[User]
Points: 30                                                                                Time: 5.0 Hours

                           Trapping Ions and Cooling Atoms
In recent decades, trapping and cooling atoms and ions has been a fascinating topic for physicists, with

several Nobel prizes awarded for work in this area. In the                           rst part of this question, we will explore a

technique for trapping ions, known as the "Paul trap". Wolfgang Paul and Hans Dehmelt received one half

of the 1989 Nobel Prize in Physics for this work. Next, we investigate the Doppler cooling technique, one

of the works cited in the press release for the 1997 Nobel Prize in Physics awarded to Steven Chu, Claude

Cohen-Tannoudji, and William Daniel Phillips "for developments of methods to cool and trap atoms with

laser light".

    A. The Paul Trap
It is known that with electrostatic                 elds, it is not possible to create a stable equilibrium for a

charged      particle.     Therefore,       creating       a     stable      equilibrium      point    for    ions     requires      more

sophisticated techniques. The Paul trap is one of these techniques.

Consider a ring of charge with a radius                        R   and a uniform positive linear charge density                       λ   . A

positive point charge        Q   with mass      m   is placed at the center of the ring.

  Figure 1 - A positively charged ring with a uniform linear charge density λ and radius R : the origin of the coordinate system is at the

                                                               center of the ring.

IPhO 2024                                                                                Page 6 of 16
Theory, English (Official)

Points: 30                                                                                Time: 5.0 Hours

              a) In cartesian coordinates (             x, y, z)   , obtain the electric       eld due to the

              charged ring in the vicinity of the ring’s center to the                     rst order in   x/R   ,

              y/R    , and   z/R    .

     A-1                                                                                1.5 pt

              b) Find the angular frequency of small oscillations of the charged particle

              around the center of the ring in the directions for which a stable

              equilibrium exists.

In order to trap the charge             Q    fully, we would like to apply alternating             elds to produce a dynamic

equilibrium. Assume that the charge density is                           λ = λ0 + u cos Ωt       in which   λ0 u,   , and   Ω are
adjustable. We shall ignore radiative e                 ects. Then the equation of motion for small displacements

from the center of the ring, along the direction perpendicular to the plane of the ring will turn out

to be:

                                              z̈ = ( +k2 + aΩ2 cos Ωt) z                 (1)

     A-2      Write   a   and   k   in terms of the known parameters.                                                     0.4 pt

We   would    like   to obtain          an    approximate        solution    to Equation       (1) by making        the   following

simplifying assumptions:            a≪1 Ω≫k   ,          , and   aΩ ≫ k
                                                                     2       2
                                                                                . With these assumptions, it can be shown

that the solution of this equation can be split into two parts:                        z(t) = p(t) + q(t)   , where       p(t)   is a

slowly varying component and                  q(t)   is a small-amplitude rapidly-varying component with a mean

value of zero. In other words,               p(t)    may be assumed constant over a few oscillations of                   q(t)   (see

Figure 2).

IPhO 2024                                                                                Page 7 of 16
Theory, English (Official)

Points: 30                                                                                Time: 5.0 Hours

  Figure 2 - A typical solution for the equation of motion of the charged particle: p(t) gives the overall motion, and q(t) represents small

                 oscillations around this trajectory. The ellipse on the right is a magni cation of a part of this trajectory.

                a) Using the approximations stated above,                       nd the equation of motion for

    A-3
                q(t)   in terms of     aΩ,   , and    p.
                                                                                1.8 pt

                b) Find the solution of this equation by considering appropriate initial

                conditions corresponding to the required properties of this function.

                a) Using the mean e            ect of the rapidly varying component and obtain an

    A-4
                e   ective equation of motion for              p(t)  .
                                                                                1.5 pt

                b) Investigate the stability of the equilibrium point and                          nd the condition

                for a stable equilibrium.

Assume that         λ0 = 8 × 10−9 C/m             and      R = 10 cm       . We would like to use this device to trap a

singly ionized atom 100 times heavier than a hydrogen atom.

                Calculate     k . Assume      a = 0. 04      and estimate the smallest frequency

    A-5         required to stabilize the motion of this ion. Use the data given at the end                                          0.4 pt

                of the question.

IPhO 2024                                                                                Page 8 of 16
Theory, English (Official)

Points: 30                                                                                Time: 5.0 Hours

    B. Doppler Cooling
It may be necessary to cool a trapped atom or ion. Assume that a trapped atom of mass                         m   , has two

energy levels with an energy di        erence of   E0 = ħωA    . Electrons in the lower level may absorb a

photon and jump to the higher level, but after a period             τthey will return to the lower level and

emit a photon with a frequency predominantly within               [ωA − Γ , ωA + Γ ]   .

    B-1        Use the Heisenberg’s uncertainty principle to          nd   Γ   .                                  0.5 pt

With a similar reasoning, when we shine a laser light on the trapped atom, if the angular

frequency of the laser,         ωL, falls in the interval     ω
                                                             [ A−     Γ , ωA + Γ ]  , the atom may absorb the

photon. Assume that the frequency           ωL   of the laser light is slightly lower than       ωA   . For a particular

device, the rate of photon absorption by an atom in the reference frame of the atom is given in

Figure 3. The absorbed photon is then re-emitted in a random direction. To make things simple, we

consider the problem in one dimension, i.e. we assume that the atoms can only move in the                                   x
                                                                                -

direction and the laser light shines on them both from the left and from the right. In the atom’s

reference frame, the light has a higher or lower frequency due to the motion of the atoms. Since

the velocity   v   of the atoms is very small, we only include terms of order              v/c       and ignore all the

higher-order terms. Moreover, we have            m ≫ ħωA/c    2
                                                                  so that the velocity of the atom nearly does

not change after absorbing the photon. Also, the change in frequency due to the Doppler e                             ect is

so small compared to           ωA − ωL   , that the function for      s   in the diagram of Figure 3 may be

approximated by the following linear function:

                                             s(ω) = sL + α(ω − ωL)
where   s   is the number of absorbed photons per unit of time,   sL           is the value of   s   for   ω = ωL   , and   α
is the slope of the line tangent to the curve at     ωL   . The frequency of the re-emitted photon is almost

equal to the frequency of the incident photon, but it is emitted with equal probability in the

positive or negative    x    -direction. In fact, up to the order considered here the two frequencies are

identical. Note that we are considering the whole process in the atom’s reference frame.

IPhO 2024                                                                                Page 9 of 16
Theory, English (Official)

Points: 30                                                                                Time: 5.0 Hours

  Figure 3 - The rate of photon absorption as a function of the frequency for a particular trap: the frequency corresponding to the energy

        di erence between the two atomic levels is indicated by ωA and the slighlty smaller frequency of the laser is indicated by ωL.

                 a) Assume that the trapped atom is moving with a velocity,                          v = vx    in the

                 lab frame. In the frame of reference of the atom, calculate the collision

                 rate of the photons, incident from each of the two directions, with the

    B-2          atoms (denoted by          s+   and   s− ) and the rate of absorption of momentum in                            1.7 pt

                 each direction (denoted by            π+   and    π−  ).

                 b) Determine the e          ective force on the atom as a function of                v kL = ωL/c
                                                                                ,                ,

                 ħ, and   α   , in the reference frame of the laboratory. Assume                    sL ≪ αωL       ,

We would like to           nd the lowest temperature that can be achieved using this technique. Assume

that the velocity of a particular atom has been reduced to zero exactly, and at this very moment it

absorbs a photon (incident from any of the two directions), and re-emits the photon randomly in

any of the two directions, with almost the same frequency. Assume that this process happens once

every   τ   units of time.

                 Considering the momentum of the atom after such a process for the two
    B-3                                                                                1.0 pt
                 possible outcomes, calculate the average power absorbed by the atom.

IPhO 2024                                                                                Page 10 of 16
Theory, English (Official)

Points: 30                                                                                Time: 5.0 Hours

              Consider the force calculated in Task B-2 and calculate the output power.

    B-4
              Then,    calculate   the   average   value   of   v2   at   equilibrium.   Using   your
                                                                                0.8 pt
              knowledge of the kinetic theory of gases estimate the temperature of the

              atoms.

              Estimate this temperature, for an atom 100 times heavier than a hydrogen
    B-5
              atom. Assume that      ωL = 2 × 1016 rad /s τ = 5 × 10−9 s
                                                                ,                  , and   α=4    .
                                                                                0.4 pt

                             mass of hydrogen atom:        mH = 1. 674 × 10−27 kg
                             charge of an electron:     e = 1. 602 × 10−19 C
                             permittivity of free space:   ε0 = 8. 854 × 10−12 F/m
                             Boltzmann constant:      kB = 1. 381 × 10−23 J/K
                             Planck constant:      ħ = 1. 055 × 10−34 J. s

IPhO 2024                                                                                Page 11 of 16
Theory, English (Official)

Solution:
S2-1
A: Paul Trap

A-1. Due to the symmetry, on the z-axis the only non-zero component of electric field is in
the z-direction. So:
                                                      dq       1            z
              EሬԦ ሺ0,0, zሻ = Ez ሺ0,0, zሻ zƸ = zƸ න                    ×            1
                                                     4πε0 ሺR 2 + z 2 ሻ ሺR 2
                                                                            + z 2 ሻ2

The element dq is equal to λRdφ where φ is the angle with the x-axis. Thus:

                                       λRdφ           z             λR        z
                    Eሺ0,0, zሻ = zƸ න                   3 = zƸ 2ε               3
                                       4πε0      2   2           0 ሺz 2 + R 2 ሻ2
                                               ሺz + R ሻ2

For z ≪ R this can be written as:

                                                   λR z     λz
                                  Ez ሺ0,0, zሻ =           =
                                                   2ε0 R 3 2ε0 R 2

Very close to the z-axis, we can write:

                                             ∂Ez              ∂Ez
            Ez ሺx, y, zሻ = Ez ሺ0,0, zሻ + x       ȁሺ0,0,zሻ + y     ȁሺ0,0,zሻ + Oሺx 2 , y 2 , z 2 ሻ
                                             ∂x               ∂y
                                                                                ∂E     ∂E
Since, there is no difference between x and −x or y and −y, it turns out that ∂xz = ∂yz = 0.
Thus, to the first order in x, y, and z we have:

                                                           λz
                                        Ez ሺx, y, zሻ =
                                                          2ε0 R 2

