                        4               GM    4    p
                               L = mvR = πρr3              R = πρr3 GM R.
                                        3                R    3
         Differentiating both sides with respect to time,
                                                             r
                                       v L        4              GM 1 dR
                                     −       Qr2 = πρr3
                                       c 4Rc      3               R 2 dt
         which simplifies to
                                               1 L    8    dR
                                             − 2   Q = πρr    .
                                              c R     3    dt

      iv. Develop an expression for the time required to remove particles of size r ≈ 1 cm and
          density ρ ≈ 1000 kg/m3 originally in circular orbits at a distance R = Rearth , and use
          the numbers below to simplify your expression.

                                                   Solution
         Integrating both sides,
                                                  L Q  4
                                              T     2
                                                      = πρrR2 .
                                                   c   3
         Since r  λ, we have Q = 1 and

                                         4 πc2
                                   T =         ρrR2 ≈ 2 × 1014 s ≈ 7 × 106 y.
                                         3L

   Some useful constants include
                                   M         =     1.989 × 1030 kg
                                   L         =     3.828 × 1026 W
                                   Rearth    =        1.5 × 1011 m
                                   λmax      =             500 nm

                  Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam      Answer Sheets                          19

                         Answer Sheets

 Following are answer sheets for some of the graphical portions of the test.

               Copyright c 2018 American Association of Physics Teachers

2018 USA Physics Olympiad Exam                       Answer Sheets                                 20

                                                                     Student AAPT ID #:

                                                                     Proctor AAPT ID #:

USAPhO 2018 Question B3


[User]
A beaker contains liquid water at its freezing point and has a big ice cube floating in
    it, also at its freezing point. If the ice cube
     (a) is solid ice,
     (b) contains a small metal ball, or
      (c) contains a lot of olive oil (which will float on the water in a thin layer),
    then how does the fluid level change when the cube melts? In all cases, neglect the density of air.
[2] Problem 3 (Moscow 1939). Consider a pair of scales with identical vessels in which there are equal
    quantities of water.

    In the left-hand vessel you suspend a very light ping-pong ball on a thin, light wire attached to the
    base of the vessel. In the right-hand vessel you suspend a ping-pong ball filled with lead, again by
    a light thin wire. Do the scales stay level, go down on the left, or go down on the right?
[2] Problem 4 (BAUPC). Two trapezoidal containers, connected by a tube as shown, hold water.

       Assume that the containers do not undergo thermal expansion.
     (a) If the water in container A is heated, causing it to expand, will water flow through the tube?
         If so, in which direction?
     (b) What if the water in container B is heated instead?
[2] Problem 5 (MPPP 85). A solid cube of volume Vi and density ρi is fastened to one end of a cord,
    the other end of which is attached to a light bucket containing water, of density ρw = ρi /10.

    The system is in equilibrium.

     (a) Find the volume Vw of the water in the bucket.

     (b) What would happen if more water were poured into the bucket?

     (c) What would happen if some or all of the water evaporated?

       Example 2

       A perfectly spherical, nonrotating planet is covered with water. Geological activity causes a
       small underwater mountain to form, made of rock that is denser than water. Does the ocean
       surface above this mountain become higher or lower?

       Solution
       Systems minimize their energy in equilibrium. This means that in hydrostatic equilibrium,
       the surface of the water is an equipotential. Since the gravitational field of the mountain
       increases the gravitational potential near it, the water surface is higher near the mountain.

       Example 3

       Robert Boyle is best known for Boyle’s law, but he also invented a remarkably simple perpetual
       motion machine, called the perpetual vase.

       Since the volume of the vase is much greater than the neck, the pressure in the neck cannot
       possibly hold up all of the water in the vase. Hence the water will flow through the neck and
       fall back into the vase, causing perpetual motion. Why doesn’t this work?

       Solution
      This is an example of the hydrostatic paradox. Most of the upward force on the water is
      not provided by the pressure in the water in the neck, but from the normal force from
      the walls; each piece of wall provides enough normal force to hold up all of the water
      above it. (Of course, ultimately each piece of the glass is held in place by internal forces
      with other pieces of the glass, which ultimately are balanced by whatever is holding the glass.)

       Thus, the water in the neck only supports the water directly above it. That’s precisely what
       is balanced by the heightened pressure in the neck, so the water doesn’t start moving. (There
       have been many more attempts at fluid-based perpetual motion, as you can see here.)

Solution:
(a) The water level does not change. The ice cube creates a “hole” in the water that
         it sits in, and when it melts it exactly fills this hole.

     (b) Initially the metal ball displaces its weight in water, which is large, but after the ice cube
         melts it falls to the bottom and only displaces its volume in water. Hence the water level goes
         down.

      (c) Initially, the oil displaces its weight in water, but after the ice cube melts it sits on top, thereby
          occupying its whole volume. Since the oil is less dense than water, the fluid level goes up.
          (Specifically, the water level goes down, and an oil layer sits on top of it, at a height greater
          than the original water level.) Note that the answer would be different if there was only a little
          olive oil, because in that case the oil would coalesce into a few drops on the water surface. In
          that case each drop displaces its weight, so that the water level stays the same.
    As a sidenote, if we accounted for the density of air, then the answer to part (a) would actually be
    that the water level goes up a tiny bit. The reason is that part of the ice cube poking out above
    the water surface experiences an extra buoyant force from the air itself, which means the ice cube
    is not as deep in the water as one would expect. So when it melts, it slightly more than fills the
    hole, causing the water level to go up.

[2] Problem 3 (Moscow 1939). Consider a pair of scales with identical vessels in which there are equal
    quantities of water.

    In the left-hand vessel you suspend a very light ping-pong ball on a thin, light wire attached to the
    base of the vessel. In the right-hand vessel you suspend a ping-pong ball filled with lead, again by
    a light thin wire. Do the scales stay level, go down on the left, or go down on the right?

    Solution. The ball on the right experiences an upward buoyant force, so it exerts a downward
    force on the water. As for the ball on the left, it has no effect whatsoever on the force on the scale,

    because this force is simply equal to the weight of all the water. Hence the scales go down on the
    right.

[2] Problem 4 (BAUPC). Two trapezoidal containers, connected by a tube as shown, hold water.

       Assume that the containers do not undergo thermal expansion.

     (a) If the water in container A is heated, causing it to expand, will water flow through the tube?
         If so, in which direction?

     (b) What if the water in container B is heated instead?

    Solution. (a) The pressure at the bottom of the containers is P = ρgh = mgh/V , where h is the
         height of the water level above the container, V the volume, and m is the mass of the water
         in the container. When A is heated and the water level in A rises, V /h (average area) will
         increase as seen by the shape of the container. Thus PA = mgh/V will decrease, and water
         will flow from B to A.

     (b) In this case, V /h for B will decrease, since when the water level rises we include a section with
         a smaller cross-sectional area. Thus PB will increase, and water will flow from B to A again.

[2] Problem 5 (MPPP 85). A solid cube of volume Vi and density ρi is fastened to one end of a cord,
    the other end of which is attached to a light bucket containing water, of density ρw = ρi /10.

    The system is in equilibrium.

     (a) Find the volume Vw of the water in the bucket.

     (b) What would happen if more water were poured into the bucket?

     (c) What would happen if some or all of the water evaporated?

    Solution. (a) There is a buoyant force of ρw Vi on the block, which pushes the block up and
         the bucket down. In equilibrium, the tension in the cord T must balance against the weight

     of the block and the buoyant force: T = (ρi Vi − ρw Vi )g. Similarly for the water/bucket,
     T = (ρw Vw + ρw Vi )g. Equating the two gets

                                          ρw Vw = ρi Vi − 2ρw Vi

                                                Vw = 8Vi

 (b) The effective weight on the left side won’t change, but the bucket will be heavier. If the
     final volume of water Vf is less than 10Vi , where Vf = 10Vi is when the system will be in
     equilibrium if the block is out of the water, then the system will be in equilibrium at some
     point with the block partially submerged. Once Vf > 10Vi , the bucket will just keep falling.

