[User]
Question A3
A ship can be thought of as a symmetric arrangement of soft iron. In the presence of an external
magnetic field, the soft iron will become magnetized, creating a second, weaker magnetic field. We
want to examine the effect of the ship’s field on the ship’s compass, which will be located in the
middle of the ship.
    Let the strength of the Earth’s magnetic field near the ship be Be , and the orientation of the
field be horizontal, pointing directly toward true north.
    The Earth’s magnetic field Be will magnetize the ship, which will then create a second magnetic
field Bs in the vicinity of the ship’s compass given by

                                  ~
                                  Bs = Be −Kb cos θ b̂ + Ks sin θ ŝ

where Kb and Ks are positive constants, θ is the angle between the heading of the ship and magnetic
north, measured clockwise, b̂ and ŝ are unit vectors pointing in the forward direction of the ship
(bow) and directly right of the forward direction (starboard), respectively.
   Because of the ship’s magnetic field, the ship’s compass will no longer necessarily point North.

  a. Derive an expression for the deviation of the compass, δθ, from north as a function of Kb ,
     Ks , and θ.

  b. Assuming that Kb and Ks are both much smaller than one, at what heading(s) θ will the
     deviation δθ be largest?

   A pair of iron balls placed in the same horizontal plane as the compass but a distance d away
can be used to help correct for the error caused by the induced magnetism of the ship.

A binnacle, protecting the ship’s compass in the center, with two soft iron spheres to help correct for errors in the
                     compass heading. The use of the spheres was suggested by Lord Kelvin.

   Just like the ship, the iron balls will become magnetic because of the Earth’s field Be . As
spheres, the balls will individually act like dipoles. A dipole can be thought of as the field produced
by two magnetic monopoles of strength ±m at two different points.
                     Copyright c 2017 American Association of Physics Teachers

2017 USA Physics Olympiad Exam                     Part A                                          5

   The magnetic field of a single pole is

                                             ~ = ±m r̂
                                             B
                                                   r2
where the positive sign is for a north pole and the negative for a south pole. The dipole magnetic
field is the sum of the two fields: a north pole at y = +a/2 and a south pole at y = −a/2, where
the y axis is horizontal and pointing north. a is a small distance much smaller than the radius of
the iron balls; in general a = Ki Be where Ki is a constant that depends on the size of the iron
sphere.

                               North

                                                                    ~
                                                                    Bi

                                       φ

                        a
                                       iron ball

                                                 ~ i from the iron a distance d  a from the center
  c. Derive an expression for the magnetic field B
     of the ball. Note that there will be a component directed radially away from the ball and a
     component directed tangent to a circle of radius d around the ball, so using polar coordinates
     is recommended.

  d. If placed directly to the right and left of the ship compass, the iron balls can be located at a
     distance d to cancel out the error in the magnetic heading for any angle(s) where δθ is largest.
     Assuming that this is done, find the resulting expression for the combined deviation δθ due
     to the ship and the balls for the magnetic heading for all angles θ.

                  Copyright c 2017 American Association of Physics Teachers

2017 USA Physics Olympiad Exam                 Part A                                               6

Solution:
Question A3
A ship can be thought of as a symmetric arrangement of soft iron. In the presence of an external
magnetic field, the soft iron will become magnetized, creating a second, weaker magnetic field. We
want to examine the effect of the ship’s field on the ship’s compass, which will be located in the
middle of the ship.
    Let the strength of the Earth’s magnetic field near the ship be Be , and the orientation of the
field be horizontal, pointing directly toward true north.
    The Earth’s magnetic field Be will magnetize the ship, which will then create a second magnetic
field Bs in the vicinity of the ship’s compass given by

                                  ~ s = Be −Kb cos θ b̂ + Ks sin θ ŝ
                                  B

where Kb and Ks are positive constants, θ is the angle between the heading of the ship and magnetic
north, measured clockwise, b̂ and ŝ are unit vectors pointing in the forward direction of the ship
(bow) and directly right of the forward direction (starboard), respectively.
   Because of the ship’s magnetic field, the ship’s compass will no longer necessarily point North.

  a. Derive an expression for the deviation of the compass, δθ, from north as a function of Kb ,
     Ks , and θ.

                                               Solution

     We add the fields to get the local field. The northward component is

                            Bnorth = Be − Be Kb cos θ cos θ − Be Ks sin θ sin θ

     while the eastward component is

                               Beast = −Be Kb sin θ cos θ + Be Ks cos θ sin θ

     The deviation is given by
                                                           sin θ cos θ
                              tan δθ = (Ks − Kb )                             .
                                                    1 − Kb cos2 θ − Ks sin2 θ
     This form is particularly nice, because as we’ll see below, Kb and Ks are small enough to
     ignore in the denominator.

  b. Assuming that Kb and Ks are both much smaller than one, at what heading(s) θ will the
     deviation δθ be largest?

                                               Solution

     By inspection, θ = 45◦ will yield the largest deviation. It’s also acceptable to list 45◦ , 135◦ ,
     225◦ , and 315◦ .

                   Copyright c 2017 American Association of Physics Teachers

2017 USA Physics Olympiad Exam                           Part A                                                     8

   A pair of iron balls placed in the same horizontal plane as the compass but a distance d away
can be used to help correct for the error caused by the induced magnetism of the ship.

A binnacle, protecting the ship’s compass in the center, with two soft iron spheres to help correct for errors in the
                     compass heading. The use of the spheres was suggested by Lord Kelvin.

   Just like the ship, the iron balls will become magnetic because of the Earth’s field Be . As
spheres, the balls will individually act like dipoles. A dipole can be thought of as the field produced
by two magnetic monopoles of strength ±m at two different points.
   The magnetic field of a single pole is

                                                   ~ = ±m r̂
                                                   B
                                                         r2
where the positive sign is for a north pole and the negative for a south pole. The dipole magnetic
field is the sum of the two fields: a north pole at y = +a/2 and a south pole at y = −a/2, where
the y axis is horizontal and pointing north. a is a small distance much smaller than the radius of
the iron balls; in general a = Ki Be where Ki is a constant that depends on the size of the iron
sphere.

                                    North

                                                                               ~i
                                                                               B

                                             φ

                            a
                                             iron ball

                                                  ~ i from the iron a distance d  a from the center
   c. Derive an expression for the magnetic field B
      of the ball. Note that there will be a component directed radially away from the ball and a
                     Copyright c 2017 American Association of Physics Teachers

2017 USA Physics Olympiad Exam                 Part A                                              9

     component directed tangent to a circle of radius d around the ball, so using polar coordinates
     is recommended.

                                               Solution

     This problem is not nearly as difficult as it looks.

                                   North

                                           φ

                            a

     Consider the colored triangle above. The black side has length a. The angle between the
     green and black sides is φ, so the length of the red side is a sin φ and the length of the green
     side is a cos φ.
     The magnetic field strength from one magnetic pole a distance d away is given by
                                                      1
                                               B = ±m 2
                                                     d
     The sum of the two fields has two components. The angular component is a measure of the
     “opening” of the triangle formed by the two vectors, and since the two vectors basically have
     the same length, we can use similar triangles to conclude
                         a sin φ   Bφ                 a           mKi
                                 ≈         ⇒   Bφ = m 3 sin φ = Be 3 sin φ.
                            d      B                 d             d
     As expected, this component vanishes for φ = 0.
     The radial component is given by the difference in the lengths of the two field vectors, or

                               1       1          m             1          m 2x
                    Br = m      2
                                  −         2
                                                = 2 1−              2
                                                                        ≈ 2 ,
                               d    (d + x)       d        (1 + x/d)       d d

     where x = a cos φ is the length of the green side, so
                                                      mKi
                                           Br = 2Be       cos φ.
                                                       d3
     That wasn’t so bad, was it?

  d. If placed directly to the right and left of the ship compass, the iron balls can be located at a
     distance d to cancel out the error in the magnetic heading for any angle(s) where δθ is largest.
     Assuming that this is done, find the resulting expression for the combined deviation δθ due
     to the ship and the balls for the magnetic heading for all angles θ.

                  Copyright c 2017 American Association of Physics Teachers

2017 USA Physics Olympiad Exam               Part A                                              10

                                             Solution

    Note that the two iron balls create a magnetic field near the compass that behaves like that
    of the ship as a whole. There is a component directed toward the bow given by

                                      Bb = −2Bθ ∝ sin φ ∝ cos θ

    and a component directed toward the starboard given by

                                       Bs = 2Br ∝ cos φ ∝ sin θ

    where the factors of 2 are because there are two balls. Note that θ is the ship heading while φ
    is the angle between North and the location of the compass relative to one of the balls. Thus,
    if the field is corrected for the maximum angles it will necessarily cancel out the induced ship
    field for all of the angles, so that
                                                δθ = 0
    for all θ. Effectively, this means placing the balls to make Kb = Ks .

                 Copyright c 2017 American Association of Physics Teachers

2017 USA Physics Olympiad Exam                Part A                                              11

USAPhO 2017 Question A3


[User]
USAPhO 2009, problem B1.

  Example 8: MPPP 42

  There are N identical tiny discs lying on a table, equally spaced along a semicircle, with total
  mass M . Another disc D of mass m is very precisely aimed to bounce off all of the discs in
  turn, then exit opposite the direction it came.

  In the limit N → ∞, what is the minimal value of M/m for this to be possible? Given this
  value, what is the ratio of the final and initial speeds of the disc?

  Solution
  The reason there is a lower bound on M is that, by problem 20(c), there is a maximal angle
  that each tiny disc can deflect the disc D. For large N , the deflection is π/N for each disc, so

                                      π         M/N   M
                                        = sin−1     ≈
                                      N          m    Nm
  which implies that M/m ≥ π.

  To see how much energy is lost in each collision, work in the center of mass frame and consider
  the first collision. In this frame, the disc D is initially approximately still, and the tiny disc
  comes in horizontally with speed v. To maximize the deflection angle in the table’s frame, the
  tiny disc should rebound vertically, as this provides the maximal vertical impulse to the disc D.

  Thus, going√back to the√table’s frame, where the disc D has speed v, the tiny disc scatters
  with speed v 2 + v 2 = 2v. By conservation of energy,

                                                 1M √ 2

                                      1   2
                                 ∆      mv = −        ( 2v) .
                                      2          2N

  This simplifies to
                                          ∆v       π
                                              =−
                                           v      N
  which means that after N collisions, we have the cute result
                                      vf     π N
                                         = 1−      ≈ e−π
                                      vi      N
  where in the last step we used a result from P1.

    Example 9: EFPhO 2003.1

    A spherical volleyball of radius r and mass m is inflated with excess pressure ∆P . If it
    is dropped from the ceiling and hits the ground, estimate how long the subsequent elastic
    collision takes.

    Solution
    Answering this question requires making a simplified physical model of how the collision
    occurs. Let’s say that when the volleyball hits the ground, it will keep going straight down,
    deforming the part that touches the ground into a flat circular face. Specifically, when the
    ball has moved a distance y into the ground, the flat face has area
                                  p             2
                           A=π      r2 − (r − y)2 = πy(2r − y) ≈ 2πry

    where we assumed that y ≪ r at all times, which is reasonable as long as the ball’s initial
    speed is not enormous. As a result, the pressure of the volleyball exerts a force

                                            F = 2πr ∆P y

    on the ground. This assumes the pressure inside the volleyball remains uniform, and that
    the rest of the volleyball stays approximately spherical, which is again reasonable as long as
    the initial speed is not huge.

