Companion to the IB-Style Practice Set · HL onlyIB 风格练习题的解析配套 · 仅 HL
Syllabus A5.1 to A5.6考纲 A5.1 至 A5.6PHYSICS HL
Train at $30\ \mathrm{m\,s^{-1}}$ east, east positive. (a) passenger $8.0\ \mathrm{m\,s^{-1}}$ east relative to train, find velocity relative to ground; (b) car $A$ $22\ \mathrm{m\,s^{-1}}$ east, car $B$ $18\ \mathrm{m\,s^{-1}}$ west, find $A$ relative to $B$ and why it fails for light.火车 $30\ \mathrm{m\,s^{-1}}$ 向东，东为正。(a) 乘客相对火车 $8.0\ \mathrm{m\,s^{-1}}$ 向东，求相对地面速度；(b) $A$ 车 $22\ \mathrm{m\,s^{-1}}$ 东、$B$ 车 $18\ \mathrm{m\,s^{-1}}$ 西，求 $A$ 相对 $B$，并说明为何对光失效。
The ground-frame velocity is the train velocity plus the passenger's velocity relative to the train: $u = u' + v = 8.0 + 30$. (M1)
$$ u = +38\ \mathrm{m\,s^{-1}}\ \text{(east)}. $$(A1)
With signed velocities $u_A = +22$ and $u_B = -18$: $u_{AB} = u_A - u_B = 22 - (-18) = +40\ \mathrm{m\,s^{-1}}$ east. (A1)
This simple subtraction fails for a pulse of light because the second postulate fixes the measured speed of light at $c$ in every inertial frame, so the speeds do not add. (R1)
地面系速度等于火车速度加上乘客相对火车的速度：$u = u' + v = 8.0 + 30$。(M1)
$$ u = +38\ \mathrm{m\,s^{-1}}\ \text{（向东）}. $$(A1)
用带符号速度 $u_A = +22$、$u_B = -18$：$u_{AB} = u_A - u_B = 22 - (-18) = +40\ \mathrm{m\,s^{-1}}$，向东。(A1)
这一简单减法对一束光脉冲失效，因为第二假设把测得的光速固定在每个惯性系中的 $c$，故速度不相加。(R1)
Spacecraft at $0.80c$ past Earth; crew measure $6.0\ \mathrm{s}$ between two beacon flashes. (a) Lorentz factor; (b) which observer measures proper time; (c) the interval an Earth observer measures.飞船以 $0.80c$ 掠过地球；船员测得信标两次闪光间隔 $6.0\ \mathrm{s}$。(a) 洛伦兹因子；(b) 哪位观察者测固有时间；(c) 地球观察者测得的间隔。
$\gamma = 1/\sqrt{1 - v^2/c^2} = 1/\sqrt{1 - 0.80^2}$. (M1)
$$ \gamma = \frac{1}{\sqrt{0.36}} = \frac{1}{0.60} \approx 1.67. $$(A1)
The crew measure the proper time $\Delta t_0$. (A1)
Both flashes happen at the same place on the ship (the beacon), so a single onboard clock is present at both events, which is the definition of proper time. (R1)
The Earth observer measures the dilated time $\Delta t = \gamma\,\Delta t_0 = 1.67 \times 6.0$. (M1)
$$ \Delta t = 10\ \mathrm{s}. $$(A1)
$\gamma = 1/\sqrt{1 - v^2/c^2} = 1/\sqrt{1 - 0.80^2}$。(M1)
$$ \gamma = \frac{1}{\sqrt{0.36}} = \frac{1}{0.60} \approx 1.67. $$(A1)
船员测得固有时间 $\Delta t_0$。(A1)
两次闪光发生在船上同一地点（信标处），故船上一只钟同时出现在两个事件处，这正是固有时间的定义。(R1)
地球观察者测得膨胀时间 $\Delta t = \gamma\,\Delta t_0 = 1.67 \times 6.0$。(M1)
