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Let $L_0=100\mathrm{m}$ be the train's proper length, $D=90\mathrm{m}$ the tunnel's length in the track reference frame, and $v=0.99c$. The Lorentz factor is $\gamma =7.0888$. At one track-frame time, length contraction makes the trainYes, the whole train fits inside at a single track-frame time. Set $t=0$ when the rear enters the near portal. Until the front reaches the far portal, the rear and front are at $vt$ and $vt+L$, respectively. Complete containment lastsa track-frame interval of approximately $256\mathrm{ns}$, counting boundary contact.

In the train's reference frame, its length is $100\mathrm{m}$, while the moving tunnel has simultaneous length $D/\gamma =12.696\mathrm{m}$. No single train-frame time has the whole train between the portals. The two answers refer to different simultaneity slices.

For an explicit check, denote the rear-entry event by $A$ and the front-exit event by $B$. In track coordinates,The Lorentz transformation givesThus the front leaves before the rear enters in train coordinates. These spatially separated events have a frame-dependent order, as allowed by relativity of simultaneity; neither answer asserts a different local event at a portal. No collision with closed doors has been stipulated.

Let the object be at rest in $S^{\prime }$, with proper length $L_0=x_2^{\prime }-x_1^{\prime }$. Measuring its length in $S$ requires two endpoint events simultaneous in $S$, so $\Delta t=0$. The Lorentz transformation then givesThis is length contraction. The simultaneous endpoint events in $S$ are not simultaneous in $S^{\prime }$; that relativity of simultaneity is essential to the measurement.

For the two rockets, take the first rocket's Earth-frame velocity as $+c/2$ and the second's as $-c/2$. The velocity-addition formula gives the second rocket's velocity in the first rocket's rest frame:Its Lorentz factor is ${\gamma }_{\mathrm{rel}}=5/3$, and therefore its measured length isHere “appear” is interpreted as a length measured using simultaneous endpoint positions in the observer's inertial frame. A photograph, involving unequal light travel times, is a different measurement.

The Michelson-Morley experiment splits coherent light into two perpendicular arms, reflects the beams and recombines them to observe interference fringes. Rotating the apparatus interchanges each arm's orientation relative to a proposed stationary luminiferous aether. The historical experiment used multiple reflections to increase the effective optical path and a rotatable support to compare orientations.^[ref]

For equal one-way effective arm lengths $\ell$, a classical stationary-aether model givesA right-angle rotation should reverse their difference, producing a fringe shift of approximately $2\ell v^2/(\lambda c^2)$. The observed orientation-dependent shift was much smaller than the stationary-aether prediction. In special relativity, the apparatus rest frame has the same vacuum speed of light along both arms, so no such orientation-dependent shift is expected. The result supports special relativity but does not, by itself, uniquely establish every postulate or exclude every modified aether theory.

For the rocket, take its tail at $x^{\prime }=0$ and nose at $x^{\prime }=L_0$, with coincident frame origins when $t=t^{\prime }=0$. Denote emission, reflection and return by $A,B,C$. In the rocket's inertial frame, their coordinates $(t^{\prime },x^{\prime })$ areThe inverse Lorentz transformation is $x=\Gamma (x^{\prime }+u{t}^{\prime })$, $t=\Gamma (t^{\prime }+u{x}^{\prime }/c^2)$, where $\Gamma =(1-u^2/c^2{)}^{-1/2}$. Thus the same spacetime events, in coordinates $(t,x)$, areThere is no requirement that emission have $t=0$ in this chosen laboratory coordinate system.

The nose-to-tail travel time is $\Delta t_{AB}=\Gamma L_0(1-u/c)/c$. If $L$ is the laboratory length measured at one laboratory time, a left-moving pulse and the right-moving tail close their initial separation at speed $c+u$. ThereforeThis derives length contraction from the pulse's travel. Comparing the positions of the two endpoints at different laboratory times would not measure the rocket's laboratory length.

Experimental evidence. Tests of special relativity check linked predictions about light propagation, clocks and four-momentum. Several independent kinds of experiment support those predictions.