    Assuming the initial velocity is not too small, gravity is negligible during the collision, so
    during the collision the force on the volleyball is effectively that of an ideal spring. The
    collision lasts for half a period, giving
                                              r        r
                                                m         πm
                                         τ =π        =         .
                                                keff     2r ∆P

    If we plug in realistic numbers, the result is of order 10 ms, which is plausible.

5    Continuous Systems
    Example 10

    As shown in M2, a hanging chain takes the form of a catenary. Suppose you pull the chain
    down in the middle. How does the center of mass of the chain move? Does the answer depend
    on how hard you pull?

    Solution
    No matter how hard you pull, or in what direction, the height of the center of mass always
    goes up! This is because this quantity measures the total gravitational potential energy of
    the chain. If you pull a chain in equilibrium, in any direction whatsoever, you will do work
    on it. So this raises its potential energy, and hence the center of mass.

        Another way of saying this is that the equilibrium position, without the extra pull you supply,
        is already in the lowest energy state, and hence already has the lowest possible center of mass.
        Changing this shape in any way raises the center of mass.

Solution:
01m   USAPhO 2009, problem B1.

       Example 8: MPPP 42

       There are N identical tiny discs lying on a table, equally spaced along a semicircle, with total
       mass M . Another disc D of mass m is very precisely aimed to bounce off all of the discs in
       turn, then exit opposite the direction it came.

  In the limit N → ∞, what is the minimal value of M/m for this to be possible? Given this
  value, what is the ratio of the final and initial speeds of the disc?

  Solution
  The reason there is a lower bound on M is that, by problem 20(c), there is a maximal angle
  that each tiny disc can deflect the disc D. For large N , the deflection is π/N for each disc, so

                                      π         M/N   M
                                        = sin−1     ≈
                                      N          m    Nm
  which implies that M/m ≥ π.

  To see how much energy is lost in each collision, work in the center of mass frame and consider
  the first collision. In this frame, the disc D is initially approximately still, and the tiny disc
  comes in horizontally with speed v. To maximize the deflection angle in the table’s frame, the
  tiny disc should rebound vertically, as this provides the maximal vertical impulse to the disc D.

  Thus, going√back to the√table’s frame, where the disc D has speed v, the tiny disc scatters
  with speed v 2 + v 2 = 2v. By conservation of energy,

                                                 1M √ 2

                                      1   2
                                 ∆      mv = −        ( 2v) .
                                      2          2N

  This simplifies to
                                          ∆v       π
                                              =−
                                           v      N
  which means that after N collisions, we have the cute result
                                      vf     π N
                                         = 1−      ≈ e−π
                                      vi      N
  where in the last step we used a result from P1.

    Example 9: EFPhO 2003.1

    A spherical volleyball of radius r and mass m is inflated with excess pressure ∆P . If it
    is dropped from the ceiling and hits the ground, estimate how long the subsequent elastic
    collision takes.

    Solution
    Answering this question requires making a simplified physical model of how the collision
    occurs. Let’s say that when the volleyball hits the ground, it will keep going straight down,
    deforming the part that touches the ground into a flat circular face. Specifically, when the
    ball has moved a distance y into the ground, the flat face has area
                                  p             2
                           A=π      r2 − (r − y)2 = πy(2r − y) ≈ 2πry

    where we assumed that y ≪ r at all times, which is reasonable as long as the ball’s initial
    speed is not enormous. As a result, the pressure of the volleyball exerts a force

                                            F = 2πr ∆P y

    on the ground. This assumes the pressure inside the volleyball remains uniform, and that
    the rest of the volleyball stays approximately spherical, which is again reasonable as long as
    the initial speed is not huge.

    Assuming the initial velocity is not too small, gravity is negligible during the collision, so
    during the collision the force on the volleyball is effectively that of an ideal spring. The
    collision lasts for half a period, giving
                                              r        r
                                                m         πm
                                         τ =π        =         .
                                                keff     2r ∆P

    If we plug in realistic numbers, the result is of order 10 ms, which is plausible.

5    Continuous Systems
    Example 10

    As shown in M2, a hanging chain takes the form of a catenary. Suppose you pull the chain
    down in the middle. How does the center of mass of the chain move? Does the answer depend
    on how hard you pull?

    Solution
    No matter how hard you pull, or in what direction, the height of the center of mass always
    goes up! This is because this quantity measures the total gravitational potential energy of
    the chain. If you pull a chain in equilibrium, in any direction whatsoever, you will do work
    on it. So this raises its potential energy, and hence the center of mass.

        Another way of saying this is that the equilibrium position, without the extra pull you supply,
        is already in the lowest energy state, and hence already has the lowest possible center of mass.
        Changing this shape in any way raises the center of mass.

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


[User]
Here is the statement of the traditional twin paradox.

          Bob is an astronaut who leaves home on a rocket with speed v. Alice stays home. After
          time T in Alice’s frame, Bob reverses direction and travels home with speed v. Who, if
          either, has aged more?

    The obvious answer is that Alice has aged more by time dilation. The trouble is explaining why we
    can’t just work in Bob’s frame and conclude that Bob has aged more by time dilation.

      (a) Draw a Minkowski diagram for Alice and Bob where Alice’s worldline is x = 0.

      (b) The reason that working in Bob’s frame is subtle is that it is not a single inertial frame. Draw
          x′ and t′ axes for Bob at several points on Bob’s worldline. Argue that when Bob turns around,
          thereby moving to a different inertial frame, Alice’s age jumps upward. (Using the results of
          chapter 11 of Morin, you can even show that the amount of aging is exactly what is needed,
          using the Minkowski diagram alone.)

    This illustrates why the situation is not symmetric between Alice and Bob. But this resolution of
    the twin paradox is a little unphysical. It does explain what goes wrong working in Bob’s frame,
    but it’s not related to what Bob actually physically sees, which is determined by when photons
    from Earth reach his eyes; nothing about that changes discontinuously when he turns around.

      (c) More physically, let us suppose that Bob continually emits radiation of frequency f (in his
          frame) towards Alice. Suppose that in Alice’s frame, Bob travels with speed v, reaches a
          maximum distance L from Alice, and accelerates quickly to return with speed v. If Alice sees
          Nb wave crests in total during Bob’s trip, then Bob has aged by Nb /f . Use the relativistic
          Doppler effect to compute Nb /f , working entirely from Alice’s perspective.

      (d) Now suppose Alice continually emits radiation of frequency f (in her frame) towards Bob. If
          Bob sees Na wave crests, use the relativistic Doppler effect to compute Na /f , working entirely
          from Bob’s perspective. If you’re careful, this should differ from the answer to (c).

 (e) [A] Now consider a trickier example. Suppose Alice and Bob live on a torus, i.e. a spacetime
     where the point (x, y, z, t) is the same as the point (x + L, y, z, t). Alice stays home, while
      Bob leaves on a rocket with velocity vx̂. After a while, Bob returns home, without having
      done any acceleration along the way! It seems like the resolution above does not apply, so
     who, if either, has aged more? Can you explain the results from Bob’s reference frame?

  Remark
  The above problem on the twin paradox is quite long. Every physics textbook that covers
  relativity mentions the twin paradox, and Morin even has a whole appendix with five
  different resolutions of it. But it’s not that hard to resolve, so why spend so much energy on it?

  The answer is that seemingly intelligent people really can get stuck on these things for
  years, or even decades. As an example, consider the honored English astronomer and
  philosopher Herbert Dingle. Dingle wrote several textbooks, served as president of the Royal
  Astronomical Society, and even wrote popular essays introducing relativity. But late in life,
  he started to believe that relativity could not explain the twin paradox.

  Dingle’s arguments contained thousands of words of vague, equation-free prose. Here’s the
  simplest mathematical formalization of what he meant. In the Lorentz transformations

               x′ = γ(x − vt),    t′ = γ(t − vx),       x = γ(x′ + vt′ ),   t = γ(t′ + vx′ ),

  we notice that if we set x = 0, then t′ = γt, while if we set x′ = 0, then t = γt′ . Therefore,
  aging in the twin paradox must be symmetric, γ = 1/γ. However, this implies γ = 1, so that
  time dilation doesn’t happen at all, and relativity collapses.

  Dingle continued pushing this for the rest of his life, writing endless letters and articles,
  and even publishing a rambling book, Science at the Crossroads, which warned of the grave
  dangers of trusting relativity. Today, it is a favorite of flat Earthers. And it’s far from the
  only example. For instance, there was a book published in Nazi Germany called A Hundred
  Authors Against Einstein, where a vast array of philosophers argued that relativity had to be
  wrong, because it contradicted the metaphysical system of the native German, 18th century
  philosopher Immanuel Kant. Kant’s ideas about space and time, they said, could be proven
  true by verbal reasoning alone, so any theory or experiment saying otherwise had to be wrong.

  If there’s a lesson to be drawn from this bizarre history, it’s that the ability to write or speak is
  not the same as the ability to think. People can churn out pages of flowing prose without ever
  having a single coherent thought. Physicists learn to think by solving well-defined problems
  mathematically. Many others never gain this skill, and spend their whole lives drunkenly
  stumbling from word to word. The real tragedy is that such people often grow to believe
  that no better method of reasoning can exist.

    4    Four-Vectors
        Idea 4

        A four-vector V µ is a set of four quantities (V 0 , V 1 , V 2 , V 3 ) that transform in the same
        manner as (ct, x, y, z). The inner product of two four-vectors is defined as

                                 V · W = V 0W 0 − V 1W 1 − V 2W 2 − V 3W 3.

        It is invariant under Lorentz transformations. By convention, V · W is also written as V µ Wµ .

Solution:
(a) Here’s the result.

 (b) Bob’s x′ and ct′ axes before and after the acceleration are also displayed. We see that as these
     axes rotate during the acceleration, Alice’s age changes extremely quickly.
    This might feel strange, but it’s really just an artifact of changing reference frames. As a
     simpler example, suppose you were a surveyor trying to measure the height of a mountain,
    which can be done by measuring the angle to its summit with respect to a horizontal level. If
     the surveyor then gets on an accelerating car, their horizontal level will tilt, causing the height
     reading to change extremely quickly. But that doesn’t mean people living on the mountain
    will be flung off! They don’t feel anything; it’s just the surveyor’s notion of horizontal that
     changed. Similarly, when Bob turns around, his definition of time changes, so that Alice’s age
    “right now” (according to Bob) suddenly changes.

 (c) The radiation that was emitted while Bob was moving away from Alice is received by Alice
     with redshifted frequency                   r
                                                    1−v
                                           fr =           f.
                                                    1+v
     The radiation that was emitted while Bob was moving towards Alice is received by Alice with
     blueshifted frequency                       r
                                                    1+v
                                           fb =           f.
                                                    1−v
     Suppose that Alice sees these frequencies for times tr and tb . Then the answer is

                                              Nb = tr fr + tb fb .

