Postulates of SR: Worked examples — Physics deep dive

First, a symbol we will lean on constantly. Everywhere below, $β$ (Greek "beta") is just a nickname:

The scenario matrix

Every relativity problem in this chapter lands in one of these cells. The examples below are labelled by cell.

Cell	What makes it special	Example
A — everyday slow speed	$β ≪ 1$ , $γ \approx 1$ ; Newton returns	Ex 1
B — moderate speed	$β \sim 0.6$ , clean $γ$	Ex 2
C — length, not time	shrink instead of stretch (Length Contraction)	Ex 3
D — velocities forward	light + rocket, does it exceed $c$ ?	Ex 4
E — velocities backward	negative $v$ , sign handling	Ex 5
F — the $v \to c$ limit	degenerate: $γ \to \infty$	Ex 6
G — real-world word problem	muon reaching the ground	Ex 7
H — exam twist / two effects	consistency of time & length together	Ex 8

Cell A — everyday slow speed

Worked example Ex 1 (Cell A) — A jet's clock

A jet flies at $v = 300 m/s$ (about Mach 0.9). It carries a clock. After the ground measures $1$ hour $= 3600 s$ , how far behind is the jet's clock?

Forecast: guess — nanoseconds? seconds? minutes?

Compute $β$ . $β = v / c = 300/ (3 \times 1 0^{8}) = 1 0^{- 6}$ . Why this step? $β$ is the only knob $γ$ depends on; get it first.
Compute $β^{2}$ . $β^{2} = 1 0^{- 12}$ . Why? $γ$ uses $β^{2}$ , and this is so tiny we can approximate next.
Approximate $γ$ . For tiny $x$ , $\frac{1}{1 - x} \approx 1 + \frac{1}{2} x$ . So $γ \approx 1 + \frac{1}{2} (1 0^{- 12}) = 1 + 5 \times 1 0^{- 13}$ . Why this step? Plugging $1 0^{- 12}$ into a calculator gives $γ = 1.000 \dots$ and you'd lose the tiny difference. The linear approximation keeps the meaningful part. This is exactly why Newton worked for 200 years — the correction hides in the 13th decimal.
Time gap. The ground time is longer by factor $γ$ , so the lag is $(γ - 1) \times Δ t_{0}$ . Here it's cleanest to say the jet's own clock reads $Δ t_{0} = Δ t / γ \approx 3600 (1 - 5 \times 1 0^{- 13})$ , so it falls behind by $3600 \times 5 \times 1 0^{- 13} = 1.8 \times 1 0^{- 9} s$ . Why? $(γ - 1)$ is the fractional slowdown; multiply by elapsed time.

Answer: about $1.8 \times 1 0^{- 9} s = 1.8 nanoseconds$ behind.

Verify: units — $[dimensionless] \times [s] = s$ ✓. Sanity — nanoseconds over a whole hour is why you never notice; relativity is always on but invisible at $β = 1 0^{- 6}$ .

Cell B — moderate speed, clean numbers

Cell C — length shrinks instead of stretching

Here is the geometry that trips people up. Time stretches ( $\times γ$ ) but length shrinks ( $\div γ$ ). The figure shows why the same $γ$ pulls two ways.

Worked example Ex 3 (Cell C) — A rod flies past

A rod is $L_{0} = 100 m$ long in its own frame (Length Contraction, proper length). It streaks past you at $v = 0.6 c$ . How long do you measure it?

Forecast: longer or shorter than $100 m$ ?

$β = 0.6$ , $β^{2} = 0.36$ , $1 - β^{2} = 0.64$ , $= 0.8$ , $γ = 1.25$ . Why? Same ladder as before; $γ$ never changes its recipe.
Divide, don't multiply. $L = L_{0} / γ = 100/1.25 = 80 m$ . Why divide here? Length in the direction of motion contracts. The proper length $L_{0}$ is the longest anyone measures; the moving observer sees it squeezed. Look at the blue bar in the figure — same $γ$ , opposite direction.

Answer: $80 m$ .

Verify: $L < L_{0}$ ✓ (contraction, not dilation). Cross-check the two effects share $γ$ : time would go $\times 1.25$ , length $\times (1/1.25) = 0.8$ — reciprocals, as the figure shows.

Cell D — velocities forward (does light beat $c$ ?)

Everyday velocities add: walk $2$ m/s on a $3$ m/s train, the ground sees $5$ . Relativity replaces that with Relativistic Velocity Addition:

Cell E — velocities backward (sign handling)

The formula holds for any signs — you just carry the minus. Backward motion is $v < 0$ or $u < 0$ .

Worked example Ex 5 (Cell E) — Two rockets, opposite directions

Rocket A moves right at $v = 0.8 c$ (relative to you). It fires a probe backwards at $u = - 0.5 c$ (in A's frame). What speed do you see for the probe?

Forecast: $0.3 c$ (naive $0.8 - 0.5$ )? Something a bit different?

Substitute with sign. $u^{'} = \frac{- 0.5 c + 0.8 c}{1 + \frac{( - 0.5 c ) ( 0.8 c )}{c ^{2}}} = \frac{0.3 c}{1 - 0.40}$ . Why keep the minus in the denominator? The product $uv$ is now negative, so the relativistic tax subtracts — you must track the sign faithfully.
Simplify. $= \frac{0.3 c}{0.60} = 0.5 c$ . Why? Arithmetic; $0.3/0.6 = 0.5$ .

Answer: $+ 0.5 c$ (still moving right, just slower than A).

Verify: result magnitude $0.5 c < c$ ✓ (never exceeds light-speed). Galilean guess was $0.3 c$ ; the relativistic answer $0.5 c$ differs — proof that even with a minus sign the formula does real work.

