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For a short pair of stationary test particles in transverse-traceless gauge, spatial coordinates stay fixed while the measured proper length changes. In background signature $+---$, let $h_{xx}=-h_{yy}=H_{+}$ and $h_{xy}=H_{\times }$. A coordinate separation of length $\Delta$ at angle $\theta$ hasThe plus polarization and cross polarization differ by a $45$-degree rotation of their principal axes. With background signature $-+++$ and the same numerical spatial perturbation, the sign reverses because the positive spatial metric is then ${\delta }_{ij}+h_{ij}$ rather than ${\delta }_{ij}-h_{ij}$. Consistently redefining the signed strain leaves the physical effect unchanged.

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.

A material standard used to compare spatial separations. In special relativity, a stationary equilibrium rod defines a proper length; measurements in another inertial frame require simultaneous endpoints in that frame. Geometry alone does not establish that arbitrary matter forms a stable length standard.

Displacement must be specified relative to an observer. In transverse-traceless gauge the free particles can have fixed coordinates, while their measured proper lengths oscillate. In a local, freely falling orthonormal frame, the geodesic deviation equation for an initial separation $\mathbf{X}=(X,Y,0)$ is, to first order in $H$,Its general solution with physical displacement $\mathbf{\xi }(0)=0$ and initial physical velocity $\mathbf{V}$ isThe source has been running indefinitely, so there is no wave-free initial state. Its initial conditions for an eternal monochromatic gravitational wave need more than positions to fix $\mathbf{V}$. The usual bounded, periodic response selects $V_i={\dot{h}}_{ij}(0,0)X^j/2$; if the particles instead start at physical rest at $t=0$, set $V_i=0$ and retain the linear term. Both solve the same linear geodesic deviation problem. The distinction between fixed TT coordinates and measured separation is explained in ^[ref].

Choose $\beta$ so the wave phase at the detector is $\Phi (t,0)=2\omega t$. For the bounded response the four initial placements then giveThese are displacements from the specified positions at $t=0$; positions are $\mathbf{X}+\mathbf{\xi }$. Equivalently, relative to mean positions the usual oscillatory displacement is ${\mathbf{\xi }}_{\mathrm{osc}}=h\mathbf{X}/2$ to first order. The response repeats after $\pi /\omega$.

Both plus polarization and cross polarization occur with equal amplitude and a quarter-cycle phase difference. Their superposition is circular gravitational-wave polarization, with handedness reversed when the viewing direction or source rotation is reversed. The principal stretching axis turns through half the wave phase. There is no leading longitudinal motion.

On large scales the appropriate background is the spatially flat FLRW metric,Here $t$ is cosmic time, the proper time of a comoving observer; $R(t)$ is the scale factor; and $(r,\theta ,\varphi )$ are dimensionless comoving coordinates. With this convention $R$ has units of length. A fixed comoving radial interval has instantaneous proper length $R(t)dr$. The cosmological principle assumes spatial homogeneity and cosmological isotropy on the scales where this background description applies.

The general FLRW metric instead has $d{r}^2/(1-k{r}^2)$ in its radial part, with normalized $k=+1,0,-1$ describing positive, zero or negative spatial curvature of an FLRW universe. The best simple description takes $k=0$; observations constrain spatial curvature to be small rather than proving it identically zero. The late accelerated expansion is encoded by the time dependence of $R$, determined by the stress-energy tensor through the Friedmann equations, rather than by an additional term in the FLRW metric. Small-scale galaxies and other inhomogeneities require perturbations of this background.

The local metric tensors do not determine the global coordinate identifications. The usual complete two-dimensional de Sitter spacetime is the hyperboloid $(X^0{)}^2-(X^1{)}^2-(X^2{)}^2=-1$ in ambient signature $(+--)$, withPulling back the ambient metric tensor gives $d{t}^2-{\cosh }^{2}td{\chi }^2$, withUnwrapping $\chi$ gives the universal cover of two-dimensional de Sitter spacetime instead. The usual embedded two-dimensional anti-de Sitter spacetime has $(X^0{)}^2+(X^1{)}^2-(X^2{)}^2=1$ in signature $(++-)$ andHere $r\in \mathbb{R}$ and $t$ is initially periodic modulo $2\pi$. The time circles are closed timelike curves. Its physically standard universal cover removes that identification:The two signs of $r$ describe two spatial ends; imposing $r\ge 0$ would retain only half of this complete model.