Consider a Gaussian surface in the shape of a symmetric cylinder around the z-axis whose
bases are parallel with the xy-plane. The cylinder’s radius is ρ and its height is 2z both of
which are small quantities. By Gauss’s law we have:

                    S1

                          S3

                   S2

                           Qin
                                                                                S2-2
                    0=         = ර EሬԦ . dSԦ = න EሬԦ . dSԦ + න EሬԦ . dSԦ + න EሬԦ . dSԦ
                           ε0                   S1            S2            S3

Integration over S1 and S2 gives:

                                                                    λz
                              න EሬԦ . dSԦ = න EሬԦ . dSԦ = πρ2 ×            .
                                S1            S2                   2ε0 R 2

Integration over S3 involves the ρ-component for which we can write the following expansion:

                                                           ∂Eρ
                      Eρ ሺz, ρ, φሻ = Eρ ሺ0, ρ, φሻ + z          ȁሺ0,ρ,φሻ + Oሺz 2 ሻ
                                                           ∂z
              ∂Eρ
We have 0 = ∂z ȁሺ0,ρ,φሻ due to symmetry between z and −z, hence, Eρ ሺz, ρ, φሻ = Eρ ሺ0, ρ, φሻ
                                                         dEρ
up to the first order. Axial symmetry also implies dφ = 0. Consequently:

                                න EሬԦ . dSԦ = Eρ ሺ0, ρ, 0ሻ × 2z × 2πρ
                                 S3

So, Gauss’s law implies:

                                                                  λz
                                     0 = Eρ × 4πzρ + 2πρ2
                                                                 2ε0 R 2

Therefore, Eρ will be:

                                                        λρ
                                            Eρ = −
                                                       4ε0 R 2

In the cylindrical coordinate we will have:

                                                     λρ           λz
                                EሬԦ ሺρ, φ, zሻ = −         2
                                                            ρො +         zƸ
                                                    4ε0 R        2ε0 R 2

In cartesian coordinates we will have:

                                                     λ
                                 EሬԦ ሺx, y, zሻ =           ሺ−x, −y, 2zሻ
                                                   4ε0 R 2

Since the ring is positively charged, the equilibrium in the x and y directions are stable, while
the equilibrium in the z-direction is unstable. The equations of motion in the x and y directions
read:

                                                            qλ
                                       mxሷ = qEx = −               x
                                                           4ε0 R 2

                                                            qλ
                                       myሷ = qEy = −               y
                                                           4ε0 R 2

Therefore, the frequencies of small oscillations are:
                                                                                S2-3
                                                             qλ
                                            ωx2 = ωy2 =
                                                           4ε0 R 2 m

 A-1 (1.5 pt)

                     −λx         −λy            λz
 (a) EሬԦ ሺx, y, zሻ = 4ε R2 xො + 4ε R2 yො + 2ε R2 zƸ
                      0           0             0
                          Qλ
 (b) ωx = ωy = √4ε R2 m
                          0

A-2.
The force in the z-direction is:

                                        Qλz        Q             Qu
                          Fz = qEz =         2
                                               =       2
                                                         λ0 z +         cos Ωt z
                                       2ε0 R     2ε0 R          2ε0 R 2

the equation of motion can thus be written as:

                                             Qλ0     Qu
                                  zሷ = ൬          +          cos Ωt൰ z
                                           2ε0 R m 2ε0 R 2 m
                                                2

Therefore:

                                  Qλ0                                        Qu
                          k=ඨ                          ,           a=
                                2ε0 R 2 m                                2ε0 R 2 mΩ2

 A-2 (0.4 pt)

           Qλ0                                                                   Qu
 k= √                                                                  ,a=
         2ε0 R 2 m                                                           2ε0 R 2 mΩ2

A.3.
              z = pሺtሻ + qሺtሻ               →         pሷ + qሷ = ሺk 2 + aΩ2 cos Ωtሻሺp + qሻ

    1. We are assuming that p is almost constant, pሷ ≃ 0.

    2. According to the assumptions k 2 ≪ aΩ2 and q ≪ p we can ignore k 2 in the first term
       on the right-hand side of the equation and q in the second term.

    hence, the equation of motion can be simplified as follows:

                                                 qሷ = paΩ2 cos Ωt.

                                                                                S2-4
   As we have assumed that p is a constant, the second derivative of q is just proportional to
   cos Ωt which gives:

                                           q = −pa cos Ωt + c1 t + c2 .

   Since q is supposed to remain small, c1 must vanish. Also c2 = 0 because the mean value
   of q is supposed to remain zero. Therefore:

                                                   q = −pa cos Ωt

 A-3 (1.8 pt)

 (a) qሷ ሺtሻ = paΩ2 cos Ωt

 (b) qሺtሻ = −pa cos Ωt

A-4. Using the final result for q the equation of motion for p reads:

                       pሷ + paΩ2 cos Ωt = ሺk 2 + aΩ2 cos Ωtሻሺp − ap cos Ωtሻ
Which gives:

                                 pሷ = k 2 p − ak 2 p cos Ωt − a2 Ω2 p cos2 Ωt

Averaging over one period, we’ll have:

                                                                              1
                                ۃcos Ωt = ۄ0              ,    ۃcos 2 Ωt= ۄ
                                                                              2
and:

                                                      2
                                                       a 2 Ω2
                                             pሷ = ቆk −        ቇ p.
                                                          2

In order for the motion to be stable, the expression inside the parentheses should be negative,
i.e.

                                     a 2 Ω2                               k
                                            > k2          →     Ω > ξ2
                                        2                                 a

 A-4 (1.5 pt)

                       a 2 Ω2
 (a) pሷ ሺtሻ = (k 2 −            )p
                         2

                k
 (b) Ω > ξ2 a

A.5. With the given data we have:

                                                                           S2-5
                                          Qλ0
                               k=ඨ                = 2 × 105 rad/s
                                        2ε0 R 2 m

                           a = 0.04      →     Ωmin = 7 × 106 rad/s

which is in the range of radio waves.

 A-5 (0.4 pt)

 k = 2 × 105 rad/s

 Ωmin = 7 × 106 rad/s

B: Doppler Cooling

B-1. From the uncertainty principle we know:
                                           ΔE × Δt ≃ ħ

Here Δt is the time τ and ΔE = ħΔω. So:

                                                            1
                              ħΔω × τ ≃ ħ →          Δω ≃     =Γ
                                                            τ

 B-1 (0.5 pt)

      1
 Γ=
      τ

B-2. We denote the forward and backward collision rates by s+ and s− respectively. Let us
                                                                           v2
proceed in the atom’s frame of reference. Ignoring the terms of the order c 2 , the Doppler
effect can be written in the following form:
                                                    v
                                         ω′ = ω (1 + )
                                                    c
Taking the atom’s velocity in the positive x-direction, we have:

                                                  v
                                                                          S2-6
                                      ω+ = ωL (1 + )
                                                  c
                                                  v
                                      ω− = ωL (1 − )
                                                  c
So:
                                           v                    v
                       s+ = sL + α (ωL (1 + ) − ωL ) = sL + αωL
                                           c                    c
                                           v                    v
                       s− = sL + α (ωL (1 − ) − ωL ) = sL − αωL
                                           c                    c
The momentum transfer per unit time from the oncoming photons to the atom is equal to:

                                     π+ = s+ × ሺ−ħk+ ሻ

For the backward photons we have:

                                     π− = s− × ሺ+ħk− ሻ
             ħω±
Where k± =         .
              c

The total momentum transferred to the atom per unit time is equal to:
                                               v            sL
                            π+ + π− = −2ħkL      ωL α ൬1 +     ൰
                                               c           αωL

Where with the approximation sL ≪ αωL , we will arrive at:
                                                     v
                                  π+ + π− = −2ħkL      ω α
                                                     c L
Note that as the atom is heavy, its velocity almost doesn’t change after the absorption of the
photon. Therefore, there will be almost no Doppler shifting in the re-emitted photon and hence,
on average there will be no momentum transfer to the atom during the re-emission process.

The above expression is, in fact, the force. Since v > 0, we have:

                                      F = −ሺ2αħkL2 ሻv

The same result holds for v < 0. This is in the atom’s reference frame. However, as we have
kept only up to the first order in v/c, the same result holds in the lab frame:

                                      F = −ሺ2αħkL2 ሻv

 B-2 (1.7 pt)
                                                                          S2-7
                  v
 s+ = sL + αωL
                  c
                  v
 s− = sL − αωL
                  c
 π+ = s+ × ሺ−ħk+ ሻ

 π− = s− × ሺ+ħk− ሻ

 F = −ሺ2αħkL2 ሻv

B-3. The atom’s momentum before the collision is zero. After the collision it will be
(assuming the photon’s momentum is in the x-direction):
                                           P1 = ħkL

After re-emitting the photon, we may have two equally likely outcomes for the final
momentum:

   1. The photon is emitted in the positive x-direction which causes the atom’s momentum
      to become zero

   2. The photon is emitted in the negative x-direction which causes the atom’s momentum
      to become: Pf = +2ħkL

Thus, the mean final energy is equal to:

                                  P2    1     1 4ħ2 kL2 ħ2 kL2
                         Eۃf  ۃ = ۄf  × = ۄ0 + ×       =
                                  2m    2     2  2m      m
This process occurs during the time τ. So, the input power (the power gained by the atom as a
result of this process) is equal to:

                                                   ħ2 kL2
                                           Pin =
                                                    mτ

 B-3 (1.0 pt)

         ħ2 kL2
 Pin =
          mτ

B.4. The output power (the power lost by the atom through collision with laser photons) can
be written as:

                                   Pout = F ∙ v = −2αħkL2 v 2
                                                                            S2-8
At equilibrium we should have:

                                       ħ2 kL2          തതത2               തതത   ħΓ
             Pout + Pin = 0    →              = 2αħkL2 v           →      v2 =
                                        mτ                                     2αm
And the temperature of this system is equal to:

                           1 തതത2 1                                 ħΓ
                             mv = kB T            →           T=
                           2      2                                2αkB

 B-4 (0.8 pt)

 Pout = −2αħkL2 v 2

 തതത    ħΓ
 v 2 = 2αm

       ħΓ
 T=
      2αkB

B-5. Considering the given data:

                                 1 055 × 10 34 J.s
                T=                                          = 2 × 10−4 K
                      2 × 4 × 1 381 × 10 23 J/K ×5 × 10 9 s

 B-5 (0.4 pt)

 T = 2 × 10−4 K

Trapping Ions and Cooling Atoms


[User] Connected in series in a closed circuit diagram is a resistor of resistance $R=\SI{100}{\ohm}$, a capacitor of capacity $C=\SI{200}{\nano\farad}$, inductive coil with a negligible active resistance and with an inductance $L=\SI{10}{\milli\henry}$ and appropriately connected ideal measuring devices. In the moment $t_0$ it was measured that the current strength through the capacitor was $I=\SI{300}{\milli\ampere}$ and that the voltage on the coil was $U=\SI{50}{\volt}$. It is known that at the moment of taking measurements the current in the coil is directed from a region with a higher potential towards a region with a lower potential. Did the capacitor or the inductor have more energy at the moment $t_0$ of taking measurements?

