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Use dimensions $\mathsf{M},\mathsf{L},\mathsf{T}$ for mass, length and time. The reduced Planck constant, gravitational constant and speed of light have dimensionsFor $l_p={\hslash }^{\alpha }{G}^{\beta }{c}^{\gamma }$ to have the dimensions of a length, dimensional analysis requiresThe first equation gives $\beta =\alpha$; the third gives $\gamma =-3\alpha$. Substitution in the second gives $2\alpha =1$. HenceWith $\hslash =1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$, $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$, the Planck length isThe proposed monomial fixes the dimensionless prefactor to one; dimensional analysis on its own cannot select an arbitrary dimensionless multiplier.

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.

The relativistic kinetic energy is the total energy minus the rest energy:Here $\gamma$ is the Lorentz factor and $c$ the speed of light. The binomial series, with $b=v/c$, givesFor $v\ll c$, the leading term is the classical kinetic energy $M{v}^2/2$. The ratio of the fourth-order term to that leading term is $3v^2/(4c^2)$. ConsequentlyThe condition concerns one term of the binomial series; it is not the condition that the entire relativistic correction be exactly $1\%$.

Continue with $m=0$. Along a characteristic $x=x_0+t$, the chain rule givesThus the initial wave packet ${\psi }_R(0,x)=f(x)$ evolves as ${\psi }_R(t,x)=f(x-t)$. Similarly, along $x=x_0-t$, the left component is constant, givingA feature initially at $x_0$ therefore appears at $x_0+t$ for the right component and $x_0-t$ for the left component. The profiles translate without changing shape at the speed of light, which is 1 in these units. For positive-energy plane waves, the right component has $E=p>0$ and the left component has $E=-p>0$, another way to identify their propagation directions.

The conclusion concerns the massless Dirac equation. For $m\ne 0$, the components couple as in c, and a positive-energy packet has group velocity $p/\sqrt{p^2+m^2}$, rather than a universal speed $\pm 1$. For example, a massive positive-energy rest solution has $p=0$ and ${\psi }_R={\psi }_L$, so neither component alone describes an independently moving light-speed packet.

The Friedmann-Lemaître-Robertson-Walker metric describes homogeneous, isotropic spatial slices. Here $t$ is cosmic time, equal to the proper time of a comoving observer; $R(t)$ is the length-valued scale factor; and $\chi ,\theta ,\varphi$ are dimensionless comoving radial, polar and azimuthal coordinates. The speed of light is $c$, andfor positive, zero and negative cosmological spatial curvature. The convention used makes $d{s}^2$ a squared time interval, with signature $(+---)$; multiplying the metric by $c^2$ gives the more usual squared-length convention. At fixed $t$, radial proper length is $R(t)d\chi$, while a sphere has area $4\pi R^2(t)S^2(\chi )$.

luminosity distance is defined by $F=L/(4\pi d_L^2)$ for bolometric radiative flux $F$ and intrinsic luminosity $L$. Radial null propagation gives $d\chi =\pm cdt/R(t)$. Applying the same path integral to two successive wave crests gives $\Delta t_0/R_0=\Delta t_e/R_e$, hence $1+z=R_0/R_e$. Each photon loses an energy factor $1+z$, and their arrival rate loses another. For transparent propagation and isotropic emission,The two redshift factors have distinct physical origins; geometrical dilution alone is insufficient.

With $H=\dot{R}/R$, differentiating $1+z=R_0/R$ yields $dt/dz=-[(1+z)H(z){]}^{-1}$. Therefore the radial comoving coordinate isThis is independent of any particular matter model; the transverse comoving distance is $R_{0}S(\chi )$, which equals $R_0\chi$ only for flat space.

For pressureless matter, the continuity equation ${\dot{\rho }}_m+3H{\rho }_m=0$ integrates to ${\rho }_m={\rho }_{m0}(R_0/R{)}^3={\rho }_{m0}(1+z{)}^3$. The cosmological constant is kept separately from this matter density. In the flat Friedmann equation, defineThe flat dust-vacuum Hubble parameter is thereforeIn the equation's convention ${\rho }_m$ is mass-equivalent energy density and $\Lambda$ has units ${\mathrm{s}}^{-2}$; the geometrical cosmological constant convention instead writes $c^2{\Lambda }_{\mathrm{geom}}$. The model contains dust and vacuum energy with negligible radiation over the relevant late epochs: “matter-dominated” here does not require matter to exceed vacuum energy today.