The observed partial-drag coefficient agrees with this first-order prediction rather than simple Galilean addition of $u$. It tests a different consequence of the Lorentz transformation from a particle-lifetime measurement.^[ref]

Atmospheric muons and the two frame descriptions. Atmospheric cosmic rays create unstable particles whose decay chains produce muons. A stopped-muon experiment measures the time between a stopping signal and the subsequent decay electron, obtaining the proper mean lifetime ${\tau }_0\simeq 2.2\mu \mathrm{s}$. The associated light-travel distance is only $c{\tau }_0\simeq 660\mathrm{m}$, yet substantial numbers of energetic atmospheric muons reach detectors after travelling several kilometres.

For an ideal selected cohort produced height $h$ above a detector and moving downward at constant speed $v$, the flight time in the terrestrial frame is $h/v$. Time dilation predicts a laboratory mean lifetime $\gamma {\tau }_0$, hence survival fractionWithout time dilation, the predicted exponent would be $-h/(v{\tau }_0)$ and the surviving population much smaller. Altitude comparisons constitute a cosmic-ray muon time-dilation experiment: select comparable momenta or energies and account for detector acceptance, atmospheric energy loss and backgrounds before interpreting the count ratio.

For illustration, not as a claimed measurement, take $h=10\mathrm{km}$ and $v=0.998c$. Then $\gamma =15.82$, the laboratory flight time is $33.4\mu \mathrm{s}$, and the predicted surviving fraction is about $0.38$. Omitting time dilation would give only about $2.5\times 10^{-7}$ on the same idealized constant-speed model.

In the muon rest frame, its decay still follows $e^{-\Delta \tau /{\tau }_0}$. The atmosphere and ground move upward. Their separation, measured simultaneously in that frame, is shortened by length contraction:The same survival fraction follows. The terrestrial description uses a longer-lived moving particle; the muon-frame description uses a shorter atmospheric flight path. They describe the same production and detection events. This consistency illustrates both time dilation and length contraction, while the survival measurement is not two independent experimental tests merely because it can be described in two frames.

Energy-momentum invariant. For a particle with rest mass $m$,Since $E=m{c}^2+K$, substitution and cancellation of $m^2{c}^4$ giveThis relativistic energy-momentum relation reduces to $K\simeq p^2/(2m)$ when $K\ll m{c}^2$.

Laboratory beam energies. If both final muons are at rest in the laboratory, their total momentum is zero and their total energy is $2m_{\mu }{c}^2$. Four-momentum conservation therefore requires equal and opposite incoming momenta. Because the electron and Positron have equal rest masses, they then have equal total energies:Subtract each particle's own rest energy:Using the usual handbook rest energies gives approximately $105.1\mathrm{MeV}$ for each beam. These are the ideal kinematic threshold values. Just above threshold, the pair has small, equal and opposite momenta in this laboratory frame.

Positron momentum. Apply the relativistic energy-momentum relation to $E_{+}=m_{\mu }{c}^2$ and use its stipulated negative-$x$ direction:The electron has the opposite momentum, $p_{e,x}=+p_0$ with $p_0=c\sqrt{m_{\mu }^2-m_e^2}$.

Boost to the electron rest frame. The electron's velocity is $v=p_{e,x}{c}^2/E_{-}=p_0/m_{\mu }>0$. Thus $S^{\prime }$ moves in the positive-$x$ direction, and its Lorentz factor satisfiesThe positron has the same laboratory total energy as the electron, so $v{E}_{+}/c^2=v{m}_{\mu }=p_0$. The supplied Lorentz transformation then givesAs a sign check, the same boost gives $p_{e,x}^{\prime }=\gamma (p_0-v{E}_{-}/c^2)=0$, as required for the electron rest frame.

Frame definitions and relativistic effects. Let $S$ be the inertial frame of Earth and the stationary asteroid, with $+x$ pointing toward the asteroid. Bob's inertial frame $S^{\prime }$ moves at $+v$ relative to $S$. For two events at Bob's fixed position, $\Delta x^{\prime }=0$, the inverse Lorentz transformation givesThe elapsed proper time on Bob's clock is $\Delta \tau =\Delta t^{\prime }$, so time dilation means $\Delta t=\gamma \Delta \tau$: the separation of his clock ticks is larger in Earth coordinates. For a rod at rest in $S^{\prime }$ with proper length $L_0$, a length measurement in $S$ uses simultaneous endpoints, $\Delta t=0$. The direct spatial Lorentz transformation gives $L_0=\Delta x^{\prime }=\gamma \Delta x$, hence length contraction:The simultaneity condition is essential: the endpoint events used to measure length are different from the two clock-tick events used to demonstrate time dilation.