Solution:
Kuna tegemist on jadaühendusega, on ka voolutugevus läbi takisti ning pooli \SI{300}{\milli\ampere}. Poolis oli seega hetkel $t_0$ energia
\[
E_L=\frac{LI^2}{2}=\SI{0.45}{\milli\joule}.
\]
Summaarne pingelang takistil ning poolil peab olema võrdne kondensaatori pingega. Antud pooli pinge suuna korral saame kondensaatori pingeks $U_C=IR+U$. Kondensaatori energia on hetkel $t_0$
\[
E_C=\frac{CU_C^2}{2}=\SI{0.64}{\milli\joule},
\]
seega oli hetkel $t_0$ rohkem energiat kondensaatoril.

Elektriahela energia

Suletud elektriahelas on jadamisi ühendatud takisti takistusega $R=\SI{100}{\ohm}$, kondensaator mahtuvusega $C=\SI{200}{\nano\farad}$, tühise aktiivtakistusega induktiivpool induktiivsusega $L=\SI{10}{\milli\henry}$ ning sobivalt ühendatud ideaalsed mõõteseadmed. Hetkel $t_0$ mõõdeti voolutugevuseks läbi kondensaatori $I=\SI{300}{\milli\ampere}$ ning pingeks poolil $U=\SI{50}{\volt}$. Teada on, et mõõtmise hetkel on vool poolis suunatud kõrgema potentsiaaliga piirkonnast madalama potentsiaaliga piirkonda. Kas mõõtmise hetkel $t_0$ oli rohkem energiat poolil või kondensaatoril?

Kuna tegemist on jadaühendusega, siis on voolutugevus läbi kõikide vooluelementide sama. Lisaks peab pingelang üle kõikide vooluelementide olema \num{0}.


[User]
During a beautiful sunny weather the atmosphere is usually adiabatic. It means that the air masses are in a continuous up and down movement. When rising the air expands and cools adiabatically; because of the continuous mixing the temperature of the rising air mass is equal to the temperatures of the air masses surrounding it at the given height. It is possible to show that during this case the temperature decreases linearly with the height, $T=T_0-\frac{\gamma-1}\gamma\frac{\mu g h}R$, where $\gamma=\SI{1.4}{	}$ is air's adiabatic index, $\mu=\SI{29}{g/mol}$ — air's average molar mass, $g=\SI{9.81}{m/s^2}$ — gravitational acceleration, $R=\SI{8.31}{J/mol.K}$ —universal gas constant and $h$ — height from the ground, the air temperature on the ground $T_0=\SI{293}K$. The balloon is made of non-stretchable but freely flexible leather that can maximally contain $V_0$ amount of gas; the amount of helium that the balloon is filled with occupies a volume $V_0/2$ on the ground. The balloon is let free and it starts to slowly rise; assume that the temperature of the helium in the balloon is always equal to the temperature of the surrounding air. Evaluate at what height $h_1$ is the balloon's buoyancy force $1\%$ less than on the ground. The evaluation's relative error is expected to be less than one tenth, moreover there is no need to prove that the error is sufficiently small.\\
\emph{Note.}  During an adiabatic process the relation $pV^\gamma = \const$ applies, where $p$ is the pressure of the gas and $V$ the volume.

Solution:
Seni, kui heelium pole võtnud enda alla veel kogu õhupalli ruumala, püsib tõstejõud konstante. Tõepoolest, $F=\rho_a g V_p$, kus $\rho_a$ on õhu tihedus ja $V_p$ --- pallis oleva gaasi ruumala. Paneme tähele, et seni kuni palli nahk ei ole pinguldunud, on pallis oleva gaasi rõhk ja temperatuur võrdsed antud hetkel palli ümbritseva õhu rõhu ja temperatuuriga. Et $\rho_a=p\mu/RT$, kus $p$ ja $T$ tähistavad rõhku ja temperatuuri antud kõrgusel, ja samal ajal $V_p=\frac {mRT}{p\mu_p}$ (kus $\mu_p$ tähistab heeliumi molaarmassi ning $m$ selle kogumassi), siis üleslükkejõud
\[
F=\rho_a V_p g=mg\frac{\mu}{\mu_p}.
\]

Üheprotsendiline vähenemine on nii väike, et me võime lugeda otsitava kõrguse võrdseks kõrgusega, kus eelpooltoodud tulemuseni jõudmiseks tehtud eeldus heeliumi ruumala kohta enam ei kehti, st see võrdsustub $V_0$-ga. Gaasi olekuvõrrand ütleb, et $V\propto T/p$ ($\propto$ tähistab võrdelisust); arvestades, et pallis oleva heeliumi ja ümbristeva õhu temperatuurid ja rõhud on võrdsed, võime järeldada, et see on kõrgus, mille juures on õhu jaoks suhe $T/p$ kasvanud 2 korda võrreldes maapinnalähedase olukorraga.

Mõttelise õhuruumala $V$ jaoks on suhe $T/p$ võrdeline $V$-ga. Seega peab mõttelise õhuruumala kerkimisel maapinnalt antud kõrguseni selle ruumala kasvama kaks korda. Et tegemist on adiabaatilise atmosfääriga, siis õhuruumala $V$ kerkimisel järgivad selle karakteristikud adiabaadiseadust $pV^\gamma=\const$; kombineerides seda ideaalse gaasi seadusega $pV/T=\const$ saame $V^{\gamma-1}T=\const$. Et $V$ peab kasvama 2 korda, siis $T$ peab kahanema $2^{\gamma-1}=2^{0.4}$ korda. Seega
\[
\frac{\gamma-1}\gamma\frac{\mu g h}R =T_0(1-2^{-0.4}),
\]
millest
\[
h=\frac\gamma{\gamma-1}\frac R{\mu g}T_0(1-2^{-0.4})\approx\SI{7250}m.
\]

Kerkiv õhupall

Ilusa päikeselise ilma korral on harilikult tegemist nn adiabaatilise atmosfääriga. See tähendab, et õhumassid on pidevas üles-alla liikumises. Kerkides õhk paisub ja jahtub adiabaatiliselt; pideva segunemise tõttu on kerkiva õhumassi temperatuur võrdne seda antud kõrgusel ümbritsevate õhumasside temperatuuriga. On võimalik näidata, et sellisel juhul kahaneb temperatuur lineaarselt kõrgusega, $T=T_0-\frac{\gamma-1}\gamma\frac{\mu g h}R$, kus $\gamma=\SI{1.4}{	}$ on õhu adiabaadinäitaja, $\mu=\SI{29}{g/mol}$ --- õhu keskmine molaarmass, $g=\SI{9.81}{m/s^2}$ --- vabalangemise kiirendus, $R=\SI{8.31}{J/mol.K}$ --- gaasikonstant, ja $h$ --- kõrgus maapinnast; õhutemperatuur maapinnal $T_0=\SI{293}K$. Venimatust kuid vabalt painduvast nahast valmistatud õhupall mahutab maksimaalselt ruumala $V_0$ jagu gaasi; see täidetakse sellise koguse heeliumiga, mis võtab maapinnal enda alla ruumala $V_0/2$. Õhupall lastakse lahti ja see hakkab aeglaselt kerkima; lugeda, et õhupallis oleva heeliumi temperatuur on kogu aeg võrdne ümbritseva õhu temperatuuriga. Hinnake, millisel kõrgusel $h_1$ on õhupalli tõstejõud $1\%$ võrra väiksem kui maapinnal. Teilt oodatakse sellist hinnangut kõrgusele, mille suhteline viga pole suurem ühest kümnendikust, kusjuures vea piisavat väiksust pole vaja tõestada.

\emph{Märkus.} Adiabaatilise protsessi korral kehtib seos $pV^\gamma = \const$, kus $p$ tähistab gaasi rõhku ja $V$ --- ruumala.

On võimalik näidata, et seni, kuni heelium pole võtnud enda alla veel kogu õhupalli ruumala, püsib tõstejõud konstante.


[User]
8. CONNECTED CHARGEs (8 points)

— Jaan
Kalda.
In the region 0 < x < L, there is an
electric field ⃗ E = E0x̂, where x̂ denotes the
unit vector parallel to the x‐axis. Two small
balls, each of mass m and carrying charge q
(qE0 > 0) are connected with a weightless
non‐stretchable string of length l. Initially,
at the moment of time t = 0, the string is
taut, the velocity of the both balls is⃗ v = v0x̂,
one of the balls, the ball A, is at x = 0 while
the other ball, the ball B is at x < 0. The
electric field created by the balls can be neg‐
lected, and it can be assumed that v0 is very
small (much smaller than
√
ELq/m).
i)(2points)Considerthecasewhenthestring
is parallel to the x‐axis, and l = L Sketch the
dependence of the velocity of the both balls
as a function of time. Will the balls collide?
If yes then when?
ii) (2 points) Now, at t = 0, the string forms
an angle of 45° with the x‐axis, l = 1.2291L.
By this string length, at the moment t = T
when the ball A reaches x = L, the string is
parallel to the x‐axis. Find T.
iii) (2 points) Under the assumptions of the
previous task, what is the speed of the ball A
at the moment t = T?
iv) (2 points) Now, at t = 0, the string forms
a very small angle φ with the x‐axis. Simil‐
arly to the previous two tasks, at the moment
t = T′
when the ball A reaches x = L, the
stringisparalleltothex‐axis. Findthestring
length l assuming that l > L. The answer to
this point should contain only L and numerical
constants.

