Unit A.5: Galilean and Special Relativity HL only单元 A.5：伽利略与狭义相对论 HL only
The HL-only capstone of Theme A "Space, time and motion". We start with classical reference frames, the Galilean transformations, and classical velocity addition, then confront the experimental fact that the speed of light $c$ is the same for every inertial observer. Einstein's two postulates force time and space themselves to depend on the observer: time dilation, length contraction, the relativity of simultaneity, the Lorentz transformations, relativistic velocity addition, spacetime (Minkowski) diagrams, and the invariant spacetime interval. Muon decay supplies the experimental confirmation.主题 A"空间、时间与运动"的 HL 专属收官单元。我们从经典参考系、伽利略变换与经典速度叠加出发，再面对一个实验事实：光速 $c$ 对每一个惯性观察者都相同。爱因斯坦的两条假设迫使时间与空间本身依赖于观察者：时间膨胀、长度收缩、同时性的相对性、洛伦兹变换、相对论速度叠加、时空（闵可夫斯基）图，以及不变的时空间隔。μ 子衰变提供了实验验证。
How to use this guide本指南使用说明
A.5 is conceptually demanding but algebraically light. Almost every numerical question reduces to one of four moves: compute the Lorentz factor $\gamma$, dilate a time, contract a length, or add velocities relativistically. The marks are lost on who measures what: proper time $\Delta t_0$ is the time between two events that happen at the same place in one frame; proper length $L_0$ is the length measured in the frame where the object is at rest. Decide whose clock and whose ruler before you reach for a formula.A.5 在概念上很难，但代数很轻。几乎每道计算题都归结为四个动作之一：算洛伦兹因子 $\gamma$、把时间膨胀、把长度收缩，或做相对论速度叠加。分数往往丢在"谁测量了什么"上：固有时间 $\Delta t_0$ 是两个在某一参考系中发生于同一地点的事件之间的时间；固有长度 $L_0$ 是在物体静止的参考系中测得的长度。动笔前先确定用谁的钟、谁的尺。
Memorise $\gamma = 1/\sqrt{1 - v^2/c^2}$, then $\Delta t = \gamma\,\Delta t_0$ (moving clocks run slow) and $L = L_0/\gamma$ (moving rulers shrink). $\gamma \ge 1$ always. Proper time and proper length are the smallest time and length any observer measures only in their home frame. The invariant interval $(\Delta s)^2 = (c\,\Delta t)^2 - (\Delta x)^2$ is the same in every frame.
背熟 $\gamma = 1/\sqrt{1 - v^2/c^2}$，再记 $\Delta t = \gamma\,\Delta t_0$（动钟变慢）与 $L = L_0/\gamma$（动尺变短）。$\gamma \ge 1$ 恒成立。固有时间与固有长度只在各自的本参考系中是"原始"的量。不变间隔 $(\Delta s)^2 = (c\,\Delta t)^2 - (\Delta x)^2$ 在每个参考系中都相同。
Be able to derive time dilation from the light-clock and the invariance of $c$, place events on a Minkowski diagram, and read off which event happens "first" in a given frame to explain the relativity of simultaneity. Know that the Lorentz transformations x' = γ(x − vt) and t' = γ(t − vx/c²) reduce to the Galilean ones when $v \ll c$, and quote muon decay as the standard evidence, framing it as dilation (ground frame) or contraction (muon frame).
能从光钟与 $c$ 的不变性推导时间膨胀，能把事件画到闵可夫斯基图上，并读出在给定参考系中哪个事件"先发生"，以解释同时性的相对性。要知道洛伦兹变换 x' = γ(x − vt) 与 t' = γ(t − vx/c²) 在 $v \ll c$ 时退化为伽利略变换，并能引用 μ 子衰变作为标准证据，从地面系（膨胀）或 μ 子系（收缩）两个角度阐述。
Reference Frames, Galilean Transformations, Velocity Addition参考系、伽利略变换、速度叠加 A.5 HL
Galilean transformation. Frame $S'$ moves at constant speed $v$ along the $x$-axis of frame $S$. Coordinates transform as: $$ x' = x - v t, \qquad y' = y, \qquad z' = z, \qquad t' = t. $$ Time is absolute and the same in both frames. Classical velocity addition. A body with velocity $u$ in $S$ has velocity in $S'$: $$ u' = u - v. $$ Key idea. In Galilean relativity the laws of mechanics are identical in all inertial frames, but each observer's $u$ differs by the simple subtraction above.
