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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 light clock measures time by counting round trips of a light pulse between mirrors. In its rest frame, mirrors separated by $h$ give a tick duration $2h/c$, where $c$ is the speed of light. Comparing this round trip in another inertial frame provides the geometric time dilation argument above.

For a relative speed $0\le v<c$, the Lorentz factor is

At $v=0.6c$, $\gamma =1/\sqrt{1-0.36}=1/0.8=1.25$. Thus a moving clock accumulates $0.8$ seconds of proper time per laboratory second, and a parallel moving rod has $0.8$ times its rest length. The Lorentz factor approaches one at small speed and grows without bound as $v$ approaches the speed of light; a massive object's rest frame never requires a boost at $v=c$.

In the observer's rest frame, $U=(c,0)$ and $P=(E^{\prime }/c,{\mathbf{p}}^{\prime })$. The Lorentz invariance of the Minkowski inner product gives $P\cdot U=E^{\prime }$ andThe relation between four-momentum and velocity is ${\mathbf{p}}^{\prime }=E^{\prime }\mathbf{v}/c^2$, soTaking the nonnegative square root provesFor a future-directed massive particle and observer, $P\cdot U>0$ and the expression lies in $[0,1)$; the massless limit, with nonzero energy, gives $v=c$.

The Polyakov action has independent worldsheet metric ${\gamma }_{ab}$ and embedding fields $X^{\mu }$. Varying the metric before imposing a gauge givesWorldsheet diffeomorphisms and Weyl transformations allow conformal gauge ${\gamma }_{ab}=e^{2\omega }{\eta }_{ab}$ locally. The Weyl factor then cancels out of the classical action, leavingThese are free two-dimensional scalar fields, but the surviving metric equations requireEquivalently, ${\dot{X}}^2+X^{\prime 2}=0$ and $\dot{X}\cdot X^{\prime }=0$. Gauge fixing does not license dropping these Virasoro constraints.

For an ordinary open string, the boundary conditions relate the two chiral sectors. With ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{k}^{\mu }$, their independent Fourier generators areThe classical oscillator Poisson brackets $\{{\alpha }_m^{\mu },{\alpha }_n^{\nu }\}=-im{\eta }^{\mu \nu }{\delta }_{m+n,0}$ give the classical Virasoro constraint algebraThus they are first-class constraints. In the quantum theory one applies normal ordering to the product and uses $[{\alpha }_m^{\mu },{\alpha }_n^{\nu }]=m{\eta }^{\mu \nu }{\delta }_{m+n,0}$. The oscillator commutator givesFor example, for $m>0$ the extra vacuum term is $(D/2){\sum }_{r=1}^{m-1}r(m-r)=D(m^3-m)/12$. This exhibits the Virasoro central extension rather than concealing it in a classical constraint equation. The matter central charge is $D$; the worldsheet ghost fields have charge $-26$. Quantum consistency sets $D=26$ and the string intercept to one.

In old covariant string quantization the physical-state Virasoro conditions for an open string arePhysical string states are then identified modulo physical null string states. One does not impose both positive and negative generators as annihilation conditions: their central extension would obstruct that prescription. The resulting null-state quotient of a string agrees with the transverse light-cone spectrum at nonzero momentum.

At $N=0$, the oscillator vacuum $|k⟩$ is a single spacetime scalar field with ${\alpha }^{\prime }{M}^2=-1$: the bosonic tachyon. At $N=1$, write $|\Psi ⟩={ϵ}_{\mu }{\alpha }_{-1}^{\mu }|k⟩$. The zero-mode equation gives $k^2=0$, and $L_1$ gives $k\cdot ϵ=0$. The null descendant $L_{-1}|k⟩=\sqrt{2{\alpha }^{\prime }}k\cdot {\alpha }_{-1}|k⟩$ identifies $ϵ\sim ϵ+\lambda k$. This leaves 24 physical polarizations of a massless gauge vector.

