Lorentz transformation — derivation: Visual walkthrough

We are relating how two observers label the same event (a flash, a click — a single point in space at a single instant). One observer sits in frame $== S ==$ (call it the lab). The other rides frame $== S^{'} ==$ , sliding along the shared $x$ -axis at steady speed $== v ==$ . We will keep the sideways directions $y, z$ aside — motion is along $x$ , so $y^{'} = y$ , $z^{'} = z$ and they never change.

Step 1 — An event lives in two grids

WHAT: We lay one event on the $S$ grid and ask for its $S^{'}$ readings. WHY: The whole derivation is one question — given $(t, x)$ , what are $(t^{'}, x^{'})$ ? We must first see that both frames describe one reality, only labelled two ways. PICTURE: Below, the black dot is a single flash. The navy grid is $S$ ; the moving observer would draw a different (tilted) grid, but for now just note: same dot, two label-sets. The origin dot at the bottom-left is the shared handshake $t = t^{'} = 0$ .

Step 2 — The naive guess and why it must break

WHAT: We draw the origin of $S^{'}$ (the point $x^{'} = 0$ ) travelling through $S$ . WHY: The moving observer sits at their own origin; in $S$ that origin is at $x = v t$ . So the shift we subtract is exactly $v t$ .

Read the guess term by term: $x^{'} = lab position x - how far S^{'} has slid by time t v t, t^{'} = t .$

PICTURE: The violet line is the moving origin's path $x = v t$ . A light pulse (orange) leaves the origin at $t = 0$ and travels at the fixed speed $c$ . With Galileo's flat clocks, the ground would see the pulse at speed $c$ but the train rider would see $c - v$ — the two orange slopes disagree. That contradicts Michelson–Morley experiment, which found light's speed never changes. So $t^{'} = t$ is the flaw.

Step 3 — Force straight lines to stay straight (linearity)

WHAT: We write the most general straight-line-preserving relation. WHY: Bend the map and a coasting particle would look accelerated to one observer — physics would differ between frames, which is forbidden. (Because of the handshake in the setup, a linear map needs no added constant — the point $0, 0$ already maps to $0, 0$ .)

$x^{'} = A x + B t, t^{'} = D x + E t .$

Term by term: $A$ stretches position, $B$ mixes in time, $D$ mixes position into the new time, $E$ stretches time. Four unknowns — track them: Step 4 fixes $B = - A v$ ; Step 6 fixes $A = γ$ ; and Step 7 solves the whole system so that $D$ and $E$ come out to $D = - γ v / c^{2}$ (the position-into-time mixing that breaks simultaneity) and $E = γ$ (the time stretch). We will name each again the moment it is pinned down.

PICTURE: Left panel — a curved map bends a straight worldline (physics breaks). Right panel — a linear map keeps it straight. That is why we must use the linear form.

Step 4 — Nail down $B$ : the moving origin rides at $v$

WHAT: Set $x^{'} = 0$ (the $S^{'}$ origin) and see where it sits in $S$ . WHY: By definition the $S^{'}$ origin moves at $v$ through $S$ , i.e. sits at $x = v t$ (this is measured from the shared handshake, so there is no extra constant). This ties $B$ to $A$ .

From $x^{'} = A x + B t = 0$ we get $x = - \frac{B}{A} t$ . Matching to $x = v t$ : $- \frac{B}{A} = v \Rightarrow B = - A v \Rightarrow x^{'} = A (x - v t) .$

Term by term: the whole bracket $(x - v t)$ is "position measured from the moving origin"; $A$ is a not-yet-known overall stretch (Galileo secretly assumed $A = 1$ ). Unknown $B$ is now named: $B = - A v$ .

PICTURE: The violet worldline $x = v t$ is exactly the set of dots the moving observer calls $x^{'} = 0$ . Everything to its right is $x^{'} > 0$ .

Step 5 — Symmetry gives the inverse for free

WHAT: Write the inverse map. WHY: If the two directions used different factors, one frame would be physically special. Symmetry forbids that. (The shared handshake again means the inverse also has no constant term.)

$x = A (x^{'} + v t^{'}) .$

Term by term: same $A$ , and $+ v$ replaces $- v$ because $S$ moves the other way as seen from $S^{'}$ . (This " $v \to - v$ " swap is exactly the sign convention noted in the setup — the inverse is not a new formula, just the same one with $v$ 's sign flipped.)