     It remains to compute tr and tb . Naively we would say tr = tb = L/v, because that’s how
     long Bob spends moving towards and away from Alice respectively, but this question is about
     what Alice sees with her eyeballs. The transition point between the two phases is when the
     radiation that Bob emitted while turning around gets to Alice. In other words,
                                              L L               L L
                                       tr =     + ,      tb =     − .
                                              v  c              v  c

     Then we have
                                           r                r     !
                          Lf                   1−v            1+v     2Lf p
                     Nb =        (v + 1)           + (1 − v)        =       1 − v2
                           v                   1+v            1−v      v

     so Bob has aged by 2L/γv, exactly as expected. Physically, Alice sees Bob aging in slow
     motion for more than half the time, and aging in fast motion for less than half the time, with
     the overall effect of Alice aging more.

 (d) Let’s define all the terms as in the previous part. Bob turns around when he is a distance
     L/γ (according to him) from Alice. The fundamental difference is that Bob starts seeing the
     higher frequency the instant he turns around, so
                                                             L
                                                 tr = tb =      .
                                                             γv
     Therefore, we have
                                           r           r         !
                                 Lf            1−v         1+v           2Lf   1
                            Na =                   +                 =       √
                                 γv            1+v         1−v            γv 1 − v 2

     so Alice has aged by 2L/v, exactly as expected. Physically, Bob sees Alice aging in slow
     motion for half the time, and aging in fast motion for half the time, with the overall effect of
     Alice aging more. Again, note that the fundamental asymmetry is due to Bob being the one
     accelerating, which is baked into how we computed the tr and tb .

 (e) This example was first considered in 1973 by this paper, and reviewed pedagogically by this
     paper. In this exotic spacetime, there really is a notion of absolute rest: we can unambiguously
     say that Bob moved and Alice didn’t, so Alice has aged more. The reason is that the torus
     itself picks out a special frame. Only in Alice’s frame is it true that when you wrap around
     the edge of the torus, you emerge on the other end at the same time. In Bob’s frame, this
     isn’t true, by loss of simultaneity. In Bob’s frame, Alice gets to the edge of the torus, then
     emerges out the other edge at a later time, which ultimately makes her older than Bob when
     she returns.
     The more general lesson is that while special relativity restricts the forms of physical laws to
     have certain symmetries, it doesn’t mean that the solutions of the corresponding equations
     must always have the same symmetry. The dynamics of salt molecules in solution obey
     perfect rotational symmetry, but when they crystallize, the faces of the crystal pick out
     special directions. Likewise, as far as we’ve ever measured, all of the dynamics in our universe
     perfectly obey the Lorentz symmetry of special relativity, but the cosmic microwave background
     radiation does provide an absolute rest frame.

  Remark
  The above problem on the twin paradox is quite long. Every physics textbook that covers
  relativity mentions the twin paradox, and Morin even has a whole appendix with five
  different resolutions of it. But it’s not that hard to resolve, so why spend so much energy on it?

        The answer is that seemingly intelligent people really can get stuck on these things for
        years, or even decades. As an example, consider the honored English astronomer and
        philosopher Herbert Dingle. Dingle wrote several textbooks, served as president of the Royal
        Astronomical Society, and even wrote popular essays introducing relativity. But late in life,
        he started to believe that relativity could not explain the twin paradox.

        Dingle’s arguments contained thousands of words of vague, equation-free prose. Here’s the
        simplest mathematical formalization of what he meant. In the Lorentz transformations

                     x′ = γ(x − vt),    t′ = γ(t − vx),    x = γ(x′ + vt′ ),   t = γ(t′ + vx′ ),

        we notice that if we set x = 0, then t′ = γt, while if we set x′ = 0, then t = γt′ . Therefore,
        aging in the twin paradox must be symmetric, γ = 1/γ. However, this implies γ = 1, so that
        time dilation doesn’t happen at all, and relativity collapses.

        Dingle continued pushing this for the rest of his life, writing endless letters and articles,
        and even publishing a rambling book, Science at the Crossroads, which warned of the grave
        dangers of trusting relativity. Today, it is a favorite of flat Earthers. And it’s far from the
        only example. For instance, there was a book published in Nazi Germany called A Hundred
        Authors Against Einstein, where a vast array of philosophers argued that relativity had to be
        wrong, because it contradicted the metaphysical system of the native German, 18th century
        philosopher Immanuel Kant. Kant’s ideas about space and time, they said, could be proven
        true by verbal reasoning alone, so any theory or experiment saying otherwise had to be wrong.

        If there’s a lesson to be drawn from this bizarre history, it’s that the ability to write or speak is
        not the same as the ability to think. People can churn out pages of flowing prose without ever
        having a single coherent thought. Physicists learn to think by solving well-defined problems
        mathematically. Many others never gain this skill, and spend their whole lives drunkenly
        stumbling from word to word. The real tragedy is that such people often grow to believe
        that no better method of reasoning can exist.

    4    Four-Vectors
        Idea 4

        A four-vector V µ is a set of four quantities (V 0 , V 1 , V 2 , V 3 ) that transform in the same
        manner as (ct, x, y, z). The inner product of two four-vectors is defined as

                                  V · W = V 0W 0 − V 1W 1 − V 2W 2 − V 3W 3.

        It is invariant under Lorentz transformations. By convention, V · W is also written as V µ Wµ .


[User] Horisontaalsel siledal dielektrilisel pinnal asuvad kaks laetud klotsi massidega $m$ ja samanimeliste laengutega $q$. Alghetkel on vahemaa nende vahel $l$. Mis tingimusel hakkavad klotsid liikuma ja kui suur on vahemaa $L$ nende vahel, kui liikumine lõppeb? Hõõrdetegur klotside ja pinna vahel on $\mu$. Klotside mõõtmeid ja liikuvate laengute elektromagnetkiirgust mitte arvestada. Pinna dielektriline läbitavus on 1.

Solution:
Klots hakkab liikuma, kui sellele mõjuv elektrostaatiline jõud ületab maksimaalsehõõrdejõu:
\[
\frac{kq^2}{l^2} > \mu mg.
\]
Klotside läbitava vahemaa leiame energia jäävuse seadusest. Punktis, milles klots peatub, on elektrostaatilise välja potentsiaalne energia väiksem kui algpunktis. Potentsiaalsete energiate vahe muundub liikumise käigus klotside kineetiliseks energiaks, mis omakorda hõõrdejõu töö tulemusena muundub soojuseks.

Kui kuulide vahemaa liikuma hakkamisel oli $l$ ning seisma jäämise hetkel $L$, siis muutus elektrostaatilises potentsiaalses energias on
\[
\Delta E = \frac{kq^2}{l} - \frac{kq^2}{L}.
\]
Hõõrdejõud $\mu mg$ mõjub kummagile klotsile vahemaa $(L-l)/2$ jooksul, seega kogu
hõõrdejõu töö on
\[
A = \mu mg (L - l).
\]
Et need energia muudud on võrdsed, saame vahemaa $L$ jaoks lihtsa võrrandi:
\[
\begin{aligned}
k q^{2}\left(\frac{1}{l}-\frac{1}{L}\right)=\mu m g(L-l) &\Rightarrow \frac{k q^{2}}{\mu m g} \frac{L-l}{l L}=L-l \Rightarrow\\
&\Rightarrow L=\frac{k q^{2}}{\mu m g l}
\end{aligned}
\]

Laetud klotsid

Horisontaalsel siledal dielektrilisel pinnal asuvad kaks laetud klotsi massidega $m$ ja samanimeliste laengutega $q$. Alghetkel on vahemaa nende vahel $l$. Mis tingimusel hakkavad klotsid liikuma ja kui suur on vahemaa $L$ nende vahel, kui liikumine lõppeb? Hõõrdetegur klotside ja pinna vahel on $\mu$. Klotside mõõtmeid ja liikuvate laengute elektromagnetkiirgust mitte arvestada. Pinna dielektriline läbitavus on 1.

Klotside läbitav vahemaa on leitav energia jäävuse seadusest. Nimelt on klotsid paigal nii alg- kui ka lõppasendis ning ainus viis soojuse eraldumiseks on hõõrdejõu kaudu, mis on omakorda avaldatav alg- ja lõppasendi potentsiaalsete energiate vahest.


[User]
1. FOUR CHARGEs (7 points)

— Päivo Simson,
Jaan Kalda.
Fouridenticalparticlesareinitiallyinthe
corners of a square, as shown in the figure
below. All particles have the same charge q,
mass m, and the same magnitude of initial
velocity v0. The directions of the initial ve‐
locities are indicated in the figure. You can
assume v << c and ignore gravity.
i)(2points)Afteralongtimehaspassed,what
isthemagnitudeofthefinalvelocityvf ofthe
particleswithrespecttothecenterofmassof
the system?
ii) (5 points) What is the angle between the
initial velocity ⃗ v0 and the final velocity ⃗ vf of
a particle?