伽利略变换（Galilean transformation）。参考系 $S'$ 沿参考系 $S$ 的 $x$ 轴以恒定速度 $v$ 运动。坐标变换为： $$ x' = x - v t, \qquad y' = y, \qquad z' = z, \qquad t' = t. $$ 时间是绝对的，在两个参考系中相同。 经典速度叠加。在 $S$ 中速度为 $u$ 的物体，在 $S'$ 中速度为： $$ u' = u - v. $$ 核心思想。在伽利略相对性中，力学定律在所有惯性系中都相同，但每个观察者测得的 $u$ 相差上面那个简单的减法。
A train moves at $30\ \mathrm{m\,s^{-1}}$ east relative to the ground. A passenger walks at $2\ \mathrm{m\,s^{-1}}$ east relative to the train. Using the Galilean transformation, find the passenger's velocity relative to the ground, and the ground's velocity relative to the train.火车相对地面以 $30\ \mathrm{m\,s^{-1}}$ 向东运动。乘客相对火车以 $2\ \mathrm{m\,s^{-1}}$ 向东行走。用伽利略变换求乘客相对地面的速度，以及地面相对火车的速度。
Identify. Let $S$ be the ground and $S'$ the train, so $v = +30\ \mathrm{m\,s^{-1}}$ (train relative to ground). Take east as positive.
识别。设 $S$ 为地面、$S'$ 为火车，故 $v = +30\ \mathrm{m\,s^{-1}}$（火车相对地面）。取东为正。
Set Up. The passenger's velocity in $S'$ is $u' = +2\ \mathrm{m\,s^{-1}}$. Inverting $u' = u - v$ gives the ground-frame velocity $u = u' + v$.
列式。乘客在 $S'$ 中的速度为 $u' = +2\ \mathrm{m\,s^{-1}}$。把 $u' = u - v$ 反解得地面系速度 $u = u' + v$。
Execute. $u = 2 + 30 = +32\ \mathrm{m\,s^{-1}}$ east. The ground's velocity relative to the train is $-v = -30\ \mathrm{m\,s^{-1}}$, i.e. $30\ \mathrm{m\,s^{-1}}$ west.
求解。$u = 2 + 30 = +32\ \mathrm{m\,s^{-1}}$，向东。地面相对火车的速度为 $-v = -30\ \mathrm{m\,s^{-1}}$，即向西 $30\ \mathrm{m\,s^{-1}}$。
Evaluate. Velocities simply add and subtract — exactly what intuition expects at everyday speeds. This breaks down only when speeds approach $c$, which is the whole point of the rest of this unit.
评估。速度只是简单地加减——这正是日常速度下直觉所预期的。只有当速度接近 $c$ 时它才失效，而这正是本单元其余部分的主题。
Going deeper: why mechanics is Galilean-invariant but light is not深入：为什么力学是伽利略不变的而光不是
Differentiate the Galilean velocity rule once more: acceleration is invariant, $a' = a$, because $v$ is constant. With invariant mass, $F = ma$ takes the same form in $S$ and $S'$ — Newtonian mechanics obeys the Galilean principle of relativity. But Maxwell's equations predict light travels at a fixed $c$, and a Galilean transformation would make light slower or faster depending on the observer's motion. Experiment (Michelson–Morley) found no such variation. Resolving this contradiction is exactly what special relativity does.
再对伽利略速度规则求一次导：因 $v$ 恒定，加速度不变，$a' = a$。在质量不变的前提下，$F = ma$ 在 $S$ 与 $S'$ 中形式相同——牛顿力学遵从伽利略相对性原理。但麦克斯韦方程预言光以固定的 $c$ 传播，而伽利略变换会使光速随观察者运动而变快或变慢。实验（迈克耳孙–莫雷）未发现这种变化。狭义相对论正是为解决这一矛盾而生。
The Two Postulates of Special Relativity狭义相对论的两条基本假设 A.5 HL
Postulate 2 (invariance of $c$). The speed of light in a vacuum, $c \approx 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$, is the same for all inertial observers, regardless of the motion of the source or the observer.
Immediate consequence. Because every observer measures the same $c$, they cannot agree on times and lengths — those must change between frames. There is no universal "now" and no preferred frame (no aether).