Solution:
8. CONNECTEDCHARGEs

—Solution
byJaanKalda,gradingschemesbyJānisCīmurs.
i) (2 points) As the ball A enters the elec‐
tric field, the force qE is exerted on it, hence
the acceleration of the system of two balls is
a1 = qE/2m; the acceleration remains con‐
stant until the ball A exits the field. This
will happen when L = a1t2
1/2, hence t1 = √
2L/a = 2
√
Lm/qE At that moment, the
acceleration of the ball A becomes zero, but
at the very same moment, ball B enters the
field. The force qE exerted on it will give it
twice as big an acceleration as before, a2 =
qE/m, and the string becomes loose. So, the
speed of the ball B is now bigger than that
of the ball A, and so they will eventually col‐
lide. The time t2 spent by the ball B in the
field can be expressed as t2 = (vf − vi)/a2,
where vi =
√
qEL/m is its initial speed, and
vf =
√
3qEL/m — the final speed. So, t2 = √
Lm/qE(
√
3−1). At the moment when the
ball B exits the field, the distance between
the balls is s = L − (vf − vi)t2/2 = vit2 =
L(
√
3 − 1), and the difference between the
velocities of the balls is ∆v = vf − vi = √
qEL/m(
√
3−1),hencethetimeremaining
until the collision t3 = s/∆v =
√
Lm/qE.
Thus, the moment of collision t = t1 + t2 +
t3 =
√
Lm/qE(
√
3 + 2).
Grading:
• Initially both charges move with the same
acceleration (0.2pts)
• When A escapes the field region A moves
with constant speed, B moves with higher
acceleration (0.2pts)
• When B escapes, both move with constant
speed (0.1pts)
• Correct a1 (0.2pts)
• Correct t1 (0.2pts)
• Correct a2 (0.1pts)
• Correct vi (0.2pts)
• Correct vf (0.2pts)
• Correct t2 (0.2pts)
• Correct s (0.2pts)
• Correct t3 and t (0.2pts)
ii) (2 points) First solution: The centre of
mass moves with constant acceleration a1 =
qE/2m, and has travelled distance sCM =
L−l/2+l/(2
√
2) = 0.8200L = a1T2
/2,hence
T = 1.811
√
mL
qE .
Grading:
• Use sCM = a1T2
/2 for centre of mass
(0.5pts)
• Correct a1 (0.5pts)
• Correct sCM (0.6pts)
• Correct T (0.4pts)
• Ifyouassumesmallangleoscillationtocal‐
culatetimenecessaryforthestringtoalign
with the x‐axis. (up to 1.4pts)
iii) (2 points) First solution: Due to the con‐
servation of y‐directional momentum, the y‐
components of the balls are always equal. At
t = T, the string is parallel to the x‐axis;
since it is not stretchable, x‐components of
the balls are equal, too, at that moment.
Hence,thespeedsareequal,andenergycon‐
servation law states EqL = mv2
so that v = √
qEL
m .
Grading:
• Using energy conservation law (1.0pts)
• Kinetic energy of both masses (0.6pts)
• Work done by electric field (0.2pts)
• Correct v (0.2pts)
Alternative solution: The time period
0 < t < T is most conveniently analysed
in the centre‐of‐mass frame. In that frame,
there is a force of inertia equal to Fi =
−ma1 = −qE/2. Hence, both balls perform
non‐linear oscillations around the centre of
mass (it is nonlinear because the amplitude
is not small). The speed of the ball at t = T
can be found from the energy conservation
law. At t = 0, the speed is zero so that the
full energy is due to the potential energy in
the homogeneous force field qE/2: ∆U =
qE
2 (l
2 − l
2
√
2
) = qEL
8 (2 −
√
2) =
mv2
y
2 , hence
v2
y = qEl
4m (2 −
√
2) = 0.1800qEL
m . This gives
us the velocity component in the y‐direction;
once we return to the laboratory frame, we
need to add the x‐component of the frame’s
velocity vx = a1T = qE
2m1.811
√
mL
qE =
0.9055
√
qEL
m so that v =
√
v2
x + v2
y =
√
qEL
m .
Grading:
• Horizontal speed vx (0.4pts)
• Useenergybalanceorsomethingsimilarto
find rotational speed vy (0.4pts)
• Correct distance travelled (0.6pts)
• Rotational speed vy (0.4pts)
• Pythagorean theorem (0.2pts)
iv) (2 points) For the string to be parallel to
the x‐axis, the system must perform n + 1
2
half‐periods, where n is an integer. Since
now the amplitude is small, the oscillations
are harmonic, and the circular frequency
ω =
√
qE
ml (we can use the formula for a
pendulum of length l/2 and equivalent grav‐
ity field g = qE/2m). So, the travel time
must be τ = π
√
ml
qE (n + 1
2). On the other
hand, we can use the result from task ii: τ = √
ml
qE
√
4λ − 2 + 2cosφ, where λ = L/l < 1
and cosφ ≈ 1. So, π(n + 1
2) =
√
4λ. Since
λ < 1, this equality can only be satisfied
with n = 0, hence λ = π2
16 = 0.6169 so that
l = L/λ = 1.6211L.
Grading:
• Correct circular frequency (0.6pts)
• If circular frequency deviates from the
right by factor
√
2 (0.2pts)
• Correct rotation time with n (0.6pts)
• Correct rotation time without n (0.2pts)
• Correct travel time from ii (0.4pts)
• Derivation of l (0.4pts)

CONNECTED CHARGEs


[User]
\begin{wrapfigure}{r}{18mm}%
\vspace{-15pt}
\includegraphics[width=\linewidth]{2012-v2g-08-rong}%
\end{wrapfigure}
Let us measure the electrical resistance between two rails on a railway, like in the drawing. A wagon of speed $v$ drives on the railway. The wagon has two pairs of wheels, one pair is at a distance $l$ from the other. Draw a plot of resistance versus time from the moment when the wagon’s first pair of wheels is at the front of the measuring point at a distance $l/2$ to the moment when the wagon’s rear wheels are behind the measuring point at the distance $l/2$.  Let the resistance of both pairs of the wheels be $r$ and a rail’s resistance per one unit of length $\rho$.

Solution:
\begin{wrapfigure}{r}{50mm}
\includegraphics[width=\linewidth]{2012-v2g-08-rong_lahendus}
\end{wrapfigure}
Ülesandes peab käsitlema kahte juhtu: a) kui mõõtepunkt ei asu vaguni rataste vahel ja b) kui mõõtepunkt asub vaguni rataste vahel. On ka näha, et ülesanne on sümmeetriline mõõtepunkti suhtes, st takistuse käitumine enne seda, kui vaguni keskpunkt möödub mõõtepunktist ja pärast seda, on üksteise peegelpildid.

Vaatleme alguses juhtu a). On näha, et sellel juhul on vagunil fikseeritud takistus, millele liitub vaguni ja mõõtepunkti vahele jäävate rööbaste takistus.
\[
R=2(l/2-vt)\rho+\frac{1}{\frac{1}{r}+\frac{1}{r+2l\rho}}.
\]

Seejärel vaatleme juhtu b). Sellel juhul moodustub takistus kahest rööpühenduses olevast osast, mille takistus muutub vaguni liikumise käigus. $R_1=(l-vt')\rho$ ja $R_2=vt'\rho$, kus $t'=t-s/v$ on aeg, mis on möödunud hetkest, kui esimene rattapaar möödus mõõtepunktist. Seega saab takistuse käitumise leida valemist:
\[
R=\frac{1}{\frac{1}{r+2R_1}+\frac{1}{r+2R_2}}=\frac{1}{\frac{1}{r+2(l-vt')\rho}+\frac{1}{r+2vt'\rho}}.
\]

Tulemuseks on graafik:

\begin{center}
\includegraphics[width=300pt]{2012-v2g-08-rong_graafik}
\end{center}

Raudtee

\begin{wrapfigure}{r}{18mm}%
\vspace{-15pt}
\includegraphics[width=\linewidth]{2012-v2g-08-rong}%
\end{wrapfigure}
Mõõdame raudteel elektritakistust kahe kõrvutise rööpa vahel nii, nagu joonisel.
Rööbastel sõidab vagun kiirusega $v$. Olgu vagunil kaks rattapaari,
mille vahekaugus on $l$. Joonistage graafik takistuse muutumisest ajas alates
hetkest, kui vaguni esimene rattapaar on mõõtepunkti ees sellest kaugusel $l/2$, kuni
ajani, kui tagumine rattapaar on mõõtepunkti taga sellest kaugusel $l/2$. Mõlema rattapaari
takistuseks olgu $r$ ja rööpa takistus pikkusühiku kohta $\rho$.

Ülesandes peab käsitlema kahte juhtu: a) kui mõõtepunkt ei asu vaguni rataste vahel ja b) kui mõõtepunkt asub vaguni rataste vahel. Mõlemal juhul käitub tekkiv elektriskeem isemoodi.


[User]
IPhO 1994, problem 1. A clean and neat relativistic dynamics problem. Print
    out the custom answer sheets before starting.

       Remark
       The setup of problem 26 is a nice model for mesons, particles composed of two quarks. And
       it’s not just something made up for an Olympiad; it is a simple version of the MIT “bag
       model”, which was one of the most important advances in the field in the 1970s. In fact, if
       you look at the original paper, which has thousands of citations, you’ll find the answer to
       the IPhO question in figure 3!

       Idea 8
       In string theory, strings carry a constant tension T , in the sense that the force F = dp/dt
       exerted on one piece of string by its neighbors is T in the momentary rest frame of that piece.
       The strings may stretch or shrink freely, and have zero mass when they have zero length.

[3] Problem 27 (Morin 12.16). A simple exercise involving relativistic string.

     (a) Two masses m are connected by a string of length ℓ and constant tension T . The masses are
         released simultaneously, and they collide and stick together. What is the mass, M , of the
         resulting blob?

     (b) Consider this scenario from the point of view of a frame moving to the left at speed v.

         The energy of the resulting blob must be γM c2 . Show that you obtain the same result by
         computing the work done on the two masses.

[3] Problem 28 (Morin 12.37). Two equal masses are connected by a relativistic string with tension
    T . The masses are constrained to move with speed v along parallel lines, as shown.

    The constraints are then removed, and the masses are drawn together. They collide and make one
    blob which continues to move to the right. Is the following reasoning correct?