 (c) The cube will fall until it hits the bottom of the bucket (the amount of evaporation doesn’t
     matter), and then the system will be stuck there since the cube can’t pass through the bucket.
  Example 2

  A perfectly spherical, nonrotating planet is covered with water. Geological activity causes a
  small underwater mountain to form, made of rock that is denser than water. Does the ocean
  surface above this mountain become higher or lower?

  Solution
  Systems minimize their energy in equilibrium. This means that in hydrostatic equilibrium,
  the surface of the water is an equipotential. Since the gravitational field of the mountain
  increases the gravitational potential near it, the water surface is higher near the mountain.

  Example 3

  Robert Boyle is best known for Boyle’s law, but he also invented a remarkably simple perpetual
  motion machine, called the perpetual vase.

  Since the volume of the vase is much greater than the neck, the pressure in the neck cannot
  possibly hold up all of the water in the vase. Hence the water will flow through the neck and
  fall back into the vase, causing perpetual motion. Why doesn’t this work?

  Solution
  This is an example of the hydrostatic paradox. Most of the upward force on the water is
  not provided by the pressure in the water in the neck, but from the normal force from
  the walls; each piece of wall provides enough normal force to hold up all of the water
  above it. (Of course, ultimately each piece of the glass is held in place by internal forces
  with other pieces of the glass, which ultimately are balanced by whatever is holding the glass.)

       Thus, the water in the neck only supports the water directly above it. That’s precisely what
       is balanced by the heightened pressure in the neck, so the water doesn’t start moving. (There
       have been many more attempts at fluid-based perpetual motion, as you can see here.)

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


[User]
Question Number 3
Theoretical Competition

25 April 2010                                                                               Page 1 of 4
__________________________________________________________________________________________

                                   Theoretical Question 3
                             Electron and Gas Bubbles in Liquids
This question deals with physics of two bubble-in-liquid systems. It has two parts:
  Part A. An electron bubble in liquid helium
  Part B. Single gas bubble in liquid

                      Part A. An Electron Bubble in Liquid Helium
   When an electron is planted inside liquid helium, it can repel atoms of liquid helium and
form what is called an electron bubble. The bubble contains nothing but the electron itself.
We shall be interested mainly in its size and stability.
  We use ∆𝑓𝑓 to denote the uncertainty of a quantity 𝑓𝑓. The components of an electron’s
position vector 𝑞𝑞⃗ = (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) and momentum vector 𝑝𝑝⃗ = (𝑝𝑝𝑥𝑥 , 𝑝𝑝𝑦𝑦 , 𝑝𝑝𝑧𝑧 ) must obey Heisenberg’s
uncertainty relations ∆𝑞𝑞𝛼𝛼 ∆𝑝𝑝𝛼𝛼 ≥ ℏ/2 , where ℏ is the Planck constant divided by 2 𝜋𝜋
and 𝛼𝛼 = 𝑥𝑥, 𝑦𝑦, 𝑧𝑧.
  We shall assume the electron bubble to be isotropic and its interface with liquid helium is a
sharp spherical surface. The liquid is kept at a constant temperature very close to 0 K with its
surface tension 𝜎𝜎 given by 3.75 × 10−4 N ∙ m−1 and its electrostatic responses to the electron
bubble may be neglected.
   Consider an electron bubble in liquid helium with an equilibrium radius 𝑅𝑅. The electron, of
mass 𝑚𝑚, moves freely inside the bubble with kinetic energy 𝐸𝐸k and exerts pressure 𝑃𝑃e on the
inner side of the bubble-liquid interface. The pressure exerted by liquid helium on the outer
side of the interface is 𝑃𝑃He .

(a) Find a relation between 𝑃𝑃He , 𝑃𝑃e , and 𝜎𝜎.                                              [0.4 point]
    Find a relation between 𝐸𝐸k and 𝑃𝑃e .                                                     [1.0 point]

(b) Denote by 𝐸𝐸0 the smallest possible value of 𝐸𝐸k consistent with Heisenberg’s uncertainty
    relations when the electron is inside the bubble of radius 𝑅𝑅. Estimate 𝐸𝐸0 as a function
    of 𝑅𝑅.                                                                         [0.8 point]

(c) Let 𝑅𝑅e be the equilibrium radius of the bubble when 𝐸𝐸k = 𝐸𝐸0 and 𝑃𝑃He = 0.
    Obtain an expression for 𝑅𝑅e and calculate its value.                                     [0.6 point]

(d) Find a condition that 𝑅𝑅 and 𝑃𝑃He must satisfy if the equilibrium at radius 𝑅𝑅 is to be
   locally stable under constant 𝑃𝑃He . Note that 𝑃𝑃He can be negative.           [0.6 point]

(e) There exists a threshold pressure 𝑃𝑃th such that equilibrium is not possible for the electron
    bubble when 𝑃𝑃He is less than 𝑃𝑃th . Find an expression for 𝑃𝑃th .                 [0.6 point]

                                                                          Question Number 3
Theoretical Competition

25 April 2010                                                                               Page 2 of 4
__________________________________________________________________________________________

        Part B. Single Gas Bubble in Liquid — Collapsing and Radiation
   In this part of the problem, we consider a normal liquid, such as water.
   When a gas bubble in a liquid is driven by an oscillating pressure, it can show dramatic
responses. For example, following a large expansion, it can collapse rapidly to a small radius
and, near the end of the collapse, emit light almost instantly. In this phenomenon, called
single-bubble sonoluminescence, the gas bubble undergoes cyclic motions which typically
consist of three stages: expansion, collapse, and multiple after-bounces. In the following we
shall focus mainly on the collapsing stage.
   We assume that, at all times, the bubble considered is spherical and its center remains
stationary in the liquid. See Fig 1. The pressure, temperature, and density are always uniform
inside the bubble as its size diminishes. The liquid containing the bubble is assumed to be
isotropic, nonviscous, incompressible, and very much larger in extent than the bubble. All
effects due to gravity and surface tension are neglected so that pressures on both sides of the
bubble-liquid interface are always equal.

 Radial motion of the bubble-liquid interface
   As the bubble’s radius 𝑅𝑅 = 𝑅𝑅(𝑡𝑡) changes with time 𝑡𝑡, the bubble-liquid interface will move
with radial velocity 𝑅𝑅̇ ≡ 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑. It follows from the equation of continuity of incompressible
fluids that the liquid’s radial velocity 𝑟𝑟̇ ≡ 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 at distance 𝑟𝑟 from the center of the bubble is
related to the rate of change of the bubble’s volume 𝑉𝑉 by
                                      𝑑𝑑𝑑𝑑
                                            = 4𝜋𝜋𝜋𝜋 2 𝑅𝑅̇ = 4𝜋𝜋𝑟𝑟 2 𝑟𝑟̇ .                             (1)
                                      𝑑𝑑𝑑𝑑
This implies that the total kinetic energy 𝐸𝐸k of the liquid with mass density 𝜌𝜌0 is
                                                                    𝑟𝑟0
                   1 𝑟𝑟0                                                1                      1 1
             𝐸𝐸k = � 𝜌𝜌0 (4𝜋𝜋𝑟𝑟 2 𝑑𝑑𝑑𝑑)𝑟𝑟̇ 2 = 2𝜋𝜋𝜌𝜌0 𝑅𝑅 4 𝑅𝑅̇ 2 � 2 𝑑𝑑𝑑𝑑 = 2𝜋𝜋𝜌𝜌0 𝑅𝑅 4 𝑅𝑅̇ 2 � − �   (2)
                   2 𝑅𝑅                                           𝑅𝑅 𝑟𝑟                        𝑅𝑅 𝑟𝑟0
where 𝑟𝑟0 is the radius of the outer surface of the liquid.