At $N=2$, use the level-two open-string polarization decompositionThe mass is $M^2=1/{\alpha }^{\prime }$. The oscillator commutators give all the nontrivial positive-mode conditions:Higher positive modes annihilate this level. To see the quotient explicitly, the 25 transverse-parameter null states $L_{-1}(\xi \cdot {\alpha }_{-1}|k⟩)$, with $k\cdot \xi =0$, shiftThe additional level-two scalar Virasoro null state $(L_{-2}+\frac{3}{2}{L}_{-1}^2)|k⟩$ shiftsSubstitution into the two physical equations verifies the null-state conditions; the second residual is $(D-26)/2$. Because $k$ is timelike, these null shifts span all components of $b$, so choose a representative with $b=0$. The physical conditions then say $k^{\mu }{h}_{\mu \nu }=0$ and $h^{\mu }{}_{\mu }=0$. In the rest frame this is a traceless second-rank tensor in 25 spatial dimensions, with $25\cdot 26/2-1=324$ polarizations. The light-cone basis independently gives $24$ states ${\alpha }_{-2}^i|k⟩$ and $24\cdot 25/2=300$ symmetric states ${\alpha }_{-1}^i{\alpha }_{-1}^j|k⟩$. Hence

For the fermionic extension, an explicit convention prevents ambiguity in the signs. Keep worldsheet signature $(-,+)$ and useTake $\psi =({\psi }_1,{\psi }_2{)}^T$ and the constant parameter $ϵ=({ϵ}_1,{ϵ}_2{)}^T$ to be real Grassmann-odd Majorana components, with $\bar{\psi }={\psi }^T{\rho }^0$. Put ${\partial }_{\pm }={\partial }_{\tau }\pm {\partial }_{\sigma }$. Apart from the common factor $1/(4\pi {\alpha }^{\prime })$, the combined Lagrangian isThe rigid transformations can be writtenor, in components,For the ${ϵ}_2$ part, the bosonic variation is $i{ϵ}_2({\partial }_{+}{\psi }_1\cdot {\partial }_{-}X+{\partial }_{+}X\cdot {\partial }_{-}{\psi }_1)$, while the fermionic variation is $-i{ϵ}_2{\partial }_{-}X\cdot {\partial }_{+}{\psi }_1+i{ϵ}_2{\psi }_1\cdot {\partial }_{+}{\partial }_{-}X$. Anticommuting the constant parameter past the fermion is essential in the second term. The first terms cancel. The other chiral component works in the same way, giving the chiral proof of rigid worldsheet supersymmetryThus the action is invariant off shell up to a boundary term. The transformation is a genuine supersymmetry: on $X$ its commutator for parameters $ϵ,\zeta$ is the translationOn $\psi$ the same translation follows using ${\partial }_{+}{\psi }_1=0$ and ${\partial }_{-}{\psi }_2=0$; the remaining cross terms are proportional to these Dirac equations. This is on-shell closure of rigid worldsheet supersymmetry. On a surface with boundary, compatible endpoint conditions must make the boundary flux vanish. On a compact worldsheet, a constant parameter must preserve the spin structure; the local identity alone does not make a constant transformation compatible with every antiperiodic sector.

The fermions contribute to the worldsheet stress-energy tensor and produce a fermionic superconformal current. In holomorphic normalization with ${\psi }^{\mu }(z){\psi }^{\nu }(w)\sim {\eta }^{\mu \nu }/(z-w)$,The free-field contractions give $G(z)G(w)\sim (2c/3)/(z-w{)}^3+2T(w)/(z-w)$ and conformal weight $3/2$ for $G$, where $c=D+D/2=3D/2$. Consequently the modes obey the N=1 super-Virasoro algebraDropping the central terms gives the classical graded constraint algebra. Integer $r$ gives the Ramond sector; half-integer $r$ gives the Neveu–Schwarz sector. In the displayed conformal-plane convention $G_0^2=L_0-c/24$ in the Ramond sector. Shifting to the Ramond vacuum-normal-ordered zero mode gives the usual Ramond supercurrent zero-mode square $G_0^2=L_0^{(R)}$.

