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Atmospheric cosmic rays generate muons whose mean lifetime at rest is about $2.2\mu \mathrm{s}$. For a selected speed $v$ and production-to-detection distance $h$ in the terrestrial frame, special relativity predicts survival fraction $e^{-h/(\gamma v{\tau }_0)}$, where $\gamma$ is the Lorentz factor. Without time dilation, the exponent would instead be $-h/(v{\tau }_0)$. Comparing altitude-dependent counts at controlled energies, allowing for acceptance and energy loss, tests these alternatives. In the muon rest frame the terrestrial distance is shortened to $h/\gamma$ by length contraction, so the proper flight time and survival fraction agree with the terrestrial calculation. This frame comparison does not turn the same decay measurement into two independent tests.

The apparent conflict between fitting a moving rod inside a shorter stationary enclosure in one frame and observing an even shorter moving enclosure in the rod's frame. Length contraction compares endpoints simultaneous in the measuring frame, and relativity of simultaneity means the two frames do not compare the same pairs of events. Transforming the enclosure-entry and enclosure-exit events resolves the apparent contradiction.

A rod of rest length $L_0$ moving parallel to its length has simultaneous laboratory length $L=L_0/\gamma$, where $\gamma$ is the Lorentz factor. Here $L_0$ is the rod's proper length measured in its rest frame. Transverse lengths are unchanged by the corresponding Lorentz transformation.

For a rod at rest in $S^{\prime }$ with endpoints $x^{\prime }=0$ and $x^{\prime }=10\mathrm{m}$, let $S^{\prime }$ move at $v=0.6c$ relative to the laboratory $S$. Record both endpoint positions at laboratory time $t=0$. The Lorentz transformation then gives $x^{\prime }=1.25x$, so their recorded positions are $0$ and $8\mathrm{m}$: the laboratory length is $10/1.25=8\mathrm{m}$. This is an operational definition: synchronized laboratory detectors register the endpoints at the same laboratory time. A photograph collects light emitted at different times and need not show this length. ^[ref].

For those same two endpoint spacetime events, $\Delta t^{\prime }=-1.25(0.6c)(8\mathrm{m})/c^2=-6\mathrm{m}/c$. Thus the front-end measurement occurs earlier in $S^{\prime }$ by about $20\mathrm{ns}$. Length contraction compares simultaneous endpoint measurements in each frame, which select different pairs of spacetime events; it does not mean that the rod has been compressed in its own rest frame.

The official notes give only “Part 1: as in textbooks” (PDF p. 13, handwritten p. 8). The requested derivation is supplied here.

Put a light source at the rod midpoint in its rest frame $S$ and emit two pulses simultaneously toward its ends, each a distance $\ell$ away. In $S$, the reflection events are $(t,x)=(\ell /c,\pm \ell )$. Invariance of light speed, the Einstein's principle of relativity and linearity give a Lorentz transformation $x^{\prime }=A(x-vt)$, $t^{\prime }=A(t-vx/c^2)$. Applying the inverse transformation with speed $-v$ gives $A^2(1-v^2/c^2)=1$, hence $A=\gamma$.

The transformed reflection events haveThese events are not simultaneous in $S^{\prime }$. Both rod ends move at $-v$ there, so extrapolating each endpoint to $t^{\prime }=0$ gives $x_{\pm }^{\prime }(0)=x_{\pm }^{\prime }+v{t}_{\pm }^{\prime }=\pm \ell /\gamma$. Measuring simultaneous endpoints therefore yields length contraction:

The Lorentz factor is $\gamma =(1-0.8^2{)}^{-1/2}=5/3$. Length contraction makes the moving javelin's length in the shed frameThis measures its two endpoints at the same shed-frame time.

The Lorentz factor for $\beta =v/c=12/13$ isThe javelin's length at simultaneous barn-frame times follows from length contraction:Its proper length remains four metres.

