Simultaneity — relativity of simultaneity: Question bank

True or false — justify

TF1. "If two events are simultaneous in one frame, they are simultaneous in every frame."

False — simultaneity holds in all frames only when

Δ x = 0

; for any nonzero separation a moving observer measures

Δ t^{'} = - γ v Δ x / c^{2} \neq = 0

TF2. "Relativity of simultaneity is caused by the finite time light takes to travel to your eyes."

False — both observers already subtract that travel time using the same

c

; the disagreement survives that correction because

c

is identical for both, so it is a real feature of spacetime, not a viewing delay.

TF3. "Two events at the exact same point in space are simultaneous for everyone."

True — with

Δ x = 0

the formula gives

Δ t^{'} = 0

regardless of

v

γ

; coincident events cannot be reordered by any observer.

TF4. "If Bob (on the train) says the front bolt struck first, then Alice must be mistaken about them being simultaneous."

False — neither is mistaken; both correctly applied the same speed of light in their own frame. There is no privileged "true now" that decides between them.

TF5. "A faster relative speed always increases the simultaneity gap for a fixed separation."

True but subtly — as

v

rises both

v

and

γ

grow, so

γ v

increases monotonically toward infinity as

v \to c

, making the disagreement larger without bound (until

v = c

, which is unreachable).

TF6. "Relativity of simultaneity can make an effect appear before its cause."

False — only spacelike-separated events (too far apart for light to connect) can swap order, and those pairs can never be cause and effect; timelike (causally linked) pairs keep their order in every frame. See Causality and the Light Cone.

TF7. "The minus sign in

Δ t^{'} = - γ v Δ x / c^{2}

is just a convention and can be dropped."

False — the sign encodes which event a given moving observer sees first (the leading clock lags). Dropping it discards the physical direction of the effect.

TF8. "If you switch which event you call 'event 1' and which is 'event 2', physics changes."

False — swapping them flips the sign of

Δ x

and hence of

Δ t^{'}

, but the physical statement ("front bolt is early in the train frame") is unchanged; only the label bookkeeping flips.

Spot the error

SE1. "

Δ t^{'} = - γ v Δ x / c^{2}

, and I'll use

Δ x

= the train's length as the train measures it (its rest length)."

Error —

Δ x

must be the separation in frame

S

where the events are simultaneous (the platform), not the rest length in

S^{'}

; mixing frames gives the wrong number. Related idea: Length Contraction.

SE2. "Bob sees the front bolt first because he is closer to the front of the train."

Error — Bob sits at the exact middle, equidistant from both ends in his own frame. He sees the front light first because he moves into it while the light travels, not because of any distance imbalance.

SE3. "Since the platform sees the strikes as simultaneous, and the train is length-contracted, the train obviously sees them simultaneous too, just squeezed."

Error — length contraction changes lengths, not the timing; the timing disagreement comes from the

- v x / c^{2}

term in the Lorentz Transformation, a separate effect that survives even if you ignore contraction.

SE4. "In the formula I plugged

v = 0.6

(as a plain number) and got nonsense units."

Error — either keep

v = 0.6 c

symbolically so the

c

's cancel with

c^{2}

, or use

v = 1.8 \times 1 0^{8} m/s

; a bare "0.6" is dimensionally meaningless here.

SE5. "Alice and Bob disagree, so at least one of their clocks is running wrong or is broken."

Error — both sets of clocks are ideal and correctly synchronized within their own frame. The point is that synchronization itself is frame-dependent; no clock is faulty. See Time Dilation for the related clock-rate effect.

SE6. "The effect is tiny, so it must be a rounding artifact that vanishes with careful measurement."

Error — the gap is proportional to

γ v Δ x / c^{2}

, an exact prediction, not noise. It is small only because

Δ x / c^{2}

is small at everyday speeds; at relativistic

v

it becomes microseconds and larger.

SE7. "Simultaneity is relative, so everything about the two events (which one is 'first') is pure opinion with no rules."

Error — the ordering is constrained: for timelike-separated events every frame agrees on the order (the light cone fixes it); only for spacelike separation is the order frame-dependent.

Why questions

WHY1. Why does the disagreement depend on spatial separation

Δ x

and not on the time-gap?