                                        𝜌𝜌0

                                            𝑃𝑃
          Fig. 1                                  𝑇𝑇    T0         𝑃𝑃0
                                       𝑅𝑅

                                                       𝑟𝑟
                                                 𝑟𝑟0

                                                                     Question Number 3
Theoretical Competition

25 April 2010                                                                          Page 3 of 4
__________________________________________________________________________________________

(f) Assume the ambient pressure 𝑃𝑃0 acting on the outer surface 𝑟𝑟 = 𝑟𝑟0 of the liquid is constant.
    Let 𝑃𝑃 = 𝑃𝑃(𝑅𝑅) be the gas pressure when the radius of the bubble is 𝑅𝑅.
    Find the amount of work 𝑑𝑑𝑑𝑑 done on the liquid when the radius of the bubble changes
    from 𝑅𝑅 to 𝑅𝑅 + 𝑑𝑑𝑑𝑑. Use 𝑃𝑃0 and 𝑃𝑃 to express 𝑑𝑑𝑑𝑑.                                [0.4 point]
    The work 𝑑𝑑𝑑𝑑 must be equal to the corresponding change in the total kinetic energy of the
    liquid. In the limit 𝑟𝑟0 → ∞, it follows that we have Bernoulli’s equation in the form
                                       1
                                    2
                                         𝜌𝜌0 𝑑𝑑�𝑅𝑅 m 𝑅𝑅̇ 2 � = (𝑃𝑃 − 𝑃𝑃0 )𝑅𝑅 n 𝑑𝑑𝑑𝑑.           (3)
   Find the exponents m and n in Eq. (3). Use dimensional arguments if necessary. [0.4 point]

 Collapsing of the gas bubble
   From here on, we consider only the collapsing stage of the bubble. The mass density of the
liquid is 𝜌𝜌0 = 1.0 × 103 kg ∙ m−3 , the temperature 𝑇𝑇0 of the liquid is 300 K and the ambient
pressure 𝑃𝑃0 is 1.01 × 105 Pa. We assume that 𝜌𝜌0 , 𝑇𝑇0 , and 𝑃𝑃0 remain constant at all times and
the bubble collapses adiabatically without any exchange of mass across the bubble-liquid
interface.
   The bubble considered is filled with an ideal gas. The ratio of specific heat at constant
pressure to that at constant volume for the gas is 𝛾𝛾 = 5/3. When under temperature 𝑇𝑇0 and
pressure 𝑃𝑃0 , the equilibrium radius of the bubble is 𝑅𝑅0 = 5.00 μm.
     Now, this bubble begins its collapsing stage at time 𝑡𝑡 = 0 with 𝑅𝑅(0) = 𝑅𝑅i = 7𝑅𝑅0 ,
𝑅𝑅̇ (0) = 0, and the gas temperature 𝑇𝑇i = 𝑇𝑇0 . Note that, because of the bubble’s expansion in
the preceding stage, 𝑅𝑅i is considerably larger than 𝑅𝑅0 and this is necessary if
sonoluminescence is to occur.

(g) Express the pressure 𝑃𝑃 ≡ 𝑃𝑃(𝑅𝑅) and temperature 𝑇𝑇 ≡ 𝑇𝑇(𝑅𝑅) of the ideal gas in the bubble as
    a function of 𝑅𝑅 during the collapsing stage, assuming quasi-equilibrium conditions hold.
    [0.6 point]
(h) Let 𝛽𝛽 ≡ 𝑅𝑅/𝑅𝑅i and 𝛽𝛽̇ = 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 . Eq. (3) implies a conservation law which takes the
   following form
                                     1
                                       𝜌𝜌 𝛽𝛽̇ 2 + 𝑈𝑈(𝛽𝛽) = 0.
                                     2 0
                                                                                (4)
   Let 𝑃𝑃i ≡ 𝑃𝑃(𝑅𝑅i ) be the gas pressure of the bubble when 𝑅𝑅 = 𝑅𝑅i . If we introduce the
   ratio 𝑄𝑄 ≡ 𝑃𝑃i /[(𝛾𝛾 − 1)𝑃𝑃0 ] , the function 𝑈𝑈(𝛽𝛽) may be expressed as
                                  𝑈𝑈(𝛽𝛽) = 𝜇𝜇𝛽𝛽 −5 [𝑄𝑄(1 − 𝛽𝛽 2 ) − 𝛽𝛽 2 (1 − 𝛽𝛽 3 )].       (5)
    Find the coefficient 𝜇𝜇 in terms of 𝑅𝑅i and 𝑃𝑃0 .                                  [0.6 point]
(i) Let 𝑅𝑅m be the minimum radius of the bubble during the collapsing stage and define
   𝛽𝛽m ≡ 𝑅𝑅m /𝑅𝑅i . For 𝑄𝑄 ≪ 1, we have 𝛽𝛽m ≈ 𝐶𝐶m �𝑄𝑄 .
    Find the constant 𝐶𝐶m .                                                             [0.4 point]

                                                                                Question Number 3
Theoretical Competition

25 April 2010                                                                                Page 4 of 4
__________________________________________________________________________________________

     Evaluate 𝑅𝑅m for 𝑅𝑅i = 7𝑅𝑅0 .                                                                   [0.3 point]
     Evaluate the temperature 𝑇𝑇m of the gas at 𝛽𝛽 = 𝛽𝛽m .                                           [0.3 point]

(j) Assume 𝑅𝑅i = 7𝑅𝑅0 . Let 𝛽𝛽𝑢𝑢 be the value of 𝛽𝛽 at which the dimensionless radial speed 𝑢𝑢 ≡
    |𝛽𝛽̇ | reaches its maximum value. The gas temperature rises rapidly for values of 𝛽𝛽 near 𝛽𝛽𝑢𝑢 .
     Give an expression and then estimate the value of 𝛽𝛽𝑢𝑢 .                                        [0.6 point]
     Let 𝑢𝑢� be the value of 𝑢𝑢 at 𝛽𝛽 = 𝛽𝛽̅ ≡ (𝛽𝛽m + 𝛽𝛽𝑢𝑢 )/2. Evaluate 𝑢𝑢�.        [0.4 point]
     Give an expression and then estimate the duration ∆𝑡𝑡m of time needed for 𝛽𝛽 to diminish
     from 𝛽𝛽𝑢𝑢 to the minimum value 𝛽𝛽m .                                           [0.6 point]

 Sonoluminescence of the collapsing bubble
   Consider the bubble to be a surface black-body radiator of constant emissivity 𝑎𝑎 so that
the effective Stefan-Boltzmann’s constant 𝜎𝜎eff = 𝑎𝑎𝜎𝜎SB . If the collapsing stage is to be
approximated as adiabatic, the emissivity must be small enough so that the power radiated by
the bubble at 𝛽𝛽 = 𝛽𝛽̅ is no more than a fraction, say 20 %, of the power 𝐸𝐸̇ supplied to it by the
driving liquid pressure.