In the javelin's rest frame, the barn moves at speed $12c/13$. Applying length contraction to its proper length givesThe lengths in the two frames use different pairs of simultaneous endpoint events.

The world lines in the train frame are $x_A=L/2$, $x_B=-L/2$ and $x_C=-vt$. In the station frame, length contraction givesThe emission event is $O=(0,0)$ in both frames. The signal world lines in $S^{\prime }$ are $x^{\prime }=u{t}^{\prime }$ and $x^{\prime }=-u{t}^{\prime }$; their intersections with the train endpoints define the reception events $R_A,R_B$. Their velocities in $S$, from the relativistic velocity-addition formula, are $(u-v)/(1-uv/c^2)$ and $-(u+v)/(1+uv/c^2)$.

The two platform-alignment events occur at $(x^{\prime },t^{\prime })=(\pm L/(2\gamma ),0)$ in $S^{\prime }$. The Lorentz transformation puts them at $(x,t)=(\pm L/2,\mp vL/(2c^2))$ in $S$, on the tilted simultaneity line $t=-vx/c^2$. The spacetime diagrams show all three observers, both signals, their receptions, and these distinct endpoint-alignment events:

Let $P$ be the platform's proper length, measured in its rest frame $S^{\prime }$. At the single station-frame time $t^{\prime }=0$, its ends coincide with train endpoints whose simultaneous station-frame separation is $L/\gamma$. ThusThe platform endpoint world lines are vertical at $x^{\prime }=\pm P/2$ in the station spacetime diagram. In the train frame they are $x=-vt\pm P/(2\gamma )$, so their separation at a common train-frame time is $\boxed{P/\gamma =L/{\gamma }^2}$. The two alignment events lie on the tilted line $t^{\prime }=0$ in the train diagram and are not simultaneous there; their separation $L$ cannot be used as the platform length measured at one train-frame time. This is the distinction enforced by length contraction.

An ideal current-carrying wire consists of opposite charge populations, one associated with the stationary lattice and one with drifting carriers. They may cancel in the lattice frame while carrying a net current, as in part (c), up to exchanging the charge signs and current direction. In a carrier-rest description their densities no longer cancel because the two populations undergo different length contractions. The wire then has both electric fields and magnetic fields.

A neutral steady current in one frame generally becomes a charged steady current in another. This illustrates how electric fields and magnetic fields are parts of one relativistic field rather than separately invariant forces. Real resistive wires also have a driving longitudinal electric field and geometry-dependent surface charges; the infinite uniform filament model isolates the transverse fields of a neutral current and does not describe every field of a complete circuit.

Write $a_y=E_{A,y}^0$, $a_z=E_{A,z}^0$, $b_y=E_{B,y}^0$ and $b_z=E_{B,z}^0$, evaluated at the same transverse coordinates $(y,z)$. A superscript $0$ labels each filament's own rest frame, rather than a single common frame in all cases. The filaments are infinite and uniform along $x$, so the change of longitudinal coordinate under a Lorentz boost creates no extra dependence in these functions. After the acceleration transients have passed, the magnetic field of each filament vanishes in its own rest frame.

For a filament moving at $+v\hat{\mathbf{x}}$, use the inverse of part (a): its transverse electric field in the laboratory is $\gamma {\mathbf{E}}^0$ and its magnetic field is $\mathbf{v}\times \mathbf{E}/c^2$. Superposition for the two co-moving filaments therefore gives, in $O$,In the common co-moving frame $O^{\prime }$ both filaments are at rest, soFor example, $B_y^{\prime }=\gamma [-\gamma v(a_z+b_z)/c^2+v\gamma (a_z+b_z)/c^2]=0$, a useful sign check. The supplied $\eta$ is the proper linear charge density. Thus each laboratory density is $\gamma \eta$ and each laboratory electric current is $\gamma \eta v$, consistently with length contraction and the four-current transformation. In $O^{\prime }$ each density is $\eta$ and both currents vanish.

