Lorentz transformation — derivation: Worked examples

Before anything, one reminder about units so no example trips you up.

The scenario matrix

Every Lorentz-transform problem is one cell of this table. The right column names the example that lands on it.

Cell class	What makes it special	Covered by
Zero velocity ( $v = 0$ )	$γ = 1$ : nothing should change	Ex. 1
Low velocity ( $v ≪ c$ )	must reduce to Galileo $x^{'} = x - v t$	Ex. 2
Event ahead ( $x > 0$ ), forward frame ( $v > 0$ )	standard positive case	Ex. 3
Event behind ( $x < 0$ ) or backward frame ( $v < 0$ )	sign of the $- v x / c^{2}$ term flips	Ex. 4
Simultaneous pair ( $Δ t = 0$ )	relativity of simultaneity, both signs	Ex. 5
Limiting speed ( $v \to c$ )	$γ \to \infty$ , degenerate squeeze	Ex. 6
Real-world word problem	muon / spaceship, pick out $x, t, v$ yourself	Ex. 7
Exam twist (inverse + invariant)	go from $S^{'}$ back to $S$ , check $s^{2}$	Ex. 8

We march through them in order. Keep the Minkowski diagram, Time dilation, Length contraction and Relativity of simultaneity pages in mind — every example quietly touches one of them.

Worked examples

Worked example 2 · Low speed → Galileo — cell:

$v ≪ c$ Statement. A car frame moves at $v = 30 m/s$ . An event is at $x = 300 m$ , $t = 2 s$ . Show $x^{'} \approx x - v t$ and $t^{'} \approx t$ . ( $c = 3 \times 1 0^{8} m/s$ .)

Forecast: Do you expect to see any relativistic difference at car speeds?

Estimate $v^{2} / c^{2}$ . $v^{2} / c^{2} = (30)^{2} / (3 \times 1 0^{8})^{2} = 900/ (9 \times 1 0^{16}) = 1 0^{- 14}$ . Why this step? This tiny number is what sits under the square root; if it is negligible, $γ \approx 1$ .
So $γ \approx 1$ to 14 decimal places. Why this step? $1 - 1 0^{- 14} \approx 1$ , so $γ$ is indistinguishable from 1.
Space: $x^{'} \approx x - v t = 300 - 30 \cdot 2 = 240 m$ — the plain Galilean answer. Why this step? With $γ = 1$ the Lorentz $x^{'}$ is the Galilean $x^{'}$ .
Time: the correction $v x / c^{2} = 30 \cdot 300/ (9 \times 1 0^{16}) = 1 0^{- 13} s$ — utterly negligible, so $t^{'} \approx t = 2 s$ . Why this step? $v x / c^{2}$ is the term that "breaks" universal time; here it is $1 0^{- 13} s$ , unmeasurable.

Verify: New theory contains the old one as a limit (Galilean transformation recovered). Units: $v x / c^{2}$ has units $(m/s) (m) / (m/s)^{2} = s$ . ✓

Worked example 3 · Event ahead, forward-moving frame — cell:

$x > 0, v > 0$ Statement. $S^{'}$ moves at $v = 0.6 c$ . An event is at $x = 10 ls$ , $t = 5 s$ . Find $(x^{'}, t^{'})$ .

Forecast: The frame chases toward $+ x$ . Will the event's primed distance be larger or smaller than 10? Will its primed time be earlier or later than 5?

Compute $γ$ . $γ = 1/ 1 - 0. 6^{2} = 1/ 1 - 0.36 = 1/ 0.64 = 1/0.8 = 1.25$ . Why this step? $γ$ multiplies both outputs.
Space. Using $c = 1$ so distances are in ls, times in s: $x^{'} = γ (x - v t) = 1.25 (10 - 0.6 \cdot 5) = 1.25 (10 - 3) = 1.25 \cdot 7 = 8.75 ls$ . Why this step? $v t = 0.6 \cdot 5 = 3$ ls is how far $S^{'}$ has slid forward; the event is now only 7 ls ahead before the $γ$ stretch.
Time. $t^{'} = γ (t - v x / c^{2}) = 1.25 (5 - 0.6 \cdot 10) = 1.25 (5 - 6) = 1.25 \cdot (- 1) = - 1.25 s$ . Why this step? With $c = 1$ , $v x / c^{2} = 0.6 \cdot 10 = 6$ ; because $x$ is large and positive, this correction is big enough to push $t^{'}$ negative — in $S^{'}$ the event already happened.