$$ \Delta t = 10\ \mathrm{s}. $$(A1)
Spaceship proper length $150\ \mathrm{m}$, past a station at $0.60c$ along its length. (a) define proper length and say who measures it; (b) the length the station measures; (c) whether a transverse flagpole appears contracted to the crew.飞船固有长度 $150\ \mathrm{m}$，以 $0.60c$ 沿长度方向掠过空间站。(a) 定义固有长度并说明谁测得；(b) 空间站测得的长度；(c) 横向旗杆对船员是否显得收缩。
Proper length $L_0$ is the length of an object measured in the inertial frame in which the object is at rest. (A1)
Here the ship is at rest relative to its own crew, so the crew measure the proper length $150\ \mathrm{m}$. (A1)
At $0.60c$, $\gamma = 1/\sqrt{1 - 0.60^2} = 1.25$. The station sees the ship contracted along its motion: $L = L_0/\gamma = 150/1.25$. (M1)
$$ L = 120\ \mathrm{m}. $$(A1)
No, the crew measure no contraction of its height. (A1)
Length contraction acts only along the direction of relative motion; the flagpole is mounted at right angles to that direction, so its measured height is unchanged. (R1)
固有长度 $L_0$ 是在物体静止的惯性系中测得的物体长度。(A1)
这里飞船相对自己的船员静止，故船员测得固有长度 $150\ \mathrm{m}$。(A1)
在 $0.60c$，$\gamma = 1/\sqrt{1 - 0.60^2} = 1.25$。空间站看到飞船沿运动方向收缩：$L = L_0/\gamma = 150/1.25$。(M1)
$$ L = 120\ \mathrm{m}. $$(A1)
不会，船员测得其高度没有收缩。(A1)
长度收缩只沿相对运动方向发生；旗杆垂直于该方向安装，故其测得高度不变。(R1)
Carriage moving past a platform; central lamp flashes toward front and rear detectors. (a) state the two postulates; (b) why the flash is simultaneous at both detectors in the carriage frame; (c) why the platform observer sees the rear first and what this shows.车厢掠过站台；中央灯向前后探测器发出闪光。(a) 写两条假设；(b) 为何车厢系中闪光同时到达两端；(c) 为何站台观察者先看到后端及其含义。
Postulate 1 (principle of relativity): the laws of physics are the same in all inertial frames. (A1)
Postulate 2 (invariance of $c$): the speed of light in a vacuum is the same, $c$, for all inertial observers, independent of the motion of source or observer. (A1)
In the carriage frame the lamp is midway between the walls, so the two detectors are equal distances from it. (M1)
By Postulate 2 the flash travels each way at the same speed $c$, so equal distances are covered in equal times and the detections are simultaneous. (R1)
In the platform frame the carriage moves, so during the flash's travel the rear wall advances toward the light while the front wall recedes from it. The light still moves at $c$ for the platform observer, so it reaches the approaching rear wall after a shorter path and therefore first. (M1)