(k) Find the power 𝐸𝐸̇ supplied to the bubble as a function of 𝛽𝛽.                                   [0.6 point]
    Give an expression and then estimate the value for an upper bound of 𝑎𝑎.                         [0.8 point]

                                                 Appendix
      𝑑𝑑 𝑛𝑛
1.        𝑥𝑥 = 𝑛𝑛𝑥𝑥 𝑛𝑛 −1
     𝑑𝑑𝑑𝑑

2. Electron mass 𝑚𝑚 = 9.11 × 10−31 kg

3. Planck constant ℎ = 2𝜋𝜋 ℏ = 2𝜋𝜋 × 1.055 × 10−34 J ∙ s

4. Stefan-Boltzmann’s constant 𝜎𝜎SB = 5.67 × 10−8 W ∙ m−2 ∙ K −4

                                                      END
----------------------------------------------------------------------------------------------------------------

Solution:
Question Number 3
Theoretical Competition

25 April 2010                                                                                Page 1 of 6
(Document Released: 14:30, 4/24)
                          __________________________________________________________________________________________

    Theoretical Question 3: Electron and Gas Bubbles in Liquids
                                                 SOLUTION
                        Part A. An Electron Bubble in Liquid Helium
(a) Consider a half of the spherical interface (see Fig. A1 below). The condition for its static
    equilibrium implies that the total force acting on it must be zero. This implies
                                  𝜋𝜋𝜋𝜋 2 (𝑃𝑃e − 𝑃𝑃He ) = 2𝜋𝜋𝜋𝜋𝜋𝜋                        (a-1)
    which leads to
                                                       2𝜎𝜎
                                        𝑃𝑃e = 𝑃𝑃He +                                                        (a-2)*
                                                       𝑅𝑅

                                                        𝑃𝑃He

                                                        𝑃𝑃e
                  Fig. A1
                                                      2𝑅𝑅
                                        𝜎𝜎                          𝜎𝜎

----------------------------------------------------------------------------------------------------------------
*An equation marked with an asterisk gives key answers to the problem.

    According to the de Broglie relation 𝑝𝑝 = ℎ/ 𝜆𝜆 ∝ 1/𝑅𝑅, the non-relativistic kinetic energy
    E k is inversely proportional to 𝑅𝑅 2 , i.e.
                                     𝑝𝑝2   const.
                               𝐸𝐸k =     =        .                                                          (a-3)
                                     2𝑚𝑚    𝑅𝑅 2
    By the work- energy theorem, we have
                                                                   const.          2
                                     −𝑃𝑃e 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝐸𝐸k = (−2)             𝑑𝑑𝑑𝑑 = －   𝐸𝐸 𝑑𝑑𝑑𝑑                 (a-4)
                                                                    𝑅𝑅 3           𝑅𝑅 k
    Thus,
                                                                 2
                                      −𝑃𝑃𝑒𝑒 (4𝜋𝜋𝑅𝑅 2 𝑑𝑑𝑑𝑑) = −     𝐸𝐸 𝑑𝑑𝑑𝑑                                   (a-5)
                                                                 𝑅𝑅 k
    or
                                                   1
                                         𝑃𝑃e =          𝐸𝐸                                                  (a-6)*
                                                 2𝜋𝜋𝑅𝑅 3 k
    [Alternative]
    The state of an electron confined in the bubble corresponds to standing waves which
    vanish on the interface. According to Part B of Question 1, these are equivalent to the
    superposition of two travelling waves moving in opposite directions and continually being
    reflected at the interface. They give rise to pressure on the interface and the relation

                                                                                Question Number 3
Theoretical Competition

25 April 2010                                                                                Page 2 of 6
(Document Released: 14:30, 4/24)
                          __________________________________________________________________________________________

    between the non-relativistic kinetic energy 𝐸𝐸k and the pressure 𝑃𝑃e for an electron inside a
    bubble is similar to that obtained from the kinetic theory of gases. Thus we have
                             2 𝐸𝐸k     2             1
                              𝑃𝑃e ==         𝐸𝐸k =        𝐸𝐸                                                (a-7)*
                             3 𝑉𝑉    4𝜋𝜋𝑅𝑅 3       2𝜋𝜋𝑅𝑅 3 k
(b) Let ℏ = ℎ/(2𝜋𝜋). From the uncertainty relations, we have
                                          1                       1                      1
                             ∆𝑥𝑥 ∆𝑝𝑝𝑥𝑥 ≥ 2 ℏ,        ∆𝑦𝑦 ∆𝑝𝑝𝑦𝑦 ≥ 2 ℏ,       ∆𝑧𝑧 ∆𝑝𝑝𝑧𝑧 ≥ 2 ℏ.                 (b-1)

    From symmetry considerations implied by isotropy, we have
                        𝑥𝑥̅ = 𝑦𝑦� = 𝑧𝑧̅ = 0, 𝑝𝑝
                                             ���𝑥𝑥 = 𝑝𝑝
                                                     ���
                                                       𝑦𝑦 = 𝑝𝑝
                                                            ���𝑧𝑧 = 0                                        (b-2)
      (∆𝑥𝑥) = 𝑥𝑥 − 𝑥𝑥̅ = 𝑥𝑥 = (∆𝑦𝑦) = (∆𝑧𝑧) , (∆𝑝𝑝𝑥𝑥 )2 = �����
           2  ���
                2     2     ���
                              2          2         2
                                                                     𝑝𝑝𝑥𝑥 2 = (∆𝑝𝑝𝑦𝑦 )2 = (∆𝑝𝑝𝑧𝑧 )2 .        (b-3)
    where 𝑓𝑓 ̅ denotes the mean value of the quantity 𝑓𝑓. Therefore, we have

                              3(∆x)2 = (∆x)2 + (∆y)2 + (∆z)2 = ���
                                                               x 2 + y���2 + z�2 = r�2 .                     (b-4)

                            3(∆𝑝𝑝𝑥𝑥 )2 = (∆𝑝𝑝𝑥𝑥 )2 + (∆𝑝𝑝𝑦𝑦 )2 + (∆𝑝𝑝𝑧𝑧 )2 = ���     ���
                                                                             𝑝𝑝𝑥𝑥2 + 𝑝𝑝 2   ���2 ���2
                                                                                𝑦𝑦 + 𝑝𝑝𝑧𝑧 = 𝑝𝑝 .      (b-5)

    Thus we obtain (cf. 18th IPhO)
                             ���  ���2 = 9(∆𝑥𝑥)2 (∆𝑝𝑝𝑥𝑥 )2 ≥ 9 ℏ2
                             𝑟𝑟 2 𝑝𝑝                                                                         (b-6)
                                                             4

    and the kinetic energy must satisfy the following inequality:
                               ���
                               𝑝𝑝2     1 9ℏ2 1
                         𝐸𝐸k =      ≥     �    �                                               (b‐7)
                               2𝑚𝑚 2𝑚𝑚 4 𝑟𝑟���2
    The smallest possible kinetic energy 𝐸𝐸0 of the electron consistent with the uncertainty
    relations is thus obtained if the mean-squared-radius 𝑟𝑟���2 is set equal to its largest possible
    value of 𝑅𝑅 2 . This gives
                                        9ℏ2
                            𝐸𝐸k ≥ 𝐸𝐸0 =                                                                     (b‐8)∗
                                       8𝑚𝑚𝑅𝑅 2
(c) If 𝐸𝐸k = 𝐸𝐸0 , it follows from Eqs. (a-2), (a-3), and (b-7) that we have
                                 1             9ℏ2       2𝜎𝜎
                         𝑃𝑃e =         𝐸𝐸0 =           =     + 𝑃𝑃He                                          (c‐1)
                               2𝜋𝜋𝑅𝑅 3       16𝑚𝑚𝜋𝜋𝜋𝜋5   𝑅𝑅
    For 𝑃𝑃He = 0, this gives the following equilibrium radius of the electron bubble:
                                                1                                                     1
                                     9ℏ2 4               9 × (1.055 × 10−34 )2       4
                            𝑅𝑅e = �          � =�                                   �
                                    32𝜋𝜋𝜋𝜋𝜋𝜋      32𝜋𝜋 × 9.11 × 10−31 × 3.75 × 10−4

                        = (2.91674 × 10−36 )1/4 m ≅ 1.31 nm                         (c‐2)*
    It might be of some interest to note that, from Eq. (c-1), the corresponding minimum
    kinetic energy is

                                                                                Question Number 3
Theoretical Competition

25 April 2010                                                                                Page 3 of 6
(Document Released: 14:30, 4/24)
                          __________________________________________________________________________________________

