Physics 410
Final Examination (Solution)
May 18 2005 8-10 am
(A figure is omitted and the underline notation for "4-vectors" is not consistent. Also, the original exam omitted wishes for a good summer to all, which are hereby supplied :)
  1. (as promised, one of 7 questions on a 3-hour exam your prof had to pass as an undergraduate)
    If the force on a particle moving in a plane is always directed toward or away from a fixed point in the plane, the magnitude of the force being a function only of the distance r from that fixed point, then:
    1. Show that the forces are conservative and that the angular momentum about the center of force is constant.
      One way would be to show that ×F = 0, which is true because in the expression for the curl the (only) radial component is differentiated only with respect to the other coordinates. But more correct and direct would be to point out that the work done along any path from r1 to r2 is just r1r2 F dr, which contains no q, hence depends only on the endpoints.
      We have dL/dt = r×F = 0 because r and F are parallel, so angular momentum is conserved.
    Now if the force is still everywhere in the radial direction but its magnitude depends on the angle q as well as on r:
    1. Show that the angular momentum is still constant, but that the force is not conservative.

      The same angular momentum argument still holds; but now we have to show that the work definitely depends on the path. Thus, consider one path that is first purely radial, then in the angular (q) direction, and another between the same points first angular and then radial. The work is given by r1r2 F dr as before, but evaluated for the initial q for the first path, and the final q for the second. Since F depends on q, it must be possible to choose beginning and end points so the two integrals are definitely different.

  2. A rigid uniform bar of mass M, length 2l, is supported in equilibrium in a horizontal position by two identical vertical springs of spring constant k, attached one at each end of the bar. The moment of inertia of the bar about its center is Ml2/3. We consider vertical motion only -- that is, the springs always remain vertical -- and small deviations from equilibrium.
    1. Choose suitable coordinates and write the Lagrangian as well as the Hamiltonian for this system.

      If we have foresight we choose the position of the CM, x, and the angle q of the bar from the horizontal as the two generalized coordinates; otherwise we choose the positions of the two endpoints, call them y+ and y-. In the latter case we still have to express the KE of the bar as the contribution of the CM + that about the CM, so we need y = x lq. Thus

      L = T - U = 1/2 M

      x
       
      2
       
      + 1/2 (Ml2/3)

      q
       
      2
       
      - 1/2 k (y+2 + y-2) = 1/2 M

      x
       
      2
       
      + 1/2 (Ml2/3)

      q
       
      2
       
      - 1/2 k (2x2 + 2l2q2)
      For the Hamiltonian we need px = Mx· and pq = (Ml2/3)q·, so H = T + U = (px2/2M) + (3pq2/Ml2) + 1/2 k (2x2 + 2l2q2).

    2. Find the normal modes and normal frequencies of the system.

      "By inspection" we see that the M- and k-matrices are already diagonal , we can write L = Lx(x, x·) + Lq(q, q·) so the normal modes consist of an excitation of x alone (with q = 0), and one of q alone (with x = 0). From Lx we get the frequency (2k/M)1/2, and from Lq we get (6k/M)1/2.

  3. Two particles having mass m and M attract each other according to the law of gravitation. Initially they are at rest an infinite distance apart.
    1. Show that their relative velocity of approach when they are a distance X apart is
        


      2G (M + m)

      X
       
      where G is the constant of gravitation.

      We can assume that the CM is at rest, then the KE is the same as that of a single particle of reduced mass m at position X, and the potential energy is -GMm/X. The total energy is that at X = , namely zero, so

      0 = 1

      2

      mM

      m+M


      X
       
      2
       
      - GMm

      X
      Solving for X· gives the required relation.
    2. Find the momentum pX conjugate to X (in terms of m, M, [X\dot]). In the phase space with coordinates X, pX sketch several orbits (including initial conditions different from those of part (a)). Restrict X to positive values, but allow pX to be of either sign.
      pX = T


      X
       
      = mM

      m+M

      X
       
      . The graphs follow from H = T + U = m+M

      2mM
      p2 - GMm

      X
      = const.
  4. In 2-dimensional space-time, (x, t) with c = 1, the four-velocity of a mirror at rest is _u = (0, 1). A photon incident on this mirror from the left has momentum-energy vector _k = (k, k) where k = hn and n is the photon's frequency. After reflection the photon has momentum-energy vector _k = (-k, k), that is, the space (momentum) component is reversed, and the time (energy) component remains the same.
    1. Consider the space-time vector1 equation
      _k = -_k - 2(_k·_u)_u
      where the dot (·) denotes the space-time scalar product. Verify that this equation correctly describes the reflection in the mirror's rest frame, and argue that it must be valid in any frame.

      It is valid in any Lorentz frame because it is a space-time vector equation.

    2. Show that the transformation _k _k is a Lorentz transformation for any vector _k (not just for null vectors). It is enough to show _k2 = _k2.

      k2 = k2 + 2·(2k·u)(u·k) + 4 (k·u)2 u2 = k2 since u2 = -1.
      The above uses the equation in (a) as intended. The question could also be interpreted in the mirror frame only, in which case you should show that the transformation of any vector k = (a, b) into k' = (-a, b) is a Lorentz transformation, that is, that k' = k. This was also counted as correct (though no one did it exactly right).

    3. Let the mirror move with speed v, so that u1 = v/{1-v2},   u4 = 1/{1-v2}. Use the equation in (a) to evaluate k and hence find the reflected frequency n in terms of the incident frequency n and the speed v.


      k4 = hn = -k4 - 2
      k1v-k4

      (1-v2)1/2


      1

      (1-v2)1/2

      = k
      v2-1 -2v + 2

      1-v2

      = hn
      1-v

      1+v

      .

Footnotes:

1usually called 4-vector, but in the present context these are 2-vectors. Nevertheless we use the subscript 4 for the time components of these vectors.


File translated from TEX by TTH, version 3.60.
On 18 May 2005, 10:29.