Two events simultaneous in the carriage frame are not simultaneous in the platform frame: simultaneity is relative, depending on the observer's frame. (R1)
假设一（相对性原理）：物理定律在所有惯性系中都相同。(A1)
假设二（$c$ 的不变性）：真空中光速对所有惯性观察者都相同，为 $c$，与光源或观察者的运动无关。(A1)
在车厢系中灯位于两壁正中，故两个探测器到它的距离相等。(M1)
由假设二，闪光朝两个方向都以相同速度 $c$ 传播，故相等距离用相等时间走完，两次探测同时发生。(R1)
在站台系中车厢在运动，故闪光传播期间后壁朝光迎去、前壁背光退离。对站台观察者光仍以 $c$ 运动，故它经更短路径先到达迎来的后壁，因而后端先被探测到。(M1)
在车厢系中同时的两个事件在站台系中不再同时：同时性是相对的，取决于观察者的参考系。(R1)
Ship $A$ at $0.60c$ east relative to a station, east positive. (a) probe fired forward at $0.50c$ relative to $A$, find its velocity relative to the station; (b) compare with the classical sum; (c) a second ship approaches head-on at $0.70c$, find $A$ relative to it.飞船 $A$ 相对空间站以 $0.60c$ 向东，东为正。(a) 向前发射相对 $A$ 为 $0.50c$ 的探测器，求其相对空间站速度；(b) 与经典求和对比；(c) 第二艘飞船以 $0.70c$ 迎面接近，求 $A$ 相对它的速率。
Take the station as frame $S$ and ship $A$ as $S'$ with $v = 0.60c$; the probe has $u' = 0.50c$ in $S'$. Invert the addition formula to find $u$ in $S$: $u = (u' + v)/(1 + u'v/c^2)$. (M1)
$$ u = \frac{0.50c + 0.60c}{1 + (0.50)(0.60)} = \frac{1.10c}{1.30}. $$(M1 for substitution)
$$ u \approx 0.846c \approx 0.85c\ \text{(east)}. $$(A1)
The Galilean prediction is $0.50c + 0.60c = 1.10c$, which exceeds $c$. (A1)
That is forbidden because no signal or object can travel faster than light; the relativistic denominator $1.30$ pulls the result down to $0.85c < c$. (R1)
In the station frame ship $A$ has $u = +0.60c$ and the second ship has velocity $v = -0.70c$ (head-on). The velocity of $A$ in the second ship's frame is $u' = (u - v)/(1 - uv/c^2)$. (M1)
$$ u' = \frac{0.60c - (-0.70c)}{1 - (0.60)(-0.70)} = \frac{1.30c}{1 + 0.42} = \frac{1.30c}{1.42}. $$(M1 for substitution)
$$ u' \approx 0.915c \approx 0.91c. $$(A1)
取空间站为 $S$、飞船 $A$ 为 $S'$，$v = 0.60c$；探测器在 $S'$ 中 $u' = 0.50c$。反解叠加公式求 $S$ 中的 $u$：$u = (u' + v)/(1 + u'v/c^2)$。(M1)
$$ u = \frac{0.50c + 0.60c}{1 + (0.50)(0.60)} = \frac{1.10c}{1.30}. $$(代入得 M1)
$$ u \approx 0.846c \approx 0.85c\ \text{（向东）}. $$(A1)
伽利略预测为 $0.50c + 0.60c = 1.10c$，超过 $c$。(A1)
这是不允许的，因为任何信号或物体都不能超过光速；相对论分母 $1.30$ 把结果压到 $0.85c < c$。(R1)
在站系中飞船 $A$ 为 $u = +0.60c$，第二艘飞船速度 $v = -0.70c$（迎面）。$A$ 在第二艘飞船系中的速度为 $u' = (u - v)/(1 - uv/c^2)$。(M1)
$$ u' = \frac{0.60c - (-0.70c)}{1 - (0.60)(-0.70)} = \frac{1.30c}{1 + 0.42} = \frac{1.30c}{1.42}. $$(代入得 M1)
$$ u' \approx 0.915c \approx 0.91c. $$(A1)