         The forces on the masses point in the y direction. Therefore, there is no change in
         the momentum of the masses in the x direction. But the mass of the resulting blob is

     greater than the sum of the initial masses (because they collide with some relative speed).
     Therefore, the speed of the resulting blob must be less than v (to keep px constant), so
     the whole apparatus slows down in the x direction.
If your answer is “no,” exactly what’s wrong about the reasoning above?

   Example 10: Right Angle Lever Paradox

   In 1909, Lewis and Tolman found one of the first relativistic paradoxes. Consider a rigid
   lever in static equilibrium, with both arms of length L, experiencing the forces shown at left.

   In a frame where the lever moves to the right with speed v, one of the lever arms will be
   contracted to L/γ, as shown at right. In addition, by the results of problem 18, the vertical
   external forces will be redshifted to F/γ. This implies a net torque of

                                                 FL
                                      τ = FL −       = F Lv 2 .
                                                 γ γ

   The paradox is, given that τ = dL/dt, why doesn’t the lever rotate?

   Solution
   The resolution is that, in the frame shown at right, the angular momentum of the lever is
   constantly increasing. The horizontal forces are continually doing equal and opposite work
   on the lever, resulting in a upward flow of energy of rate F v in the vertical arm. As explained
   below problem 1, in relativity, energy flow is equal to momentum density, so the total upward
   momentum in the vertical arm is F Lv. Therefore,
                                      dL   dx
                                         =    (F Lv) = F Lv 2
                                      dt   dt
   exactly as expected.

   Remark: Relativistic Torque

   The resolution of the right angle lever paradox is very controversial, with dozens of papers
   written on the subject, so we should discuss what it even means to “resolve” a paradox. As
   long as we believe relativity is self-consistent, we already know what’s going to happen: the
   lever won’t rotate. Everything the lever does is determined by F = dp/dt alone, so if it looks
   like angular momentum considerations give a different answer, that just means we haven’t
   formulated the latter correctly. The reason there are so many different resolutions out there

  is just that people choose different ways to define torque and angular momentum.

  The solution above is the standard one, and its implicit definition of angular momentum can
  be motivated by Noether’s theorem. That’s a reasonable choice, since it’s a specific output
  of a useful and general theorem, and we thereby know for sure that it’s conserved for isolated
  systems. Unfortunately, explaining the definition takes some advanced math.

  We define the angular momentum density tensor

                                M µνρ (x) = xµ T νρ (x) − xν T µρ (x)

  where the right-hand side contains the stress-energy tensor, from the solution to problem 17.
  The total angular momentum is an antisymmetric rank 2 tensor,
                                              Z
                                   J (t) = dx M µν0 (x).
                                     µν

  Noether’s theorem states that it is this quantity that is conserved for an isolated system, due
  to symmetry under rotations and boosts. More specifically, the three spatial components
  J xy , J yz , and J zx just make up ordinary angular momentum, e.g. for a single point particle
  they would assemble into the vector r × p = r × (γmv). And the other components J 0x , J 0y
  and J 0z have to do with the center of mass motion.

  If there is an external four-force per unit proper volume f µ (x), which in terms of the stress-
  energy tensor implies ∂µ T µν = f ν , the rate of change of angular momentum is

                           dJ µν
                                                  Z
                                 = τ , τ = dx xµ f ν (x) − xν f α (x)
                                     µν     µν
                            dt
  which looks quite similar to the Newtonian expression. The component of this equation
  relevant to this paradox is dJ xy /dt = τ xy , where
                          Z                            X
                    J xy = dx xT y0 − yT x0 , τ xy =     x(k) Fy(k) − y (k) Fx(k)
                                                          k

  where the index k sums over the four forces, and the T i0 stand for the density of momentum
  in the i direction. From this point on, the solution proceeds as above.

  There is something a bit strange here, though. In the lever’s rest frame, the angular momen-
  tum is zero, so if J µν were a tensor, it would have to be zero in all frames, but instead it
  rises to arbitrarily high values in the other frame. The reason is that when there are external
  torques, J µν isn’t a tensor at all, just like how the four-momentum wasn’t a four-vector in the
  solution to problem 17. That’s one of the reasons there’s a controversy: there just doesn’t
  exist any definition that has all the nice properties one might want.

Solution:
(a) The total work done on the masses is ℓT , so by energy conservation this must
         manifest as rest energy in the final blob, M = 2m + ℓT /c2 .

     (b) Let c = 1. The initial energy is 2γm, so we need to show that the work done is γℓT .
         At first glance, this is puzzling, because the
                                                     R initial distance between the masses in this frame
         is ℓ/γ. Therefore, naively applying W = F dx, we have
                                      Z          Z             Z
                                 W = T dx1 − T dx2 = T dx1 − dx2 = T ℓ/γ

         which is wrong. The resolution is that we have assumed the masses are released simultaneously
         in the original frame, which means they aren’t released simultaneously in this frame.
         The mass on the left will start accelerating first, and after some time, the mass on the right
         will accelerate. In the original frame, these two events have ∆x = ℓ and ∆t = 0. Thus,
         applying the Lorentz transformation,

                                                ∆x′ = γ∆x = γℓ.

         Suppose that after it starts experiencing the tension, the left mass moves a distance x0 before
         it collides with the right mass. Then the above calculation shows that after the right mass
         starts experiencing the tension, it moves a distance x0 − ∆x′ until collision. Thus,

                                        W = T (x0 − (x0 − ∆x′ )) = γℓT

         as desired.

[3] Problem 28 (Morin 12.37). Two equal masses are connected by a relativistic string with tension
    T . The masses are constrained to move with speed v along parallel lines, as shown.

    The constraints are then removed, and the masses are drawn together. They collide and make one
    blob which continues to move to the right. Is the following reasoning correct?

         The forces on the masses point in the y direction. Therefore, there is no change in
         the momentum of the masses in the x direction. But the mass of the resulting blob is
         greater than the sum of the initial masses (because they collide with some relative speed).
         Therefore, the speed of the resulting blob must be less than v (to keep px constant), so
         the whole apparatus slows down in the x direction.

    If your answer is “no,” exactly what’s wrong about the reasoning above?

    Solution. The reasoning is incorrect. To see this, we can consider working in the initial rest frame
    of the system. In this frame, the masses just approach each other and collide, ending up at rest. So
    in the original frame, the whole apparatus must keep going at the same speed as before.

   There are two ways to see what’s going on. First, consider just the top mass, and work throughout
in the original frame. Then the incorrect statement is the very first sentence: the three-force on the
top mass is not always in the y direction. Recall the relativistic transformation of the three-force
derived in problem 18. This tells us that if we align the x′ axis with the instantaneous motion of
the particle, then
                                      F = (Fx′ ′ , Fy′ ′ /γ, Fz′ ′ /γ).
Once the top mass gets moving, it has velocity components along both x and y, so the x′ axis must
be tilted accordingly. Upon applying this formula (i.e. redshifting the y ′ component of the force),
we end up with a nonzero x component of the force, so the logic above fails.
    Alternatively, we can consider the entire system, of the masses and string. In this case, the
statement that fails is the second parenthetical, “to keep px constant”. The issue here is that the
string itself has a linear mass density of T /c2 , due to the energy stored in it in the stretching process,
and hence also carries momentum. This needs to be accounted for in the momentum conservation
equation, and gives the “missing” momentum we need. Note that this is totally compatible with
the previous paragraph; the force discussed there is precisely how this string momentum ends up
transferred to the masses.

   Example 10: Right Angle Lever Paradox

   In 1909, Lewis and Tolman found one of the first relativistic paradoxes. Consider a rigid
   lever in static equilibrium, with both arms of length L, experiencing the forces shown at left.

   In a frame where the lever moves to the right with speed v, one of the lever arms will be
   contracted to L/γ, as shown at right. In addition, by the results of problem 18, the vertical
   external forces will be redshifted to F/γ. This implies a net torque of

                                                    FL
                                        τ = FL −        = F Lv 2 .
                                                    γ γ

   The paradox is, given that τ = dL/dt, why doesn’t the lever rotate?

   Solution
   The resolution is that, in the frame shown at right, the angular momentum of the lever is
   constantly increasing. The horizontal forces are continually doing equal and opposite work
   on the lever, resulting in a upward flow of energy of rate F v in the vertical arm. As explained
   below problem 1, in relativity, energy flow is equal to momentum density, so the total upward

  momentum in the vertical arm is F Lv. Therefore,
                                     dL   dx
                                        =    (F Lv) = F Lv 2
                                     dt   dt
  exactly as expected.

  Remark: Relativistic Torque

  The resolution of the right angle lever paradox is very controversial, with dozens of papers
  written on the subject, so we should discuss what it even means to “resolve” a paradox. As
  long as we believe relativity is self-consistent, we already know what’s going to happen: the
  lever won’t rotate. Everything the lever does is determined by F = dp/dt alone, so if it looks
  like angular momentum considerations give a different answer, that just means we haven’t
  formulated the latter correctly. The reason there are so many different resolutions out there
  is just that people choose different ways to define torque and angular momentum.

  The solution above is the standard one, and its implicit definition of angular momentum can
  be motivated by Noether’s theorem. That’s a reasonable choice, since it’s a specific output
  of a useful and general theorem, and we thereby know for sure that it’s conserved for isolated
  systems. Unfortunately, explaining the definition takes some advanced math.

  We define the angular momentum density tensor

                                M µνρ (x) = xµ T νρ (x) − xν T µρ (x)

  where the right-hand side contains the stress-energy tensor, from the solution to problem 17.
  The total angular momentum is an antisymmetric rank 2 tensor,
                                              Z
                                   J (t) = dx M µν0 (x).
                                     µν

  Noether’s theorem states that it is this quantity that is conserved for an isolated system, due
  to symmetry under rotations and boosts. More specifically, the three spatial components
  J xy , J yz , and J zx just make up ordinary angular momentum, e.g. for a single point particle
  they would assemble into the vector r × p = r × (γmv). And the other components J 0x , J 0y
  and J 0z have to do with the center of mass motion.

  If there is an external four-force per unit proper volume f µ (x), which in terms of the stress-
  energy tensor implies ∂µ T µν = f ν , the rate of change of angular momentum is
                           dJ µν
                                                  Z
                                 = τ µν , τ µν = dx xµ f ν (x) − xν f α (x)
                            dt
  which looks quite similar to the Newtonian expression. The component of this equation
  relevant to this paradox is dJ xy /dt = τ xy , where
                          Z                            X
                    J = dx xT y0 − yT x0 , τ xy =
                      xy
                                                         x(k) Fy(k) − y (k) Fx(k)
                                                          k

  where the index k sums over the four forces, and the T i0 stand for the density of momentum
  in the i direction. From this point on, the solution proceeds as above.

