Relativity of Simultaneity

Derivation

Relativity of Simultaneity

D-092 Home PU-105 Threads light · symmetry Depends on Lorentz Transformation from the Two Postulates

Statement

In two inertial frames \(S\) and \(S'\) in standard configuration (relative speed \(v\) along the common \(x\)-axis), two events that are simultaneous in \(S\) but separated by a spatial interval \(\Delta x\) along the direction of motion have, in \(S'\), a nonzero time separation \(\Delta t' = -\gamma v\,\Delta x/c^2\). Equivalently, two clocks synchronised in their common rest frame and separated there by proper distance \(L_0\) along the motion are, in any frame through which they move at speed \(v\), out of synchronisation: the leading clock lags the trailing clock by the leading-clocks-lag relation \(\Delta t = L_0 v/c^2\).

Why it matters

Simultaneity is the hidden premise behind almost every classical intuition about "now". Once it is frame-dependent, the notions of a universal present, of absolute time ordering for spacelike-separated events, and of a rigid extended body all dissolve. Every relativistic "paradox" (the ladder-in-the-barn, the train-and-lightning, the twin asymmetry at turnaround) is, at heart, an accounting error about which events one frame counts as simultaneous.

Quantitatively, the leading-clocks-lag term \(L_0 v/c^2\) is the piece most often dropped in careless bookkeeping. It is the reason length contraction and time dilation combine consistently, and it is the offset that GPS and other precision-timing systems must respect when comparing clocks in relative motion.

Assumptions

The Lorentz transformation holds between the two frames.If the Galilean transformation is used instead, \(\Delta t' = \Delta t\) always and simultaneity is absolute; the entire effect vanishes.

Both frames are inertial (unaccelerated, gravity-free).In a non-inertial or gravitating frame there is no single global time coordinate; simultaneity can only be defined locally and becomes convention- and path-dependent.

Clocks are synchronised by the Einstein convention in their own rest frame.If a different synchronisation convention is chosen, the numerical offset changes; the coordinate-dependent "lag" is only physical relative to a stated synchronisation rule.

The two clocks (or events) are separated along the boost direction.Only the component of separation parallel to \(\vec v\) enters; a purely transverse separation gives \(\Delta t'=0\), so ignoring the projection over-counts the effect.

Derivation

\[ \Delta t' = \gamma\!\left(\Delta t - \frac{v\,\Delta x}{c^2}\right), \qquad \Delta x' = \gamma\left(\Delta x - v\,\Delta t\right), \qquad \gamma = \frac{1}{\sqrt{1-v^2/c^2}} \]

Lorentz transformation for coordinate differences between the same pair of events, taken from the prior result lorentz-transformation-from-postulates. The map is linear, so differences transform like coordinates. A

\[ \Delta t = 0 \quad\Longrightarrow\quad \Delta t' = -\,\gamma\,\frac{v\,\Delta x}{c^2} \]

Impose simultaneity in \(S\) (the two events carry the same \(S\)-time) and read off the first transformation equation. A

\[ \Delta t' \neq 0 \iff \Delta x \neq 0 \ \text{and}\ v \neq 0 \]

Since \(\gamma>0\), the offset vanishes only for coincident events or zero relative velocity: simultaneity is not preserved. This is the relativity of simultaneity as a theorem, not an example. B

\[ \text{Two clocks at rest in } S',\ \text{proper separation } \Delta x' = L_0 \ \text{along } x. \]

Specialise to physical clocks so the abstract offset becomes a readable "lag". Their rest-frame separation is the proper length \(L_0\). B

\[ \Delta t = 0:\qquad \Delta x' = \gamma\,\Delta x \ \Longrightarrow\ \Delta x = \frac{L_0}{\gamma} \]

Sample both clocks at one instant of lab time (\(\Delta t=0\)) and use the second transformation equation. This is length contraction, derived inline: the lab separation is \(L_0/\gamma\). C

\[ \Delta t' = -\,\gamma\,\frac{v\,\Delta x}{c^2} = -\,\gamma\,\frac{v}{c^2}\cdot\frac{L_0}{\gamma} = -\,\frac{v L_0}{c^2} \]

Substitute the lab separation from step 5 into step 2. The factor \(\gamma\) cancels exactly against length contraction, leaving a \(\gamma\)-free reading difference. C

\[ \Delta t' = t'_{\text{lead}} - t'_{\text{trail}} = -\,\frac{v L_0}{c^2} < 0 \]