Solution:
1. FOUR CHARGEs

— Solution by
PäivoSimson,gradingschemesbyPäivoSimson,
Oleg Košik and Lasse Franti.
i) (2 points) Assuming v << c and ignoring
gravity, the initial total energy of the system
is the sum of classical kinetic and potential
energies
Etotal = Ekin + Eelectric =
Ekin +
1
2
Σi̸=j k
qiqj
rij
=
4
mv2
0
2
+ 4
kq2
L
+ 2
kq2
√
2L
=
= 4
mv2
0
2
+
kq2
L
4 +
√
2
4
!
.
After the particles have moved infinitely far
from each other, the potential energy be‐
comes zero and the total energy is only kin‐
etic:
Etotal = 4
mv2
f
2
.
Since the total energy does not change we
have from the equality of these two expres‐
sions
vf =
s
v2
0 +
kq2
Lm
4 +
√
2
2
.
Grading:
• Idea of using energy conservation (0.2pts)
• Ideaofsymmetryandequalityofquantities
(0.2pts)
• Idea of total energy as a sum of kinetic and
electrostatic (0.2pts)
• Formula for electrostatic energy including
the two different distances (0.3pts)
• Factor 1
2 from pairings (0.4pts)
• Final energy is purely kinetic (0.4pts)
• Correct final answer (0.3pts)
• If only dimensionless factors missing and
final answer is reasonable (0.2pts)
ii) (5 points)
Let the required angle be φ.
The expression for the total energy re‐
veals that the particles behave effectively in‐
dependent of each other as if they are only
subjecttoacentralfieldgeneratedbyasingle
effective charge
Qe = q
2
√
2 + 1
4
≈ 0.96q,
fixed at the center of mass of the system,
and initially at a distance L/
√
2 from each
particle. This observation allows us to re‐
duce the initial four‐body problem into four
independent (and identical) two‐body prob‐
lemsinvolvingonlythespatiallyfixedcharge
Qe and one moving charge q.The trajectories
are therefore hyperbolas with one focus loc‐
ated at the position of the effective charge.
Since the force is repulsive, the effective
charge is at the focus F1 as shown in the fol‐
lowing figure.
Let’s now calculate the asymptote angle
α, since the required angle φ is just π/2 − α.
A hyperbola is defined as a set of points
P, such that the absolute difference of the
distances from P to two fixed points F1 and
F2 (the foci) is constant. Let the distances be
r1 and r2. From the figure it is easy to see
that the constant is equal to 2a: if we take
the point P on the x‐axis, then r1 − r2 =
c + a − (c − a) = 2a.
To determine the asymptote angle α, let
us look at the triangle F1PF2. From the law
of cosines, we have
r2
2 = r2
1 + 4c2
− 4cr1 cosβ.
Solving for cosβ and using r2 = r1 − 2a we
get
cosβ =
r2
1 − r2
2 + 4c2
4cr1
=
a
c
+
c2
− a2
cr1
.
As r1 → ∞, then β → α, and we get the
expression for the cosine of the asymptote
angle:
cosα =
a
c
.
Now using c = L/
√
2 − a and knowing that
E =
kqQe
2a
=
mv2
0
2
+
√
2kqQe
L
(the vis‐viva equation for the hyperbola),
from which
1
a
=
mv2
0
kqQe
+
2
√
2
L
,
we finally get
sinφ = cosα =
a
c
=
1
L √
2a
− 1
=
=
1
1 +
Lmv2
0 √
2kqQe
=
1
1 +
4Lmv2
0
kq2(4+
√
2)
.
Grading:
• Effective repulsive central force F = A
r2
pointingfromtheCOM(explicitexpression
or statement is required) (1pts)
• Realising hyperbolic motion (0.5pts)
• Central charge is located at the correct fo‐
cus (0.5pts)
• Expression for the angle φ or α in terms of
geometrical parameters e.g. cosα = a/c,
tanα = b/a (1pts)
• Vis‐viva equation OR angular momentum
conservation (1pts)
• Deriving final answer (1pts)
Subpoints for angular momentum conservation:
• Mentionedthatangularmomentumiscon‐
served (0.3pts)
• Writes initial angular momentum around
COM using L and v0 (0.3pts)
• Writes final angular momentum around
COM using vf (0.4pts)
Variation 1.
If the student doesn’t know the vis‐viva
equation for the hyperbola
E =
kqQe
2a
,
which differs from the corresponding equa‐
tionfortheellipseonlybythesignofthetotal
energy, it can be derived as follows.
From the asymptote angle formula
cosα =
a
c
the geometric relationships shown in the fol‐
lowing figure follow immediately.
We see that the two right triangles on the
figure are identical.
From the conservation of angular mo‐
mentum (with respect to the point F1) we
have
m(c + a)v0 = mbvf =⇒ vf = v0
c + a
b
,
and from the conservation of energy using
b2
= c2
− a2
kqQe
a + c
=
mv2
f
2
−
mv2
0
2
=
=
mv2
0
2

(c + a)2
c2 − a2
− 1

=
mv2
0
2
2a
c − a
,
from which
mv2
0
2
=
c − a
2a
·
kqQe
c + a
.
Inserting this into the total energy expres‐
sion we finally get
E = kqQe

1
2a
·
c − a
c + a
+
1
c + a

=
kqQe
2a
.
Variation 2.
Using the algebraic equation for the hy‐
perbola
x2
a2
−
y2
b2
= 1,
and knowing that b2
= c2
− a2
, we get
tanα = lim
x→∞
y
x
= lim
x→∞
b
r
1
a2
−
1
x2
=
b
a
.
Variation 3.
The vis‐viva equation step can be done
using conservation of angular momentum
around COM, as all the forces are radial.
L = mv0
L
√
2
= mvfb
since we already know a + c = L √
2
, c2
=
a2
+b2
, we can solve for the hyperbolas para‐
meters.

FOUR CHARGEs


[User]
Veepüstoliga (vt joonist) tekitatakse veejuga, surudes vett läbi kitsa silindrilise suudme, mille sisediameeter on $d_2 = \SI{1}{mm}$. Päästik on ühendatud kolviga, mis saab tihedalt liikuda silindrilises torus diameetriga $d_1 = \SI{1}{cm}$. Oletagem, et sõrmed suruvad päästikule jõuga $F = \SI{20}{N}$ (jõu rakenduspunkt ja suund on näidatud joonisel). Kui suure kiirusega väljub veejuga püstolist? Vee liikumise võib lugeda laminaarseks, vee viskoossust ja püstoli liikuvatele osadele mõjuvaid hõõrdejõude võib ignoreerida. Vee tihedus on $\rho = \SI{1000}{kg/m^3}$

\begin{center}
	\includegraphics[width=0.6\linewidth]{2006-v3g-04-yl}
\end{center}

Solution:
Kuna me ei arvesta dissipatiivseid efekte, siis peab kehtima mehaanilise energia jäävus. Liikugu kolb jõu $F$ toimel kiirusega $v_1$ ja olgu vee kiirus suudmes $v_2$. Kuna vee koguruumala ei muutu, siis $v_1S_1 = v_2S_2$. Ajavahemiku $t$ vältel
läbib kolb vahemaa $x = v_1t$ tehes tööd $A = F x$. Vaatame, milline on sellele liikumisele vastav vee summaarne kineetilise energia muut. Ühelt poolt kolvi eest \enquote{kaob ära} veehulk $m = S_1x\rho$, mille kiirus oli $v_1$, teiselt poolt ilmub suudmesse sama kogus vett liikudes kiirusega $v_2$. Energia jäävus annab
\[
A+\frac{m v_{1}^{2}}{2}=\frac{m v_{2}^{2}}{2}
\]
ehk peale asendamist ja $v_1$ elimineerimist
\[
\frac{F}{S_{1}}+\frac{\rho v_{2}^{2}}{2} \frac{S_{2}^{2}}{S_{1}^{2}}=\frac{\rho v_{2}^{2}}{2}.
\]
Aga $d_2 \ll d_1$ ja $S \propto d^2$, seega ammugi $S_2^2 \ll S_1^2$ ning teise liikme vasakul pool võrdusmärki võib ära jätta. Avaldame $v_2$:
\[
v_{2}=\frac{1}{d_{1}} \sqrt{\frac{8 F}{\pi \rho}} \approx \SI{22,6}{m/s}.
\]

\emph{Märkus:} Bernoulli seadusele tuginedes oleks saanud ülesande lahendada ka lühemalt:
\[
p+\frac{\rho v^{2}}{2}=\const,
\]
mille me lahendamise käigus sisuliselt tuletasime.

Veepüstol

Veepüstoliga (vt joonist) tekitatakse veejuga, surudes vett läbi kitsa silindrilise suudme, mille sisediameeter on $d_2 = \SI{1}{mm}$. Päästik on ühendatud kolviga, mis saab tihedalt liikuda silindrilises torus diameetriga $d_1 = \SI{1}{cm}$. Oletagem, et sõrmed suruvad päästikule jõuga $F = \SI{20}{N}$ (jõu rakenduspunkt ja suund on näidatud joonisel). Kui suure kiirusega väljub veejuga püstolist? Vee liikumise võib lugeda laminaarseks, vee viskoossust ja püstoli liikuvatele osadele mõjuvaid hõõrdejõude võib ignoreerida. Vee tihedus on $\rho = \SI{1000}{kg/m^3}$

\begin{center}
	\includegraphics[width=0.6\linewidth]{2006-v3g-04-yl}
\end{center}

Kolbile mõjuv jõud tekitab kolvi sees lisarõhu $F/S$. Kuna tegu on laminaarse vooga, kehtib Bernoulli seadus. Alternatiivselt võib rakendada energia jäävust kolvi ees ja suudme juures.


[User]
8. PHAsE sPıRAL (9 points)

— Taavet Kalda.
Here we shall study the motion of Milky Way
stars in the Solar neighbourhood in the dir‐
ection of the z‐axis, i.e. perpendicular to
the galactic plane. For our purposes, we
can model the galactic gravity field as be‐
ing created by a continuous mass density ρ
(that accounts for the masses of stars, dark
matter, gas, interstellar dust, etc), and as‐
sumethatthismassformsaninfinitemirror‐
symmetric plate, i.e. ρ ≡ ρ(z) and ρ(z) =
ρ(−z) is independent of x and y. Through‐
out the problem, you may assume that each
star’s total energy is conserved over the en‐
tire considered time period. Gravitational
constant G = 6.67 × 10−11
m3
kg−1
s−2
=
4.30 × 10−3
pcM−1
⊙ (km/s)2
.
i) (1 point) Assuming that the mass density
is constant over the plate’s thickness. i.e.
ρ(z) = ρ0, what is the acceleration az of a
star at a distance z from the mid‐plane?
ii) (0.5 points)Consider a star that starts with
zero velocity at a distance of z = a from the
mid‐plane. With what period does it start os‐
cillating around the mid‐plane?
Inreality, densitydecreaseswithgrowing
|z|. Measuring density has been a great chal‐
lenge because of contributions from dark
and other difficult‐to‐see matter. Here, we
consider a breakthrough method of doing
it. Consider the distributions of the stars in
our neighbourhood on the z–vz phase plane,
where each star is a dot with coordinates
(vz,z); vz denotes the z‐component the star’s
velocity, and z – the vertical coordinate. Ini‐
tially, these dots were distributed nearly ho‐
mogeneously, but some time ago, the Milky
Way was perturbed externally, probably by
a passing‐by dwarf galaxy; this shuffled the
positions and velocities of stars, creating a
bar‐shaped overdensity region. When mov‐
ing within that bar‐shaped region from the
centre to the periphery, the total energy per
mass of stars increased monotonously. Over
time, this overdensity region started “wind‐
ing up”, due to the oscillation periods of stars
intheverticalplanedependingontheiroscil‐
lationamplitudezm,andevolvedintoaspiral
pictured below (Antoja et al. 2018, Nature
561, 360). An observation that you need to
exploit below is that the ordering of stars by
energies along the spiral today remains the
same as it was at the time of perturbation.
Theoscillationperiodofstarsdependson
the amplitude zm because the gravitational
potential (the potential energy per mass)
Φ(z) is not parabolic. In such a case, the
periodcanbeapproximatelyfoundbysubsti‐
tuting the real Φ(z) with a kz2
matching Φ(z)
at z = zm, i.e. with k = Φ(zm)/z2
m.
−1.5 −1.0 −0.5 0.0 0.5 1.0 1.5
z (kpc)
−60
−40
−20
0
20
40
60
v z (km/s)
iii) (2.5 points) At the intersection points of
the spiral with vz = 0, calculate Φ(z) by in‐
terpolating data linearly where appropriate;
plot your results (this follows the analysis of
Guo et al. 2024, ApJ, 960, 133).
iv) (1 point) Assuming that the mass density
is almost constant for |z| ≤ 0.3kpc, what is
the mass density near z = 0?
v)(2points)Darkmatterisan“invisible”form
of matter that only interacts by gravity. In
general, it is found that dark matter forms
halos that extend significantly farther than
visible matter structures. By assuming that
the dark matter density doesn’t vary signific‐
antlywithinthevolumeofinterestandthatit
starts dominating far away from the galactic
plane, from around z = 0.7kpc, estimate the
local dark matter density ρDM.
vi) (2 points)How long ago did the perturba‐
tion occur?

