[User]
7th Asian Physics Olympiad                                           Theoretical Question 3   Page 1 /2

                                     Theoretical Question 3

                             This problem consists of four not related parts.

A. [2.5 points] The Mariana Abyss in the Pacific Ocean has a depth of H = 10920 m .
    Density of salted water at the surface of the ocean is ρ 0 = 1025 kg / m 3 , bulk modulus is
                                                              2
     К = 2,1 ⋅ 10 9 Pa , acceleration of gravity is g=9.81m/s . Neglect the change in the
    temperature and in the acceleration of gravity with the depth, and also neglect the
    atmospheric pressure.
            А1) Find the relation between the density ρ (x) and pressure P (x) at the depth
                    of х.
            А2) Find the numerical value of the pressure P(H ) at the bottom of the
                    Mariana Abyss. You may use iterative methods to solve this part.

    Note: The fluids have very small compressibility. Compressibility coefficient is
    defined as
                                                 1 ⎛ dV ⎞
                                            κ =− ⎜      ⎟
                                                 V ⎝ dP ⎠ T =const
    Bulk modulus K is the inverse of κ: K=1/ κ.

B. [2.5 points] Light mobile piston separates the vessel into two parts. The vessel is
    isolated from the environment. One part of the vessel contains m1 = 3g of hydrogen
    at the temperature of Т10 = 300 К, and the other part contains т2 = 16 g of oxygen at
    the temperature of Т20 = 400 К. Molar masses of hydrogen and oxygen are
     µ1 = 2 g / mole and µ 2 = 32 g / mole respectively, and R = 8.31J /( K ⋅ mole ). The piston
    weakly conducts heat between oxygen and hydrogen, and eventually the temperature
    in the system equilibrates. All the processes are quasi stationary.
            В1) What is the final temperature of the system T?
            В2) What is the ratio between final pressure Pf and initial pressure Pi ?
            В3) What is the total amount of heat Q, transferred from oxygen to hydrogen?

C. [2.5 points] Two identical conducting plates α and β with charges –Q and +q
    respectively (Q > q > 0) are located parallel to each other at a small distance.
    Another identical plate γ with mass m and charge +Q is situated parallel to the
    original plates at distance d from the plate β (see fig 1). Surface area of the plates is
    S. The plate γ is released and can move freely, while the plates α and β are kept
    fixed. Assume that the collision between the plates β and γ is elastic, and neglect the
    gravitational force and the boundary effects. Assume that the charge has enough time
    to redistribute between plates β and γ during the collision.

7th Asian Physics Olympiad                                      Theoretical Question 3   Page 2 /2

            С1) What is the electric field E1 acting on the plate γ before the collision with
                the plate β ?
            С2) What are the charges of the plates Qβ and Qγ after the collision?
            С3) What is the velocity υ of the plate γ after the collision at the distance d
                from the plate β ?
                                α         β          γ

                                                 d
                               -Q        +q          +Q

                                        Fig. 1

D. [2.5 points] Two thin lenses with lens powers D1 and D2 are located at distance
    L = 25cm from each other, and their main optical axes coincide. This system creates a
    direct real image of the object, located at the main optical axis closer to lens D1, with
    the magnification Γ'= 1 . If the positions of the two lenses are exchanged, the system
    again produces a direct real image, with the magnification Γ' ' = 4.
           D1) What are the types of the lenses? On the answer sheet you should mark
                 the gathering lens as «+», and the diverging lens as «-» .
           D2) What is the difference between the lens powers ∆D = D1 − D2 ?

Solution:
7th Asian Physics Olympiad                                                    Theoretical Question 3 /Solutions   Page 1/6

                                     Theoretical Question 3
                                                     Solutions

                                                    A
    The change of the pressure is related to the change in the density via

                                                        ∆V    ∆ρ
                                           ∆P = − K        =K    ,                                                (1)
                                                        V     ρ0

where ρ 0 is the density of water at the surface.

                                                        ⎛       ∆ρ ⎞              ∆P ⎞
                                 ρ = ρ 0 + ∆ρ = ρ 0 ⎜⎜1 +          ⎟⎟ = ρ 0 ⎛⎜1 +    ⎟,                           (2)
                                                        ⎝       ρ0 ⎠         ⎝    K ⎠

where ∆P ≈ P (we neglect the atmospheric pressure). Then

                                                            ⎛     P( x) ⎞
(A1)                                         ρ ( x ) = ρ 0 ⎜1 +         ⎟.                                        (3)
                                                            ⎝      K ⎠

    The change of the hydrostatic pressure with the depth equals

                                                      dP                          P ( x)
                             dP = g ⋅ ρ ( x ) dx ,       = gρ ( x ) = gρ 0 + gρ 0        ,                        (4)
                                                      dx                           K

                                           dP ( x) gρ 0
                                                  −     P ( x ) = gρ 0 .                                          (5)
                                            dx      K

    The solution of this differential equation with boundary condition P (0) = 0 is

                                                       ⎛    gρ      ⎞
                                             P( x) = K ⎜ exp 0 x − 1⎟.                                            (6)
                                                       ⎝     K      ⎠

            gρ 0
    Since        H << 1 , we can use the expansion
             K
                                                                  z2
                                             exp z ≈ 1 + z +         + ...,                                       (7)
                                                                  2!
    thus
                                                                 1
                                           P ( x ) ≅ gρ 0 x +      (gρ 0 x )2                                     (8)
                                                                2K

    The last formula can be simply derived using the method of successive iterations. First, the
pressure can be estimated without compressibility taken into account:

                                                      P0 ( x) = gρ 0 x.                                           (9)

7th Asian Physics Olympiad                                                                Theoretical Question 3 /Solutions   Page 2/6

    Correction to the density in the first approximation can be obtained using P0 ( x ) :

                                                                                gρ 0 x
                                                           ρ1 ( x) = ρ o (1 +            ).                                   (10)
                                                                                K

    Now, correction to pressure can be obtained using ρ1 ( x) :

                                                      H
                                                                                1
                                          P1 ( H ) = ∫ ρ1 ( x) gdx = gρ 0 x +              (gρ 0 x )2 ,                       (11)
                                                      0
                                                                                2K
as obtained earlier.
    Putting in the numerical values, we get

(A2)                                  P( H ) = (1098 ⋅10 5 + 28,7 ⋅10 5 ) Па ≈ 1,13 ⋅108 Па.                                  (12)

                                                                     B
    The total work done by the gases is zero. Thus at any moment the total internal energy equals the
original value:

                                      m1              m2                 m1               m2
                                            CV T1 +        CV T2 =             CV T10 +        CV T20,                        (13)
                                      µ1              µ2                 µ1               µ2

where µ1 = 2 g / mole and µ 2 = 32 g / mole are molar masses of hydrogen and oxygen, and СV=5R/2
is the molar heat capacity of diatomic gas. The final temperature of the system is

                                                      m1            m2
                                                           T10 +         T20
                                                      µ1            µ2
(B1)                                          T=                               = 325 K .                                      (14)
                                                           m1       m2
                                                                +
                                                           µ1       µ2

   The temperature of oxygen decreases, and the amount of heat Q is transferred to hydrogen by heat
conduction. The piston will move in the direction of the oxygen, thus the hydrogen does a positive
work А>0, and the change of the internal energy of oxygen is ∆U = A − Q . On the other hand,

                                                           m2 5
                                                ∆U =            R (T − T20 ) = −779 J .                                       (15)
                                                           µ2 2

    To find А, let us prove that the pressure P doesn’t change. Differentiating the equations of the state
for each gas, we get

                                     µ1                                            µ2
                             ∆T1 =          (P∆V + V1∆P ) ,              ∆T2 =            (− P∆V + V2 ∆P ).                   (16)
                                     m1 R                                         m2 R

where Vi are the gas volumes, and ∆V = ∆V1 = −∆V2 is the change of the volume of the hydrogen.
Differentiating (1), we get

7th Asian Physics Olympiad                                                 Theoretical Question 3 /Solutions   Page 3/6

                                          m1           m2
                                               ∆T1 +        ∆T2 = 0.                                           (17)
                                          µ1           µ2

Substituting (2) into (3), we obtain (V1+V2)·∆P = 0, thus

(B2)                                           Pf /Pi= 1.                                                      (18)

Then the work done by the hydrogen is

                                               m2               m2
                              A = P ⋅ ∆V = −        R ⋅ ∆T2 =          R (T20 − T ) = 312 J .                  (19)
                                               µ2                µ2

The total amount of heat transferred to hydrogen is

(B3)                                      Q = A − ∆U = 1091J .                                                 (20)

                                                        C

The electric field acting on the plate γ before the collision is

                                                            Q−q
(C1)                                                E1 =           .                                           (21)
                                                            2ε 0 S

The force acting on the plate is

                                             F1 = E1Q =
                                                             (Q − q )Q .                                       (22)
                                                               2ε 0 S

The work done by the electric field before the collision is

                                             A1 = F1d =
                                                            (Q − q )Qd .                                       (23)
                                                               2ε 0 S

The charge will get redistributed between two touching conducting plates during the collision. The
values of the charges can be obtained from the condition that the electric field between the touching
plates vanishes. If one assumes that the plate γ is on the right side, the left surface of the combined
plate will have the charge

(С2a)                                            Qβ = Q + q / 2,                                               (24)

and the right surface will have the charge

(C2b)                                                  Qγ = q / 2 .                                            (25)

7th Asian Physics Olympiad                                                Theoretical Question 3 /Solutions   Page 4/6