假设二（$c$ 的不变性）。真空中的光速 $c \approx 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$ 对所有惯性观察者都相同，与光源或观察者的运动无关。
直接后果。因为每个观察者测得相同的 $c$，他们无法在时间与长度上达成一致——这些量必须在不同参考系间发生变化。不存在普适的"此刻"，也不存在优先参考系（没有以太）。
A spaceship travels at $0.60c$ toward a star and switches on a headlamp. (a) At what speed does the ship's crew measure the emitted light moving away from them? (b) At what speed does an observer on the star measure that same light approaching? Justify with the postulates.一艘飞船以 $0.60c$ 朝某恒星飞行并打开前灯。(a) 飞船船员测得发出的光以多大速度离开他们？(b) 恒星上的观察者测得同一束光以多大速度接近？用基本假设说明理由。
Identify. The unknowns are two measured light speeds, in two different inertial frames (ship and star).
识别。未知量是两个测得的光速，分属两个不同的惯性系（飞船与恒星）。
Set Up. Apply Postulate 2 directly: the vacuum speed of light is $c$ for every inertial observer, independent of source or observer motion.
列式。直接用假设二：真空光速对每个惯性观察者都是 $c$，与光源或观察者的运动无关。
Execute. (a) The crew measure the light leaving at $c$. (b) The star observer also measures it approaching at $c$ — not $c + 0.60c$.
求解。(a) 船员测得光以 $c$ 离开。(b) 恒星观察者同样测得它以 $c$ 接近——不是 $c + 0.60c$。
Evaluate. Classical (Galilean) velocity addition would give $1.60c$ for the star observer, which contradicts the postulate and all experiment. Reconciling "everyone gets $c$" is exactly why time dilation and length contraction must exist.
评估。经典（伽利略）速度叠加会给出 $1.60c$，这与假设及所有实验相矛盾。要使"人人都测得 $c$"自洽，正是时间膨胀与长度收缩必须存在的原因。
Going deeper: the aether and Michelson–Morley深入：以太与迈克耳孙–莫雷
Nineteenth-century physics assumed light travelled through a medium, the "luminiferous aether", which would single out a preferred rest frame. The Michelson–Morley interferometer experiment (1887) tried to detect Earth's motion through the aether by comparing light speeds along perpendicular arms. It found no difference at any time of year. Einstein's response in 1905 was radical: drop the aether entirely and elevate the constancy of $c$ to a postulate. The "null result" of Michelson–Morley is the experimental backbone of Postulate 2.
十九世纪物理学假设光通过一种介质——"发光以太"——传播，这将确立一个优先的静止参考系。迈克耳孙–莫雷干涉实验（1887）试图通过比较两条垂直臂上的光速来检测地球穿过以太的运动，却在一年中任何时候都未发现差异。爱因斯坦 1905 年的回应很激进：彻底抛弃以太，把 $c$ 的恒定性提升为一条假设。迈克耳孙–莫雷的"零结果"是假设二的实验支柱。
The Lorentz Factor and Time Dilation洛伦兹因子与时间膨胀 A.5 HL
γ = 1/√(1 − v²/c²).)
Proper time. $\Delta t_0$ is the time between two events measured by a clock present at both events (they occur at the same place in that frame). It is the shortest time any frame records. Time dilation. Any other observer, moving at $v$, measures a longer time: $$ \Delta t = \gamma\,\Delta t_{0}. $$ "Moving clocks run slow." (Data booklet:
Δt = γ Δt₀.)
γ = 1/√(1 − v²/c²)。）
固有时间（proper time）。$\Delta t_0$ 是由同时出现在两个事件处的钟所测得的两事件间时间（这两事件在该参考系中发生于同一地点）。它是任何参考系所记录的最短时间。 时间膨胀（time dilation）。任何其他以 $v$ 运动的观察者测得更长的时间： $$ \Delta t = \gamma\,\Delta t_{0}. $$ "动钟变慢。"（数据手册：
Δt = γ Δt₀。）
A spacecraft moves past Earth at $v = 0.80c$. The crew measure a $5.0\ \mathrm{s}$ interval between two flashes of an onboard beacon. How long is that interval as measured by an observer on Earth?一艘飞船以 $v = 0.80c$ 掠过地球。船员测得船上信标两次闪光之间的间隔为 $5.0\ \mathrm{s}$。地球上的观察者测得这段间隔为多长？
Identify. The two flashes occur at the same place on the ship, so the crew's $5.0\ \mathrm{s}$ is the proper time $\Delta t_0$. The Earth observer's time $\Delta t$ is the unknown.
