[User]
In this problem, you will analyze the normal modes of the double pendulum,
    which consists of a pendulum of length ℓ and mass m attached to the bottom of another pendulum,
    of length ℓ and mass m. To solve this problem directly, one has to compute the tension forces in
    the two strings, which are quite complicated. A much easier method is to use energy.

 (a) Parametrize the position of the pendulum in terms of the angle θ1 the top string makes with
     the vertical, and the angle θ2 the bottom string makes with the vertical. Write out the kinetic
     energy K and the potential energy V to second order in the θi and θ˙i .

 (b) The Euler–Lagrange equations for the system are

                                             d ∂K       ∂V
                                                     =−     .
                                                   ˙
                                             dt ∂ θi    ∂θi

     Using the results of part (a), write these equations in the form

                                           θ¨1

                                                   g    θ1
                                               = −   A
                                           θ¨2     L    θ2

     where A is a 2 × 2 matrix. This is a generalization of θ̈ = −gθ/L for a single pendulum.

 (c) Find the normal modes and their angular frequencies, using the general method in section 4.5
     of Morin.

For larger deviations, the double pendulum can become chaotic. For a beautiful visualization of
both the chaotic behavior and the “islands of stability” within, see this video.

   Remark
   We mostly considered examples with two or three masses, but the techniques above work for
   systems with arbitrarily many degrees of freedom. However, this quickly becomes intractable
   unless the setup is highly symmetric, as in problem 23. Without such symmetry, a computer
   is generally necessary, so this sort of question won’t appear on standard Olympiads. However,
   if you’re curious, see ITPO 2016, problem 1 and Physics Cup 2021, problem 3 for examples.

5 [A] Adiabatic Change
   Idea 5
   When a problem contains two widely separate timescales, such as a fast oscillation superposed
   on a slow overall motion, one can solve for the fast motion while neglecting the slow motion,
   then solve for the slow motion by replacing the fast motion with an appropriate average.

   Example 9: MPPP 21

   A small smooth pearl is threaded onto a rigid, smooth, vertical rod, which is pivoted at
   its base. Initially, the pearl rests on a small circular disc that is concentric with the rod,
   and attached to it a distance d from the rotational axis. The rod starts executing simple
   harmonic motion around its original position with small angular amplitude θ0 .

  What angular frequency of oscillation is required for the pearl to leave the rod?

  Solution
  The reason the pearl leaves the rod is that the normal force rapidly varies in direction, with
  an average upward component. If this average upward force is greater than gravity, the pearl
  accelerates upward and leaves the rod.

  In this case, the fast motion is the oscillation of the rod, while the slow motion is the rate of
  change of the pearl’s distance from the pivot, which can be neglected during one oscillation.
  The pearl has horizontal displacement and acceleration

                  x(t) = −d sin θ ≈ −dθ(t) = −θ0 d sin ωt,    ax (t) = θω 2 d sin ωt.

  This is supplied by the horizontal component of the normal force. The vertical component is

                        Ny = Nx tan θ(t) ≈ max (t)θ(t) = mθ02 ω 2 d sin2 ωt.

  Now we average over the fast motion to understand the slow motion. Since the average value
  of sin2 (ωt) is 1/2, the condition for the pearl to go up is
                                          1 2 2
                                           mθ ω d > mg
                                          2 0
  which gives                                      r
                                              1        2g
                                           ω>             .
                                              θ0       d

  Example 10

  A mass m oscillates on a spring with spring constant k = k0 with amplitude A0 . Over a very
  long period of time, the spring smoothly and continuously weakens until its spring constant
  becomes k = k0 /2. Find the new amplitude of oscillation.

   Solution
   In this case the fast motion is the oscillation of the mass, while the slow motion is the
   weakening of the spring. We can solve the problem by considering how the energy changes
   in each oscillation, due to the slight decrease in k.

   Suppose that the spring constant drops in one instant by a factor of 1 − ϵ. Then the kinetic
   energy stays the same, while the potential energy drops by a factor of 1 − ϵ. Since the kinetic
   and potential energy are equal on average, this means that if the spring constant gradually
   decreases by a factor of 1 − x over a full cycle, with x ≪ 1, then the energy decreases by a
   factor of 1 − x/2.

