Special relativity is the topic IB Physics HL students fear most — and the one that most often surprises them in the exam, because once you have the right mental model the formulas are fewer and simpler than in any other HL topic. Topic A.5 takes you from Galilean transformations (your everyday intuition) to Einstein's two postulates, then derives the Lorentz factor, time dilation, length contraction, the full Lorentz transformations, relativistic velocity addition, and the invariant space-time interval. It finishes with space-time diagrams and the muon-decay experiment that confirms it all.
This cheatsheet condenses every formula, trick and trap from Topic A.5 Galilean and Special Relativity into one page. It covers Galilean vs Lorentz transformations, the Lorentz factor (with memorable values for $v = 0.6c$ and $0.8c$), proper time and proper length, simultaneity, world lines and tilted axes on space-time diagrams, and how to argue the muon-decay paradox from both reference frames — the standard IB full-marks question. Scroll to the bottom for the printable PDF and the gated full library.
§1 — Galilean Relativity A.5 HL
Galilean transformations
If frame S$'$ moves at $+v$ relative to S in the $x$-direction:
Inertial frames & the principle
- Inertial frame: a non-accelerating reference frame. Newton's laws hold unchanged.
- Galilean Relativity: the laws of mechanics are the same in every inertial frame. No mechanical experiment can detect uniform motion.
§2 — The Lorentz Factor A.5 HL
$$\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}} \;\geq\; 1$$
| $v / c$ | $\gamma$ |
|---|---|
| 0.6 | 1.25 |
| 0.8 | $5/3 \approx 1.667$ |
| 0.9 | 2.29 |
| 0.99 | 7.09 |
Time dilation
$$\Delta t = \gamma \,\Delta t_0 \quad (\Delta t \geq \Delta t_0)$$
- $\Delta t_0$ — proper time: measured by a clock present at both events (shortest time).
- $\Delta t$ — dilated time measured by an observer for whom the events occur at different places.
- Moving clocks run slow.
§3 — Length Contraction & Lorentz Transforms A.5 HL
Length contraction
$$L = \dfrac{L_0}{\gamma} \quad (L \leq L_0)$$
- $L_0$ — proper length: measured in the rest frame of the object (longest length).
- Only the dimension parallel to motion contracts. $y$ and $z$ are unchanged.
Lorentz transformations
S$'$ moves at $+v$ relative to S:
Inverse (S$' \to$ S):
§4 — Velocity Addition & Space-Time Interval A.5 HL
Relativistic velocity addition
An object moves at $u$ in S; S$'$ moves at $v$ relative to S. Then:
$$u' = \dfrac{u - v}{1 - u v / c^2}$$
Inverse: $\;u = \dfrac{u' + v}{1 + u' v / c^2}$.
Sanity check: if $u = c$, then $u' = c$ as well — the speed of light is the same in both frames. ✓
Space-time interval (invariant)
$$(\Delta s)^2 = (c\,\Delta t)^2 - (\Delta x)^2$$
Same value in all inertial frames (invariant under Lorentz transforms).
- $(\Delta s)^2 > 0$ — timelike separation: a causal link is possible.
- $(\Delta s)^2 = 0$ — lightlike: events connected by a light signal.
- $(\Delta s)^2 < 0$ — spacelike: no causal connection possible.
§5 — Space-Time Diagrams A.5 HL
World lines
- Axes: horizontal $= x$; vertical $= ct$ (so both axes have units of distance).
- Stationary object: vertical world line.
- Object moving at $v$: straight line with $\tan\theta = v/c$ from the $ct$-axis.
- Light: 45° line ($\tan\theta = 1$, $v = c$).
- Every world line of a physical object lies between vertical and the 45° line.
Drawing S$'$ axes on an S diagram
If S$'$ moves at $+v$ relative to S:
- The $ct'$ axis tilts towards the 45° light line by angle $\theta$, where $\tan\theta = v/c$.
- The $x'$ axis tilts towards the 45° light line by the same angle (symmetric — the light line bisects the primed axes).
- Scales on the primed axes are not equal to the unprimed ones — unit intervals are larger. Lines of constant $(\Delta s)^2$ (hyperbolas) calibrate the axes.