识别。两次闪光发生在船上同一地点，故船员的 $5.0\ \mathrm{s}$ 是固有时间 $\Delta t_0$。地球观察者的时间 $\Delta t$ 为未知量。
Set Up. Compute $\gamma$, then apply $\Delta t = \gamma\,\Delta t_0$.
列式。先算 $\gamma$，再用 $\Delta t = \gamma\,\Delta t_0$。
$$ \gamma = \frac{1}{\sqrt{1 - 0.80^{2}}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.60} \approx 1.67. $$Execute.
求解。
$$ \Delta t = \gamma\,\Delta t_0 = 1.67 \times 5.0 \approx 8.3\ \mathrm{s}. $$Evaluate. The Earth observer records a longer interval — the moving ship's clock appears to run slow. The answer exceeds $5.0\ \mathrm{s}$, as it must whenever $\gamma > 1$.
评估。地球观察者记录到更长的间隔——运动飞船的钟看起来走得慢。结果大于 $5.0\ \mathrm{s}$，只要 $\gamma > 1$ 就必然如此。
Going deeper: deriving time dilation from the light clock深入：用光钟推导时间膨胀
A light clock bounces a pulse between two mirrors a distance $d$ apart. In the clock's rest frame the round trip takes $\Delta t_0 = 2d/c$. Now view the clock moving at $v$ perpendicular to $d$: the pulse travels a longer diagonal path of length $c\,\Delta t$, while the clock advances $v\,\Delta t$ horizontally. Pythagoras on half the trip gives
光钟让一束脉冲在相距 $d$ 的两面镜子间往返。在钟的静止系中往返用时 $\Delta t_0 = 2d/c$。现在看这只钟以垂直于 $d$ 的方向、速度 $v$ 运动：脉冲沿更长的斜线走过长度 $c\,\Delta t$，而钟在水平方向前进 $v\,\Delta t$。对半个行程用勾股定理：
$$ \left(\tfrac{c\,\Delta t}{2}\right)^{2} = d^{2} + \left(\tfrac{v\,\Delta t}{2}\right)^{2}. $$Substituting $d = c\,\Delta t_0 / 2$ and solving for $\Delta t$ gives $\Delta t = \Delta t_0 / \sqrt{1 - v^2/c^2} = \gamma\,\Delta t_0$. The result follows purely from the invariance of $c$: the only thing assumed is that the pulse moves at $c$ in both frames.
代入 $d = c\,\Delta t_0 / 2$ 并解出 $\Delta t$，得 $\Delta t = \Delta t_0 / \sqrt{1 - v^2/c^2} = \gamma\,\Delta t_0$。这一结果完全源自 $c$ 的不变性：唯一的假设是脉冲在两个参考系中都以 $c$ 运动。
Length Contraction and the Relativity of Simultaneity长度收缩与同时性的相对性 A.5 HL
L = L₀/γ.)
Relativity of simultaneity. Two events that are simultaneous in one frame are generally not simultaneous in a frame moving relative to it. There is no universal "now"; ordering can differ only for events with a spacelike separation.
L = L₀/γ。）
同时性的相对性（relativity of simultaneity）。在某一参考系中同时发生的两个事件，在相对它运动的参考系中通常不同时。不存在普适的"此刻"；只有类空间隔的事件，其先后次序才可能不同。
A spaceship has a proper length of $120\ \mathrm{m}$ and flies past Earth at $0.80c$. (a) How long is the ship as measured by an Earth observer? (b) The Earth observer says a $90\ \mathrm{m}$-long landing strip lies along the flight path; how long is that strip in the ship's frame?一艘飞船固有长度为 $120\ \mathrm{m}$，以 $0.80c$ 掠过地球。(a) 地球观察者测得飞船有多长？(b) 地球观察者说飞行路径上有一条 $90\ \mathrm{m}$ 长的跑道；在飞船系中这条跑道有多长？
Identify. From A5.3, $\gamma = 1/\sqrt{1 - 0.80^2} \approx 1.67$. The ship's $120\ \mathrm{m}$ is proper length $L_0$ (ship at rest in its own frame); the strip's $90\ \mathrm{m}$ is proper length in the Earth frame (strip at rest on Earth).
识别。由 A5.3，$\gamma = 1/\sqrt{1 - 0.80^2} \approx 1.67$。飞船的 $120\ \mathrm{m}$ 是固有长度 $L_0$（飞船在自身系中静止）；跑道的 $90\ \mathrm{m}$ 是地球系中的固有长度（跑道在地面静止）。
Set Up. Each observer applies $L = L_0/\gamma$ to the object that is moving relative to them.