Both spacetimes have constant curvature and are geodesically complete. In the paper's convention their Ricci scalars are respectively $+2$ and $-2$. Their embeddings show that geodesics are intersections with planes through the ambient origin. They nevertheless have very different causal structures.

For the geodesics of two-dimensional de Sitter spacetime, use an affine parameter $\lambda$, normalization $\kappa ={\dot{t}}^2-{\cosh }^{2}t{\dot{\chi }}^2\in \{1,0,-1\}$ and conserved momentum $L={\cosh }^{2}t\dot{\chi }$. ThenTimelike geodesics extend to both infinite proper times; $\chi =\text{constant}$ is a simple example. Null geodesics obey $\sinh t=\pm |L|(\lambda -{\lambda }_0)$, so their affine parameter is also infinite at either temporal end. Spacelike geodesics have $|L|\ge 1$, obey $\sinh t=\sqrt{L^2-1}\sin (\lambda -{\lambda }_0)$ and are closed curves on the embedded cylinder; $t=0$ is its simplest spacelike geodesic circle. Unwrapping the spatial circle removes closure but not completeness.

For the geodesics of two-dimensional anti-de Sitter spacetime, the timelike Killing vector ${\partial }_t$ gives $E={\cosh }^{2}r\dot{t}$. The normalization is $\kappa ={\cosh }^{2}r{\dot{t}}^2-{\dot{r}}^2$, henceChoose a future-directed timelike geodesic, so $E\ge 1$ andFor $E>1$, it oscillates radially with proper period $2\pi$; $E=1$ gives the central geodesic $r=0$. In both cases, global time advances by $2\pi$ over a proper time interval $2\pi$. The curve is closed on the original hyperboloid, while its lift on the universal cover is nonclosed and future-directed. Null geodesics have $\sinh r=\pm E(\lambda -{\lambda }_0)$ and reach either spatial end only at infinite affine parameter. Spacelike geodesics satisfy $\sinh r=\sqrt{E^2+1}\sinh (\lambda -{\lambda }_0)$ and also have infinite proper length toward either end. Thus the finite coordinate time discussed below does not imply physical geodesic incompleteness.

The conformal cylinder of two-dimensional de Sitter spacetime follows by setting $\eta =\arctan (\sinh t)$:Its past and future conformal boundaries are spacelike circles. Null geodesics have $d\chi /d\eta =\pm 1$ and can travel only a finite angular distance over the entire infinite proper-time history. For the complete observer $\chi =0$, an event can send a signal to that observer precisely when its shortest angular distance $d(\chi ,0)$ is less than $\pi /2-\eta$. The observer horizons in two-dimensional de Sitter spacetime are therefore the null curves $d(\chi ,0)=\pi /2-\eta$; its past signal horizon has $d(\chi ,0)=\eta +\pi /2$. These are observer horizons, without curvature singularities. Their intersection bounds a static patch with metric tensorThe horizons at $\rho =\pm 1$ are regular Killing horizons beyond which this static chart fails, while the global coordinates remain smooth.

For the conformal strip of two-dimensional anti-de Sitter spacetime, set $\psi =\arctan (\sinh r)$. On its universal cover,The two conformal boundaries are timelike. A null ray from the center approaches a boundary after coordinate time $\pi /2$, though its affine parameter diverges. Every interior event can signal to the complete central static observer in finite global time, so that observer has no event horizon. A Poincaré horizon concerns a restricted coordinate patch, and accelerated observers can have different causal horizons.

Finally, the global de Sitter cylinder and its spatial cover are globally hyperbolic: constant-time slices are Cauchy hypersurfaces. The anti-de Sitter universal cover is not globally hyperbolic, because timelike infinity admits incoming signals within finite global time. A field's evolution therefore requires boundary conditions at its two conformal boundaries in addition to initial data. The opposite placement of the conformal boundaries explains why de Sitter observers have cosmological horizons despite global hyperbolicity, while the central anti-de Sitter observer has no event horizon despite the failure of global hyperbolicity.