The cosmic microwave background supplies a standard ruler through its acoustic scale: the angular positions of the Cosmic microwave background acoustic peaks constrain cosmological spatial curvature, while their relative heights constrain matter and baryon densities. Together with independent distance information, these favour a nearly flat Lambda-CDM model containing substantial cold dark matter and vacuum energy. The cosmic microwave background alone has a geometrical degeneracy between parameters affecting distance to last scattering; its analysis also includes early radiation, which the late-time formula above omits. The nine-year WMAP results were already available as a December 2012 preprint.^[ref]

A standardized Type Ia supernova at appreciable redshift is fainter, hence farther in luminosity distance, than in a matter-only Einstein-de Sitter universe with the same local expansion scale. Fitting the supernova Hubble diagram supports late acceleration; in flat dust plus vacuum, $q_0={\Omega }_{m0}/2-{\Omega }_{\Lambda 0}$ is negative for ${\Omega }_{\Lambda 0}>1/3$. Light-curve and colour corrections, extinction, calibration, selection bias and possible population evolution must be controlled. The supernova evidence and cosmic microwave background constraints are complementary; together they favour this model without proving that every alternative dark-energy model is impossible.^[ref][ref]

For the final comparison, assume $z>0$, a common $H_0$ and $0<{\Omega }_{\Lambda 0}<1$ in model (a). ThenThus $H_c<H_a<H_b$ and the integrands obey $1/H_b<1/H_a<1/H_c$. Because all three models are flat, compare the normalization-independent distance $R_0\chi =c\int d{z}^{\prime }/H(z^{\prime })$, rather than separately normalized coordinates. The luminosity-distance ordering of flat dust-vacuum universes is $d_{L,b}<d_{L,a}<d_{L,c}$, henceModel (b) gives the largest received flux, model (c) the smallest. Equalities replace the strict ordering at the endpoint values of the vacuum fraction; all have the same leading local Hubble law as $z\to 0$.

Corotation reaches the speed of light at $R_c=c/\Omega$. Extrapolating a dipole field gives $B_c\sim B_{\ast }(R_{\ast }/R_c{)}^3$ and magnetic energy density ${ϵ}_c\sim B_c^2/{\mu }_0$, suppressing order-one factors. An outward electromagnetic energy flux of scale ${ϵ}_{c}c$ through area of scale $R_c^2$ therefore givesThis assumes a dipole-dominated field tied to the rotating star and an effective outgoing Poynting flux near the light cylinder. The dipole extrapolation, angular pattern and surface-field convention affect the coefficient. A strictly aligned vacuum dipole would not radiate, whereas an oblique vacuum dipole does. A plasma-filled magnetosphere opens field lines and can spin down even an aligned rotator, so the light-cylinder estimate is a scaling argument, not an exact vacuum emission calculation or literal rigid motion at $c$. Force-free calculations explicitly demonstrate this distinction.^[ref]

The energy comes primarily from stellar rotation. With constant moment of inertia $I$, $E_{\mathrm{rot}}=I{\Omega }^2/2$, and conservation of energy gives $I\Omega \dot{\Omega }=-P$. ThusFor constant $K$, differentiating gives $\ddot{\Omega }=-3K{\Omega }^2\dot{\Omega }$, hence $n=\Omega \ddot{\Omega }/{\dot{\Omega }}^2=3$. Evolving field strength, inclination, moment of inertia or additional torques invalidate the constant-$K$ prediction; generally $n=3+(\Omega /\dot{\Omega })(\dot{K}/K)$ in this parametrization.

For radial inflow onto mass $M$, take $v_{\mathrm{ff}}\sim (GM/r{)}^{1/2}$ and $\rho \sim \dot{M}/(4\pi r^2{v}_{\mathrm{ff}})$. magnetic pressure decreases as $B_{\ast }^2{R}_{\ast }^6/({\mu }_0{r}^6)$, while ram pressure scales as $\rho v_{\mathrm{ff}}^2\propto \dot{M}(GM{)}^{1/2}{r}^{-5/2}$. Balancing them yieldsNumerical factors and disc geometry are not determined by this spherical estimate. The displayed proportionality holds $M$ fixed and requires a magnetospheric boundary outside the stellar surface.