列式。每个观察者对相对自己运动的物体用 $L = L_0/\gamma$。
Execute. (a) Earth sees the ship contracted: $L = 120 / 1.67 \approx 72\ \mathrm{m}$. (b) The ship sees the strip contracted: $L = 90 / 1.67 \approx 54\ \mathrm{m}$.
求解。(a) 地球看到飞船收缩：$L = 120 / 1.67 \approx 72\ \mathrm{m}$。(b) 飞船看到跑道收缩：$L = 90 / 1.67 \approx 54\ \mathrm{m}$。
Evaluate. Each frame finds the other's object shortened — the situation is symmetric, with no contradiction. Always contract the proper length measured in the rest frame of the object in question.
评估。每个参考系都发现对方的物体变短——情形是对称的，没有矛盾。永远对所讨论物体静止系中测得的固有长度做收缩。
Going deeper: the train-and-platform simultaneity thought experiment深入：火车—站台同时性思想实验
A lightning bolt strikes each end of a moving train. In the platform (ground) frame the two strikes are simultaneous, and their light reaches the platform mid-point together. But an observer at the centre of the train is moving toward the light from the front strike and away from the rear strike. Since light travels at $c$ in the train frame too, the train observer receives the front flash first and concludes the front strike happened earlier. Simultaneity is frame-dependent. Mathematically this comes straight from the $-vx/c^2$ term in the Lorentz time transformation (next section): events separated in $x$ are separated in $t'$ even if $\Delta t = 0$.
闪电击中一列运动火车的两端。在站台（地面）系中两次击打同时发生，其光同时到达站台中点。但火车中央的观察者正朝前端击打的光迎去、背离后端击打的光。由于光在火车系中也以 $c$ 传播，火车观察者先收到前端闪光，从而判定前端击打更早发生。同时性依赖于参考系。从数学上看，这直接来自洛伦兹时间变换中的 $-vx/c^2$ 项（见下一节）：即使 $\Delta t = 0$，在 $x$ 上分开的事件在 $t'$ 上也会分开。
Lorentz Transformations, Velocity Addition, Spacetime Diagrams洛伦兹变换、速度叠加、时空图 A.5 HL
x' = γ(x − vt), t' = γ(t − vx/c²).)
Relativistic velocity addition. A body with velocity $u$ in $S$ has velocity in $S'$: $$ u' = \frac{u - v}{1 - \dfrac{u v}{c^{2}}}. $$ The denominator keeps the result $\le c$. (Data booklet:
u' = (u − v)/(1 − uv/c²).)
Spacetime (Minkowski) diagram. Plot $ct$ (vertical) against $x$ (horizontal). A stationary object is a vertical worldline; light travels along $45^{\circ}$ lines. A moving observer's $ct'$ and $x'$ axes are tilted symmetrically toward the light line.
x' = γ(x − vt)、t' = γ(t − vx/c²)。）
相对论速度叠加。在 $S$ 中速度为 $u$ 的物体，在 $S'$ 中速度为： $$ u' = \frac{u - v}{1 - \dfrac{u v}{c^{2}}}. $$ 分母使结果保持 $\le c$。（数据手册：
u' = (u − v)/(1 − uv/c²)。）
时空（闵可夫斯基）图。以 $ct$（纵轴）对 $x$（横轴）作图。静止物体是竖直世界线；光沿 $45^{\circ}$ 线传播。运动观察者的 $ct'$ 与 $x'$ 轴对称地朝光线倾斜。
Spaceship $A$ moves at $0.70c$ east relative to a space station. It fires a probe forward at $0.50c$ east relative to itself. Find the probe's velocity relative to the station, and contrast with the classical (Galilean) answer.飞船 $A$ 相对空间站以 $0.70c$ 向东运动。它向前发射一个探测器，相对自身以 $0.50c$ 向东。求探测器相对空间站的速度，并与经典（伽利略）结果对比。
Identify. Let $S$ be the station and $S'$ the ship, with $v = 0.70c$. The probe's velocity in $S'$ is $u' = 0.50c$; we want its velocity $u$ in $S$.
识别。设 $S$ 为空间站、$S'$ 为飞船，$v = 0.70c$。探测器在 $S'$ 中的速度 $u' = 0.50c$；求它在 $S$ 中的速度 $u$。
Set Up. Invert the addition formula to solve for $u$ given $u'$:
列式。把叠加公式反解，由 $u'$ 求 $u$：
$$ u = \frac{u' + v}{1 + \dfrac{u' v}{c^{2}}}. $$Execute.