§6 — Muon Decay & Experimental Evidence A.5 HL
The argument from both frames
- Earth frame: the muon travels at $\sim 0.98c$. Its lifetime is dilated to $\gamma \tau_0$. Distance travelled $= v \cdot \gamma\tau_0 \gg v\tau_0$ — long enough to reach sea level.
- Muon frame: the muon is at rest. The atmosphere is length-contracted to $L_0 / \gamma$, so the distance to Earth is short enough to traverse in one lifetime $\tau_0$.
- Both frames agree on the same physical outcome: the muon reaches sea level.
Summary: which is "proper"?
| Quantity | Proper | Measured by |
|---|---|---|
| Time | $\Delta t_0$ | Single clock present at both events |
| Length | $L_0$ | Rest frame of the object |
Proper time is the shortest time. Proper length is the longest length.
§7 — Exam Attack Plan A.5 — All sections
| Question trigger | Reach for |
|---|---|
| "State the postulates of SR" | Both — laws of physics in every inertial frame; $c$ is invariant |
| "Calculate $\gamma$ for $v = 0.8c$" | $\gamma = 1/\sqrt{1 - 0.64} = 5/3$ |
| "Find the dilated time" | $\Delta t = \gamma \Delta t_0$ — identify proper time first |
| "Find contracted length" | $L = L_0 / \gamma$ — identify proper length first |
| "Two events, find $x'$, $t'$ in moving frame" | Lorentz transforms; sign of $v$ matters |
| "Spaceship A fires probe; find probe speed in lab" | $u = (u' + v)/(1 + u'v/c^2)$ |
| "Are these events causally linked?" | $(c\Delta t)^2 - (\Delta x)^2$: $> 0$ ⇒ timelike (causal) |
| "Draw S$'$ axes on S diagram" | Both $ct'$ and $x'$ tilt by $\tan\theta = v/c$ towards 45° line |
| "Explain muon decay" | Earth frame: time dilation. Muon frame: length contraction. |
| "Are these two events simultaneous in S$'$?" | Use $t' = \gamma(t - vx/c^2)$ — simultaneity is frame-dependent |
Worked Example — IB-Style Time Dilation & Length Contraction
Question (HL Paper 2 style — 6 marks)
A spaceship travels from Earth to a star 4.0 ly (light-years) away, as measured in the Earth frame, at a constant speed of $0.80c$. Calculate (a) the time taken for the journey as measured by an observer on Earth, (b) the time taken for the journey as measured by the astronaut on the spaceship, and (c) the distance between Earth and the star as measured by the astronaut.
Solution
- For $v = 0.80c$: $\gamma = \dfrac{1}{\sqrt{1 - 0.64}} = \dfrac{1}{0.6} = \dfrac{5}{3} \approx 1.667$. (M1)
- Earth-frame journey time: $\Delta t = \dfrac{L_0}{v} = \dfrac{4.0\;\text{ly}}{0.80c} = 5.0$ years. (A1) [part (a)]
- The astronaut's clock measures the proper time $\Delta t_0$ — the spaceship is at both events (departure and arrival from its own perspective). So $\Delta t = \gamma\,\Delta t_0$. (R1)
- $\Delta t_0 = \dfrac{\Delta t}{\gamma} = \dfrac{5.0}{5/3} = 3.0$ years. (A1) [part (b)]
- From the astronaut's frame the Earth–star distance is length-contracted ($L_0 = 4.0$ ly is the Earth-frame proper length of the Earth–star separation):
$L = L_0 / \gamma = 4.0 / (5/3) = 2.4$ ly. (M1) - Cross-check: in the astronaut's frame, the star moves towards the ship at $0.80c$, so the journey takes $L/v = 2.4 / 0.80 = 3.0$ years — matches (b). ✓ (A1) [part (c)]
Examiner's note: Students often mix up which observer measures the proper time. Here the astronaut is the only observer for whom departure and arrival happen at the same place (right next to him in the cockpit) — so the astronaut measures the proper time $\Delta t_0 = 3.0$ years. The Earth observer sees two events at different places, so measures the dilated time $5.0$ years. The two cross-check via $L/v$ in the astronaut's frame.
Common Student Questions
Which observer measures the proper time?
What are the two postulates of special relativity?
How do I explain muon decay using both reference frames?
Which length is the proper length?
Why don't velocities just add the way Galileo said they do?
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