Muons at $0.98c$, rest lifetime $\tau = 2.2\ \mathrm{\mu s}$, $N = N_0 e^{-t/\tau}$ in the muon frame; detectors $2000\ \mathrm{m}$ apart vertically. (a) $\gamma$; (b) ground-frame travel time; (c) muon-frame time; (d) surviving fraction; (e) classical prediction and what the comparison shows.μ 子以 $0.98c$、静止寿命 $\tau = 2.2\ \mathrm{\mu s}$，μ 子系中 $N = N_0 e^{-t/\tau}$；探测器竖直相距 $2000\ \mathrm{m}$。(a) $\gamma$；(b) 地面系行进时间；(c) μ 子系时间；(d) 存活比例；(e) 经典预测及对比含义。
$\gamma = 1/\sqrt{1 - 0.98^2} = 1/\sqrt{1 - 0.9604} = 1/\sqrt{0.0396}$. (M1)
$$ \gamma \approx 5.0. $$(A1)
In the ground frame the muons cross $2000\ \mathrm{m}$ at $0.98c$: $t = d/v = 2000/(0.98 \times 3.00\times 10^8)$. (M1)
$$ t \approx 6.8\times 10^{-6}\ \mathrm{s} = 6.8\ \mathrm{\mu s}. $$(A1)
The ground-frame time is dilated, so the proper time in the muon frame is $t_0 = t/\gamma = 6.8/5.0$. (M1)
$$ t_0 \approx 1.4\ \mathrm{\mu s}. $$(A1)
The decay clock runs in the muon frame, so use $t_0$ in $N/N_0 = e^{-t_0/\tau}$. (M1)
$$ \frac{N}{N_0} = e^{-1.35/2.2} = e^{-0.615}. $$(M1 for the exponent)
$$ \frac{N}{N_0} \approx 0.54. $$(A1)
Ignoring dilation, the classical physicist substitutes the ground-frame time directly: $N/N_0 = e^{-t/\tau} = e^{-6.8/2.2} = e^{-3.1}$. (M1)
$$ \frac{N}{N_0} \approx 0.045. $$(A1)
The measured survival is far closer to $0.54$ than to $0.045$, so the large observed muon flux at the ground can only be explained if the muon clock runs slow: this is direct evidence for time dilation. (R1)
$\gamma = 1/\sqrt{1 - 0.98^2} = 1/\sqrt{1 - 0.9604} = 1/\sqrt{0.0396}$。(M1)
$$ \gamma \approx 5.0. $$(A1)
在地面系中 μ 子以 $0.98c$ 穿过 $2000\ \mathrm{m}$：$t = d/v = 2000/(0.98 \times 3.00\times 10^8)$。(M1)
$$ t \approx 6.8\times 10^{-6}\ \mathrm{s} = 6.8\ \mathrm{\mu s}. $$(A1)
地面系时间被膨胀，故 μ 子系中的固有时间为 $t_0 = t/\gamma = 6.8/5.0$。(M1)
$$ t_0 \approx 1.4\ \mathrm{\mu s}. $$(A1)
衰变钟在 μ 子系中运行，故在 $N/N_0 = e^{-t_0/\tau}$ 中用 $t_0$。(M1)
$$ \frac{N}{N_0} = e^{-1.35/2.2} = e^{-0.615}. $$(指数得 M1)
$$ \frac{N}{N_0} \approx 0.54. $$(A1)
忽略膨胀，经典物理学家直接代入地面系时间：$N/N_0 = e^{-t/\tau} = e^{-6.8/2.2} = e^{-3.1}$。(M1)
$$ \frac{N}{N_0} \approx 0.045. $$(A1)
实测存活比例远接近 $0.54$ 而非 $0.045$，故地面处观测到的大量 μ 子通量只有在 μ 子钟变慢时才能解释：这是时间膨胀的直接证据。(R1)
