[User]
An infinite grounded conducting plane at z = 0 is deformed with a hemispherical
    bump of radius R centered at the origin, as shown. A charge q is placed at z = a as shown.

    Can the method of images be used to find the potential in the region with the charge? If so, specify
    the image charges; if not, explain why not.

[2] Problem 4 (Purcell 3.50). A point charge q is located a distance b > r from the center of a
    nongrounded conducting spherical shell of radius r, which also has charge q. When b is close to r,
    the charge is attracted to the shell because it induces negative charge; when b is large the charge is
    clearly repelled. Find the value of b so that the point charge is in equilibrium. (Hint: you should
    have to solve a difficult polynomial equation. You can either use a computer or calculator, or use
    the fact that it contains a factor of 1 − x − x2 .)

Solution:
It can be done with three image charges, all on the z-axis:

      • A charge −q at z = −a.

      • A charge −qR/a at z = R2 /a.

      • A charge qR/a at z = −R2 /a.

    These ensure that the voltage vanishes on both the whole plane z = 0 and on the sphere r = R.

[2] Problem 4 (Purcell 3.50). A point charge q is located a distance b > r from the center of a
    nongrounded conducting spherical shell of radius r, which also has charge q. When b is close to r,
    the charge is attracted to the shell because it induces negative charge; when b is large the charge is
    clearly repelled. Find the value of b so that the point charge is in equilibrium. (Hint: you should
    have to solve a difficult polynomial equation. You can either use a computer or calculator, or use
    the fact that it contains a factor of 1 − x − x2 .)

    Solution. The image charge has value −q ′ = −qr/b, and is at position r2 /b. As of now, the
    spherical shell has total charge −q ′ (by Gauss’s law), so to compensate, we add a charge of value
    q + q ′ at the center. Thus, to balance forces we have
                                                 q′        q + q′
                                                         =        .
                                            (b − r2 /b)2     b2
    Defining x = r/b, this simplifies to

                                            x = (1 + x)3 (1 − x)2

    which is a rather intimidating quintic equation. At this point you can pull out your calculator, but
    amazingly, it factors as
                                       (1 − x − x2 )(1 + x − x3 ) = 0.
                                                                √
    The only root with 0 < x < 1 is from the quadratic, x = ( 5 − 1)/2, giving
                                                        √
                                                    1+ 5
                                               b=           r.


[User]
Silindriline veetünn, milles hoitakse muutumatut veetaset kõrgusega $h$, on tõstetud horisontaalsele platvormile, mille kõrgus maapinnast on $H$ (vt joonist). Tünni seina kõrgusele $a$ selle põhjast puuritakse auk. Väljuv veejuga puudutab maapinda kaugusel $L$ platvormi jalamist. Graafikul on kujutatud kauguse $L$ sõltuvus augu kõrgusest $x = a/h$. Määrake tünni kõrgus $h$ ning aluse kõrgus $H$ eeldusel, et $h > H$.

\begin{center}
	\includegraphics[width=\linewidth]{2005-v3g-05-yl}
\end{center}

Solution:
Leiame esmalt veejoa väljumise kiiruse. Kiiruse avaldis on tuntud Torricelli seadusena, kuid selle leidmiseks võime arutleda nõnda: avast väljuv juga omab kineetilist energiat $mv^2/2$, teisalt peab see olema võrdne potentsiaalse energiaga, mis saadakse vee ülemiselt pinnalt kuni auguni langedes: $mg(1 - x)h$. Seega väljumiskiirus avaldub
\[
v = \sqrt{2g(1 - x)h}.
\]
\emph{Märkus:} impulsi jäävusest saaksime (võrreldes tünni vasakule ja paremale seinale mõjuvate rõhumisjõudude vahe avaldist ning veejoa impulssi) tulemuse $v = \sqrt{g(1 - x)h}$. See avaldis kehtib siis, kui vee liikumine tünnis pole laminaarne ning energia ei säili (läheb veekeeristesse). Laminaarse (energiakadudeta) voolu korral tuleks rõhumisjõudude vahe leidmisel arvestada Bernoulli seadusest tingitud rõhu muutust, mistõttu impulsi jäävusest tuletatud vastus ei kehti. Kuivõrd antud ülesandes on voolu laminaarsuse küsimus jäetud täpsustamata, siis on mõlemad meetodid korrektsed.

Et veejoal vertikaalset kiiruskomponenti esialgu pole, kulub langemiseks aeg
\[
\tau = \sqrt{\frac{2(H+xh)}{g}}.
\]
Selle ajaga liigub aga veejuga horisontaalsihis kaugusele $L = v\tau$ ehk:
\[
L=2 \sqrt{(1-x) h(H+x h)},
\]
mida aga ongi antud graafikul kujutatud.

Seega piisaks $H$ ja $h$ leidmiseks kahe joone punkti koordinaatide määramisest ning tekkiva võrrandisüsteemi lahendamisest. Et aga võimalikult lihtsalt tulemuseni jõuda, kasutame tähelepanekut, et $x = 0$ korral $L = 2\sqrt{hH}$. Võtame graafikult lugemi punktis $x = 0$ ning saame esimese võrrandi: $hH = \SI{9}{m^2}$.

Nagu öeldud, võib teise võrrandi saada suvalise punkti abil, kuid uurime natuke ekstreemumtingimust. Tähistame esmalt $\alpha = H/h$, $L$ avaldub seega kui
\[
L=2 h \sqrt{(1-x)(\alpha+x)}.
\]
Kui võtame $L$-ist $x$-i järgi tuletise, näeme, et $L$ omab ekstreemumväärtust $x = (1 - \alpha )/2$ korral. Selle tulemiseni võib jõuda ka arutledes nõnda: $y = (1-x)(\alpha +x)$ kujutab endast allapoole suunatud parabooli nullkohtadega $1$ ja $-\alpha$, ekstreemumväärtus on seega nende vahel ehk kohal $x = (1 - \alpha )/2$. Asendame selle $L$-i avaldisse:
\[
L=2 h \sqrt{\left(1-\frac{1-\alpha}{2}\right)\left(\alpha+\frac{1-\alpha}{2}\right)}=2 h \frac{1+\alpha}{2}=h+H.
\]
Seega saame teise võrrandi $L$-i maksimumväärtust kasutades:
\[
H + h = \SI{10}{m}.
\]
Nendest kahest lihtsast võrrandist tekkiva võrrandisüsteemi lahendamisel (viies need näiteks ruutvõrrandi kujule) leiame väärtused: $h = \SI{9}{m}$ ja $H = \SI{1}{m}$.

Veetünn

Silindriline veetünn, milles hoitakse muutumatut veetaset kõrgusega $h$, on tõstetud horisontaalsele platvormile, mille kõrgus maapinnast on $H$ (vt joonist). Tünni seina kõrgusele $a$ selle põhjast puuritakse auk. Väljuv veejuga puudutab maapinda kaugusel $L$ platvormi jalamist. Graafikul on kujutatud kauguse $L$ sõltuvus augu kõrgusest $x = a/h$. Määrake tünni kõrgus $h$ ning aluse kõrgus $H$ eeldusel, et $h > H$.

\begin{center}
	\includegraphics[width=\linewidth]{2005-v3g-05-yl}
\end{center}

Veejoa väljumise kiirus on leitav energia jäävuse seadusest või alternatiivselt impulsi jäävusest. Mõlemad lähenemised on korrektsed, aga annavad numbrilise konstandi võrra erineva vastuse, mis on tingitud jäävusseaduste kehtivuse eelduste erinevusest. Tünni ja aluse kõrgused on leitavad valides graafikult kaks (või vajadusel rohkem) punkti ja lahendades tekkinud võrrandisüsteemi.


[User]
In most common materials, µ ≈ µ0 while ϵ depends on frequency. We’ll investigate
    the origin of this frequency dependence below.
     (a) Model an electron in an atom as a mass m with charge q attached to a spring, with natural
         angular frequency ω0 and a damping force −mγv, in an electric field E0 e−iωt . Write down
         the equation of motion for the electron.
     (b) The atomic polarizability α is defined by p = αE. Show that
                                                        q 2 /m
                                              α=                    .
                                                   −ω 2 + ω02 − iγω

    Now we restrict to a gas with small number density n, so that nα ≪ ϵ0 . For simplicity, you may
    also assume that the damping is weak, γ ≪ ω0 . Now, the Clausius–Mossotti formula reduces to
                                                ϵ = ϵ0 + nα
    and α is a complex number, so we learn that ϵ is also cmmplex.