Bob's speed. Food consumption measures Bob's own proper time, ${\tau }_B=4\text{ months}=1/3\mathrm{yr}$. The Earth-frame distance is $D=1/4\mathrm{ly}$ and the Earth travel time is $D/v$. ConsequentlyPutting $\beta =v/c$, ${\beta }^2/(1-{\beta }^2)=9/16$, soThe Earth travel time is $5/12\mathrm{yr}=5$ months. Equivalently Bob measures the contracted Earth-asteroid separation $D/\gamma =0.20\mathrm{ly}$ and sees the asteroid approach at $0.60c$, taking $1/3\mathrm{yr}$ to reach him.

Events A and B. Set both coordinate origins at Event A. Alice reaches the asteroid at Event B, travelling at $u=2c/3$ in $S$, soThe Lorentz transformations into Bob's frame giveHere $c(1\mathrm{yr})=1\mathrm{ly}$ has been used consistently. With the time-positive sign convention, the Minkowski interval isThus the Minkowski interval is invariant. It is timelike, as expected for two events on Alice's subluminal worldline. Using the opposite metric sign convention would change both signs together, without changing invariance.

Relative velocity and spacetime diagram. Divide the differential spatial and temporal Lorentz transformations for the same trajectory:For Alice,The spacetime diagram uses horizontal $x^{\prime }$ and vertical $c{t}^{\prime }$. In these coordinates Bob's worldline is $x^{\prime }=0$, Alice's is $x^{\prime }=c{t}^{\prime }/9$, and the asteroid's isThus Alice has slope $d(c{t}^{\prime })/d{x}^{\prime }=9$, while the asteroid has slope $-5/3$. Their intersection is exactly B, $(x^{\prime },c{t}^{\prime })=(1/32,9/32)\mathrm{ly}$. The asteroid later reaches Bob at $(0,1/3)\mathrm{ly}$. These intercepts and gradients show the relativistic velocity addition geometrically.

The separation simultaneous in the mothership frame at launch is the length contraction of the Earth-frame simultaneous separation:In that frame Earth approaches the launch position at $0.6c$ while the lander travels toward Earth at $0.8c$. Their coordinate separation closes at $1.4c$, soA closing rate between two objects can exceed $c$; it is not either object's speed in the other's rest frame. Simultaneity matters: simply transforming the Earth event at Earth-frame launch time would not give the simultaneous mothership-frame separation.

The speed of one rocket in the other's rest frame is obtained by relativistic velocity addition:The Lorentz factor is $\gamma =(1-u^2/c^2{)}^{-1/2}=5/3$. By length contraction, the spatial length measured at simultaneous times in that rocket's rest frame isThis is the inertial-frame length measurement; a visual image assembled from light arriving at one instant is a different measurement.

Use the inverse standard Lorentz boost, $t={\gamma }_v(t^{\prime }+v{x}^{\prime }/c^2)$ and $x={\gamma }_v(x^{\prime }+v{t}^{\prime })$. Label two rod endpoints by a separation $\Delta x^{\prime }=L$; their world lines satisfy $y^{\prime }=u{t}^{\prime }$. A rod's orientation in $S$ must compare endpoint events simultaneous in $S$, not in $S^{\prime }$. ConsequentlyThenThe signed counterclockwise angle therefore satisfiesFor positive $u,v$, the rod slopes downward to the right, so the acute clockwise angle is $\arctan ({\gamma }_{v}uv/c^2)$. This is a relativity of simultaneity effect in addition to the longitudinal length contraction; the rod is not being mechanically bent by the coordinate change.

Let starred quantities denote the capacitor's rest inertial frame, $u_0={ϵ}_0{E}^2/2$, $\beta =v/c$, and $\gamma =(1-{\beta }^2{)}^{-1/2}$. Work in the uniform interplate region of an ideal parallel-plate capacitor, neglecting its fringing field of a capacitor. To transform from that rest inertial frame to a laboratory in which the capacitor moves along $+x$, use the positive-sign Lorentz transformationFor a purely electric field, the spatial block of the electromagnetic stress-energy tensor isThis is the negative of the Maxwell stress tensor: it has $-u_0$ along the electric field and $+u_0$ perpendicular to it. The stress contribution to energy transport in a moving capacitor is essential to the Lorentz transformation of the energy density and flux.