Events $P, Q$ in frame $S$: $c\,\Delta t = 5.0$, $\Delta x = 4.0$ (light-microseconds); $S'$ moves at $0.60c$ along $x$, $\gamma = 1.25$. (a) define the invariant interval; (b) compute $(\Delta s)^2$ and classify; (c) find $c\,\Delta t'$ and $\Delta x'$; (d) show $(\Delta s')^2 = (\Delta s)^2$ and name the principle.$S$ 系中事件 $P, Q$：$c\,\Delta t = 5.0$、$\Delta x = 4.0$（光微秒）；$S'$ 沿 $x$ 以 $0.60c$ 运动，$\gamma = 1.25$。(a) 定义不变间隔；(b) 算 $(\Delta s)^2$ 并分类；(c) 求 $c\,\Delta t'$ 与 $\Delta x'$；(d) 证 $(\Delta s')^2 = (\Delta s)^2$ 并命名原理。
The spacetime interval between two events is the combination $(\Delta s)^2 = (c\,\Delta t)^2 - (\Delta x)^2$. (A1)
Although $\Delta t$ and $\Delta x$ differ between inertial frames, this combination has the same value in every inertial frame, hence "invariant". (A1)
Substitute the frame-$S$ data: $(\Delta s)^2 = 5.0^2 - 4.0^2$. (M1)
$$ (\Delta s)^2 = 25 - 16 = 9.0 \ (\mathrm{l.\mu s})^2. $$(A1)
Since $(\Delta s)^2 > 0$ the separation is timelike: the events can be causally linked and a single clock can be present at both. (A1)
Apply the Lorentz transformations with $\gamma = 1.25$, $\beta = 0.60$, using $c\,\Delta t$ and $\Delta x$ in the same unit: $c\,\Delta t' = \gamma(c\,\Delta t - \beta\,\Delta x)$ and $\Delta x' = \gamma(\Delta x - \beta\,c\,\Delta t)$. (M1)
$$ c\,\Delta t' = 1.25\,(5.0 - 0.60\times 4.0) = 1.25\,(2.6) = 3.25. $$ $$ \Delta x' = 1.25\,(4.0 - 0.60\times 5.0) = 1.25\,(1.0) = 1.25. $$(M1 for both substitutions)
So $c\,\Delta t' = 3.25$ and $\Delta x' = 1.25$ (light-microseconds). (A1)
$(\Delta s')^2 = (c\,\Delta t')^2 - (\Delta x')^2 = 3.25^2 - 1.25^2 = 10.5625 - 1.5625 = 9.0$. (A1)
This equals the value found in $S$, confirming the invariance of the spacetime interval under a Lorentz transformation. (A1)
两事件间的时空间隔是组合量 $(\Delta s)^2 = (c\,\Delta t)^2 - (\Delta x)^2$。(A1)
尽管 $\Delta t$ 与 $\Delta x$ 在不同惯性系间不同，该组合在每个惯性系中取值相同，故称"不变"。(A1)
代入 $S$ 系数据：$(\Delta s)^2 = 5.0^2 - 4.0^2$。(M1)
$$ (\Delta s)^2 = 25 - 16 = 9.0 \ (\mathrm{l.\mu s})^2. $$(A1)
因 $(\Delta s)^2 > 0$，间隔为类时：事件可有因果关联，且可有一只钟同时出现在两处。(A1)
用 $\gamma = 1.25$、$\beta = 0.60$ 施加洛伦兹变换，$c\,\Delta t$ 与 $\Delta x$ 取同一单位：$c\,\Delta t' = \gamma(c\,\Delta t - \beta\,\Delta x)$，$\Delta x' = \gamma(\Delta x - \beta\,c\,\Delta t)$。(M1)
$$ c\,\Delta t' = 1.25\,(5.0 - 0.60\times 4.0) = 1.25\,(2.6) = 3.25. $$ $$ \Delta x' = 1.25\,(4.0 - 0.60\times 5.0) = 1.25\,(1.0) = 1.25. $$(两处代入得 M1)
故 $c\,\Delta t' = 3.25$、$\Delta x' = 1.25$（光微秒）。(A1)
$(\Delta s')^2 = (c\,\Delta t')^2 - (\Delta x')^2 = 3.25^2 - 1.25^2 = 10.5625 - 1.5625 = 9.0$。(A1)