These charges remain on the plates after the collision is over. Now the force acting on the plate γ
equals F2 = E2 q / 2 , where E2 = (q / 2) / (2ε 0 S ) . The work done by field E2 is

                                                              q2d
                                                A2 = F2 d =          .                                        (26)
                                                              8ε 0 S

The total work done by the electric fields is

                                                                            2
                                                           d ⎛       q⎞
                                        A = A1 + A2 =            ⎜Q − ⎟ .                                     (27)
                                                          2ε 0 S ⎝   2⎠

Velocity at the distance d can be calculated using the following relation:

                                                      mυ 2
                                                           = A.                                               (28)
                                                       2

Substituting (5) into (6), we finally get

                                                  ⎛      q⎞   d
(С3)                                         υ = ⎜Q − ⎟             .                                         (29)
                                                  ⎝      2 ⎠ mε 0 S

                                                         D
    First one has to determine the types of the lenses. If both lenses are negative, one always obtains a
direct imaginary image. If one lens is positive and the other is negative, three variants are possible: an
inversed real image, a direct imaginary image or an inversed imaginary image, all contradicting the
conditions of the problem. Only the last variant is left – two positive lenses. The first lens creates an
inversed real image, and the second one inverts in once more, creating the direct real image. Using the
lens equations, the magnifications of the lenses can be written as

                                                F1                     F2
                                      Γ1 =           ;       Γ2 =            ,                                  (30)
                                             d1 − F1                d 2 − F2

where d1 is the distance from the object to the first lens, d 2 = L − f1 is the distance from the image of
the first lens to the second lens, and f1 is the distance from the first lens to the first image. The total
magnification of the system is Γ′ = Γ1 ⋅ Γ 2 . Using the expression for d2, inverted magnification
coefficient can be written as

                                      1 d1 [L − ( F1 + F2 )] L
                                         =                  −    + 1.                                           (31)
                                      Γ'       F1 F2          F2

One notices from this expression that if two lenses are exchanged, the first term stays invariant, and
only the second term changes. Thus the expression for the inverted magnification in the second case is:

7th Asian Physics Olympiad                                       Theoretical Question 3 /Solutions   Page 5/6

                                        1 d1 [L − ( F1 + F2 )] L
                                             =                − + 1.                                 (32)
                                        Γ' '     F1 F2         F1

Subtracting these two formulas, we get:

                                      1 1     ⎛1 1 ⎞
                                        − = L ⎜ − ⎟ = L( D1 − D2 );                                  (33)
                                      Γ′ Γ′′  ⎝ F1 F2 ⎠

                                    1⎛ 1 1 ⎞       1 ⎛ 1⎞     1 3
                        D1 − D2 =     ⎜ ′ − ′′ ⎟ =   ⎜1 − ⎟ =   ⋅ = 3 diopters.                      (34)
                                    L ⎝ Γ Γ ⎠ 0, 25 ⎝ 4 ⎠ 0, 25 4

7th Asian Physics Olympiad                                     Theoretical Question 3 /Solutions   Page 6/6

                                         Mark Distribution

           Total       Partial
 No                                                            Contents
            Pt           Pt
  A         2.5         0.5                                    Equation
                        1.5        Differential   1.0              Equation (4) or (5)
                                    equation
                                                  0.5                    Equation (6)
                                     method
                                    Iterative     0.5                         P0 ( x )
                                     method       0.5                         ρ1 ( x)
                                                  0.5                         P1 ( H )
                             0.5                        Numerical value of P(H)
  B         2.5              0.5                         Numerical value of T
                             1.0                          Constant pressure
                             0.3                        Numerical value of ∆U
                             0.3                         Numerical value of A
                             0.4                         Numerical value of Q
  C         2.5              0.5                                   E1
                             0.5                                   Qβ
                             0.5                                  Qγ
                             0.5                  Total work by electric fields
                             0.5                    Answer for final velocity
  D         2.5              1.0                         Signs of lenses
                             0.5           Expression for the magnification coefficient
                             1.0                    Final answer for D1 − D2

APhO 2006 Theory T3


[User]
7. WAVEs (4 points)

— Janis Huns, Jaan
Kalda. The dispersion relation (i.e. the de‐
pendence of the circular frequency ω on the
wave vector k =
2π
λ
) of capillary‐gravity
waves is
ω2
= gkα
+
σ
ρ
kβ
,
where σ denotes the surface tension, g =
9.81ms−2
, and ρ = 1000kgm−3
.
i) (1 point)determine the values of the expo‐
nents α and β.
F. Moisy, M. Rabaud, PRE 90, 023009 (2014)
ii) (3 points) In the image above, we can see
how an object moving with a constant speed
U = 60cms−1
generates a wake — a set
of waves of different wavelengths. Pay at‐
tention to the short‐wavelength waves whose
wavecrestextendsfromtheobjectalmostup
to the edges of the photo: the presence of
a very long wavefront testifies that for these
particular waves, the phase and group velo‐
cities are equal. Determine the surface ten‐
sion of water. You can take measurements
from the photo. Note that while phase speed
is the speed of a constant phase of the wave,
the group speed vg = dω
dk is the speed of a
wave packet (a train of waves).

Solution:
7. WAVEs

— Solution by Tudor
Plopeanu.
i) (1 point) Both gkα
and σ
ρ kβ
’s dimension‐
alities are s−2
, while k’s dimensionality is
m−1
. The surface tension’s dimensionality is
kg /s2
. As such, α = 1 and β = 3.
Grading: (preliminary)
1. Dimensional analysis (or similar tech‐
nique). (0.2pts)
2. Units of σ (0.2pts)
3. α = 1. (0.3pts)
4. β = 3. (0.3pts)
ii) (3 points)
The phase and group velocities of the
wake are equal, i.e. dω
dk = ω
k . We can find
ω
k from the Mach angle formed by the wave
front in the picture: sinμ = ω
Uk. Differenti‐
ating the dispersion relation,
2ω
dω
dk
= (g +3
σ
ρ
k2
) =⇒ 2ω2
= k(g +3
σ
ρ
k2
).
Substituting our dispersion relation, we ob‐
tain g = σ
ρ k2
and ω2
= 2gk. Substituting
g
1
2 = (σ
ρ )
1
2 k, we reach ω2
k2 = 2(gσ
ρ )
1
2 . Thus,
we can find
σ =
ρU4
sin4
μ
4g
≈ 60gs−2
.
for measured μ ≈ 21.5°.
Grading: (preliminary)
1. Identifying dω/dk = ω/k. (0.6pts)
2. Correct angle i.e. sin(μ) = ω/(Uk)
(0.6pts)
3. Measuring μ from the picture (0.4pts)
4. Value μ ∈ [20◦
,30◦
] (0.3pts)
5. Calculatethederivativedω/dk (0.4pts)
6. Correct expression for σ (0.4pts)
7. Value σ ≈ 60gs−2
(0.3pts)

WAVEs


[User]
The general, turn-the-crank method to find the time evolution of an arbitrary
    wave on a string of length L is Fourier series. In this method, we write the initial shape y0 (x) of
    the wave as a combination of standing waves,
                                          X          πnx
                                 y0 (x) =     cn sin     ,    0 ≤ x ≤ L.
                                           n
                                                      L
    We know how each standing wave oscillates in time, so by linearity, the entire wave evolves as
                                            X          πnx
                                 y(x, t) =      cn sin     cos(ωn t)
                                             n
                                                        L

    where ωn is the angular frequency of the nth harmonic.
      (a) The coefficients cn can be extracted by integrating y0 (x) against another sine,
                                                 Z L
                                                                   πnx
                                            cn ∝     dx y0 (x) sin     .
                                                  0                 L
          Explain why this works, and find the constant of proportionality.

      (b) Now let’s consider the plucked string considered in part (a) of problem 19. If the pluck is
          centered at the middle of the string and has height h, find the coefficients cn . (If you’re so
          inclined, you can use a computer to see how the resulting y(x, t) approaches the answer to
          problem 19 as more terms are included.)
      (c) Argue that in general, we have
                                             Z L
                                                                  LX
                                                   y02 (x) dx =       |cn |2 .
                                              0                   2 n
          By applying this result to the plucked string, show that the Riemann zeta function has value
                                                           ∞
                                                           X 1             π4
                                                  ζ(4) =             4
                                                                       =      .
                                                                 n         90
                                                           n=1

          In fact, this is one of the simplest ways to compute ζ(4).

    We’ll use the idea of Fourier series to illustrate some conceptual points in W2.

       Idea 7
       More generally, the relation between the incoming and reflected waves may depend on the
       exact form of the incoming wave. In this case, it’s useful to consider sinusoidal solution. Let

                                             yin (x, t) = ei(kx−ωt) .

       Almost all boundary conditions will state that something at the boundary is constant in time,
       which is only possible if the reflected wave has the same frequency. So in general we have

                                            yr (x, t) = rei(−kx−ωt)

       where r is the reflection coefficient. If the medium exists for x > 0, there is also a transmitted
       wave there, of the form
                                                                ′
                                               yt (x, t) = tei(k x−ωt)
       where k ′ might differ from k, and t is the transmission coefficient. In general, both r and t
       may depend on k as well as the boundary conditions. Note that the phases of r and t depend
       on the conventions we used to define yr (x, t) and yt (x, t), though the magnitudes don’t.