With \(\Delta x = x_{\text{lead}} - x_{\text{trail}} > 0\) (leading clock farther along \(+x\)), the negative sign shows the leading clock reads the smaller time: it lags. B

Result

\[ \boxed{\;\Delta t_{\text{lag}} = \frac{L_0\, v}{c^2}\;}\qquad\qquad \Delta t'\big|_{\Delta t=0} = -\,\frac{\gamma\, v\, \Delta x}{c^2} \]

Reading. Clocks synchronised and separated by proper distance \(L_0\) along the motion do not stay synchronised in a frame through which they move at speed \(v\): the clock that is ahead in the direction of motion (the "leading" clock) reads behind the trailing clock by \(L_0 v/c^2\). More generally, events simultaneous in \(S\) and separated by \(\Delta x\) along the boost acquire a time separation \(-\gamma v\,\Delta x/c^2\) in \(S'\); simultaneity is a property of a frame, not of the events themselves.

Units check. \(\dfrac{[\text{m}]\,[\text{m s}^{-1}]}{[\text{m}^2\,\text{s}^{-2}]} = \dfrac{\text{m}^2\,\text{s}^{-1}}{\text{m}^2\,\text{s}^{-2}} = \text{s}\), a time, as required. \(\gamma\) is dimensionless, so the general form carries the same units.

Limiting cases

\(v \ll c\): \(L_0 v/c^2 \to 0\); Newtonian absolute simultaneity is recovered, which is why the effect went unnoticed pre-1905.
\(L_0 \to 0\): \(\Delta t_{\text{lag}} \to 0\). Events at the same point are simultaneous in every frame — local simultaneity is invariant.
\(v \to c\): the leading-clocks-lag \(L_0 v/c^2 \to L_0/c\), the light-crossing time of the separation (finite), while the general offset \(\gamma v\,\Delta x/c^2\) diverges.
Transverse separation (\(\Delta x = 0\), only \(\Delta y,\Delta z\)): \(\Delta t' = 0\); separation perpendicular to \(\vec v\) never breaks simultaneity.

Breaks when

Accelerating or rotating frames. The global Lorentz transformation no longer applies; simultaneity becomes path-dependent (e.g. the Sagnac effect, where a light beam circling a rotating ring cannot be globally synchronised).
Curved spacetime / strong gravity. No global inertial frame exists, so there is no single \(\Delta t=0\) surface to transform; simultaneity is defined only locally and one must use general relativity.
Separation not along the boost. The formula uses the component of \(\Delta x\) parallel to \(\vec v\); applying \(L_0 v/c^2\) with a total (including transverse) separation over-counts the lag.
Timelike-separated (causally connected) events. If the events can be linked by a signal with \(|\Delta x| < c\,|\Delta t|\), their time order is invariant and no frame can reverse it; the "relativity" applies only to spacelike separations.

Failure modes

Wrong clock lags. Asserting the trailing clock lags. It is the leading clock (ahead in the direction of motion) that reads behind; the rear clock is ahead by \(L_0 v/c^2\).
Length mix-up. Using the lab (contracted) separation \(L_0/\gamma\) inside \(\gamma v\,\Delta x/c^2\) and then also keeping \(\gamma\) — double counting. The reading-difference form \(L_0 v/c^2\) already has \(\gamma\) cancelled.
Signal-delay confusion. Believing the offset is a light-travel-time artifact ("we just see the far clock late"). It is a genuine coordinate desynchronisation after correcting for propagation, not an optical illusion.
Conflating with time dilation. Treating \(L_0 v/c^2\) as a rate (clocks running slow). It is a constant offset in synchronisation, independent of elapsed time; dilation is the separate factor \(\gamma\).
Dropping the term in paradoxes. Solving the ladder/barn or twin problem with length contraction and time dilation alone, omitting the simultaneity offset — the source of nearly every apparent contradiction.

Discussion

The relativity of simultaneity is the most conceptually radical of the kinematic effects, because it attacks the structure of time itself rather than merely rescaling intervals. Time dilation and length contraction are single numbers (a stretched duration, a shrunk length); simultaneity is about the entire foliation of spacetime into "instants". Different inertial observers slice the same four-dimensional spacetime into stacks of simultaneity surfaces that are tilted relative to one another. Two events on one observer's horizontal slice of constant \(t\) lie on a tilted line for another.