它等于 $S$ 系中所得的值，确证时空间隔在洛伦兹变换下不变。(A1)
Muons made $4.5\ \mathrm{km}$ up, $v = 0.995c$, $\gamma = 10.0$, proper lifetime $2.2\ \mathrm{\mu s}$. (a) dilated lifetime; (b) ground-frame distance in one lifetime and whether they reach the ground; (c) contracted atmosphere thickness in the muon frame; (d) muon-frame crossing time and how it confirms (b); (e) the key observation and the two effects used.μ 子在 $4.5\ \mathrm{km}$ 高处产生，$v = 0.995c$、$\gamma = 10.0$、固有寿命 $2.2\ \mathrm{\mu s}$。(a) 膨胀寿命；(b) 一寿命内地面系距离及能否抵地；(c) μ 子系中大气收缩厚度；(d) μ 子系穿越时间及如何印证 (b)；(e) 关键观测与所用两效应。
The rest lifetime is the proper time $\Delta t_0$; the ground frame measures $\Delta t = \gamma\,\Delta t_0 = 10.0 \times 2.2$. (M1)
$$ \Delta t = 22\ \mathrm{\mu s}. $$(A1)
Distance covered in one dilated lifetime: $d = v\,\Delta t = (0.995)(3.00\times 10^8)(22\times 10^{-6})$. (M1)
$$ d \approx 6.6\times 10^3\ \mathrm{m} = 6.6\ \mathrm{km}. $$(A1)
Since $6.6\ \mathrm{km}$ exceeds the $4.5\ \mathrm{km}$ of atmosphere, a large fraction of muons survive to the ground. (R1)
In the muon frame the $4.5\ \mathrm{km}$ is the proper length of the atmosphere (it is at rest in the ground frame), so it is contracted: $L = L_0/\gamma$. (M1)
$$ L = \frac{4.5\ \mathrm{km}}{10.0}. $$(M1 for substitution)
$$ L \approx 0.45\ \mathrm{km} = 450\ \mathrm{m}. $$(A1)
In its own frame the muon crosses the contracted distance at $0.995c$: $t = L/v = 450/(0.995 \times 3.00\times 10^8)$. (M1)
$$ t \approx 1.5\ \mathrm{\mu s}. $$This is less than the $2.2\ \mathrm{\mu s}$ rest lifetime, so most muons survive, the same conclusion as (b) reached by length contraction instead of time dilation. (A1)
The key observation is that far more muons are detected at ground level than the un-dilated $2.2\ \mathrm{\mu s}$ lifetime would allow. (B1)
The ground-frame explanation uses time dilation; the muon-frame explanation uses length contraction. (B1)
静止寿命是固有时间 $\Delta t_0$；地面系测得 $\Delta t = \gamma\,\Delta t_0 = 10.0 \times 2.2$。(M1)
$$ \Delta t = 22\ \mathrm{\mu s}. $$(A1)
一个膨胀寿命内走过的距离：$d = v\,\Delta t = (0.995)(3.00\times 10^8)(22\times 10^{-6})$。(M1)
$$ d \approx 6.6\times 10^3\ \mathrm{m} = 6.6\ \mathrm{km}. $$(A1)
由于 $6.6\ \mathrm{km}$ 超过 $4.5\ \mathrm{km}$ 的大气层，相当大比例的 μ 子能存活到地面。(R1)
在 μ 子系中 $4.5\ \mathrm{km}$ 是大气层的固有长度（它在地面系中静止），故被收缩：$L = L_0/\gamma$。(M1)
$$ L = \frac{4.5\ \mathrm{km}}{10.0}. $$(代入得 M1)