求解。
$$ u = \frac{0.50c + 0.70c}{1 + (0.50)(0.70)} = \frac{1.20c}{1.35} \approx 0.89c. $$Evaluate. The classical sum would be $0.50c + 0.70c = 1.20c$, faster than light — forbidden. The relativistic denominator $1.35$ pulls the result down to $0.89c < c$. No combination of sub-$c$ speeds ever exceeds $c$.
评估。经典求和会得 $0.50c + 0.70c = 1.20c$，比光还快——这是不允许的。相对论分母 $1.35$ 把结果压到 $0.89c < c$。任何低于 $c$ 的速度组合都不会超过 $c$。
Going deeper: reading worldlines off a Minkowski diagram深入：从闵可夫斯基图读出世界线
On a spacetime diagram with $ct$ up and $x$ across, the slope of a worldline is $c/v$ (steeper means slower). Light sits at $45^{\circ}$; nothing physical can have a worldline flatter than $45^{\circ}$ (that would mean $v > c$). For an observer moving at $v$, the $ct'$ axis is the line $x = vt$ (their own worldline) and the $x'$ axis is its mirror image across the light line; both tilt by the same angle, so the light line bisects them — a geometric statement of the invariance of $c$. Simultaneity in $S'$ means "parallel to the $x'$ axis", which is why simultaneous events in $S'$ are not simultaneous in $S$.
在 $ct$ 向上、$x$ 向右的时空图上，世界线的斜率为 $c/v$（越陡越慢）。光位于 $45^{\circ}$；任何实物的世界线都不能比 $45^{\circ}$ 更平（否则意味着 $v > c$）。对以 $v$ 运动的观察者，$ct'$ 轴是直线 $x = vt$（其自身世界线），$x'$ 轴是它相对光线的镜像；两者倾斜相同角度，故光线平分二者——这是 $c$ 不变性的几何表述。$S'$ 中的"同时"意味着"平行于 $x'$ 轴"，这正是 $S'$ 中同时的事件在 $S$ 中不同时的原因。
The Invariant Spacetime Interval and Experimental Evidence不变时空间隔与实验证据 A.5 HL
(Δs)² = (cΔt)² − (Δx)².)
- $(\Delta s)^2 > 0$: timelike — events can be causally linked; a clock can be present at both.
- $(\Delta s)^2 = 0$: lightlike — connected by a light signal.
- $(\Delta s)^2 < 0$: spacelike — no causal link; their time-order is frame-dependent.
(Δs)² = (cΔt)² − (Δx)²。）
- $(\Delta s)^2 > 0$：类时——事件可有因果关联；可有一只钟同时出现在两处。
- $(\Delta s)^2 = 0$：类光——由光信号相连。
- $(\Delta s)^2 < 0$：类空——无因果关联；其时间次序依赖于参考系。
Muons are created $4.5\ \mathrm{km}$ above the ground and travel down at $v = 0.995c$ ($\gamma \approx 10$), with a proper lifetime of $2.2\ \mathrm{\mu s}$. Show, from the ground frame, that a large fraction survive to the surface; then confirm the same result from the muon's frame.μ 子在地面上方 $4.5\ \mathrm{km}$ 处产生，以 $v = 0.995c$（$\gamma \approx 10$）向下运动，固有寿命 $2.2\ \mathrm{\mu s}$。从地面系证明有相当一部分能存活到地表；再从 μ 子系确认同一结果。
Identify. Take $c = 3.0 \times 10^{8}\ \mathrm{m\,s^{-1}}$. Compare the muon's available lifetime with the travel time over $4.5\ \mathrm{km}$.
识别。取 $c = 3.0 \times 10^{8}\ \mathrm{m\,s^{-1}}$。比较 μ 子可用寿命与穿过 $4.5\ \mathrm{km}$ 所需时间。
Ground frame (time dilation). The dilated lifetime is $\Delta t = \gamma\,\Delta t_0 = 10 \times 2.2 = 22\ \mathrm{\mu s}$. In that time the muon covers $d = v\,\Delta t = (0.995)(3.0 \times 10^{8})(22 \times 10^{-6}) \approx 6.6\ \mathrm{km}$, comfortably more than $4.5\ \mathrm{km}$.