In a disc, inflowing matter near $R_M$ has specific angular momentum approximately $\sqrt{GM{R}_M}$. If it is transferred to a slower star, the material torque is $N\sim \dot{M}\sqrt{GM{R}_M}$ and $I\dot{\Omega }\sim N>0$. The corotation radius is $R_{\mathrm{co}}=(GM/{\Omega }^2{)}^{1/3}$, distinct from the light-cylinder radius. As stellar spin approaches the inner disc's Keplerian frequency, net spin-up weakens. If $R_M$ lies outside corotation, the field rotates faster than the disc and can centrifugally expel matter in the propeller regime. A characteristic equilibrium isfor fixed mass and magnetic-coupling convention. A complete magnetic torque model can change the coefficient; a star also cannot rotate past its mass-shedding limit, of order $\sqrt{GM/R_{\ast }^3}$.

For opacity per unit mass $\kappa$, an isotropic radiative force balances gravity at $L_E=4\pi GMc/\kappa$. Efficient release of surface gravitational binding energy gives $L_{\mathrm{acc}}\sim GM\dot{M}/R_{\ast }$ and therefore the surface accretion Eddington rateThe radius dependence matters: it uses efficiency $\eta \sim GM/(R_{\ast }{c}^2)$, not an arbitrarily fixed reference efficiency. Substituting this maximal rate into the equilibrium period givesmass, opacity and geometry factors are held fixed. This is the classical radiatively efficient Eddington-limited scaling, not a universal upper bound on every mass supply rate: beaming, advection, outflows and opacity changes can alter it. For sufficiently small field, $R_M\lesssim R_{\ast }$ and the magnetic scaling must be replaced by surface-disc and mass-shedding physics.

The metric uses signature $(+---)$ and a time-valued interval $ds$: a comoving clock has $ds=dt$. Here $t$ is cosmic proper time, $R(t)$ a scale factor with dimensions of length, $c$ the speed of light, $\chi$ a dimensionless comoving radial coordinate, and $\theta ,\varphi$ the usual polar and azimuthal angles. With curvature normalized to $k=+1,0,-1$, $S(\chi )=\sin \chi ,\chi ,\sinh \chi$, respectively. Homogeneity and isotropy apply to each constant-$t$ spatial slice. A comoving object's fixed coordinates do not imply a constant proper separation: spatial lengths scale with $R$.

Radial null propagation obeys $d\chi =cdt/R(t)$ in magnitude. Two neighbouring wave crests between the same comoving emitter and observer have the same radial integral, giving $\delta t_0/R_0=\delta t_e/R_e$. The wavelength redshift is consequentlyPeculiar Doppler shifts are excluded by the comoving assumption. Define angular diameter distance by $d_{\theta }={\ell }_{\perp }/\delta \theta$, where ${\ell }_{\perp }$ is proper transverse size at emission. The transverse metric gives ${\ell }_{\perp }=R_{e}S(\chi )\delta \theta$, henceThis distance is neither the photon travel time times $c$ nor the object's present proper distance.

The comoving particle horizon at $t_1$, with the initial boundary at the Big Bang, is ${\chi }_p(t_1)={\int }_0^{t_1}cdt/R(t)$. Its proper radial distance is $D_p=R(t_1){\chi }_p$; for a power-law scale factor $R=A{t}^p$, $0<p<1$,Thus a matter-only flat Einstein-de Sitter universe, $p=2/3$, has $D_p=3c{t}_1$, and the flat radiation-only case, $p=1/2$, has $D_p=2c{t}_1$. Strictly, Einstein-de Sitter denotes the pressureless matter model; the radiation calculation uses the flat zero-$\Lambda$ radiation analogue requested here.

For the source seen at time $t_1=t_{\mathrm{rec}}$, the radial coordinate is the integral from recombination to today, not the earlier horizon integral. In these flat models,Taking the linear horizon scale in the question to be $D_p$, its small angular extent isThereforeUsing the supplied times gives ${\theta }_{p,m}=0.02684\mathrm{rad}=1.54^{\circ }$ and ${\theta }_{p,r}=0.004244\mathrm{rad}=0.243^{\circ }$. There is a genuine radius/diameter convention to state: the supplied $3c{t}_1$ is a radial horizon distance. If angular diameter means the full diameter of a spherical horizon patch of radius $D_p$, the two answers are instead $2{\theta }_{p,m}=3.08^{\circ }$ and $2{\theta }_{p,r}=0.486^{\circ }$. Both conventions follow from the same distance calculation; a factor of two should not be hidden in the word horizon.

Matter-only evolution is the better of these two approximations because the universe is already matter dominated around recombination and evolves mostly through a subsequent matter era, rather than remaining radiation dominated to the present. Neither is exact: pre-recombination radiation changes the particle horizon, and late-time dark energy changes the distance to the last-scattering surface. No supplied ages have been replaced by later best-fit values.

