2023年泛珠三角及中华名校物理奥林匹克邀请赛（Word版含解析）

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1
Pan Pearl River Delta Physics Olympiad 2022
2023 年泛珠三角及中华名校物理奥林匹克邀请赛
Sponsored by Institute for Advanced Study, HKUST
香港科技大学高等研究院赞助
Simplified Chinese Part-1 (Total 4 Problems, 40 Points) 简体版卷-1 (共4题， 40分) (9:30 am – 12:00 pm, 29th Jan 2023)
Please fill in your final answers to all problems on the answer sheet. 请在答题纸上填上各题的最后答案。 At the end ofthe competition, please submit the answer sheet only. Question papers and working sheets will not be collected. 比赛结束时，请只交回答题纸，题目纸和草稿纸将不会收回。
1. [10 points] Consider a satellite that has a shape of a plate of surface area A and thickness d VA. The satellite can convert solar energy into electrical energy and charge the onboard batteries making use of the temperature difference. The solar energy flux density [solar power per unit area] at the position of the satellite is S and the satellite is facing towards Sun. Assuming that the emissivity of both sides of the satellite is e and the temperatures of the satellite on two sides are T1 and T2 respectively and is the Stefan-Boltzmann constant.
1. [ 10 分] 考虑一个人造卫星，其形状为表面积为 A 且厚度为d VA 的平板。卫星可以利用温差将太阳能转化为电能，为星载电池充电。人造卫星所在位置的太阳能通量密度[单位面积的太阳能功率]为S ，卫星正对着太阳。假设整颗卫星的发射率 (又称辐
射率) 为e ，卫星两侧的温度分别为 T1 和 T2 ，是 Stefan-Boltzmann 常数。
(a) [2] What is the net heat flux [energy per second] absorbed by the bright side (the side facing towards Sun) of the satellite
(a) [2]卫星的亮面(面向太阳的一面) 吸收的净热通量 [每秒能量] 是多少？
(b) [1] What is the net heat flux [energy per second] released from the dark side of the satellite
(b) [1] 从卫星暗面释放的净热通量 [每秒能量] 是多少？
(c) [1] What is value of the emissivity e to get the theoretically maximal charging power Pmax
(c) [1]可得到理论上最大充电功率 Pmax 的发射率 e 的值是多少？
(d) [3] Find a condition for the temperature T1 in order to get the theoretically maximal charging power Pmax provided by the satellite. Express the condition in term of the dimensionless variable x = . You don’t need to solve the equation in this part.
(d) [3] 求温度T1 的条件，使得卫星能達到理论上最大充电功率 Pmax 。用无量纲变量 x = 表示你的答案。你不需要解这部分的方程。
(e) [3] What is the theoretically maximal charging power Pmax provided by the satellite Calculate the numerical value of
(
AS
.
Your
answer
should
be
correct
to
at
least
2
significant
figures
.
)Pma%
(e) [3] 卫星可提供的理论上最大充电功率Pmax 是多少？计算的数值。你的答案至少应正确到 2 位有效数字。
1
1
2. [10 points] Binary-sun solar system: Consider a binary pair of identical suns of mass M orbiting in the x y plane in an orbit centred at the origin. The gravitational constant is G . Now add a planet of mass m with an initial condition on the z axis above the center of mass of the two suns and with a velocity along the z direction. By the symmetry of the system, the small planet will remain on the z axis, suns will have equal z coordinates and the center of mass of two suns will also remain on the z axis.
2. [10 分]双星太阳系：考虑两个质量均为 M 的相同太阳在 x
y 平中以原点为中的轨道运。万有引常数为 G 。现在
添加个质量 m 的星，其初始条件位于两个太阳质上的
z 轴上，速度在 z 向上。通过系统的对称性，星将保持
在 z 轴上，两个太阳的 z 坐标相同，其质也维持在 z 轴上。
We use the Cartesian coordinates to describe the dynamics of the system: the coordinate of the planet (0,0, u), and the coordinates of two suns are (±R cose , ±R sine , Z).
我们笛卡尔坐标来描述系统的动学：星坐标为(0,0, u) ，两个太阳的坐标为(±R cose , ±R sine , Z) 。 (a) [0.5] What is the total kinetic energy, T, of the system Express the answer in terms of R, e, Z, u and their time derivative.
(a) [0.5] 系统的总动能 T 是多少？ R, e , Z, u 及其时间导数表答案。
(b) [0.5] What is the total potential energy, V, of the system Express the answer in terms of R, e, Z, u and their time derivative.
(b) [0.5] 系统的总势能 V 是多少？ R, e , Z, u 及其时间导数表答案。
(c) [0.5] What is the total linear momentum, P, of the system Express the answer in terms of R, e, Z, u and their time derivative.
(c) [0.5] 系统的总线性动量 P 是多少？ R, e , Z, u 及其时间导数表答案。
(d) [0.5] What is the total angular momentum, L, about the z axis of the system Express the answer in terms of R, e, Z, u and their time derivative.
(d) [0.5] 关于系统 z 轴的总动量 L 是多少？ R, e , Z, u 及其时间导数表答案。
From now on, we introduce the dynamical variables q(t) = u Z and the center of mass coordinate of the system Q(t) = .
从现在开始，我们引入动学变量q(t) = u Z 和系统的质坐标 Q(t) = . 。
(e) [2] Find the equation of motion for q(t). Express your answer in terms of , q, R and given physical parameters. (e) [2] 找出 q(t) 的运动程。 , q, R 和给定的物理参数表达你的答案。
(f) [2] Find the equation of motion for R (t). Express your answer in terms of , R, q, L and given physics parameters. (f) [2] 求出 R (t) 的运动程。 , R, q, L 和给定的物理参数表达你的答案。
(g) [2] In the limit of small planetary mass m M we can ignore the effect of the planet on the motion of the suns. Find the explicit solution for the motion of the suns R(t) for orbits with small eccentricity e 1. Write your solution as circular motion plus a term proportional to e, i.e. R (t) = R2 + eR1 (t) where R2 is the radius of the circular orbit. You can assume the initial condition R (0) = R2 (1 + e).
(g) [2] 在星质量 m M 的极限下，我们可以忽略星对太阳运动的影响。对于偏率 e 1 的轨道，找到太阳运动 R (t) 的显式解。将你的答案写成圆周运动加上与e 成比例的项，即 R (t) = R2 + eR1 (t) ，其中 R2 是
圆形轨道的半径。你可以假设初始條件 R (0) = R2 (1 + e) 。
(h) [2] Obtain the equation of motion for q(t) in the limit m M . You can see that q(t) is a nonlinear oscillator driven by a nonlinear force term.
(h) [2] 求 q(t) 在极限m M 时的运动程。你可以看到q(t)是个由非线性项驱动的非线性振荡器。
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1
3. [10 points] A massless rod can rotate without friction about the pivot point at its center. Light, propagating as a plane wave, propagates from left to right, along the x axis. The electric field of the light is given by
3. [10 分] 根质量的杆可以绕其中的枢轴点摩擦地旋转。平光波沿x 轴从左向右传播。光的电场由下式给出
(x, t) = E2 cos(kx 业t)
where = k and E2 is a real number. The angle between the
rod and is denoted by e.
其中 = k 且 E2 为实数。杆和之间的度 e 表。
Cantered at the ends of the rod are disks, each with one side perfectly mirror with 100% reflection and the other side with 100% absorbing. The disks are oriented so that light in the upper part of the rob (above the pivot) always strikes an absorptive surface, while in the lower part, it strikes a reflective surface. Each of the disks have mass m and radius T . Assume that the distance R from the pivot to the center of each disk satisfies R T .
在杆的两端是圆盘，每个圆盘的侧是 100% 反射的完美镜，另侧 100% 吸收。圆盘的向使得杆上部(枢轴上 )的光总是照射到吸收，在下部，它照射到反射。每个圆盘的质量为 m ，半径为 T 。假设枢轴点到每个圆盘中的距离R 满 R T 。
The Poynting vector, which describes the energy flux density (i.e. the energy per unit area per unit time) is given by 描述能量通量密度(即每单位时间每单位积的能量)的坡印亭量由下式给出
= U × X,
where u2 is the vacuum permeability. The momentum density carried by the EM wave is
其中 u2 是真空磁导率。光波携带的动量密度为
= ,
where c is the speed of light in vacuum.
其中 c 是真空中的光速。
(a) [ 1] What are the frequency f, wavelength 入, and magnetic field (x, t) of the light
(a) [1] 光的频率 f 、波入和磁场 (x, t) 分别是什么？
(b) [ 1] What is the time-averaged Poynting vector of the incident light
(b) [1] 入射光的时间平均坡印廷量是什么？
(c) [6] What is the total torque which is delivered by the light to the system of rod plus disks around the pivot point at a given angle e What is the average torque over a full rotation of the rod
(c) [6] 在给定度 e 下，光传递到围绕枢轴点的杆加圆盘系统的总扭矩是多少？杆旋转圈的平均扭矩是多少？
(d) [2] Find the average angular acceleration of the rod over a revolution.
(d) [2] 求杆旋转圈的平均加速度。
3
1
4. [10 points] A vertical, insulated and sealed cylinder with a cross-sectional area A , and an insulating piston of mass m inside, whose thickness is negligible compared with the length of the cylinder . At the beginning, the piston is fixed in the center of the cylinder, which divides the cylinder into two air chambers with the same length l , as shown in the figure . Assume that the upper and lower gas chambers of the cylinder each contain n moles of monatomic ideal gas with temperature T2 . In the following problems, it can be assumed that there is little friction between the piston and the cylinder wall, and that l is much larger than the distance traveled by the piston ( z) .
4. [10 分] 有垂直豎立的密閉絕熱圓筒，截積為 A ，內裝有質量為m的
絕熱活塞，其厚度和圓筒的長度相比，可忽略不計。起始時，活塞被固定在
圓筒的中央，將圓筒分隔成兩個長度同為 l 的氣室，如圖所。設圓筒的上
下兩氣室各含有溫度為T2 和n 摩耳的單原分理想氣體。在下列的問題
中，可以假定活塞與圓筒壁之間的摩擦很少，且 l 遠於活塞所移動的距離 (l z) 。
(a) [4] Release the piston from rest at time t = 0 so that it can move freely up and down . Find the trajectory of the piston z(t) . You can neglect the friction between the piston and the cylinder in the part .
(a) [4] 在時間 t = 0時從静中釋放活塞，使其能由上下運動，找出活塞的軌跡 z(t) 。在这部分，你可以忽略活塞和缸之间的摩擦。
(b) [2] How does the temperature of the upper and lower chambers in the cylinder change with the position of the piston, z
(b) [2] 圓筒內上下两氣室的溫度如何隨活塞位置 z 的變動改變
(c) [4] Although there is little friction between the piston and the cylinder, the piston will eventually come to rest after a long time . Find the position of the piston , zf , when it rests and the temperature, Tf , of the gas in the cylinder at that time. . We can assume that the heat capacity of the cylinder and the piston is negligible, the temperature of the upper and lower chambers will eventually come to the same because of the movement of the piston and all heat lost due to friction will transfer into the internal energy of the gas .
(c) [4] 活塞與圓筒壁間的摩擦雖然很少，但經過段長時間後，终會使活塞靜下來。試求活塞最後靜時的位置 zf 和其時筒內氣體的温度 Tf？我们假設圓筒壁和活塞的熱容量可忽略不計，且由於活塞的運動，使上下氣室溫度最後趨於致，且由于摩擦损失的所有热量都将转化为体的内能。
~ End of Part 1 卷-1 完 ~
42
Pan Pearl River Delta Physics Olympiad 2023
2023 年泛珠三角及中华名校物理奥林匹克邀请赛
Sponsored by Institute for Advanced Study, HKUST
香港科技大学高等研究院赞助
Simplified Chinese Part-2 (Total 2 Problems, 60 Points) 简体版卷-2 (共2题，60分)
(1:30 pm – 5:00 pm, 29 January 2023)
All final answers should be written in the answer sheet. 所有最后答案要写在答题纸上。 All detailed answers should be written in the answer book. 所有详细答案要写在答题簿上。 There are 2 problems. Please answer each problem starting on a new page. 共有 2 题，每答 1 题，须采用新一页纸。 Please answer on each page using a single column. Do not use two columns on a single page. 每页纸请用单一直列的方式答题。不可以在一页纸上以双直列方式答题。 Please answer on only one page of each sheet. Do not use both pages of the same sheet. 每张纸单页作答。不可以双页作答。 Rough work can be written in the answer book. Please cross out the rough work after answering the questions. No working sheets for rough work will be distributed. 草稿可以写在答题簿上，答题后要在草稿上划上交叉，不会另发草稿纸。 If the answer book is not enough for your work, you can raise your hand. Extra answer books will be provided. Your name and examination number should be written on all answer books. 考试中答题簿不够可以举手要，所有答题簿都要写下姓名和考号。 At the end of the competition, please put the question paper and answer sheet inside the answer book. If you have extra answer books, they should also be put inside the first answer book. 比赛结束时，请把考卷和答题纸夹在答题簿里面，如有额外的答题簿也要夹在第一本答题簿里面。
1
2
Problem 1: Vacuum bubbles (28 points)
问题 1: 真空泡泡 (28 分)
Is our vacuum stable We don't know. It's possible that we do not live in the true vacuum. Rather, we live in a false vacuum which can decay into true vacuum by emerging and expanding bubbles. To describe such a possibility, we will make use of a space-time dependent "scalar field" (t, x, y, z), which takes a real value at every space-time point. (Similar to height on a map, which takes a real value at every point on the x-y plane, while a scalar field takes a real value for any given t, x, y, z . Also, in a full quantum theory, we have to distinguish operators and numbers, but here we will assume the scalar field only take real number values.)
我们的真空是稳定的吗？我们不知道。有可能我们并不是生活在一个稳定的真真空(既真的“真空”)里面。我们可能生活在一个假真空里，而假真空可以通过自发产生和膨胀的泡泡衰变到真真空。为了描述这种可能，我们将使用一个依赖于时空点到“标量场”(t, x, y, z) ：它在每个时空点上取一个实数。(这有点像地图上的高度，在 x-y 平面的每个点上取一个实数。不过标量场是在每个t, x, y, z 时空点上取一个实数。另外，在完整的量子理论中，我们需要考虑算符和数的区别，但这里我们假设标量场的取值仅是普通的实数。)
(
a
2
p
a
2
p
a
2
p
a
2
p
dV (p)
) (
density
of
the
field
,
and we will call it potential for short in this problem. The energy density of the scalar field is
1
2
/
d
d
p
t
0
2
+
/
0
2
+
/
0
2
+
/
0
2
+
V(
)
.
标量场满足运动方
程