Solution:
8. PHAsE sPıRAL

— Taavet Kalda.
i) (1 point) Gravitational acceleration obeys
Gauss’ law, i.e., the number of field lines
passing through a closed surface is propor‐
tional to the enclosed mass. We can see from
the example of a point mass M that
∫
gdA =
4πGM. Applied for the case of an infinite
plane with constant density with a cuboid of
area A and half‐thickness z, we get −2azA =
4πG2Azρ0 so
az = −4πGρ0z.
Grading: (preliminary)
• Idea of using Gauss’ law 0.3pts
• Formularelatingthemassinsidewithgrav‐
itational flux 0.3pts
• Application of Gauss’ law on a cuboid
0.2pts
• Final result 0.2pts
Alternative solution:
• Finding the acceleration of a thin disk by
integrating over its surface, of which:
0.5pts
– Writing down the integral 0.3pts
– Correct evaluation, including finding
that the acceleration is independent of
the displacement from the surface
0.2pts
• Inferring that only the surface density in‐
side −a < z < a contributes to the final
acceleration 0.3pts
• Final result 0.2pts
ii) (0.5 points) The acceleration is propor‐
tional to displacement and therefore corres‐
ponds to a harmonic oscillator. The period
of oscillation is thus
T =
2π
√
4πGρ0
=
√
π
Gρ0
.
Grading: (preliminary)
• Noticingthatthemovementisthatofahar‐
monic oscillator 0.3pts
• Expressionfortheoscillationperiod0.2pts
iii) (2.5 points) If we follow the trajectory of
a single star that lies on the spiral, we would
find it oscillating around the mid‐plane with
some period T(z) that decreases with in‐
creasing z. Over the course of an orbit, the
energy per unit mass is conserved and is
givenbyE = v2
z/2+Φ(z). Weknowthatnear
the mid‐plane, the gravitational potential is
minimal and equal to zero, so vz is maximal
and the kinetic energy is equal to the total
energy. Hence, we know the total energy of
stars at the seven intersection points of the
spiral with z = 0. Similarly, when vz = 0, the
kinetic energy is minimal and equal to zero,
so the total energy is equal to Φ(z) at those
points.
As one traces the various intersection
points with vz = 0 and z = 0 along the
spiral, the maximal extent of the orbit keeps
increasing. To the first order, if we assume
that the maximal extent increases linearly
with each crossing, we can find the potential
energy of the crossings of vz = 0 as the aver‐
agebetweenthekineticenergiesoftheprevi‐
ousandsubsequentcrossingoverz = 0. This
allows us to determine the potential energy
at all the crossings with vz = 0, as tabulated
and plotted below.
i zi (kpc) Φ(zi) (km2
/s2
)
1 0.27 180
2 0.39 330
3 0.54 530
4 0.72 800
5 0.97 1200
6 1.34 1800
Grading: (preliminary)
• Making use of the total energy at z = 0 in‐
tersection points being known (either ex‐
plicitly or implicitly) 0.7pts
• Interpolating the values at vz = 0 from
neighbouring z = 0 crossovers. This
should be explicitly mentioned 0.8pts
• Tabulating the potential, of which: 0.7pts
– using six points 0.7/0.7pts
– using four to five points 0.4/0.7pts
– using one to three points 0.1/0.7pts
• Φ(z) vs z correctly plotted 0.3pts
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4
z (kpc)
0
250
500
750
1000
1250
1500
1750
2000
Φ(z) (km 2
/s 2
)
iv) (1point)For a harmonic oscillator, the po‐
tentialenergywouldgrowasz2
. Basedonthe
plotofpotentialenergyweobtained,itseems
to grow roughly quadratically in the begin‐
ning, and then transition into a more lin‐
ear regime, implying that smaller values of z
have more uniform ρ. Taking the first value
of Φ(z1) = 180km2
/s2
that we obtained, we
can estimate the mid‐plane density. From
the first part, Φ(z) =
∫
azdz = 2πGρ0z2
and
so
ρ0 =
Φ(z1)
2πGz2
1
= 6.1 × 10−21
kg/m3
= 0.090M⊙/pc3
.
Grading: (preliminary)
• ConnectingΦ(z1)withρ0 byassumingcon‐
stant mass density. 0.8pts
• If the first data point is not used −0.2pts
• Final expression for ρ0 0.1pts
• Numerical value within 10% 0.1pts
v) (2 points) We can compute the enclosed
surface density Σ(z) between 0 < z by us‐
ing the previous harmonic oscillator estim‐
ate. With the constant density approxima‐
tion, the surface density is Σ(z) = ρ0z =
Φ(z)/(2πGz)
Ofcourse,here,ρ0 isaplaceholdervariablewhile
using the constant profile approximation to sim
plify the calculus. The final result is expected to
deviatefromthetruevaluebyanumericalfactor
that’s close to unity.
Assuming that dark matter density dom‐
inates far away, we can use the difference
betweenthefarthesttwodatapointsatz5 and
z6 to estimate the dark matter density via
Σ(z6) − Σ(z5) = ρDM(z6 − z5)
=
1
2πG
(
Φ(z6)
z6
−
Φ(z5)
z5
)
.
Thus,
ρDM ≈
1
2πG(z6 − z5)
(
Φ(z6)
z6
−
Φ(z5)
z5
)
= 7.7 × 10−22
kg/m3
= 0.011M⊙/pc3
.
Dark matter therefore makes up around
ρDM/ρ0 = 13% of the total local matter
budget.
Grading: (preliminary)
• Obtaining an expression for the total mass
contained within z 0.7pts
• Taking the difference between the total
mass within z6 and z5 for calculating the
dark matter content 0.9pts
• Final expression for ρDM based on z6 and
z5 0.3pts
• Numerical value within 10% 0.1pts
Alternative scheme:
• Using the previous expression for ρ based
onΦ(z)toexpress(z6−z5)ρDM = z6ρ(z6)−
z5ρ(z5) 1.6pts
• Final expression for ρDM based on z6 and
z5 0.3pts
• Numerical value within 10% 0.1pts
vi) (2 points) We can estimate the time of
the perturbation by using the winding rate
between two points on the spiral and how
many full turns around the origin they have
made relative to each other. Using the har‐
monic estimate, firstly, ρ0 = Φ(z)/(2πGz2
)
and secondly the angular frequency is
ω(z) =
√
4πGρ0 =
√
2Φ(z)/z2.
By, for example, picking points z1 and z6 and
seeing that they have 2.5 full turns between
them, we can express how long ago the per‐
turbation happened:
T0 = 2.5
2π
ω(z6) − ω(z5)
= 5π
(√
2Φ(z1)/z2
1 −
√
2Φ(z6)/z2
6
)−1
= 1.9 × 1016
s = 620Myr.
The timescale is relatively long, but com‐
paredtothelifespanoftheMilkyWay, which
is around 13.6 billion years, it’s relatively re‐
cent.
Grading: (preliminary)
• Idea of using differences in the winding
rate between two points on the spiral
1.0pts
• Expression for angular frequency ω in
terms of Φ(z) by assuming a harmonic os‐
cillator 0.5pts
• Picking two points and connecting the age
of the spiral, ω and the winding amount
0.3pts
• Numerical value within 10% of the solu‐
tion value 0.2pts

PHAsE sPıRAL


[User]
2. Consider a particle of mass m confined to a one dimensional box of length L. The particle moves
   in the box with momentum p colliding elastically with the walls. We consider the quantum
   mechanics of this system. As far as possible, express your answers in terms of α = h2 /8m.
   (a) At each energy state, the particle may be represented by a standing wave given by the de         [1]
       Broglie hypothesis. Express its wavelengths λdB in terms of L in the nth energy state.

         λdB =

INPhO 2019 Page 2          Questions & Summary Answers                    Last four digits of Roll No.:
   (b) Write the energy of the nth energy state, En .                                                       [1]

         En =

    (c) Let there be N (mass m) electrons in this box where N is an even number. Obtain the                 [2]
        expression for the lowest possible total energy U0 of the system (e.g., the ground state energy
        of this N -particle system). Neglect coulombic interaction between the electrons.

         U0 =

   (d) Express the total energy U1 in terms of U0 and relevant quantities when the system is in the       [31/2]
       first excited state. Also express the total energy U2 in terms of U0 and relevant quantities
       when the system is in the second excited state.

         U1 =

         U2 =

                                     C1                 9
    (e) When the system is in the ground state, let the length of the box change slowly from L to           [1]
        L − ∆L. Obtain the magnitude of the force F on each wall in terms of U0 , when ∆L  L.

         F =

                HB
    (f) Assuming N is large (N  1) obtain the ratio r of dU0 /dN to the energy level of the highest
        occupied ground state.
                                                                                [1]

         r=

   (g) We assume once again that N is large. Consider the possibility of the electrons forming a          [11/2]
       uniform continuum of length L with constant linear density. Using dimensional analysis,
       calculate the gravitational energy of this system UG assuming that it depends on its total
       mass, universal gravitational constant G and L. Equate this (attractive) energy to the
       (repulsive) energy U0 (N ). Obtain L in terms of N and related quantities.

         UG =

         L=

                                 Detailed answers can be found on page numbers:

INPhO 2019 Page 3          Questions & Summary Answers

Solution:
2. Consider a particle of mass m confined to a one dimensional box of length L. The particle moves
   in the box with momentum p colliding elastically with the walls. We consider the quantum
   mechanics of this system. As far as possible, express your answers in terms of α = h2 /8m.
   (a) At each energy state, the particle may be represented by a standing wave given by the de           [1]
       Broglie hypothesis. Express its wavelengths λdB in terms of L in the nth energy state.

         λdB =

                            2L
          Solution: λdB =
                             n

   (b) Write the energy of the nth energy state, En .                                                     [1]

         En =

          Solution:

                                              C1
                                            En =
                                                   p2
                                                   2m
                                                      =
                                                        n2 h2
                                                        8mL2
                                                              9
                                                              =
                                                                αn2
                                                                L2
                                                                                (2.1)

          where we have used p = h/λdB .

    (c) Let there be N (mass m) electrons in this box where N is an even number. Obtain the ex-           [2]

                      HB
        pression for the lowest possible total energy U0 of the system (e.g., the ground state energy
        of this N -particle system). Neglect coulombic interaction between the electrons.

         U0 =

          Solution: Highest occupied level is nmax = N/2 and each level is occupied by the two
          electrons.
                                     N/2            N/2
                                     X            2α X 2 αN (N + 1)(N + 2)
                              U0 =         En =          n =                                  (2.2)
                                     n=1
                                                  L2 n=1       12L2

   (d) Express the total energy U1 in terms of U0 and relevant quantities when the system is in the     [31/2]
       first excited state. Also express the total energy U2 in terms of U0 and relevant quantities
       when the system is in the second excited state.