PICTURE: Two mirrored observers pointing at each other; each sees the other recede at speed $v$ , and each uses the identical factor $A$ . The setup is perfectly symmetric.

Step 6 — The light postulate solves for $A = γ$

WHAT: Send one light pulse from the shared origin at $t = t^{'} = 0$ (the handshake). It obeys $x = c t$ in $S$ and $x^{'} = c t^{'}$ in $S^{'}$ — same speed $c$ in both. WHY: This is the one experimental demand (Michelson–Morley experiment) that repairs Galileo. It is what makes the derivation relativistic.

Put $x = c t$ into $x^{'} = A (x - v t)$ : $x^{'} = A t (c - v)$ . Put $x^{'} = c t^{'}$ into $x = A (x^{'} + v t^{'})$ : $c t = A t^{'} (c + v)$ . Multiply the two equations (using $x^{'} = c t^{'}$ so the left sides give $c^{2} t t^{'}$ ), then cancel $t t^{'}$ :

$c^{2} = A^{2} (c - v) (c + v) = A^{2} (c^{2} - v^{2}) .$

Solve (taking the positive root, since $A > 0$ : the grids are not flipped): $A^{2} = \frac{c ^{2}}{c ^{2} - v ^{2}} = \frac{1}{1 - v ^{2} / c ^{2}} \Rightarrow A = \frac{1}{1 - v ^{2} / c ^{2}} \equiv γ .$

Term by term inside $γ$ : $v^{2} / c^{2}$ is "how close to light-speed, squared"; subtract it from 1, take the root, flip it. Unknown $A$ is now named: $A = γ$ . See Time dilation.

PICTURE: Both orange light lines have the same slope (speed $c$ ) in each grid — the single requirement that forces $A$ to be $γ$ , not 1. The curve of $γ$ versus $v / c$ shows it hugging 1 for slow speeds and rocketing up near $v = c$ .

Step 7 — The time equation drops out (this names $D$ and $E$ )

WHAT: Feed $x^{'} = γ (x - v t)$ into the inverse $x = γ (x^{'} + v t^{'})$ and solve for $t^{'}$ . WHY: We already have the space rule; the same two relations, combined, force the time rule. No new assumption is needed — and this is where the last two unknowns $D$ and $E$ from Step 3 finally get their values.

$x = γ (γ (x - v t) + v t^{'}) \Rightarrow γ v t^{'} = x (1 - γ^{2}) + γ^{2} v t .$

For $v \neq = 0$ , use $1 - γ^{2} = - γ^{2} v^{2} / c^{2}$ and divide by $γ v$ : $t^{'} = γ (t - \frac{v x}{c ^{2}}) = D (- \frac{γ v}{c ^{2}}) x + E γ t .$ (And at $v = 0$ this same formula gives $t^{'} = t$ , so it covers both cases.)

Term by term: $t$ is the lab time; $\frac{v x}{c ^{2}}$ is a position-dependent time offset — the $1/ c^{2}$ converts a distance $x$ into a time. Reading off Step 3's coefficients: $D = - γ v / c^{2}$ (position mixing into the new time — the term that breaks simultaneity) and $E = γ$ (the time stretch). All four unknowns $A, B, D, E$ are now determined.

PICTURE: For a fixed lab time $t$ , the $S^{'}$ -time depends on where the event is (its $x$ ): far-right events (magenta) get a bigger negative offset than left events (violet). Same $t$ , different $t^{'}$ — that is Relativity of simultaneity.

Step 8 — The invariant hiding inside: $c^{2} t^{2} - x^{2}$

WHAT: Take our two formulas and compute $c^{2} t^{'2} - x^{'2}$ . WHY: It ties the linearity/light arguments together: the transform is exactly the shear that leaves this combination fixed — the geometric heart of relativity, and the thing the summary picture will draw.

$c^{2} t^{'2} - x^{'2} = c^{2} γ^{2} (t - \frac{v x}{c ^{2}})^{2} - γ^{2} (x - v t)^{2} = c^{2} t^{2} - x^{2} .$

Term by term: the cross terms of the form $v x t$ cancel; the leftover $γ^{2} (1 - v^{2} / c^{2}) = 1$ collapses the rest back to $c^{2} t^{2} - x^{2}$ . See the Verify block for the full cancellation.

PICTURE: The dotted navy curve is the set of all events with the same interval — a hyperbola. The tilted $S^{'}$ grid slides events along this curve but never off it: the interval is frame-independent.