+
= 0
，其中
V(
)
是场的势能密度，我们将简称它为标量场的势能。标量
)The scalar field satisfies the following equation of motion: at2 ax2 ay2 az2 + dp = 0, where V() is the potential
场的能量密度是 2 (1) /d (d)t (p)02 + 2 (1) /d (d)x (p)02 + 2 (1) /d (d)y (p)02 + 2 (1) /d (d)z (p)02 + V()。
We consider the following potential: the false vacuum has field value = + where V(+ ) = 0, and the true vacuum has field value = - , where V(- ) is slightly negative. In the left panel of the following figure, we plot the shape of the potential. The right panel is an example of the false and true vacuum in position space.
我们考虑如下势能：假真空处标量场取值是 = + ，满足 V(+ ) = 0 ；真真空处标量场取值是 = -， V(- ) 取一个接近 0 的负值。下图(左)是势能的函数形式，下图(右)是位置空间中假真空和真真空的一个例子。
In this problem, we will use natural units and set the speed of light in vacuum c = 1 (by redefining the time unit as the time that light traveled over unit length). In this unit, when an object is at rest, its energy equals its mass by the famous formula E =
2
mc = m .
本题中，我们将使用自然单位并定义真空中的光速 c = 1 (既定义时间的单位为光穿越单位长度的时间)。在自然单位制中，物体的静止能量等于它的质量，这就是著名的公式 E = mc2 = m。
(
A
.
DOMAIN
WALL
畴壁
)
Before coming to the asymmetric potential which generates vacuum bubbles, let us consider a symmetric potential as follows: 在我们研究非对称的势能以及它生成的真空泡泡前，我们先考虑如下一个对称的势能：
2
2
Let's find a static solution which is homogeneous along the y and z directions, known as a domain wall. The potential of the domain wall V() = VD () is illustrated above, with two minima VD (± ) = 0. The domain wall can be used as the local approximation of the bubble wall.
我们将找一种在 y 和 z 方向均匀的“畴壁”解。上图是畴壁解对应的势能 V() = VD ()，它具有两个最小值 VD (± ) = 0 。畴壁可以作为泡泡壁的局域近似。
A1. SIMPLIFY THE EQUATION OF MOTION 化简运动方程
A1 Given the static and homogeneity (independency of y and z) conditions, write down the simplified equation of motion for . 根据静态，以及均匀(不依赖于y 和 z 方向)的条件，写出化简后的运动方程。 2 points 2 分
A2. THE DOMAIN WALL PROFILE 畴壁上标量场取值的空间变化
A2 (
d
"
dx
)Express in terms of VD (). 请把用 VD () 表达出来。 3 points 3 分
A3. THE DOMAIN WALL TENSION 畴壁的张力 (2')
A3 The tension of the domain wall (energy density of the wall for unit area in the y and z directions) is = ∫ f()d . Find f() in terms of VD (). 畴壁的张力(既在 y 、z 方向单位面积上，畴壁的能量密度)是 = ∫f()d。请把 f()用 VD () 表达出来。 2 points 2 分
Note: to avoid propagation of possible errors, in the later part of the questions, please still use the domain wall tension where applicable, instead of using the integral expression that you obtain.
注：为避免潜在的错误传播，在本题后面的部分中，当用到畴壁张力时，请仍使用符号，而不是这里你求出的积分表达式。
(
B
.
BUBBLE
WALL
泡泡壁
)
If we only look at a small part, a bubble wall can be approximated as a domain wall. But globally, the bubble can be approximated to be spherical with radius R . Let's assume that R is large enough, such that the thickness of the bubble wall is much smaller than R (thin wall approximation). Inside and outside the bubble, → ± exponentially quickly.
如果我们只看一泡泡壁的一小部分的话，泡泡壁上的一小块可以用畴壁来近似。但是整体上，真空泡泡可以近似为球形
的，具有半径 R。假设 R 足够大，泡泡壁的厚度远远小于 R (薄壁近似)。在泡泡壁的内部和泡泡壁的外部，指数快地趋向于 ± 。
At the moment when a bubble is nucleated, the bubble is static, and the bubble nucleation and motion create negligible amount of radiation or other dissipations.
在真空泡泡产生的时刻，泡泡是静止的。泡泡产生的过程带来的辐射或其它耗散可以忽略。
3
(
(2) The real-time and imaginary time field configurations are related by
p(t = 0,x,y,Z) =
p
(
T
= 0,
x
,
y
,
Z
)
.
) (
aT
2
ax
2
ay
2
az
2

dp
)2
B1 . THE ENERGY ON THE BUBBLE WALL 泡泡壁的能量 (1')
B1 At the momentum of bubble nucleation, calculate the energy Ew carried by the bubble wall using R and the bubble tension G. 在真空泡泡产生的时刻，利用 R 和泡泡壁的张力 G 计算泡泡壁的能量 Ew。 1 point 1 分
B2 . FALSE AND TRUE VACUA 假真空和真真空
B2 (
lowest order in Taylor expansion which contain
s
e
).
Calculate
e
using
G
and
R
.
) (
阶
效应
(
既泰勒展开中含有
e
的最低阶
)
。
利用
G
和
R
计算
e
。
)For a spherical bubble to appear, there must be an energy density difference between V(p±). Thus, to (
(
p

p
+
)
. In the thin wall approximation, we are only interested in leading order results in
e
(the
)write down a potential to model bubble nucleation, we consider the potential V(p) = VD (p) + (
势
能
模型，我们考虑势能
V(p) = V
D
(p) +
(p p
+
)
。在薄壁近似下，我们只感兴趣
e
零头
)为了让球形真空泡泡能够出现， V(p±) 的取值必须不同。所以，为了对真空泡泡的产生过程建立 (
2
)points 2 分
B3. BUBBLE MOTION 泡泡的运动
B3 At the moment of bubble nucleation, calculate the acceleration a of the bubble wall in terms of G and R. 在真空泡泡产生的瞬间，利用 G 和 R 计算泡泡的加速度 a。 (
2
)points 2 分
(
B
4.
BEYOND
NEWTONIAN
MECHANICS
超越牛顿力
学
)
B4 (
the speed of the b
ubble
wall
to
reach
0.6
.
当真空
泡
泡的速度接近光速，
牛顿力学不再适用，我们应该使用狭义相对论。在狭义相对论里，
)When the speed of the bubble wall is close to the speed of light, Newtonian mechanics breaks down and (
E
K
= (y
1)m
, where
y ≡
.. Calculate the time needed from the nucleation of the bubbl
e
to
that
)special relativity should be used instead. In special relativity, the kinetic energy of a moving object is (
运
需
动
要
物
的
体
时
的
间
动
。
能为
E
K
= (y 1)m
，其中
y
≡
。
计算真
空泡泡从产生到泡泡壁速度达到
0.6
所
d
(
x
2
-1
x
dx
(
x
2
-1
.
)Hint: you may need the mathematical relation = (
dx
(
x
2
-1
。
)提示：你可能需要数学关系d(x2-1 = x 4 points 4 分
C . NUCLEATION RATE OF THE BUBBLE 泡泡的产生率
What's the probability for a bubble to appear It can be shown that the nucleation rate Γ of the bubble, i.e., the probability for a bubble to appear in unit volume and during unit time, can be written as Γ Ae-SE/i, where A and are constants, and SE is a "Euclidean action", which can be calculated with the following procedure:
真空泡泡产生的概率是多少？可以证明，泡泡的产生率 Γ，即单位时间单位体积，一个泡泡产生的概率，可以由 Γ
Ae-SE/i 计算，其中 A 和是常数， SE 是一个“欧氏作用量”，由以下步骤计算：
(
(1)
将
物理时间
t
“
旋转
”
到欧氏时间
T
=
it
(
其中
i
2
= 1
)
。
)(1) We "rotate" our physical time t to "Euclidean time" T = it (where i2 = 1).
(2) 实数时间和虚数时间到场位形由 p(t = 0,x,y,Z) = p(T = 0,x,y,Z) 联系起来。
(3) Given the above time boundary condition, find a 4-dimensional rotational symmetric solution of the Euclidean equation of
motion a2p + a2p + a2p + a2p dV(p) = 0.
(3) 给定上述时间边界条件，找到欧氏运动方程 + + + = 0 的四维旋转对称的解。
4
(
于
p
(
这
就
是四维转动不变的含义
)
，一般的欧氏运动方程可以写成
+
f
(
p
)

= 0
。
) (

) (
为了
寻找运动方程的四维转动不变解，使用
p
=
PT
2
+
x
2
+
y
2
+
z
2
作为解方程的变量比较方便。假设
p
=
p
(
p
)
只依赖
)2
(4) Insert the solution to the Euclidean action SE = ∫ dt d3x I /02 + /02 + /02 + /02 + V(p)J to find Γ .
(4) 将找到的解带入欧氏作用量 SE = ∫ dt d3x I /02 + /02 + /02 + /02 + V(p)J 来求出 Γ。
Let us do this calculation in this part. Note that a 4-dimensional unit ball with radius r has "volume" 2 (冗)2 r4 and surface "area"
2几2r3.
让我们在这一部分中做上述计算。注意，四维球的“体积”为 2 (冗)2 r4 ，球面面积为 2几2r3。
(
C
1
.
THE
EUCLIDEAN
EQUATION
OF
MOTION
欧氏运动方程
)
Since we are to look for a 4-dimentional rotational symmetric solution of the Euclidean equation of motion, it is convenient to
(
dimensional rotational symmetric), the
general Euclid
ean equation of m
otion can be w
ritten
as
+
f
(
p
)

= 0
.
)use p = PT2 + x2 + y2 + z2 as the variable for equation solving. Assuming that p = p(p) only depends on p (i.e., 4-
C1 Find the expression of f(p). 求 f(p)。 Hint: if the formal calculation needs too much calculus, you can consider an example: by tuning the form of V(p), one can obtain a solution p = p2. Then f(p) can be solved by this example (and this form f(p) will apply for all forms of V(p), not limited to this special form of solution). 提示：如果进行普适的计算需要太多微积分，你可以考虑一个例子：通过调整 V(p) 的形式，我们得到一个解 p = p2。这时， f(p) 可以从这个例子里解出来(之后这个 f(p) 的形式对所有 V(p) 都适用，不仅限于这个特殊解)。 2 points 2 分
C2. THE EUCLIDEAN ACTION 欧氏作用量
C2 Write the Euclidean action SE as an integral of p from p = 0 to p → ∞. 以对 p 的积分(从p = 0 积到 p → ∞)的形式写出欧氏作用量 SE 。 (
2
)points 2 分
(
C
3
.
QUALITATIVE
INSPECTION
定性讨论
(4')
)
Before trying to solve the Euclidean equation of motion, let us first see how it behaves. If you are not familiar with this Euclidean equation of motion, you can consider the following analogy: consider p as the position of a particle, and p as an effective time
variable. In this case, answer the following questions (choose one from the options):
在解欧氏运动方程之前，我们先看看方程的性质。如果你不熟悉欧氏运动方程，你可以用如下类比来理解：把 p 类比为一个粒子的位置，把 p 类比为适用于这个粒子的有效时间变量。在这种情况下，回答以下问题(单项选择题)：
C3- 1 What's the nature of f(p) 以下哪项是 f(p) 的性质？ (A) friction (i.e. decelerate the particle) 阻力(即让粒子减速) (B) anti-friction (i.e. accelerate the particle) 推力(即让粒子加速) (C) friction for > 0 and anti-friction otherwise 在 > 0 情况下是阻力，否则是推力 (D) friction for < 0 and anti-friction otherwise 在 < 0 情况下是阻力，否则是推力 1 point 1 分
C3-2 What's the force that drives the motion of the "particle" p 驱动粒子 p 运动的力是哪个？ (A) V (B) V (C) dV/dp (D) dV/dp 1 point 1 分
5
(

)2
C3-3 Where is the "starting point" of p at p = 0 (Here 6 is extremely small but finite) 在 p = 0， p 的“起始位置”在哪里？(其中6 是一个非常小但有限的数) (A) p. |0(6)| (B) p. (C) p. + |0(6)| (D) p+ |0(6)| (E) p+ (F) p+ + |0(6)| 1 point 1 分
C3-4 Where is the "end point" of p at p → ∞ 在 p → ∞， p 的“最终位置”在哪里？ (A) p. |0(6)| (B) p. (C) p. + |0(6)| (D) p+ |0(6)| (E) p+ (F) p+ + |0(6)| 1 point 1 分
(
C
4.
THE
) (
BUBBLE
NUCLEATION
RATE
泡泡产生
率
)
C3-4 Express SE in terms of R and G . 用 R 和 G 写出 SE 。 4 points 4 分
Problem 2: Lorentz reciprocity (32 points)
问题 2: 洛伦兹互易性 (32 分)
Lorentz reciprocity is a fundamental principle in electromagnetism with important application in antenna design theory. It states that the receiving and transmitting capabilities of an antenna are identical. On the other hand, reciprocity can be broken by using magnetic materials under an external magnetic field with strong magneto-optical effect. The study of Lorentz reciprocity can be extended to nearly zero frequency at magnetostatics, shown in the figure below, with two current coils and an arbitrary object fixed in locations. When one current coil works in transmitting mode, another one works in receiving mode. If we model the two current coils as magnetic dipole moments m1 and m2 at locations r1 and r2 , generating magnetic flux densities B1 (r) and B2 (r)
in transmitting mode, respectively. Then, the reciprocity relationship can be expressed as
(
洛伦兹互易性是电磁学的基本原理，在天线设计理论中具有重要应用。
它指出天线的接收和发射能力是相同的。
另
一
方
面，在外磁场下，使用具有强磁光效应的磁性材料可以破坏互易性
。
洛伦兹互易的研究可以扩展到静磁学，如下图所
示，
有两个电流环和一个固定在某个位置的物体。
当一个电流环在发射模式工作时，另一个电流环在接收模式工作。
如
果我
们将两个电流环设为位置
r
1
和
r
2
处的磁偶极子
m
1
和
m
2
，分别在发射模式下产生磁通密度
B
1
(r)
和
B
2
(r)
。
那么，
互
易
关系
可以表示为
)m1 B2 (r1 ) = m2 B1 (r2 )
m1 B2 (r1 ) = m2 B1 (r2 )
6
2
7
m! at r1
Transmitting
Object
B!(r2)
Receiving
B2(r!)
Receiving
Object
m2 at r2
Transmitting
The magnetic field B(r) generated by a magnetic dipole mi located at ri is given by,
在位置 ri 处的磁偶极子 mi 所产生的磁场 Bi(r) 由下式给出，
Bi(r) = 4 (u)几 (@) ]3(r i)r mi |r r (i)i |3_
In this problem, we will first establish the reciprocity relationship and will investigate how it can be broken by imposing a constant velocity on the object.
本题中，我们将首先建立互易关系，并研究如何通过对物体施加恒定速度来打破它。
(
A
.
ESTABLISHING
MAGNETOSTATIC
RECIPROCITY
AND
NON
-
RECIPROCITY
建立静磁互易性和非互易性
)
For magnetostatics, we have vector potential A, magnetic flux density B, magnetic field strength H and impressed current density J. These fields satisfy the Ampère’s law
× H =J
with material response
× A = B = uH
where u(r) is the isotropic magnetic permeability profile for the material, i.e. the object in the figure.
对于静磁学，我们有矢量势 A、磁通密度 B、磁场强度 H 和外加电流密度 J。这些场满足安培定律
× H =J
关于物质反应的表达式为
× A = B = uH
其中 u(r) 是材料的各向同性磁导率分布，即图中的物体。
We want to establish the reciprocity relationship for magnetostatics when we have a magnetic dipole moment m1 at location r1
(
respectively. The two cases, labeled by
i = 1,2
, have the current density
J
i
= × M
i
and magnetization
M
i
= m
i
6(r
r
i
)
for
)in one case and a magnetic dipole m2 at r2 in another case. The two dipoles generate magnetic fields B1(r) and B2(r),
the two dipoles.
(
∫