Rigid symmetry of matter already supplies these currents, but promoting their vanishing to new gauge constraints requires coupling to worldsheet supergravity and varying its worldsheet gravitino. In superconformal gauge one retains $T=G=0$. The fermionic current constraints and their physical-state conditions extend the bosonic Virasoro conditions. The ghost charges are $-26+11=-15$, so cancellation with $3D/2$ gives the critical dimension of the RNS superstring $D=10$; the NS and Ramond intercepts are respectively $1/2$ and zero. This distinguishes the global supersymmetry established above from its locally gauged string completion.

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.

With negligible cold-dark-matter velocity, the synchronous perfect-fluid density equation gives ${\delta }_C^{\prime }=-h^{\prime }/2$. Differentiate and use the trace Einstein equation, taking the cold matter sound speed to vanish and its density to dominate:In a spatially flat matter-dominated universe, $a\propto {\tau }^2$, $\mathcal{H}=2/\tau$, and the conformal Friedmann equation gives $4\pi G{a}^2{\rho }_C=3{\mathcal{H}}^2/2=6/{\tau }^2$. HencePut ${\delta }_C\propto {\tau }^s$. The indicial equation is $s(s-1)+2s-6=(s-2)(s+3)=0$. These two independent modes span the solution space, soEquivalently the growing mode is proportional to $a$ and the decaying mode to $a^{-3/2}$. The coefficients are set by initial data. Since synchronous gauge has residual coordinate modes, interpreting the density contrast physically requires fixing its residual freedom consistently, for example by the cold-matter rest frame and metric tensor integration convention; the mode equation alone is not a gauge-invariant observable.

In the pion rest frame, conservation of three-momentum makes the Electron and massless neutrino momenta equal and opposite, of common magnitude $p$. Their energies obeySquaring $E_e=m_{\pi }{c}^2-pc$ and cancelling $p^2{c}^2$ givesThe Electron four-momentum satisfies $p=E_{e}v/c^2$. Thus, for the physically allowed decay $m_{\pi }>m_e$,

If one clock measures a proper time interval $\Delta \tau$ while moving at constant speed $v$ in an inertial frame, that frame measures $\Delta t=\gamma \Delta \tau$, where $\gamma$ is the Lorentz factor. The two spacetime events must lie on that clock's world line.

Suppose a clock moving at $0.6c$ advances by $4\mu \mathrm{s}$ between two ticks. Set $t=x=t^{\prime }=x^{\prime }=0$ at the first tick. The laboratory's synchronized clocks assign $\Delta t=1.25\times 4=5\mu \mathrm{s}$. During that interval the clock travels $\Delta x=0.6c\times 5\mu \mathrm{s}=3c\mu \mathrm{s}$, approximately $899.38\mathrm{m}$. The Minkowski interval checks the result:Both frames therefore recover the same $4\mu \mathrm{s}$ of proper time. This calculation is useful whenever a process is timed on a moving apparatus but described using laboratory clocks.

A light clock makes the mechanism visible. Two mirrors separated by $h$ perpendicular to the motion exchange a pulse; define one tick as its complete return to the lower mirror. In the clock's rest frame the tick takes $\Delta \tau =2h/c$. In the laboratory each half-tick lasts $T$ and the light traverses a diagonal, so the Pythagorean theorem givesFor $h=1\mathrm{m}$ and $v=0.6c$, each leg covers $1.25\mathrm{m}$ while the mirrors move $0.75\mathrm{m}$. The return tick lasts $2.5\mathrm{m}/c$ instead of $2\mathrm{m}/c$: the light has the same speed of light in both accounts. ^[ref].

Reciprocal time dilation is consistent: each inertial frame times the other frame's single moving clock using its own synchronized clock network. These comparisons use different simultaneous spacetime events. Comparing reunited clocks instead requires their complete world lines, as in the twin paradox.