  There is something a bit strange here, though. In the lever’s rest frame, the angular momen-
  tum is zero, so if J µν were a tensor, it would have to be zero in all frames, but instead it
  rises to arbitrarily high values in the other frame. The reason is that when there are external
  torques, J µν isn’t a tensor at all, just like how the four-momentum wasn’t a four-vector in the
  solution to problem 17. That’s one of the reasons there’s a controversy: there just doesn’t
  exist any definition that has all the nice properties one might want.

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


[User]
A block of mass M and velocity v0 to the right approaches a stationary puck of
    mass m ≪ M . There is a wall a distance L to the right of the puck.

      (a) Assuming all collisions are elastic, find the minimum distance between the block and the wall
          by explicitly analyzing each collision. (Note that it does not suffice to just use the adiabatic
          theorem, because it applies to slow change, while the collisions are sharp. Nonetheless, you
          should find a quantity that is approximately conserved after many collisions have occurred.)

     (b) Approximately how many collisions occur before the block reaches this minimum distance?

      (c) The adiabatic index γ is defined so that P V γ is conserved during an adiabatic process. In one
          dimension, the volume V is simply the length, and P is the average force. Using the adiabatic
          theorem, infer the value of γ for a one-dimensional monatomic gas.

Solution:
(a) Let the speeds of the block and puck be v and w. Every collision, w increases by 2v.
        If the block is a distance x from the wall, then a collision happens in time 2x/w. Therefore,
        we have
                                            ∆w        2v          w
                                               =            =− .
                                            ∆x     −2xv/w         x
         Because m ≪ M , many collisions happen. After many collisions have happened, w will be
         very large, so in the next collision, ∆x will be small compared to x, and ∆w will be small

     compared to w. In this case, we can approximate the finite differences with a derivative,
     giving
                                             dw      w
                                                 ≈− .
                                             dx       x
     Separating and integrating shows that wx is conserved. We could also have arrived at this by
     the adiabatic theorem,             I
                                     I=      p dx = mw(2x) ∝ wx.

     However, in the earlier collisions (∆w)/w and (∆x)/x aren’t small, so this reasoning is invalid.
     For instance, wx is zero before the first collision and nonzero right after it. Thus, we must
     treat the first few collisions manually. Right before the second collision, we have

                                          w ≈ 2v0 ,     x ≈ L/3

     by one-dimensional kinematics. Right before the third collision we have

                                          w ≈ 4v0 ,     x ≈ L/5

    where for these early few collisions we are treating v as constant since m ≪ M . It is not hard
    to show that right before collision n + 1, we have w ≈ 2nv0 and x ≈ L/(2n + 1), which means
    that after many (but not too many collisions) we have wx ≈ Lv0 . Then, for future collisions,
    wx stays at this value.
    The block turns around when the puck has all its energy, so
                                             1        1
                                               M v02 = mw2 .
                                             2        2
                                                                     p
     Plugging in wx = Lv0 and solving for x gives the solution, x = L m/M .

 (b) At each collision we have ∆w = 2v, and energy conservation gives
                                                    m 2
                                             v2 +     w = v02 .
                                                    M
    Therefore, the number of collisions is approximately
           Z v0 √M/m          Z √                              r Z              r
                      dw    1 v0 M/m               dw         1 M 1    dx      π M
       n≈                =                 p                =       √        =     .
             0        2v    2 0              v02 − (m/M )w2   2 m 0   1 − x2   4 m

     Note that we didn’t need to separate out the first few collisions here, even though the approx-
     imation as an integral technically doesn’t work, because they’re just that not important for
     calculating the total number of collisions, which is large. The appearance of π in this result
     has a nice geometric interpretation, as explained here.

 (c) The analogue of pressure in one dimension is just force. The average force exerted by the
     puck, which we now think of as a gas molecule, is
                                             ∆p   2mw    mw2
                                       F =      =      =     .
                                             ∆t   2x/w    x
     Meanwhile, the analogue of volume is one dimension is simply x. Then the conservation of
     wx says that F x3 is conserved, which means γ = 3. This is exactly what we would expect for
     a one-dimensional gas, where Cv = kT /2 and Cp = 3kT /2.


[User]
01^ Solve F = ma 2018 problems A12, A25, B19, and B25, and F = ma 2019
    problems A16, B18, and B25. Make sure to strictly adhere to the total time. Since these are
    F = ma problems, you don’t have to produce a writeup. If you find these questions difficult to
    finish in the allotted time, go back and review the earlier material!

Solution:
01^  Solve F = ma 2018 problems A12, A25, B19, and B25, and F = ma 2019
    problems A16, B18, and B25. Make sure to strictly adhere to the total time. Since these are
    F = ma problems, you don’t have to produce a writeup. If you find these questions difficult to
    finish in the allotted time, go back and review the earlier material!


[User]
\begin{wrapfigure}[5]{r}{0.45\linewidth}
	\vspace{-5pt}
\begin{tikzpicture}[scale=0.4]
% Nõlv
\draw [thick] (0,0) -- (160:5);
\draw [thick] (0,0) -- (-10:7);

% Nurgad
\draw [dashed] (-4.6,0) -- (7,0);
\draw (160:3.5) arc (160:180:3.5) node at (170:3) {\(\alpha\)};
\draw (0:6.5) arc (0:-10:6.5) node at (-5:6) {\small\(\beta\)};

% Suurused
\draw [<->] (-5,0) -- (-5,1.7) node[pos=0.5, left] {\(h\)};
\draw [<->, yshift=-0.5cm] (0,0) -- (-10:7) node[pos=0.5, below] {\(s\)};
\end{tikzpicture}
%\caption{Künka läbilõige} \label{liug}
\end{wrapfigure}
Juss found a hill for Fastelavn (a carnival tradition and holiday in Northern Europe) sliding. The hill’s cross-cut consists of two line segments, as shown in the figure. After the hill there is a horizontal ground. The height of the hill’s first part is \(h=\SI{2}{\meter}\) and its inclination is \(\alpha=\ang{20}\). The length of the second part is \(s=\SI{20}{\meter}\) and inclination \(\beta=\ang{5}\). The mass of Juss and the sledge is \(m=\SI{47}{\kilogram}\) and coefficient of friction between the snow and the sledge is \(\mu=\num{0.08}\), gravitational acceleration is \(g=\SI{9.8}{\meter\per\second\squared}\). Find how long is Juss’s slide.

Solution:
Hõõrdejõu tõttu kulutatud energia sõltub läbitud teepikkusest ja selle kaldest, potentsiaalse energia vahe sõltub kõrguse muudust. Esimesel lõigul vabaneb potentsiaalne energia \(E_1 = mgh\) ja hõõrdejõu mõjul liikumiseks kaotatakse energia
\[
A_1 = \mu \cdot mg\cos\alpha \cdot \frac{h}{\sin\alpha} = \mu mgh \cot\alpha.
\]
Teisel lõigul vabaneb potentsiaalne energia \(E_2 = mgs \sin\beta\) ja hõõrdejõu tõttu kaotatakse energia
\[
A_2 = \mu \cdot mg \cos \beta \cdot s = \mu mgs \cos \beta.
\]
Seega nõlva lõppedes on kelgutajal alles kineetiline energia
\begin{align*}
\Delta E &= E_1 - A_1 + E_2 - A_2 = \\
&= mgh(1-\mu\cot\alpha) + mgs(\sin\beta - \mu\cos\beta) = \SI{787.4}{\joule} \, .
\end{align*}
Tasase maa peal enam potentsiaalset energiat ei vabane, aga kineetiline energia väheneb hõõrdumise tõttu. Kelk libiseb tasase maa peal
\begin{align*}
l = \frac{\Delta E}{F_h} = \frac{\Delta E}{\mu mg} = \SI{21.4}{\meter} \, .
\end{align*}
Seega kogu vastlaliu pikkus on
\[
\frac{h}{\sin\alpha} + s + l = \SI{47.2}{\meter}.
\]

Vastlaliug

\begin{wrapfigure}[5]{r}{0.45\linewidth}
	\vspace{-5pt}
\begin{tikzpicture}[scale=0.4]
% Nõlv
\draw [thick] (0,0) -- (160:5);
\draw [thick] (0,0) -- (-10:7);

% Nurgad
\draw [dashed] (-4.6,0) -- (7,0);
\draw (160:3.5) arc (160:180:3.5) node at (170:3) {\(\alpha\)};
\draw (0:6.5) arc (0:-10:6.5) node at (-5:6) {\small\(\beta\)};

% Suurused
\draw [<->] (-5,0) -- (-5,1.7) node[pos=0.5, left] {\(h\)};
\draw [<->, yshift=-0.5cm] (0,0) -- (-10:7) node[pos=0.5, below] {\(s\)};
\end{tikzpicture}
%\caption{Künka läbilõige} \label{liug}
\end{wrapfigure}

Juss leidis vastlaliu laskmiseks künka, mille läbilõige koosneb kahest sirglõigust, nagu näha joonisel, edasi on horisontaalne maa. Esimese nõlvaosa kõrgus on \(h=\SI{2}{\meter}\) ja selle kalle \(\alpha=\ang{20}\), teise osa pikkus on \(s=\SI{20}{\meter}\) ja kalle \(\beta=\ang{5}\). Jussi mass koos kelguga on \(m=\SI{47}{\kilogram}\) ja hõõrdetegur lume ja kelgu vahel on \(\mu=\num{0.08}\), raskuskiirendus \(g=\SI{9.8}{\meter\per\second\squared}\). Leidke, kui pikk on Jussi vastlaliug.


[User]
If you look down a large body of water before sunset, the sun’s reflection will appear
    very long. Why?

Solution:
Because of the large length scales involved, and the absence of dispersion, we can tell
    that this is a geometrical optics effect. The explanation is as follows. If the water’s surface was
    perfectly flat, then the sun’s reflection would be at a point. Because there are waves, different parts
    of the surface are tilted with respect to the vertical by a small angle, so that other light rays can
    also make it to your eye.
       Because the sun is low in the sky, this effect is much more dramatic in the forward/backward
    direction. For instance, suppose the ideal reflection occurs a distance d away from you, and your
    height is h ≪ d, so that the light ray enters your eye at a low angle θ ≈ h/d to the horizontal. Then
    sunlight can also hit the water a distance 2d away from you, bounced off a tilted water surface,
    and enter your eye at an angle h/(2d) = θ/2. Just a small change θ/2 in the direction of reflection
    can double the distance of the reflection! Such an enhancement does not occur in the left/right
    direction, so the reflection of the sun looks much longer than it is wide.