×
AdV
= ∫
A
×
da
) (
∫

AdV
= ∫
A

da
) (
∫
B
×
AdV
= ∫
A
×
BdV
+ ∫
A
×
B

da
)2
(
在
一
种情况下
r
1
处有一个磁偶极子
m
1
并产生磁场
B
1
(
r
)
而
在
另
一种情况下
r
2
处有一个磁偶极子
m
2
并产生磁场
B
2
(
r
)
。
) (
的
互易关系。
)对于由 i = 1,2 索引的两个偶极子，我们有电流密度 Ji = × Mi 和磁化强度 Mi = mi 6(r ri ) 。以下我们想要建立静磁学
Here, we also give some formulas for these differential operators:
在这里，我们提供一些关于微分算子的公式：
(

×
A
=
ga
y
A
z

a
z
A
y
i
+
(
a
z
A
x

a
x
A
z
)
+
ga
x
A
y

a
y
A
x
i
) (
(
A
×
B
) = ( ×
A
)
B

A
×
B
) A = ax Ax + ayAy + azAz
and Kronecker delta function 6(r) is defined by
6(r) = l
which satisfies ∫ 6(r)dV = 1 when we integrate a volume V enclosing the origin. a is defined as the closed area enclosing volume
(
V
).
当我们对包含原点的一个体积 V 进行积分时， Kronecker delta 函数 6(r) 定義為:
6(r) = w 0 (∞) 除 ( i)此 (f r)以 (=)外 (0)
而且满足∫ 6(r)dV = 1 。 a 定义为包围体积 V 的封闭区域。
(
Prove
the
reciprocity
relationship
m
1

B
2
(
r
1
) =
m
2

B
1
(
r
2
)
.
Hint
:
you
may
consider
(
H
1
×
A
2
)
.
)A1. PROVING THE RECIPROCITY RELATIONSHIP 证明互易关系
A1 (
证明互易关系
m
1

B
2
(
r
1
) =
m
2

B
1
(r
2
)
。
提示
：你可以考虑
(H
1
× A
2
)
。
) (
3
)points 3 分
If the material conducts electricity with an electrical conductivity a, we have to add an additional term to the current density J due to the free current through
J → J + aE
in the Ampère’s law stated previously. Suppose now we move the conductor by a constant velocity v. There will be a Lorentz force on the free charges proportional to E + v × B. It further updates the additional term in the current density through
J → J + a(E + v × B),
which may upset the reciprocity relationship.
如果材料以电导率 σ 导电，由于通过的自由电流，我们必须为在此前安培定律中的电流密度 J 添加一个附加项
J → J + aE.
假设现在我们以恒定速度 v 移动导体。自由电荷上将存在与 E + v × B成比例的洛伦兹力，这进一步更新电流密度中的附加项以使破坏互易关系变的可能：
J → J + a(E + v × B).
8
(

) (

) (

) (

) (
You
can
complete
part
B
and
C
without
part
A
.
你可以在没有
A
部分的情况下完成
B
部分和
C
部
分。
) (
We
further
simplify
the
problem
by
removing
the
conductor at the moment.
) (
A
2
Express
the
possibly
non
-
zero
m
1
B
2
(
r
1
)
m
2
B
1
(
r
2
)
as
a
volume
integral
in
terms
of
the
vector potentials
A
1
and
A
2
and
the
conductor
velocity
v
.
将可能非零的
m
1
B
2
(
r
1
)
m
2
B
1
(
r
2
)
表示为根据矢量势
A
1
和
A
2
以及导体速度
v
的体积积分。
3
p
oints
3
分
) (
v
) (
Conducting
Object
) (
y
z
'
x
)2
(
A
2.
BREAKING
RECIPROCITY
RELATIONSHIP
打破互易关
系
)
(
B
.
MAGNETIC
DIPOLES
ON
A
MOVING
PERFECT
CONDUCTOR
运动理想导体上的磁偶极子
)
In this part, we first obtain the magnetic field of a single magnetic dipole on a moving conductor, which is a perfect conductor (i.e. the electrical conductivity G = ∞) here, occupying z ≤ 0 and is moving with a velocity v in the positive x-direction. This is defined
(
as
the
laboratory
frame
,
as
shown
in
the
figure
below
.
The
magnetic dipole, situated at
(X,y,z) = (0,0,z
@
)
on top of the conductor,
) (
在这一部分中，我们首先得到运动导体上
单个磁偶极子的磁场，这里是理想导体
(
即电导率
G = ∞
)
，
占据
z
≤
0
并且在正
x
方向上
以
速度
v
运动。
这被定义为实验室坐标系，如下图所示
。
位于导体顶部
(X,y,z) = (0,0,z
@
)
处的磁偶极子写为
)has a magnetic moment m = mx + mz with zero component in the y-direction.
m = mx + mz ，在 y 方向上没有分量。
z
m = mx + m&
(
giving us a convenience to find the
magnetic fields generated by the magnetic dip
ole
.
The
coordinates
in
the
moving
frame
,
denoted
)In the moving frame at a velocity v = v with respect to the laboratory frame, the object is simply a perfect conductor at rest,
as (XC,yC, zC, tC), is transformed from the coordinates in the laboratory frame (X,y, z, t) through the Lorentz transformation:
(
生的磁场。
移动坐标系中的坐标，表示为
(
X
C
,
y
C
,
z
C
,
t
C
)
，
由实验室坐标系
(
X
,
y
,
z
,
t
)
中的坐标通过洛伦兹变换转换
而
来：
X
C
=
y(X vt), y
C
= y, z′ = z, ct′ = y(ct vX/c)
)在相对于实验室坐标系以速度 v = v 运动的坐标系中，物体只是一个静止的理想导体，使我们更方便地找到磁偶极子产
(
where
y
= 1/
P
1
v
2
/
c
2
and
c
is
the
speed
of
light
.
)其中 y = 1/P1 v2/c2， c 是光速。
The magnetic and electric fields (B′and EC ) in the moving frame are related to the fields (B and E) in the laboratory frame by the Lorentz transformation,
移动坐标系中的磁场和电场(B′和 E′) 与实验室坐标系中的场(B 和 E) 相关，可由洛伦兹变换给出:
(
}
)
C
x
C
y
C
z
=
}
)
+
}
y
)
,
)} )CxCyCz = }) vy } y)
9
(
x
)2
我们现在通过移除导体来进一步简化问题。
B1 . MAGNETIC FIELD OF A MOVING MAGNETIC DIPOLE IN FREESPACE 自由空间中运动磁偶极子的磁场
B1 (
to
the
laboratory
frame

In
this
part
,
we
only
consider
a
magnetic
dipole
m
=
m
$
pointing
in
the
x
-
) (
direction
and
there
is
no
conductor
below
the
dipole
yet
.
Please
express
your
answer
in
the
coordinates of
the
moving
frame
.
请找出相对于实验室
坐标系以速度
v
移动的观察者的偶极子磁场
B
C
(X′, y′, z′, t′)
。
在这一部分中，
)What is the magnetic field BC (X′, y′, z′, t′) from the dipole for an observer moving at velocity v with respect (
我们只考虑指向
x
方向的磁偶极子
m
=
m
$
并且偶极子下方还没
有导体。
请用移动坐标系的坐标
)表达你的答案。 3 point 3 分
Next, we introduce the perfect conductor below the dipole. The perfect metal has a planar interface at z = 0 while the dipole is placed at a distance z@ above the metal. The metal is being moved at a velocity v in the positive x-direction.
接下来，我们引入偶极子下方的理想导体。理想导体在 z=0 处有一个平面界面，而偶极子位于金属上方 z@ 处。金属在正 x 方向上以速度 v 移动。
(
Please
express
the
magnetic
field
of
the
magnetic
dipole
m
=
m
$
on
top
of
a
moving
perfect
conductor
)B2 . MAGNETIC FIELD OF A MAGNETIC DIPOLE ON A MOVING CONDUCTOR (I) 运动导体上磁偶极子的磁场 ( 1 )
B2 (
(
i
.
e
B
(
X
,
y
, 0
-
)
)
in
terms
of
the
laboratory
frame
coordinate
.
Please
also
verify
the
boundary
condition
of the
magnetic
field
in
the
moving
frame
at
which
the
conductor
at
rest
.
Hint
:
adopt
the
method
of
images
and
assume
both
electric
and
magnetic
fields
inside
a
perfect
conductor
is
zero
at
a
nearly
-
zero
frequency
.
) (
请用实验室坐标系表示在移动中的理想导体上磁偶极
子
m
=
m
$
的磁场
(
即
B
(
X
,
y
, 0
-
)
)
。
还请验
) (
证导体静止时移动坐标系中的磁
场边界条件。
提示：采用镜像法
。假定在接近零频率时完美导体
里电场和磁场为零。
) (
4
)points 4 分
B3. MAGNETIC FIELD OF A MAGNETIC DIPOLE ON A MOVING CONDUCTOR (2) 运动导体上磁偶极子的磁场 ( 2 )
B3 (
如果运动导体上方的磁偶极
子
转为
m = m
&
，指向正
z
方向，
那么请找出实验室坐标系中磁偶极子
) (
在移
动中的理想导体上的磁场
(
即
B(X, y, 0
-
)
)
？
)What is the magnetic field from the magnetic dipole on top of a moving perfect conductor (i.e (
m
=
m
&
,
pointing
in
the
positive
z
direction
)B(X, y, 0- )) in the laboratory frame if the magnetic dipole above the moving conductor is changed to (
3
)points 3 分
Now, we consider the two situations: one with magnetic dipole moment m+ at r+ = (0,0, z@ ) and another magnetic dipole moment m2 at r2 = (6, 0, z@ ). When one of them is turned on, another one is turned off.
现在，我们考虑两种情况：一种是 r+ = (0,0, z@ ) 处的磁偶极矩 m+ ，另一种是 r2 = (6, 0, z@) 处的磁偶极矩 m2 。当其中一个打开时，另一个被关闭。
z
(
m
1
=
m
&
)m2 = mz
6
y
(
z
)'
(
Conducting
Object
v
)
10
(
值
定义为

= gB
1z
(6) B
2x
( 6)i/gB
1z
(6) + B
2x
( 6)i
。
) (
对
于第二个问题，假设我们将
m
1
的指向更改为正
x
方向而大小保持不变。
m
2
方向和大小不变
。
在这种情况下，
互易功
) (
When

= 0
,
reciprocity
is
satisfied
.
Reciprocity
is
broken
when

deviates
from
value
zero
.
) (
3
) (
B
4
.
RECIPROCITY
MERIT
FOR
DIPOLES
ON
A
PERFECT
CONDUCTOR
理想导体上偶极子的互易功值
B
4
Find
the
reciprocity
merit

for
the
two
defined
problems
about
magnetic
dipoles
on
the
moving
perfect conductor
.
求出关于以上两个问题运动理想导体上
磁偶极子的互易功值

。
p
oints
3
分
C
.
MAGNETIC
DIPOLES
ON
A
MOVING
CONDUCTOR
WITH
FINITE
CONDUCTIVITY
具有有限电导率的移动导体上
的磁偶极
子
) (
施加到
r
1
= (x
1
, 0, z
0
)
上。
然后我们定义互易功值
) (
current
case
of
m
=
m
z
,
we
put
an
image
magnetic
dipole
with
given
form
of
magnetic
moment
m
r
)
+
m
r
)
and
electric
)2
Now, we define the two problems to solve next. For the first problem, suppose the two magnetic dipoles moments m1 and m2
(
point
in
the
z
-
direction
with
same
size
m
z
,
as
shown
in
the
above
figure
.
Dipole
moment
m
1
imposes
a
magnetic
field
of
z
-
) (
function
B
z
(
x