Solution:
(a) This works because all the other terms in y0 (x) will cancel out, as seen here:
                 Z L                                          Z L
                                 πnx     πmx   1                           πx               πx
                        dx sin       sin     =                       dx cos    (m − n) − cos    (m + n)
                    0             L       L    2                0           L                L

     which will always equal to 0 when m ̸= n, since the arguments of sine (the anti-derivative of
     cosine) will always be an integer multiple of π. To find the constant of proportionality, we
     only need to look at the nth term of the expansion of y0 (x):
                               Z L                                       Z L
                                                     πnx                                     πnx
                                       dx y0 (x) sin     = cn                  dx sin2           = cn L/2
                                   0                  L                   0                   L

     Thus the constant of proportionality is 2/L.

 (b) The equation of the plucked string is

                                        y0 (x) = h − 2h|x − L/2|/L,                     0 < x < L.

     By symmetry, if n is even, cn = 0 since y0 (x) is even about x = L/2 and sin(πnx/L) is odd
     about x = L/2, so the integral will be 0. Thus we will consider the case where n is odd,
                              Z L/2                                       Z L
                         2             2hx      πnx       2                                              πnx
                    cn =                   sin         dx +                         (2h − 2hx/L) sin                dx
                         L     0        L         L         L                 L/2                           L

     Using symmetry again, the two integrals above are equal, so we only need to evaluate the
     first. We have
                                         Z L/2               !
                                                                4h L
                                                                   Z
            4h    L     πnx  L/2    L             πnx                    x      πnx
     cn = 2 − x cos                +           cos         dx +          1−     sin         dx
            L     πn      L     0    πn 0             L          L L/2      L          L

     Considering only odd n, the first term will vanish since cos(πn/2) = 0 for odd n. Also, by
     symmetry the integral that goes from L/2 to L should be equal to the one that goes from 0
     to L/2. Thus, for odd n,
                                               8h       πn
                                         cn = 2 2 sin
                                             π n          2
     while cn = 0 for even n.

 (c) Again, integrating sinusoids with different values of n will get 0, and the same value will get
     L/2. Thus when representing y0 (x) as a sum of sinusoids and having the integral of all the
     cross terms go to 0, we get that
               Z L                     Z L                                              Z L
                                                  X               πnx X 2                               πnx   LX 2
                     y0 (x) dx =             dx        c2n sin2      =   cn                   dx sin2       =     c .
                0                       0         n
                                                                   L   n                 0               L    2 n n

     Using y0 (x) = h − 2h|x − L/2|/L, and symmetry about x = L/2, we get
                         Z L                          Z L/2                     Z L/2
                                                                                4h2 x2     1
                               y02 (x) dx = 2                 y02 (x) dx = 2                    2
                                                                                dx = h2 L.
                          0                            0                            0         L         3

          To find the sum of cn , we use our answer above and consider the odd terms:
                              LX           LX     64h2        32h2 L X     1
                                  |cn |2 =      4         4
                                                            =     4
                              2 n          2 n π (2n + 1)       π    n
                                                                       (2n + 1)4

          To relate that sum to ζ(4), define the sums for the even and odd numbers as E and O so that
                                              X       1    1                              15
                       ζ(4) = E + O,    E=                = ζ(4),       O = ζ(4) − E =       ζ(4)
                                               n
                                                    (2n)4  16                             16

          Now equating our expressions will get
                                               1 2    32h2 L 15
                                                 h L=           ζ(4)
                                               3        π 4 16
          from which we conclude
                                                               π4
                                                      ζ(4) =      .

       Idea 7
       More generally, the relation between the incoming and reflected waves may depend on the
       exact form of the incoming wave. In this case, it’s useful to consider sinusoidal solution. Let

                                             yin (x, t) = ei(kx−ωt) .

       Almost all boundary conditions will state that something at the boundary is constant in time,
       which is only possible if the reflected wave has the same frequency. So in general we have

                                            yr (x, t) = rei(−kx−ωt)

       where r is the reflection coefficient. If the medium exists for x > 0, there is also a transmitted
       wave there, of the form
                                                                ′
                                               yt (x, t) = tei(k x−ωt)
       where k ′ might differ from k, and t is the transmission coefficient. In general, both r and t
       may depend on k as well as the boundary conditions. Note that the phases of r and t depend
       on the conventions we used to define yr (x, t) and yt (x, t), though the magnitudes don’t.

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


[User]
Problem 2

                  Magnetic field with a V-shaped wire

Among the first successes of the interpretation by Ampère of magnetic phenomena, we have the
computation of the magnetic field B generated by wires carrying an electric current, as compared
to early assumptions originally made by Biot and Savart.

A particularly interesting case is that of a very long thin wire, carrying a constant current i, made
out of two rectilinear sections and bent in the form of a "V", with angular half-span 1 α (see figure).
According to Ampère's computations, the magnitude B of the magnetic field in a given point P
lying on the axis of the "V", outside of it and at a distance d from its vertex, is proportional to
      α
tan      . Ampère's work was later embodied in Maxwell's electromagnetic theory, and is
     2
universally accepted.

                                                                    α

Using our contemporary knowledge of electromagnetism,

1.        Find the direction of the field B in P.   [1 point]
                                                            α
2.        Knowing that the field is proportional to tan         , find the proportionality factor k in
                                                            2
                          α
          B(P) = k tan       .     [1.5 points]
                           2
3.        Compute the field B in a point P* symmetric to P with respect to the vertex, i.e. along the
          axis and at the same distance d, but inside the "V" (see figure). [2 points]

1
 Throughout this problem α is expressed in radians
Final

Problem 2                                                                                Page 2

4.      In order to measure the magnetic field, we place in P a small magnetic needle with moment
        of inertia I and magnetic dipole moment μ; it oscillates around a fixed point in a plane
        containing the direction of B. Compute the period of small oscillations of this needle as a
        function of B.    [2.5 points]

        In the same conditions Biot and Savart had instead assumed that the magnetic field in P
                                                                iμ α
might have been (we use here the modern notation) B(P) = 20 , where μ0 is the magnetic
                                                                 π d
permeability of vacuum. In fact they attempted to decide with an experiment between the two
interpretations (Ampère's and Biot and Savart's) by measuring the oscillation period of the magnetic
needle as a function of the "V" span. For some α values, however, the differences are too small to
be easily measurable.

5.      If, in order to distinguish experimentally between the two predictions for the magnetic
        needle oscillation period T in P, we need a difference by at least 10%, namely T1 > 1.10 T2
        (T1 being the Ampere prediction and T2 the Biot-Savart prediction) state in which range,
        approximately, we must choose the "V" half-span α for being able to decide between the
        two interpretations. [3 points]

Hint

Depending on which path you follow in your solution, the following trigonometric equation might
               α     sin α
be useful: tan   =
               2   1 + cos α

Final

                               NAME____________________________                 Problem         2
                               TEAM_____________________________                 Page n.        A

                               CODE______________________________
                                                                               Page total

                                        Answer sheet
In this problem write the requested results as analytic expressions, not as numerical values and
units, unless explicitly indicated.

1.      Using the following sketch draw the direction of the B field (the length of the vector is not
        important). The sketch is a spatial perspective view.

2.      Proportionality factor k ............................

3.      Absolute value of the magnetic field intensity at the point P*, as described in the

        text...............................................

        Draw the direction of the B field in the above sketch

4.      Period of the small angle oscillations of the magnet .................................

5.      Write for which range of α values (indicating here the numerical values of the range limits)
        the ratio between the oscillation periods, as predicted by Ampère and by Biot and Savart, is
        larger than 1.10:

           ......................... ............................

Final

Solution:
Problem 2 – Solution                                                                              Page 1

      Solution
1.    The contribution to B given by each leg of the "V" has the same direction as that of a corresponding
      infinite wire and therefore - if the current proceeds as indicated by the arrow - the magnetic field is
      orthogonal to the wire plane taken as the x-y plane. If we use a right-handed reference frame as
      indicated in the figure, B(P) is along the positive z axis.

      For symmetry reasons, the total field is twice that generated by each leg and has still the same
      direction.

2A.   When α=π/2 the "V" becomes a straight infinite wire. In this case the magnitude of the field B(P) is
                          i       iμ 0                                         i μ0
      known to be B =           =      , and since tan(π/4)=1, the factor k is      .
                      2π ε 0 c d 2π d
                              2
                                                                               2πd

      The following solution is equally acceptable:
                                                    μ 0i cosθ1 − cosθ 2
2B.   If the student is aware of the equation B =                       for a finite stretch of wire lying on a
                                                    4π         h
      straight line at a distance h from point P and whose ends are seen from P under the angles θ1 and θ2,
                                                                        μ i 1 − cosα
      he can find that the two legs of the “V” both produce fields 0                    and therefore the total
                                                                        4π d sinα
                     iμ 0 1 − cos α     iμ 0     α
      field is B =                   =       tan     . This is a more complete solution since it also proves
                    2π d sin α         2π d       2
      the angular dependence, but it is not required.

3A.   In order to compute B(P*) we may consider the "V" as equivalent to two crossed infinite wires (a
      and b in the following figure) plus another "V", symmetrical to the first one, shown in the figure as
      V', carrying the same current i, in opposite direction.

      Problem 2 – Solution                                                                             Page 2

      Then B (P * ) = Ba (P * ) + Bb (P * ) + BV' (P * ) . The individual contributions are:
                               iμ 0
      Ba (P* ) = Bb (P* ) =            , along the negative z axis;
                            2π d sin α
                   iμ 0      α
      BV' (P* ) =       tan     , along the positive z axis.
                  2π d       2
                                       iμ 0  2        α              1 + cosα         α
      Therefore we have B(P* ) =                − tan           =k            = k cot   , and the field is
                                      2π d sinα       2                sinα           2
      along the negative z axis.