                                                                −N x = 1/2. At this point, the
   The process finishes after N oscillations, where (1 − x)N ≈ e√
                                               N
   energy has dropped by√a factor of (1 − x/2) ≈ e   −N x/2 = 1/ 2. Since the energy is kA2 /2,
   the new amplitude is 4 2A0 .

Amazingly, the question can also be solved in one step using a subtle conserved quantity.

   Solution
   Sinusoidal motion is just a projection of circular motion. In particular, it’s equivalent
   to think of the mass as being tied to a spring of zero rest length attached to the origin,
   and performing a circular orbit about the origin, with the “actual” oscillation being the x
   component. (This is special to zero-length springs obeying Hooke’s law, and occurs because
   the spring force −kx = −k(x, y) has its x-component independent of y, and vice versa.)

   Since the spring constant is changed gradually, the orbit has to remain circular. Then angular
   momentum is conserved, and we have
                                                        √
                                      L ∝ vr = ωA2 ∝ kA2 .
                                 √
   Then the final amplitude is 4 2A0 as before.

Both of these approaches are tricky. The energy argument is very easy to get wrong, while the
angular momentum argument seems to come out of nowhere and is inapplicable to other situations.
But the formal angular momentum here turns out to be a special case of a more general conserved
quantity, which is useful in a wide range of similar problems.

   Idea 6: Adiabatic Theorem
   If a particle performs a periodic motion in one dimension in a potential that changes very
   slowly, then the “adiabatic invariant”
                                                I
                                           I = p dx

   is conserved. This integral is the area of the orbit in phase space, an abstract space whose
   axes are position and momentum.

      Solution
      Since the potential changes slowly, the energy is roughly conserved in each oscillation cycle,

                                                   p2  1
                                             E=       + kx2 .
                                                   2m 2
      Thus, within one oscillation cycle, the curve p(x)
                                                     √ traces out  pan approximate ellipse in phase
      space, with semimajor and semiminor axes of 2mE and 2E/k. Over the course of many
      oscillations, the energy changes, but the area of this ellipse is the adiabatic invariant,
                                          √                                  √
                              I                                  r
                                                p                   m
                          I = p dx = π 2mE 2E/k = 2πE                   ∝ A2 km.
                                                                    k

      Thus, A ∝ k −1/4 in an adiabatic change of k, recovering the answer found earlier.

      Remark
      The existence of the adiabatic invariant is hard to see in pure Newtonian mechanics, but
      it falls naturally out of Hamiltonian mechanics, which is built on phase space. In fact,
      Hamiltonian mechanics makes a lot of useful facts easier to see, which is why it’s the most
      commonly used foundation for introducing quantum mechanics. It is commonly introduced
      at the end of an undergraduate upper-division mechanics course, and therefore beyond the
      Olympiad syllabus. If you’d like to learn more about Hamiltonian mechanics, or just see
      how the adiabatic theorem is derived, see David Tong’s lecture notes.

      The conservation of the adiabatic invariant has important consequences throughout physics.
      As we’ll see in problem 25 and in X1, it ensures that the conditions which determine
      energy levels in quantum mechanics remain true as a system is changed. As we’ll discuss
      in R3, the adiabatic invariant is also useful to analyze the motion of charges in magnetic fields.

      It’s also closely connected to adiabatic processes in thermodynamics. You’ve probably heard
      that an adiabatic thermodynamic process has to be fast, so that no heat exchange can
      happen. But the more fundamental definition is that it’s slow, relative to the dynamics of the
      particles. In this case, the conservation of the adiabatic invariant for each particle implies the
      conservation of the entropy of the gas. That’s because, as we’ll discuss in T2, the entropy
      fundamentally measures the volume of phase space that the system can occupy.

Solution:
(a) To second order, the horizontal displacements of the masses are

                                         x1 = ℓθ1 ,        x2 = ℓ(θ1 + θ2 )

     which gives a kinetic energy of

                                             mℓ2 ˙ 2
                                       K=       ((θ1 ) + (θ˙1 + θ˙2 )2 ).