x
i
)
. We then define a reciprocity figure-of-merit
)component Bz (6) on r2 = (x2, 0, z0 ) and dipole m2 imposes magnetic field Bz ( 6) on r1 = (x1, 0, z0 ) according to the same
(
B
z
(6) + B
z
( 6)
.
) = Bz (6) Bz ( 6)
(
如上图所示
。
偶极子
m
1
根据函数
B
z
(x x
i
)
将
z
分量
B
z
(6)
的磁场施加到
r
2
= (x
2
, 0, z
0
)
上，
偶极子
m
2
将磁场
B
z
( 6)
)现在，我们定义接下来要解决的两个问题。对于第一个问题，假设两个磁偶极子m1 和m2 都指向 z 方向，大小mz 相同，
(
B
z
(6) + B
z
( 6)
.
) = Bz (6) Bz ( 6)
当 = 0时，满足互易性。当偏离零值时，互易性被打破。
For the second problem, suppose we change the pointing direction for m1 to the positive x-direction with magnitude remaining
(
gB
1
z
(6)
B
2
x
( 6)i/gB
1z
(6) + B
2x
( 6)i
.
)the same. The magnitude and direction of m2 are not changed. In this case, the reciprocity merit is defined as =
In this part, we move to a more realistic situation that the conductor is a metal. It has a large but finite conductivity G (in unit of Ω- 1m- 1 ), deviating from the perfect conductor condition. Current density in the conductor is given by J = GE. We also assume that the current on the conductor surface is confined by a skin depth d of small thickness so that the electric and magnetic fields cannot penetrate beyond the skin depth from the conductor surface. Then, the surface current density can be written as (Gd)E. We further take the approximation that d is just a constant. We only consider the two dipoles pointing in the positive z-direction with same size mz in this part.
在这一部分中，我们转向一个更现实的情况，即金属导体具有较大但有限的电导率 σ (单位为 Ω- 1m- 1 )，偏离理想导体条件。导体中的电流密度由 J = GE 给出。我们还假设电流只在导体表面厚度较小的趋肤深度 d 内流动，电场和磁场不能从导体表面穿透超过趋肤深度。因此表面电流密度可以写成 (Gd)E 。我们进一步假定 d 为一个常数。在这部分我们只考虑两个偶极子都指向正 z 方向，大小 mz 相同。
Again, we need to solve the magnetic field from only one dipole at (0,0, z0 ) first. In fact, the surface current profile generated on the surface of conductor cannot be easily solved without adopting a numerical solver. Instead, we can approach the problem by
(
position
(
0,0,
z
0
)
in
order
to
give
as
closely
as
possible
the
same
reflected
field
generated
from
the
surface
current
.
For
the
)extending the method of image as an approximation. In this case, we would like to have a point-like multipolar source at the image
11
(
分
量，
同时忽略高阶多极子。
这些偶极矩的大小有待确定。
) (
子设置在
(0,0,
z
0
)
处，它
拥
有给定的表达式：
磁矩为
m
r)
+ m
r)
、电矩为
p
r)
。
镜像磁偶极子现在同时具有磁和电
)2
(
components
while
neglecting
the
higher
order
multipoles
.
The size of these dipole moments are yet to be determined.
同样，我们需要先求解
(
0,0, z
0
)
处只有一个偶极子的磁场。
实际上，
导体表面产生的表面电流分布不采用数值求解器是
)moment pr) at the same location (0,0, z0 ). The mirrored magnetic dipole is now relaxed to have both magnetic and electric
(
有一个点状多极源，以便尽可能地提供相同的从表面电流生成的反射场。
对于当前
m =
m
z
的情况，我们将镜像磁偶极
)无法精确求解的。这里，我们尝试通过电像法来近似解决这个问题。在这种情况下，我们希望在镜像位置 (0,0, z0) 处
C1. GENERALIZED METHOD OF IMAGES 广义电像法
C1 Find the magnetic field B, (x′, y′, 0+) and electric field E, (x′, y′, 0+ ) on the conductor surface in the moving frame. Express your answer in mz , m mr) and p You can use either the moving frame or laboratory frame coordinates. Do not need to solve the mirrored dipole moments yet. 求移动坐标系导体表面的磁场B′(x′, y′, 0+ ) 和电场E, (x′, y′, 0+ ) 。请用mz 、m mr) 和pr) 表达你的答案。可以使用移动坐标或实验室坐标表达你的答案。暂时不需要求解镜像偶极矩。 (
5
)points 5 分
(
C
2
.
FINDING
THE
MIRRORED
DIPOLE
MOMENTS
计算镜像偶极矩
)
C2 Find m mr) and pr) in response to a given mz . The mirror dipole gives the same reflected field generated by the surface current on the conductor. As approximation, only apply the boundary condition (in the moving frame) on the surface current along the y-direction, which is the dominant current than the one along the x-direction. It may be useful to express the answers in term of the dimensionless parameter K = u0 vyGd 1. 用给定的mz 表示 m mr) 和 p 镜像偶极子给出了与导体表面电流产生的相同的反射场。作为近似，请仅将边界条件(在移动坐标系中)应用于沿 y 方向的表面电流，该电流比沿 x 方向的电流占主导地位。可以考虑用无量纲参数K = u0 vyGd 1 表达答案。 (
5
)points 5 分
C3. RECIPROCITY MERIT FOR DIPOLES ON A CONDUCTOR OF FINITE CONDUCTIVITY 有限电导率导体上偶极子的互易功值
C3 Find the reciprocity merit for the two identical dipoles mz displaced by 6 (with r1 = (0,0, z0 ), r2 = (6, 0, z0 ))in the x-direction. 找出在 x 方向上位移6 (r1 = (0,0, z0 ), r2 = (6, 0, z0))) 的两个相同偶极子 mz 的互易功价值。 3 point 3 分
~~ END of Paper 2 卷 2 完 ~~
12Pan Pearl River Delta Physics Olympiad 2023
2023 年泛珠三角及中华名校物理奥林匹克邀请赛
Sponsored by Institute for Advanced Study, HKUST
香港科技大学高等研究院赞助
Simplified Chinese Part-2 (Total 2 Problems, 60 Points)
简体版卷-2 (共2题，60分)
(1:30 pm – 5:00 pm, 29 January 2023)
All final answers should be written in the answer sheet. 所有最后答案要写在答题纸上。 All detailed answers should be written in the answer book. 所有详细答案要写在答题簿上。 There are 2 problems. Please answer each problem starting on a new page. 共有 2 题，每答 1 题，须采用新一页纸。 Please answer on each page using a single column. Do not use two columns on a single page. 每页纸请用单一直列的方式答题。不可以在一页纸上以双直列方式答题。 Please answer on only one page of each sheet. Do not use both pages of the same sheet. 每张纸单页作答。不可以双页作答。 Rough work can be written in the answer book. Please cross out the rough work after answering the questions. No working sheets for rough work will be distributed. 草稿可以写在答题簿上，答题后要在草稿上划上交叉，不会另发草稿纸。 If the answer book is not enough for your work, you can raise your hand. Extra answer books will be provided. Your name and examination number should be written on all answer books. 考试中答题簿不够可以举手要，所有答题簿都要写下姓名和考号。 At the end of the competition, please put the question paper and answer sheet inside the answer book. If you have extra answer books, they should also be put inside the first answer book. 比赛结束时，请把考卷和答题纸夹在答题簿里面，如有额外的答题簿也要夹在第一本答题簿里面。
Problem 1: Vacuum bubbles (28 points)
问题 1: 真空泡泡 (28 分)
Is our vacuum stable We don't know. It's possible that we do not live in the true vacuum. Rather, we live in a false vacuum which can decay into true vacuum by emerging and expanding bubbles. To describe such a possibility, we will make use of a space-time dependent "scalar field" (t, x, y, z), which takes a real value at every space-time point. (Similar to height on a map, which takes a real value at every point on the x-y plane, while a scalar field takes a real value for any given t, x, y, z. Also, in a full quantum theory, we have to distinguish operators and numbers, but here we will assume the scalar field only take real number values.)
我们的真空是稳定的吗？我们不知道。有可能我们并不是生活在一个稳定的真真空(既真的“真空”)里面。我们可能生活在一个假真空里，而假真空可以通过自发产生和膨胀的泡泡衰变到真真空。为了描述这种可能，我们将使用一个依赖于时空点到“标量场”(t, x, y, z) ：它在每个时空点上取一个实数。(这有点像地图上的高度，在 x-y 平面的每个点上取一个实数。不过标量场是在每个t, x, y, z 时空点上取一个实数。另外，在完整的量子理论中，我们需要考虑算符和数的区别，但这里我们假设标量场的取值仅是普通的实数。)
(
a
2
p
a
2
p
a
2
p
a
2
p
dV (p)
) (
density
of
the
field
,
and
we
will
call
it
potential
for
short
in
this
problem
.
The
energy
density
of
the
scalar
field
is
1
2
/
d
d
p
t
0
2
+
/
0
2
+
/
0
2
+
/
0
2
+ V(
)
.
标
量场满足运动方程

+
=
0
，其中
V
(
)
是场的势能密度，我们将简称它为标量场的势能。标
)The scalar field satisfies the following equation of motion: at2 ax2 ay2 az2 + dp = 0, where V() is the potential
量场的能量密度是 2 (1) /d (d)t (p)02 + 2 (1) /d (d)x (p)02 + 2 (1) /d (d)y (p)02 + 2 (1) /d (d)z (p)02 + V()。
We consider the following potential: the false vacuum has field value = + where V(+ ) = 0, and the true vacuum has field value = - , where V(- ) is slightly negative. In the left panel of the following figure, we plot the shape of the potential. The right panel is an example of the false and true vacuum in position space.
我们考虑如下势能：假真空处标量场取值是 = + ，满足 V(+ ) = 0 ；真真空处标量场取值是 = -， V(- ) 取一个接近 0 的负值。下图(左)是势能的函数形式，下图(右)是位置空间中假真空和真真空的一个例子。
In this problem, we will use natural units and set the speed of light in vacuum c = 1 (by redefining the time unit as the time that light traveled over unit length). In this unit, when an object is at rest, its energy equals its mass by the famous formula E =
2
mc = m.
本题中，我们将使用自然单位并定义真空中的光速 c = 1 (既定义时间的单位为光穿越单位长度的时间)。在自然单位制中，物体的静止能量等于它的质量，这就是著名的公式 E = mc2 = m。
(
A
.
DOMAIN
WALL
畴壁
)
Before coming to the asymmetric potential which generates vacuum bubbles, let us consider a symmetric potential as follows: 在我们研究非对称的势能以及它生成的真空泡泡前，我们先考虑如下一个对称的势能：
Let's find a static solution which is homogeneous along the y and z directions, known as a domain wall. The potential of the domain wall V() = VD () is illustrated above, with two minima VD (± ) = 0. The domain wall can be used as the local approximation of the bubble wall.
我们将找一种在 y 和 z 方向均匀的“畴壁”解。上图是畴壁解对应的势能 V() = VD ()，它具有两个最小值 VD (± ) = 0 。畴壁可以作为泡泡壁的局域近似。
A1. SIMPLIFY THE EQUATION OF MOTION 化简运动方程
A1 Given the static and homogeneity (independency of y and z) conditions, write down the simplified equation of motion for . 根据静态，以及均匀(不依赖于y 和 z 方向)的条件，写出化简后的运动方程。 (
2
)points 2 分
Solution: = . [Note, if the student did not convert to d, we also consider it as correct. Also, since in Part A, V = VD , both are correct.]
A2. THE DOMAIN WALL PROFILE 畴壁上标量场取值的空间变化
A2 (
dp
dx
)Express in terms of VD (). 请把用 VD () 表达出来。 (
3
)points 3 分
(
Solution:
=
=
0
and
V
D
=
0
, we have
/
0
2
,
thus
6/
0
2
7 =
.
Since
far
away
from the domain wall, we have boundary condition
=
=
2V
D
.
)
A3. THE DOMAIN WALL TENSION 畴壁的张力 (2')
A3 The tension of the domain wall (energy density of the wall for unit area in the y and z directions) is G = ∫ f()d . Find f() in terms of VD (). 畴壁的张力(既在 y、z 方向单位面积上，畴壁的能量密度)是 G = ∫f()d。请把 f()用 VD () 表达出来。 2 points 2 分
Note: to avoid propagation of possible errors, in the later part of the questions, please still use the domain wall tension G where
applicable, instead of using the integral expression that you obtain.
注：为避免潜在的错误传播，在本题后面的部分中，当用到畴壁张力时，请仍使用符号 G ，而不是这里你求出的积分表达式。
(
Solution
:
G
=
伪
伪
>
/
0
2
+ V
D
(
) dx =
∫
[2V
D
(
)]
d
=
∫
[2V
D
(
)]
1/2
d
f
(
) = B2V
D
(
)
)
B . BUBBLE WALL 泡泡壁
If we only look at a small part, a bubble wall can be approximated as a domain wall. But globally, the bubble can be approximated to be spherical with radius R. Let's assume that R is large enough, such that the thickness of the bubble wall is much smaller than R (thin wall approximation). Inside and outside the bubble, p → p± exponentially quickly. 如果我们只看一泡泡壁的一小部分的话，泡泡壁上的一小块可以用畴壁来近似。但是整体上，真空泡泡可以近似为球形的，具有半径 R。假设 R 足够大，泡泡壁的厚度远远小于 R (薄壁近似)。在泡泡壁的内部和泡泡壁的外部， p 指数快地趋向于 p± 。 At the moment when a bubble is nucleated, the bubble is static, and the bubble nucleation and motion create negligible amount of radiation or other dissipations. 在真空泡泡产生的时刻，泡泡是静止的。泡泡产生的过程带来的辐射或其它耗散可以忽略。
B1 . THE ENERGY ON THE BUBBLE WALL 泡泡壁的能量 (1')
B1 At the momentum of bubble nucleation, calculate the energy Ew carried by the bubble wall using R and the bubble tension a. 在真空泡泡产生的时刻，利用 R 和泡泡壁的张力 a 计算泡泡壁的能量 Ew 。 1 point 1 分
(
Solution
:
E
w
=
4
几
R
2
a
.
)
B2 . FALSE AND TRUE VACUA 假真空和真真空
B2 (
(p p
+
)
. In the thin wall approximation, w
e
are
only
interested
in
leading
order
results
in
e
(
the
) (
lowest order in Taylor expansion which cont
ains
e
).
Calculate
e
using
a
and
R
.
) (
阶效应
(
既泰勒展开中含有
e
的最低阶
)
。
利用
a
和
R
计算
e
。
)For a spherical bubble to appear, there must be an energy density difference between V(p± ). Thus, to write down a potential to model bubble nucleation, we consider the potential V(p) = VD (p) + (
势能模型，我们考虑势能
V(p) = V
D
(p) +
(p p
+
)
。在薄壁近似下，我们只感兴趣
e
零头
)为了让球形真空泡泡能够出现， V(p± ) 的取值必须不同。所以，为了对真空泡泡的产生过程建立 (
2
)points 2 分
(
Solution
:
The
energy
density
inside
the
bub
b
le
is

e
.
From energy conse
rvation
,
E
w
=
4
3
几
R
3
e
,
thus
e =
.
Note
:
though
the
introduction
of
the
linear
term
modifies
the
minima
of
the
potential
a
little
,
but the modification multiplying the
energy
density
will
be
0
(e
2
)
and thus neglected.
)
B3. BUBBLE MOTION 泡泡的运动
B3 At the moment of bubble nucleation, calculate the acceleration a of the bubble wall in terms of a and R. 在真空泡泡产生的瞬间，利用 a 和 R 计算泡泡的加速度 a。 (
2
)points 2 分
(
Solution
:
Consider
bubble
expansion
R → R + 6R
.
Since volume expands faster than surface, the energy obtained by the bubble wall from this expansion is
6E
w
=
3
4
几
(R
+
6R)
3
e

4
几
(R
+
6R)
2
a
=
4
几
Ra
6R
)
(
(1)
将物理时间
t
“
旋转
”
到欧氏时间
T = i
t
(
其中
i
2
= 1
)
。
) (
(2)
The
real
-
time and imaginary time field configurations are related by
p(t = 0,x,y,z) = p(T = 0,x,y,z)
.
) (
Since
the
force
F
=
= ma = E
w
a
(in the unit
c =
1
)
Acceleration
a
=
6
E
%
/6
R
=
1
E
%
R
.
)
B4. BEYOND NEWTONIAN MECHANICS 超越牛顿力学
B4 (
the speed of the bubb
le
wall
to
reach
0.6
.
当真空
泡
泡的速度接近光速，
牛顿力学不再适用，我们应该使用狭义相对论。在狭义相对论里，
)When the speed of the bubble wall is close to the speed of light, Newtonian mechanics breaks down and (
E
K
= (
y
1)
m
,
where
y ≡
.. Calculate the time needed from the nucleation of the bubble to that
)special relativity should be used instead. In special relativity, the kinetic energy of a moving object is (
运
需
动
要
物
的
体
时
的
间
动
。
能为
E
K
=
(y

1)
m
，其中
y
≡
。
计算真
空泡泡从产生到泡泡壁速度达到
0.6
所
d
(
x
2
-1
x
dx
(
x
2
-1
.
)Hint: you may need the mathematical relation = (
dx
(
x
2
-1
。
)提示：你可能需要数学关系d(x2-1 = x (
4
)points 4 分
(
Solution
:
The
rest
+
kinetic
energy
of
the
bubble
is
E
w
=
ym
=
4
几
r
2
G
,
where
r
=
r
(
t
)
is
the
time-dependent radius of the bubble
(
to
be
distinguished
from
the
initial
radius
of
the
bubble
R
.
Thus, energy conservation at time
t
yields
1
4
几
r
2
G =
4
几
r
3
e
,
i
.
e
.,
=
, i.e.,
=
, i.e.,
= d√
r
2

R
2
= dt
.
Given
the
initial
condition
r
(
t
= 0)
= R
, we have
r
2

R
2
= t
2
.
(
Note
,
after
a
complicated
calculation
(
it
'
s
more
complicated
if
you use relativistic force), you get a surprisingly simple result. This is
not
a
coincidence. In fact, this is an analytical continuation of a Euclidean 4-sphere. We will see a little bit of this in Part C.)
v
=
=
=
0.6
,
t
=
0.75R
.
(
1-
v
2
3
)
(
C
.
NUCLEATION
RATE
OF
THE
BUBBLE
泡泡的产生率
)
What's the probability for a bubble to appear It can be shown that the nucleation rate Γ of the bubble, i.e., the probability for a bubble to appear in unit volume and during unit time, can be written as Γ Ae-SE/i, where A and are constants, and SE is a "Euclidean action", which can be calculated with the following procedure:
真空泡泡产生的概率是多少？可以证明，泡泡的产生率 Γ，即单位时间单位体积，一个泡泡产生的概率，可以由 Γ
(
(1)
We
"
rotate
"
our
physical
time
t
to
"
Euclidean
time
"
T = it
(where
i
2
= 1
).
(2)
实数时间和虚数时间到场位形由
p(t = 0,x,y,z) = p(T = 0,x,y,
z
)
联系起来。
)Ae-SE/i 计算，其中 A 和是常数， SE 是一个“欧氏作用量”，由以下步骤计算：
(
于
p
(
这就是四维转动不变的含义
)
，一般的欧氏运动方程可以写成
+ f(p)