                                                   𝜎𝜎 1/2
                                   𝐸𝐸0 = 4𝜋𝜋𝑅𝑅02 𝜎𝜎 = 3ℎ �
                                                       � = 0.100 eV                     (c‐3)
                                                 2𝜋𝜋𝜋𝜋
(d) The condition for stable local equilibrium of the electron bubble at radius 𝑅𝑅 is that
   when 𝑅𝑅 is increased by a small amount 𝑑𝑑𝑑𝑑 > 0, the inward force pushing on the interface
   must be greater than the outward force so as to decrease the radius. Thus, from Eq. (c-1),
   we obtain
                              2𝜎𝜎                     9ℏ2
                                      + 𝑃𝑃He >                                         (d‐1)
                          (𝑅𝑅 + 𝑑𝑑𝑑𝑑)          16𝑚𝑚𝑚𝑚(𝑅𝑅 + 𝑑𝑑𝑑𝑑)5
    By keeping only terms linear in 𝑑𝑑𝑑𝑑 after both sides of the inequality are expanded as a
    power series and making use of Eq. (c-1) to eliminate 𝑃𝑃He , we obtain
                                  2𝜎𝜎             9ℏ2
                                  (−1) > (−5)                                             (d‐2)
                                  𝑅𝑅 2         16𝑚𝑚𝑚𝑚𝑅𝑅 6
    Note that the same inequality is obtained if we consider a small change 𝑑𝑑𝑑𝑑 < 0. Using Eq.
    (c-1), we may express Eq. (d-2) in terms of 𝑃𝑃He as
                                     2𝜎𝜎            2𝜎𝜎
                                         < 5 �𝑃𝑃He + �                                                       (d‐3)
                                     𝑅𝑅             𝑅𝑅
    or equivalently,
                                           8𝜎𝜎
                                      𝑃𝑃He > − �
                                               �                                                           (d-4)*
                                          5𝑅𝑅
(e) From Eqs. (a-2), (a-6), and (b-8), we have
                                      2𝜎𝜎                 𝐸𝐸k     𝐸𝐸0      9ℏ2
                                          + 𝑃𝑃He = 𝑃𝑃e =        ≥       =                                    (e‐1)
                                      𝑅𝑅                 2𝜋𝜋𝑅𝑅 3 2𝜋𝜋𝑅𝑅 3 16𝑚𝑚𝜋𝜋𝜋𝜋 5
    or equivalently,
                                       9ℏ2       2𝜎𝜎
                                      𝑃𝑃He ≥ 5
                                               −                                       (e‐2)
                                    16𝑚𝑚𝜋𝜋𝜋𝜋      𝑅𝑅
    The minimum of the right-hand side of the inequality occurs when its derivative vanishes,
    i.e.
                                       −45ℏ2      2𝜎𝜎
                                               6
                                                 + 2 =0                                                      (e‐3)
                                      16𝑚𝑚𝜋𝜋𝜋𝜋    𝑅𝑅
    or
                                                                  1/4
                                                 45ℏ2
                                   𝑅𝑅 = 𝑅𝑅th = �         �                                                   (e‐4)
                                                32𝑚𝑚𝑚𝑚𝑚𝑚
    Substituting the last result back into Eq. (e-2), we obtain

                           9ℏ2       2𝜎𝜎   1      2σ −8σ       −16σ 2𝑚𝑚𝑚𝑚𝑚𝑚 1/4
           𝑃𝑃He ≥ 𝑃𝑃th ≡         5 − 𝑅𝑅 = �5 − 1� 𝑅𝑅 = 5𝑅𝑅   =     �       �                                (e-5)*
                         16𝑚𝑚𝑚𝑚𝑅𝑅th    th           th    th    5    45ℏ2

                                                                                Question Number 3
Theoretical Competition

25 April 2010                                                                                Page 4 of 6
(Document Released: 14:30, 4/24)
                          __________________________________________________________________________________________

    For PHe < Pth , no equilibrium is possible for the electron bubble.

         Part B. Single Gas Bubble in Liquid — Collapsing and Radiation
(f) When the bubble’s radius 𝑅𝑅 changes by 𝑑𝑑𝑑𝑑, the volume of the liquid displaced by the
   interface is 𝑑𝑑𝑑𝑑 = 4𝜋𝜋𝑅𝑅 2 𝑑𝑑𝑑𝑑. But the total volume of the incompressible liquid cannot
   change, so the change of the volume at the outer surface of the liquid must also be 𝑑𝑑𝑑𝑑.
   Thus the amount of work done on the liquid is
                                    𝑑𝑑𝑑𝑑 = 𝑃𝑃𝑃𝑃𝑃𝑃 − 𝑃𝑃0 𝑑𝑑𝑑𝑑 = (𝑃𝑃 − 𝑃𝑃0 )4𝜋𝜋𝑅𝑅 2 𝑑𝑑𝑑𝑑 (f-1)*
    From Eq.(2), the change in total kinetic energy of the liquid is, in the limit 𝑟𝑟0 → ∞,
                                                              1 1
                               𝑑𝑑𝐸𝐸k = 𝑑𝑑 �2𝜋𝜋𝜌𝜌0 𝑅𝑅 4 𝑅𝑅 ̇2 � − �� = 2𝜋𝜋𝜌𝜌0 𝑑𝑑�𝑅𝑅 3 𝑅𝑅 ̇2 �.                �f‐2�
                                                              𝑅𝑅 𝑟𝑟0
    Since 𝑑𝑑𝐸𝐸k = 𝑑𝑑𝑑𝑑, we obtain
                                       1
                                         𝜌𝜌 𝑑𝑑�𝑅𝑅 m 𝑅𝑅 ̇2 � = (𝑃𝑃 − 𝑃𝑃0 )𝑅𝑅 n 𝑑𝑑𝑑𝑑.                          �f‐3�
                                       2 0
    with
                                    m = 3,    n = 2.                                       (f-4)*
(g) The initial gas temperature is 𝑇𝑇0 . According to the ideal gas law, the initial gas pressure
    𝑃𝑃i = 𝑃𝑃(𝑅𝑅i ) is thus given by
                              𝑃𝑃i 𝑅𝑅i3 = 𝑃𝑃0 𝑅𝑅03 .                                                        (g-1)*
    Since the process is adiabatic, the radial dependence of the gas pressure P is
                                                            𝛾𝛾
                                             𝑅𝑅i3          𝑅𝑅i 5      𝑅𝑅i 5 𝑅𝑅0 3
                              𝑃𝑃 ≡ 𝑃𝑃(𝑅𝑅) = � 3 � 𝑃𝑃i = � � 𝑃𝑃i = � � 𝑃𝑃0 � � .                             (g-2)*
                                             𝑅𝑅            𝑅𝑅         𝑅𝑅    𝑅𝑅i
    and the temperature 𝑇𝑇 corresponding to the radius 𝑅𝑅 is given by
                                                            (𝛾𝛾−1)
                                            𝑅𝑅i3                            𝑅𝑅i 2
                             𝑇𝑇 ≡ 𝑇𝑇(𝑅𝑅) = � 3 �                     𝑇𝑇0 = � � 𝑇𝑇0 .                       (g-3)*
                                            𝑅𝑅                              𝑅𝑅
(h) From Eqs. (3) and (g-2), we have
                               1 𝑑𝑑         3 ̇2
                                                          𝑃𝑃 − 𝑃𝑃0 𝑃𝑃0 𝑃𝑃𝑖𝑖 𝑅𝑅𝑖𝑖 3𝛾𝛾
                                         �𝑅𝑅 𝑅𝑅 � =                 = � � � − 1�                            (h-1)
                              2𝑅𝑅 2 𝑑𝑑𝑑𝑑                     𝜌𝜌0        𝜌𝜌0 𝑃𝑃0 𝑅𝑅
    In terms of 𝛽𝛽 = 𝑅𝑅/𝑅𝑅i and 𝛽𝛽̇ = 𝑅𝑅̇ /𝑅𝑅i , the last equation may be rewritten as
                                1 𝑑𝑑                         𝑃𝑃0 𝑃𝑃i −3𝛾𝛾
                                          �𝛽𝛽 3 𝛽𝛽 ̇2 � =          � 𝛽𝛽        − 1� .                        (h-2)
                                   2
                               2𝛽𝛽 𝑑𝑑𝑑𝑑                    𝜌𝜌0 𝑅𝑅i2 𝑃𝑃0
    This may be integrated to give
                     1 3 ̇2 𝑃𝑃0 𝛽𝛽 𝑃𝑃i 2−3𝛾𝛾     2
                                                          𝑃𝑃0 𝑃𝑃i 𝛽𝛽 3−3𝛾𝛾 − 1 𝛽𝛽 3 − 1
                       𝛽𝛽 𝛽𝛽 = � � 𝑦𝑦        − 𝑦𝑦 � 𝑑𝑑𝑑𝑑 = �� �               −         �
                     2        𝜌𝜌0 1 𝑃𝑃0                   𝜌𝜌0 𝑃𝑃0 3(1 − 𝛾𝛾)        3