     The vertical displacements are

                          y1 = ℓ(1 − cos(θ1 )),       y2 = ℓ(2 − cos(θ1 ) − cos(θ2 ))

     and expanding the cosines to second order gives
                                           ℓ                   ℓ
                                       y1 = θ12 ,          y2 = (θ12 + θ22 )
                                           2                   2
     which gives a potential energy of
                                                  mgℓ 2
                                            V =      (2θ1 + θ22 ).

 (b) The resulting Euler–Lagrange equations are
                                                 2g                         g
                                   2θ¨1 + θ¨2 = − θ1 ,         θ¨1 + θ¨2 = − θ2 .
                                                  ℓ                         ℓ
     Solving the system, we find
                                                            2 −1
                                              A=
                                                           −2 2
     straightforwardly.

 (c) We must find the eigenvalues of the matrix, which obey the equation

                                           (2 − λ)2 − 2 = 0
                            √
    which implies λ = 2 ±   2. The normal mode amplitudes are

                                            √                      1
                         high frequency :       , low frequency : √
                                           − 2                      2

                                     2 = (g/L)(2 ±
                                                      √
    and the angular frequencies are ω±                    2).

  Remark
  We mostly considered examples with two or three masses, but the techniques above work for
  systems with arbitrarily many degrees of freedom. However, this quickly becomes intractable
  unless the setup is highly symmetric, as in problem 23. Without such symmetry, a computer
  is generally necessary, so this sort of question won’t appear on standard Olympiads. However,
  if you’re curious, see ITPO 2016, problem 1 and Physics Cup 2021, problem 3 for examples.

5 [A] Adiabatic Change
  Idea 5
  When a problem contains two widely separate timescales, such as a fast oscillation superposed
  on a slow overall motion, one can solve for the fast motion while neglecting the slow motion,
  then solve for the slow motion by replacing the fast motion with an appropriate average.

  Example 9: MPPP 21

  A small smooth pearl is threaded onto a rigid, smooth, vertical rod, which is pivoted at
  its base. Initially, the pearl rests on a small circular disc that is concentric with the rod,
  and attached to it a distance d from the rotational axis. The rod starts executing simple
  harmonic motion around its original position with small angular amplitude θ0 .

  What angular frequency of oscillation is required for the pearl to leave the rod?

   Solution
   The reason the pearl leaves the rod is that the normal force rapidly varies in direction, with
   an average upward component. If this average upward force is greater than gravity, the pearl
   accelerates upward and leaves the rod.

   In this case, the fast motion is the oscillation of the rod, while the slow motion is the rate of
   change of the pearl’s distance from the pivot, which can be neglected during one oscillation.
   The pearl has horizontal displacement and acceleration

                   x(t) = −d sin θ ≈ −dθ(t) = −θ0 d sin ωt,    ax (t) = θω 2 d sin ωt.

   This is supplied by the horizontal component of the normal force. The vertical component is

                         Ny = Nx tan θ(t) ≈ max (t)θ(t) = mθ02 ω 2 d sin2 ωt.

   Now we average over the fast motion to understand the slow motion. Since the average value
   of sin2 (ωt) is 1/2, the condition for the pearl to go up is
                                           1 2 2
                                            mθ ω d > mg
                                           2 0
   which gives                                      r
                                               1        2g
                                            ω>             .
                                               θ0       d

   Example 10

   A mass m oscillates on a spring with spring constant k = k0 with amplitude A0 . Over a very
   long period of time, the spring smoothly and continuously weakens until its spring constant
   becomes k = k0 /2. Find the new amplitude of oscillation.

   Solution
   In this case the fast motion is the oscillation of the mass, while the slow motion is the
   weakening of the spring. We can solve the problem by considering how the energy changes
   in each oscillation, due to the slight decrease in k.

   Suppose that the spring constant drops in one instant by a factor of 1 − ϵ. Then the kinetic
   energy stays the same, while the potential energy drops by a factor of 1 − ϵ. Since the kinetic
   and potential energy are equal on average, this means that if the spring constant gradually
   decreases by a factor of 1 − x over a full cycle, with x ≪ 1, then the energy decreases by a
   factor of 1 − x/2.

                                                                −N x = 1/2. At this point, the
   The process finishes after N oscillations, where (1 − x)N ≈ e√
                                               N
   energy has dropped by√a factor of (1 − x/2) ≈ e   −N x/2 = 1/ 2. Since the energy is kA2 /2,

   the new amplitude is 2A0 .