The information in the original question is insufficient to determine a unique spin-induced change of charge density. It specifies ideal dielectric filaments with a prescribed excess linear charge density, but not the charge carrier species, their spin state, charge carrier mobility, binding potential or relaxation mechanism. A classical line charge has no spin variable at all. The following conditional conclusions are supported by the fields already calculated.

For a carrier moving with its filament, the local magnetic field in its instantaneous rest frame is, for the transverse fields here,where $\mathbf{u}$ is that carrier's laboratory velocity. Exclude its filament's singular self-field and first retain only the prescribed straight convective motion.

In Case 1 there is a common rest frame in which both filaments and the excess charges are stationary and the interfilament magnetic field vanishes. Equivalently, in the laboratory ${\mathbf{B}}_A=\mathbf{v}\times {\mathbf{E}}_A/c^2$, so B's co-moving charges have ${\mathbf{B}}_{A,\mathrm{rest}}=0$. Thus uniform common translation alone creates no extra mutual rest-field spin splitting. This does not exclude intrinsic spin-orbit coupling of electrons moving within their atomic binding potentials; that material physics was never specified.

In Case 2 there is relative motion and no frame in which both source distributions are stationary. Put $E_{\ast }=\eta /(2\pi {ϵ}_{0}d)$ as in part (c). At the respective filament axes, the external fields experienced in each carrier distribution's own rest frame areFor the first expression B is at rest in $O$; for the second A is at rest in $O^{\prime }$ and the field comes from B moving at $-v\hat{\mathbf{x}}$. For a carrier with magnetic dipole moment ${\mathbf{\mu }}_s$, the interaction $H_s=-{\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}}$ distinguishes opposite transverse spin orientations. A field gradient could exert a spin-dependent force $\nabla ({\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}})$ on mobile carriers; spin relaxation could instead produce a spin polarization. These are possible responses, not consequences of charge density alone. In B's frame the displayed interaction is directly a magnetic-moment coupling to the moving source's field; treating the moving carrier's electrostatic orbital term in another frame also requires consistent binding and acceleration dynamics, including any applicable Thomas precession.

For example, if $\eta >0$ and mobile electron spins can relax, the lower magnetic-energy orientation has the electron's magnetic moment toward $+\hat{\mathbf{y}}$ and its spin toward $-\hat{\mathbf{y}}$. Oppositely oriented moments would experience opposite gradient forces. The sign changes if the source charge changes sign. None of these statements fixes a net charge redistribution without a confinement and response model, and the stipulated zero transverse size supplies no resolved “sides” on which charge could accumulate.

Therefore Case 1 has no new mutual spin effect from common bulk motion alone; Case 2 permits motion-induced spin-dependent interactions, but neither case determines an additional charge-density profile from the supplied data. In particular, the factor $\gamma$ relating proper and laboratory linear charge densities in part (b) follows from length contraction and is not an effect of spin-orbit coupling. If the charges remain pinned in an ideal dielectric, their imposed density remains the prescribed density in either case.

Spacetime geometry gives an acceptable structural explanation of length contraction, provided the relation between material rods, measurement and the metric tensor is included. It is not a force that squeezes a rod, and the mere existence of a quadratic form cannot guarantee how arbitrary matter behaves. The explanatory issue is whether a common geometrical constraint is sufficient for the phenomenon at issue, or whether a constructive theory of the rod is also required.

For an equilibrium rod with proper length $L_0$, use its rest frame $S^{\prime }$, where its endpoint histories are $x_A^{\prime }=0$ and $x_B^{\prime }=L_0$. In a laboratory frame useA laboratory measurement compares endpoints at the same laboratory time. Subtract the two endpoint equations at $\Delta t=0$ to obtainThe two frames measure different pairs of spacetime events. There is no contradiction in each assigning the other's moving rods smaller lengths; their simultaneity relations differ. This is neither an optical distortion nor an invariant compression diagnosed by an attached strain gauge. A photograph also involves unequal emission times and propagation of light, so it is not this simultaneous endpoint measurement.

