Postulates of SR: Exercises — Physics deep dive

Definition Symbols used on this whole page (build them once, use them everywhere)

$c = 3 \times 1 0^{8} m/s$ — the speed of light in vacuum. By Postulate 2 it is the same number for every inertial (non-accelerating) observer.
$v$ — the relative speed between two inertial frames (e.g. how fast a ship moves past Earth). Always $v < c$ .
$Δ t_{0}$ — the proper time: the time between two events measured in the frame where they happen at the same place (e.g. on the moving clock itself). It is the shortest possible time for those events.
$Δ t$ — the dilated time: the time between the same two events measured by any other inertial observer. Always $Δ t \geq Δ t_{0}$ .
$γ$ — the Lorentz factor, the "stretch factor" linking the two: $γ = \frac{1}{1 - v ^{2} / c ^{2}} \geq 1, Δ t = γ Δ t_{0} .$ Because $v < c$ we have $v^{2} / c^{2} < 1$ , so the square root is below $1$ and its reciprocal $γ$ is at least $1$ . At everyday speeds $γ \to 1$ and Newton returns.
$u$ — the speed of an object as measured inside a moving frame (e.g. a ball thrown inside a rocket, measured by someone in that rocket).
$u^{'}$ — the speed of the same object as measured by the outside observer (e.g. Earth). The Relativistic Velocity Addition rule links them: $u^{'} = \frac{u + v}{1 + uv / c ^{2}}$ .

Level 1 — Recognition

Recall Solution

(b) and (d).

(a) is the opposite of Postulate 2 — light's speed does not depend on the source. Wrong.
(b) is Postulate 1 (Principle of Relativity). ✅
(c) is exactly the Newtonian idea Einstein threw away. Wrong.
(d) is Postulate 2 (constancy of $c$ ). ✅

Recall Solution

The ground lab is (approximately) inertial: a free object left alone stays put or moves in a straight line at constant speed — Newton's first law holds, no pseudo-forces. The braking car is not inertial: it is accelerating, so a free object (a ball on the seat) lurches forward on its own — a pseudo-force appears. SR's Postulate 1 does not cover accelerating frames.

Level 2 — Application

Recall Solution

Step 1 — compute $γ$ . With $v^{2} / c^{2} = 0.64$ : $γ = \frac{1}{1 - 0.64} = \frac{1}{0.36} = \frac{1}{0.6} = 1.6 \overline{6} .$ Step 2 — apply $Δ t = γ Δ t_{0}$ (here $Δ t_{0} = 5 s$ is the ship's proper time, $Δ t$ is Earth's measurement): $Δ t = 1.6 \overline{6} \times 5 = 8.33 s .$ Earth measures the moving clock running slow (longer elapsed time). See Time Dilation.

Recall Solution

$c$ exactly ( $3 \times 1 0^{8} m/s$ ), not $1.95 c$ . Postulate 2: light's speed is frame-independent. Check with the Relativistic Velocity Addition formula. Define its symbols first:

$u$ = the object's speed as measured inside the rocket's frame (here the laser light, so $u = c$ );
$v$ = the speed of the rocket's frame relative to the planet observer ( $v = 0.95 c$ );
$u^{'}$ = the object's speed as measured by the planet observer (what we want). $u^{'} = \frac{u + v}{1 + uv / c ^{2}} = \frac{c + 0.95 c}{1 + ( c ) ( 0.95 c ) / c ^{2}} = \frac{1.95 c}{1 + 0.95} = \frac{1.95 c}{1.95} = c .$ The formula forces light back to $c$ .

Recall Solution

$v / c = 300/ (3 \times 1 0^{8}) = 1 0^{- 6}$ , so $v^{2} / c^{2} = 1 0^{- 12}$ . Now use the small-quantity expansion. Let $x \equiv v^{2} / c^{2}$ — here $x = 1 0^{- 12}$ , a tiny number. For any tiny $x$ , $\frac{1}{1 - x} \approx 1 + \frac{1}{2} x$ , so $γ - 1 \approx \frac{1}{2} x = \frac{1}{2} (1 0^{- 12}) = 5 \times 1 0^{- 13} .$ Utterly negligible — this is why Newtonian physics works in daily life.