地面系（时间膨胀）。膨胀后寿命 $\Delta t = \gamma\,\Delta t_0 = 10 \times 2.2 = 22\ \mathrm{\mu s}$。这段时间内 μ 子走过 $d = v\,\Delta t = (0.995)(3.0 \times 10^{8})(22 \times 10^{-6}) \approx 6.6\ \mathrm{km}$，远超 $4.5\ \mathrm{km}$。
Muon frame (length contraction). The atmosphere rushes up at $0.995c$, so the $4.5\ \mathrm{km}$ proper distance contracts to $L = L_0/\gamma = 4.5 / 10 = 0.45\ \mathrm{km}$. Crossing it takes $t = L/v = (450)/(0.995 \times 3.0 \times 10^{8}) \approx 1.5\ \mathrm{\mu s}$, less than the $2.2\ \mathrm{\mu s}$ rest lifetime.
μ 子系（长度收缩）。大气以 $0.995c$ 迎面冲上，故 $4.5\ \mathrm{km}$ 的固有距离收缩为 $L = L_0/\gamma = 4.5 / 10 = 0.45\ \mathrm{km}$。穿越它用时 $t = L/v = (450)/(0.995 \times 3.0 \times 10^{8}) \approx 1.5\ \mathrm{\mu s}$，小于 $2.2\ \mathrm{\mu s}$ 的静止寿命。
Evaluate. Both frames agree the muon reaches the ground — one invokes time dilation, the other length contraction. This dual explanation of the same fact is the classic exam framing of muon-decay evidence.
评估。两个参考系都认为 μ 子能抵达地面——一个用时间膨胀，另一个用长度收缩。对同一事实的这种双重解释正是 μ 子衰变证据的经典考法。
Going deeper: why the interval is invariant, and $E = \gamma m c^2$深入：为什么间隔不变，以及 $E = \gamma m c^2$
Substituting the Lorentz transformations into $(c\,\Delta t')^2 - (\Delta x')^2$ and simplifying, all the $\gamma$ and $v$ terms cancel, leaving $(c\,\Delta t)^2 - (\Delta x)^2$. The interval is the spacetime analogue of a length: rotations in space leave $x^2 + y^2$ fixed, while Lorentz "rotations" leave $(c\,\Delta t)^2 - (\Delta x)^2$ fixed. The proper time between timelike events is $c\,\Delta\tau = \Delta s$, independent of frame.
把洛伦兹变换代入 $(c\,\Delta t')^2 - (\Delta x')^2$ 并化简，所有 $\gamma$ 与 $v$ 项相消，余下 $(c\,\Delta t)^2 - (\Delta x)^2$。间隔是长度的时空类比：空间转动保持 $x^2 + y^2$ 不变，而洛伦兹"转动"保持 $(c\,\Delta t)^2 - (\Delta x)^2$ 不变。类时事件间的固有时间满足 $c\,\Delta\tau = \Delta s$，与参考系无关。
The total relativistic energy $E = \gamma m c^2$ reduces, for $v \ll c$, to $E \approx mc^2 + \tfrac{1}{2}mv^2$ — the rest energy plus the familiar Newtonian kinetic term. As $v \to c$, $\gamma \to \infty$, so accelerating a massive object to $c$ would require infinite energy. This is the energetic reason $c$ is an ultimate speed limit, complementing the velocity-addition argument of A5.5.
总相对论能量 $E = \gamma m c^2$ 在 $v \ll c$ 时退化为 $E \approx mc^2 + \tfrac{1}{2}mv^2$——静能加上熟悉的牛顿动能项。当 $v \to c$ 时 $\gamma \to \infty$，故把有质量物体加速到 $c$ 需无限能量。这正是 $c$ 为终极速度上限的能量学理由，与 A5.5 的速度叠加论证互为补充。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Proper time $\Delta t_0$ is measured where the two events happen at the same place; multiply by $\gamma$ to dilate. Identify whose clock is present at both events before computing.
- 固有时间 $\Delta t_0$ 在两事件发生于同一地点处测得；乘以 $\gamma$ 得膨胀时间。计算前先确定谁的钟同时出现在两事件处。
- Proper length $L_0$ is measured in the object's rest frame; divide by $\gamma$ to contract. The contracted answer is always shorter than $L_0$.
- 固有长度 $L_0$ 在物体静止系中测得；除以 $\gamma$ 得收缩长度。收缩后的答案总比 $L_0$ 短。
- Times get longer ($\times \gamma$), lengths get shorter ($\div \gamma$). A quick sanity check: $\gamma \ge 1$, so dilated time $>$ proper time and contracted length $<$ proper length.