                                                                                Question Number 3
Theoretical Competition

25 April 2010                                                                                Page 5 of 6
(Document Released: 14:30, 4/24)
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                                   𝑃𝑃0         𝑃𝑃i     1
                             =             �− � �           (1 − 𝛽𝛽 2 ) + 𝛽𝛽 2 (1 − 𝛽𝛽 3 )�.                �h‐3�
                                 3𝜌𝜌0 𝛽𝛽 2     𝑃𝑃0 (𝛾𝛾 − 1)
    Since 𝑄𝑄 ≡ 𝑃𝑃i /[(𝛾𝛾 − 1)𝑃𝑃0 ] and 𝛾𝛾 = 5/3, the last equation leads to
                             1                        −𝑃𝑃0
                               𝜌𝜌0 𝛽𝛽 2̇ = −𝑈𝑈(𝛽𝛽) ≡ 2 5 [𝑄𝑄(1 − 𝛽𝛽 2 ) − 𝛽𝛽 2 (1 − 𝛽𝛽 3 )]
                             2                      3𝑅𝑅i 𝛽𝛽
                                             −𝑃𝑃0 (1 − 𝛽𝛽 2 )       𝛽𝛽 2 (1 − 𝛽𝛽 3 )
                                           =                  �𝑄𝑄 −                  �.                     �h‐4�
                                                3𝑅𝑅i2 𝛽𝛽 5            (1 − 𝛽𝛽 2 )
    Thus we obtain
                                  𝑃𝑃0
                                    𝜇𝜇 =
                                       .                                                (h-5)*
                                 3𝑅𝑅i2
(i) The radius of the bubble reaches its minimum value when 𝑅𝑅̇ = 𝑅𝑅i 𝛽𝛽̇ = 0. Thus, from Eq.
    (h-4), we obtain
                                            2                             2
                                          𝛽𝛽m          3       2
                                                                        𝛽𝛽m
                                   𝑄𝑄 =       2
                                                (1 − 𝛽𝛽m ) = 𝛽𝛽m �1 +         �.                             (i-1)
                                        1 − 𝛽𝛽m                       1 + 𝛽𝛽m
    The last equality shows that 𝛽𝛽m must be very small in order that 𝑄𝑄 ≪ 1. Thus
                                   2
                            𝑄𝑄 ≈ 𝛽𝛽m , or 𝛽𝛽m ≈ �𝑄𝑄 , i.e. 𝐶𝐶m = 1                                          (i-2)*
    For 𝑅𝑅i = 7𝑅𝑅0 = 35.0 μm, we have, from Eq. (g-1),
                                              𝑃𝑃i        1    𝑅𝑅0 3 3 1 3
                                 𝑄𝑄 =               =        � � = � � = 0.00437.                            (i-3)
                                        𝑃𝑃0 (𝛾𝛾 − 1) (𝛾𝛾 − 1) 𝑅𝑅i   2 7
    Therefore
                                 𝛽𝛽m = �𝑄𝑄 = 0.0661,                                                         (i-4)
                            𝑅𝑅m = 𝛽𝛽m 𝑅𝑅𝑖𝑖 = �𝑄𝑄𝑅𝑅𝑖𝑖 = 0.0661 × 35 μm = 2.31 μm,                            (i-5)*
    and from Eq. (g-3), the corresponding temperature 𝑇𝑇m is
                                           1 2                       1         2
                               𝑇𝑇m = � � 𝑇𝑇0 = �                            � × 300 K = 6.86 × 104 K. (i-6)*
                                          𝛽𝛽m                    0.0661
(j) From Eq. (h-4), the maximum value of the radial speed 𝑢𝑢 ≡ �𝛽𝛽̇ � occurs at 𝛽𝛽 = 𝛽𝛽𝑢𝑢
    where −𝑈𝑈(𝛽𝛽) is also at its maximum, i.e. the derivative of 𝑈𝑈(𝛽𝛽) with respect to 𝛽𝛽 must
    vanish at 𝛽𝛽 = 𝛽𝛽𝑢𝑢 . Since
                                                  𝑃𝑃0            1       1            1
                                  𝑈𝑈(𝛽𝛽) =               �𝑄𝑄  �      −       �   − �      − 1�� ,      (j-1)
                                                3𝑅𝑅i2           𝛽𝛽 5 𝛽𝛽 3            𝛽𝛽 3
    we have
                                   𝑑𝑑𝑑𝑑                   −𝑃𝑃0             5        3        3
                                        �          =        2       �𝑄𝑄 � 5 − 3 � − 3 � = 0.           (j-2)
                                   𝑑𝑑𝑑𝑑 𝛽𝛽 =𝛽𝛽         3𝑅𝑅  i 𝛽𝛽 𝑢𝑢       𝛽𝛽 𝑢𝑢    𝛽𝛽𝑢𝑢     𝛽𝛽𝑢𝑢
                                               𝑢𝑢
    Thus
                                                  3𝛽𝛽𝑢𝑢2                   2
                                                                                5       𝑄𝑄
                                    𝑄𝑄 =                      ,   or    𝛽𝛽𝑢𝑢  =    �         �.        (j-3)
                                           (5 − 3𝛽𝛽𝑢𝑢2 )                         3 1 + 𝑄𝑄

                                                                                Question Number 3
Theoretical Competition

25 April 2010                                                                                Page 6 of 6
(Document Released: 14:30, 4/24)
                          __________________________________________________________________________________________

    which implies
                                          5   𝑄𝑄
                                  𝛽𝛽𝑢𝑢 = � �       � = 0.0852.                                              (j-4)*
                                          3 1 + 𝑄𝑄
    The radius midway between 𝛽𝛽𝑢𝑢 (corresponding to maximum speed) and 𝛽𝛽m
    (corresponding to zero speed) is given by
                               1                1
                                 𝛽𝛽̅ ≡
                                 (𝛽𝛽m + 𝛽𝛽𝑢𝑢 ) ≅ (0.0661 + 0.0852) = 0.0757                                   (j-5)
                               2                2
    From Eq. (h-4), the dimensionless radial speed at radius 𝛽𝛽̅ is
                                                      −2
                                 𝑢𝑢� = −𝛽𝛽̇ �𝛽𝛽̅ � = � 𝑈𝑈(𝛽𝛽̅ )
                                                      𝜌𝜌0

