Lorentz transformation — derivation: Question bank

Before we start, the symbols used everywhere below:

$S$ = lab frame, $S^{'}$ = frame moving at speed $v$ along the shared $x$ -axis.
$c$ = speed of light (same in every frame — the whole point).
$γ = \frac{1}{1 - v ^{2} / c ^{2}}$ = the Lorentz factor, always $\geq 1$ .
An event = a point in spacetime, coordinates $(t, x)$ in $S$ and $(t^{'}, x^{'})$ in $S^{'}$ .

True or false — justify

Is the transform

x^{'} = γ (x - v t)

valid for any relative speed

v

Only for

∣ v ∣ < c

. At

v = c

the root

1 - v^{2} / c^{2}

is zero, so

γ

blows up and the formula breaks — no massive frame can reach

c

Time dilation and length contraction are two independent postulates.

False. Both are consequences of the single pair

x^{'} = γ (x - v t), t^{'} = γ (t - v x / c^{2})

; you derive them, you don't assume them.

The Lorentz transformation reduces to the Galilean transformation when

v ≪ c

True. Then

v^{2} / c^{2} \to 0

γ \to 1

and the term

v x / c^{2} \to 0

, leaving

x^{'} = x - v t, t^{'} = t

. The old theory survives as a limit.

γ

can be less than 1 for some speeds.

False. For

0 \leq v < c

we have

0 \leq v^{2} / c^{2} < 1

, so the root is

< 1

and its reciprocal

γ

> 1

always. Moving clocks run slow, never fast.

Two events simultaneous in

S

are simultaneous in every frame.

False. If

t_{1} = t_{2}

but

x_{1} \neq = x_{2}

, then

t^{'} = γ (t - v x / c^{2})

gives different

t^{'}

for each — this is Relativity of simultaneity.

The spacetime interval

s^{2} = c^{2} t^{2} - x^{2}

has the same value in

S

and

S^{'}

True. Direct substitution shows the cross terms cancel and

c^{2} t^{'2} - x^{'2} = c^{2} t^{2} - x^{2}

; see Spacetime interval.

The transverse coordinates change under a boost along

x

False. Motion is along

x

only, so

y^{'} = y

and

z^{'} = z

— nothing squeezes perpendicular to the motion.

Length contraction means the rod physically compresses like a spring.

False. Nothing pushes on it; it's a geometric result of measuring both ends at the same lab time. In the rod's own frame it has full length.

The inverse transform uses a different formula from the forward one.

False (in structure). It's the same formula with

v \to - v

, because

S

moves at

- v

as seen from

S^{'}

— neither frame is special.

Spot the error

A student writes

t^{'} = t - v x / c^{2}

(forgetting

γ

The whole time equation must be scaled by

γ

t^{'} = γ (t - v x / c^{2})

. Dropping

γ

violates the light-speed condition and the interval invariance.

A student adds velocities as

u^{'} = u - v

for a beam of light

u = c

That gives

c - v \neq = c

, breaking the postulate. The correct Relativistic velocity addition

u^{'} = \frac{u - v}{1 - uv / c ^{2}}

returns exactly

c

when

u = c

A student says "the moving clock in

S^{'}

runs fast because it gains the

γ

factor."

Backwards. The moving clock runs slow: a proper time interval

Δ τ

S^{'}

appears as

γ Δ τ > Δ τ

in the lab, so more lab time passes per tick.

A student measures a rod's length by noting its two ends at different lab times.

Invalid. Length requires both ends recorded at the same

t

(set

Δ t = 0

). Different times let the rod move between readings, corrupting the measurement.

A student claims the derivation needs a curved (nonlinear) map to fit the data.

No. Homogeneity of space and uniformity of time force the map to be linear — straight-line (uniform) motion must map to straight-line motion, which only linear maps do.

A student writes the interval as

s^{2} = c^{2} t^{2} + x^{2}

(plus sign).

Wrong sign. The invariant is

c^{2} t^{2} - x^{2}

; the minus is what distinguishes spacetime geometry from ordinary Euclidean distance.

A student concludes

x^{'} = γ (x - v t)

implies

x^{'} > x

for all events.

Not general. The sign and size of

x^{'}

depend on the specific

x, t

; for many events

x^{'}

is smaller or negative.

γ

scales, but

(x - v t)

can be any sign.

Why questions

Why must the transform be linear?

Because space is homogeneous and time uniform, a free particle moving uniformly in

S

must move uniformly in

S^{'}

; only linear maps send straight lines to straight lines.

Why is the same factor

γ

used in both the forward and inverse transforms?

The principle of relativity says neither frame is privileged, so the inverse must look identical with

v \to - v

, forcing one common stretch factor.

Why does the term

v x / c^{2}

appear in the time equation and not in Galileo's?

Because the light-speed demand mixes space into time; the

v x / c^{2}

term is exactly what makes distant, spatially-separated events shift in time between frames.

Why divide the two light-cone equations to find

γ

Each alone tangles

t

and

t^{'}

; multiplying and dividing lets the

t t^{'}

cancel, isolating

A^{2} = c^{2} / (c^{2} - v^{2})

cleanly.

Why did the Michelson–Morley experiment force this whole rebuild?

It found light's speed unchanged regardless of Earth's motion, killing the additive-velocity picture and demanding a transform that keeps

c

fixed in every frame.

Why does the factor

1/ c^{2}

sit next to

x

rather than

t

Because

x / c^{2}

has units of time per length squared... more precisely

v x / c^{2}

carries units of time, converting a spatial offset into the time shift the equation needs.

Edge cases

What happens to

γ

v \to 0

γ \to 1

: the transform becomes the Galilean

x^{'} = x, t^{'} = t

— no relativistic effects at rest.

What happens to

γ

v \to c

γ \to \infty

: lengths contract toward zero and time dilation grows without bound, which is why

v = c

is unreachable for massive frames.

For a light pulse itself, what is the spacetime interval

s^{2}

Exactly zero: on the path

x = c t

s^{2} = c^{2} t^{2} - (c t)^{2} = 0

. Light lies on the "light cone," a null interval in every frame.

v

is negative (frame moving in

- x

direction), do the formulas still hold?

Yes.

γ

depends on

v^{2}

so it's unchanged, and the signed

v

inside

(x - v t)

and

(t - v x / c^{2})

automatically handles the reversed direction.

For two events at the same place in

S

(

Δ x = 0

), what does

Δ t^{'}

become?

Δ t^{'} = γ Δ t

— pure Time dilation with no simultaneity mixing, since the

v x / c^{2}

term drops out.

Is there any frame where two timelike-separated events swap order?

No. If

c^{2} Δ t^{2} > Δ x^{2}

(timelike), the interval's sign guarantees the time-ordering is preserved in all frames, protecting cause and effect.

Recall Final self-test

Name the one demand that fixes $γ$ . ::: The speed of light equals $c$ in both frames.
Name the quantity invariant under the boost. ::: The spacetime interval $s^{2} = c^{2} t^{2} - x^{2}$ .
Where do time dilation and length contraction come from? ::: Both are consequences of the two Lorentz equations, not separate postulates.