    The following solutions are equally acceptable:
3B. The point P* inside a "V" with half-span α can be treated as if it would be on the outside of a "V"
    with half-span π-α carrying the same current but in an opposite way, therefore the field is
                   π −α             π α             α
    B(P* ) = k tan          = k tan   −     = k cot      ; the direction is still that of the z axis but it is
                     2              2 2              2
    along the negative axis because the current flows in the opposite way as previously discussed.

3C. If the student follows the procedure outlined under 2B., he/she may also find the field value in P*
    by the same method.

4.    The mechanical moment M acting on the magnetic needle placed in point P is given by M = μ ∧ B
      (where the symbol ∧ is used for vector product). If the needle is displaced from its equilibrium
      position by an angle β small enough to approximate sinβ with β, the angular momentum theorem
                                  d β
                                    2
                           dL
      gives M = - μBβ =       = I 2 , where there is a minus sign because the mechanical momentum is
                           dt      dt
      always opposite to the displacement from equilibrium. The period T of the small oscillations is
                               2π          I
      therefore given by T =       = 2π        .
                               ω          μB
      Writing the differential equation, however, is not required: the student should recognise the same
      situation as with a harmonic oscillator.

5.    If we label with subscript A the computations based on Ampère's interpretation, and with subscript
      BS those based on the other hypothesis by Biot and Savart, we have

Problem 2 – Solution                                                                         Page 3
        iμ 0     α                            iμ 0
BA =         tan( )                  B BS =        α
        2πd      2                            π 2d
                  2πId                            π 2 Id
TA = 2π                              TBS = 2π
                        α                         μ 0 μiα
             μ 0 μi tan( )
                         2
TA             2α
    =
TBS             α
          π tan( )
                    2

                                                                         2
For α = π/2 (maximum possible value) TA = TBS; and for α → 0 TA →            TBS ≈ 1.128 TBS . Since
                                                                         π
                   tan(α / 2)                                               T
within this range             is a monotonically growing function of α, A is a monotonically
                      α /2                                                 TBS
decreasing function of α; in an experiment it is therefore not possible to distinguish between the
two interpretations if the value of α is larger than the value for which TA = 1.10 TBS (10%
                                α       4 α         α
difference), namely when tan        =         = 1.05 . By looking into the trigonometry tables or
                                2     1.21π 2        2
using a calculator we see that this condition is well approximated when α/2 = 0.38 rad; α must
therefore be smaller than 0.77 rad ≈ 44°.
A graphical solution of the equation for α is acceptable but somewhat lengthy. A series
development, on the contrary, is not acceptable.

Problem 3 – Solution                                                                           Page 1

Grading guidelines

1.    1     for recognising that each leg gives the same contribution
      0.5   for a correct sketch

2.    0.5  for recognising that α=π/2 for a straight wire, or for knowledge of the equation given in
           2B.
      0.25 for correct field equation (infinite or finite)
      0.25 for value of k

3.    0.7 for recognising that the V is equivalent to two infinite wires plus another V
      0.3 for correct field equation for an infinite wire
      0.5 for correct result for the intensity of the required field
      0.5 for correct field direction
alternatively
      0.8 for describing the point as outside a V with π-α half-amplitude and opposite current
      0.7 for correct analytic result
      0.5 for correct field direction
alternatively
      0.5 for correctly using equation under 2B
      1      for correct analytic result
      0.5 for correct field direction

4.    0.5   for correct equation for mechanical moment M
      0.5   for doing small angle approximation sin β ≈ β
      1     for correct equation of motion, including sign, or for recognizing analogy with
            harmonic oscillator
      0.5   for correct analytic result for T

5.    0.3   for correct formulas of the two periods
      0.3   for recognising the limiting values for α
      0.4   for correct ratio between the periods
      1     for finding the relationship between α and tangent
      0.5   for suitable approximate value of α
      0.5   for final explicit limiting value of α

For the numerical values a full score cannot be given if the number of digits is incorrect (more than
one digit more or less than those given in the solution) or if the units are incorrect or missing

Magnetic field with a V-shaped wire


[User]
A narrow tube is formed in the shape of ring of radius R. Initially, it is stationary
    and horizontal in the lab frame. Then, it is quickly spun by 180◦ about its east-west diameter.

      (a) Suppose the tube contains water, and the Earth’s rotational velocity ω makes an angle ϕ to
          the vertical in the lab frame. What is the speed of the water afterward?

      (b) Suppose the tube is a conductor with self-inductance L, and the Earth’s magnetic field B
          makes an angle ϕ to the vertical in the lab frame. What is the current in the tube afterward?

Solution:
(a) This is called the Compton generator. It was invented by Compton while he was
         still an undergraduate to measure the Coriolis force, and he found agreement to within 3%.
          We first compute the Coriolis impulse on a small piece of the water in the tube, with mass
          dm, as the ring spins around. We have
                                     Z                  Z
                          dJc = (dm) 2 ω × v dt = (dm) 2 ω × dr = (dm) 2 ω × ∆r

          where ∆r is the total displacement of that piece of water. Since the rotation is about the
          east-west axis, all the displacements are north-south, which means that only the vertical
          component of ω matters. If we let θ = 0 at the easternmost point of the ring, then the
          component of the impulse on this fluid element along the ring is

                                              dJc = (2ω cos ϕ dm) ∆r sin θ

          Next, we integrate over the ring, letting θ = 0 at the easternmost point, so that
                                         Z 2π
                                                  dm                               m
                           Jc = 2ω cos ϕ       dθ     (2R sin θ) sin θ = 4ωR cos ϕ    π.
                                          0       dθ                               2π

          The final velocity is given by Jc = mvf , so that

                                                    vf = 2ωR cos ϕ.

     (b) This is called an Earth inductor, or Delzenne’s circle. We simply apply Faraday’s law, using
         the fact that the change in magnetic flux is 2πR2 B cos ϕ, along with
                                                    Z
                                              ∆Φ = E dt = LIf

          to conclude that
                                                    2πR2
                                                If =     B cos ϕ.
                                                      L
         This is similar in form to the answer in part (a), and the reason is that the magnetic force
         v × B and the Coriolis force 2v × ω are similar. Indeed, we could have solved part (a) much
         faster by thinking like the magnetic case, and computing a change in the “flux” of ω. Of
         course, the analogy isn’t perfect. The fluid motion is dominated by kinetic energy, while, as
         we mentioned in E5, in a typical circuit the kinetic energy of the charges is negligible, and
         field energy dominates instead. Also, in a typical circuit the density of electrons is almost
         perfectly uniform, while in a mechanical system the mass density can be arbitrary.
          However, it can sometimes be helpful to keep this analogy in mind. If the force is the only
          thing that matters, then we can often exchange magnetic and Coriolis force effects. For
          example, as we discussed in E5, a superconductor in a magnetic field will produce currents
          that expel that magnetic field. But since the Coriolis force has the same form, if you just
          rotate a superconductor in a lab on the Earth, it will also produce currents, because the
          electrons respond to the Coriolis force in the same way! This neat effect is called the London
          moment.


[User]
This problem is about some neat atmospheric phenomena. For some parts, it will
    be useful to use Babinet’s principle.

     (a) On a foggy night, there are many tiny water droplets in the air. On such nights one can see a
         ring around the moon, called a lunar corona, shown at left above. The ring is usually reddish
         in color. If one looks very carefully on a good night, one can see a blue ring outside the red
         ring and a blueish-white region inside the red ring. On other nights, one can only see a white
         haze around the moon. Explain these observations.

     (b) The size of the corona depends on the atmospheric conditions. Estimate the diameter of the
         water droplets in the air if the first red ring around the moon appears to have a diameter 4
         times that of the moon. The angular diameter of the moon in the sky is 0.5◦ .

        (c) On a cold night, there are many thin hexagonal ice crystals in the air. On such nights one
            can see a much larger, sharper ring around the moon, called a 22◦ halo, shown at right above.
            The size of the halo does not depend on the size of the crystals. Explain these observations.

     (d) In the photo used in part (c), the moon is shaped like an octagon. Why?

      (e) On a cold and exceptionally calm night, the results will be different.

          Instead of a circle, one will see two “moon dogs”, bright spots displaced about 22◦ from
          the moon horizontally. In addition, lights on the ground will produce vertical “light pillars”.
          Explain these observations.

[2] Problem 21 (Povey). Consider a reflective metal tube, such as a length of copper pipe, with length
    L and radius r. If you place a diffuse light source at one end of the tube, on the axis of symmetry,
    and look at it from the other end, with your eye also on the axis of symmetry, then you will see
    both the light source and bright circular rings around it. Why does this happen? Assuming the
    light has wavelength λ, calculate the angles of the bright rings.

Solution:
(a) Fog is made of small water droplets dispersed throughout the air, which leads to
     single slit diffraction by Babinet’s principle. The red wavelengths are spread out more, so the
     bluer regions are seen inside the red rings.
     Note that the colors are only pronounced if the droplets are small (to give a large diffraction
     angle), and nearly uniform in size. A wide range of droplet sizes will wash out the diffraction
     features, giving a white haze.

 (b) This light was deflected by θ = 1◦ , and from the one-dimensional single slit diffraction pattern,
     we can estimate a drop diameter a = λ/θ. (The numeric factor is actually different, since it’s
     two-dimensional diffraction, but we’re just doing a rough estimate here.) Taking λ = 650 nm
     for red light gives a = 4 × 10−5 m. Anything within a factor of a few is acceptable.