Verify: Sign check with the Minkowski diagram below: a far-ahead event, seen by a forward observer, is dragged to an earlier primed time. Positive $x$ , positive $v$ ⇒ subtract ⇒ time can go negative. ✓

Worked example 4 · Event behind / backward frame — cell:

$x < 0$ or $v < 0$ Statement. Same speed magnitude but now $S^{'}$ moves the other way, $v = - 0.6 c$ , and the event is behind the origin at $x = - 10 ls$ , $t = 5 s$ . Find $(x^{'}, t^{'})$ .

Forecast: Two sign flips ( $v$ and $x$ ). Do they cancel or double up in the time equation?

$γ$ is unchanged. $γ = 1/ 1 - (- 0.6)^{2} = 1/ 0.64 = 1.25$ . Why this step? $γ$ depends on $v^{2}$ , so the sign of $v$ never affects it — a moving clock slows the same whichever way it moves.
Space. $x^{'} = γ (x - v t) = 1.25 (- 10 - (- 0.6) (5)) = 1.25 (- 10 + 3) = 1.25 (- 7) = - 8.75 ls$ . Why this step? $- v t = + 3$ now (double negative); the event stays behind, at $- 8.75$ ls.
Time. $t^{'} = γ (t - v x / c^{2}) = 1.25 (5 - (- 0.6) (- 10)) = 1.25 (5 - 6) = - 1.25 s$ . Why this step? $(- 0.6) (- 10) = + 6$ : the two minus signs multiply to a plus, so the correction is identical to Ex. 3. This is the key lesson — it is the product $v x$ , not either sign alone, that sets the time shift.

Verify: By pure mirror symmetry the answer is $(- 8.75 ls, - 1.25 s)$ : the whole picture of Ex. 3 flipped left-to-right. Sign discipline confirmed. ✓

Worked example 5 · Simultaneous in

$S$ , not in $S^{'}$ — cell: $Δ t = 0$ , both signs of $x$ Statement. In $S$ , two flashes happen at the same time $t = 0$ : flash A at $x_{A} = + 8 ls$ , flash B at $x_{B} = - 8 ls$ . Frame $S^{'}$ moves at $v = 0.6 c$ . Which flash does $S^{'}$ see first?

Forecast: They are simultaneous in $S$ . Guess: does $S^{'}$ still call them simultaneous? If not, which comes first — the one ahead or the one behind?

$γ = 1.25$ as before. Why this step? Same speed, same factor.
Flash A ( $x = + 8$ ): $t_{A}^{'} = γ (0 - v x_{A} / c^{2}) = 1.25 (0 - 0.6 \cdot 8) = 1.25 (- 4.8) = - 6 s$ . Why this step? Only the $- v x / c^{2}$ term survives ( $t = 0$ ); a positive $x$ gives a negative primed time.
Flash B ( $x = - 8$ ): $t_{B}^{'} = γ (0 - v x_{B} / c^{2}) = 1.25 (0 - 0.6 \cdot (- 8)) = 1.25 (+ 4.8) = + 6 s$ . Why this step? Negative $x$ flips the sign, giving a positive primed time.
Compare. $t_{A}^{'} = - 6 < t_{B}^{'} = + 6$ , so in $S^{'}$ flash A (ahead) happens first, a full 12 s before B. Why this step? The observer moving toward $+ x$ meets the event on that side "earlier" in her time coordinate.

Verify: $Δ t = 0$ in $S$ but $Δ t^{'} = t_{B}^{'} - t_{A}^{'} = 12 s \neq = 0$ in $S^{'}$ . This is exactly Relativity of simultaneity — simultaneity is not absolute, and the ordering is set by the sign of $x$ . ✓

Worked example 6 · The limit

$v \to c$ — cell: degenerate / limiting speed Statement. Track $γ$ and the transform as $v$ climbs: $v = 0.8 c, 0.99 c, 0.9999 c$ . What happens to $γ$ , and to an event at $x = 1 ls, t = 0$ ?

Forecast: As $v$ approaches $c$ , does $γ$ approach some finite ceiling or blow up?

$v = 0.8 c$ : $γ = 1/ 1 - 0.64 = 1/ 0.36 = 1/0.6 = 1.6 \overline{6}$ . Why this step? Establish the baseline before pushing higher.
$v = 0.99 c$ : $γ = 1/ 1 - 0.9801 = 1/ 0.0199 \approx 7.089$ . Why this step? The number under the root is shrinking toward 0, so $γ$ shoots up.
$v = 0.9999 c$ : $γ = 1/ 1 - 0.99980001 = 1/ 0.00019999 \approx 70.71$ . Why this step? Each step closer to $c$ makes $1 - v^{2} / c^{2}$ ten-fold smaller, $γ$ ten-fold... no, $\approx 10$ -fold bigger. It diverges: $γ \to \infty$ as $v \to c$ .
The event's time $t^{'} = γ (0 - v x / c^{2}) = - γ v x / c^{2}$ grows without bound in magnitude. Why this step? Nothing with mass can reach $v = c$ — the transform becomes singular (division-by-zero inside $γ$ ), which is physics saying " $c$ is a wall."