Because the events start simultaneous (

Δ t = 0

), so the entire

S^{'}

time difference comes from the position-dependent term

- γ v x / c^{2}

; with equal times, only differing positions can produce differing

t^{'}

WHY2. Why does the constancy of

c

force simultaneity to be relative?

If everyone measures light at

c

, then a moving observer travelling toward one flash and away from another must attribute the unequal arrival to unequal emission times, since he cannot blame it on unequal light speeds.

WHY3. Why is it always the leading clock (front, in the direction of motion) that lags?

The

- v x / c^{2}

term is more negative for larger

x

(the front), so the front clock's

t^{'}

reading is pushed further behind — the direction of motion picks out the front as the "early to strike, behind on the clock" end.

WHY4. Why can't we just define one universal master clock and settle all disputes?

Synchronizing distant clocks requires sending a signal (fastest is light), and the "same instant" that synchronization defines depends on your motion; there is no frame-independent procedure, so no universal clock exists.

WHY5. Why doesn't relativity of simultaneity break cause and effect?

Cause-effect pairs are timelike-separated (a signal at

\leq c

can link them), and timelike order is invariant; the reorderable pairs are spacelike, precisely those no signal can connect, so no cause is ever seen after its effect.

WHY6. Why does the effect vanish when

v = 0

even if

Δ x

is huge?

With no relative motion the observer is not moving into either light beam, so there is nothing to distinguish the arrivals; algebraically

v = 0

kills the whole

- γ v Δ x / c^{2}

term.

WHY7. Why do we say simultaneity is "the first casualty" of the light postulate rather than time dilation or length contraction?

Because both dilation and contraction can be derived from the loss of a shared "now"; once clocks at different places disagree on synchronization, mismatched clock rates and shrunken lengths follow. It is logically upstream.

Edge cases

EC1. What happens to

Δ t^{'}

v \to c

for a fixed nonzero

Δ x

γ \to \infty

Δ t^{'} \to \infty

— the disagreement grows without bound, but

v = c

itself is forbidden for massive observers, so the divergence is a limit never reached.

EC2. What is

Δ t^{'}

when

Δ x = 0

but

v

is huge (say

0.999 c

Exactly zero — a large

γ v

multiplied by

Δ x = 0

is still zero. Coincident events are absolutely simultaneous no matter how fast the observer moves.

EC3. If the two events are separated perpendicular to the direction of motion (along

y

, with equal

x

), what does a moving observer see?

They remain simultaneous — the formula uses

Δ x

along the motion; a purely transverse separation has

Δ x = 0

, so

Δ t^{'} = 0

EC4. Two events are simultaneous in

S

; a frame

S^{'}

moves in the

+ x

direction and another frame

S^{''}

moves in the

- x

direction, both at speed

v

. What differs?

The magnitude of the gap is equal but the sign flips:

S^{'}

and

S^{''}

disagree about which event is first, because reversing

v

reverses the sign of

Δ t^{'}

EC5. Consider events with

Δ x \neq = 0

that are simultaneous in

S

. Is there a frame where they occur at the same

S^{'}

-time yet a different

S^{'}

-place?

Only

S

(and frames with

v = 0

relative to it) keeps them simultaneous; any

v \neq = 0

along

x

gives

Δ t^{'} \neq = 0

, so no distinct moving frame preserves their simultaneity. Draw it on a spacetime diagram: tilting the axes always splits equal-time events.

EC6. Two events are simultaneous and co-located in

S

. Are they actually one event?

Effectively yes — same time and same place means the same spacetime point; every observer agrees they coincide, and no reordering is possible.

EC7. If

Δ x

is spacelike-large enough that no light can connect the events, can a frame reverse their order?

Yes — spacelike-separated events have frames where either one is first, or where they are simultaneous; this is exactly the reorderable regime, and it is safe because no causal signal links them.

Recall One-line self-test before you leave

Which single quantity in $Δ t^{'} = - γ v Δ x / c^{2}$ is the trigger for all disagreement, and in which frame is it measured? Trigger and its frame ::: The spatial separation $Δ x$ , measured in the frame $S$ where the events are simultaneous; if it is zero, everyone agrees.