 (c) The fact that you always get the same angle, regardless of crystal size, is a tipoff that this
     isn’t actually a diffraction effect. Instead, it’s a geometric optics effect. The hexagonal ice
     crystals refract and reflect the light. The resulting angular deflection has a critical point at
     22◦ , which causes a lot of light to come out with that deflection. The reason that the halo
     has a red to blue gradient is simply because the index of refraction depends slightly on the
     wavelength of light.
     In XRev, you’ll carry out this kind of calculation explicitly for a spherical drop of water,
     which leads to the familiar rainbow.

 (d) The octagon is just showing the aperture shape of the camera, i.e. the shape of the opening
     that the light goes through. The reason it’s apparent here is because the photograph is taken
     in low light conditions, requiring a wide aperture, the moon itself is quite small, and the

          camera evidently was not focused properly. In bokeh photography, this sort of effect is done
          intentionally. We’ll discuss cameras further in W3.

      (e) In the absence of wind, the hexagonal ice crystals will lie flat in the air, so light arriving
          from the top can just pass right through them without much deflection. We only see a 22◦
          deflection when light enters the hexagon horizontally, as shown.

          This yields the two “moon dogs”, and if the same thing happens in the daytime, one can
          instead see “sun dogs”.
          As for the light pillars, they simply occur when light reflects directly off the flat bottom faces
          of the hexagons to reach the viewer, as shown here.

[2] Problem 21 (Povey). Consider a reflective metal tube, such as a length of copper pipe, with length
    L and radius r. If you place a diffuse light source at one end of the tube, on the axis of symmetry,
    and look at it from the other end, with your eye also on the axis of symmetry, then you will see
    both the light source and bright circular rings around it. Why does this happen? Assuming the
    light has wavelength λ, calculate the angles of the bright rings.

    Solution. You can tell that a diffraction explanation is implausible, because the copper pipe is too
    big for diffraction effects to be prominent, and a diffuse light source probably isn’t coherent, yet
    the result is sharp. Instead, the effect is due to geometric optics.
       The closest ring is due to light that bounces off the inner surface of the pipe, then arrives back at
    the middle. By basic trigonometry, the angle is θ = tan−1 (2r/L). Higher rings are due to multiple
    bounces, and arrive at angles tan−1 (2nr/L). There are infinitely many rings, though they blur
    together at high n.


[User]
01m
                     IPhO 2023, problem 1, parts A, B, and D.3. A short derivation of the key features
    of Brownian motion. It requires only the ideas of this problem set, and some basic mechanics.

    4      Data Analysis
         Idea 5
         All data analysis for the USAPhO and IPhO can be done using extremely basic methods.
         Sometimes, it suffices to just calculate a value based on a single data point, or by cleverly
         using a pair of data points. When this isn’t enough, you’ll have to do graphical data analysis,
         which will usually correspond to drawing a line and measuring its slope and intercept. This
         is quite limited compared to modern statistical tools, but also can be surprisingly powerful.

         Example 5

         The activity of a radioactive substance obeys A(t) = A0 e−t/τ . Using measurements of t and
         A(t), plot a line to find A0 and τ .

         Solution
         To handle exponential relationships, take the logarithm of both sides for

                                            log A(t) = log A0 − t/τ.

         Then a plot of log A(t) vs. t has slope −1/τ and y-intercept log A0 .

Solution:
IPhO 2023, problem 1, parts A, B, and D.3. A short derivation of the key features
    of Brownian motion. It requires only the ideas of this problem set, and some basic mechanics.

    4      Data Analysis
         Idea 5
         All data analysis for the USAPhO and IPhO can be done using extremely basic methods.
         Sometimes, it suffices to just calculate a value based on a single data point, or by cleverly
         using a pair of data points. When this isn’t enough, you’ll have to do graphical data analysis,
         which will usually correspond to drawing a line and measuring its slope and intercept. This
         is quite limited compared to modern statistical tools, but also can be surprisingly powerful.

       Example 5

       The activity of a radioactive substance obeys A(t) = A0 e−t/τ . Using measurements of t and
       A(t), plot a line to find A0 and τ .

       Solution
       To handle exponential relationships, take the logarithm of both sides for

                                           log A(t) = log A0 − t/τ.

       Then a plot of log A(t) vs. t has slope −1/τ and y-intercept log A0 .


[User]
6. Truck (7 points)

1)Aropeisputoverapolesothattheplaneoftheropeisperpendicular
totheaxesofthepole,andthelengthofthatsegmentoftherope,which
touches the pole is l, much shorter than the radius of the pole R, see fi-
gure (a). To the one end of the rope, a force T is applied; the sliding of
the rope can be preventedby applying a force T1 to the other end ofthe
rope. Express the ratio T1/T via l, R, and μ, where μ is the coefficient
of friction between the rope and the pole.
2)Answerthefirstquestion,iflisnotsmall(i.e.withouttheassumption
l fi R).
Remark: you may use equality
lim
x→0
(1 + nx)1/x
= en
.
3)Theropemakesexactlyn = 2windsaroundthepole. One endofthe
ropeis attachedtoatruckstandingonaslope(slantingangleφ = 10◦
);
the mass of the truck m = 20 t, see figure (b). Find the force F, needed
to apply to the other end of the rope, in order to keep the truck at rest.
Use the numerical value μ = 0,3. All the other friction forces acting
upon the truck can be neglected.
4) How does the answer change, if the cross-section of the pole is not
circular, but instead,egg-shaped? Motivate youranswer.
T T1
(a)
(b)

Solution:
6. Truck

1) First, since l is small, T1 ≈ T. From the radial force balance, N ≈
Tα = Tl/R, where α is the angle, by which the direction of the rope is
changed. From the tangential balance, T1 = T − μN = T(1 − μα).
2) If the angle is not small, we divide the touching segment into M
smallsegments andusetheprevious result: Ti+1 = Ti(1−μα/M). So,
TM = T(1 − μα/M)M
→ Te−μα
(asM → ∞). So, T0
= Te−μl/R
.
3) Using the previous formula and the provided numerical values (α =
2π),we obtain
T = Mg sinφe−μα
≈ 800N.
4) If we re-examine the solution to the second question, we notice that
we haven’t used the assumption of cylindrical shape. What matters, is
just the rotation angle of the tangent to the rope. So, the answer does
not change for egg-like cross-section.

Truck


[User]
T2: Floating table

A table is made by fastening a metal frame to a mas-
sive uniform plate (so they form a rigid body) and at-
taching it with chains to another frame that is fixed
on the horizontal ground. The motion of the table is
limited to the plane of the side view (right picture).
a) (4 pts) Show that in the configuration on the side
view, the table is in a stable equilibrium.
b) (6 pts) Find the period T of the small oscillations.
The masses of the chains and the frame can be
neglected. The chains are frictionless, inextensible,
and remain tensioned in oscillations. The grid step
is a = 0.100m, the acceleration of gravity g = 9.81m/s2
.
a
g
a