= 0
。
) (
为了寻找运动方程的四维转动不变解，使用
p = BT
2
+ x
2
+ y
2
+ z
2
作为解方程的变量比较方便。假设
p = p(
p
)
只依赖
)
(
(3)
Given
the
above
time
boundary condition, find a 4-dimensional rotational symmetric solution of the Euclidean equation of
motion
a
2
p
+
a
2
p
+
a
2
p
+
a
2
p dV
(p)
= 0
.
(3)
给定上述时间边
界
条件，找到欧氏运动方程
a
a
2
T
p
2
+
+
+

=
0
的四维旋转
对称的解。
(4)
Insert
the
solution
to
the
Euclidean
action
S
E
= ∫ dt d
3
x >
/
d
d
p
T
0
2
+
/
d
d
p
x
0
2
+
/
d
d
p
y
0
2
+
/
d
d
p
z
0
2
+ V(p)
to find
Γ
.
(4
)
将找到的解带入欧氏作用量
S
E
= ∫ dt d
3
x >
/
0
2
+
/
0
2
+
/
0
2
+
/
0
2
+ V(p)
来求出
Γ
。
)aT2 ax2 ay2 az2 dp
Let us do this calculation in this part. Note that a 4-dimensional unit ball with radius r has "volume" r4 and surface "area"
2几2r3 .
让我们在这一部分中做上述计算。注意，四维球的“体积”为 r4 ，球面面积为 2几2r3 。
(
C
1
.
THE
EUCLIDEAN
EQUATION
OF
MOTION
欧氏运动方程
)
Since we are to look for a 4-dimentional rotational symmetric solution of the Euclidean equation of motion, it is convenient to
(
dimensional rotational symmetri
c
),
the
general
Euclid
ean
equation
of
m
otion
can
be
written
as
+
f
(
p
)

= 0
.
)use p = BT2 + x2 + y2 + z2 as the variable for equation solving. Assuming that p = p(p) only depends on p (i.e., 4-
C1 Find the expression of f(p). 求 f(p)。 Hint: if the formal calculation needs too much calculus, you can consider an example: by tuning the form of V(p), one can obtain a solution p = p2 . Then f(p) can be solved by this example (and this form f(p) will apply for all forms of V(p), not limited to this special form of solution). 提示：如果进行普适的计算需要太多微积分，你可以考虑一个例子：通过调整 V(p) 的形式，我们得到一个解 p = p2 。这时， f(p) 可以从这个例子里解出来(之后这个 f(p) 的形式对所有 V(p) 都适用，不仅限于这个特殊解)。 2 points 2 分
(
Solution
:
Since the above equati
on
is
a
general
one
,
we
can
design
a
potential
,
such
that
the
solution
of
the
Euclidean
equation
is
p
=
p
2
= T
2
+ x
2
+ y
2
+ z
2
. We have
=
=
=
= 2
On
the
other
hand
,
= 2, = 2p
and
8
=
+
+
+
=
+ f(p)
= 2 + 2pf(p)
Then
2p × f(p) = 6, f(p) = 3/
p
.
d
2
p
dp
dp
2
dp
)
(
C
2.
THE
EUCLIDEAN
ACTION
欧氏作用
量
)
C2 Write the Euclidean action SE as an integral of p from p = 0 to p → ∞ . 以对 p 的积分(从p = 0 积到 p → ∞ )的形式写出欧氏作用量 SE 。 2 points 2 分
(
Solution
:
For
4
-
dimensional space, we have
dTdxdydz = 2
几
2
p
3
dp
and
)
(
S
E
=
2
几
2
伪
p
3
6
/
0
2
+
V7
dp
.
)
(
C
3
.
QUALITATIVE
INSPECTION
定性讨论
(4')
)
Before trying to solve the Euclidean equation of motion, let us first see how it behaves. If you are not familiar with this Euclidean equation of motion, you can consider the following analogy: consider p as the position of a particle, and p as an effective time
variable. In this case, answer the following questions (choose one from the options):
在解欧氏运动方程之前，我们先看看方程的性质。如果你不熟悉欧氏运动方程，你可以用如下类比来理解：把 p 类比为一个粒子的位置，把 p 类比为适用于这个粒子的有效时间变量。在这种情况下，回答以下问题(单项选择题)：
C3- 1 What's the nature of f(p) 以下哪项是 f(p) 的性质？ (A) friction (i.e. decelerate the particle) 阻力(即让粒子减速) (B) anti-friction (i.e. accelerate the particle) 推力(即让粒子加速) (C) friction for > 0 and anti-friction otherwise 在 > 0 情况下是阻力，否则是推力 (D) friction for < 0 and anti-friction otherwise 在 < 0 情况下是阻力，否则是推力 1 point 1 分
C3-2 (
(B)

V
(D)

dV/
d
p
)What's the force that drives the motion of the "particle" p 驱动粒子 p 运动的力是哪个？ (A) V (C) dV/dp 1 point 1 分
(
C
3-3
Where
is
the
"
starting
point
"
of
p
at
p
= 0
(
Here
6
is
extremely
small
but
finite
)
在
p
= 0
，
p
的
“
起始位置
”
在哪里？
(
其中
6
是一个非常小但有限的数
)
(
B
)
p
-
(
C
)
p
-
+ |0(6)|
(
E
)
p
+
(
F
)
p
+
+ |0(6)|
1
point
1
分
) (
(A)
p
-

|0(6)|
)(D) p+ |0(6)|
(
(A)
p
-

|0(6)|
) (
C
3-4
Where
is
the
"
end
point
"
of
p
at
p → ∞

在
p → ∞
，
p
的
“
最终位置
”
在哪里？
(
B
)
p
-
(
C
)
p
-
+ |0(6)|
(
E
)
p
+
(
F
)
p
+
+ |0(6)|
1
point
1
分
)(D) p+ |0(6)|
(
Solution
: 1.
A
; 2.
C
; 3.
C
; 4.
E
Note
:
For
(
C
3-3)
the
solution
is C instead of B. One can see it from two ways: First, consider x=y=z=0. In this case,
p = 0
denotes
the
bubble
center
after
bubble
nucleation
.
Then
from continuity of
p
, the field value at the bubble center should have a slightly
(
by
an
exponentially
small
margin) larger field value than
p
-
. The second way is considering the effective particle and the friction
force. F
or (C3-4), at
infinity
,
consider
T
→ 0
,
then
it
corresponds
to
spatial
infinity
,
thus
p
=
p
+
(
spatial
infinity
makes
it
exact
).
)
(
C
4
.
THE
BUBBLE
NUCLEATION
RATE
泡泡产生率
)
C3-4 Express SE in terms of R and a. 用 R 和 a 写出 SE 。 4 points 4 分
(
Solution
:
For
the integral
S
E
= 2
几
2
伪
p
3
6
/
d
d
p
p
0
2
+ V7 dp
Note
that
at
p
→
∞
, the contribution vanishes. Thus, the integral breaks up to
)
(
(
1)
The
contribution
of
the
domain
wall
(
replacing
x
with
p
):
(
surface
area
)
x
(
tension
) =
2
几
2
R
3
.
Note: you m
ay
notice
that
the
dimension
of
the
problem
here
is
different
from
the
problem
above
.
But
at
the
domain
wall
,
p
is
a constant
and
all
the
above
calculation
of
the
domain wall tension carries over to here.
(
2)
The
contribution
of
the
vacuum
energy
inside
the
bubble
.
(volume) x (
vacuum
energy
density
)
=
几
2
R
4
e
Summing
the
two
terms
up
and
inserting
e
=
37
R
,
we
get
S
E
=
1
几
2
R
3
.
Reference
: "
Fate of the false vacuum: Semiclassical theory", by Sidney Coleman, Phys. Rev. D
15
, 2929 (1977).
2
)
(
洛伦兹互易性是电磁
学
的基本原理，在天线设计理论中具有重要应用。
它指出天线的接收和发射能力是相同的。
另一方
面，在外磁场下，使用具有
强
磁光效应的磁性材料可以破坏互易性
。
洛伦兹互易的研究可以扩展到静磁学，如下图所示，
有两个电流环和一
个
固定在某个位置的物体。
当一个电流环在发射模式工作时，另一个电流环在接收模式工作。
如果我
们将两个电
流
环设为位置
r
1
和
r
2
处的磁偶极子
m
1
和
m
2
，分别在发射模式下产生磁通密度
B
1
(r)
和
B
2
(r)
。
那么，
互
)
Problem 2: (32 points)
问题 1: (32 分)
Lorentz reciprocity is a fundamental principle in electromagnetism with important application in antenna design theory. It states that the receiving and transmitting capabilities of an antenna are identical. On the other hand, reciprocity can be broken by using magnetic materials under an external magnetic field with strong magneto-optical effect. The study of Lorentz reciprocity can be extended to nearly zero frequency at magnetostatics, shown in the figure below, with two current coils and an arbitrary object fixed in locations. When one current coil works in transmitting mode, another one works in receiving mode. If we model the two current coils as magnetic dipole moments m1 and m2 at locations r1 and r2 , generating magnetic flux densities B1 (r) and B2 (r)
(
m
1

B
2
(
r
1
) =
m
2

B
1
(
r
2
)
)in transmitting mode, respectively. Then, the reciprocity relationship can be expressed as
(
m
1

B
2
(
r
1
) =
m
2

B
1
(
r
2
)
)易关系可以表示为
m! at r1
Transmitting
Object
B! (r2 )
Receiving
B2 (r! )
Receiving
Object
m2 at r2
Transmitting
The magnetic field B(r) generated by a magnetic dipole mi located at ri is given by,
(
B
i
(
r
)
=
j
3(
r
i
)
r

m
i

k
)在位置 ri 处的磁偶极子 mi 所产生的磁场 Bi (r) 由下式给出，
(
In this problem, we will first establish the reciprocity relationship and will in
vestigate
how
it
can
be
broken
by
imposing
a
constant velocity
on
the
object
.
)本题中，我们将首先建立互易关系，并研究如何通过对物体施加恒定速度来打破它。
(
A
.
ESTABLISHING
MAGNETOSTATIC
RECIPROCITY
AND
NON
-
RECIPROCITY
建立静磁互易性和非互易性
)
(

(A × B) = ( × A) B A × B
) (
∫ AdV = ∫ A d
a
) (
∫
B × AdV = ∫ A × BdV + ∫ A × B da
) (
∫
× AdV = ∫ A × da
) (

A = a
x
A
x
+ a
y
A
y
+ a
z
A
z
)
For magnetostatics, we have vector potential A, magnetic flux density B, magnetic field strength H and impressed current density J. These fields satisfy the Ampère’s law
× H = J
with material response
× A = B = uH
where u(r) is the isotropic magnetic permeability profile for the material, i.e. the object in the figure. 对于静磁学，我们有矢量势 A 、磁通密度 B 、磁场强度 H 和外加电流密度 J。这些场满足安培定律
× H = J
关于物质反应的表达式为
× A = B = uH
其中 u(r) 是材料的各向同性磁导率分布，即图中的物体。
(
respectively
.
The
two
cases
,
labeled
by
i = 1,2
, have the current density
J
i
=
× M
i
and magnetization
M
i
= m
i
6(r r
i
)
for
)We want to establish the reciprocity relationship for magnetostatics when we have a magnetic dipole moment m1 at location r1
in one case and a magnetic dipole m2 at r2 in another case. The two dipoles generate magnetic fields B1 (r) and B2 (r) ,
the two dipoles.
(
对于由
i = 1,2
索引的两个偶极子，我们有电流密度
J
i
= × M
i
和磁化强度
M
i
= m
i
6(r r
i
)
。以下我们想要建立静磁
)在一种情况下r1 处有一个磁偶极子 m1 并产生磁场 B1 (r)而在另一种情况下r2 处有一个磁偶极子 m2 并产生磁场 B2 (r)。
学的互易关系。
Here, we also give some formulas for these differential operators:
(

× A =
ra
y
A
z

a
z
A
y
s +
(a
z
A
x

a
x
A
z
) +
ra
x
A
y

a
y
A
x
s
)在这里，我们提供一些关于微分算子的公式：
and Kronecker delta function 6(r) is defined by
6(r) = v
which satisfies ∫ 6(r)dV = 1 when we integrate a volume V enclosing the origin. a is defined as the closed area enclosing volume
(
V
).
当我们对包含原点的一个体积 V 进行积分时， Kronecker delta 函数 6(r) 定義為:
6(r) = 0 (∞) 除 ( i)此 (f r)以 (=)外 (0)
而且满足∫ 6(r)dV = 1 。 a 定义为包围体积 V 的封闭区域。
(
Prove the reciprocity relation
ship
m
1

B
2
(
r
1
) =
m
2

B
1
(
r
2
)
.
Hint
:
you
may
consider
(
H
1
×
A
2
)
.
)A1. PROVING THE RECIPROCITY RELATIONSHIP 证明互易关系
A1 (
证明互易关系
m
1

B
2
(r
1
) = m
2

B
1
(r
2
)
。
提示
：你可以考虑
(H
1
× A
2
)
。
) 3 points 3 分
(
1
+
1
) (
B
i
(
r
)
=
u
4
D
几
k
) (
∫

(H
1
×
A
2

H
2
×
A
1
)dV
) (
j
3(r

i
)
r

m
i

|
r
i
r
i
|
3
) (
∫
M
1

B
2
dV

∫
M
2

B
1
dV
=
∫
M
1
×
A
2

da
+
∫
M
2
×
A
1

da
magnetization
M
1
and
M
2
,
we
arrive
m
1

B
2
(r
1
) = m
2

B
1
(r
2
)
Localized
sources
imply
the
right
hand
side
goes
to
zero
.
Finally
,
from
two
point
dipoles
m
1
at
r
1
and
m
2
at
r
2
for
the
two
) (
Soluti
on
:
In magnetosta
tics
,
we
have

×
A
=
B
=
uH

×
H
=J
where
the
free
currents
are
from
impressed
current coils which can be described as
magnetization
M
through
J
=

×
M
For
two
different
sets
of
fields
and
currents
labeled
by
subscript
1
and
2 (
with
same
medium
u
),
we
have

(H
1
×
A
2

H
2
×
A
1
)
) (
A
H
B
A
H
B
) (
As
integration
volume
grows
to
infinity
,
the
integral
(with Gauss’s law) decays to zero by
) (
|H|
∝
1/r
3
,
|A|
∝
1/r
2
,
|H
×
A|
∝
1/r
5
) (
If the material conduc
ts
electricity
with
an
electrical
conductivity
G
,
we
have
to
add
an
additional
term
to
the
current
density
J
due t
o
the
free
current
through
)
(
=
J
1