Verify: Sequence $γ \in {1.667, 7.089, 70.71}$ is monotone increasing and unbounded. The $v = c$ case is genuinely undefined ( $0$ in the denominator), the correct degenerate behaviour. ✓

Worked example 7 · Real-world word problem — the muon — cell: word problem

Statement. A muon is created high in the atmosphere and moves straight down at $v = 0.98 c$ . In its own frame it lives $τ = 2.2 μ s$ before decaying. Using the transform, how far does the muon travel in the ground frame before decaying? ( $c = 3 \times 1 0^{8} m/s$ .)

Forecast: In $2.2 μ s$ at $0.98 c$ a Newtonian would get $\approx 647 m$ . Guess: will the ground see it go further or less far?

Identify frames. Let $S^{'}$ = muon (it is at rest there), $S$ = ground. Two events: birth $(x^{'} = 0, t^{'} = 0)$ and death $(x^{'} = 0, t^{'} = τ = 2.2 \times 1 0^{- 6} s)$ — same place in $S^{'}$ because the muon sits still in its own frame. Why this step? Choosing the frame where the muon is at rest makes its lifetime the clean proper time $τ$ .
$γ$ . $γ = 1/ 1 - 0.9 8^{2} = 1/ 1 - 0.9604 = 1/ 0.0396 \approx 5.025$ . Why this step? This factor stretches the muon's short life into ground time.
Ground time of death via the inverse $t = γ (t^{'} + v x^{'} / c^{2})$ with $x^{'} = 0$ : $t = γ τ = 5.025 \times 2.2 \times 1 0^{- 6} \approx 1.106 \times 1 0^{- 5} s = 11.06 μ s$ . Why this step? We go from primed → unprimed, so we use the inverse transform; the moving muon's clock ran slow, so the ground clock reads longer (Time dilation).
Ground distance. In $S$ the muon travels $d = v t = 0.98 \times (3 \times 1 0^{8}) \times 1.106 \times 1 0^{- 5} \approx 3251 m \approx 3.25 km$ . Why this step? Distance is speed × ground time; using the dilated time gives the real experimental range.

Verify: Naive answer $647 m$ vs relativistic $\approx 3251 m$ — the factor is $γ \approx 5.03$ , and indeed $647 \times 5.025 \approx 3251$ . This extra reach is exactly why muons reach the ground and get detected. Units: $(m/s) (s) = m$ . ✓

Worked example 8 · Exam twist — inverse transform + invariant interval — cell: inverse &

$s^{2}$ Statement. You are given the primed coordinates of an event: $x^{'} = 3 ls$ , $t^{'} = 5 s$ , with $v = 0.6 c$ . (a) Find $(x, t)$ in $S$ using the inverse transform. (b) Confirm the Spacetime interval $s^{2} = c^{2} t^{2} - x^{2}$ matches in both frames.

Forecast: Two independent checks. Will the interval computed from primed values equal the one from unprimed values?

$γ = 1.25$ (from $v = 0.6 c$ ). Why this step? Same speed as our earlier examples; reuse it.
Inverse space $x = γ (x^{'} + v t^{'}) = 1.25 (3 + 0.6 \cdot 5) = 1.25 (3 + 3) = 1.25 \cdot 6 = 7.5 ls$ . Why this step? Going from $S^{'}$ back to $S$ flips the sign of $v$ , turning the minus into a plus — the inverse transform.
Inverse time $t = γ (t^{'} + v x^{'} / c^{2}) = 1.25 (5 + 0.6 \cdot 3) = 1.25 (5 + 1.8) = 1.25 \cdot 6.8 = 8.5 s$ . Why this step? Same sign-flip logic for the time equation.
Interval in $S$ ( $c = 1$ ): $s^{2} = t^{2} - x^{2} = 8. 5^{2} - 7. 5^{2} = 72.25 - 56.25 = 16$ . Why this step? Compute the candidate invariant from the unprimed values.
Interval in $S^{'}$ : $s^{'2} = t^{'2} - x^{'2} = 5^{2} - 3^{2} = 25 - 9 = 16$ . Why this step? Compute the same quantity from the primed values.

Verify: $s^{2} = s^{'2} = 16$ — the spacetime interval is frame-independent, the geometric heart the parent note promised. Both the inverse coordinates and the invariance check out. ✓