Solution:
T2: Floating table

Solution with forces First, let us consider the
forces and torques and show that in the configuration
shown in the figure, the table is indeed in equilib-
rium. By the word “table”, we understand the rigid
body formed by the plate and the frame attached to
it. In the following considerations and descriptions
of the table position we often use positions of points
A, B and C, which belong to the rigid table: distances
between them do not change. Since in the problem
statement, the motion is limited to the side view only,
the pair of short chains will always have the same
position when viewed from the side. So we will con-
sider them as a single chain (chain 1) with the tension
doubled. The same is true for the pair of long chains
(chain 2).
2a
4a
m® g
® F1
® F2
A
B C
Figure 3: Scheme of forces for the initial state.
There are 3 forces acting on a table (see Fig. 3):
plate weight m® g (at the center of mass of the table
C), and 2 tension forces from the chains: ® F1 (at point
A, chain 1) and ® F2 (at point B, chain 2). Since all the
forces only have vertical components different from
0, their horizontal components are balanced and for
their vertical components, the balance equation is:
−mg + F1 − F2 = 0. (8)
Finally, consider the torque balance around the
point B:
mg · 4a − F1 · 2a = 0. (9)
Solving the system of equations (8-9), we get F1 =
2mg and F2 = mg. Since both F1,F2 > 0, we conclude
that the chains are indeed tensioned and the system
is in equilibrium.
Now let’s analyze how the table rotates, when it
is displaced. Since the chains are tensioned and in-
extensible, the point B can only travel along the cir-
cle cb. Similarly, the distance between the point A0
and any point inside the circle ca is smaller than the
length of the short chain (4a); the distance between
the point A0 and any point outside of the circle ca is
larger than 4a.
Since the table is rigid, if we fix the point B and ro-
tate the table around it, point A would travel along
the circle ct. From the way how ct intersects with
ca, we can conclude that if the table is rotated clock-
wise, the distance A0A decreases. And if it is rotated
counter-clockwise, A0A increases.
Let’s imagine we translate (move without rotation)
the table to the left, so the point B travels to the point
B0. Point A would move to the point A0, as shown in
Fig. 4. Since A0 is outside of the circle ca, A0A0 > A0A
and the table has to rotate clockwise (around B0) to
compensate for this change and return the point A0
back to the circle ca (point A00) to satisfy the constant
chain length constraint.
B
A
C
A00
B0
A0
A0
B0
cb
ca
ct
A
A00
A0
Figure 4: Translation to the left and clockwise rota-
tion of the table
If we translate the table in the opposite direction,
following similar arguments, we find that the table
rotates clockwise in this case too, see Fig. 5.
B B0
B0
C
A0
A
A00
A0
cb
ca
ct
Figure 5: Translation to the right and clockwise ro-
tation of the table
Theoretical Problems - Solutions
Since the table rotates in the same direction for
both positive and negative horizontal displacements,
the Taylor expansion of the angle of rotation φ(x)
does not contain a linear term in x: φ(x) ∼ x2
. This
means that for small horizontal displacements, we
can neglect the table rotation.
Let’s look at what happens when the table is dis-
placed infinitesimally from its initial position in the
horizontal direction (it will also move in the vertical
direction, but the displacement in the vertical direc-
tion is an order of magnitude smaller and we will ne-
glect it). The vertical components of the forces in the
chains do not change significantly, but as you can see
in the Fig. 6, the horizontal components of the forces
in both chains appear in the direction opposite to the
displacement. This means that the equilibrium is sta-
ble.
m® g
® F1
® F2
B C
A
Figure 6: Horizontal displacement of the table
Since we have already drawn a figure showing the
displaced position of the table, let us write the equa-
tion of motion for the table plate for small horizon-
tal displacements x, where θ1 and θ2 are small an-
gles between the chains 1 and 2 respectively and the
vertical. Since the table displacement is small, ten-
sion forces F1 and F2 cannot change significantly (the
force change is of the same order as the displace-
ment). Thus, we can use the previously found values:
m¥ x = −F1 sinθ1 − F2 sinθ2 = −F1θ1 − F2θ2 =
= −2mg
x
4a
− mg
x
6a
= −mg
2x
3a
,
where we have also considered the lengths of the
chains 4a (chain 1) and 6a (chain 2), neglected the
rotation of the plate and used small angle approxi-
mations. Rearranging this equation, we get:
¥ x +
2g
3a
x = 0,
which describes horizontal oscillations of the table
plate with ω =
√
2g/(3a) = 8.09s−1
, frequency ν = 1.29Hz
and period
T = 777ms.
Solution with energies Let’s denote the small dis-
placement of the centre of mass from equilibrium po-
sition C to C0 as (x, y), where x, y fi a, and the small
tilt of the table as φ, as shown in Fig. 7.
5a
2a
4a
4a
A0
B0
C0
φ
φ
φ
B
A
C
Figure 7: Infinitesimal displacement of the table
Given the small angle approximation φ fi 1 (sin φ ≈
φ and cos φ ≈ 1), the displacement of points from equi-
librium positions A and B to A0 and B0 are:
ΔAx = x − 5asin φ − 2a(1 − cos φ) ≈ x − 5aφ,
ΔAy = y − 2asin φ + 5a(1 − cos φ) ≈ y − 2aφ,
ΔBx = x − 4a(1 − cos φ) ≈ x,
ΔBy = y − 4asin φ ≈ y − 4aφ.
Now we can write a condition that the squared
length of the small chain remains the same:
(x − 5aφ)2
+ (4a − y + 2aφ)2
= (4a)2
x2
− 10axφ + 25a2
φ2
− 8a(y − 2aφ) + (y − 2aφ)2
= 0,
Let’s notice that the largest term in x is x2
and in
y and aφ, it’s 8a(y + 2aφ). Thus, we can omit all other
terms, since they are of the higher order (and there-
fore smaller):
x2
− 8a(y − 2aφ) = 0,
x2
8a
− y + 2aφ = 0. (10)
Doing the same for the long chain we get:
x2
+ (6a + y − 4aφ)2
= (6a)2
x2
+ 12a(y − 4aφ) + (y − 4aφ)2
= 0,
Omitting higher order terms we get:
x2
+ 12a(y − 4aφ) = 0,
x2
12a
+ y − 4aφ = 0. (11)
Solving (10) and (11) as linear equations on y and φ,
we get:
y =
x2
3a
, φ =
5x2
48a2
.
Since the potential energy of the table
U = mg y =
2
3a
mg
x2
2
, (12)
Theoretical Problems - Solutions
doesn’t have a term linear in x, we conclude that the
system is in equilibrium. And since the coefficient
for x2
is positive, the equilibrium is stable.
The kinetic energy of the table is:
E = m
¤ x2
+ ¤ y2
2
+ I
¤ φ2
2
≈ m
¤ x2
2
, (13)
where I is a moment of inertia with respect to rota-
tion around the axis through the centre of mass C.
Since both y and φ depend on x2
, their squared
derivatives will be proportional to x2 ¤ x2
and can be
neglected when compared to ¤ x2
term.
Finally, we can find the period of oscillations from
the ratio of coefficients of (13) and (12):
T = 2π
√
m
/
2mg
3a
= 2π
√
3a
2g
= 0.777 s.
Solution with the curvature We begin with the
observation that at the initial horizontal position,
points A and B in Fig. 3 move horizontally. There-
fore, the instantaneous centre of table’s rotation is
at infinity (found as the intersection of lines perpen-
dicular to the velocity vectors, which are parallel in
this case), or in other words there is no rotation. An-
other way to see it is to consider the projections of
the velocities vA and vB to the line AB which should
be the same in a rigid body. Thus vA = vB = v and all
the points of the table move with the same velocity,
i.e. there is no immediate rotation.
® vB
® vA
A
B C
Figure 8: Immediate velocities given the constraints
Consequently, no rotational kinetic energy is in-
volved, and the motion of the table can be treated
as the motion of a point mass located at the centre of
mass, point C. Since at the given configuration, the
vC can only be horizontal, we conclude that initially
table is at equilibrium. All that remains is to find the
radius of curvature R of its trajectory and calculate
T = 2π
√
R/g.
To find the radius of curvature, we note that for an
arbitrary point P on the table, the curvature ® c = ˆ n/R
C
B
A
wB
wAy
wAx α
α
α
α
α
α
wC
Figure 9: Immediate accelerations
of its trajectory is a linear function of its coordinates,
where ˆ n denotes a unit vector pointing towards the
centre of curvature. Indeed, using a non-inertial ref-
erence frame where the rigid body remains at rest
with origin at O, the acceleration of point P is the su-
perposition of: the acceleration of O, the centripetal
acceleration − ® OPω2
, and the tangential acceleration
® ε × ® OP. All of these are linear functions of the coordi-
nates of point P, where ® ε denotes the angular accel-
eration and ω the angular speed. Hence, in the lab
frame, the acceleration of point P is a linear function
of coordinates, as must be its centripetal accelera-
tion v2 ® c. Since the angular speed of the table is zero,
all points move with the same speed, so ® c must be a
linear function of coordinates.
The remaining calculation is straightforward: due
to the translational motion of the body, curvatures
depend only on the horizontal coordinate x. The cur-
vature of B at x = 4a is cB = −1/(6a), with the minus sign
indicating that the vector points downwards. The
curvature of A at x = 2a is cA = 1/(4a). Hence, the
curvature of C is 1/(4a) + (1/(4a) + 1/(6a)) = 2/(3a), giv-
ing R = 1.5dm and resulting in T = 777ms. Positive
curvature means that the position is a stable equilib-
rium.
Another way to see it, is to consider normal accel-
eration wC of point C. For 2 points of a rigid body C
and B, the projections of their accelerations wC, wB to
the segment BC differ by the value ω2
|BC|. However,
in our case since there is no rotation, the projections
of acceleration are equal (like velocities). This means
that the accelaration wB is also normal and equal to
wB = v2
/(6a). Point A has both normal and tangential
acceleration. Comparing projections of points A, B,
Theoretical Problems - Solutions
and A, C, we get:
−wB cosα = wAy cosα − wAy sinα,
wC cosα = wAy cosα + wAy sinα,
Thus, we find that
wC = wB + 2wAy =
v2
6a
+ 2
v2
4a
=
v2
3a/2
,
so radius of curvature is R = 3a/2.
You can visualize the motion of the table using a
GeoGebra tool, following the web link or the qr code
https://www.geogebra.org/m/wbqwp3tf
Marking scheme T2
A Table is in equilibrium
Partial points:
– (1pt) Sketch of forces or eqn for forces
2.0
B The equilibrium is stable
Partial points:
– (1pt) Sketch with returning forces, but
there is no proof that ω0 = 0
– (1pt) There is a statement that the
stability comes from y = kx2
if k > 0, but
k is not found correctly (and thus could
potentially be negative)
2.0
C Table can rotate:
whether it is clear from the sketch or
from the equations, i.e. φ is introduced
1.0
D Table does not have immediate rotation
whether it is proved separately or follows
naturally from the results of calculations:
– geometry considered ⇒ φ ∼ x2
– eqn for contraints ⇒ φ ∼ x2
– immediate velocities ⇒ no rotation
1.0
E The plan how to find the period of oscil-
lation:
– 2nd Newton’s law ¥ x ∼ −x
– idea to find kinetic and potential ener-
gies
– using the curvature of the trajectory
1.0
F Finding all the elements for the plan
above:
– small horizontal forces
– energies to the correct approximations
– accelerations/curvatures
Partial points:
– (1pt) if not all elements are found or
they are found with mistakes
– (1pt) if there is no proof ω0 = 0 (or no
consideration about the rotation at all),
but other than that, everything is correct
– (0pts) if some elements are there, but
the plan was different (e.g. the student
was looking for energies, but has the
expression for forces)
2.0
G Formula and number for T
Partial points:
– (0.5pts) only formula or a number
– (0.5pts) simple mistake (like inverse
formula under the root)
1.0
Total 10.0
– Only a single partial point is applied (whatever is
larger)
– If there is a single simple arithemetic error/typo in
calculations in item F, which leads to a wrong for-
mula value in item G, then F and G receive 1 pt each
(no propagation of error). This doesn’t apply to more
profound errors in physics/geometry
Theoretical Problems - Solutions

Floating table


[User]
In this problem, we’ll go through Laplace’s slick derivation of Kepler’s first law.
    Throughout, we assume the orbit takes place in the xy plane, with the Sun at the origin.
     (a) Show that
                                                    γx          γy
                                                ẍ = −3
                                                        , ÿ = − 3
                                                    r           r
           where γ is a constant that depends on the parameters.
     (b) Show that
                                         d 3                  d 3
                                            (r ẍ) = −γ ẋ,      (r ÿ) = −γ ẏ.
                                         dt                   dt
        (c) Show that
                                                     d 3
                                                        (r r̈) = −γ ṙ.
                                                     dt
           (Hint: this can get messy. As a first step, try showing the left-hand side is equal to
           (r2 /2) d3 (r2 )/dt3 . You will have to switch variables to x and y and then switch back; for
           these purposes it’s useful to use the results of part (a), and the definition r2 = x2 + y 2 .)
     (d) Define ψ(t) = r(t)3 . In parts (b) and (c), we have shown that the differential equation

                                              d         du
                                                   ψ(t)      = −γu
                                              dt        dt
           has three solutions, namely ẋ, ẏ, and ṙ. Any second-order linear differential equations only
           has two independent solutions. If ẋ and ẏ are not independent, the orbit is simply a line,
           which is trivial. Assuming that doesn’t happen, they are independent, so ṙ must be a linear
           combination of them,
                                                     ṙ = Aẋ + B ẏ.
           Use this result to argue that the orbit is a conic section.