2

1

2

J
2

2

) (
=J
1

A
2

J
2

A
1

uH
1

H
2
+uH
2

H
1
) (
There
are
only
close
-
loop
currents
,
i
.
e
.
made
of
magnetic
dipoles
.
For
each
magnetic
dipole
,
we
have
) ∫ (H1 × A2 H2 × A1)dV = ∫ J1 A2dV ∫ J2 A1dV
Then
(
By substituting
J = × M
, we hav
e
)∫ J1 A2dV = ∫ J2 A1dV
J → J + GE
in the Ampère’s law stated previously. Suppose now we move the conductor by a constant velocity v. There will be a Lorentz force on the free charges proportional to E + v × B. It further updates the additional term in the current density through
J → J + G(E + v × B),
which may upset the reciprocity relationship.
如果材料以电导率 σ 导电，由于通过的自由电流，我们必须为在此前安培定律中的电流密度 J 添加一个附加项
J → J + GE.
假设现在我们以恒定速度 v 移动导体。自由电荷上将存在与 E + v × B成比例的洛伦兹力，这进一步更新电流密度中的附加项以使破坏互易关系变的可能：
(

) (

) (
A
2
Express
the
possibly
non
-
zero
m
1
B
2
(
r
1
)
m
2
B
1
(
r
2
)
as
a
volume
integral
in
terms
of
the
vector
potentials
A
1
and
A
2
and
the
conductor
velocity
v
.
将可能非零的
m
1
B
2
(r
1
) m
2
B
1
(r
2
)
表示为根据矢量势
A
1
和
A
2
以及导体速度
v
的体积积分。
3
p
oints
3
分
)
J → J + G(E + v × B).
(
A
2.
BREAKING
RECIPROCITY
RELATIONSHIP
打破互易关
系
)
(
Solution
:
The
Ampère’s law becomes
Then
,
the
additional
part
to
(
H
1
×
A
2

H
2
× A
1
)
is
Gv (B
1
× A
2

B
2
×
A
1
)
T
herefore
=
Gv

r( × A
1
) × A
2

( × A
2
) × A
1
s
m
1

B
2
(r
1
) m
2

B
1
(r
2
) = ∫ Gv rA
2
× ( × A
1
) A
1
× ( × A
2
)sd
V

×
H = J + G(E + v × B)
G
(
E
1

A
2

E
2

A
1
) + G(v × B
1

A
2

v × B
2

A
1
)
The volume integration of the first term goes to zero for
× E
i
= 0
and decay o
f
the
boundary
terms
as
T
approaching
infinity
.
The
second term is
)
You can complete part B and C without part A.
你可以在没有 A 部分的情况下完成 B 部分和 C 部分。
(
B
.
MAGNETIC
DIPOLES
ON
A
MOVING
PERFECT
CONDUCTOR
运动理想导体上的磁偶极子
)
In this part, we first obtain the magnetic field of a single magnetic dipole on a moving conductor, which is a perfect conductor (i.e. the electrical conductivity G = ∞ ) here, occupying Z ≤ 0 and is moving with a velocity v in the positive x-direction. This is defined
(
has a magne
tic
moment
m
=
m
x
+
m
&
with
zero
component
in
the
y
-
direction
.
)as the laboratory frame, as shown in the figure below. The magnetic dipole, situated at (X, y, Z) = (0,0, ZD ) on top of the conductor,
(
在这
一部分中，我们首先得到运动导体上单个磁偶极子的磁场，这里是理想导体
(
即电导率
G = ∞
)
，
占据
z
≤
0
并且在正
x
方向上
以速度
v
运动。
这被定义为实验室坐标系，如下图所示
。
位于导体顶部
(X, y, Z) = (0,0, Z
D
)
处的磁偶极子写为
)m = mx + m& ，在 y 方向上没有分量。
(
其
中
y = 1/B1 v
2
/c
2
，
c
是光速。
) (
v
) (
y
z
'
x
) (
The
magnetic
and
electric
fields
(
B
′
and
E
J
)
in
the
moving
frame
are related to the fields (
B
and
E
) in the laboratory frame by the
Lorentz transfor
mation
,
) (
where
y
= 1/
B1 v
2
/
c
2
and
c
is the speed of light.
) (
Conducting
Object
)
z
m = mx + m&
(
giving
us
a
convenience
to
find
the
magnetic
fields
generated
by
the
magnetic
dipole
.
The
coordinates
in
the
moving
frame
,
denoted
as
(
x
J
,
y
J
,
z
J
,
t
J
)
,
is
transformed
from
the
coordinates
in
the
laboratory frame
(x,y,z,t)
through the Lorentz
transformation
:
)In the moving frame at a velocity v = v with respect to the laboratory frame, the object is simply a perfect conductor at rest,
(
在相对于实验室坐标系以速度
v
=
v
运动的坐标系中，物体只是一个静止的
理想导体，
使我们更方便地找到磁偶极子
) (
x
J
=
y(x

vt),
y
J
=
y,
z′
=
z,
ct′
=
y(ct

vx/c)
)产生的磁场。移动坐标系中的坐标，表示为 (xJ,yJ,zJ,tJ)，由实验室坐标系(x,y,z,t) 中的坐标通过洛伦兹变换转换而来：
移动坐标系中的磁场和电场(B′和 E′) 与实验室坐标系中的场(B 和 E) 相关，可由洛伦兹变换给出:
(
à

J
$
J
y
J
&
=
à

+
à
y

,
) (
We
further
simplify
the
problem
by
removing
the
conductor at the moment.
我们现
在通过移除导体来进一步简化问题。
)à J$JyJ& = à vy à y
(
B
1
.
MAGNETIC
FIELD
OF
A
MOVING
MAGNETIC
DIPOLE
IN
FREESPACE
自由空间中运动磁偶极子的磁场
)
B1 (
to
the
laboratory
frame

In
this
part
,
we
only
consider
a
magnetic
dipole
m
=
m
$
pointing
in
the
x
-
) (
direction
and
there
is
no
conductor
below
the
dipole
yet
.
Please
express
your
answer
in
the
coordinates
of
the
moving
frame
.
请
找出相对于实验室坐标系以速度
v
移动的观察者的偶极子磁场
B
J
(x′,y′,z′,t′)
。
在这一部分中，
)What is the magnetic field BJ (x′,y′,z′,t′)from the dipole for an observer moving at velocity v with respect (
我们只考虑指向
x
方向的磁偶极子
m
=
m
$
并且偶极子下方
还没有导体。
请用移动坐标系的坐标
)表达你的答案。 3 point 3 分
(
Soluti
on
:
u
4
D
几
j
3(r

z
|
D
z
r
r
z
D
)

m

|
r
D
|
3
k
The
B-field for a magnetic dipole
m = m
$
h
e
l
d
at
z
D
is
)
(
For
m
x
dipole
,
we
have
In
the
moving
frame
,
the
fields
tr
à
a

or
=
m
4
t
o
几
z
D
)
2
)
5
/
2
à
2x
2
y
x
2
(
3
z
z
z
D
z
D
)
2

à

J
x
J
y
J
&
=
à

where
x
=
y(x
J
+ vt
J
), y = y
J
, z = z
J
, ct = y(ct
J
+ vx
J
/c)
with
y = 1/B
1 v
2
/c
2
.
We also have
x
J
= vt
J
+ x/y
. Full
marks
if
the
answer
is
given
in
non
-
relativistic
limit
y
→ 1
.
=
4
几
z
D
)
2
)
5/2
z
z
J
D
z
D
)
2

)
Next, we introduce the perfect conductor below the dipole. The perfect metal has a planar interface at z = 0 while the dipole is placed at a distance zD above the metal. The metal is being moved at a velocity v in the positive x-direction.
接下来，我们引入偶极子下方的理想导体。理想导体在 z=0 处有一个平面界面，而偶极子位于金属上方 zD 处。金属在
正 x 方向上以速度 v 移动。
B2 . MAGNETIC FIELD OF A MAGNETIC DIPOLE ON A MOVING CONDUCTOR (I) 运动导体上磁偶极子的磁场 ( 1 )
B2 (
(i.e
B(x, y, 0
+
)
) in terms of the laboratory
frame coordinate. Pleas
e
also
verify
the
boundary
condition
of the
magnetic
field
in
the
moving
frame
at
which
the
conductor
at
rest
.
Hint
:
adopt
the
method
of
images
) (
证导体静止时移动坐标系中的磁
场边界条件。
提示：采用镜像法
。假定在接近零频率时完美导体
里电场和磁场为零。
)Please express the magnetic field of the magnetic dipole m = mx on top of a moving perfect conductor (
请用实验室坐标系表示
在
移动中的理想导体上磁偶极子
m = m
x
的磁场
(
即
B(x, y, 0
+
)
)
。
还请验
)and assume both electric and magnetic fields inside a perfect conductor is zero at a nearly-zero frequency. (
4
)points 4 分
(
Solution
:
We
adopt
the
method
of
images
by
having
a
magnetic
dipole
m
x
at
mirror
location
at

z
D
.
In
the
moving
frame
with
velocity
v
In
the
laboratory
frame
(
dipole
at
rest
),
the
total
field
is
à

=
u
D
4
m
几
z
D
)
2
)
5
/
2
à
2x
2
y
x
2
(
3
z
z
z
D
z
D
)
2

+
u
D
4
m
几
x
à
2x
2
y
x
2
(
3
z

The
total field in the moving frame is
with respect to the laboratory frame, the conductor is at rest, and we can app
ly
the
normal
boundary
condition
B
J
&
= 0
on
the
surface
of
the
perfect
conductor
.
for
z ≥ 0
.
)
(
à

J
x
J
y
J
z
=
à

=
(y
2
(x
J
+ vt
J
)
2
+
+ (z
J

z
D
)
2
)
5/2
z
z
J
D
z
D
)
2

for
z
J
≥ 0
.
At
z
J
= 0
+
,
we
have
à
J
x
J
y
J
z
=
u
D
2
m
冗
x
(y
2
(x
J
+ vt
J
)
2
1
+y
J2
+z
2
D
)
5/2
à
2

z
2
D

satisfying boundary condition on surface of conductor
B
J
z
= 0
in jus
tifying
the
solution
from
the
method
of
image
.
The
magnetic
field
in
the
laboratory
frame
(
for
the
case
of
no
electric
fields
in
the
laboratory
frame
)
is
à

=
à
B
B
B
J
y
J
z
J
x
/
/
y
y

=
u
D
2
m
冗
x

(2x
2

y
3
y

z
2
D
)/y
ó
Full
marks
are
given
if
answers
are
expressed
in
either
laboratory
frame
or
moving
frame
coordinates
.
For
a
perfect
metal
,
the
mirror
dipole
of
rm
x
,
m
y
,
m
z
s
is
rm
x
,
m
y
,
m
z
s
by
definition
.
However
,
it
is
required
to
verify
the
+
4
冗
(y
2
(
x
J
+ vt
J
)
2
+y
J2
+ (
z
J
+ z
D
)
2
)
5/2

boundary
condition in the moving frame as stated.
)
B3. MAGNETIC FIELD OF A MAGNETIC DIPOLE ON A MOVING CONDUCTOR (2) 运动导体上磁偶极子的磁场 ( 2 )
B3 (
m = m
z
, pointing in the positive
z
directi
on

) (
如果运动导体上方的磁偶极子转为
m = m
z
z
，指向正
z
方向，
那么请找出实验室坐标系中磁偶极子
在移动中的
理
想导体上的磁场
(
即
B(x,y,0
+
)
)
？
)What is the magnetic field from the magnetic dipole on top of a moving perfect conductor (i.e (

)B(x,y, 0+)) in the laboratory frame if the magnetic dipole above the moving conductor is changed to (
3
)points 3 分
(
Solution
:
For
a
m
z
dipole
,
we
have
the
magnetic
field
in
the
laboratory
frame
as
Putting
a
mirror
dipole

m
z
results
the
total
field
as
At
z = 0
, we have
à

=
4
冗

z
D
)
2
)
5/2
à
x
2

3
3
(
(
2

z
D
)
2

4
冗
à
x
2

3
3
(
(
2
+
z
D
)
2

à

=
4
冗

z
D
)
2
)
5/2
à
x
2

3
3
(
(
2

z
D
)
2

)
(

=
B
z
(6) B
z
( 6)
) (
值定义为

= rB
1z
(6) B
2x
( 6)s/rB
1z
(6) + B
2x
( 6)
s
。
) (
'
) (
B
4
.
RECIPROCITY
MERIT
FOR
DIPOLES
ON
A
PERFECT
CONDUCTOR
理想导体上偶极子的互易功值
) (
Conducting
Object
v
) (
施加到
r
1
=
(x
1
, 0, z
D
)
上。
然后我们定义互易功值
) (
function
B
z
(x x
i
)
. We th
en
define
a
reciprocity
figure
-
of
-
merit
) (
z
)
(
à

=
j
k
In
principle
,
the
fields
are
transformed
to
the
moving
frame
at
which
the
boundary
condition
is verified, as we have done in part B
.2
, but we have skipped here for brevity.
)
Now, we consider the two situations: one with magnetic dipole moment m1 at r1 = (0,0, zD ) and another magnetic dipole moment m2 at r2 = (6, 0, zD ). When one of them is turned on, another one is turned off.
现在，我们考虑两种情况：一种是 r1 = (0,0, zD ) 处的磁偶极矩 m1 ，另一种是 r2 = (6, 0, zD) 处的磁偶极矩 m2 。当其中一个打开时，另一个被关闭。
z
(
m
1
=
m
&
)m2 = mz
6
y
x
Now, we define the two problems to solve next. For the first problem, suppose the two magnetic dipoles moments m1 and m2
(
component
B
z
(6)
on
r
2
= (
x
2
, 0,
z
D
)
and
dipole
m
2
imposes
magnetic
field
B
z
( 6)
on
r
1
= (x
1
, 0, z
D
)
according to the
same
)point in the z-direction with same size mz , as shown in the above figure. Dipole moment m1 imposes a magnetic field of z-
(

=
B
z
(6) B
z
( 6)
) (
When

= 0
, reciprocity is satisfied. Reciprocit
y
is
broken
when

deviates
from
value
zero
.
)Bz (6) + Bz ( 6).
(
如
上图所示
。
偶极子
m
1
根据函数
B
z
(x x
i
)
将
z
分量
B
z
(6)
的磁场施加到
r
2
= (x
2
, 0, z
D
)
上，
偶极子
m
2
将磁场
B
z
( 6)
)现在，我们定义接下来要解决的两个问题。对于第一个问题，假设两个磁偶极子m1 和m2 都指向 z 方向，大小mz 相同，
(
B
z
(
6) + B
z
( 6)
.
)当 = 0时，满足互易性。当偏离零值时，互易性被打破。
For the second problem, suppose we change the pointing direction for m1 to the positive x-direction with magnitude remaining
(
the
same
.
The
magnitude
and
direction
of
m
2
are
not
changed
.
In
this
case
,
the
reciprocity
merit
is
defined
as

=
) (
对于第二个问题，假设我们将
m
1
的指向更改为正
x
方向而大小保持不变。
m
2
方向和大小不变
。
在这种情况下，
互易功
)rB1z (6) B2x ( 6)s/rB1z (6) + B2x ( 6)s.
(
moment
p
r)
at the same location
(0,0, z
0
)
. The
mirrored
magnetic
dipole
is
now
relaxed
to
have
both
magnetic
and
electric
) (
components
while
neglecting
the
higher
order
multipoles
.
The size of these dipole moments are yet to be determined.
) (
current
case
of
m
=
m
z
,
we
put
an
image
magnetic
dipole
with
given
form
of
magnetic
moment
m
r
)
+
m
r
)
and
electric
)
B4 Find the reciprocity merit for the two defined problems about magnetic dipoles on the moving perfect conductor. 求出关于以上两个问题运动理想导体上磁偶极子的互易功值。 3 points 3 分
(
Solution
:
For t
he
first
case
,
the
magnetic
field
from
dipole
1
is
B
z
(
x,0,z
0
)
=
u
0
4
冗
x
2