                                        2𝑃𝑃0 (1 − 𝛽𝛽̅ 2 )       𝑄𝑄       𝛽𝛽̅ 2
                                     =�                   �1 −       +               � = 5.52 × 106 .       (j-6)*
                                                 2 ̅3
                                          3𝜌𝜌0 𝑅𝑅i 𝛽𝛽          𝛽𝛽̅ 2   (1 +    𝛽𝛽̅ )

    Thus an estimate of the duration ∆𝑡𝑡m for the radius of the bubble to diminish from 𝛽𝛽𝑢𝑢 to
    the minimum value 𝛽𝛽m is
                               (𝛽𝛽𝑢𝑢 − 𝛽𝛽m ) (0.0852 − 0.0661)
                          ∆𝑡𝑡m =              =                             = 3.45 × 10−9 s. (j-7)*
                                     𝑢𝑢�               5.52 × 106
(k) Suppose the bubble is a surface radiator with emissivity 𝑎𝑎. By making use of Eq. (g-3), the
    radiant power 𝑊𝑊r of the bubble at temperature 𝑇𝑇 can be written as a function of 𝛽𝛽, i.e.
                                                                             1
                           𝑊𝑊r = 𝑎𝑎(𝜎𝜎SB 𝑇𝑇 4 )4𝜋𝜋𝑅𝑅 2 = 4𝜋𝜋𝑅𝑅i2 𝑎𝑎𝜎𝜎SB 𝑇𝑇04 6 ,             (k-1)
                                                                            𝛽𝛽
    where 𝜎𝜎SB is the Stefan-Boltzmann constant. The power supplied to the bubble is
                 4T

                                       𝑑𝑑𝑑𝑑         𝑉𝑉i 𝛾𝛾 𝑑𝑑𝑑𝑑                𝛽𝛽̇
                                ̇
                              𝐸𝐸 = −𝑃𝑃      = −𝑃𝑃i � �          = −4𝜋𝜋𝑅𝑅i3 𝑃𝑃i 3        (k-2)*
                                       𝑑𝑑𝑑𝑑         𝑉𝑉 𝑑𝑑𝑑𝑑                   𝛽𝛽
    The assumption of an adiabatic collapsing of the bubble is deemed reasonable when the
    radiant power is less than 20 % of the power supplied to the bubble at 𝛽𝛽 = 𝛽𝛽� . Thus we
    have
                                                         1            3      𝑢𝑢�
                                  4𝜋𝜋𝑅𝑅i2 𝑎𝑎𝜎𝜎SB 𝑇𝑇04         ≤ 4𝜋𝜋𝑅𝑅i  𝑃𝑃i       × 20 %                     (k-3)
                                                        𝛽𝛽̅ 6               𝛽𝛽̅ 3
    or
                                            𝑃𝑃i 𝑅𝑅i              𝑃𝑃 𝑅𝑅     𝑅𝑅 3
                                   𝑎𝑎 ≤              𝛽𝛽̅ 3 𝑢𝑢� = 0 i � 0 � 𝛽𝛽̅ 3 𝑢𝑢� = 0.0107              (k-4)*
                                          5𝜎𝜎SB 𝑇𝑇04            5𝜎𝜎SB 𝑇𝑇04 𝑅𝑅i

APhO 2010 Theory T3


[User]
NBPhO 2016, problem 4. A rather complicated problem involving several exotic
    circuit elements.
   Next, we consider some qualitatively new behavior that can emerge from less familiar circuit elements,
   such as amplification, hysteresis, and instability.

      Idea 1
      Tunnel diodes are a variant of diodes, whose I(V ) rises, falls, and rises again. That is, they
      have a region with negative differential resistance, dI/dV < 0. This allows them to amplify
      signals, as we’ll see below, and also can make them unstable.

      Example 1: EFPhO 2003

      The circuit below, containing a tunnel diode, acts as a simple amplifier.

      Here, R = 10 Ω and E = 0.25 V. If a small signal voltage Vin (t) is applied across the input,
      then an amplified and shifted version of the signal appears across the output. Find the
      amplification factor.

      Solution
      When a constant emf E is applied, Kirchoff’s laws give

                                             E = I0 R + V (I0 )

      where V (I) is the voltage characteristic of the diode.

      By plotting E − IR on the graph above, we find I0 at the intersection. (Notice the x-axis
      label: it is common to write a decimal point as a comma in Eastern Europe.) Now consider
      the effect of applying the signal voltage, which changes the current by ∆I,

                                   E + Vin = (I0 + ∆I)R + V (I0 + ∆I).

      Since the signal voltage is small, we can Taylor expand the voltage characteristic, giving

                                          Vin = ∆I(R + V ′ (I0 )).

      This in turn tells us that

                                                      0          R
                                Vout = (I0 + ∆I)R = Vout +                 Vin .
                                                             R + V ′ (I0 )

      In other words, the change in Vout is just Vin , times the amplification factor
                                           R           10       5
                                             ′
                                                   =         =−
                                       R + V (I0 )   10 − 16    3

      where we read V ′ (I0 ) off the graph by drawing a tangent. The intuition here is that the
      circuit is like a voltage divider, but the tunnel diode acts like a negative resistance. If we
      had V ′ (I0 ) close to −R, for example, the amplification factor would have been huge. Since
      V ′ (I0 ) is more negative than −R, the signal ends up flipped.

Solution:
See the official solutions here.

    Next, we consider some qualitatively new behavior that can emerge from less familiar circuit elements,
    such as amplification, hysteresis, and instability.

       Idea 1
       Tunnel diodes are a variant of diodes, whose I(V ) rises, falls, and rises again. That is, they
       have a region with negative differential resistance, dI/dV < 0. This allows them to amplify
       signals, as we’ll see below, and also can make them unstable.

       Example 1: EFPhO 2003

       The circuit below, containing a tunnel diode, acts as a simple amplifier.

  Here, R = 10 Ω and E = 0.25 V. If a small signal voltage Vin (t) is applied across the input,
  then an amplified and shifted version of the signal appears across the output. Find the
  amplification factor.

  Solution
  When a constant emf E is applied, Kirchoff’s laws give

                                         E = I0 R + V (I0 )

  where V (I) is the voltage characteristic of the diode.

  By plotting E − IR on the graph above, we find I0 at the intersection. (Notice the x-axis
  label: it is common to write a decimal point as a comma in Eastern Europe.) Now consider
  the effect of applying the signal voltage, which changes the current by ∆I,

                               E + Vin = (I0 + ∆I)R + V (I0 + ∆I).

  Since the signal voltage is small, we can Taylor expand the voltage characteristic, giving

                                      Vin = ∆I(R + V ′ (I0 )).

  This in turn tells us that
                                                 0           R
                           Vout = (I0 + ∆I)R = Vout +                  Vin .
                                                         R + V ′ (I0 )

       In other words, the change in Vout is just Vin , times the amplification factor
                                             R             10       5
                                                       =         =−
                                         R + V ′ (I0 )   10 − 16    3

       where we read V ′ (I0 ) off the graph by drawing a tangent. The intuition here is that the
       circuit is like a voltage divider, but the tunnel diode acts like a negative resistance. If we
       had V ′ (I0 ) close to −R, for example, the amplification factor would have been huge. Since
       V ′ (I0 ) is more negative than −R, the signal ends up flipped.


[User]
01W
                     USAPhO 2017, problem A4. However, to make it a little harder, solve part (a)
    without assuming Eb is small.

    2    Optimal Collisions
    These collision problems are conceptually simple, but somewhat more mathematically challenging.

        Idea 3
        The minimum energy configuration of a system of particles with fixed total momentum is
        the one where they all move with the same velocity. This is easiest to show by boosting to

  the center of mass frame (i.e. the frame with zero total momentum) and then boosting back.

  Example 4: KK 14.3

  A high energy photon (γ ray) collides with a proton at rest. A neutral pi meson is produced
  according to the reaction
                                        γ + p → p + π0.
  What is the minimum energy the γ ray must have for this reaction to occur? The rest mass
  of a proton is 938 MeV and the rest mass of a neutral pion is 135 MeV.