Amazingly, the question can also be solved in one step using a subtle conserved quantity.

   Solution
   Sinusoidal motion is just a projection of circular motion. In particular, it’s equivalent
   to think of the mass as being tied to a spring of zero rest length attached to the origin,
   and performing a circular orbit about the origin, with the “actual” oscillation being the x
   component. (This is special to zero-length springs obeying Hooke’s law, and occurs because
   the spring force −kx = −k(x, y) has its x-component independent of y, and vice versa.)

  Since the spring constant is changed gradually, the orbit has to remain circular. Then angular
  momentum is conserved, and we have
                                                       √
                                     L ∝ vr = ωA2 ∝ kA2 .
                                √
  Then the final amplitude is 4 2A0 as before.

Both of these approaches are tricky. The energy argument is very easy to get wrong, while the
angular momentum argument seems to come out of nowhere and is inapplicable to other situations.
But the formal angular momentum here turns out to be a special case of a more general conserved
quantity, which is useful in a wide range of similar problems.

   Idea 6: Adiabatic Theorem
   If a particle performs a periodic motion in one dimension in a potential that changes very
   slowly, then the “adiabatic invariant”
                                                I
                                           I = p dx

   is conserved. This integral is the area of the orbit in phase space, an abstract space whose
   axes are position and momentum.

   Solution
   Since the potential changes slowly, the energy is roughly conserved in each oscillation cycle,

                                              p2  1
                                         E=      + kx2 .
                                              2m 2
   Thus, within one oscillation cycle, the curve p(x)
                                                  √ traces out  pan approximate ellipse in phase
   space, with semimajor and semiminor axes of 2mE and 2E/k. Over the course of many
   oscillations, the energy changes, but the area of this ellipse is the adiabatic invariant,
                                       √                                  √
                           I                                  r
                                             p                   m
                       I = p dx = π 2mE 2E/k = 2πE                   ∝ A2 km.
                                                                 k

   Thus, A ∝ k −1/4 in an adiabatic change of k, recovering the answer found earlier.

      Remark
      The existence of the adiabatic invariant is hard to see in pure Newtonian mechanics, but
      it falls naturally out of Hamiltonian mechanics, which is built on phase space. In fact,
      Hamiltonian mechanics makes a lot of useful facts easier to see, which is why it’s the most
      commonly used foundation for introducing quantum mechanics. It is commonly introduced
      at the end of an undergraduate upper-division mechanics course, and therefore beyond the
      Olympiad syllabus. If you’d like to learn more about Hamiltonian mechanics, or just see
      how the adiabatic theorem is derived, see David Tong’s lecture notes.

      The conservation of the adiabatic invariant has important consequences throughout physics.
      As we’ll see in problem 25 and in X1, it ensures that the conditions which determine
      energy levels in quantum mechanics remain true as a system is changed. As we’ll discuss
      in R3, the adiabatic invariant is also useful to analyze the motion of charges in magnetic fields.

      It’s also closely connected to adiabatic processes in thermodynamics. You’ve probably heard
      that an adiabatic thermodynamic process has to be fast, so that no heat exchange can
      happen. But the more fundamental definition is that it’s slow, relative to the dynamics of the
      particles. In this case, the conservation of the adiabatic invariant for each particle implies the
      conservation of the entropy of the gas. That’s because, as we’ll discuss in T2, the entropy
      fundamentally measures the volume of phase space that the system can occupy.


[User]
With a little vector calculus, the results above can be generalized to an arbitrary
    number of dimensions. For example, ideal waves in three dimensions obey
                                         2
                               ∂2ψ                ∂2ψ ∂2ψ

                                       2 ∂ ψ
                                   =v          +        +        = v 2 ∇2 ψ
                               ∂t2         ∂x2     ∂y 2   ∂z 2

    where the function ψ(r, t) could stand for a variety of things, such as the pressure, density, or
    temperature (for a sound wave) or the electric or magnetic field (for an electromagnetic wave).

     (a) For simplicity, let’s restrict to waves which have spherical symmetry, so that ψ only depends
         on r and t. Plug such a spherical wave into the wave equation, and simplify until you get an
         equation only in terms of the partial derivatives of ψ(r, t).