Level 3 — Analysis

Worked example L3.1 — Rebuild the light-clock diagonal

A light-clock (mirror separation $L$ ) is watched from Earth's frame, in which the clock moves right at speed $v$ . In this Earth frame the photon travels a diagonal each half-tick because the top mirror slides sideways while the light is in flight. (In the clock's own frame the light would go straight up and down — that shorter, purely-vertical trip is the proper time $Δ t_{0}$ ; the diagonal trip we analyse here is the dilated time $Δ t$ .) Using the figure below, write the length of ONE diagonal in terms of $L$ , $v$ , and the full Earth-frame tick $Δ t$ — then explain WHY the speed used is $c$ and not $c + v$ .

Figure — Earth-frame view of a moving light-clock. The two gray horizontal bars are the bottom and top mirrors, drawn at three moments as the clock slides right. A blue double-arrow marks the fixed vertical separation $L$ . An orange double-arrow along the bottom marks the sideways shift $v Δ t /2$ the clock makes in half a tick. The red line is the photon's actual slanted path (up then down) — the "diagonal" $d$ we are solving for.

Recall Solution

During one full tick $Δ t$ the mirror moves right by $v Δ t$ ; during half a tick it moves $v Δ t /2$ (the orange base of each triangle). The vertical side is $L$ (blue). By Pythagoras, one diagonal (red) is $d = L^{2} + (\frac{v Δ t}{2})^{2} .$ Why speed $c$ ? Postulate 2: the photon's speed is $c$ in every inertial frame, including Earth's frame where the clock moves. The path got longer, but the speed did not increase to compensate — so the tick must take longer, i.e. $c Δ t = 2 d$ . That extra time is exactly time dilation. (Full algebra → Time Dilation.)

Recall Solution

Square both sides. Note $(2 L^{2} + (v Δ t /2)^{2})^{2} = 4 (L^{2} + \frac{v ^{2} Δ t ^{2}}{4}) = 4 L^{2} + v^{2} Δ t^{2}$ : $c^{2} Δ t^{2} = 4 L^{2} + v^{2} Δ t^{2} .$ Group the $Δ t^{2}$ terms: $Δ t^{2} (c^{2} - v^{2}) = 4 L^{2} .$ Take the square root. The equation gives two roots, $Δ t = \pm 2 L / c^{2} - v^{2}$ ; we keep only the positive root because $Δ t$ is an elapsed time and cannot be negative (the negative root is unphysical here). So $Δ t = \frac{2 L}{c ^{2} - v ^{2}} .$ Factor $c$ out of the root: $c^{2} - v^{2} = c 1 - v^{2} / c^{2}$ , so $Δ t = \frac{2 L}{c 1 - v ^{2} / c ^{2}} = \frac{2 L / c}{1 - v ^{2} / c ^{2}} .$ Now substitute $Δ t_{0} = 2 L / c$ : $Δ t = \frac{Δ t _{0}}{1 - v ^{2} / c ^{2}} = γ Δ t_{0} . ■$

Recall Solution

(a) $v^{2} / c^{2} = 0.9604$ , so $1 - v^{2} / c^{2} = 0.0396$ and $γ = 1/ 0.0396 = 5.025$ . $Δ t = γ Δ t_{0} = 5.025 \times 2.2 μ s = 11.06 μ s .$ (b) Distance in Earth's frame $= v Δ t = 0.98 \times (3 \times 1 0^{8}) \times 11.06 \times 1 0^{- 6}$ $= 2.94 \times 1 0^{8} \times 1.106 \times 1 0^{- 5} = 3251 m \approx 3.25 km .$ Without dilation the muon would cover only $0.98 c \times 2.2 μ s \approx 647 m$ — it would never reach the ground. Yet muons do reach detectors: experimental proof of time dilation. (From the muon's own view, it's Length Contraction that shrinks the atmosphere instead — two frames, same physics.)

Level 4 — Synthesis

Recall Solution

Using the symbols from the global box: $u = 0.6 c$ (the ball's speed measured inside frame $B$ ), $v = 0.6 c$ (speed of $B$ relative to Earth), and $u^{'}$ = Earth's measurement. Galilean (wrong here): $u + v = 0.6 c + 0.6 c = 1.2 c$ — impossible, faster than light. Relativistic Relativistic Velocity Addition: $u^{'} = \frac{u + v}{1 + uv / c ^{2}} = \frac{1.2 c}{1 + 0.36} = \frac{1.2 c}{1.36} = 0.882 c .$ Still below $c$ — relativity keeps everything sub-luminal. The correction term $uv / c^{2} = 0.36$ in the denominator is exactly what "eats" the excess.