The cosmic microwave background originates near last scattering and samples many horizon-sized regions. In an always-decelerating hot-Big-Bang model, widely separated last-scattering patches lack overlapping early causal histories, motivating the horizon problem. The measured acoustic angular scale uses the smaller sound horizon, not automatically this light-travel particle horizon; baryon-photon pressure changes the propagation speed. Inflation supplies an earlier accelerated expansion during which the comoving Hubble radius shrinks, allowing a patch with a common earlier causal history to expand to the size of the observable universe. Quantum perturbations can then seed later acoustic and density fluctuations. This does not require local signals to propagate faster than light, and the Hubble radius must not be equated indiscriminately with the particle horizon. CMB temperature and lensing spectra test the resulting perturbation and expansion model.^[ref]

Definitions and pressure convention. $R(t)$ is the scale factor, $t$ cosmic time and dots time derivatives. $G$ is the gravitational constant, $c$ the speed of light, $\rho$ the mass-equivalent density of the material fluid and $k=+1,0,-1$ the spatial-curvature sign in this normalization. $\Lambda$ here has units of inverse time squared, rather than inverse length squared. Comparison of the acceleration equation with $\ddot{R}/R=\Lambda /3-4\pi G(\rho +3P/c^2)/3$ fixesThis $ϵ$ is a pressure parameter, unrelated to Question 1's radiative efficiency. pressureless matter has $P=0$ and $ϵ=0$. When solving the generalized model below, $ϵ$ is assumed constant.

Evidence for a positive vacuum component. Standardized Type Ia supernovae measure the luminosity-distance–redshift relation. Their corrected apparent magnitudes at cosmological distances imply greater distances than a decelerating matter-dominated model of comparable parameters, favoring late accelerated expansion. Light-curve standardization, interstellar extinction, selection bias and possible population evolution must be controlled rather than treating every supernova as an identical candle.^[ref][ref]

The cosmic microwave background and baryon acoustic oscillation measurements constrain geometry and matter density in combination. Near spatial flatness together with a subcritical matter density leaves room for an additional smooth component; cosmic ages and structure growth supply consistency tests. A positive cosmological constant fits these observations within the standard framework. They do not establish that every allowed dark-energy theory must be a strictly constant vacuum energy; the inference depends on the cosmological and gravitational model.

pressureless matter evolution. Set $k=0$ and $ϵ=0$. Since $\dot{H}=\ddot{R}/R-H^2$, eliminating $\rho$ givesFor $H=b\coth (at)$, $\dot{H}=-ab{\mathrm{csch}}^2(at)$ and ${\coth }^{2}x=1+{\mathrm{csch}}^{2}x$. ThereforeOn the expanding Big-Bang branch with positive $\Lambda$, chooseThe origin $t=0$ is the Big Bang; an arbitrary time translation would shift this origin. Integrating gives $R\propto {\sinh }^{2/3}(at)$. At early times $H\sim 2/(3t)$, while at late times $H\to \sqrt{\Lambda /3}$.

Age and the density fractions. Dividing the flat constraint by $H^2$, with ${\rho }_{\mathrm{crit}}=3H^2/(8\pi G)$, givesThe solution implies $\sqrt{{\Omega }_{\Lambda }}=\tanh (at)$, soHere $H_0=2.3331\times 10^{-18}{\mathrm{s}}^{-1}$, using the printed megaparsec conversion. Taking one year as $365.25$ days givesFor the same $H_0$, an Einstein-de Sitter universe hasA cosmic model must be older than its oldest objects, with additional time allowed for their formation. Old stellar populations are dated through stellar-evolution fitting, particularly the globular cluster main-sequence turnoff, with uncertainties from distances, composition and input stellar physics. A pre-examination globular-cluster age constraint on cosmology obtained a $95\%$ lower cosmic-age limit of $11\mathrm{Gyr}$ after allowing for formation time. This conflicts with the $9.05\mathrm{Gyr}$ model but is compatible with the $13.09\mathrm{Gyr}$ calculation.^[ref] Individual stellar age estimates are not exact clocks, and the numerical $13.09\mathrm{Gyr}$ is the result for the specified $H_0$ and matter fraction, not a claim to be a contemporary best-fit age.

Constant nonzero pressure parameter. Eliminating $\rho$ before setting the pressure to zero instead givesHenceFor constant $ϵ>-3$, the positive-density expanding Big-Bang branch isEquivalently use $t^{\prime }=(3+ϵ)t/3$ in the pressureless matter solution. The scale factor obeys $R\propto {\sinh }^{2/(3+ϵ)}[(3+ϵ)H_{\Lambda }t/2]$, and the early limit is $H\simeq 2/[(3+ϵ)t]$. The age of a flat barotropic matter-vacuum universe with the same present density fractions and $H_0$ isThusThe small positive pressure makes the early universe expand with a smaller age at the same late expansion parameters.