Step 9 — Every case: slow, fast, and the $v \to c$ wall

$v = 0$ (frames at rest together): $γ = 1$ , offset $v x / c^{2} = 0$ → $x^{'} = x$ , $t^{'} = t$ . Identity — as it must be (this is the $v = 0$ branch of Step 7).
$v ≪ c$ (everyday cars): $v^{2} / c^{2} \approx 0$ , $γ \approx 1$ → $x^{'} \approx x - v t$ , $t^{'} \approx t$ . We recover Galilean transformation. The new theory contains the old.
$v < 0$ (S' moves left): identical formulas, $γ$ unchanged (depends on $v^{2}$ ); only the signs of the $v t$ and $v x / c^{2}$ terms flip. No new physics.
$v$ close to $c$ (e.g. $0.9 c$ ): $γ$ blows up; space and time mix heavily.
$v \to c$ : $γ \to \infty$ — the grid shears until the time and space axes fold onto the light line. Nothing with mass reaches this wall.
$v \geq c$ : $1 - v^{2} / c^{2} \leq 0$ , the root is imaginary or zero → no real transform. This is why $c$ is a speed limit.

WHAT / WHY: Checking the limits proves the formula is well-behaved everywhere allowed and forbids the everywhere it shouldn't reach. PICTURE: The $S^{'}$ axes (violet = line of constant $x^{'}$ , magenta = line of constant $t^{'}$ ) tilt toward the orange light line as $v$ grows from $0.3 c$ to $0.9 c$ . They meet the light line only in the impossible limit $v = c$ .

The one-picture summary

Everything above lives in one diagram — the Minkowski diagram. The navy grid is $S$ . The tilted violet/magenta grid is $S^{'}$ . The orange 45° line is light: it keeps the same slope in both grids — the one demand that set $A = γ$ . Read any event's $(x, t)$ off the straight navy grid and its $(x^{'}, t^{'})$ off the tilted grid. The bracket $(x - v t)$ is the tilt of space; the offset $v x / c^{2}$ is the tilt of time; $γ$ scales both. The quantity $c^{2} t^{2} - x^{2}$ (the Spacetime interval from Step 8) is the same number in either grid — the hyperbola that no shear can change.

Recall Feynman retelling — the whole walkthrough in plain words

Two people label the same flash of light, and they agree on one starting handshake: the instant their origins pass, both clocks read zero. Newton said: just slide your ruler sideways and keep the same clock. But light refuses to change speed, so that story breaks. To fix it we insist the map be "straight-lines-stay-straight" (linear), then we pin its knobs one at a time: one knob says the moving origin drifts at $v$ ; symmetry says the return trip uses the same knob; and the light rule — light is $c$ for everyone — squeezes that knob to a precise number we name $γ$ . Once $γ$ is fixed, the time rule is not a new idea; it falls out automatically (with a tiny care that if the two frames aren't moving at all, we don't divide by zero — we just get "same time"), and it carries a sneaky $v x / c^{2}$ term meaning far-away events don't stay "at the same time" when you move. Deeper still, one combination $c^{2} t^{2} - x^{2}$ never changes at all — that's the real bedrock. Turn the speed down and everything melts back into Newton; turn it up toward light and the grid shears until it can't shear any more — that wall is the speed limit $c$ .

Recall Active recall

What single demand fixes the stretch factor $A = γ$ ? ::: The speed of light equals $c$ in both frames.
Why must the map be linear? ::: So constant-velocity (straight) worldlines stay straight in both frames.
Which term breaks simultaneity, and why? ::: The $- v x / c^{2}$ in $t^{'}$ — it makes the new time depend on position $x$ .
What must we check before dividing by $v$ in Step 7? ::: That $v \neq = 0$ ; the $v = 0$ case gives the identity $x^{'} = x, t^{'} = t$ directly.
What quantity is the same in every frame? ::: The spacetime interval $c^{2} t^{2} - x^{2}$ .
What happens to $γ$ as $v \to c$ ? ::: It diverges to $\infty$ ; at $v \geq c$ no real transform exists.
What are the four linear coefficients $A, B, D, E$ once pinned down? ::: $A = γ$ , $B = - γ v$ , $D = - γ v / c^{2}$ , $E = γ$ .
How do we handle $S^{'}$ moving in the negative $x$ -direction? ::: Use $v < 0$ ; the same formulas hold and $γ$ is unchanged since it depends on $v^{2}$ .