Cell F — the degenerate limit $v \to c$

What happens at light-speed? The figure plots $γ$ climbing toward the wall.

Worked example Ex 6 (Cell F) — Pushing toward light-speed

Compute $γ$ at $β = 0.9$ , then $0.99$ , then $0.999$ . What is the pattern as $β \to 1$ ?

Forecast: does $γ$ level off, or blow up?

$β = 0.9$ : $1 - 0.81 = 0.19$ , $= 0.436$ , $γ = 2.29$ . Why? Same ladder; watch the root shrink.
$β = 0.99$ : $1 - 0.9801 = 0.0199$ , $= 0.141$ , $γ = 7.09$ . Why? The root just got tiny, so its reciprocal jumps.
$β = 0.999$ : $1 - 0.998 = 0.001999$ , $= 0.0447$ , $γ = 22.4$ . Why? As $β^{2} \to 1$ , the number under the root $\to 0$ , so $γ \to \infty$ .

Answer: $γ \to \infty$ as $β \to 1$ . Time dilation becomes infinite; you'd need infinite energy (Mass-Energy Equivalence) to reach $c$ . That's why massive objects never touch light-speed — the wall at $β = 1$ is unreachable. See the red asymptote in the figure.

Verify: at $β = 1$ exactly, $1 - β^{2} = 0$ and $γ = 1/0$ is undefined — the degenerate case is genuinely forbidden, consistent with " $<$ " in the postulates.

Cell G — real-world word problem

Worked example Ex 7 (Cell G) — The muon that shouldn't survive

Muons form $\sim 10 km$ up in the atmosphere, travel down at $v = 0.98 c$ , and live only $Δ t_{0} = 2.2 μ s$ in their own frame. Without relativity, how far do they get? With time dilation, do they reach the ground?

Forecast: will they make it the full $10$ km?

Naive distance (no relativity). $d = v Δ t_{0} = 0.98 \times (3 \times 1 0^{8}) \times (2.2 \times 1 0^{- 6}) \approx 647 m$ . Why? Classical distance = speed × lifetime. Only $0.65$ km — they'd die high up.
Compute $γ$ . $β = 0.98$ , $β^{2} = 0.9604$ , $1 - β^{2} = 0.0396$ , $= 0.199$ , $γ \approx 5.03$ . Why? The lab sees the muon's clock run slow, extending its life.
Dilated lifetime in our frame. $Δ t = γ Δ t_{0} = 5.03 \times 2.2 = 11.06 μ s$ . Why $\times$ ? $2.2 μ s$ is proper time; the ground frame measures longer.
Distance with dilation. $d = v Δ t = 0.98 \times (3 \times 1 0^{8}) \times (11.06 \times 1 0^{- 6}) \approx 3250 m \approx 3.25 km$ . Why? Same distance formula, but with the dilated lifetime.

Answer: classically $\approx 0.65$ km (fails); relativistically $\approx 3.25$ km — many do reach detectors, and this is measured. Real experimental proof of time dilation.

Verify: $3250/647 = γ$ ? $3250/647 \approx 5.02 \approx γ$ ✓ — the extra reach is exactly the $γ$ factor.

Cell H — exam twist: two effects must agree

Worked example Ex 8 (Cell H) — Same journey, two viewpoints

A ship crosses a $L_{0} = 6 \times 1 0^{9} m$ lane (measured by ground) at $v = 0.6 c$ . (a) Ground clock: how long does the crossing take? (b) Ship crew's clock: how long for them? Show the two views are consistent.

Forecast: which clock reads less — ground or ship?

Ground time. $t = L_{0} / v = \frac{6 \times 1 0 ^{9}}{0.6 \times 3 \times 1 0 ^{8}} = \frac{6 \times 1 0 ^{9}}{1.8 \times 1 0 ^{8}} = 33.3 s$ . Why? Ground measures full lane length and its own time — plain distance/speed.
$γ$ . $β = 0.6 \Rightarrow γ = 1.25$ (from Ex 3). Why? Needed to convert between frames.
Ship time via dilation. The crew is one clock at one place → their crossing time is proper time: $t_{ship} = t / γ = 33.3/1.25 = 26.7 s$ . Why divide? The moving crew's clock runs slow as seen by ground; equivalently their own elapsed time is shorter.
Cross-check via length contraction. The crew see the lane contracted: $L = L_{0} / γ = 6 \times 1 0^{9} /1.25 = 4.8 \times 1 0^{9} m$ . Their time = $L / v = 4.8 \times 1 0^{9} / (1.8 \times 1 0^{8}) = 26.7 s$ . Why do this? Two different effects (dilation vs contraction) must give the same ship time. They do — that's the internal consistency of relativity.

Answer: ground $33.3 s$ , ship $26.7 s$ ; both methods agree.

Verify: $33.3/26.7 = 1.25 = γ$ ✓, and the length- and time-based ship computations match to $26.7 s$ ✓.

Recall Quick self-test

A rod is $50$ m proper length at $β = 0.6$ ; what length do you measure? ::: $50/1.25 = 40$ m A clock reads $4 μ s$ proper at $β = 0.8$ ; what does the lab read? ::: $γ = 5/3$ , so $6.67 μ s$ Light fired from a $0.99 c$ ship — its speed to you? ::: exactly $c$ As $β \to 1$ , $γ \to$ ? ::: $\infty$

Connections

Parent: Postulates of SR
Time Dilation · Length Contraction · Lorentz Transformations · Relativistic Velocity Addition · Michelson-Morley Experiment · Mass-Energy Equivalence · Galilean Relativity