         U1 =

INPhO 2019 Page 4                Questions & Answers                       Last four digits of Roll No.:

         U2 =

         Solution: 1st excited state configuration is
                                     Energy level     Configuration

                                        N/2 + 1

                                          N/2

                                            α N 2     α             N    2

                                  U1 = U0 −         + 2               +1                         (2.3)
                                           L2 2       L             2
                                            α
                                     = U0 + 2 (N + 1)                                            (2.4)
                                           L

         Solution: 2nd excited state configuration is
                                    Energy level

                                    C1  N/2 + 1

                                          N/2
                                                        9
                                                      Configuration

                                        N/2 − 1

                                                   2                         2

                HB              U2 = U0 −

                                         L
                                          α N
                                         L2 2
                                          α
                                   = U0 + 2 2N
                                               −

                                                 1
                                                       α
                                                     + 2
                                                      L

                                                                        N
                                                                        2
                                                                          +1

    (e) When the system is in the ground state, let the length of the box change slowly from L
        to L − ∆L. Obtain the magnitude of the force F on each wall in terms of U0 , when ∆L  L.
                                                                                (2.5)

                                                                                (2.6)

                                                                                [1]

         F =

         Solution: From the work-energy theorem,

                             F ∆L = Ufinal − Uinitial                                            (2.7)
                                    αN (N + 1)(N + 2)        1     1

                                  =                              −                               (2.8)
                                              12        (L − ∆L)2 L2
                                    αN (N + 1)(N + 2)2     2U0
                                F ≈                     =                                        (2.9)
                                             12L3           L

    (f) Assuming N is large (N  1) obtain the ratio r of dU0 /dN to the energy level of the highest       [1]
        occupied ground state.

INPhO 2019 Page 5                Questions & Answers

         r=

         Solution:
                                    dU0    α d
                                        =          (N 3 + 3N 2 + 2N )                        (2.10)
                                    dN    12L2 dN
                                           α
                                        ≈      3N 2                                          (2.11)
                                          12L2
         Highest occupied ground state corresponds to N/2, for which the energy level is αN 2 /4L2 .
         Thus

                                                   r=1                                       (2.12)

   (g) We assume once again that N is large. Consider the possibility of the electrons forming a       [11/2]
       uniform continuum of length L with constant linear density. Using dimensional analysis,
       calculate the gravitational energy of this system UG assuming that it depends on its total
       mass, universal gravitational constant G and L. Equate this (attractive) energy to the

                                                              9
       (repulsive) energy U0 (N ). Obtain L in terms of N and related quantities.

         UG =

                                           C1
         L=

                      HB
         Solution: [UG ] =
                             GM 2
                              L
                                  =
                                    GN 2 m2
                                      L
                                            or −
                                                 GN 2 m2

         Solution: For large N , U0 ≈ αN 3 /12L2 .
                                                   L

                                      GN 2 m2   αN 3
                                              =                                              (2.13)
                                        L       12L2
                                                 αN 3     N h2
                                            L=         =                                     (2.14)
                                                12GM 2   96Gm3

                                Detailed answers can be found on page numbers:

INPhO 2019 Problem 2


[User]
IPhO 2004, problem 1. A nice question on the dynamics of a multi-part system.

    2    The Lorentz Force
        Idea 2
        Some questions below will involve special relativity. The Lorentz force law as written in
        idea 1 is still valid as long as F is interpreted as dp/dt, where the relativistic momentum is

                                               p = γmv,    γ=p            .
                                                              1 − v 2 /c2
        The relativistic energy is also modified to

                                      E = γmc2 = mc2 + mv 2 + . . . .

        We will return to this subject in more detail in R2, but for now this is all you need.

  Example 3: Kalda 163

  A beam of electrons, of mass m and charge q, is emitted with a speed v almost parallel to
  a uniform magnetic field B. The initial velocities of the electrons have an angular spread
  of α ≪ 1, but after a distance L the electrons converge again. Neglecting the interaction
  between the electrons, what is L?

  Solution
  Consider an electron initially traveling at an angle α to the magnetic field. This electron
  has a speed v∥ = v cos α ≈ v parallel to the field, which is unchanged by the magnetic force.
  Instead, the magnetic force causes the perpendicular component v⊥ to rotate about the B
  axis, so that the electron spirals along the field lines.

  The acceleration of the electron has magnitude
                                                F   qv⊥ B
                                         a⊥ =     =
                                                m     m
  and v⊥ goes through a circle in velocity space of circumference 2πv⊥ . After one such circle,
  the total perpendicular displacement is zero, so the beam refocuses. Thus we have
                                           2πv⊥      2πmv
                                      L=        v∥ ≈      .
                                             a        qB
  Since the quantity α dropped out, this setup acts like a magnetic “lens”.

  Example 4: Griffiths 7.50

  In a “betatron”, electrons move in circles in a magnetic field. When the magnetic field is
  slowly increased, the accompanying electric field will impart tangential acceleration.

  Suppose the field always has the same spatial profile B(r, t) = B0 (r)f (t). For what B0 (r) is
  it possible for an electron to start at rest in zero magnetic field, and then move in a circle of
  constant radius as the field is increased?

       Solution
       The electrons experience a tangential force

                                                        Φ̇B  qr
                                          ṗ = qE = q       = Ḃav
                                                        2πr  2
       where Bav is the average field over the orbit. Since the particles start from rest in zero field,
       we can integrate this to find
                                                     qr
                                                 p = Bav .

       On the other hand, the standard result for cyclotron motion is p = qrB, which means we
       must have B = Bav /2, i.e. the field at any radius is half the average magnetic field inside,
                                                    Z r
                                              1 1
                                     B(r) =             B(r′ )(2πr′ ) dr′ .
                                              2 πr2 0

       This rearranges slightly to give
                                                     Z r

                                          r B(r) =          r′ B(r′ ) dr′ .

       Differentiating both sides with respect to r, we have

                                          2rB(r) + r2 B ′ (r) = rB(r)

       which simplifies to
                                                dB       dr
                                                    =−
                                                B        r
       which means the field profile should be B0 (r) ∝ 1/r. (Of course, a real betatron might differ
       since it only needs to obey B = Bav /2 at the radii where electrons will be orbiting.)

[3] Problem 8 (Griffiths 5.17). A large parallel plate capacitor with uniform surface charge σ on the
    upper plate and −σ on the lower is moving with a constant speed v as shown.

     (a) Find the magnetic field between the plates and also above and below them.

     (b) Find the magnetic force per unit area on the upper plate, including its direction.

      (c) What happens to the net force between the plates in the limit v → c? Explain your result
          using some basic ideas from special relativity.

Solution:
(a) Let x̂ be the direction of the velocity. Let ŷ point into the page, and let ẑ point
         up. The magnetic field due to the top plane is − 12 µ0 σvŷ above the above plane and 12 µ0 σvŷ
         below the above plane. Similarly for the lower plane, we have 12 µ0 σvŷ above and − 12 µ0 σvŷ
         below. Thus, the magnetic field is µ0 σvŷ between the plates, and zero outside.

     (b) The force per unit area (i.e. pressure) is σvx̂ × 12 µ0 σvŷ = 12 µ0 σ 2 v 2 ẑ. The factor of 1/2 is there
         since it only feels a force due to the contribution of the other plate; this is the essentially the
         same 1/2 as we found for the pressure on a conductor in E1.

     (c) The forces balance when v = c,
                                                     1              1 2
                                                       µ0 σ 2 c2 =     σ
                                                     2             2ϵ0
         because c2 = 1/ϵ0 µ0 . Thus, as v increases to approach c, the attractive force between the
         plates gets smaller and smaller. If we invoke special relativity, this makes perfect sense. In
         the rest frame of the plates, there is only the attractive electrostatic force, so the plates move
         together. This implies that in the lab frame, the plates also have to move together, so the
         force must be attractive. But for high v, there’s a lot of time dilation, so the plates move
         together more slowly. (This is partially due to the force decreasing, as derived here, and
         partially due to the higher “transverse mass” due to the plates’ relativistic momentum, as
         we’ll see in R2.)


[User]
Two cars are driving after one another with speed $v=\SI{50}{km/h}$. The first car brakes maximally, seeing that the first car brakes the car in the back also brakes maximally. The brakes of the first car are put to use at the exact same time when the brake lights turn on. It takes the rear car $t=\SI{1,5}{s}$ to apply the brakes from the moment when the front car’s brake lights turn on. Coefficient of friction is $\mu=1$ and gravitational acceleration $g=\SI{9,8}{m/s^2}$.\\
a) How big should the distance be between the two cars be while driving so that when braking there would be no collision?\\
b) If the distance between the cars is $l=\SI{5}{m}$ before the braking then how big is the speed difference between the cars at the moment of the collision?

Solution:
\osa Kuna autod pidurdavad maksimaalselt, siis on nende aeglustused võrdsed ning pidurdamise teepikkused on sama pikad. Seega, kui tagumise auto nina on pidurite rakendumisel samas kohas, kus eesmise auto saba oli piduritulede süttides, siis sellel piirjuhul veel otsasõitu ei toimu. Leiame vahemaa, millal see täpselt nii on:
\[
s=vt=\SI{50}{km/h}\cdot\SI{1,5}{s}=\SI{20,8}{m}.
\]
\osa Vaatleme liikumist taustsüsteemis, mis liigub kiirusega $v$ autodega samas suunas. Selles taustsüsteemis on autode esialgne kiirus null ja esimese auto pidurdamisel hakkab ta selle taustsüsteemi suhtes ühtlaselt kiirenema kiirendusega, mis on leitav seosest $F=ma$, kus $F=\mu mg$, seega $a=\mu g$. Esmalt tuleb kindlaks teha, kas kokkupõrge leiab aset enne või pärast tagumise auto pidurite rakendumist. Kui kokkupõrge toimuks enne tagumise auto pidurdama hakkamist, siis kehtiks kokkupõrke ajal $l=at^2/2$, millest
\[
t=\sqrt{2l/ug}=\SI{1.0}{s}.
\]
Kuna see on väiksem kui \num{1,5} sekundit, siis toimub autode kokkupõrge enne teise auto pidurite rakendumist autodevahelise kiirusega $\Delta v=at=\SI{36}{km/h}$.

Pidurdus

Kaks autot sõidavad teineteise järel kiirustega $v=\SI{50}{km/h}$. Esimene auto pidurdab maksimaalselt, mida nähes tagumise auto juht samuti pidurdab maksimaalselt. Esimese auto pidurid rakenduvad samal hetkel, kui süttivad pidurituled. Tagumise auto juhil kulub eesmise auto piduritulede süttimisest kuni oma auto pidurite rakendumiseni $t=\SI{1,5}{s}$. Teekatte hõõrdetegur $\mu=1$ ning raskuskiirendus $g=\SI{9,8}{m/s^2}$.\\
\osa Kui suur peaks olema autodevaheline vahemaa sõidu ajal, et pidurdamisel ei toimuks tagant otsasõitu?\\
\osa Kui autodevaheline vahemaa enne pidurdamist on $l=\SI{5}{m}$, siis kui suur on autode kiirus üksteise suhtes kokkupõrke hetkel?

\osa Kõige suurem kokkupõrke oht on siis, kui esimene on jõudnud täielikult seiskuda (selles saab veenduda, kui liikuda tagumise autoga kaasa liikuvasse taustsüsteemi).\\
\osa Kahe liikuva objekti suhtelist liikumist on kasulik uurida liikuvas taustsüsteemis.


[User]
USAPhO 2013, problem B1. This problem is quite tricky! Once you’re done,
    carefully read the official solution, which describes how center of mass work is applied.