The three kinematic effects are not independent: they are three projections of the single Lorentz transformation and must be used together. In particular the leading-clocks-lag term is the "glue" that makes length contraction self-consistent. When a moving rod is measured, its two ends must be located at the same lab instant; because the rod's own end-clocks are desynchronised by \(L_0 v/c^2\), the measurement that looks simultaneous in the lab is not simultaneous in the rod frame, and this is precisely what reconciles the two frames' disagreement about the rod's length.

Causally, the effect is tightly constrained: only spacelike-separated events (those that cannot influence one another, \(|\Delta x| > c|\Delta t|\)) can have their time order reversed by a change of frame. For timelike or lightlike separations the ordering is absolute, which is exactly what protects causality — no observer ever sees an effect precede its cause.

Geometrically, a plane of simultaneity for an observer with four-velocity \(u^\mu\) is the set of events whose separation \(\Delta x^\nu\) satisfies \(\eta_{\mu\nu}\,u^\mu\,\Delta x^\nu = 0\), i.e. the spacelike hyperplane orthogonal (in the Minkowski metric) to the worldline. Boosting the observer rotates \(u^\mu\) in the \(t\)-\(x\) plane as a hyperbolic rotation of rapidity \(\phi\) with \(\tanh\phi = v/c\), and the orthogonal simultaneity hyperplane rotates by the same hyperbolic angle in the opposite sense — the two "close" toward the light cone. This is why the light cone (\(\phi \to \infty\)) is the invariant limiting surface every observer agrees on, and why the leading-clocks-lag saturates at \(L_0/c\).

Common misconceptions. The disagreement about simultaneity is symmetric and real, not a bookkeeping quirk of one privileged frame; neither observer is "wrong". It is not caused by the finite speed of light delaying signals — it survives after all propagation delays are removed. And it does not permit sending information into one's own past: only the ordering of events that could never have communicated is frame-dependent.

Worked examples

A train of proper length \(L_0 = 300\ \text{m}\) has clocks at its front and rear, synchronised in the train frame, and moves at \(v = 0.60c\) past a platform. Find the desynchronisation seen on the platform and say which clock lags.

Setup: identify \(L_0\) and \(v\), then apply the leading-clocks-lag relation. A

\[ \Delta t_{\text{lag}} = \frac{L_0 v}{c^2} = \frac{L_0\,(0.60c)}{c^2} = \frac{0.60\,L_0}{c} \]

Symbols first: cancel one power of \(c\) using \(v = 0.60c\). A

\[ \Delta t_{\text{lag}} = \frac{0.60 \times 300\ \text{m}}{3.0\times10^{8}\ \text{m s}^{-1}} = \frac{180}{3.0\times10^{8}}\ \text{s} = 6.0\times10^{-7}\ \text{s} \]

Insert numbers with units; \(c = 3.0\times10^8\ \text{m s}^{-1}\). A

\[ \Delta t_{\text{lag}} = 0.60\ \mu\text{s}\quad\text{(front clock lags; rear clock reads ahead).} \]

Reading. On the platform the front (leading) clock shows a time \(0.60\ \mu\text{s}\) behind the rear clock, even though they are perfectly synchronised aboard the train.

Two firecrackers explode simultaneously in frame \(S\), separated by \(\Delta x = 100\ \text{m}\) along \(x\). A frame \(S'\) moves at \(v = 0.80c\) along \(+x\). Find their time separation in \(S'\).

Setup: this is the general (event) form, so \(\gamma\) is retained; \(\Delta t = 0\) in \(S\). A

\[ \gamma = \frac{1}{\sqrt{1-(0.80)^2}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.60} = 1.667 \]

Evaluate the Lorentz factor before substituting. A

\[ \Delta t' = -\frac{\gamma v\,\Delta x}{c^2} = -\frac{1.667\,(0.80c)(100\ \text{m})}{c^2} = -\frac{1.667\times 0.80 \times 100}{3.0\times10^{8}}\ \text{s} \]

Symbols rearranged, then numbers; one power of \(c\) cancels via \(v=0.80c\). A

\[ \Delta t' = -\frac{133.3}{3.0\times10^{8}}\ \text{s} = -4.4\times10^{-7}\ \text{s} \]

Arithmetic. A

\[ \Delta t' = -0.44\ \mu\text{s} \]

Reading. Events simultaneous in \(S\) are separated by \(0.44\ \mu\text{s}\) in \(S'\); the one farther along the direction of \(S'\)'s motion happens first. Simultaneity did not survive the boost.