    6     Mechanics and Geometry
    For dessert, we’ll consider a few cute problems that relate statics to geometry.

         Example 8

         Given a triangle ABC, the Fermat point is the point X that minimizes AX + BX + CX.
         Design a machine that finds the Fermat point.

         Solution
         We take a horizontal plane and drill holes at points A, B, and C. A mass M on a rope is
         fed through each hole, and the three ends of the rope are tied together at point X. The
         gravitational potential energy is proportional to AX + BX + CX, so in equilibrium X lies
         on the Fermat point. Moreover, since the tensions in each rope are all equal to M g, force
         balance requires ∠AXB = ∠BXC = ∠CXA = 120◦ .

Solution:
(a) This just follows from F = ma. In terms of the usual parameters, γ = GM .

 (b) This immediately follows from clearing denominators in the results of part (a) and differenti-
     ating both sides.

    (c) Following the hint, we have
                             d 3            ...           1 d3            d2
                                (r r̈) = r3 r + 3r2 ṙr̈ = r2 3 (r2 ) = r2 2 (rṙ).
                             dt                           2 dt            dt
       At this point, we switch back to x and y. By differentiating r2 = x2 + y 2 ,

                                                 rṙ = xẋ + y ẏ.

       Plugging this in gives
                           d 3           d2                  d
                              (r r̈) = r2 2 (xẋ + y ẏ) = r2 (xẍ + y ÿ + ẋ2 + ẏ 2 ).
                           dt            dt                  dt
       We see that we’ll have a lot of factors involving ẍ and ÿ, but we know how to handle these
       using part (a). Using part (a) several times, we have
                                                γ                  γ           γ ṙ
                                ẋẍ + ẏ ÿ = − 3 (xẋ + y ẏ) = − 3 (rṙ) = − 2
                                                r                  r           r
       and
                                                      γ                   γ
                                       xẍ + y ÿ = − 3 (x2 + y 2 ) = − .
                                                      r                   r
       Plugging these results in, we find

                                  d 3         2      d −γ          2γ ṙ
                                    (r r̈) = r −              − 2 = −γ ṙ
                                 dt                 dt    r         r
       just as desired.

 (d) Integrating both sides,
                                               r = Ax + By + C.
       But then squaring both sides shows that the equation of the orbit is just a quadratic in
       x and y, which is precisely the form of a conic section in Cartesian coordinates. You can
       also show that the focus is at the origin, though this requires a bit more knowledge about conics.

This question was inspired by this paper, which has a few more derivations of Kepler’s first law.

6      Mechanics and Geometry
For dessert, we’ll consider a few cute problems that relate statics to geometry.

       Example 8

       Given a triangle ABC, the Fermat point is the point X that minimizes AX + BX + CX.
       Design a machine that finds the Fermat point.

       Solution
       We take a horizontal plane and drill holes at points A, B, and C. A mass M on a rope is
       fed through each hole, and the three ends of the rope are tied together at point X. The
       gravitational potential energy is proportional to AX + BX + CX, so in equilibrium X lies
       on the Fermat point. Moreover, since the tensions in each rope are all equal to M g, force
       balance requires ∠AXB = ∠BXC = ∠CXA = 120◦ .


[User]
In physics, whenever we have an equality relation, both sides of the equation should be
of the same type i.e. they must have the same dimensions. For example you cannot have a
situation where the quantity on the right-hand side of the equation represents a length and
the quantity on the left-hand side represents a time interval. Using this fact, sometimes one
can nearly deduce the form of a physical relation without solving the problem analytically.
For example if we were asked to find the time it takes for an object to fall from a height of
 h under the influence of a constant gravitational acceleration g , we could argue that one
only needs to build a quantity representing a time interval, using the quantities g and h and
the only possible way of doing this is T = a (h / g )1 / 2 . Notice that this solution includes an
as yet undetermined coefficient a which is dimensionless and thus cannot be determined,
using this method. This coefficient can be a number such as 1 , 1 2 , 3 , π , or any other
real number. This method of deducing physical relations is called dimensional analysis. In
dimensional analysis the dimensionless coefficients are not important and we do not need
to write them. Fortunately in most physical problems these coefficients are of the order of 1
and eliminating them does not change the order of magnitude of the physical quantities.
Therefore, by applying the dimensional analysis to the above problem, one
obtains T = (h / g )1 / 2 .

    Generally, the dimensions of a physical quantity are written in terms of the dimensions
of four fundamental quantities: M (mass), L (length), T (time), and K (temperature).
The dimensions of an arbitrary quantity, x is denoted by [x] . As an example, to express
the dimensions of velocity v , kinetic energy Ek , and heat capacity                    CV     we
                  −1               2     −2             2   −2   −1
write: [v] = LT        , [ Ek ] = ML T        , [CV ] = ML T K .

1     Fundamental Constants and Dimensional Analysis

         Find the dimensions of the fundamental constants, i.e. the Planck's
         constant, h , the speed of light, c , the universal constant of gravitation, G ,
1.1                                                                                0.8
         and the Boltzmann constant, k B , in terms of the dimensions of length, mass,
         time, and temperature.
The Stefan-Boltzmann law states that the black body emissive power which is the total
                                                                                4
energy radiated per unit surface area of a black body in unit time is equal to σθ            where
σ is the Stefan-Boltzmann's constant and θ is the absolute temperature of the black body.
         Determine the dimensions of the Stefan-Boltzmann's constant in terms of the
1.2                                                                                0.5
         dimensions of length, mass, time, and temperature.
The Stefan-Boltzmann's constant is not a fundamental constant and one can write it in
                                                                        δ
terms of fundamental constants i.e. one can write σ = a hα c β G γ k B . In this relation a is a
dimensionless parameter of the order of 1. As mentioned before, the exact value of a is not
significant from our viewpoint, so we will set it equal to 1.
1.3      Find α , β , γ , and δ using dimensional analysis.                                          1.0

                                                        1

2     Physics of Black Holes

     In this part of the problem, we would like to find out some properties of black holes
using dimensional analysis. According to a certain theorem in physics known as the no hair
theorem, all the characteristics of the black hole which we are considering in this problem
depend only on the mass of the black hole. One characteristic of a black hole is the area of
its event horizon. Roughly speaking, the event horizon is the boundary of the black hole.
Inside this boundary, the gravity is so strong that even light cannot emerge from the region
enclosed by the boundary.
     We would like to find a relation between the mass of a black hole, m , and the area of
its event horizon, A . This area depends on the mass of the black hole, the speed of light,
and the universal constant of gravitation. As in 1.3 we shall write A = G α c β m γ .

2.1       Use dimensional analysis to find α , β , and γ .                                        0.8
          From the result of 2.1 it becomes clear that the area of the event horizon of a black
      hole increases with its mass. From a classical point of view, nothing comes out of a
      black hole and therefore in all physical processes the area of the event horizon can only
      increase. In analogy with the second law of thermodynamics, Bekenstein proposed to
      assign entropy, S , to a black hole, proportional to the area of its event horizon i.e.
       S = η A . This conjecture has been made more plausible using other arguments.

          Use the thermodynamic definition of entropy dS = dQ θ to find the dimensions
2.2       of entropy.                                                                             0.2
          dQ is the exchanged heat and θ is the absolute temperature of the system.

          As in 1.3, express the dimensioned constant η as a function of the fundamental
2.3                                                                                1.1
          constants h , c , G , and k B .

Do not use dimensional analysis for the rest of problem, but you may use the results you
have obtained in previous sections.

3     Hawking Radiation

    With a semi-quantum mechanical approach, Hawking argued that contrary to the
classical point of view, black holes emit radiation similar to the radiation of a black body at
a temperature which is called the Hawking temperature.

          Use E = mc 2 , which gives the energy of the black hole in terms of its mass,
3.1       and the laws of thermodynamics to express the Hawking temperature θ H of                0.8
          a black hole in terms of its mass and the fundamental constants. Assume that
          the black hole does no work on its surroundings.
          The mass of an isolated black hole will thus change because of the Hawking
          radiation. Use Stefan-Boltzmann's law to find the dependence of this rate of
3.2                                                                                0.7
          change on the Hawking temperature of the black hole, θ H and express it in
          terms of mass of the black hole and the fundamental constants.

                                                  2

3.3      Find the time t * , that it takes an isolated black hole of mass m to evaporate       1.1
         completely i.e. to lose all its mass.