[User] 01W   USAPhO 2013, problem A1. An elegant conduction question.

Solution: 01W   USAPhO 2013, problem A1. An elegant conduction question.


[User] Let us study an hourglass model. The hourglass consists of a cylindrical tube of length $L$, the tube is separated with a plate in the middle and the plate is evenly covered with holes. The sand can flow through these holes. In a good approximation the mass flow rate $w$ of the sand flowing through the holes does not depend on the amount of sand in the upper part of the tube. The hourglass is put on a scale while the sand is flowing and again when the sand has flown down. What is the difference between the scale readings? The sand’s density is $\rho$ and the area of the tube’s cross section is $S$. Assume that the mass of the falling sand is insignificant compared to the sand’s whole mass.

Solution:
Olgu kogu liivahulga kõrgus $H$ ja liiva kõrgus alumises anumas $h$. Alumisse anumasse voolab liiv masskiirusega $w$.

Paneme kirja liivakella masskeskme koordinaadi:
\[x=\frac{\frac{h}{2}hS\rho+(\frac{L}{2}+\frac{H-h}{2})(H-h)S\rho}{HS\rho}=\frac{2h^2+LH-Lh+H^2-2Hh}{2H}.\]
Kui liivakella masskese omab kiirendust, peab masskeskme koordinaadi teine tuletis aja järgi olema nullist erinev. Järeldusena Newtoni II seadusest võib õelda, et liivakellale peab mõjuma masskeskme kiirendusele vastav lisajõud -- see ongi kaalu näidu muutus, mida otsime.
Võtame nüüd masskeskme koordinaadist kaks korda järjest ajalist tuletist, pidades meeles, et liivataseme kõrguse $h$ tuletis aja järgi (muutumiskiirus) on seotud otseselt liiva voolu masskiirusega $v$ järgnevalt:
\[w=S\rho\dot{h}.\]
Täpiga suuruse peal tähistame vastavat ajalist tuletist. Peame meeles, et $h$ muutumise kiirendus (teine tuletis aja järgi) peab võrduma nulliga ning et $L$ ja $H$ on konstantsed suurused, mistõttu nende tuletised on nullid.
\[\dot{x}=\frac{4h\dot{h}-L\dot{h}-2H\dot{h}}{2H}.\]
Võtame veelkord tuletist:
\[a=\ddot{x}=\frac{4h\ddot{h}+4\dot{h}^2-L\ddot{h}-2H\ddot{h}}{2H}=\frac{2\dot{h}^2}{H}=\frac{2w^2}{HS^2\rho^2}.\]
Kogu liivakellas oleva liiva mass on $m=SH\rho$.
Liivakellale mõjuv täiendav jõud (kaalu näidu muutus):
\[F=ma=\frac{2w^2}{S\rho}.\]

Liivakell

Uurime liivakella mudelit. Liivakell koosneb silindrilisest torust pikkusega
$L$, mis on keskelt eraldatud ühtlaselt aukudega läbistatud plaadiga, millest
liiv saab läbi voolata. Heas lähenduses ei sõltu liiva aukude läbimise
masskiirus $w$ ülemises anumas olevast liivahulgast. Liivakell asetatakse
kaalule töörežiimis (kui liiv voolab) ja siis, kui kogu liiv on alla voolanud.
Milline on kaalunäitude vahe? Liiva tihedus on $\rho$ ja liivakella
ristlõikepindala on $S$. Eeldage, et hetkel kukkuva liiva mass on tühine
võrreldes liiva kogumassiga.

Juhul, kui süsteemi massikeskme kõrgus on $x(t)$, siis Newtoni II seaduse kohaselt $M\ddot{x}(t) = F - Mg$, kus $M$ on süsteemi kogumass ja $F$ kaalu näit. Seega taandub ülesanne $x(t)$ leidmisele liivakella töörežiimis.


[User]
Gaasitermomeeter koosneb mõõteampullist 1 ja manomeetrist 2, mis on omavahel ühenduses peenikese kapillaari 3 kaudu (vt joonist). Manomeetri ja mõõteampulli ruumalade suhe on $\alpha = 30$. Kapillaari ruumala võib lugeda tühiselt väikeseks. Seade täidetakse toatemperatuuril oleva gaasiga rõhuni $p_0 = \SI{1}{atm}$. Gaasi võib lugeda ideaalseks. Manomeetrit hoitakse toatemperatuuril $T_0 = \SI{293}{K}$, mõõteampull asetatakse keskkonda, mille temperatuuri on tarvis määrata. Leidke keskkonna temperatuur, kui manomeetri näit on $p = \SI{0,7}{atm}$.

\begin{center}
	\includegraphics[width=0.25\linewidth]{2006-v2g-05-yl}
\end{center}

Solution:
Olgu mõõteampulli ruumala $V$ ning manomeetri ruumala $V_m$. Kui toatemperatuuril $T_0$ täideti seade $n$ mooli gaasiga, siis ideaalse gaasi olekuvõrrandi põhjal
\[
\frac{p_0V}{T_0} + \frac{p_0V_m}{T_0} = nR.
\]
Kuna manomeeter ja mõõteampull on kapillaari kaudu ühenduses, siis nende gaasirõhud on isegi temperatuuride erinedes ühesugused. Kui mõõteampull on temperatuuril $T$, siis (gaasi koguhulk jääb samaks)
\[
\frac{pV}{T} + \frac{pV_m}{T_0} = nR.
\]
Elimineerides $n$ ja asendades $V_m/V = \alpha$, saame
\[
T=\frac{p T_{0}}{p_{0}+\left(p_{0}-p\right) \alpha} \approx \SI{20,5}{K}.
\]

Gaasitermomeeter

Gaasitermomeeter koosneb mõõteampullist 1 ja manomeetrist 2, mis on omavahel ühenduses peenikese kapillaari 3 kaudu (vt joonist). Manomeetri ja mõõteampulli ruumalade suhe on $\alpha = 30$. Kapillaari ruumala võib lugeda tühiselt väikeseks. Seade täidetakse toatemperatuuril oleva gaasiga rõhuni $p_0 = \SI{1}{atm}$. Gaasi võib lugeda ideaalseks. Manomeetrit hoitakse toatemperatuuril $T_0 = \SI{293}{K}$, mõõteampull asetatakse keskkonda, mille temperatuuri on tarvis määrata. Leidke keskkonna temperatuur, kui manomeetri näit on $p = \SI{0,7}{atm}$.

\begin{center}
	\includegraphics[width=0.25\linewidth]{2006-v2g-05-yl}
\end{center}


[User]
The vectors E and B cannot go into four-vectors, as they transform among each other,
    but rather fit together into an antisymmetric rank two tensor. As a result, there is a different set
    of associated invariant quantities.

 (a) Show that under the relativistic field transformations, the quantities E · B and E 2 − B 2 are
     both invariant. (Hint: this can be done using vector notation, using E⊥ · E∥ = B⊥ · B∥ = 0.)
     These are the two basic invariants, out of which all other invariants can be constructed.
 (b) Suppose that in an inertial frame, E is zero at a given point and B is nonzero. Is it possible
     to find an inertial frame where B is zero at that point?
 (c) Recall from E7 that, in units where ϵ0 = µ0 = 1, the energy density of the electromagnetic
     field is E = E 2 /2 + B 2 /2, and the Poynting vector is S = E × B. Show that E 2 − |S|2 is
     invariant. (Hint: don’t use the field transformations for this part.)
  Remark: Is Magnetism Real?

  Purcell’s electromagnetism textbook is exceptional because it shows that a force like
  magnetism must exist, if one believes Coulomb’s law and relativity. The idea is simple. We
  know how forces transform between frames, and given some reasonable assumptions, can
  also deduce how electric fields transform between frames. If electric fields were all there
  were, then electric forces would have to transform just like three-forces, but they don’t. So
  there must be some other force to make up the difference, and it turns out to be precisely
  the magnetic force. We saw an example of this in problem 2.

  It is important not to misunderstand this beautiful idea. Many people, upon reading such
  arguments, believe that magnetism “doesn’t exist” because it’s “all just electric fields”.
  Sometimes people even say that magnetic forces are a “mistake” caused by “forgetting
  about” relativistic corrections. This is all totally backwards. Sometimes time dilation in one
  frame can be explained in terms of length contraction in another, but that doesn’t mean
  that length contraction doesn’t exist, or is a mistake – it’s perfectly real in that particular
  frame. (Furthermore, while you can always get rid of the magnetic force on one particle at
  one moment by going to that particle’s rest frame, there are plenty of situations where you
  can’t remove the magnetic field, as we saw in problem 8!)

  The real lesson of relativity isn’t that magnetic fields are a mistake, it’s that electric and
  magnetic fields are as intertwined as space and time, as you can see from their transformation
  properties. Just as space and time combine into a four-vector, electric and magnetic fields
  combine, in an equal footing, into the electromagnetic field tensor.

  Remark: Electromagnetism in Covariant Form

  Problem 5 is a first step to showing that J µ = (ρ, J) is a four-vector, where ρ is the charge
  density and J is the current density. Note that the continuity equation for charge, as
  mentioned in T2, can be simply written in four-vector notation as

                                            ∂µ J µ = 0.

  As another example, you can show that the four-current of a single charged particle q is
  J µ = quµ . We can go even further and write the whole of electromagnetism in terms of
  four-vectors and tensors. Maxwell’s equations can be written as

                                          ∂µ F µν = J ν .

      The invariant quantities found in problem 8 can be written in terms of the field strength
      tensor as Fµν F µν and ϵµνρσ F µν F ρσ where ϵµνρσ is the Levi–Civita symbol. These are the
      only two ways to “contract all the indices” to get a scalar.