Plates parallel to $xy$. Choose ${\mathbf{E}}_{\ast }=E\hat{\mathbf{z}}$ and ${\mathbf{B}}_{\ast }=0$. The electromagnetic field tensor and electromagnetic stress-energy tensor areThe transformed nonzero independent electromagnetic field tensor components are $F^{03}=\gamma E/c$ and $F^{13}=\gamma \beta E/c$. Thus $\mathbf{E}=\gamma E\hat{\mathbf{z}}$ and $\mathbf{B}=-\gamma vE\hat{\mathbf{y}}/c^2$, consistent with $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$. Transforming the electromagnetic stress-energy tensor givesIn particular, the laboratory Poynting vector is $\mathbf{S}=2u_0{\gamma }^{2}v\hat{\mathbf{x}}$, and the electromagnetic energy density is $u=u_0{\gamma }^2(1+{\beta }^2)$. The ratio $S_x/u=2v/(1+{\beta }^2)$ is subluminal for $v<c$, but differs from the capacitor's velocity. The difference from a simple convective flux isThe extra term is $v{T}_{\ast }^{xx}$, the rest-frame stress multiplied by the velocity. This shows why treating stored field energy as an unstressed parcel moving rigidly with the plates misses part of the energy transport.

Plates parallel to $yz$. Now ${\mathbf{E}}_{\ast }=E\hat{\mathbf{x}}$. The rest tensors areThe electric field is parallel to the Lorentz boost, so $F^{01}=E/c$ remains unchanged and $\mathbf{B}=0$. The cancellation between $T_{\ast }^{00}$ and $T_{\ast }^{11}$ givesThere is no Poynting flux in the ideal interior in this orientation, although the region containing the electric field moves. The longitudinal negative stress contribution cancels the convective term: $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})=0$. Field energy is removed at the receding boundary and supplied at the advancing boundary by work on the charged plates; it need not flow through the interior to accompany the moving region. At edges, omitted fringing fields provide additional local structure.

For a finite idealized interplate rest volume $V_0$, length contraction gives $V=V_0/\gamma$. The field-only integrated laboratory energies would consequently be $U_{xy}=\gamma (1+{\beta }^2)U_0$ and $U_{yz}=U_0/\gamma$, with $U_0=u_0{V}_0$. Their different transformation laws are not paradoxical: the field alone is not a closed system. Include the plates and the supports holding them apart, whose mechanical stresses balance the Maxwell stress tensor. The total closed-system four-momentum then has the ordinary Lorentz transformation law. This is the same distinction between a subsystem and a conserved whole that appears in b.

Take the divergence of the Ampère-Maxwell equation and use that the divergence of a curl is zero:The last step uses Gauss's law. Thus the charge continuity equation is exactlyFor a direct justification that this four-current is a four-vector, take one locally comoving population of electric charges, with rest charge density ${\rho }_0$ and four-velocity $U^{\mu }=\gamma (c,\mathbf{v})$. Invariance of electric charge and length contraction give $\rho =\gamma {\rho }_0$ and $\mathbf{J}=\rho \mathbf{v}$, hence $J^{\mu }={\rho }_0{U}^{\mu }$. Summing populations preserves the four-vector rule. In particular, $c{\rho }^{\prime }=\gamma (c\rho -\beta J_x)$ and $J_x^{\prime }=\gamma (J_x-\beta c\rho )$. The Lorentz invariance of its four-divergence then explains why charge conservation holds in every inertial frame. A single zero-divergence condition alone would not determine a field's transformation law; the physical charge density and current density provide the additional content.

For a particle instantaneously at rest in a primed frame moving at $v\hat{\mathbf{x}}$ relative to the laboratory, the ordinary three-force transforms as $F_{\parallel }=F_{\parallel }^{\prime }$ and $F_{\perp }=F_{\perp }^{\prime }/\gamma$. The spatial components of the four-force are ${\gamma }_u\mathbf{F}$, whereas its time component is ${\gamma }_u\mathbf{F}\cdot \mathbf{u}/c$. For a co-moving material segment, length contraction also changes its laboratory length, so transforming total force is not the same as transforming force per unit length.