Solution:
As usual, see the official solutions. Recently, this Veritasium video reignited the debate
    over the Blackbird, leading to this followup video, which cites the USAPhO solution. If you’re still
    confused about how the Blackbird works, I recommend watching the second video, which shows
    various arguments and a mechanical model.


[User]
Diffraction limits the resolution of optical instruments.

 (a) Suppose that light of wavelength λ enters through an aperture of width D. As a result, the
     light diffracts, which causes the angle of the light’s propagation to pick up an additional spread
     of order θ. Estimate θ.

 (b) The diameter of a human pupil is about 3 mm. Estimate the size of the smallest text that a
     human being could read from 5 m away.

 (c) A typical amateur telescope has an aperture of order 10 cm. The Sun has a radius of 7 × 108 m.
     Estimate the furthest possible distance, in light years, that such a telescope could resolve a
     Sun-sized star. (Stars further away than this will just show up as blurry points.)

  Remark
  The effect described in problem 17 motivates astronomers to build ever larger telescopes. The
  largest examples are radio telescopes, such as the Arecibo observatory that collapsed in 2020,
  because diffraction effects are more significant the larger the wavelength is. However, the
  telescope doesn’t have to be one big piece. An array of telescopes can effectively be combined
  into a telescope whose radius, for the purposes of the diffraction limit, is the total width of
  the array, as in the Very Large Array. The Event Horizon Telescope was able to resolve a
  black hole 5 × 107 ly away because it combined telescopes spaced around the entire Earth.

  Example 9

  How close does a Sun-like star have to be in order to see it with the naked eye in daylight?

  Solution
  Let the distance to the Sun be d, and the distance to the star be D. Then the ratio of
  intensities of the two is naively
                                               2
                                      Istar    d
                                            =       .
                                      ISun     D
  This suggests the star is hard to see if D > d, which is always true. But this is too pessimistic,
  because the light from the Sun comes from all directions in the sky, while the light from the
  star comes from only a single direction. The actual ratio we want to calculate is
                                                             2
                                     Istar /Ωstar         d         2π
                                                  =
                                      ISun /2π            D        Ωstar

  where Ωstar is the apparent solid angle of the star in the sky.

  This in turn is given by the diffraction limit: if your pupils have radius r, then

                                      Ωstar ∼ (∆θ)2 ∼ (λ/r)2 .

  The star should be visible above daylight if the ratio above is at least one or so, which means
  the maximum distance is
                               r     3 mm
                         D∼      d∼        (1 AU) ∼ 5 × 103 AU ∼ 0.1 ly.
                               λ    600 nm

         This is still closer than the closest other star, so you would need a telescope to see any.

         Notice how this differs from a microscope! Microscopes are used to resolve finer details on a
         small object. But most telescopes can’t resolve any of the details of any but the nearest stars.
         Increasing the size of the telescope has two benefits: increasing the amount of light that goes
         through, and improving the contrast due to decreasing the blurring due to diffraction.

[3] Problem 18 (PPP 126). A compact disc contains approximately 650 MB of information. Estimate
    the size of one bit on a CD using an ordinary ruler. Confirm your estimate using a laser pointer.
    (If you can’t find a CD, a DVD will also work.)

Solution:
(a) Whenever diffraction occurs, it creates an angular spread of order θ ∼ λ/D. For
         example, the angular width of the central maximum for diffraction through a circular aperture
         is sin−1 (1.22λ/D), as noted earlier.
      (b) For concreteness, taking λ = 500 nm, the formula above gives an angular spread of θ ∼
          1.7 × 10−4 rad. At a distance of 5 m, this corresponds to a distance of ∼ 1 mm, so a letter
          smaller than this will just get blurred into a single blob.
          And indeed, the letters on the bottom row of a standard eye chart are about 2 mm tall. Of
          course, most people can’t see this well because of nearsightedness, but that’s a consequence
          of geometric optics (i.e. the eye not focusing light properly) rather than diffraction.
      (c) Repeating the reasoning of part (b) yields a distance of around 0.02 ly, so no stars can be
          resolved by such a telescope at all. Even the biggest optical telescopes ever built can resolve
          essentially no stars.

       Remark
       The effect described in problem 17 motivates astronomers to build ever larger telescopes. The
       largest examples are radio telescopes, such as the Arecibo observatory that collapsed in 2020,
       because diffraction effects are more significant the larger the wavelength is. However, the
       telescope doesn’t have to be one big piece. An array of telescopes can effectively be combined
       into a telescope whose radius, for the purposes of the diffraction limit, is the total width of
       the array, as in the Very Large Array. The Event Horizon Telescope was able to resolve a
       black hole 5 × 107 ly away because it combined telescopes spaced around the entire Earth.

       Example 9

       How close does a Sun-like star have to be in order to see it with the naked eye in daylight?

       Solution
       Let the distance to the Sun be d, and the distance to the star be D. Then the ratio of
       intensities of the two is naively
                                                    2
                                           Istar    d
                                                 =       .
                                           ISun     D
       This suggests the star is hard to see if D > d, which is always true. But this is too pessimistic,
       because the light from the Sun comes from all directions in the sky, while the light from the
       star comes from only a single direction. The actual ratio we want to calculate is
                                                                  2
                                          Istar /Ωstar         d         2π
                                                       =
                                           ISun /2π            D        Ωstar

       where Ωstar is the apparent solid angle of the star in the sky.

       This in turn is given by the diffraction limit: if your pupils have radius r, then

                                           Ωstar ∼ (∆θ)2 ∼ (λ/r)2 .

       The star should be visible above daylight if the ratio above is at least one or so, which means
       the maximum distance is
                                   r     3 mm
                              D∼     d∼        (1 AU) ∼ 5 × 103 AU ∼ 0.1 ly.
                                   λ    600 nm
       This is still closer than the closest other star, so you would need a telescope to see any.

       Notice how this differs from a microscope! Microscopes are used to resolve finer details on a
       small object. But most telescopes can’t resolve any of the details of any but the nearest stars.
       Increasing the size of the telescope has two benefits: increasing the amount of light that goes
       through, and improving the contrast due to decreasing the blurring due to diffraction.

[3] Problem 18 (PPP 126). A compact disc contains approximately 650 MB of information. Estimate
    the size of one bit on a CD using an ordinary ruler. Confirm your estimate using a laser pointer.
    (If you can’t find a CD, a DVD will also work.)

   Solution. CDs have a radius of around 6 cm and an inner radius of around 2.5 cm, giving a
   surface area of around 0.01 m2 . Then the area of 1 bit can be found by dividing the total area
   by 650 × 106 × 8 (since there are 8 bits in a byte), and the square root of that would give the
   approximate size of a bit as 1 micrometer.
      The bits are arranged in concentric rings, so a laser pointer hitting part of the CD will effectively
   see a reflective diffraction grating, with the slits parallel to the tangential direction on the CD. The
   resulting diffraction peaks can be used to find the ring spacing, as you can try at home!


[User]
Use similar reasoning to find the acceleration of the Atwood’s machine. (We will
    show an even easier way to do this, using “generalized coordinates”, in M4.)
[2] Problem 27 (Morin 1.6). A person throws a ball (at an angle of her choosing, to achieve the
    maximum distance) with speed v from the edge of a cliff of height h. Which of the below could be
    an expression for the maximal range?
                               s           r
                   gh2    v2     v2h    v2             v2                    v 2 /g

                                                2gh               2gh
                       ,     ,       ,      1 +     ,        1 +        ,             .
                    v2    g       g      g       v2     g          v2     1 − 2gh/v 2

    If desired, try Morin problems 1.13, 1.14, and 1.15 for additional practice.

Solution:
We know from dimensional analysis that the acceleration is gf (m, M ) where f (m, M )
    is dimensionless. Thus it should be a fraction.
        If either of the masses is much more massive than the other mass, then the acceleration should
    be g. Thus the coefficients of m, M should be ±1. If the masses are equal, then the acceleration
    should be 0. This leads to a M − m term in the numerator. Since the denominator should be
    different but still have factors of ±1, a reasonable answer is
                                                     m−M
                                               a=        g.
                                                     m+M
    which is indeed the real answer.

[2] Problem 27 (Morin 1.6). A person throws a ball (at an angle of her choosing, to achieve the
    maximum distance) with speed v from the edge of a cliff of height h. Which of the below could be
    an expression for the maximal range?
                                s          r
                      2     2     v2h   v2             v2                    v 2 /g

                   gh     v                     2gh               2gh
                        ,     ,       ,     1 +     ,        1 +        ,             .
                    v2    g        g     g       v2     g          v2     1 − 2gh/v 2

    If desired, try Morin problems 1.13, 1.14, and 1.15 for additional practice.

    Solution. First check if they’re all dimensionally correct (they are). When h = 0, the maximum
    range as found above with sin(2θ) = 1 is v 2 /g. Also the maximum range obviously depends on the
    height of the edge of the cliff, and there shouldn’t be a case of a finite height or velocity where the
    range becomes infinite. This leaves 2 options:
                                           r
                                        v2               v2

                                                 2gh              2gh
                                             1+ 2 ,          1+ 2
                                         g        v      g         v

    When h is small, the extra distance at the end of the trajectory from dipping down a vertical
    distance h can be found with binomial theorem: h, and 2h respectively. Since the trajectory is
    symmetric, when h ≈ 0 (to be more concise, h ≪ v 2 /g) the optimal launch angle is 45 deg, so by
    geometry the extra distance should also be h. Thus the correct formula is
                                                 r
                                              v2      2gh
                                                  1+ 2 .
                                              g        v


[User] Üldrelatiivsusteooria ennustab, et mustast august möödumisel kaldub valguskiir gravitatsiooni tõttu kõrvale oma esialgsest liikumissuunast nurga $\varphi = 4GM/c^2r$ võrra, kus $M$ on musta augu mass ning $r$ trajektoori lähima punkti kaugus selleni. Oletame, et must auk satub täpselt vaatleja ja tähe vahele nii, et kaugus vaatlejast musta auguni on $L_1$ ning mustast august täheni $L_2$. Missugune on tähe kujutis vaatleja jaoks (põhjendage oma vastust kiirte käigu visandi abil) ning kui suur on kujutise nurkläbimõõt? Kuna vaatlejani jõudvate kiirte jaoks on $r$ palju väiksem tähe kaugusest, võib kasutada väikeste nurkade lähendust $\sin \alpha \approx \tan \alpha \approx \alpha$.

Solution:
Tähest väljunud kiired kõverduvad musta augu lähiümbruses (vt joonist).

\begin{center}
	\includegraphics[width=0.9\textwidth]{2007-v2g-10-lah1}
\end{center}

Kujutise konstrueerimisel aga eeldatakse, et kiired on kogu tee otse liikunud. Et kiired jõuavad vaatlejani kõikjalt ümber musta augu, on kujutiseks ringjoon (eeldusel, et täht on punkt).

Et silmani jõudvate kiirte jaoks kehtib $r \ll L$ ning trajektoori kõverdumine toimub tähe lähiümbruses, võib vaadelda kiire teekonda lihtsustatult: sirge liikumine musta auguni, hetkeline nurga muutus musta augu juures ning edasi sirge tee vaatlejani (vt joonist). Joonisel arusaadavalt on vertikaalskaala võrreldes horisontaalskaalaga oluliselt välja venitatud.