     (c) The wavevector and angular frequency are related by k 2 = µϵω 2 . Explain why the fact that ϵ
         is complex indicates that waves can be absorbed.
     (d) What value of ω maximizes the absorption rate of the electromagnetic waves? Roughly how
         many wavelengths does such a wave propagate before being mostly absorbed?
     (e) What value of ω maximizes the speed of the electromagnetic waves, and what is that speed?
     (f) Transparent objects such as glass can be modeled as having a very high resonant frequency,
         much higher than that of visible light. Does blue light or red light refract more when passing
         from air to glass?

   The intuitive reason that these electrons can affect the propagation speed of light is because they
   emit secondary electromagnetic waves that are out of phase with the original wave; this “pushes” the
   phase of the composite wave forward or backward, affecting the phase velocity. A nice explanation

                   01h
   of this can be found in chapter I.31 of the Feynman lectures.

                   01h

Solution:
(a) We have
                                   mr̈ = −mω02 r − mγv + qE0 e−iωt .

 (b) Suppose r = r0 eiωt where r0 is potentially complex. Then, we see that E0 ∥ r0 and

                                −mω 2 r = −mω02 r − mγiωr + q(E0 /r0 )r.

     Thus,
                                                 m(ω02 − ω 2 + iγω)
                                      E0 /r0 =                      .
                                                         q
     We have that p = −r0 q/E0 which yields the result.

 (c) If ϵ is complex, then with µ ≈ µ0 and ω 2 being real, then k 2 = µϵω 2 will also be complex.
     Thus with a complex wavevector k, the field of E0 ei(k·x−ωt) will exponentially decay.
                                                                          √        √        nα
 (d) The absorption arises from the imaginary part of of kx. With k = ω µϵ ≈ ω µ0 ϵ0 (1 + 2ϵ  0
                                                                                ),
     the absorption rate is maximized when the imaginary part of k is maximized, and

                                                               q 2 /m

                                         ωn        ωn
                       β ≡ Im(k) = Im        α =                             (γω)
                                        2cϵ0      2cϵ0 (ω02 − ω 2 )2 + (ωγ)2

                                           q 2 γn        ω2
                                      =
                                          2mcϵ0 (ω02 − ω 2 )2 + γ 2 ω 2

        The maximum value of this occurs when
                            dβ
                                 ∝ (ω02 − ω 2 )2 + γ 2 ω 2 − ω 2 (2(ω 2 − ω02 ) + γ 2 ) = 0

                            dω 2

        which simplifies to yield
                                                  ω04 − ω 4 = 0.
         So an electromagnetic wave with angular frequency ω = ω0 has the maximum absorption rate.
                                                                              q n      2
        The electric field will have a factor of e−βx , and at ω = ω0 , β = 2γmcϵ 0
                                                                                . The value of the real
        wavevector Re k will be close to (note that Re(α) = 0 at ω = ω0 ):

                                                 ω0            nα         ω0
                                       Re(k) =       1 + Re            =
                                                  c            2ϵ0         c
        Then for the wave to fall off by a factor of e, the wave will need to travel a distance of β1 ,
                  1    k
        which is βλ = 2πβ wavelengths. Thus,
                                                      k    ω0 γmϵ0
                                                         =
                                                     2πβ    πq 2 n
         is the number of wavelengths it will travel before the amplitude gets reduced by a factor of e.
     (e) The phase velocity is maximized when Reω k , or Re √1µϵ is maximized.

                                                                cq 2 n     ω 2 − ω02

                                1              1 nα
                      vp = Re √ ≈ c 1 − Re               =c+
                                 µϵ            2 ϵ0             2mϵ0 (ω 2 − ω02 )2 + (γω)2
         Differentiating with respect to ω 2 and finding where it’s zero yields
                               (ω 2 − ω02 )2 + γ 2 ω 2 − (ω 2 − ω02 )(2(ω 2 − ω02 ) + γ 2 ) = 0
                                                   (ω 2 − ω02 )2 = ω02 γ 2
                                                      ω 2 = ω02 ± ω0 γ
         Looking at the original, the smaller solution yields the minimum velocity, and the larger
         solution yields the maximum velocity (which happens to be greater than c). The maximum
         phase velocity is
                                              cq 2 n         ω0 γ
                                   vmax = c +
                                              2mϵ0 (ω0 γ)2 + γ 2 (ω02 + ω0 γ)
     (f) From the previous part, we have
                                                     cq 2 n     ω02 − ω 2
                                          vp = c −
                                                     2mϵ0 (ω 2 − ω02 )2 + (γω)2
         and now we know that ω0 ≫ ω, so
                          vp      q2n        ω02 − ω 2              q2n
                             ≈1−                             ≈ 1 −          (1 + ω 2 /ω02 ).
                           c     2mϵ0 ω04 − 2ω02 ω 2 + (γω)2       2mϵ0 ω02
        Thus, increasing the frequency would decrease vp and increase the index of refraction, so blue

                   01h
        light would refract more.

                   01h

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


[User]
Libedale klaaspulgale on pehmest traadist tihedasti keritud solenoid pikkusega $\ell$, keerdude arvuga $N$ ja ristlõikepindalaga $S$. Selles hoitakse konstantset voolu tugevusega $I$. Millist jõudu $F$ oleks vaja rakendada pooli otstele südamiku sihis, et venitada seda pisutki pikemaks, kui kehtiks eeldus, et venitamisel suurenevad kõigi naaberkeerdude vahekaugused võrdselt? Võite lugeda, et klaasi magnetiline läbitavus $\mu = 1$.

\emph{Vihje:} tiheda solenoidi südamikus on homogeenne magnetinduktsioon$ B = \mu_0IN/\ell$.

Solution:
Venitame pooli väikese $\Delta \ell$ võrra pikemaks ja avaldame tehtud töö ($A$) kahel eri viisil. Ühelt poolt $A = F_1\Delta \ell$. Samas salvestub osa kulutatud energiat ($\Delta E_m$) magnetväljas ja ülejäänu ($\Delta E_v$) kas eraldub vooluallika sisetakistusel (see peab nt lühise korral jääva voolu hoidmiseks alati olemas olema) või, kui $A < \Delta E_m$, täiendatakse vooluallika tööga. Igal juhul:
\[
A = \Delta E_m + \Delta E_v.
\]
$\ell$ suurenedes $B$ väheneb, mistõttu ilmselt $\Delta E_m < 0$ ja vooluallika sisetakistusel eraldub energiat:
\[
A > 0 \Longrightarrow \Delta E_v > 0.
\]
Vajalikud energiamuudud võime leida mitmel eri viisil.

\emph{Esimene meetod}. Olgu kogu magnetvälja energia $E_m$. Selle energia ruumtihedus:
\[
w=\frac{B^{2}}{2 \mu_{0}} \Longrightarrow E_{m}=w \ell S=\frac{\ell S B^{2}}{2 \mu_{0}} \Longrightarrow \Delta E_{m}=\frac{S \Delta\left(\ell B^{2}\right)}{2 \mu_{0}}=\frac{I N S \cdot \Delta B}{2}.
\]
Toimugu pooli pikenemine ajaga $\Delta t$ ja indutseerigu magnetvoo muutus poolil elektromotoorjõu absoluutväärtusega $E$. Faraday induktsiooniseadusest:
\[
\mathcal{E}=N \frac{|\Delta B| \cdot S}{\Delta t}.
\]
Siit saamegi $\Delta Ev$:
\[
\Delta E_v = I\mathcal{E}\Delta t = INS|\Delta B|.
\]