Use geometrized units with $\hslash =k_B=1$ initially, and write $f(r)=1-2M/r$, $h(r)=1-Q^2/(Mr)$. The given nonextremality assumption makes $h_H=h(2M)=1-Q^2/(2M^2)>0$. Thus the static Killing vector field has a simple zero of its norm atThe exterior $r>2M$ is static. For radial null geodesics, $dt/dr=\pm 1/f$, independently of $h$. With $v=t+\int dr/f$, the metric iswhich is regular at $r_H$ since $h_H>0$. The outgoing null equation is $dr/dv=f/2$. Inside the future horizon these rays move to smaller radius, while outside they can escape to infinity. Hence the zero is the event horizon. The inner zero of $h$ is not another regular Killing horizon: the Killing norm there diverges rather than vanishes. The strict charge bound matters when making the regular extension.

For the Euclidean black-hole regularity condition, continue $t=i{t}_E$, consistent with the sign of the given quantum time-translation operator. Set $x=r-2M$. To leading order $f=x/(2M)$ and $h=h_H$, so the radial–time part of the Euclidean metric isIntroduce the radial proper length coordinate $\rho$ by $x=h_H{\rho }^2/(8M)$. ThenThe first two terms are a polar plane with angle $t_E/(4M)$. Smoothness at its origin requiresA different period would make a conical singularity. The time coordinate is normalized to proper time at infinity, so no further asymptotic redshift factor is needed. The cancellation of $h_H$ proves the charge independence of the Euclidean horizon period under a regular radial conformal factor; this argument would fail at $h_H=0$.

To justify the thermal interpretation, analytic continuation of the time-evolution operator $U(t)=e^{iHt}$ gives $U(i{t}_E)=e^{-H{t}_E}$. Its matrix element is an imaginary-time propagation kernel. Identifying the initial and final configurations and integrating over them takes its trace, giving the Euclidean path integral $Z=\mathrm{Tr}{e}^{-\beta H}$. In an energy eigenbasis each state has weight $e^{-\beta E}$, exactly the Boltzmann factor of a canonical ensemble. Thus the geometrically required period is the inverse Hawking temperature. With dimensions restored, an imaginary time period $\Delta t_E$ gives $\beta =\Delta t_E/\hslash =1/(k_{B}T)$. ConsequentlyHere $\mathcal{M}$ is mass in ordinary units; for a neutral hole $M=G\mathcal{M}/c^2$ and the physical imaginary-time period is $8\pi G\mathcal{M}/c^3$. Bosonic thermal fields are periodic and fermionic fields antiperiodic over that same Euclidean period. The relation to the canonical density operator, not just dimensional reasoning, identifies the temperature.

For an evaporation estimate take a neutral, isolated Schwarzschild black hole and approximate its radiated power with the Stefan–Boltzmann law. Its area is $A=16\pi G^2{\mathcal{M}}^2/c^4$, and the photon Stefan-Boltzmann constant is ${\sigma }_{SB}={\pi }^2{k}_B^4/(60{\hslash }^3{c}^2)$. Substituting the temperature givesThe black-hole evaporation speeds up as the mass decreases. Integrating, rather than dividing the initial rest energy by the initial power, givesThe corresponding initial Hawking temperature is about $6\times 10^{-8}$K. This is the Stefan-Boltzmann estimate of Schwarzschild evaporation time at the level of accuracy requested. Greybody factors and other emitted species alter the numerical coefficient, but not the leading cubic scaling of a large isolated hole. The estimate assumes negligible accretion or ambient radiation absorption; a stellar-mass hole in the present cosmic radiation bath is not currently losing mass at that isolated-hole rate. Finally the semiclassical formula does not describe the Planck-scale endpoint: extrapolating it to zero mass is an estimate of the macroscopic lifetime, not a proof of what the final quantum object is.

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.

Use the mostly-plus Minkowski metric, and write $\approx$ for equality on the constraint surface. Assume that the mechanical constraints are locally independent. They are first-class constraints whenThus their Poisson brackets vanish on the constraint surface, and their Hamiltonian flows preserve that surface. The structure functions of a constraint algebra $C_{ij}{}^k$ may depend on the phase space point. The finite real span of the constraints is a Lie algebra if it closes with constant structure coefficients, in a suitable choice of generators. The Jacobi identity then gives the usual conditions on the structure constants of a Lie algebra. With general structure functions the finite real span need not close, even though the Poisson bracket of all smooth functions is itself a Lie bracket.