Problems

Two clocks are at rest in a rocket, separated by a proper distance \(L_0 = 6.0\times10^{8}\ \text{m}\) along the direction of flight and synchronised in the rocket frame. The rocket moves at \(v = 0.50c\). By how much does the leading clock lag in the ground frame?
Solution
\(\Delta t = L_0 v/c^2 = (6.0\times10^8)(0.50c)/c^2 = (0.50)(6.0\times10^8)/(3.0\times10^8) = 3.0\times10^8/3.0\times10^8 = 1.0\ \text{s}\). The front clock lags the rear by \(1.0\ \text{s}\).
Two events are simultaneous in frame \(S\), separated by \(\Delta x = 1.5\times10^{11}\ \text{m}\) (about one astronomical unit) along \(x\). Frame \(S'\) moves at \(v = 0.10c\). Find the magnitude of their time separation in \(S'\).
Solution
\(\gamma = 1/\sqrt{1-0.01} = 1/\sqrt{0.99} = 1.005\). \(|\Delta t'| = \gamma v\,\Delta x/c^2 = 1.005\,(0.10c)(1.5\times10^{11})/c^2 = 1.005\,(0.10)(1.5\times10^{11})/(3.0\times10^8) = 1.005 \times (1.5\times10^{10}/3.0\times10^8) = 1.005 \times 50\ \text{s} = 50.3\ \text{s}\).
A rod of proper length \(2.0\ \text{m}\), with synchronised end-clocks, moves at \(v = 0.99c\). Find the leading-clock lag in the lab frame, and compare it with the light-crossing time \(L_0/c\).
Solution
\(\Delta t = L_0 v/c^2 = (2.0)(0.99c)/c^2 = (0.99)(2.0)/(3.0\times10^8) = 1.98/3.0\times10^8 = 6.6\times10^{-9}\ \text{s} = 6.6\ \text{ns}\). The light-crossing time is \(L_0/c = 2.0/(3.0\times10^8) = 6.67\ \text{ns}\); the lag is \(0.99\) of it, approaching the \(v\to c\) ceiling of \(L_0/c\).
Two firecrackers explode simultaneously in the ground frame \(S\), \(300\ \text{m}\) apart along \(x\). An observer in \(S'\) moves at \(v = 0.60c\) in the \(+x\) direction. Which explosion does \(S'\) record as first, and by what time interval?
Solution
\(\gamma = 1/\sqrt{1-0.36} = 1/\sqrt{0.64} = 1.25\). Label the event at larger \(x\) as B: \(\Delta t' = t'_B - t'_A = -\gamma v\,\Delta x/c^2 = -1.25\,(0.60c)(300)/c^2 = -1.25(0.60)(300)/(3.0\times10^8) = -1.25 \times 180/3.0\times10^8 = -225/3.0\times10^8 = -7.5\times10^{-7}\ \text{s}\). Since \(\Delta t' < 0\), event B (the one farther along \(+x\), the direction of \(S'\)'s motion) occurs first, by \(0.75\ \mu\text{s}\).
Two clocks separated by proper distance \(L_0 = 300\ \text{m}\) are synchronised in their rest frame \(S'\) (so the two "read zero" events are simultaneous in \(S'\)). The frame moves at \(v = 0.60c\) relative to the lab \(S\). (a) Find the lab time separation of the two "read zero" events. (b) Find the leading-clock lag (the reading difference at a single lab instant). (c) Show the two answers differ by exactly \(\gamma\), and explain physically.
Solution
\(\gamma = 1.25\). (a) The "read zero" events are simultaneous in \(S'\) (\(\Delta t'=0\), \(\Delta x' = L_0\)); by the inverse transform \(\Delta t = \gamma(\Delta t' + v\,\Delta x'/c^2) = \gamma v L_0/c^2 = 1.25(0.60)(300)/(3.0\times10^8) = 225/3.0\times10^8 = 7.5\times10^{-7}\ \text{s} = 0.75\ \mu\text{s}\). (b) The reading difference at one lab instant is \(\Delta t_{\text{lag}} = L_0 v/c^2 = (0.60)(300)/(3.0\times10^8) = 180/3.0\times10^8 = 6.0\times10^{-7}\ \text{s} = 0.60\ \mu\text{s}\). (c) Ratio \(0.75/0.60 = 1.25 = \gamma\). Physically: (a) compares two events along a tilted \(S'\)-simultaneity slice, picking up the full \(\gamma v L_0/c^2\); (b) samples both worldlines at one lab time, where the clocks' lab separation is contracted to \(L_0/\gamma\), cancelling one factor of \(\gamma\).