$$ L \approx 0.45\ \mathrm{km} = 450\ \mathrm{m}. $$(A1)
在自身系中 μ 子以 $0.995c$ 穿过收缩距离：$t = L/v = 450/(0.995 \times 3.00\times 10^8)$。(M1)
$$ t \approx 1.5\ \mathrm{\mu s}. $$它小于 $2.2\ \mathrm{\mu s}$ 的静止寿命，故大多数 μ 子存活，与 (b) 结论相同，但用的是长度收缩而非时间膨胀。(A1)
关键观测是：在地面探测到的 μ 子远多于未膨胀的 $2.2\ \mathrm{\mu s}$ 寿命所允许的数量。(B1)
地面系的解释用时间膨胀；μ 子系的解释用长度收缩。(B1)
Event $E$ in $S$ at $x = 900\ \mathrm{m}$, $t = 2.0\ \mathrm{\mu s}$; $S'$ at $0.60c$ along $+x$, $\gamma = 1.25$, origins coincide at $t = t' = 0$. (a) find $x'$ and $t'$; (b) slope of a light worldline and of the $S'$ origin worldline on a $ct$-$x$ diagram; (c) for two events simultaneous in $S$ separated by $\Delta x = 600\ \mathrm{m}$, find $\Delta t'$ and explain.$S$ 中事件 $E$ 在 $x = 900\ \mathrm{m}$、$t = 2.0\ \mathrm{\mu s}$；$S'$ 沿 $+x$ 以 $0.60c$，$\gamma = 1.25$，$t = t' = 0$ 时原点重合。(a) 求 $x'$、$t'$；(b) $ct$-$x$ 图上光世界线与 $S'$ 原点世界线的斜率；(c) 对 $S$ 中相隔 $\Delta x = 600\ \mathrm{m}$ 的两同时事件求 $\Delta t'$ 并解释。
Use $x' = \gamma(x - vt)$ and $t' = \gamma(t - vx/c^2)$ with $v = 0.60c$. First $vt = 0.60(3.00\times 10^8)(2.0\times 10^{-6}) = 360\ \mathrm{m}$. (M1)
$$ x' = 1.25\,(900 - 360) = 1.25\,(540) = 675\ \mathrm{m}. $$(A1)
For the time, $vx/c^2 = (0.60)(3.00\times 10^8)(900)/(3.00\times 10^8)^2 = 1.8\times 10^{-6}\ \mathrm{s} = 1.8\ \mathrm{\mu s}$. (M1)
$$ t' = \gamma\!\left(t - \frac{vx}{c^2}\right) = 1.25\,(2.0 - 1.8)\ \mathrm{\mu s} = 1.25\,(0.20)\ \mathrm{\mu s} = 0.25\ \mathrm{\mu s}. $$(A1)
A light pulse obeys $x = ct$, so on a $ct$-versus-$x$ diagram its worldline has slope $1$, i.e. $45^{\circ}$. (A1)
The origin of $S'$ moves at $v$, tracing $x = vt$, so its worldline has slope $c/v = 1/0.60 \approx 1.67$ (steeper than the light line). (A1)
For simultaneous events in $S$, $\Delta t = 0$, so $\Delta t' = \gamma(\Delta t - v\,\Delta x/c^2) = -\gamma\,v\,\Delta x/c^2$. With $\Delta x = 600\ \mathrm{m}$: $\Delta t' = -1.25\,(0.60)(600)/(3.00\times 10^8)$. (M1)
$$ \Delta t' = -1.5\times 10^{-6}\ \mathrm{s} = -1.5\ \mathrm{\mu s}. $$The non-zero $\Delta t'$ shows that events simultaneous in $S$ are not simultaneous in $S'$: this is the relativity of simultaneity, driven by the $-vx/c^2$ term. (A1)
用 $x' = \gamma(x - vt)$ 与 $t' = \gamma(t - vx/c^2)$，$v = 0.60c$。先算 $vt = 0.60(3.00\times 10^8)(2.0\times 10^{-6}) = 360\ \mathrm{m}$。(M1)
$$ x' = 1.25\,(900 - 360) = 1.25\,(540) = 675\ \mathrm{m}. $$(A1)