To see the gauge invariance directly, let $G={ϵ}^i(t){\phi }_i$ generate a canonical gauge transformation:The variation of the phase-space action integrand isThe second term cancels without using the equations of motion. Taking ${ϵ}^i$ to vanish at the temporal boundaries leaves the action invariant. Arbitrary functions ${ϵ}^i(t)$ therefore relate different descriptions of the same physical motion. This reasoning also works with structure functions; constant structure coefficients are only needed for the finite-dimensional Lie algebra claim.

For a closed string, choose $0\le \sigma <2\pi$ and periodic fields. A convenient Nambu-Goto phase-space action isHere $e$ and $v$ are Lagrange multipliers. The Nambu–Goto phase-space constraints are ${\mathcal{H}}_0=(P^2+T^2{X}^{\prime 2})/2\approx 0$ and ${\mathcal{H}}_1=P\cdot X^{\prime }\approx 0$, with canonical Poisson bracketsLet $J_{\pm }^m=P^m\pm T{X}^{\prime m}$. Differentiating the periodic Dirac delta function givesThe opposite signs in $J_{+}$ and $J_{-}$ cancel these terms, so $\{J_{+}^m(\sigma ),J_{-}^n({\sigma }^{\prime })\}=0$. Replace the original constraints by the equivalent chiral densitiesTheir mixed Poisson brackets vanish. Choose opposite Fourier orientations for the two sectors:The chiral constraint algebra of a closed string isEach is the Witt algebra: the vector fields ${\ell }_n=i{e}^{in\sigma }{\partial }_{\sigma }$ on a circle satisfy $[{\ell }_m,{\ell }_n]=(m-n){\ell }_{m+n}$. Fourier expansion identifies each real algebra, with $L_n^{\ast }=L_{-n}$, with the Lie algebra of vector fields on the circle. The two commuting copies give ${\mathrm{Diff}}_1\oplus {\mathrm{Diff}}_1$, not a quantum central extension.

For an open string, allowed boundary conditions must remove the endpoint term in the variation of the action, consistently with the allowed endpoint variations. The spatial boundary term isIt expresses the open-string endpoint momentum flux. In the temporal gauge for a string $X^0=t$, take a boundary-adapted parametrization with $v=0$ at the ends. Fixing $X^i(t,a)=0$ gives $\delta X^i(t,a)=0$, a Dirichlet boundary condition. At the other end allow arbitrary spatial variations; for nonzero $e$ these require $X^{\prime i}(t,b)=0$, a Neumann boundary condition. Also $X^{\prime 0}=0$ in this temporal gauge for a string, so $X^{\prime m}(t,b)=0$. The constraint at this free-end string boundary condition reduces to $P^2=0$. Hamilton's equation $\dot{X}=eP+v{X}^{\prime }$ consequently gives ${\dot{X}}^2=0$ there. Since ${\dot{X}}^0=1$, the free endpoint has spatial speed one. This is the null motion of a free string endpoint.

A straight rotating string with one fixed endpoint supplies the required solution in at least two spatial dimensions. Set $e=1/T$, $v=0$, andTake $P_m=T{\dot{X}}_m$. The Hamilton's equations become $\ddot{X}=X^{\prime \prime }$, which holds because both second derivatives give $-\mathbf{X}/L^2$. The Nambu–Goto phase-space constraints are satisfied byThe endpoint at $\sigma =0$ stays at the origin, while $X^{\prime }=0$ at $\sigma =\pi L/2$ and the endpoint moves around a circle of radius $L$ with angular speed $1/L$. At each time the whole string lies on a straight radial segment. Its spatial proper length isThe velocity is everywhere perpendicular to the segment, so this also equals the sum of local rest-frame lengths. The induced worldsheet metric becomes degenerate at the null free endpoint, as expected for the limiting free-end solution.

Use Minkowski spacetime signature $(-,+,+,+)$ and a closed spatial parameter of period $2\pi$. Choose the future-directed branch with positive lapse $e$.

Canonical dynamics and gauge freedom. Variation of momentum and embedding in the Nambu-Goto phase-space action givesThe Lagrange multipliers impose the Nambu–Goto phase-space constraints $P^2+T^2{X}^{\prime 2}=0$ and $P\cdot X^{\prime }=0$. These two first-class constraints reflect freedom to relabel time and space on the same string worldsheet. The canonical Hamiltonian is a linear combination of constraints, with arbitrary multiplier functions. These functions specify a coordinate description rather than additional propagating fields. No explicit gauge transformation or Poisson-bracket calculation is required.