\emph{Teine meetod}. Leiame tiheda pooli induktiivsuse L:
\[
N B S=L I \Longrightarrow L=\frac{N B S}{I}.
\]
Magnetvälja energia:
\[
E_{m}=\frac{L I^{2}}{2}=\frac{N B S I}{2} \Longrightarrow \Delta E_{m}=\frac{I N S \cdot \Delta B}{2}.
\]
Toimugu pooli pikenemine ajaga $\Delta t$ ja indutseerigu magnetvoo muutus poolil elektromotoorjõu absoluutväärtusega $\mathcal{E}$. Eneseinduktsioonielektromotoorjõud tuleneb Faraday induktsiooniseadusest:
\[
\mathcal{E}=\frac{|\Delta(N B S)|}{\Delta t}=\frac{|\Delta(L I)|}{\Delta t}=\frac{N S|\Delta B|}{\Delta t},
\]
kust saame:
\[
\Delta E_{v}=I \mathcal{E} \Delta t=I N S|\Delta B|.
\]

\emph{Ühine osa mõlemale lahendusele}. Leiame $\Delta B$, eeldades, et $\Delta \ell$ on väike:
\[
\Delta B=\mu_{0} I N \Delta\left(\frac{1}{\ell}\right)=\mu_{0} I N\left(\frac{1}{\ell+\Delta \ell}-\frac{1}{\ell}\right)=-\mu_{0} I N \frac{\Delta \ell}{(\ell+\Delta \ell) \ell} \approx-\mu_{0} I N \frac{\Delta \ell}{\ell^{2}}.
\]
Lõpuks:
\[
F=\frac{A}{\Delta \ell}=\left(-\frac{I N S}{2}+I N S\right) \frac{|\Delta B|}{\Delta \ell}=\frac{I N S|\Delta B|}{2 \Delta \ell}=\frac{\mu_{0} I^{2} N^{2} S}{2 \ell^{2}}.
\]

\emph{Märkus}. Paar sõna ülesande tekstis tehtud eelduse kohta, et kõik keerud hakkavad otstest tõmbamisel eemalduma võrdse vahemaa võrra. Tegelikult see ei kehti, otstest tõmbamisel hakkaksid kõigepealt lahti hargnema otsmised keerud (ja märksa väiksema jõu juures). Ilmselt oleks võimalik keerdude ühtlane eemaldumine spetsiaalse mehaanilise konstruktsiooni abil, kui keerud poleks mitte klaaspulgal, vaid vastava raamistiku peal. Aga selgub, et antud ülesande vastus realiseerub lihtsamas olukorras ka. Kui võtta kinni poolist kahest lähestikku asuvas kohas keskpaiga läheduses, siis allpool leitav jõud on ühtlasi selline jõud, millega tõmbamisel saab natuke eemaldada pika poolikeskkohast vasakule- ja paremale poole jäävaid keerde (seda väidet me siinkohas tõestama ei hakka).

Pool

Libedale klaaspulgale on pehmest traadist tihedasti keritud solenoid pikkusega $\ell$, keerdude arvuga $N$ ja ristlõikepindalaga $S$. Selles hoitakse konstantset voolu tugevusega $I$. Millist jõudu $F$ oleks vaja rakendada pooli otstele südamiku sihis, et venitada seda pisutki pikemaks, kui kehtiks eeldus, et venitamisel suurenevad kõigi naaberkeerdude vahekaugused võrdselt? Võite lugeda, et klaasi magnetiline läbitavus $\mu = 1$.

\emph{Vihje:} tiheda solenoidi südamikus on homogeenne magnetinduktsioon$ B = \mu_0IN/\ell$.

Selleks, et venitada pooli väikse vahemaa $\Delta \ell$ võrra pikemaks, on vaja teha tööd, millest osa kulub magnetväljas salvestatud energia suurendamiseks ning teine osa kulub vooluallika poolt tehtud töö kompenseerimiseks. Magnetväljas salvestatud energia leidmiseks võib kasutada kas pooli koguenergia valemit $\frac{LI^2}{2}$ või magnetvälja energiatihedust $w = \frac{B^2}{2\mu_0}$.


[User]
5. MAGNETIC BILLIARD (9 points)

— Jaan
Kalda. Consider two absolutely elastic dielec-
tric balls of radius r and mass m one of which
carries isotropically distributed charge −q, and
the other — +q. There is so strong homogen-
eous magnetic field B, parallel to the axis z, that
electrostatic interaction of the two charges can
be neglected; neglect also gravity and friction
forces. The first ball (negatively charged) moves
with speed v and collides with the second ball
which had been resting at the origin. The colli-
sion is central, and immediately before the im-
pact, the velocity of the first ball was parallel to
the x-axis.
i) (1 point)What is the speed of the second ball
immediately after the collision?
ii)(2points)Sketchthetrajectoriesofthecentres
of the both balls during the subsequent motion.
iii) (3 points)What is the average velocity (mag-
nitude and direction) of the balls during their
subsequent motion?
iv) (3 points)Consider the same situation as be-
fore, except that the there are three differences
in the assumptions: the both balls have now
identical positive charge +q; electrostatic repul-
sion between the balls is no longer negligible;
the collision is not necessarily central (but the
balls move still at the same value of z so that
the collision will not induce any motion in the
direction of the z-axis). Let Pi denote the point
where the surfaces of the two balls are in contact
during the i-th collision. What is the maximal
distance between Pi and Pj (maximize over all
the values i,j =1,...∞, and over all the impact
parameters of the collisions for fixed values of B,
m, and q)?

Solution:
5. MAGNETIC BILLIARD

i) After the first collision, let the velocities of
the first and second ball be v1 and v2 respect-
ively. Applying the conservation of energy gives
mv2
2 =
mv2
1
2 +
mv2
2
2 or v2
= v2
1 + v2
2. Conserva-
tion of momentum yields mv = mv1 +mv2 or
v= v1+v2. Combining the two equations gives
v2
= v2
2+(v−v2)2
= v2
−2vv2+2v2
2 and v2 =0;v.
Since the first solution corresponds to the case
when the collision doesn’t happen, the speed of
the second ball must be v2 = v.
ii) The balls experience Lorentz force due to the
external magnetic field. Since the Lorentz force
is perpendicular to the line of motion and con-
stant in magnitude, the balls move along a circu-
lar orbit. Equating the Lorentz force with cent-
rifugal force gives mv2
R = qvB. Thus, R = mv
qB
and ω = v
R =
qB
m . This means that one of the
charges moves along the orbit clockwise and the
other anticlockwise.
After each collision, one of the balls moves
at speed v while the other one is at rest. The
moving ball travels a part of the full cyclotron
period (either clockwise or anticlockwise, de-
pending on the charge) before making a head-
on collision with the ball at rest. The mo-
mentum is given over to the first ball and the
previously moving ball stays at rest and the mo-
tion starts once again. During the subsequent
collisions, the balls start drifting in one direc-
tion as can be seen in the figure.
α
α
R
r r
+q -q
v
α
iii) The average velocity of the balls is equal to
the average speed of the collision points. From
the figure, it can be seen that the direction of
the average velocity is π − α clockwise from
the initial direction of the incoming ball, where
α = arctan 2r
R . The collision point moves by
d = rcosα between two subsequent collisions.
In between the two collisions, one of the balls
moves 2π−2α along a cyclotron orbit. The time
taken is then t = 2π−2α
ω
= 2m
qB
(
π−arctan 2r
R
)
and the average velocity is
vavg =
d
t
=
rωcosα
2(π−α)
=
vrR
R
p
4r2 +R2 (π−α)
=
=
v
√
4+ R2
r2
(
π−arctan 2r
R
).
iv) Let the velocities of the two balls at any mo-
ment of time be⃗ v1 and⃗ v2. The velocity of the
centre of mass is then ⃗ vCM = ⃗ v1+⃗ v2
2 . The equa-
tion of motion of the system is
q⃗ v1 ×⃗ B−⃗ F +q⃗ v2 ×⃗ B+⃗ F = m ̇ ⃗ v1 +m ̇ ⃗ v2,
where ⃗ F is the force between the two balls,
either the elastic forces during a collision or the
electrostatic forces. Then
q(⃗ v1 +⃗ v2)×⃗ B = m
d
dt
(⃗ v1 +⃗ v2),
q⃗ vCM ×⃗ B = m
d
dt
⃗ vCM.
This means that the centre of mass of the sys-
tem undergoes cyclotronic motion with a radius
of R = mv
qB . Because every collision point can
only be located where the center of mass is, the
collision points must also be limited to the same
circle. Thus, the maximum distance between
any two collisions is 2R = 2mv
qB

MAGNETIC BILLIARD


[User]
Consider a pendulum whose length adiabatically changes from L to L/2.