     (b) Because the area of a sphere goes as r2 , we expect the energy density of a spherical wave
         to fall as 1/r2 , and therefore expect the amplitude to fall as 1/r. Therefore, it is useful to
         consider the quantity rψ, which has this falloff factored out. By considering the differential
         equation that rψ obeys, find the general solution for ψ(r, t).

    2    Standing Waves
        Idea 3
        A standing wave is a solution to the wave equation of the form

                                           y(x, t) = f (x) cos(ωt).

        Typically, only discrete values of ω are possible, with the allowed values depending on the
        boundary conditions. If the setup is translationally symmetric, then f (x) will be sinusoidal.
        If you want to get some intuition, try playing with this PhET simulation.

Solution:
(a) To do this, we need to simplify the partial derivatives with respect to x, y, and z.
     We have
                                                 ∂ψ    ∂ψ ∂r
                                                    =
                                                 ∂x    ∂r ∂x
     where these partial derivatives are all keeping the other spatial variables y and z constant. To
     evaluate ∂r/∂x, we note that
                                             r 2 = x2 + y 2 + z 2
     and take ∂/∂x of both sides, giving
                                                      ∂r
                                                 2r      = 2x
                                                      ∂x
     from which we conclude
                                                   ∂r  x
                                                      = .
                                                   ∂x  r
     Plugging this back in, we conclude
                                            ∂ψ     x ∂ψ
                                                =        .
                                             ∂x    r ∂r
     Of course, we actually want the second spatial derivative, which is

                       ∂2ψ   1 ∂ψ x2 ∂ψ x ∂ ∂ψ        1 − x2 /r2 ∂ψ x2 ∂ 2 ψ
                           =      −       +         =               + 2 2
                       ∂x2   r ∂r   r3 ∂r   r ∂x ∂r       r      ∂r  r ∂r
     where we used the product rule. By similar logic for the other components, we conclude

          ∂2ψ              2     2   2    2       x2 + y 2 + z 2 ∂ 2 ψ                ∂2ψ

                  2 3 − (x + y + z )/r ∂ψ                                   2 2 ∂ψ
               =v                              +                         =v         +       .
           ∂t2                 r            ∂r         r2        ∂r2           r ∂r   ∂r2

     Of course, a shortcut to this result would be to just look up the formula for the Laplacian in
     spherical coordinates, but it’s good to go through this once in your life.

 (b) We notice that
                                        ∂2          2 ∂

                                           (rψ) = v       (rψ)
                                       ∂t2           ∂r2
     by the product rule. That is, the quantity rψ obeys the ordinary, one-dimensional wave
     equation, for which we already know the general solution! We thus conclude

                                                 f (r − vt) + g(r + vt)
                                     ψ(r, t) =                          .
                                                            r
     If we also want the wave not to blow up at r = 0, we additionally need f (−vt) + g(vt) = 0.

2    Standing Waves

       Idea 3
       A standing wave is a solution to the wave equation of the form

                                          y(x, t) = f (x) cos(ωt).

       Typically, only discrete values of ω are possible, with the allowed values depending on the
       boundary conditions. If the setup is translationally symmetric, then f (x) will be sinusoidal.
       If you want to get some intuition, try playing with this PhET simulation.

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


[User]
An infinite cylindrical rod of radius R has a fixed, uniform magnetization M.

     (a) If M is parallel to the rod’s axis, find the magnetic field everywhere.

     (b) If M is perpendicular to the rod’s axis, describe B outside, and find the value of B inside.

Solution:
Let ẑ point along the axis of the rods.

     (a) The surface bound current density is Kb = M θ̂, which is simply that of an infinite solenoid.
         So the magnetic field is µ0 M inside the rod, and zero outside.

     (b) Let’s say that M is parallel to x̂. Then in cylindrical coordinates, we have Kb = M sin θ ẑ. This
         current density is equivalent to superposing two cylinders carrying uniform current density
         ±J ẑ, separated by d along the ŷ direction, where Jd = M .

         Outside the rod, the cylinders can be replaced with wires carrying current I = πR2 J, and the
         corresponding magnetic field was found in a problem in E3. Inside the rod, superposing the
         magnetic fields of the cylinders yields B = µ0 M/2.