Eliminating auxiliary variables. Let $V=\dot{X}-u{X}^{\prime }$. Eliminating momentum using $P=V/e$ leavesOn a patch with spacelike spatial tangent, variation of $u$ gives $V\cdot X^{\prime }=0$, hence $u=(\dot{X}\cdot X^{\prime })/X^{\prime 2}$. For the induced worldsheet metric $g_{ab}={\partial }_{a}X\cdot {\partial }_{b}X$, this gives $V^2=\det g/X^{\prime 2}$. Variation of $e$ then gives $e=\sqrt{-\det g}/(T{X}^{\prime 2})$ on the positive branch. Substitution yieldsThis is minus string tension times Lorentzian worldsheet area. Vary the area using $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{ab}\delta g_{ab}$ and integrate by parts. The Nambu–Goto equations of motion areThe auxiliary-variable elimination and this metric form apply on nondegenerate timelike patches.

Circular motion, length and energy. For the circular embedding, direct differentiation givesAway from collapse, $\sqrt{-g}{g}^{ab}=\mathrm{diag}(-1,1)$, so the equations become ${\ddot{X}}^m-X^{\prime \prime m}=0$. The time and out-of-plane coordinates satisfy them immediately. For $Z=X^1+i{X}^2$, both $\ddot{Z}$ and $Z^{\prime \prime }$ are $-Z$. The relations $\dot{X}\cdot X^{\prime }=0$ and ${\dot{X}}^2+X^{\prime 2}=0$ verify the Virasoro constraints. Equivalently, $e=1/T$, $u=0$, $P=T\dot{X}$ solve the phase-space equations and constraints at all times.

The ring stays in a fixed plane, with radius $R|\cos t|$. It contracts to a point and re-expands. Each labeled point moves radially with speed $|\sin t|$ in target time $X^0=Rt$. The geometric ring repeats after target-time interval $\pi R$, although the labels have then shifted by half a circumference. It is not rigidly rotating. Velocity is perpendicular to the tangent, so simultaneous spatial arclength is also the local proper length along the string. Its length and conserved energy areAway from collapse, the same energy follows from $T\int dl/\sqrt{1-v^2}$. Increasing kinetic energy compensates the shrinking length. At collapse the induced metric degenerates; the area-form equation alone is undefined there. The regular phase-space solution supplies the continuation and the limiting constant energy. This is a pulsating circular string.

Endpoint variation. For an open string, integration by parts produces the boundary termIts coefficient is the open-string endpoint momentum flux. Evaluate this ungauge-fixed variation in a boundary-adapted temporal gauge for a string $X^0=t$. Then $X^{\prime 0}=0$ and the time equation gives $P^0=1/e\ne 0$. The free variation of time forces $u{P}_0=0$, hence $u=0$ at each end. Since $e$ is nonzero, the remaining spatial term requiresFree variation in every spatial direction gives ${\vec{X}}^{\prime }=0$, and together with $X^{\prime 0}=0$ gives the free-end string boundary condition $X^{\prime m}=0$. The general starting condition is the flux condition; the shift term should not be silently discarded before choosing the gauge.

Target-space charges. The Noether charges for translations and Lorentz transformations are the target-space Noether charges of a stringWriting $F_m=e{T}^2{X}_m^{\prime }+u{P}_m$, the canonical equations implyFor the second identity, antisymmetry cancels the $e{P}_m{P}_n$ terms. Integration by parts then cancels the $u{X}^{\prime }P$ terms and the symmetric $e{T}^2{X}_m^{\prime }{X}_n^{\prime }$ terms. For free ends $F_m=0$ individually, so both charges are constant. In conformal gauge, the same result follows from $u=0$ and $X^{\prime m}=0$. Target-space energy is $-{\mathcal{P}}_0$.

Other possibilities fix selected spatial directions by Dirichlet boundary conditions, while leaving Neumann boundary conditions in the permitted tangent directions. These mixed conditions describe endpoints on a D-brane; the two ends can lie on different branes. Time stays free under the given assumption. Fixed supports may absorb momentum in their Dirichlet directions and break corresponding translations or Lorentz symmetries. More general force-law conditions require an additional boundary interaction.