     (a) If the initial (small) amplitude was θ0 , find the final amplitude using the adiabatic theorem.

     (b) Give a physical interpretation of the adiabatic invariant.

     (c) When quantum mechanics was being invented, it was proposed that the energy in a pendulum’s
         oscillation was always a multiple of ℏω, where ω is the angular frequency. At the first Solvay
         conference of 1911, Lorentz asked whether this condition would be preserved upon slow changes
         in the length of the pendulum, and Einstein said yes. Reproduce Einstein’s analysis.

Solution:
(a) Using the small angle approximation, we have
                                                   1      1
                                                E = mv 2 + mgLθ2
                                                   2      2
         and the adiabatic invariant is
                                          I              I               I
                                              p dx = L       p dθ = mL       v dθ.

         On the other hand, from
                              p conservation
                                          p of energy, we know that v(θ) is an ellipse with semimajor
         and semiminor axes 2E/m and 2E/mgL, so
                                 I                                   s
                                               p      p                L
                                    p dx ∝ mL E/m E/mgL = E               .
                                                                       g

         The total energy is E = mgLθ02 /2, so
                                             I
                                                   p dx ∝ θ02 L3/2 g 1/2

         which implies that when L halves, the amplitude becomes 23/4 θ0 . Since we kept track of
         factors of g, this derivation also tells us what happens to the amplitude if g is slowly changed.
         The most famous literary example of a pendulum with changing length appears in Edgar Allan
         Poe’s short story, The Pit and the Pendulum. In the story, the narrator is strapped to a table,
         and sees a pendulum above him slowly moving and lengthening, bringing its razor edge toward
         him. Poe describes the pendulum’s amplitude as initially small, but “increasing inexorably”.
         As you can see from the solution to this problem, precisely the opposite happens. When the
         length goes up, the amplitude goes down, so it would just gently land on the narrator’s chest.

     (b) As for the case of a mass on a spring, we can add a third dimension and let the pendulum
         oscillate in a horizontal circle. Then the adiabatic invariant is simply
                                              I
                                                 Lz dθ = 2πLz ∝ Lz

         which is the angular momentum in the z-direction.

     (c) Given the way we did part (a), this is immediate. The adiabatic invariant is
                                                   s
                                                     L    E
                                                 E     = .
                                                     g    ω

         Therefore, E/ω remains an integer multiple of ℏ under adiabatic change.
         This is admittedly a rather contrived example. However, the conservation of E/ω does have
         practical applications. For example, in fluid mechanics, a relative of this quantity is called
         the wave action. The conservation law derived in this problem can be used to see how the
         amplitudes of water waves change as they travel through the ocean, which has varying depth.


[User]
Suppose x has uncertainty ∆x and y has uncertainty ∆y, where x and y are indepen-
    dent. Explain why the uncertainty of x + y is
                                                  p
                                     ∆(x + y) = (∆x)2 + (∆y)2 .

   This is called “addition in quadrature”. What is the uncertainty of x − y? How about x + x?

        Remark
        Note how this differs from “high school” uncertainty analysis. In school, you might be told
        to show uncertainty using significant figures, and when adding two things, to keep only the
        figures that are significant in both of them. That corresponds to

                                         ∆(x + y) = max(∆x, ∆y)

        which is an underestimate. Or, you might be told that the uncertainty needs to encapsulate
        all the possible values, which implies that

                                           ∆(x + y) = ∆x + ∆y

        which is an overestimate, since the uncertainties could cancel.

       Example 2: F = ma 2016 25

       Three students make measurements of the length of a 1.50 m rod. Each reports an uncertainty
       estimate representing an independent random error applicable to the measurement.

         • Alice performs a single measurement using a 2.0 m tape measure, to within 2 mm.

         • Bob performs two measurements using a wooden meter stick, each to within 2 mm, which
           he adds together.

         • Christina performs two measurements using a machinist’s meter rule, each to within
           1 mm, which she adds together.

       Rank the measurements in order of their uncertainty.

       Solution
                                                                                √
       The uncertainty in Alice’s measurement is 2 mm. The     √ uncertainty in Bob’s is 2  2 mm by
       quadrature, while the uncertainty in Christina’s is 2 mm by quadrature. So the lowest
       uncertainty is Christina’s, followed by Alice’s, followed by Bob’s.

Solution:
For independent variables, var(X
                                           p 1 + X2 ) = var(X1 ) + var(X2 ). Since our uncertainties
    represent the standard deviation, σX = var(X), we have
                                                p
                                      ∆(x + y) = (∆x)2 + (∆y)2 .

    For x − y = x + (−y), and ∆(−y) = ∆y, we get that ∆(x − y) = ∆(x + y). Finally, by linearity we
    clearly have ∆(x + x) = 2∆x. (The formula above doesn’t apply, because x isn’t independent of x.)

       Remark
       Note how this differs from “high school” uncertainty analysis. In school, you might be told
       to show uncertainty using significant figures, and when adding two things, to keep only the
       figures that are significant in both of them. That corresponds to

                                         ∆(x + y) = max(∆x, ∆y)

       which is an underestimate. Or, you might be told that the uncertainty needs to encapsulate
       all the possible values, which implies that

                                           ∆(x + y) = ∆x + ∆y

       which is an overestimate, since the uncertainties could cancel.

       Example 2: F = ma 2016 25

       Three students make measurements of the length of a 1.50 m rod. Each reports an uncertainty
       estimate representing an independent random error applicable to the measurement.

         • Alice performs a single measurement using a 2.0 m tape measure, to within 2 mm.

         • Bob performs two measurements using a wooden meter stick, each to within 2 mm, which
           he adds together.

         • Christina performs two measurements using a machinist’s meter rule, each to within
           1 mm, which she adds together.

       Rank the measurements in order of their uncertainty.

       Solution
                                                                                √
       The uncertainty in Alice’s measurement is 2 mm. The     √ uncertainty in Bob’s is 2 2 mm by
       quadrature, while the uncertainty in Christina’s is 2 mm by quadrature. So the lowest
       uncertainty is Christina’s, followed by Alice’s, followed by Bob’s.


[User]
Two stars of mass M orbit each other in a circle. The separation between them is
    2R, and their angular velocity about their center is ω. Work in the frame rotating with the stars.

      (a) In this frame, how many places can a third object of negligible mass stay at rest? Qualitatively
          indicate where all of them are, and when possible, analytically solve for their locations.

      (b) If we ignored the Coriolis force, how many of these locations would be stable equilibria?

       Remark: Lagrange Points

       In part (b) above, you should have found that none of the locations are stable equilibria.
       This is a consequence of Earnshaw’s theorem, which is usually stated in the context of
       electrostatics. In that context, suppose that in the presence of electric charges, a point P
       outside of the charges is an equilibrium point, i.e. one where the electric field vanishes. We
       then draw a small Gaussian surface S about P . For P to be a stable equilibrium point, we
       would need the electric field to point inward everywhere on S. But this is impossible: since
       there is no charge inside S, Gauss’s law implies that the electric flux through it must be zero.

       The same argument applies to gravitational fields, as they satisfy ∇ · g = 0 away from other
       masses, and therefore obey the same Gauss’s law constraint. In the above problem, there was
       also a centrifugal acceleration ac , so that the relevant quantity was atot = g + ac . However,
       ∇ · ac is positive, so it tends to make equilibrium points even more unstable, as you saw.

       However, to determine stability correctly, we have to account for the Coriolis force, which
       tends to deflect things sideways. To do this, we expand Newton’s second law at first order
       about the equilibrium point. The resulting differential equations are still linear, even with
       the Coriolis force added, and the equilibrium point is unstable if there is a normal mode
       solution that grows exponentially.

       This more general analysis is carried out here, and the result is that all five equilibrium
       points are still unstable. (This situation was considered in IPhO 2011, problem 1, but there
       the Coriolis force was ignored and an unphysical assumption was introduced, leading to
       the incorrect conclusion that some of the points were stable. I don’t recommend that problem.)

       More generally, if the stars have masses M1 > M2 , there are still five equilibrium points.
       Two of them are still at the vertex of an equilateral triangle, and they are stable if
                                                       √
                                          M1     25 + 3 69
                                              >              ≈ 25.
                                          M2         2

       For a proof of this remarkable statement, see this answer. Since MSun /MEarth = 3 × 105 , the
       corresponding Lagrange points for the Earth-Sun system are stable.