The geometrical explanation has genuine unifying content. Once all suitable equilibrium rods measure the spatial distances induced by the same Minkowski metric, the result is independent of chemical composition. It relates length contraction to time dilation and relativity of simultaneity, rather than postulating separate mechanisms for every device. A structural explanation need not be a temporal causal story: the result compares descriptions of an existing world tube, not successive states of a rod being accelerated.

The main objection asks what warrants the italicized measurement premise. Matter on a mathematical Minkowski spacetime could obey equations involving an additional preferred vector field, as in 3. Such matter need not form devices with the usual transformation behaviour. Harvey R. Brown and Oliver Pooley argue that a constructive understanding must attend to the Lorentz-invariant dynamics of matter.^[ref] This objection has force against treating a bare metric tensor as an independently acting agent. It does not invalidate the endpoint derivation once the physical metric interpretation is established.

Acceleration introduces a different explanatory task. A rod pushed at one end can undergo stresses, heating and oscillations; its final proper length can change. A dynamical calculation is needed to determine whether it settles into an equivalent equilibrium state. The comparison $L=L_0/\gamma$ assumes a specified equilibrium proper length, and does not assert that every acceleration history preserves it. Geometry constrains the measurement of that state; material dynamics explain its formation and stability. Those explanatory roles can coexist without making either empty.

There are two distinct explanatory questions. One asks why a proposed system of equations has a particular dynamical symmetry. The other asks why the common spacetime geometry is the right geometry for the behavior of physical rods, clocks and fields. A geometric account and a dynamical account give different directions of explanation, and their merits cannot be decided by a change of notation alone.

For the geometric account, Minkowski spacetime has a fixed Lorentzian metric; its spacetime isometries form the Poincaré group. If the matter equations use that metric tensor and no additional symmetry-breaking background structure, with couplings and boundary assumptions appropriate to the transformation, these spacetime isometries take solutions into solutions. For example, the Klein-Gordon equationin metric signature $(-,+,+,+)$ is invariant under the Poincaré group. This provides one common reason for the same transformation rules appearing in many laws. An individual solution, such as a particular plane wave, need not possess every symmetry of the equation.

Geometry alone is insufficient unless the assumed coupling to geometry is included. In the covariant equation with a preferred timelike vector,the fixed timelike $n$ selects a preferred rest frame. All the quantities can be transformed consistently under a coordinate transformation, but boosts that change $n$ are not symmetries with the physical background fixed. Thus general covariance, tensor notation and the mere presence of the Minkowski metric do not by themselves establish Einstein's principle of relativity.

A dynamical account instead starts with the common invariance of the actual interactions and asks why their composite rods and clocks measure a common interval. If their equations and suitable equilibrium solutions transform consistently under Lorentz transformations, their moving configurations can exhibit length contraction and their periodic processes time dilation. This is the explanatory strategy defended by Harvey R. Brown and Oliver Pooley, who deny that merely redescribing these effects geometrically supplies their microscopic explanation.^[ref] Such an account must still explain why independent interactions have the same relevant symmetry and why a particular clock is a good clock; it cannot simply infer those facts from the word “dynamics”.

The positions therefore need not disagree about experimental predictions. The geometric presentation makes a shared constraint explicit and permits economical deductions. The constructive presentation investigates why material systems realize that constraint. Defining geometry from universal dynamical behavior is a substantive proposal; deriving behavior from geometry also requires substantive matter-coupling assumptions. Neither direction is made compulsory by the formal equivalence of the descriptions.