Järgnevalt leiame tähe nurkdiameetri. Lihtsast geomeetriast järeldub, et $\alpha + \beta = \varphi$. Kaugused on suured ning nurgad väikesed, seega võime kasutada ligikaudseid valemeid $\alpha = r/L_1$ ja $\beta = r/L_2$. Niisiis

\begin{center}
	\includegraphics[width=0.9\textwidth]{2007-v2g-10-lah2}
\end{center}

\begin{center}
	\includegraphics[width=\textwidth]{2007-v2g-10-lah3}
\end{center}

\[
\frac{r}{L_{1}}+\frac{r}{L_{2}}=\frac{4 G M}{c^{2} r} \Rightarrow r=\sqrt{\frac{4 G M L_{1} L_{2}}{c^{2}\left(L_{1}+L_{2}\right)}}.
\]
Tähe kujutise nurkdiameeter on
\[
\gamma=\frac{2 r}{L_{1}}=\sqrt{\frac{4 G M L_{2}}{c^{2} L_{1}\left(L_{1}+L_{2}\right)}}.
\]

Gravitatsioonilääts

Üldrelatiivsusteooria ennustab, et mustast august möödumisel kaldub valguskiir gravitatsiooni tõttu kõrvale oma esialgsest liikumissuunast nurga $\varphi = 4GM/c^2r$ võrra, kus $M$ on musta augu mass ning $r$ trajektoori lähima punkti kaugus selleni. Oletame, et must auk satub täpselt vaatleja ja tähe vahele nii, et kaugus vaatlejast musta auguni on $L_1$ ning mustast august täheni $L_2$. Missugune on tähe kujutis vaatleja jaoks (põhjendage oma vastust kiirte käigu visandi abil) ning kui suur on kujutise nurkläbimõõt? Kuna vaatlejani jõudvate kiirte jaoks on $r$ palju väiksem tähe kaugusest, võib kasutada väikeste nurkade lähendust $\sin \alpha \approx \tan \alpha \approx \alpha$.

Kuna silmani jõudvate kiirte jaoks kehtib $r \ll L$ ning trajektoori kõverdumine toimub musta augu lähiümbruses, võib vaadelda kiire teekonda lihtsustatult: kiire liikumine musta auguni, hetkeline nurga muutus musta augu juures ning edasi sirge tee vaatlejani.


[User] An electrically insulated metal ball of mass $M$ and charge $Q>0$ is hanging at the end of a vertical spring of stiffness $k$ in a gravitational field $g$, the ball is in its equilibrium position. Now a vertical electrical field of strength $E$ is created. At the first moment it is directed downwards and afterwards always at the same direction as the ball’s velocity. Assume that the electrical field changes instantaneously. Find the ball’s distance from its initial position at a moment of time $t=7\pi \sqrt{\frac{M}{k}}$.

Solution:
Kuna liikumine toimub vaid vertikaalsihis, võtame $x$-teljeks vertikaalsihi suunaga alla. Algsest jõudude tasakaalust saame
$$Mg = kx_0,$$
kust kuuli algkõrgus on $x_0 = \frac{Mg}{k}$. Teame, et vedru otsa asetatud kuuli pool võnkeperioodi on
$$\tau = \pi \sqrt{\frac{M}{k}}.$$
On selge, et see ei sõltu gravitatsioonivälja tugevusest. Kuna nii gravitatsioonivälja kui ka elektrivälja poolt tekitatud jõud käituvad analoogselt, võime järeldada, et võnkeperiood ei sõltu ka elektrivälja tugevusest ning jääb konstantseks. Seega muudab elektrivälja jõud vaid kuuli tasakaaluasendit iga kord, kui kuul muudab oma liikumissuunda (sest ülesandes on öeldud, et elektriväli on alati kuuli liikumise suunas) ehk iga poole võnkeperioodi $\tau$ järel. Alla liikumisel on elektriväli suunatud alla ning saame kirja panna uue jõudude tasakaalu
$$Mg + Eq = kx_\downarrow.$$
Seega on tasakaaluasend alla liikumisel
$$x_\downarrow = \frac{Mg}{k} + \frac{Eq}{k} = x_0 + \Delta x.$$
Analoogselt saame leida tasakaaluasendi üles liikumisel, kuid nüüd on elekriväli suunatud üles:
$$x_\uparrow = \frac{Mg}{k} - \frac{Eq}{k} = x_0 - \Delta x.$$
On selge, et algselt elektrivälja sisselülitamisel, kui elektriväli on alla suunatud, muutub tasakaaluasend hetkeliselt vastavaks, kuid kuuli asukoht jääb algselt samaks. Seega pole kuul enam tasakaalus, vaid uuest tasakaaluasendist kaugusel $\Delta x$ ning hakkab võnkuma amplituudiga $A_1 = \Delta x$ tasakaaluasendi $x_\downarrow$ ümber. Aja $\tau$ järel on kuul jõudnud kaugusele $x_0 + 2\Delta x$. Kuna kuul hakkab nüüd üles liikuma, muutub ka kuuli tasakaaluasend koos elektriväljaga vastavaks ning üles minnes võngub kuul juba amplituudiga $A_2 = 3 \Delta x$ tasakaaluasendi $x_\uparrow$ ümber.
On selge, et iga tasakaaluasendi vahetumisega kasvab kuuli amplituud edaspidi $2 \Delta x$ võrra. Tasakaaluasend vahetub aga iga ajavahemiku $\tau$ järel.
Nüüd leiame kuuli asukoha ülesandes antud ajahetkel, milleks on $7 \tau$. Selle aja jooksul liigub kuul korra algasendist alla, ning teeb siis $3$ täisvõnget, jõudes nende järel uuesti alla. Pärast algasendist alla jõudmist on kuuli asukoht $x_0 + 2 \Delta x $. Seejärel liigub kuul üles asendisse $x_0 - 4 \Delta x$ ja uuesti alla asendisse $x_0 + 6 \Delta x$. Kauguse aboluutväärtus algasendist suureneb seega iga $\tau$ järel $2 \Delta x$ võrra. $7 \tau$ järel on kaugus tasakaaluasendist seega $7 \cdot 2 \Delta x = 14 \Delta x$.
Nagu enne selgeks tegime, on kuul seega küsitud ajahetkel $7 \tau$ algasendist $x_0$ $14 \Delta x = 14 \frac{Eq}{k}$ võrra all pool.

Pendel

Elektriliselt isoleeritud metallkuul massiga $M$ ja laenguga $Q>0$ ripub vertikaalse vedru otsas jäikusega $k$ tasakaaluasendis gravitatsiooniväljas $g$. Nüüd tekitatakse vertikaalne elektriväli tugevusega $E$, mis on esialgu suunatud alla ning edaspidi alati kuuli liikumise suunas. Eeldada, et elektriväli muutub hetkeliselt. Leida kuuli kaugus algsest asukohast ajahetkel $t=7\pi \sqrt{\frac{M}{k}}$.

Elektrivälja lisamisel nihkub kuuli tasakaaluasend kas üles või alla, sõltuvalt elektrivälja suunast. Seega on igat poolperioodi mõistlik eraldi vaadata, sest selle raames liigub kuul ümber fikseeritud tasakaaluasendi.


[User]
Rong liikus kiirusega $v=\SI{54}{km/h}$ läbi pika horisontaalse silindrikujulise tunneli.
Kui palju tõusis tunnelis asuva õhu temperatuur? Tunneli läbimõõt oli $d=\SI{5}{m}$.
Rongi elektrimootor tarbis tunnelit läbides võimsust $P=\SI{800}{kW}$.
Õhu molaarmass on $M=\SI{29}{g/mol}$, õhurõhk tunnelis $p=\SI{100}{kPa}$ ja algtemperatuur $t_0=\SI{17}{\degreeCelsius}$.
Õhk lugeda kaheaatomiliseks ideaalseks gaasiks. Eeldada, et rongi liikumisest tekkinud õhuvoolude liikumisest tulenev alarõhk on atmosfäärirõhuga võrreldes tühine\\
\emph{Märkus:} kaheaatomilise gaasi siseenergia ühe molekuli kohta on $5/3$ korda suurem kui samal temperatuuril oleval üheaatomilisel gaasil.

Solution:
Õhu temperatuur tunnelis kasvab, kuna mootor soojendab tunneli läbimisel selles olevat õhku. Vaatleme rongi liikumist ajavahemiku $\Delta t$ jooksul. Selle ajaga läbib rong vahemaa $s = v \Delta t$ ja rongist mööduva õhu ruumala on $\Delta V = \pi d^2 s /4$. Õhu mass on $m = \Delta V \rho$ ja moolide arv on
\[
N = {m \over M}={\Delta V \rho \over M}={\pi d^2 v \Delta t\rho\over 4M}.
\]
Ideaalse gaasi olekuvõrrandist $pV={m\over M}RT$ saame avaldada $\rho={m\over V}={pM\over RT}.$

Rongi mootoris eraldub samal ajal soojushulk $Q_1 = P \Delta t$.

Üheaatomilise gaasi erisoojus jääval ruumalal on $C_1 = 3/2 R$. Seega on kaheaatomilise gaasi erisoojus $C_2 = 5/3 \cdot 3/2R = 5/2 R$. Gaasi erisoojus jääval rõhul on seega $C = 5/2R + R = 7/2 R$. Õhu soojendamiseks $\Delta T$ võrra kulub soojushulk $Q_2 = N C \Delta T$. Võrdsustame soojushulgad $Q_1$ ja $Q_2$.
\[
N C \Delta T = P \Delta t.
\]
Asendades leitud avaldised $N$ ja $C$ jaoks saame pärast teisendusi
\[
\Delta T = {8 P T \over 7\pi d^2 v p } = \SI{2.25}{K}.
\]

Rong tunnelis

Rong liikus kiirusega $v=\SI{54}{km/h}$ läbi pika horisontaalse silindrikujulise tunneli.
Kui palju tõusis tunnelis asuva õhu temperatuur? Tunneli läbimõõt oli $d=\SI{5}{m}$.
Rongi elektrimootor tarbis tunnelit läbides võimsust $P=\SI{800}{kW}$.
Õhu molaarmass on $M=\SI{29}{g/mol}$, õhurõhk tunnelis $p=\SI{100}{kPa}$ ja algtemperatuur $t_0=\SI{17}{\degreeCelsius}$.
Õhk lugeda kaheaatomiliseks ideaalseks gaasiks. Eeldada, et rongi liikumisest tekkinud õhuvoolude liikumisest tulenev alarõhk on atmosfäärirõhuga võrreldes tühine\\
\emph{Märkus:} kaheaatomilise gaasi siseenergia ühe molekuli kohta on $5/3$ korda suurem kui samal temperatuuril oleval üheaatomilisel gaasil.

Ülesannet on mugav lahendada uurides ajavahemikku $\Delta t$ ning vaadeldes, kuidas rongist eraldunud võimsus soojendab rongist möödunud õhu molekule.