                       01E
                    01^

Solution:
(a) There are 5 such locations. Let’s place the stars at (−R, 0) and (R, 0). The angular
     velocity obeys
                                          GM 2
                                                 = M ω2R
                                          (2R)2
     from which we conclude that the centrifugal acceleration is
                                                           GM r
                                            ac = ω 2 r =        .
                                                           4R3
     The desired points are the ones where g cancels the centrifugal acceleration.

       • Clearly, both vanish at (0, 0) by symmetry.
       • There are also other points on the x-axis where they cancel. Clearly, such points must lie
         at |x| > R. For the case x > R, we must solve
                                        GM x     GM         GM

                                             =        2
                                                        +
                                        4R     (x − R)    (x + R)2
         which, after clearing denominators, gives a quintic equation. It’s not practically possible
         to solve it analytically, but the answer is roughly x ≈ 2.4 R. Similarly, there’s another
         point at negative x.
       • There are also points on the y-axis where they cancel. The relevant equation is
                                           GM y         2Gmy

                                                  =
                                            4R      (R + y 2 )3/2

                                                           √
         which can be straightforwardly solved to get y = ± 3 R. At these points, the three bodies
         form an equilateral triangle.

 (b) Without the Coriolis force, none of them are stable. For (0, 0), a displacement towards either
     mass would just make its gravitational attraction stronger. (The centrifugal force adds to this
     effect, making the point even more unstable.) The four other equilibrium points actually all lie
     at maxima of the potential energy in the plane, and therefore are unstable upon displacement
     in any direction in the plane. For example, in all four cases, a displacement away from the
     origin would make the gravitational attraction weaker but the centrifugal force stronger, so
     that the particle would keep moving away.

   Remark: Lagrange Points

   In part (b) above, you should have found that none of the locations are stable equilibria.
   This is a consequence of Earnshaw’s theorem, which is usually stated in the context of
   electrostatics. In that context, suppose that in the presence of electric charges, a point P
   outside of the charges is an equilibrium point, i.e. one where the electric field vanishes. We
   then draw a small Gaussian surface S about P . For P to be a stable equilibrium point, we
   would need the electric field to point inward everywhere on S. But this is impossible: since
   there is no charge inside S, Gauss’s law implies that the electric flux through it must be zero.

       The same argument applies to gravitational fields, as they satisfy ∇ · g = 0 away from other
       masses, and therefore obey the same Gauss’s law constraint. In the above problem, there was
       also a centrifugal acceleration ac , so that the relevant quantity was atot = g + ac . However,
       ∇ · ac is positive, so it tends to make equilibrium points even more unstable, as you saw.

       However, to determine stability correctly, we have to account for the Coriolis force, which
       tends to deflect things sideways. To do this, we expand Newton’s second law at first order
       about the equilibrium point. The resulting differential equations are still linear, even with
       the Coriolis force added, and the equilibrium point is unstable if there is a normal mode
       solution that grows exponentially.

       This more general analysis is carried out here, and the result is that all five equilibrium
       points are still unstable. (This situation was considered in IPhO 2011, problem 1, but there
       the Coriolis force was ignored and an unphysical assumption was introduced, leading to
       the incorrect conclusion that some of the points were stable. I don’t recommend that problem.)

       More generally, if the stars have masses M1 > M2 , there are still five equilibrium points.
       Two of them are still at the vertex of an equilateral triangle, and they are stable if
                                                       √
                                          M1     25 + 3 69
                                              >              ≈ 25.
                                          M2         2

       For a proof of this remarkable statement, see this answer. Since MSun /MEarth = 3 × 105 , the
       corresponding Lagrange points for the Earth-Sun system are stable.

                       01E
                    01^


[User] Teritamata pliiatsi telg ühtib koondava läätse peateljega. Mitu korda on pliiatsi kujutise pikkus tema enda pikkusest erinev, kui pliiatsi ühe otsa kujutise diameetri ja pliiatsi diameetri suhe on $k_1$ ning teise otsa jaoks on see suhe $k_2$? Pliiatsi mõlemad otsad asuvad läätsest kaugemal kui on läätse fookuskaugus.

Solution:
Olgu pliiatsi otste kaugused läätsest $d_1 = |BO|$ ja $d_2 = |DO|$ ning pliiatsi kujutise otste kaugused läätsest vastavalt $f_1 = |B'O|$ ja $f_2 = |D'O|$ (vt joonist).

\begin{center}
	\includegraphics[width=\linewidth]{2007-lahg-09-lah}
\end{center}

Paneme kirja valemid suurendustegurite jaoks:
\[
k=\frac{L^{\prime}}{L}, \quad k_{1}=\frac{|A^{\prime}B^{\prime}|}{|AB|}, \quad k_{2}=\frac{|C^{\prime}D^{\prime}|}{|CD|}.
\]
Sarnastest kolmnurkadest $F'B'A'$ ja $FOE$ leiame, et
\[
\frac{|A^{\prime}B^{\prime}|}{|EO|}=\frac{|F^{\prime}B^{\prime}|}{|FO|}=\frac{f_{1}-f}{f}.
\]
Analoogiliselt, sarnastest kolmnurkadest $F'D'C'$ ja $FOE$ leiame, et
\[
\frac{|C^{\prime}D^{\prime}|}{|EO|}=\frac{|F^{\prime}D^{\prime}|}{|FO|}=\frac{f_{2}-f}{f}.
\]
Kuna $|EO| = |AB| = |CD|$, siis
\[
k=\frac{f_{2}-f_{1}}{d_{1}-d_{2}}, \quad k_{1}=\frac{f_{1}-f}{f}, \quad k_{2}=\frac{f_{2}-f}{f}.
\]
Kasutades läätse valemit mõlema otsa jaoks:
\[
\frac{1}{d_{1}}+\frac{1}{f_{1}}=\frac{1}{f} \quad \operatorname{ning} \quad \frac{1}{d_{2}}+\frac{1}{f_{2}}=\frac{1}{f}
\]
saame avaldada pikkused $d_1$ ja $d_2$ pikkuste $f_1$ ja $f_2$ kaudu:
\[
d_{1}=\frac{f_{1} f}{f_{1}-f}, \quad d_{2}=\frac{f_{2} f}{f_{2}-f}.
\]
Asendame saadud väärtused suurendusteguri $k$ avaldisse:
\[
\begin{aligned}
	k&=\frac{f_{2}-f_{1}}{f_{1}-f}-\frac{f_{2} f}{f_{2}-f}=\frac{f_{2}-f_{1}}{f} \frac{\left(f_{1}-f\right)\left(f_{2}-f\right)}{f_{1}\left(f_{2}-f\right)-f_{2}\left(f_{1}-f\right)} \\
	&=\frac{f_{2}-f_{1}}{f} \frac{\left(f_{1}-f\right)\left(f_{2}-f\right)}{f_{1} f_{2}-f_{1} f-f_{2} f_{1}+f_{2} f}=\frac{f_{2}-f_{1}}{f} \frac{\left(f_{1}-f\right)\left(f_{2}-f\right)}{f_{2} f-f_{1} f} \\ &=\frac{f_{2}-f_{1}}{f} \frac{\left(f_{1}-f\right)\left(f_{2}-f\right)}{f\left(f_{2}-f_{1}\right)}=\frac{\left(f_{1}-f\right)\left(f_{2}-f\right)}{f^{2}}.
\end{aligned}
\]
Võrreldes saadud avaldist varem saadud avaldistega $k_1$ ja $k_2$ jaoks, on lihtne näha, et
\[
k=\frac{f_{1}-f}{f} \frac{f_{2}-f}{f}=k_{1} k_{2}.
\]
\emph{Märkus:} mugavuse kaalutlusel vaatleme joonisel tegelikult vaid poolt pliiatsit, aga on ilmne, et lahenduskäik kehtib ka terve pliiatsi kohta (joonisel on pliiatsi teine pool näidatud punktiirjoonega).