For the approaching ship, ${\beta }_s=0.85$ and ${\gamma }_s=(1-0.85^2{)}^{-1/2}$. Its station-frame length contraction gives $L_s=100/{\gamma }_s\mathrm{m}$. Adding the two station-frame clearances givesThis is the berth's proper length, since the berth is at rest in the station. It is a simultaneous station-frame allocation, not a ship-frame guarantee of a safe docking maneuver.

The berth moves at $0.85c$ in the ship's initial rest frame, so it too is subject to length contraction:The ship remains $100\mathrm{m}$ long in this frame. The two length comparisons use different simultaneous endpoint events; this is the same relativity of simultaneity that resolves the ladder paradox. Physical docking also requires deceleration, whose later effects are outside these initial inertial measurements.

If $S^{\prime }$ moves at $+v$ relative to $S$, the Lorentz transformations areA rod at rest in $S$ has length $L$. To measure it in $S^{\prime }$, select endpoint events simultaneous there: $\Delta t^{\prime }=0$, so $\Delta t=vL/c^2$ and $\Delta x^{\prime }=\gamma (L-v\Delta t)=L/\gamma$. This is length contraction.

Now let a clock at rest in $S^{\prime }$ cross the two rod endpoints. These are different events, with $\Delta x=L$, $\Delta t=L/v$ and $\Delta x^{\prime }=0$. Their transformed time interval isThusThe moving clock measures the proper time $\Delta t^{\prime }$. The rod's contracted length divided by its speed in that clock's frame gives exactly the same crossing time; time dilation and length contraction are consistent consequences of the same transformation. Measuring a length and measuring a traversal interval require different simultaneity conditions.

For $\beta =0.99$, the Lorentz factor is $\gamma =1/\sqrt{1-{\beta }^2}=7.0888$. In the tracks' rest frame the train undergoes length contraction to $100/\gamma =14.1067\mathrm{m}$, which fits in the $90\mathrm{m}$ tunnel. There is a time interval during which it is entirely inside. In the train's rest frame the tunnel instead has length $90/\gamma =12.6961\mathrm{m}$, so it never simultaneously contains the whole $100\mathrm{m}$ train:These refer to different sets of simultaneous events, as required by relativity of simultaneity. Quantitatively let $A$ be the rear entering the near portal, with $(x_A,t_A)=(0,0)$, and $B$ the front reaching the far portal. In the track frame,The train is wholly inside between these events. Transforming their time separation givesThus the front leaves before the rear enters in the train frame. The two events are spacelike separated, so their time order may reverse without a causal contradiction. No doors or forced acceleration need be added to the problem.

At that simultaneous entrance-clock event, $x_G=0$ and $t_G=-vL/c^2$. Its runner-frame position isThe runner is at $x^{\prime }=0$, so the entrance is still ahead by distance $\gamma {\beta }^{2}L$ at the runner-frame emission time. The simultaneously measured distance between the two tunnel ends is $x_E^{\prime }-x_G^{\prime }=\gamma L(1-{\beta }^2)=L/\gamma$, consistent with length contraction.

For a Lorentz boost along $x$, the Lorentz transformation isTo calculate a photograph, compare photons that arrive together, which were generally emitted at different times. Put the lamina in the $xy$ plane, moving along $+x$, and take the distant camera to lie along negative $y$. Let $(X,Y)$ be a material point's coordinates in the lamina's rest frame. At a laboratory time $t$, its position satisfies $x=X/\gamma +vt$, $y=Y$.

The distant-camera approximation makes the variable part of the light travel time $Y/c$. Equal arrival time therefore requires $t+Y/c=\text{constant}$. Choose that constant to be zero. The point's photographed transverse coordinate becomesA stationary lamina rotated in its plane by $\theta$ would instead have the projected coordinate $X\cos \theta -Y\sin \theta$. The two projections agree whenThis is Terrell rotation. The shape defined by simultaneous laboratory positions still has length contraction; the photographic rotation comes from combining that contraction with unequal emission times. Reversing the viewing or motion direction reverses the apparent rotation's sign.