Spatially separated simultaneous events in one inertial frame are generally not simultaneous in a relatively moving frame. The Lorentz transformation explicitly mixes position into time. This is essential when comparing distances between moving light pulses: one must compare their positions at the same time in the measuring frame.

At $v=0.6c$, the rod's simultaneous laboratory endpoint measurements in length contraction have $(ct,x)=(0,0)$ and $(0,8\mathrm{m})$. They become $(c{t}^{\prime },x^{\prime })=(0,0)$ and $(-6\mathrm{m},10\mathrm{m})$. In contrast, simultaneous rod-frame measurements at $t^{\prime }=0$ use $(ct,x)=(0,0)$ and $(7.5\mathrm{m},12.5\mathrm{m})$. The latter front-end event occurs later in the laboratory, after the front has moved. The sloping $t^{\prime }=0$ line in the figure is a set of simultaneous spacetime events, not a signal's trajectory. ^[ref].

Special relativity describes physics in Minkowski spacetime, where inertial frames assign different space and time coordinates to the same spacetime events. Its two postulates are Einstein's principle of relativity and the same vacuum speed of light $c$ in every inertial frame, independent of the source's motion. Together they replace the Galilean transformation with the Lorentz transformation, coupling space and time rather than providing a universal clock. Einstein's 1905 paper developed this account through measuring rods, clocks, and light signals. ^[ref].

A spacetime event is an occurrence at a definite place and time: a pulse reaching a detector, for example. An inertial frame supplies a Cartesian coordinate system and clocks at rest throughout it. Einstein synchronisation sets a distant clock's reflection reading to the midpoint between a light signal's departure and return readings at the sending clock. The resulting coordinate time records the local synchronized clock at each spacetime event; it differs from the time when a distant observer receives its image. This distinction makes time dilation, length contraction, and the relativity of simultaneity precise measurements. ^[ref].

The beauty of special relativity is that these different measurements fit one geometry: all inertial frames agree on the Minkowski interval. It lets us translate an experiment between a laboratory and a moving apparatus while preserving the light-signal relations between spacetime events.

The apparent rotation of a moving object arises when length contraction is combined with different light emission times for photons arriving at a camera together. For an edge-on plane lamina moving along x, let X,Y be its rest coordinates and put a distant camera on the negative y axis. Simultaneous reception requires $t+Y/c=\text{constant}$. Since $x=X/\gamma +vt$, its transverse image coordinate is $X/\gamma -\beta Y$. This is the projection of a stationary lamina rotated through $\theta$ with $\cos \theta =1/\gamma$, $\sin \theta =\beta$. It describes a photographic projection; simultaneous laboratory positions retain length contraction. The simple angle formula assumes the transverse distant-camera geometry.

Special relativity connects measurable clock rates, length contraction, light propagation, and four-momentum through the Lorentz transformation. Tests probe particular predictions rather than measuring a coordinate transformation in isolation. Examples include cosmic-ray muon time-dilation experiments, muon storage-ring time-dilation experiments, the Michelson-Morley experiment, the Fizeau experiment, and particle-collision energy and momentum balances.

A runner enters a stationary tunnel of length $L$ at the tunnel-frame instant a photon is emitted towards them from the far end. The interception fraction is $f=\beta /(1+\beta )$, where $\beta =v/c$. In the runner frame, with entry at the origin, emission is at $(x^{\prime },t^{\prime })=(\gamma L,-\gamma vL/c^2)$ and the entrance clock simultaneously reads $-vL/c^2$. The entrance is then ahead by $\gamma {\beta }^{2}L$, so the photon travels $\gamma L$, not the contracted tunnel length $L/\gamma$. Subtracting that initial entrance gap from the entrance's motion during the photon flight gives the same interception fraction. This is a concrete application of relativity of simultaneity, length contraction and the Lorentz transformation.

The region of spacetime occupied by an extended object through its history. Different simultaneous slices intersect a moving tube in different endpoint events, explaining why length contraction compares different sections rather than successive physical compressions.