- 时间变长（$\times \gamma$），长度变短（$\div \gamma$）。快速自检：$\gamma \ge 1$，故膨胀时间 $>$ 固有时间、收缩长度 $<$ 固有长度。
- If your dilated time came out shorter, or your contracted length came out longer, you inverted the factor. Fix it before moving on.
- 若你的膨胀时间反而更短，或收缩长度反而更长，说明你把因子用反了。先改正再继续。
- Never add velocities classically near $c$; always use $u' = (u - v)/(1 - uv/c^2)$. The result is guaranteed $\le c$ for sub-light inputs.
- 接近 $c$ 时绝不用经典相加；始终用 $u' = (u - v)/(1 - uv/c^2)$。对低于光速的输入，结果必 $\le c$。
- Track signs carefully for head-on (opposite-direction) cases. A subtraction of a negative velocity in the numerator increases the relative speed.
- 迎面（反向）情形要小心符号。分子中减去一个负速度会增大相对速率。
- State the frame first, then name the single effect that applies in it. Ground frame uses time dilation of the muon's lifetime; muon frame uses length contraction of the ground distance.
- 先说明参考系，再点明该系中适用的那一个效应。地面系用 μ 子寿命的时间膨胀；μ 子系用地面距离的长度收缩。
- Both frames must reach the same physical conclusion (the muon arrives). Markschemes reward showing the two viewpoints agree.
- 两个参考系必须得出相同的物理结论（μ 子到达）。评分会奖励展示两种视角一致。
Flashcards闪卡
$$t' = t$$
$$t' = \gamma\!\left(t - \frac{vx}{c^{2}}\right)$$
Unit A.5 Practice Quiz单元 A.5 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- HL Apply the Galilean transformation and classical velocity addition $u' = u - v$ in 1D在一维中应用伽利略变换与经典速度叠加 $u' = u - v$
- HL State both postulates of special relativity and explain the role of the Michelson–Morley null result陈述狭义相对论两条假设，并解释迈克耳孙–莫雷零结果的作用
- HL Compute the Lorentz factor $\gamma = 1/\sqrt{1 - v^2/c^2}$ for any sub-light speed对任意低于光速求洛伦兹因子 $\gamma = 1/\sqrt{1 - v^2/c^2}$
- HL Identify the proper time $\Delta t_0$ and apply time dilation $\Delta t = \gamma\,\Delta t_0$辨认固有时间 $\Delta t_0$ 并应用时间膨胀 $\Delta t = \gamma\,\Delta t_0$
- HL Derive time dilation from the light-clock and the invariance of $c$由光钟与 $c$ 的不变性推导时间膨胀
- HL Identify the proper length $L_0$ and apply length contraction $L = L_0/\gamma$辨认固有长度 $L_0$ 并应用长度收缩 $L = L_0/\gamma$
- HL Explain the relativity of simultaneity with the train-and-platform thought experiment用火车—站台思想实验解释同时性的相对性
- HL Apply the Lorentz transformations $x' = \gamma(x - vt)$ and $t' = \gamma(t - vx/c^2)$应用洛伦兹变换 $x' = \gamma(x - vt)$ 与 $t' = \gamma(t - vx/c^2)$
- HL Use relativistic velocity addition $u' = (u - v)/(1 - uv/c^2)$ and confirm the result stays $\le c$使用相对论速度叠加 $u' = (u - v)/(1 - uv/c^2)$ 并确认结果保持 $\le c$
- HL Plot worldlines on a spacetime (Minkowski) diagram and locate the $45^{\circ}$ light line在时空（闵可夫斯基）图上画世界线并定出 $45^{\circ}$ 光线
- HL Compute the invariant interval $(\Delta s)^2 = (c\Delta t)^2 - (\Delta x)^2$ and classify it计算不变间隔 $(\Delta s)^2 = (c\Delta t)^2 - (\Delta x)^2$ 并对其分类
- HL Explain muon-decay evidence from both the ground frame (dilation) and muon frame (contraction)从地面系（膨胀）与 μ 子系（收缩）两方面解释 μ 子衰变证据
IB Paper-Style PracticeIB 试卷风格练习
A.5 Practice and Solutions are on the roadmap, to ship under Practice Questions/Unit_A5_*.html with the bilingual built-in pattern.
A.5 配套 Practice 与 Solutions 在排期，上线后位于 Practice Questions/Unit_A5_*.html。