      From the viewpoint of thermodynamics, black holes exhibit certain exotic behaviors.
      For example the heat capacity of a black hole is negative.
3.4      Find the heat capacity of a black hole of mass m .                                    0.6

4     Black Holes and the Cosmic Background Radiation

   Consider a black hole exposed to the cosmic background radiation. The cosmic
background radiation is a black body radiation with a temperature θ B which fills the entire
                                                                                4
universe. An object with a total area A will thus receive an energy equal to σ θ B × A per
unit time. A black hole, therefore, loses energy through Hawking radiation and gains
energy from the cosmic background radiation.
         Find the rate of change of a black hole's mass, in terms of the mass of the
4.1      black hole, the temperature of the cosmic background radiation, and the               0.8
         fundamental constants.

         At a certain mass, m * , this rate of change will vanish. Find m * and express it
4.2                                                                                0.4
         in terms of θ B and the fundamental constants.

         Use your answer to 4.2 to substitute for θ B in your answer to part 4.1 and
4.3      express the rate of change of the mass of a black hole in terms of m , m * ,          0.2
         and the fundamental constants.
         Find the Hawking temperature of a black hole at thermal equilibrium with
4.4                                                                                0.4
         cosmic background radiation.
         Is the equilibrium stable or unstable? Why? (Express your answer
4.5                                                                                0.6
         mathematically)

                                                   3

Solution:
1.1) One may use any reasonable equation to obtain the dimension of the questioned
quantities.

I) The Planck relation is hν = E ⇒ [h][ν ] = [ E ] ⇒ [h] = [ E ][ν ]−1 = ML2T −1
(0.2)
II) [c] = LT −1                                                                (0.2)
           Gmm
III) F =          ⇒ [G ] = [ F ][r 2 ][m]−2 = M −1 L3T −2                      (0.2)
            r2
IV) E = K Bθ ⇒ [ K B ] = [θ ]−1[ E ] = ML2T −2 K −1                            (0.2)

1.2) Using the Stefan-Boltzmann's law,
 Power
         = σ θ 4 , or any equivalent relation, one obtains:
  Area
        (0.3)
[σ ]K = [ E ]L−2T −1 ⇒ [σ ] = MT −3 K −4 .
      4
                                                                                (0.2)

1.3) The Stefan-Boltzmann's constant, up to a numerical coefficient, equals
σ = hα c β G γ k B δ , where α , β , γ , δ can be determined by dimensional analysis. Indeed,
[σ ] = [h]α [c]β [G ]γ [k B ]δ , where e.g. [σ ] = MT −3 K −4 .

             (
MT − 3 K − 4 = ML2T −1   ) (LT ) (M L T ) (ML T K ) = M
                         α   −1 β   −1 3   −2 γ   2   −2   −1 δ   α −γ +δ
                                                                            L2α + β + 3γ + 2δ T −α − β − 2γ − 2δ K −δ ,
                                                                                (0.2)
The above equality is satisfied if,

  α − γ + δ = 1,                                             α = −3,
  2α + β + 3γ + 2δ = 0,                                      β = −2,
                                                             
⇒                                  (Each one (0.1))         ⇒        (Each one (0.1))
  − α − β − 2γ − 2δ = −3,                                     γ = 0,
  − δ = −4,                                                 δ = 4.
               4
            kB
⇒      σ = 2 3.
            ch

2.1) Since A , the area of the event horizon, is to be calculated in terms of m from a
classical theory of relativistic gravity, e.g. the General Relativity, it is a combination of
c , characteristic of special relativity, and G characteristic of gravity. Especially, it is

independent of the Planck constant h which is characteristic of quantum mechanical
phenomena.

A = G α c β mγ

Exploiting dimensional analysis,
⇒ [ A] = [G ]α [c]β [m]γ ⇒ L2 = M −1L3T − 2    (              ) (LT ) M = M
                                                              α       −1 β    γ         −α + γ
                                                                                L3α + β T − 2α − β
                                                                                (0.2)
The above equality is satisfied if,

   − α + γ = 0,                                                   α = 2,
                                                                 
⇒  3α + β = 2, (Each one (0.1))                       ⇒          β = −4,        (Each one (0.1)) ⇒
  − 2α − β = 0,                                                   γ = 2,
                                                                 
                 m 2G 2
           A=           .
                  c4

2.2)

                                                   dQ
From the definition of entropy dS =                     , one obtains [ S ] = [ E ][θ ]−1 = ML2T −2 K −1 (0.2)
                                                   θ

2.3) Noting η = S A , one verifies that,

[η ] = [ S ][ A]−1 = MT −2 K −1 ,

                                                                                (0.2)
            α    β   γ     δ      −α + β +δ 3α + 2 β +γ + 2δ − 2α − β −γ − 2δ
[η ] = [G ] [h] [c] [k B ] = M             L                T                 K −δ ,
Using the same scheme as above,

           − α + β + δ = 1,                                                   α = −1,
           3α + 2β + γ + 2δ = 0,                                              β = −1,
                                                                              
          ⇒                          (Each one (0.1))                        ⇒                        (Each one (0.1))
            − 2α  − β − γ − 2δ = −2,                                          γ = 3,
           δ = 1,                                                            δ = 1,
                c3k B
thus,     η=          .                                                                                (0.1)
                Gh

3.1)

The first law of thermodynamics is dE = dQ + dW . By assumption, d W = 0 . Using the
                            dQ
definition of entropy, dS =    , one obtains,
                              θ
               dE = θ H dS + 0,                      (0.2) + (0.1), for setting d W = 0 .
            G kB 2
       S =         m ,
Using,       ch        [(0.1) for S]
        E = mc 2 ,
       
                                  −1            −1
                     dE  dS    2  dS 
one obtains, θ H =     =     =c                                  (0.2)
                     dS  dE       dm 
                 1 c h 1
                         3
Therefore, θ H =           .                               (0.1)+(0.1) (for the coefficient)
                  2  GkB m

3.2) The Stefan-Boltzmann's law gives the rate of energy radiation per unit area. Noting
that E = mc 2 we have:

dE / dt = −σθ H 4 A,            (0.2)
          4
       kB
σ = c 2 h 3 ,
                                                              4
                                           k B  c 3 h 1  m 2G 2
                                              4
                                2 dm
                         ⇒c           = − 2 3              4
                                                                  , (0.2)
 A =   m 2 2
           G                      d t     c  h   2G k B m   c
        c4
 
  E = mc 2,
          dm        1 c4h 1
⇒               =−            . (0.1) (for simplification) + (0.2) (for the minus sign)
           dt      16 G 2 m 2

3.3)
By integration:
dm       1 c4h 1                             c 4h
     =−            .     ⇒ ∫ m 2 dm = − ∫         dt (0.3)
 dt     16 G 2 m 2                          16G 2
                       3c 4 h
⇒ m3 (t ) − m3 (0) = −        t , (0.2) + (0.2) (Integration and correct boundary values)
                       16G 2
At t = t * the black hole evaporates completely:
                                  16G 2
m(t * ) = 0 (0.1)        ⇒ t * = 4 m3 (0.2)+(0.1) (for the coefficient)
                                  3c h

3.4) CV measures the change in E with respect to variation of θ .

       dE          (0.2)
CV = dθ ,
                                                  2G k B 2
  E = mc ,
           2        (0.2)          ⇒       CV = −         m . 0.1)+(0.1) (for the coefficient)
        3
                                                     ch
 θ = c h 1
     2G k B m

4.1)      Again the Stefan-Boltzmann's law gives the rate of energy loss per unit area of the
          black hole. A similar relation can be used to obtain the energy gained by the black
          hole due to the background radiation. To justify it, note that in the thermal
          equilibrium, the total change in the energy is vanishing. The blackbody radiation
          is given by the Stefan-Boltzmann's law. Therefore the rate of energy gain is given
          by the same formula.

                         (0.1) + (0.4) (For the first and the second terms respectively)
dE                  4
    = −σθ 4 A + σθ B A                    dm     hc 4 1    G2
 dt                                 ⇒        =−      2 2
                                                          +  8 3
                                                                 (kB θ B )4 m 2            (0.3)
 E = mc 2 ,                              dt    16G m      ch

4.2)
          dm
Setting         = 0 , we have:
          dt
    hc 4 1          G2              2
                 + 8 3 (k B θ B ) m* = 0
                                 4
 −      2      2
                                                                  (0.2)
   16G m     *     ch
and consequently,
        c3h 1
m* =                                                                               (0.2)
      2G k B θ B

4.3)
      c 3h 1                dm     hc 4 1        m4 
θB =                 ⇒         =−             1−                                         (0.2)
     2G k B m*              dt    16G 2 m 2  m* 4 

4.4) Use the solution to 4.2,
   *   c 3h 1                                         *  c 3h 1
 m =                (0.2) and 3.1 to obtain,        θ =           =θ B     (0.2)
      2G k B θ B                                        2G k B m*
One may also argue that m* corresponds to thermal equilibrium. Thus for m = m* the
black hole temperature equals θ B .

Or one may set
               dE
                dt
                            4
                             (      4
                                       )
                    = −σ θ * − θ B A = 0 to get θ * = θ B .

4.5) Considering the solution to 4.3, one verifies that it will go away from the
equilibrium.                                                    (0.6)
                                                     dm
                                  m > m*     ⇒            >0
  dm         4
          hc 1       m 4
                                                     dt
      = − 2 2 1 − * 4  ⇒ 
   dt     G m  m              m < m*               dm
                                            ⇒            <0
                                                       dt

of the same type i.e. they must have the same dimensions. For example you cannot