Recall Solution

$γ = 1/ 1 - 0.36 = 1/0.8 = 1.25$ for both — the situation is symmetric. Each measures the other's moving clock as ticking slow by $\times 1.25$ (i.e. for every 1 s of the other's clock, $1.25$ s pass on their own). No contradiction because "moving clock is slow" is a statement about two events measured in one frame vs the proper time in the other. Alice and Bob use different pairs of synchronised clocks and different notions of simultaneity (the classical idea Einstein sacrificed). To settle "who is truly younger" you'd need to bring them back together — which requires acceleration (a turnaround), breaking the symmetry. See Lorentz Transformations.

Recall Solution

Maxwell's Equations predict one fixed light speed $c$ . They contain the constants of electricity and magnetism, and out pops a single wave speed $c$ — with no mention of who is watching.
Postulate 1 says those equations must look the same in every inertial frame. Why? Postulate 1 states the laws of physics are identical in all inertial frames; Maxwell's equations are laws of physics, so no inertial observer may write them differently. Hence every inertial observer, applying the same equations, must get the same $c$ .
If light instead needed a medium (the "ether"), Earth's motion through it should produce a direction-dependent "ether wind" — $c$ would vary with direction.
Michelson–Morley (1887) searched for that wind and found none: $c$ was the same in every direction, at every season.
Conclusion: there is no preferred medium and no special frame. $c$ is the same for all inertial observers — that is Postulate 2.

Level 5 — Mastery

Recall Solution

Ship's proper time is $Δ t_{0} = Δ t / γ$ , so the deficit is $Δ t - Δ t_{0} = Δ t (1 - \frac{1}{γ}) = Δ t (1 - 1 - v^{2} / c^{2}) .$ For small $v / c$ : with $x \equiv v^{2} / c^{2}$ , $1 - x \approx 1 - \frac{1}{2} x$ , so $1 - 1 - x \approx \frac{1}{2} x = \frac{1}{2} v^{2} / c^{2}$ . Set the deficit to $1 s$ : $86400 \times \frac{1}{2} \frac{v ^{2}}{c ^{2}} = 1 \Rightarrow \frac{v ^{2}}{c ^{2}} = \frac{2}{86400} = 2.315 \times 1 0^{- 5} .$ $\frac{v}{c} = 2.315 \times 1 0^{- 5} = 4.81 \times 1 0^{- 3} .$ So $v \approx 0.00481 c \approx 1.44 \times 1 0^{6} m/s$ . Since $v / c ≪ 1$ , the approximation is safe (exact $γ$ differs only in the 6th decimal).

Recall Solution

(a) Each leg covers $6$ light-years at $0.6 c$ , so it takes $6/0.6 = 10$ Earth-years. Out and back gives $Δ t = 2 \times 10 = 20$ Earth-years. (b) $γ = 1/ 1 - 0.36 = 1/0.8 = 1.25$ . The traveller's clock records the proper time for the journey (their own frame), which is shorter: $Δ t_{0} = \frac{Δ t}{γ} = \frac{20}{1.25} = 16 years .$ (c) The traveller is younger by $20 - 16 = 4$ years. The apparent symmetry (each could claim the other moves) is broken because the traveller accelerates at the turnaround — they change inertial frames, while the Earth twin stays in one inertial frame the entire time. That frame-switch is physically real; the Lorentz Transformations confirm the traveller's total elapsed time is the shorter one. (This is the famous twin paradox — see Time Dilation.)

Recall Solution

Total energy: $E = γ m c^{2} = 5.025 \times 105.7 MeV = 531.1 MeV .$ Kinetic energy: $K = (γ - 1) m c^{2} = 4.025 \times 105.7 MeV = 425.4 MeV .$ The same $γ$ that stretches the muon's lifetime also multiplies its energy — one factor, two consequences. See Mass-Energy Equivalence.

Connections

Postulates of SR — the parent note this page drills.
Time Dilation, Length Contraction, Lorentz Transformations, Relativistic Velocity Addition, Mass-Energy Equivalence — the machinery used above.
Michelson-Morley Experiment, Maxwell's Equations, Galilean Relativity — the historical/logical roots.

Recall Self-test checklist

Can you state both postulates without looking? ::: Same laws (all inertial frames) + same light speed (all inertial observers). Multiply or divide by $γ$ to get the other frame's time? ::: Multiply: $Δ t = γ Δ t_{0}$ , always longer. Why does the light-clock's diagonal NOT mean light exceeds $c$ ? ::: Longer path is paid for by longer time, since speed is locked at $c$ . What breaks the twin symmetry? ::: The turnaround acceleration — only one twin changes inertial frames.