Lääts

Teritamata pliiatsi telg ühtib koondava läätse peateljega. Mitu korda on pliiatsi kujutise pikkus tema enda pikkusest erinev, kui pliiatsi ühe otsa kujutise diameetri ja pliiatsi diameetri suhe on $k_1$ ning teise otsa jaoks on see suhe $k_2$? Pliiatsi mõlemad otsad asuvad läätsest kaugemal kui on läätse fookuskaugus.

Suhteliselt kindel meetod on avaldada kõik ülesandes antud ja otsitavad muutujad võimalikult mugavate jooniselt leitavate suuruste kaudu ning loota, et saadud võrranditest on lihtne näha, kuidas lõppvastus avaldub $k_1$ ja $k_2$ kaudu. Kehvemal juhul tuleb lahendada kolmest võrrandist koosnev võrrandisüsteem ($k_1$, $k_2$ ning otsitava suhte jaoks). Silmas peab pidama, et jooniselt valitud muutujad peavad ülesande geomeetria üheselt ära defineerima. Selleks sobivad näiteks läätse fookuskaugus ning pliiatsi mõlema otsa kaugused läätsest.


[User]
7. ELECTRONs ıN MAGNETıC FıELD (9 points)

— Kaarel Hänni, Jaan Kalda.
Inwhatfollowsweconsidertwoelectrons
(of mass m and charge −e) moving in homo‐
geneous magnetic field of flux density B in
suchawaythatthedistancebetweenthetwo
electrons remains always constant; different
subtasks explore different possibilities.
i) (2 points) The distance between the elec‐
trons is so large that their electrostatic repul‐
sion can be neglected. At a certain moment,
the angle between the velocity vectors of the
electrons is α ̸= 0, and one of them moves
in the direction of the other with speed v.
Sketch the trajectories of the both electrons.
What is the speed of the other electron?
ii) (1 point)The electrostatic repulsion is still
to be neglected. Now, the trajectories of the
two electrons intersect, and at a certain mo‐
ment, the velocity of one of them is ⃗ v. What
can be said about the velocity of the other
electron at the same moment? Sketch the
both trajectories.
iii) (2 points) What is the distance l between
the electrons if both of them perform a peri‐
odic motion of period 6πm
Be while moving
with a constant speed?
iv) (2 points)The figure below depicts the tra‐
jectoryofoneoftheelectronsforacertainset
of initial conditions. Where is the other elec‐
tron at the moment when the first electron is
at the position of the small circle? What is
the period of this motion?
v) (2 points)The figure below depicts the tra‐
jectoryofoneoftheelectronsforacertainset
ofinitialconditions. Whatisthespeedofthe
other electron at the moment when the first
electron is at the position of the small circle?