  Solution
  The total four-momentum is (E + mp , E) where E is the energy of the γ ray in the lab frame.
  This four-momentum has norm 2Emp + m2p . Crucially, the norms of four-momenta don’t
  change upon changing frames, so the total four-momentum in the center of mass frame is
                                    q
                                        2Emp + m2p , 0

  because the total spatial momentum vanishes by definition. On the other hand, we also know
  that the reaction can just barely happen when both the proton and pion are produced at rest
  in the center of mass frame, with a final four-momentum of (mp + mπ , 0). Hence we have
                                    q
                                      2Emp + m2p = mp + mπ

  and plugging in the numbers gives E = 145 MeV. As expected, this is a little bit more than
  the mass-energy of the pion, because the final system inevitably has some kinetic energy too.

  Example 5

  Two photons of angular frequencies ω1 and ω2 collide head-on. Under what conditions can
  an electron-positron pair be created?

  Solution
  The naive answer is to say the energy present must exceed the rest energy,

                                       ℏω1 + ℏω2 ≥ 2me .

  However, this is incorrect because the electron and positron will inevitably have kinetic
  energy, since the photons initially have a net momentum. The lowest total kinetic energy
  is achieved when the electron and positron come out with the same velocity, which is the
  velocity of the center of mass frame of the photons.

  The total four-momentum of the photons is

                                    (ℏ(ω1 + ω2 ), ℏ(ω1 − ω2 ))

        in the lab frame, and (Ecm , 0) in the center of mass frame. Therefore,

                                Ecm = ℏ2 ((ω1 + ω2 )2 − (ω1 − ω2 )2 ) = 4ℏ2 ω1 ω2 .

        In the center of mass frame, the electron and positron can be produced at rest, so the
        condition is Ecm ≥ 2me , which means
                                                  √
                                                 ℏ ω1 ω2 ≥ m e .

Solution:
USAPhO 2017, problem A4. However, to make it a little harder, solve part (a)
    without assuming Eb is small.

2    Optimal Collisions
These collision problems are conceptually simple, but somewhat more mathematically challenging.

    Idea 3
    The minimum energy configuration of a system of particles with fixed total momentum is
    the one where they all move with the same velocity. This is easiest to show by boosting to
    the center of mass frame (i.e. the frame with zero total momentum) and then boosting back.

    Example 4: KK 14.3

    A high energy photon (γ ray) collides with a proton at rest. A neutral pi meson is produced
    according to the reaction
                                          γ + p → p + π0.
    What is the minimum energy the γ ray must have for this reaction to occur? The rest mass
    of a proton is 938 MeV and the rest mass of a neutral pion is 135 MeV.

    Solution
    The total four-momentum is (E + mp , E) where E is the energy of the γ ray in the lab frame.
    This four-momentum has norm 2Emp + m2p . Crucially, the norms of four-momenta don’t
    change upon changing frames, so the total four-momentum in the center of mass frame is
                                      q
                                          2Emp + m2p , 0

    because the total spatial momentum vanishes by definition. On the other hand, we also know
    that the reaction can just barely happen when both the proton and pion are produced at rest
    in the center of mass frame, with a final four-momentum of (mp + mπ , 0). Hence we have
                                      q
                                        2Emp + m2p = mp + mπ

    and plugging in the numbers gives E = 145 MeV. As expected, this is a little bit more than
    the mass-energy of the pion, because the final system inevitably has some kinetic energy too.

    Example 5

    Two photons of angular frequencies ω1 and ω2 collide head-on. Under what conditions can
    an electron-positron pair be created?

    Solution
    The naive answer is to say the energy present must exceed the rest energy,

                                         ℏω1 + ℏω2 ≥ 2me .

    However, this is incorrect because the electron and positron will inevitably have kinetic
    energy, since the photons initially have a net momentum. The lowest total kinetic energy

       is achieved when the electron and positron come out with the same velocity, which is the
       velocity of the center of mass frame of the photons.

       The total four-momentum of the photons is

                                         (ℏ(ω1 + ω2 ), ℏ(ω1 − ω2 ))

       in the lab frame, and (Ecm , 0) in the center of mass frame. Therefore,

                             Ecm = ℏ2 ((ω1 + ω2 )2 − (ω1 − ω2 )2 ) = 4ℏ2 ω1 ω2 .

       In the center of mass frame, the electron and positron can be produced at rest, so the
       condition is Ecm ≥ 2me , which means
                                               √
                                              ℏ ω1 ω2 ≥ m e .


[User] Kahte suurt paralleelset metallplaati hoitakse horisontaalselt vastastikku. Üks plaatidest on temperatuuril $T_1 = \SI{-20}{\degreeCelsius}$ ja teine temperatuuril $T_2 = \SI{20}{\degreeCelsius}$. Metallplaatide vahel on vesi. Ilmselgelt on külma plaadi läheduses vesi tahkes olekus. On teada, et vee tahke ja vedela kihi paksuste suhe on \num{4}. Millisele temperatuurile tuleb soojendada teine metallplaat, et vedela kihi paksus saaks võrdseks tahke kihi paksusega?

Solution:
Soojusvoog läbi vee kihi on määrtud valemiga
\[
Q = D\frac{S\Delta T}{l},
\]
kus $S$ on kihi pindala, $l$ selle paksus, $\Delta T$ kihi ülemise ja alumise pinna temperatuuride vahe ning $D$ vastava vee faasi soojusjuhtivuskoefitsient. Vaatleme vedela ja tahke faasi piirpinda. Antud pinna temperatuur on \SI{0}{\degreeCelsius}. Tahkest poolest tulev soojusvoo võimsus on
$Q_t = D_tST_1/l_t$
ja vedelast faasist tuleva voo võimsus on
$Q_v = D_vST_2/l_v$.
Statsionaarses olukorras, kus piirpinna asukoht ei muutu, tasakaalustavad antud vood üksteist ära. Arvestades, et mõlema faasi pindalad on võrdsed, saame
\[
\frac{D_tT_1}{l_t} = -\frac{D_vT_2}{l_v}.
\]
Kasutame ära asjaolu, et tahke ja vedela kihi paksuste suhe oli alguses \num{4}:
\[
\frac{D_t}{D_v} = -\frac{T_{2} l_{t}}{T_1 l_{v1}}=-\frac{\num{-20}\cdot \num{4}}{\num{20}\cdot \num{1}} = \num{4},
\]
Teisel juhul
\[
T_{2}' = -\frac{D_{t}}{D_{v}}\frac{l_{v}'}{l_{t}'} T_1 = \num{-4}\frac{l_v'}{l_t'} T_2 = \SI{80}{\degreeCelsius}.
\]

Vesi ja jää

Kahte suurt paralleelset metallplaati hoitakse horisontaalselt vastastikku. Üks plaatidest on temperatuuril $T_1 = \SI{-20}{\degreeCelsius}$ ja teine temperatuuril $T_2 = \SI{20}{\degreeCelsius}$. Metallplaatide vahel on vesi. Ilmselgelt on külma plaadi läheduses vesi tahkes olekus. On teada, et vee tahke ja vedela kihi paksuste suhe on \num{4}. Millisele temperatuurile tuleb soojendada teine metallplaat, et vedela kihi paksus saaks võrdseks tahke kihi paksusega?

Soojusvoog läbi kihi paksusega $d$, pindalaga $S$, temperatuuride vahega $\Delta T$ ja soojusjuhtivuskoefitsiendiga $D$ on $Q = D\frac{S\Delta T}{d}$. Statsionaarses olekus tasakaalustab jää ja vee eralduspinnal ühe plaadi poolt tulev soojusvoog teiselt poolt tuleva soojusvoo täielikult ära.