Use the Minkowski metric ${\eta }_{mn}=\mathrm{diag}(-1,1,\dots ,1)$ and units with speed of light one. The contractions are $P^2={\eta }^{mn}{P}_m{P}_n$ and $(T{X}^{\prime }{)}^2=T^2{\eta }_{mn}(X^m{)}^{\prime }(X^n{)}^{\prime }$; $(X^m{)}^{\prime }{P}_m$ and ${\dot{X}}^m{P}_m$ pair a vector with a covector without another metric. The Lagrange multipliers impose the two first-class constraintsThey generate worldsheet diffeomorphisms, so the phase space contains both constrained directions and gauge redundancy. Two first-class constraints remove two canonical pairs, leaving $D-2$ physical degrees of freedom per point.

In Monge gauge, $X^0=t$ and $X^1=\sigma$. Write the transverse canonical variables as $\mathbf{X}=(X^2,\dots ,X^{D-1})$ and $\mathbf{P}=(P_2,\dots ,P_{D-1})$. Solving the first-class constraints givesThe negative root selects positive energy. Substitution into the phase-space action gives the Hamiltonian reductionFor a static segment, $\mathbf{P}=0$, its proper length element is $d\ell =\sqrt{1+|{\mathbf{X}}^{\prime }{|}^2}d\sigma$, and $\mathcal{E}d\sigma =Td\ell$. Thus the string tension is the rest energy per unit proper length. In particular a straight resting segment has $\mathcal{E}=T$. Monge gauge is a local choice on a string embedding map for which $X^1$ is a valid coordinate; it need not cover folded strings or all endpoint configurations.

Let $q=\sigma /L$, with $L>0$, and let $Z=X^1+i{X}^2$. Direct differentiation gives $\ddot{Z}=Z^{\prime \prime }=-Z/L^2$, while the time coordinate also satisfies the wave equation. Using the Minkowski metric,Thus both Virasoro constraints hold, and the induced worldsheet metric is $g_{\mu \nu }={\cos }^{2}q\mathrm{diag}(-1,1)$. The configuration solves the Nambu–Goto action equations in its interior.

Place the fixed end at $\sigma =0$. The first free endpoint has $Z^{\prime }=e^{it/L}\cos q=0$ at $q=\pi /2$. For the usual single unfolded rotating Nambu–Goto string,This parameter interval need not equal the earlier interval $[0,\pi ]$, since that interval was a coordinate convention. The proper length is measured on a constant-$X^0$ slice; the local motion is perpendicular to the string, so there is no longitudinal Lorentz contraction.

All points have the same phase $t/L$, hence the segment rotates rigidly with angular velocity $\Omega =1/L$. At radius $r=L\sin q$, its speed is $v=r/L$. Therefore the free tip moves at the speed of light. This is consistent with its Neumann boundary condition: $X^{\prime }=0$ and the Virasoro constraints imply ${\dot{X}}^2=0$ at the free tip. The null endpoint is a boundary limit, not a nondegenerate interior point.

In this conformal gauge, $P_m=T{\dot{X}}_m$, so the energy density per $\sigma$ is $-P_0=T$. Equivalently, the energy density per proper length is $T/\sqrt{1-r^2/L^2}$. Thusand the rotational kinetic energy isThe angular momentum about the fixed point isTherefore the Regge trajectory isThe endpoint conditions alone also admit folded extensions with parameter length $(n+\frac{1}{2})\pi L$, $n\ge 0$. Their proper length, energy and angular momentum are $(2n+1)$ times the values above, and $\beta =1/[(2n+1)\pi T]$. The displayed answer uses the implicit unfolded-segment convention.

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.

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.

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.

In the stipulated constant-speed model, the Earth-frame speed is $v=D/t=3.844\times 10^8/(76\times 3600)\simeq 1.405\times 10^3$ m/s. The Earth and Moon are stationary endpoints in this model, so their Earth-frame separation $D$ is the proper length. The spacecraft-frame separation, measured simultaneously in that frame, is $D^{\prime }=D/\gamma =D\sqrt{1-v^2/c^2}$. ThusThe spacecraft-frame distance is shorter by about 4.2 mm. This compares simultaneous endpoint measurements in two frames, not the coordinate difference between a launch event and an arrival event. The small-speed length contraction estimate also avoids numerical cancellation when evaluating the tiny contraction.

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.