Solution:
7. ELECTRONs ıN MAGNETıC FıELD

— Solution by Taavet Kalda.
i) (2 points) Due to Lorentz force not acting
along the direction that’s parallel to B, the
condition for electrons to stay the same dis‐
tance along that axis becomes simply that
theirvelocitycomponentsalongB areequal.
We will henceforth only consider the motion
on the plane that’s perpendicular to B.
Neglecting electrostatic interactions,
electrons move on circular trajectories of
frequency ω given by the right hand rule,
found from the centrifugal and Lorentzforce
balance:
mv2
R
= mvω = veB,
hence ω0 = eB/m. In other words, the angu‐
lar frequency for both electrons is the same
and independent of their speeds. We also get
an expression for the radius of the circular
trajectory: R = v/ω0 = mv/(eB).
An important consequence is that the
angle between the velocities of the two elec‐
trons stays constant in time. Hence, the re‐
lative velocity between the two electrons is
also constant in magnitude (and not zero!)
androtatingwithω0. Fortherelativevelocity
to keep the distance between electrons con‐
stant, the distance vector between the two
electrons must be perpendicular to the rel‐
ative velocity and hence also rotate with an‐
gular speed ω0. This is enough to conclude
that the only way this is satisfied is when the
two electrons move on concentric circular
trajectories.
The condition for the relative velocity to
be perpendicular to the distance vector gives
us that the speed of the other electron is u =
v/cosα. The radii of the two circles are then
R1 = v/ω0 and R2 = v/(cosαω0). The tra‐
jectories are illustrated below.
Grading:
• Findingthecyclotronfrequencyandradius
of curvature from force balance (0.5pts)
• Showing that the two orbits must be con‐
centric, of which
– deducing that the angle between velocit‐
ies is constant (0.3pts)
– showing that the displacement traject‐
ory rotates with ω0 on a circular traject‐
ory (0.4pts)
– concludingthattheorbitsareconcentric
(0.2pts)
• Correct sketch of the trajectories (0.3pts)
• Correct expression for the speed of the
other electron (0.3pts)
ii) (1 point) The class of solutions we found
in the previous part does not cover the case
where the two trajectories intersect. How‐
ever, the previous part assumed that α ̸= 0.
Hence, for the trajectories to intersect, we
need α = 0 and u = v. In other words, ⃗ u = ⃗ v.
This leaves us complete freedom in the loca‐
tions of the centres of the two trajectories, as
long as they intersect. A sketch of a potential
trajectory is shown below
Grading:
• Deducing that α = 0 and u = v (0.4pts)
• Concluding that ⃗ u = ⃗ v (0.3pts)
• Sketch (0.3pts)
iii) (2 points) The frequency of the periodic
motion is 2π/T = eB/(3m) = ω0/3, i.e.
a third of the cyclotron frequency. Now,
it’s clear that a simple solution that can sat‐
isfy the electrons being equidistant and with
equal speeds is simply the one where they
move on the same circular trajectory, but in
the opposite phase. We proceed to reason
why this is the only potential solution.
Thenetforceonanyoftheelectronsmust
be perpendicular to the velocity of the said
electron (otherwise v ̸= const). The Lorentz
force is automatically perpendicular to the
velocity, and hence we need the Coulomb
force to also be perpendicular to the velo‐
city. This means that the displacement vec‐
tor has to be perpendicular to the velocities
and this, combined with the electrons being
equidistant, gives us the aforementioned cir‐
cular solution.
The radial force balance equation reads
mω2
l
2
= evB −
ke
l2
,
where negative sign points outward (Lorentz
force must be positive signed for the elec‐
trons not to repulse each‐other). Using v =
ωl/2 and substituting ω, we get
e2
B2
l
18m
=
e2
B2
l
2m
−
ke2
l2
and so
l =
3
√
9km
4B2
.
Grading:
• Showing that the electrons must be orbit‐
ing each‐other on a circular orbit (0.7pts)
• Radial force balance (0.8pts)
• Final answer (0.5pts)
iv) (2 points)In order to analyse the trajector‐
ies given on the figure, we need some basic
understanding of their underlying dynam‐
ics. As compared to the previous task, the
condition for the centre of mass to be at rest
is now relaxed. We start in the most general
formandwritedowntheforcesactingonthe
two electrons, 1 and 2:
m ̇ ⃗ v1 = −
ke2
l2
ˆ l + e⃗ v1 × ⃗ B
m ̇ ⃗ v2 =
ke2
l2
ˆ l + e⃗ v2 × ⃗ B,
where⃗ l = ⃗ r2 − ⃗ r1 is the displacement vector
from 1 to 2 and ˆ l is the corresponding unit
vector. We can cancel out the Coulomb force
by adding the two equations together:
m( ̇ ⃗ v1 +  ̇ ⃗ v2) = e(⃗ v1 +⃗ v2) × ⃗ B.
Thiscanbefurthersimplifiedbysubstituting
thecentreofmassvelocity⃗ vCM = (⃗ v1+⃗ v2)/2:
m ̇ ⃗ vCM = e⃗ vCM × ⃗ B.
This mirrors exactly the equation of motion
of a single electron, meaning the centre of
mass moves on a circular trajectory with fre‐
quency ω0, even in the presence of the Cou‐
lomb force.
Weproceedwithasimilaranalysisforthe
difference of velocities:
m( ̇ ⃗ v1 −  ̇ ⃗ v2) = −2
ke2
l2
ˆ l + e(⃗ v1 −⃗ v2) × ⃗ B.
Substituting in the velocity of 1 w.r.t. CM
∆⃗ v1 = ⃗ v1 −⃗ vCM = (⃗ v1 −⃗ v2)/2 we get
m∆ ̇ ⃗ v1 = −
ke2
l2
ˆ l + e∆⃗ v1 × ⃗ B
which, once again, mirrors the equation of
motion of the previous sub‐task. Explicitly,
the EoM of the previous part was
m ̇ ⃗ v = −
ke2
l2
ˆ l + e⃗ v × ⃗ B.
Therefore, ∆⃗ v1 rotates with an angular fre‐
quency ω that’s smaller than ω0.
In conclusion, the motion of an electron
is the superposition of the circular trajectory
of the centre of mass of radius |⃗ R1| = R1 and
frequency ω0 and the circular motion of the
electron of radius |⃗ R2| = R2 and frequency
ω < ω0 around the centre of mass. The other
electronisdiametricallyoppositearoundthe
centre of mass, separated by a distance R2
from the first electron. R1, R2 and ω are free
parameters.
We finally continue with the graph given
in the statement. Since the trajectory makes
one hoop in a full period, ω0 must be an in‐
teger multiple of ω. If this wasn’t the case,
both ⃗ R1 and ⃗ R2 must make more than one
full rotation in a period and the result would
be a hoop that goes around itself more than
once (possibly self‐intersecting in the pro‐
cess). Thus, ω0 = Nω, where N ∈ {2,3,...}
(N = 1 gives a circle).
For the next step, let’s consider the ef‐
fect of radius of curvature, an easily observ‐
able property of the trajectory. In general,
the bigger the speed, the bigger the radius
of curvature (Coulomb force gives a constant
contribution to the radial force, and Lorentz
force has a monotonously increasing rela‐
tion between R and v). Hence, the point
on the trajectory with the biggest radius of
curvature has the highest speed (marked on
the figure with B), and vice‐versa (marked
with A). Further, the biggest speed happens
when ⃗ R1 and ⃗ R2 are parallel, and the smal‐
lest when they’re antiparallel (having speeds
ω0R1 + ωR2 and |ω0R1 − ωR2| respectively).
Now, in a full period, ⃗ R2 goes around once,
and ⃗ R1 goes around N times. Therefore, ⃗ R1
overtakes ⃗ R2 a total of N − 1 times and we
expect to see N − 1 occurrences of maximal
speed, i.e. maximal radius of curvature. Be‐
cause we see this happen once, N = 2. This
immediately gives the period of the motion
to be T = 2π/ω = 4π/ω0 = 4πm/(eB).
From the previous theory, |AB| = 2R2,
and the centre point O1 between A and B
is where ⃗ R1 intersects with the axis of sym‐
metry. The other point where ⃗ R1 intersects
with the axis is when ⃗ R1 and ⃗ R2 are per‐
pendicular. This can be found by finding
the points (using a ruler) which are |AB|/2
away from the principal axis (we mark these
by C and D). There are two solutions, the
right one of which is unphysical (as can be
seenbythefollowingconstructionsbreaking
down). Having found the other intersection
pointO2 (whichhappenstocoincidewithA),
we can fully reconstruct ⃗ R1 (colored blue)
and from there it’s easy to find the centre
of mass corresponding to the marked elec‐
tron (by finding the point on ⃗ R1 which is a
distance R2 from the marked point using a
compass)andhencethelocationoftheother
electron (by mirroring the marked electron
w.r.t. the centre of mass), marked cyan.
Alternative solution.
An alternative solution with simpler geo‐
metric operations, but more complex al‐
gebra would follow a similar line of reas‐
oning until N = 2. After that, one
can show that the two electrons follow
the same orbit by expressing the locations
of the two electrons via complex numbers
z± = R1 exp(2iωt) ± R2 exp(iωt). This is
a commonly deployed method to simplify
vector operations. The real part of the
complex number is the x‐coordinate, and
the imaginary the y‐coordinate. Now, if
we apply a phaseshift of π to ωt, z+ be‐
comes z− and vice‐versa. This is because
exp(2i(ωt + π)) = exp(2iωt) · exp(2πi) =
exp(2iωt) and exp(iωt + iπ) = −exp(iωt).
Hence, z+ and z− follow the same trajectory
butwithaπ phaseshift,exactlyaswewanted.
Now, the location of the other electron
can be simply found by drawing a circle of
radius |AB| from the first electron and see‐
ing where it intersects with the trajectory.
Grading:
• Analysis of the dynamics of the system, of
which:
– Equations of motions for both electrons
(0.4pts)
– Deducing that the centre of mass moves
on a circular orbit of frequency ω0
(0.3pts)
– Deducing that the electrons orbit centre
of mass with frequency ω < ω0. (0.4pts)
If the student implicitly assumes ω > ω0
but does the rest correctly (including the
constructions), award a maximum of 1.5
points.
• Analysing the trajectory to show that ω =
ω0/2. (0.3pts)
• Using points A and B to get 2R2 (0.3pts)
• UsingC andDtofindR1 andhencetheloc‐
ationoftheotherelectron(alternativelyus‐
ing |AB| and reasoning that the electrons
must share a trajectory) (0.3pts)
v) (2 points) Based on the previous reason‐
ing, we see that the trajectory has two points
withmaximalradiusofcurvature,andhence
N = 3, i.e. the same as in part iii). Now, the
first electron is at the inflection point where
it’s momentarily at rest (before being pushed
into motion by Coulomb force). Also, since
it’s the point with highest curvature, we have
v1 = |ω0R1 − ωR2| = 0 so ω0R1 = ωR2. Be‐
cause the other electron is diametrically op‐
positew.r.t. centreofmass, it’satthepointof
highest speed, i.e. v2 = ω0R1 +ωR2 = 2ωR2.
Now, because the equation of motion de‐
fining R2 is the same as in part iii) (as high‐
lightedintheprevioussubtask),wecanreuse
the result form that part to get
R2 =
l
2
=
3
√
9km
32B2
and so
v2 = 2ωR2 =
2eB
3m
R2 = e
3
√
kB
12m2
.
Grading:
• Deducingthatω = ω0/3(thetoolsetforthis
should’ve been developed in the previous
part) (0.3pts)
• Finding that the highlighted electron is at
rest (0.3pts)
• Rest condition ω0R1 = ωR2 (0.3pts)
• Expressing the speed of the other electron
as 2ωR2 (0.3pts)
• Solving the force balance equation for
R2 (or reusing the result from part iii))
(0.7pts)
• Final expression (0.1pts)

ELECTRONs ıN MAGNETıC FıELD


[User]
A painter of mass M stands on a platform of mass m as shown.

    He pulls each rope down with force F , and accelerates upward with acceleration a. Find a.

Solution:
A painter of mass M stands on a platform of mass m as shown.

    He pulls each rope down with force F , and accelerates upward with acceleration a. Find a.

    Answer. −g + 4F/(M + m)


[User]
A multimeter is the most complicated piece of equipment that ever shows up on
    Olympiad labs. Get a cheap multimeter – something like this will do. By reading the manual
    and experimenting, teach yourself what every setting on the multimeter does, and how to read the
    display. Test your knowledge by performing a variety of measurements on cheap circuit components,
    such as batteries, resistors, capacitors, and diodes. Be careful not to burn out the ammeter.

Solution:
A multimeter is the most complicated piece of equipment that ever shows up on
    Olympiad labs. Get a cheap multimeter – something like this will do. By reading the manual
    and experimenting, teach yourself what every setting on the multimeter does, and how to read the
    display. Test your knowledge by performing a variety of measurements on cheap circuit components,
    such as batteries, resistors, capacitors, and diodes. Be careful not to burn out the ammeter.


[User]
oscillations.
                    01T   IPhO 2004, problem 3. A practical problem which also reviews damped/driven

Solution:
oscillations.
                    01T   IPhO 2004, problem 3. A practical problem which also reviews damped/driven


[User]
A comet passes by Sun as shown, in a parabolic path.

    How long, in years, does the comet take to get from point A to point B? (Hint: if you apply Kepler’s
    laws and properties of conics, this problem can be done with almost no computation.)

Solution:
A comet passes by Sun as shown, in a parabolic path.

    How long, in years, does the comet take to get from point A to point B? (Hint: if you apply Kepler’s
    laws and properties of conics, this problem can be done with almost no computation.)
               √
    Answer. 2 2/3π years


[User]
Consider an electric charge q placed at x = 0 and a charge −q placed at x = d. The
    electric field along the x axis is then

                                                q    1     1
                                        E(x) =         −          .
                                               4πϵ0 x2 (x − d)2

    For large x, use the binomial theorem to approximate the field.