4
冗
2
0
)
5/2
(

x
2
+ 8z
2
0
)
which
is
even
in
x
. Then, we have

= 0
.
For
the
second
case
,
the
magnetic
field
from
dipole
1
on
dipole
2
is
B
1
z
(6) =
u
0
4
m
冗
x
The
magnetic field from dipole 2 on dipole 1 is
B
2
x
(

6) =
u
0
4
m
冗
z
with
m
=
m
They have the same values and
hence

= 0
.
In
fact
,
for
a
moving
perfect
metal
,
although
breaking
the
time
-
reversal
symmetry
,
is
still not able to generate the non-reciprocal coupling
.
We
need
dissipation
with
a
finite conductivity.
z
x
.
)
(
C
.
MAGNETIC
DIPOLES
ON
A
MOVING
CONDUCTOR
WITH
FINITE
CONDUCTIVITY
具有有限电导率的移动导体上
的磁偶极
子
)
In this part, we move to a more realistic situation that the conductor is a metal. It has a large but finite conductivity a (in unit of
Ω_1m_1), deviating from the perfect conductor condition. Current density in the conductor is given by J = aE. We also assume that the current on the conductor surface is confined by a skin depth d of small thickness so that the electric and magnetic fields cannot penetrate beyond the skin depth from the conductor surface. Then, the surface current density can be written as (ad)E. We further take the approximation that d is just a constant. We only consider the two dipoles pointing in the positive z-direction with same size mz in this part.
在这一部分中，我们转向一个更现实的情况，即金属导体具有较大但有限的电导率σ (单位为 Ω_1m_1 )，偏离理想导体条件。导体中的电流密度由 J = aE 给出。我们还假设电流只在导体表面厚度较小的趋肤深度 d 内流动，电场和磁场不能从导体表面穿透超过趋肤深度。因此表面电流密度可以写成 (ad)E。我们进一步假定 d 为一个常数。在这部分我们只考虑两个偶极子都指向正 z 方向，大小 mz相同。
Again, we need to solve the magnetic field from only one dipole at (0,0,z0) first. In fact, the surface current profile generated on the surface of conductor cannot be easily solved without adopting a numerical solver. Instead, we can approach the problem by
(
position
(0
,0, z
0
)
in order to give as closely as possible the same
reflected field generated from the surface current. For the
)extending the method of image as an approximation. In this case, we would like to have a point-like multipolar source at the image
(
电
分
量，
同时忽略高阶多极子。
这些偶极矩的大小有待确定。
) (
à
y

) (
à
= à

vy
à
y

) (
极子设置在
(0,0, z
0
)
处，它拥有给定的表达式：
磁矩为
m
r)
+ m
r)
、电矩为
p
r
)
。
镜像磁偶极子现在同时具有磁和
)
(
同样，我
们
需要先求解
(0,0, z
0
)
处只有一个偶极子的磁场。
实际上，
导体表面产生的表面电流分布不采用数值求解器是
) (
有一个
点
状多极源，以便尽可能地提供相同的从表面电流生成的反射场。
对于当前
m = m
z
的情况，我们将镜像磁偶
)无法精确求解的。这里，我们尝试通过电像法来近似解决这个问题。在这种情况下，我们希望在镜像位置 (0,0, z0) 处
(
C
1
.
GENERALIZED
METHOD
OF
IMAGES
广义电像法
)
C1 Find the magnetic field B, (x′, y′, 0+) and electric field E, (x′, y′, 0+) on the conductor surface in the moving frame. Express your answer in mz , m mr) and p You can use either the moving frame or laboratory frame coordinates. Do not need to solve the mirrored dipole moments yet. 求移动坐标系导体表面的磁场B′(x′, y′, 0+) 和电场E, (x′, y′, 0+)。请用mz 、m mr) 和pr) 表达你的答案。可以使用移动坐标或实验室坐标表达你的答案。暂时不需要求解镜像偶极矩。 5 points 5 分
(
Solution
:
The
magnetic
field
comes
from
the
magnetic dipole
m
z
at
(0,0, z
0
)
, magnetic dipole
m
r)
+ m
r)
at
(0,0, z
0
)
and a small
contribution from electric dip
ole
p
r
)
at
(0,0,
z
0
)
due
to
the
change
of
reference
frame
.
The dipole fields on metal surface in the laboratory frame
are
à

=
4
几
(
x
2
+
x
0
2
2
0
4
z
几
0
)
u
2
r
z
)
+
z
0
)
2
)
5/2
à
x
2

3
3
(
(
2
+
z
0
)
2

at
z
=
0
, giving
)
We also have à = 4几0 (2))5/2 r)z20r (s)yxs)0m (m)r (r)ú
(
à
=
4
几
)
z
2
0
)
5
/
2
à

x
2

z
2
0

)
(
In
the
moving
frame
,
the
fields
are
transformed
by
)
(
à

= à
+
)
(
Then
,
)
(
à

/
4
几
2
0
)
5
/
2
=

(2z
2
0

x
2

y
2
3
xz
3y
yr
0
r
z
0
m
m
yr
(r)
z
)
2

z
2
0
)
vy
2
p
r)
ú
à

/
4
几
(
x
2
+
u
y
0
2
+z
2
0
)
5
/
2
=
( x
2
+ 2y
2

z
r)
2
p
p
y(
x
2
(r)
z
r
v
+
m
z
s
s 3xz
0
vym
r)
Full
marks
will be given for fields expressed either in lab frame or moving frame coordinates.
)
C2. FINDING THE MIRRORED DIPOLE MOMENTS 计算镜像偶极矩
C2 Find m mr) and pr) in response to a given mz . The mirror dipole gives the same reflected field generated by the surface current on the conductor. As approximation, only apply the boundary condition (in the moving frame) on the surface current along the y-direction, which is the dominant current than the one along the x-direction. It may be useful to express the answers in term of the dimensionless parameter x = u0vyGd 1. 用给定的mz表示 m mr) 和 p 镜像偶极子给出了与导体表面电流产生的相同的反射场。作为近似，请仅将边界条件(在移动坐标系中)应用于沿 y 方向的表面电流，该电流比沿 x 方向的电流占主导地位。可以考虑用无量纲参数x = u0vyGd 1 表达答案。 (
5
)points 5 分
(
Solution
:
The
surface
current
is
given
by
3
xz
0
rm
r)

m
z
s + (2x
2

y
2

z
2
0
)m
r)
and
hence
r
=
m
x
z
x
+
m
z
+ 2c
2
p
r)
/vs
x
m
z
1 +
x
2
x
z
p
y
v
2
v
2
m
z
c
2
1 + x
2
c
2
x
2
where
=
x
/(
x
2
+ 2y
2

z
2
0
)c
2
p
r)
/v (2z
2
0

x
2
y
2
)rm
r)
+m
z
s 3xz
0
m
r)
0
where
x
=
u
0
vyGd
.
Equating
coefficients
for
the
different
powers
of
x
and
y
give
2
m
r
)
=
xrm
r
)
+
m
z

c
2
p
r
)
/
vs
J
,
=
d
G
E
,
=
×
B
,
/u
0
at
z
,
= 0
. The second equality comes from Ampère’s l
aw
with
the
assumption
that
the
field
decays
to
negligible
values
beyond
skin depth. The dominant current is along the y-direction. We h
a
ve
B
,
=
u
0
d
G
E
,
:
=

1
+
m
(
r
)
1
x
2
2
m
z
1 +
x
2
x
2
(
r)
=

m
(
r
)
2
x
2
=
)
x = u0vyad. We have assumed a large x limit so that the material deviates a bit from the perfect metal.
C3. RECIPROCITY MERIT FOR DIPOLES ON A CONDUCTOR OF FINITE CONDUCTIVITY 有限电导率导体上偶极子的互易功值
C3 Find the reciprocity merit for the two identical dipoles mz displaced by 6 (with r1 = (0,0,z0), r2 = (6, 0,z0))in the x-direction. 找出在 x 方向上位移6 (r1 = (0,0,z0), r2 = (6, 0,z0))) 的两个相同偶极子 mz 的互易功价值。 3 point 3 分
(
Solution
:
The
magnetic
field
has
contributions
from
the
original
m
z
dipole
and
the
mirror
magnetic
dipole
:
B
z
(
x
,
y
,
z
)
=
4
几
(
x
2
+y
2
u
z
z
z
r
)
2
)
5
/
2
(

x
2

y
2
+ 2(z
z
0
)
2
)
+
4
几
(
x
2
+y
u
2
r
z
)
+ z
0
)
2
)
5
/
2
(

x
2

y
2
+ 2(z + z
0
)
2
)
=
4
几
2
0
)
5/2

+
(8z
2
0

6
2
)
+
6z
0
6ó
Then

=
B
z
(6,0,z
0
) B
z
( 6,0,z
0
)
m

sign(6)
x
By furt
her
substituting
last
part
’
s
answer
,
we
obtain
sign
(
6) 12|6/z
0
|
4
Full
marks
can be given
without the final substitution.
Reference
:
J
.
Prat-Camps, P. Maurer, G. Kichmair, and O. Romero-Isart, Phys. Rev. Lett. 121, 213903 (2018).
When
the
material
is
approaching
a
perfect
metal
,
we
have
m
r
)

m
z
to
be
substituted
into
the
denominator
,
giving
x
6|6/z
0
|
4
=
sign(6)

(4
+
|6/
z
0
|
2
)
5
/
(8

|6/
z
0
|
2
)|6/
z
0
|
3

x
(4
+
|6/
z
0
|
2
)
5/2
+
(8

|6/
z
0
|
2
)|6/
z
0
|
3
m
(
r)
6|6/z
0
|
4
m
z
(4
+
|6/z
0
|
2
)
5
/
2
+
(8

|6/z
0
|
2
)|6/z
0
|
3
B
z
(6
,0,z
0
)
+
B
z
(

6,0
,z
0
)
+
4
几
/2
r3x(z
+
z
0
)s
B
z
(6
,0,z
0
)
(
r)
)1
Pan Pearl River Delta Physics Olympiad 2022
2023 年泛珠三角及中华名校物理奥林匹克邀请赛
Sponsored by Institute for Advanced Study, HKUST
香港科技大学高等研究院赞助
Simplified Chinese Part-1 (Total 4 Problems, 40 Points) 简体版卷-1 (共4题， 40分) (9:30 am – 12:00 pm, 29th Jan 2023)
Please fill in your final answers to all problems on the answer sheet. 请在答题纸上填上各题的最后答案。 At the end of the competition, please submit the answer sheet only. Question papers and working sheets will not be collected. 比赛结束时，请只交回答题纸，题目纸和草稿纸将不会收回。
1. [10 points] Consider a satellite that has a shape of a plate of surface area A and thickness d VA. The satellite can convert solar energy into electrical energy and charge the onboard batteries making use of the temperature difference. The solar energy flux density [solar power per unit area] at the position of the satellite is S and the satellite is facing towards Sun. Assuming that the emissivity of both sides of the satellite is e and the temperatures of the satellite on two sides are T1 and T2 respectively and G is the Stefan-Boltzmann constant.
1. [10 分] 考虑一个人造卫星，其形状为表面积为 A 且厚度为d VA 的平板。卫星可以利用温差将太阳能转化为电能，为星载电池充电。人造卫星所在位置的太阳能通量密度[单位面积的太阳能功率]为S ，卫星正对着太阳。假设整颗卫星的发射率 (又称辐射率) 为e ，卫星两侧的温度分别为 T1 和 T2 ，G 是 Stefan- Boltzmann 常数。
(a) [2] What is the net heat flux [energy per second] absorbed by the bright side (the side facing towards Sun) of the satellite
(a) [2]卫星的亮面(面向太阳的一面) 吸收的净热通量 [每秒能量] 是多少？
(b) [1] What is the net heat flux [energy per second] released from the dark side of the satellite
(b) [1] 从卫星暗面释放的净热通量 [每秒能量] 是多少？
(c) [1] What is value of the emissivity e to get the theoretically maximal charging power Pmax
(c) [1] 可得到理论上最大充电功率 Pmax 的发射率 e 的值是多少？
(d) [3] Find a condition for the temperature T1 in order to get the theoretically maximal charging power Pmax provided by the satellite. Express the condition in term of the dimensionless variable X = . You don’t need to solve the equation in this part.
(d) [3] 求温度T1 的条件，使得卫星能達到理论上最大充电功率 Pmax 。用无量纲变量 X = 表示你的答案。你不需要解这部分的方程。
(e) [3] What is the theoretically maximal charging power Pmax provided by the satellite Calculate the numerical value of . Your answer should be correct to at least 2 significant figures.
(e) [3] 卫星可提供的理论上最大充电功率Pmax 是多少？计算的数值。你的答案至少应正确到 2 位有效数字。
(
Q

1
= eA(S GT
4
1
)
Solution
:
(
a
)
)
1
(
Q

1
Q

2
) (

) (
=
) (
T
1
) (

T
3
2
=
) (
Sub
.
Into
eqtn
(1)
,
) (
P
=
eA
(
S aT
4
1

aT )
4
2
= eA 4S aT
4
1

a 5
T
3
1
6
4/3
7
) (
(
1 X)
(1 X)
) (
+
) (
T
1
T
2
) (
For
maximal
power
,
the
satellite
should
be
a
Carnot
engine
,
) (
Obviously
,
we
can
get
maximal
power
if
e
= 1
.
) (

T
3
1
)1
(
(b)
) (
P
=
Q

1

Q

2
= eA(S aT
4
1

aT
4
2
)
(1)
)(c) The power is given by Q 2 = eaAT2 (4)
(
eA
(
S aT
4
1
)
)=
(d) Define X = , we have
(
= 1 X 8
1
X
9
= 1
X