对时间，$vx/c^2 = (0.60)(3.00\times 10^8)(900)/(3.00\times 10^8)^2 = 1.8\times 10^{-6}\ \mathrm{s} = 1.8\ \mathrm{\mu s}$。(M1)
$$ t' = \gamma\!\left(t - \frac{vx}{c^2}\right) = 1.25\,(2.0 - 1.8)\ \mathrm{\mu s} = 1.25\,(0.20)\ \mathrm{\mu s} = 0.25\ \mathrm{\mu s}. $$(A1)
光脉冲满足 $x = ct$，故在 $ct$ 对 $x$ 的图上其世界线斜率为 $1$，即 $45^{\circ}$。(A1)
$S'$ 原点以 $v$ 运动，描出 $x = vt$，故其世界线斜率为 $c/v = 1/0.60 \approx 1.67$（比光线更陡）。(A1)
对 $S$ 中的同时事件 $\Delta t = 0$，故 $\Delta t' = \gamma(\Delta t - v\,\Delta x/c^2) = -\gamma\,v\,\Delta x/c^2$。代 $\Delta x = 600\ \mathrm{m}$：$\Delta t' = -1.25\,(0.60)(600)/(3.00\times 10^8)$。(M1)
$$ \Delta t' = -1.5\times 10^{-6}\ \mathrm{s} = -1.5\ \mathrm{\mu s}. $$非零的 $\Delta t'$ 表明在 $S$ 中同时的事件在 $S'$ 中不再同时：这就是同时性的相对性，由 $-vx/c^2$ 项驱动。(A1)
Spacecraft to a star $6.0$ light-years away and back at $0.60c$ each leg ($\gamma = 1.25$), distances in the Earth frame. (a) Earth round-trip time; (b) which time is proper, with reason; (c) crew round-trip time; (d) Earth-to-star distance the crew measure and consistency check.飞船往返 $6.0$ 光年外恒星，每段 $0.60c$（$\gamma = 1.25$），距离以地球系给出。(a) 地球往返时间；(b) 哪个是固有时间及理由；(c) 船员往返时间；(d) 船员测得的地球-恒星距离及一致性检验。
In the Earth frame the total distance is $2 \times 6.0 = 12\ \mathrm{ly}$ at $0.60c$: $t = d/v = 12/0.60$. (M1)
$$ t_{\text{Earth}} = 20\ \mathrm{yr}. $$(A1)
The crew's time is the proper time $\Delta t_0$. (A1)
Departure and return happen at the same place in the crew's frame (right where they are aboard the ship), so one onboard clock is present at both events, which defines proper time. (R1)
The Earth time is dilated relative to the proper time: $\Delta t_0 = t_{\text{Earth}}/\gamma = 20/1.25$. (M1)
$$ \Delta t_0 = 16\ \mathrm{yr}. $$(A1)
In the crew frame the Earth-to-star distance is contracted: $L = L_0/\gamma = 6.0/1.25 = 4.8\ \mathrm{ly}$ each way. (M1)
Crossing $2 \times 4.8 = 9.6\ \mathrm{ly}$ at $0.60c$ takes $9.6/0.60 = 16\ \mathrm{yr}$, matching the crew time in (c). (A1)
在地球系中总距离为 $2 \times 6.0 = 12\ \mathrm{ly}$，速度 $0.60c$：$t = d/v = 12/0.60$。(M1)
$$ t_{\text{地球}} = 20\ \mathrm{yr}. $$(A1)
船员的时间是固有时间 $\Delta t_0$。(A1)
出发与返回在船员系中发生于同一地点（就在船上他们所在处），故船上一只钟同时出现在两个事件处，这定义了固有时间。(R1)
地球时间相对固有时间被膨胀：$\Delta t_0 = t_{\text{地球}}/\gamma = 20/1.25$。(M1)
$$ \Delta t_0 = 16\ \mathrm{yr}. $$(A1)
在船员系中地球到恒星的距离被收缩：$L = L_0/\gamma = 6.0/1.25 = 4.8\ \mathrm{ly}$，单程。(M1)
以 $0.60c$ 穿过 $2 \times 4.8 = 9.6\ \mathrm{ly}$ 用时 $9.6/0.60 = 16\ \mathrm{yr}$，与 (c) 的船员时间相符。(A1)