Solution:
Use the binomial theorem with d/x ≪ 1 to get

                                         1       1       2d
                                               =     1 +      .
                                      (x − d)2   x2      x

    Then
                                                    2qd          qd
                                        E(x) = −          3
                                                            =−         .
                                                   4πϵ0 x      2πϵ0 x3
    This is the on-axis field of an electric dipole.


[User]
7. Mass-spectrometer (9 points)

In the figure below, a simplified scheme of a
mass-spectrometer is given. It is a device for
measuring the masses of molecules. The sub-
stance under investigation is ionised by heat-
ing up to a temperature T on a hot filament
(molecules undergo a single-electron ionisation).
The ions are accelerated using voltage U. At
first, let us neglect the thermal energy of the
ions (eU ≫ kT, where e is elementary charge
and k the Boltzmann's constant). A nar-
row beam of accelerated ions enters a region
with magnetic field. For the sake of simpli-
city, let us assume that the region has a rect-
angular shape, and the magnetic field is ho-
mogeneous inside it. The magnetic fields de-
flects the ions and depending on their mass, they
may hit the detector. Let us assume that those
ions which hit the centre of the detector enter
and exit the region with magnetic field perpen-
dicularly with its boundary, and the distance
between the entry and exit points is l (see figure)
.
1) Express the mass M of those ions which hit
the centre of the detector via the quantities B,
l, U and e.
2) The entrance to the detector is a circle with
radius r (r ≪ l). Because of the finite size of the
detector entrance, ions within a mass range from
M − ∆M to M + ∆M will enter the detector;
find the range width ∆M.
3) Under the assumptions of the previous ques-
tion, what is the width of the range of the exit
angles ∆φ of those ions which can still hit the
detector?
4) Now let us assume that r is negligibly small,
but we cannot neglect the thermal eects (still
eU ≫ kT). Due to dierence in energies, the
detector can be hit by ions of dierent masses,
within the range from M − δM to M + δM.
What is the resolving power δM of the mass-
spectrometer?

Solution:
7. Mass-spectrometer

1) The trajectory of a charged particle in the magne-
tic field is circle of radius R = l/
√
2. Lorenz force
is responsible for the acceleration, Bev = Mv2
/R,
hence BeR = p. Substituting p2
= 2MUe =
B2
e2
R2
, we obtain
M = B2
l2
e/4U.
2)Now, the radius can be R±r. Approximate calcu-
lus yields ∆R/R = r
√
2/l ≈ ∆M/2M, hence
∆M ≈ Mr2
√
2/l, i.e.
∆M = B2
lre/
√
2U.
3) Ion leaves the magnetic field at the distance r be-
fore (or after) performing a quarter of the circle. So,
∆φ ≈ r/R = r
√
2/l.
4)CertaininitialenergykT implies thattheterminal
energyUe+kT = e(U+kT/e);this isequivalent to
the change of the voltageby δU = kT/e. Using app-
roximate calculus and the result of the first question,
we obtain: δM = dM
dU
kT/e,i.e.
δM = B2
l2
kT/4U2
.

Mass-spectrometer


[User]
A frictionless tube of length R is rotated with fixed angular velocity ω about one of
    its ends. A package is placed in the tube at a distance r0 from the axis of rotation, with no initial
    radial velocity. When the package flies out the other end of the tube, what is its speed?

[2] Problem 21 (Cahn). A pendulum is designed for use on a gravity-free spacecraft. The pendulum
    consists of a mass at the end of a rod of length ℓ. The pivot at the other end of the rod is forced to
    move in a circle of radius R with fixed angular frequency ω. Let θ be the angle the rod makes with
    the radial direction.

    Show this system behaves exactly like a pendulum of length ℓ in a uniform gravitational field
    g = ω 2 R. That is, show that θ(t) is a solution for one system if and only if it is for the other.

Solution:
In the frame rotating with the tube, there is only a centrifugal force, which produces a
    radial acceleration a = ω 2 r. The solution to this equation is

                                             r(t) = Aeωt + Be−ωt

    and fitting to the initial conditions r(0) = r0 and v(0) = 0 gives

                                  r(t) = r0 cosh(ωt),    v(t) = ωr0 sinh(ωt).

    When the package flies out the end, we have R = r0 cosh(ωt0 ), which implies
                                                    q
                                          v(t0 ) = ω R2 − r02 .

    Of course, we have to go back to the lab frame, where there’s a tangential speed ωR, giving
                                                 q
                                            v = ω 2R2 − r02 .

[2] Problem 21 (Cahn). A pendulum is designed for use on a gravity-free spacecraft. The pendulum
    consists of a mass at the end of a rod of length ℓ. The pivot at the other end of the rod is forced to
    move in a circle of radius R with fixed angular frequency ω. Let θ be the angle the rod makes with
    the radial direction.

    Show this system behaves exactly like a pendulum of length ℓ in a uniform gravitational field
    g = ω 2 R. That is, show that θ(t) is a solution for one system if and only if it is for the other.
    Solution. This system experiences no gravitational force, but instead experiences a Coriolis and
    centrifugal force. The Coriolis force plays no role, because it is always perpendicular to the velocity
    of the mass and the angular velocity, which implies it is directed along the rigid rod; it merely
    changes the tension in the rod.
         The centrifugal acceleration a is directed away from the origin; the relevant part of it is the
    component a⊥ perpendicular to the rod. Referring to the below diagram, we see that a⊥ =
    ω 2 (R sin θ).

    This is exactly the same a⊥ as for a pendulum in gravity g = ω 2 R, so the systems are equivalent.


[User]
Staircase
The equilibrium shape of bodies in zero-gravity is determined by the minimum of their surface
energy. Thus, for example, the equilibrium shape of a water droplet turns out to be spherical: the
sphere has the smallest surface area among bodies of the same volume. At low temperature, the
equilibrium shape of crystals may have flat facets. The parts of the crystal surface that have a
small angle φ with the facet are in fact staircases of rare steps on this facet. The height of such
steps is equal to the period of the crystal lattice h.

Equilibrium surface profile y (x) of a certain crystal and the corresponding microscopic staircase
are shown schematically in the figure, where n denotes the step number, counting from x = 0.
                                                                     3
The profile shape at x > 0 can be approximated as y (x) = − λx 2 h , where λ = 45 μm and
h = 0.3 nm.

a) Express the distance dn between two adjacent steps as a function of n for n fl 1.

b) The interaction energy E of two steps depend on the distance d between them as

                                             E (d) = μdν

   where μ is a constant. Assume that only adjacent steps interact. Find the numerical value for
   the exponent ν.

Solution:
Solution - Staircase
   A Since n = −y/h = (x/λ)2/3 , x(n) = n2/3 λ. The distance between the steps is:

                                                         dx  2
                          dn = x(n + 1) − x(n) =≈           = λn−1/3 = n−1/3 · 30 μm.
                                                         dn  3
   B Equilibrium energy value, being minimum, must be stable against small perturbations of the
     crystal shape. Allowed are perturbations which conserve the total volume of the crystal. In
     other words a small horizontal displacement of one step must be accompanied by an equal
     and opposite displacement of another step.
       The energy change εn (δ) associated with a small horizontal displacement δ of the n-th step
       is:

                    εn (δ) = μ (dn + δ)ν − dνn + (dn+1 + δ)ν − dνn+1 ≈ μν(dν−1 − dν−1

                                                                           n      n+1 )δ.

       In order for εn (δ) + εm (−δ) to be zero for arbitrary n and m it is necessary to require that
       the factor in the parentheses does not depend on n:

                                               dν−1
                                                n      n+1 = const.
                                                    − dν−1

       Substituting d ∝ n−1/3 , we get1 :

                                                               1 − ν 1−ν −1
                                n(1−ν)/3 − (n + 1)(1−ν)/3 ≈         n 3     = const,
                                                                 3
                                            1−ν
                                                − 1 = 0 =⇒ ν = −2.
                                             3

       The interaction energy corresponds to that of two dipoles in 2D:

                                                              1
                                                     E(d) ∝      .
                                                              d2

   1
    Trivial solutions ν = 0 and ν = 1 imply that the total energy within given constraints does not depend on the
shape of the crystal.

Staircase