      Remark: Elegant Notation

      Sometimes people dislike the index notation above because of all the little Greek letters
      floating around. If you only want to deal with vectors, vector notation is often better. It
      hides all the indices, at the cost of requiring you to introduce special symbols like · and × to
      specify the vector operations you want to do. The reason we don’t use a vector-like notation
      for tensors is because there are too many operations you can do with them (e.g. “contract the
      3rd index of a rank 4 tensor with the 1st index of a rank 2 tensor”) to define separate symbols
      for each one; indices are just more efficient. On the other hand, if you only work with totally
      antisymmetric tensors, then there are only a few possible operations, and one can use the
      elegant, index-free “differential form” notation. In this notation, Maxwell’s equations are

                                                d⋆F =J

      where d is called the exterior derivative, ⋆ is the Hodge dual, and the fact that the electro-
      magnetic fields are derivatives of potentials is expressed as

                                                 F = dA.

      So is this the best, most true formulation of Maxwell’s equations? Well, as Feynman once
      pointed out, you can easily do better. For example, you can define the “unworldliness”

                                 U = |F − ma|2 + (∇ · E − ρ/ϵ0 )2 + . . . .

      Then all physical laws can be expressed in terms of the amazingly simple equation

                                                  U = 0.

      But this doesn’t actually help, because to use the equation for anything, you need to plug in
      the definition of U , and then you’re back to where you were before. In general, more elegant
      notation is often more brittle: it only works well in a smaller set of situations. (For example,
      with differential form notation, you just can’t write down the stress-energy tensor of the
      electromagnetic field, because that’s symmetric rather than antisymmetric.) Index notation
      is great because it works as long as indices are contracted in pairs, which holds as long as
      you’re dealing with laws that are independent of coordinate system. In general, there’s no
      need to be ideological about notation; it’s just a tool, and we should use the best tool for
      each job. If anyone tells you that their preferred alternative notation for vectors or tensors
      will revolutionize physics, keep your hand on your wallet.

Solution:
(a) Setting c = 1 for convenience as usual, we have

                    E′ · B′ = E′⊥ · B′⊥ + E∥′ B∥′ = γ 2 (E⊥ + v × B⊥ ) · (B⊥ − v × E⊥ ) + E∥ B∥ .

          The cross terms vanish (try an explicit example if you don’t see why), which leaves

                                   E′ · B′ = γ 2 (1 − v 2 )E⊥ · E⊥ + E∥ B∥ = E · B

     as desired. As for the other quantity, we have
                              2                   2
             E ′2 − B ′2 = E ′ ∥ + E′⊥ · E′⊥ − B ′ ∥ − B′⊥ · B′⊥
                        = E∥2 + γ 2 (E⊥ + v × B⊥ )2 − B∥2 − γ 2 (B⊥ − v × E⊥ )2
                        = E∥2 + γ 2 (E⊥ · E⊥ + v 2 B⊥ · B⊥ ) − B∥2 − γ 2 (B⊥ · B⊥ + v 2 E⊥ · E⊥ )
                        = E∥2 + E⊥ · E⊥ − B∥2 − B⊥ · B⊥
                        = E2 − B2

     as desired, where in the third equality the cross-terms canceled.

 (b) This is impossible, because then E 2 − B 2 would have different signs in the two frames.

 (c) If we plugged in the field transformations, the algebra would get extremely messy. Instead,
     we use the hint that this invariant can be constructed out of the ones found in part (a), so

                                          1 2 1 2 2

                              2     2
                            E − |S| =       E + B        − |E × B|2
                                          2      2
                                          1 2 1 2 2

                                      =     E + B        + (E · B)2 − E 2 B 2
                                          2      2
                                          1 2 1 2 2

                                      =     E − B        + (E · B)2 .
                                          2      2
     This is constructed out of the invariants in part (a), so it is invariant as well.
  Remark: Is Magnetism Real?

  Purcell’s electromagnetism textbook is exceptional because it shows that a force like
  magnetism must exist, if one believes Coulomb’s law and relativity. The idea is simple. We
  know how forces transform between frames, and given some reasonable assumptions, can
  also deduce how electric fields transform between frames. If electric fields were all there
  were, then electric forces would have to transform just like three-forces, but they don’t. So
  there must be some other force to make up the difference, and it turns out to be precisely
  the magnetic force. We saw an example of this in problem 2.

  It is important not to misunderstand this beautiful idea. Many people, upon reading such
  arguments, believe that magnetism “doesn’t exist” because it’s “all just electric fields”.
  Sometimes people even say that magnetic forces are a “mistake” caused by “forgetting
  about” relativistic corrections. This is all totally backwards. Sometimes time dilation in one
  frame can be explained in terms of length contraction in another, but that doesn’t mean
  that length contraction doesn’t exist, or is a mistake – it’s perfectly real in that particular
  frame. (Furthermore, while you can always get rid of the magnetic force on one particle at
  one moment by going to that particle’s rest frame, there are plenty of situations where you
  can’t remove the magnetic field, as we saw in problem 8!)

  The real lesson of relativity isn’t that magnetic fields are a mistake, it’s that electric and
  magnetic fields are as intertwined as space and time, as you can see from their transformation
  properties. Just as space and time combine into a four-vector, electric and magnetic fields

  combine, in an equal footing, into the electromagnetic field tensor.

  Remark: Electromagnetism in Covariant Form

  Problem 5 is a first step to showing that J µ = (ρ, J) is a four-vector, where ρ is the charge
  density and J is the current density. Note that the continuity equation for charge, as
  mentioned in T2, can be simply written in four-vector notation as

                                            ∂µ J µ = 0.

  As another example, you can show that the four-current of a single charged particle q is
  J µ = quµ . We can go even further and write the whole of electromagnetism in terms of
  four-vectors and tensors. Maxwell’s equations can be written as

                                           ∂µ F µν = J ν .

  The invariant quantities found in problem 8 can be written in terms of the field strength
  tensor as Fµν F µν and ϵµνρσ F µν F ρσ where ϵµνρσ is the Levi–Civita symbol. These are the
  only two ways to “contract all the indices” to get a scalar.

  Remark: Elegant Notation

  Sometimes people dislike the index notation above because of all the little Greek letters
  floating around. If you only want to deal with vectors, vector notation is often better. It
  hides all the indices, at the cost of requiring you to introduce special symbols like · and × to
  specify the vector operations you want to do. The reason we don’t use a vector-like notation
  for tensors is because there are too many operations you can do with them (e.g. “contract the
  3rd index of a rank 4 tensor with the 1st index of a rank 2 tensor”) to define separate symbols
  for each one; indices are just more efficient. On the other hand, if you only work with totally
  antisymmetric tensors, then there are only a few possible operations, and one can use the
  elegant, index-free “differential form” notation. In this notation, Maxwell’s equations are

                                            d⋆F =J

  where d is called the exterior derivative, ⋆ is the Hodge dual, and the fact that the electro-
  magnetic fields are derivatives of potentials is expressed as

                                             F = dA.

  So is this the best, most true formulation of Maxwell’s equations? Well, as Feynman once
  pointed out, you can easily do better. For example, you can define the “unworldliness”

                             U = |F − ma|2 + (∇ · E − ρ/ϵ0 )2 + . . . .

  Then all physical laws can be expressed in terms of the amazingly simple equation

                                              U = 0.

      But this doesn’t actually help, because to use the equation for anything, you need to plug in
      the definition of U , and then you’re back to where you were before. In general, more elegant
      notation is often more brittle: it only works well in a smaller set of situations. (For example,
      with differential form notation, you just can’t write down the stress-energy tensor of the
      electromagnetic field, because that’s symmetric rather than antisymmetric.) Index notation
      is great because it works as long as indices are contracted in pairs, which holds as long as
      you’re dealing with laws that are independent of coordinate system. In general, there’s no
      need to be ideological about notation; it’s just a tool, and we should use the best tool for
      each job. If anyone tells you that their preferred alternative notation for vectors or tensors
      will revolutionize physics, keep your hand on your wallet.


[User] A geostationary orbit is such an orbit that if it has a satellite on it, the satellite does not move with respect to the Earth. How big is the area of ground that can be observed from such a satellite? Give the area’s diameter measured along the surface of the Earth as an answer. The gravitational constant is $G=\SI{6.7e-11}{N \cdot m^2/kg^2}$, the Earth’s mass $M=\SI{6,0e24}{kg}$, the Earth’s radius $r=\SI{6400}{km}$, the Earth’s rotation period $t=\SI{24}{h}$.

Solution:
Satelliidile orbiidil mõjuv kesktõmbejõud on Maa poolt satelliidile avaldatav gravitatsioonijõud. Saame $$\frac{mv^2}{R}=G\frac{Mm}{R^2},$$kus $m$ on satelliidi mass ja $R$ orbiidi raadius. Kuna geostatsionaarne satelliit Maa suhtes ei liigu, peab selle tiirlemisperiood olema samuti 24 h. Saame $v=\frac{2\pi R}{t}$. Neist võrranditest saame $$4\pi^2 R^3=GM \Rightarrow R=\sqrt[3]{\frac{GMt^2}{4\pi^2}}=\SI{42400}{km}.$$ Maa keskpunkt, satelliit ning satelliidilt nähtava maa-ala serval asetsev suvaline punkt moodustavad täisnurkse kolmnurga, mille hüpotenuusiks on satelliidi orbiidi raadius ning üheks kaatetiks Maa raadius. Maa keskmes asuvaks nurgaks saame $\alpha=\arccos{\frac{r}{R}}$, kuid kuna meid huvitab satelliidilt nähtava ala läbimõõt, peame leidma nurga $2\alpha$. Sellele nurgale vastav kaare pikkus Maa pinnal on $d=2r\alpha$ (kui nurk on radiaanides). Lõppvastuseks saame
$$d=2r\arccos \left(\frac{r}{\sqrt[3]{\frac{GMt^2}{4\pi^2}}}\right) =\SI{18000}{km}.$$

Satelliit

Geostatsionaarseks orbiidiks nimetatakse sellist orbiiti, millel asuv satelliit
Maa suhtes ei liigu. Kui suur on maa-ala, mida sellisel orbiidil olevalt
satelliidilt jälgida saab? Vastuseks esitage selle maa-ala läbimõõt mõõdetuna
mööda Maa pinda. Gravitatsioonikonstant $G=\SI{6.7e-11}{N \cdot m^2/kg^2}$, Maa
mass $M=\SI{6,0e24}{kg}$, Maa raadius $r=\SI{6400}{km}$, Maa
pöörlemisperiood $t=\SI{24}{h}$.

Geostatsionaarse orbiidi raadius on leitav jõudude tasakaalust ning orbiidi perioodist.