(1
)
(
We
can
get
max
.
power
P
if
X
satisfies
the
following
equation
:
)
(
5
6 = 1
X
1/3
3
X
4/3
= 0
)
(
(
e
)
We
can
solve
the
equation
numerically
and
we
get
:
)
X ≈ 0.751
The corresponding value of power is
(
P
) max
= 0.077
AS
(
2. [10 points] Binary-sun solar system: Consider a binar
y
pair
of
identical
suns
of
mass
M
orbiting
in
the
X

y
) (
center
of
mass
of
two suns
will also remain on the
z
axis.
)plane in an orbit centred at the origin. The gravitational constant is G . Now add a planet of mass m with an initial condition on the z axis above the center of mass of the two suns and with a velocity along the z direction. By the symmetry of the system, the small planet will remain on the z axis, suns will have equal z coordinates and the
(
2. [10
分
]
双星太阳系：考
虑两个质量均为
M
的相同
太阳在
X y
平中以原点为中的轨道运。万有引
) 常数为 G 。现在添加个质量 m 的星，其初始条件位于两个太阳质上的 z 轴上，速度在 z 向上。通过系统的对称性，星将保持在 z 轴上，两个太阳的 z 坐标相同，其质也维持在 z 轴上。
2
1
We use the Cartesian coordinates to describe the dynamics of the system: the coordinate of the planet (0,0, u), and the coordinates of two suns are (±R cose , ±R sine , Z).
我们笛卡尔坐标来描述系统的动学：星坐标为(0,0, u) ，两个太阳的坐标为(±R cose , ±R sine , Z) 。
(a) [0.5] What is the total kinetic energy, T, of the system Express the answer in terms of R, e, Z, u and their time derivative.
(a) [0.5] 系统的总动能 T 是多少？ R, e , Z, u 及其时间导数表答案。
(b) [0.5] What is the total potential energy, V, of the system Express the answer in terms of R, e, Z, u and their time derivative.
(b) [0.5] 系统的总势能 V 是多少？ R, e , Z, u 及其时间导数表答案。
(c) [0.5] What is the total linear momentum, P, of the system Express the answer in terms of R, e, Z, u and their time derivative.
(c) [0.5] 系统的总线性动量 P 是多少？ R, e , Z, u 及其时间导数表答案。
(d) [0.5] What is the total angular momentum, L, about the z axis of the system Express the answer in terms of R, e, Z, u and their time derivative.
(d) [0.5] 关于系统 z 轴的总动量 L 是多少？ R, e , Z, u 及其时间导数表答案。
(
Q
(
t
) =
mu
+
2
MZ
) (
m
+2
M
.
)From now on, we introduce the dynamical variables q(t) = u Z and the center of mass coordinate of the system
(
m
+
2
M
) (
(e) [2] Find the equation of motion for
q(t).
Express
your answer in
terms
of
,
q
,
R
and
given
physical
parameters
.
)从现在开始，我们引入动学变量q(t) = u Z 和系统的质坐标 Q(t) = mu+2MZ . 。
(
(f) [2] Find the equation of motion for
R(t).
Express your answer in t
erms
of
,
R
,
q
,
L
and
given
physics
)(e) [2] 找出 q(t) 的运动程。 , q, R 和给定的物理参数表达你的答案。
(
parameters
.
)(f) [2] 求出 R(t) 的运动程。 , R, q, L 和给定的物理参数表达你的答案。
(g) [2] In the limit of small planetary mass m M we can ignore the effect of the planet on the motion of the suns. Find the explicit solution for the motion of the suns R(t) for orbits with small eccentricity e 1. Write your
3
1
solution as circular motion plus a term proportional to e, i.e. R(t) = R5 + eR1 (t) where R5 is the radius of the circular orbit. You can assume the initial condition R(0) = R5 (1 + e).
(g) [2] 在星质量 m M 的极限下，我们可以忽略星对太阳运动的影响。对于偏率 e 1 的轨道，找到太阳运动 R(t) 的显式解。将你的答案写成圆周运动加上与 e 成比例的项，即 R(t) = R5 + eR1 (t) ，其中 R5 是圆形轨道的半径。你可以假设初始條件R(0) = R5 (1 + e) 。
(h) [2] Obtain the equation of motion for q(t) in the limit m M. You can see that q(t) is a nonlinear oscillator driven by a nonlinear force term.
(h) [2] 求 q(t) 在极限m M 时的运动程。你可以看到 q(t) 是个由非线性项驱动的非线性振荡器。
(
Solution
:
(
a
)
The
total
kinetic
energy
is
T
=
MR

2
+
MZ

2
+ MR
2
e

2
+
mu

2
(
b
)
The
total
potential
energy
is
V =

Y
Z
)
2
P
=
mu

+ 2MZ

L
=
2MR
2
e

(
e
)
We
can
rewrite
u
and
Z
in
term
of
q
and
Q
,
u
=
Q +
q
Z = Q
q
The
EOM
for
the
planet
is
m
=
(R
2
+ q
2
)
3/2
q
Since
u = Q +
q
=
2
mM
2GMm
(
c
)
The
total
linear
momentum
is
(d)
The
total
angular
momentum
is

m
+
2M
=

(R
2
+ q
2
)
3/2
q
2
GMm

u
=
(R
2
+ q
2
)
3/2
q
2
M
2
M
m
+
2M
m + 2
M
2
GMm
)
4
(
1
)
where the reduced mass of the system u is defined by:
= +
(f) The total energy can be rewritten as,
E = T + V = MR 2 + MZ 2 + MR2 e 2 + mu 2 Y Z)2
E = MR 2 + M \Q q ]2 + MR2 e 2 + m \Q + q ]2
(
1
Mm
GM
2
2GMm
2
m
+
2M 2R
YR
2
+ q
2
)E = MR 2 + (m + 2M)Q 2 + q 2 + MR2 e 2
E = MR 2 + (m + 2M)Q 2 + uq 2 + MR2 e 2
E = MR 2 + + uq 2 + MR2 562
(
P
2
GM
2
L
2
1
2GMm
2(
m
+
2M)
2R 4MR
2
2
YR
2
+ q
2
) E = + MR 2 + + uq 2
where the momentum and angular momentum
(
L
=
2MR
2
e

)p = mu + 2MZ = m 5Q + q 6 + 2M 5Q q 6 = (m + 2M)Q
are conserved.
The constant energy gives
= 0 2MR + R R + uq + _2RR + +2qq ` = 0
42M + + 7 R + 4u + 7 q = 0
2M + + = 0
(g) In the limit → 0,
2M + + = 0
5
(
1
)
+ + ≈ + = 0
The orbit, R(t) = R5 + eR1 (t),
R 1 51 6 + 51 6 = 0
L2 G L2
(
2
M
3
R
3
5
2
R
2
5
GM
3
) = R5 =
And
R 1 = + = 47 R1
(
G
M
4R
3
5
) R 1 = R1
R1 = A cos bc t + pe
The radial coordinate of the suns are
R1 (t) = R5 b1 + Ae cos bc t + pee = R5 b1 + e cos bc tee
(h) In the limit m M, u ≈ m and
(
3
)(R2 + q2)< = fR5 (2) b1 + e cos bc tee2 + q2 g<2 ≈ (R5 (2) + q2 )< b1 + e cos bc tee< ≈ e cos bc te
Finally, we obtain the EOM for q(t),
m + q cos bc te = 0
The dynamics of this system given by the EOM for q(t) turns out to be complex and chaotic even in the limit m M. If you want to learn more about the complex dynamic (chaos) of this system, you can search the “Slitnikov Problem” in the internet.
3. [10 points] A massless rod can rotate without friction about the pivot point at its center. Light, propagating as a plane wave, propagates from left to right, along the x axis. The electric field of the light is given by
6
1
3. [10 分] 根质量的杆可以绕其中的枢轴点摩擦地旋转。平光波沿 x 轴从左向右传播。光的
电场由下式给出
(
(
x,
t)
= E
5
cos
(kx

业
t
)
) (
其中
=
k
且
E
5
为实数。杆和
之间的度
e
表
。
Cantered
at
the
ends
of
the
rod
are
disks
,
each
with
one
side
perfectly
mirror
with
100%
reflection
and
the
other side
with
100%
absorbing
.
The
disks
are
oriented
so
that
light
in
the
upper
part
of
the
rob
(
above
the
pivot
)
always strikes
an
absorptive
surface
,
while
in
the
lower
part
,
it
strikes
a
reflective
surface
.
Each
of
the
disks
have
mass
m
and
radius
T
.
Assume
that
the
distance
R
from
the
pivot
to
the
center
of
each
disk
satisfies
R

T
.
)where = k and E5 is a real number. The angle between the rod and is denoted by e.
在杆的两端是圆盘，每个圆盘的侧是 100% 反射的完美镜，另侧 100% 吸收。圆盘的向使得杆上部(枢轴上 )的光总是照射到吸收，在下部，它照射到反射。每个圆盘的质量为 m ，半径为 T 。假设枢轴点到每个圆盘中的距离R 满 R T 。
The Poynting vector, which describes the energy flux density (i.e. the energy per unit area per unit time) is given by 描述能量通量密度(即每单位时间每单位积的能量)的坡印亭量由下式给出
= _ × `,
where u5 is the vacuum permeability. The momentum density carried by the EM wave is
其中 u5 是真空磁导率。光波携带的动量密度为
= ,
where c is the speed of light in vacuum.
其中 c 是真空中的光速。
(
(
a
)
[1]
光的频率
f
、波
入
和磁场
(x, t)
分别是什么？
)(a) [ 1] What are the frequency f, wavelength 入, and magnetic field (x, t) of the light
(b) [ 1] What is the time-averaged Poynting vector of the incident light
(b) [1] 入射光的时间平均坡印廷量是什么？
(c) [6] What is the total torque which is delivered by the light to the system of rod plus disks around the pivot point at a given angle e What is the average torque over a full rotation of the rod
(c) [6] 在给定度 e 下，光传递到围绕枢轴点的杆加圆盘系统的总扭矩是多少？杆旋转圈的平均扭矩是多少？
(d) [2] Find the average angular acceleration of the rod over a revolution.
(d) [2] 求杆旋转圈的平均加速度。
7
1
(
(
a
)
f
=
业
2
冗
,
入
=
=
2
冗
k
and
(x, t) =
cOs (kx
业
t
)
y
z
=
(
c
)
The
force
exerted
on
the
disk
in
the
direction
normal
to
its
surface is given by the total momentum per second
transferred
by
the
light
in
this
direction
.
The momentum transferred per unit area per second is
p
n
c cos e =
pc cos
2
e
, where
p
n
is the component o
f
momentum
density
of
the
wave
along
the
normal
to
the
disk
surface
.
The total
momentum transfer per second to the absorbing disk (i.e. the force) is then
F
abs
= (pc cos
2
e)(
几
r
2
)
=
几
r
2
cos
2
e =
E
2
0
几
r
2
cos
2
e
For
the
reflecting
surface, the corresponding total momentum transfer is twice as much,
F
ref
=
E
2
0
几
r
2
cos
2
e
(d) Th
e
torque
is
given
by
the
sum
of
×
.
The
net
torque
is
=
E
2
0
几
r
2
R cos
2
e
Taking
time
average
over
one
full
rotation
in
e
,

=
E
2
0
几
r
2
R
(e) Since
T = Ia
where
I
is the moment of in
ertia
of
the
system
about
the
pivot
.
In
the
limit
R

r
,
we
have
I
=
2
mR
2
and
(
b
)
and
are
perpendicular
and
in
phase
.
The
time
average of
cos
2
业
t
is
1
2
, so that
Solu
tion
:
)
8
(
1
) (
y
z
=
8
E
52
几
m
Rc
)
4. [10 points] A vertical, insulated and sealed cylinder with a cross-sectional area A, and an insulating piston of mass m inside, whose thickness is negligible compared with the length of the cylinder . At the beginning, the piston is fixed in the center of the cylinder, which divides the cylinder into two air chambers with the same length l, as shown in the figure . Assume that the upper and lower gas chambers of the cylinder each contain n moles of monatomic ideal gas with temperature T5 . In the following problems, it can be assumed that there is little friction between the piston and the cylinder wall, and that l is much larger than the distance traveled by the piston ( z) .
4. [10 分] 有垂直豎立的密閉絕熱圓筒，截積為 A ，內裝有質量為m的絕熱活塞，其厚度和圓筒的長度相比，可忽略不計。起始時，活塞被固定在圓筒的中央，將圓筒分隔成兩個長度同為 l 的氣室，如圖所。設圓筒的上下兩氣室各含有溫度為T5 和n 摩耳的單原分理想氣體。在下列的問題中，可以假定活塞與圓筒壁之間的摩擦很少，且 l 遠於活塞所移動的距離 (l z) 。
(a) [4] Release the piston from rest at time t = 0 so that it can move freely up and down . Find the trajectory of the piston z(t). You can neglect the friction between the piston and the cylinder in the part .
(a) [4] 在時間 t = 0時從静中釋放活塞，使其能由上下運動，找出活塞的軌跡 z(t) 。在这部分，你可以忽略活塞和缸之间的摩擦。
(b) [2] How does the temperature of the upper and lower chambers in the cylinder change with the position of the piston , z
(b) [2] 圓筒內上下两氣室的溫度如何隨活塞位置 z 的變動改變
(c) [4] Although there is little friction between the piston and the cylinder, the piston will eventually come to rest after a long time . Find the position of the piston , zf , when it rests and the temperature, Tf , of the gas in the cylinder at that time . . We can assume that the heat capacity of the cylinder and the piston is negligible, the temperature of the upper and lower chambers will eventually come to the same because of the movement of the piston and all heat lost due to friction will transfer into the internal energy of the gas .
(c) [4] 活塞與圓筒壁間的摩擦雖然很少，但經過段長時間後，终會使活塞靜下來。試求活塞最後靜時的位置 zf 和其時筒內氣體的温度 Tf？我们假設圓筒壁和活塞的熱容量可忽略不計，且由於活塞的運動，使上下氣室溫度最後趨於致，且由于摩擦损失的所有热量都将转化为体的内能。
9
(
1
)
Solution:
(a) The process is adiabatic and we have PVy = C where y = = .
(
From
the
figure
,
we have
)
P1 V1y = P2 V2y = P0 V0y
P1 (l + z)y = P2 (l z)y = P0 ly
P1 = P0 bey ≈ P0 \1 ]
P2 ≈ P0 \1 + ] = P0 51 + 6
Newton’s 2nd law gives
P0A \1 + ] + P0A \ 1 ] mg = z mg = m
= z g
= 业2 \z + 业 (g)2 ]
where 业 = à . The general solution of z (t) is:
(
From
the
initial
conditions
z (0) = z

(0) = 0,
)z(t) = A cos(业t + p) 业 (g)2
10
(
1
)
(

A
业
sin
p = 0
) A cos p 业 (g)2 = 0
(
g
) (
业
) z(t) = 2 (1 cos业t)
(b)
T! V!I T! = T5 beIT2 = T5 41 + (y 1)z7 = T5 51 + 6
(c) At equilibrium z5 , both chambers have the same temperature T and the pressure satisfies
(P! P2 )A = mg
(

= (1)
) (
1
1 mg
) 5A(l (n)z (T)5 ) A(l (n)z (T)5 )6 A = mg
(
By
energy
conservation, we have
)l + z5 l z5 nRT
2 × 5nRT5 6 = 2 × 5nRT6 + mgz5
T = T5 z5 (2)
From equations (1) and (2), we can eliminate T and get
l z5 = l2 5z5 (2) = = nR \T5 R (g) z5 ]
l2 z5 (2) = z5 \T5 z5 ]
(

z
2
5

z
5
+ l
2
= 0
)6nRT5 3
5mg 5
z5 = 56 ± c562 +
Since z5 < 0, we get
11
1
(
Z
0
= 5
6

c
5
6
2
+
Note
:
In
the
limit
when
\
]
2

\
i
.
e
.
l],
we have
Z
0
= 5
6 b1 c1 +
5
6
2
e ≈
5
6
5
6
2
=
=
This
result
is
justified
if
|
Z
0
|
=
l l
~
i
s
valid
at
high
temperature
T
0

and
Z

l
.
And
the temperature is
T
=
T
0
+
c
5
6
2
+
5
6
2
= T
0
b
+
c
1 +
5
6
2
e
≈ T
0
41 +
7
)
~ End of Part 1 卷-1 完 ~
12

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2023年泛珠三角及中华名校物理奥林匹克邀请赛（Word版含解析）

2023年泛珠三角及中华名校物理奥林匹克邀请赛（Word版含解析）