The acceleration transition. Normalize the present scale factor to one. Constant $ϵ$ and the continuity equation give $\rho ={\rho }_0(1+z{)}^{3+ϵ}$. Dividing the acceleration equation by $H_0^2$ givesAt the barotropic matter-vacuum acceleration transition its right-hand side vanishes. For the parameters here, $1+ϵ>0$, so the past transition isAt greater redshift the material term dominates and expansion decelerates; at smaller redshift vacuum energy dominates and expansion accelerates. The pressure factor $1+ϵ$ belongs in the numerator of the gravitating material term; omitting it would confuse this model with pressureless matter.

The required invariant is obtained by tensor contraction, rather than by an ordinary sum of squares of the entries of the electromagnetic field tensor. With ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, raising both indices givesThe electromagnetic field tensor is an antisymmetric second-rank tensor, so every independent off-diagonal entry occurs twice in its contraction. The three electric contributions each acquire a minus sign, whereas the magnetic contributions retain their sign. ConsequentlyAll indices are contracted, so this quantity is a Lorentz scalar. Since the speed of light is invariant,is Lorentz invariant. Both terms have the units of squared electric field, ${\mathrm{V}}^2{\mathrm{m}}^{-2}$ in SI units.

Write ${\Gamma }_a(v)=(1-v^2/a^2{)}^{-1/2}$. The first proposed Lorentz transformation isFor the next transformation, write ${\Gamma }_b(u)=(1-u^2/b^2{)}^{-1/2}$ and substitute the first pair into $x^{\prime \prime }={\Gamma }_b(u)(x^{\prime }-u{t}^{\prime })$ and $t^{\prime \prime }={\Gamma }_b(u)(t^{\prime }-u{x}^{\prime }/b^2)$. The result isThe direct transformation between $S$ and $S^{\prime \prime }$ must belong to the same standard boost family with the same calibration of space and time units. Such a transformation has one common factor multiplying the coefficient of $x$ in $x^{\prime \prime }$ and that of $t$ in $t^{\prime \prime }$. Equality of these coefficients for generic $uv\ne 0$ requiresThe case $uv=0$ alone cannot constrain the scales, but a transformation law must work for arbitrary successive boosts. Once the scales agree, their composed velocity is $V=(u+v)/(1+uv/a^2)$ and the common factor is ${\Gamma }_a(V)$. This establishes the universality of an invariant speed under boost composition within the proposed family.

Finiteness needs physical input. Sending $a^2$ to infinity gives the consistent Galilean transformation $x^{\prime }=x-vt$, $t^{\prime }=t$. Its boosts compose with $V=u+v$ and satisfy a relativity principle. Thus the composition calculation itself does not prove that $a^2$ cannot be infinite. The finite propagation speed of light in vacuum and its invariance, supplied by the second of the postulates of special relativity, rule out this limit for physical spacetime. Under a Galilean transformation, a ray with laboratory speed $c$ would have speed $c-v$ in the moving frame, in conflict with that postulate.

Finally, the common invariant scale must be identified with the physical speed of light, rather than an arbitrary speed measured in one frame. Requiring a ray $x=ct$ also to obey $x^{\prime }=c{t}^{\prime }$ givesHence the physical scale is $a=c$, taking the positive speed. The postulate about light identifies and requires a finite invariant scale; closure alone only establishes its universality. This distinction is necessary for a logically valid answer to the question's request to exclude infinity.

The two postulates of special relativity are: all nongravitational laws have the same form in every inertial frame, and light in vacuum has the same finite speed of light $c$ in every inertial frame, independently of the motion of its source. The first is Einstein's principle of relativity; the second supplies an experimentally identified invariant speed.

For the stated standard configuration, the Lorentz transformation isTake differentials along the particle's worldline, rather than comparing coordinate displacements at equal times in the two frames. Writing $u_i=d{x}^i/dt$, one obtainsDividing by the transformed time interval proves the velocity-addition formula:For a massive particle, $|\mathbf{u}|<c$ and $|v|<c$, so the denominator $1-v{u}_x/c^2$ is positive.

Let $\Psi =k_{\alpha }{x}^{\alpha }$. For a nonzero symmetric amplitude, substituting the plane wave into the wave equation gives $-(k_{\sigma }{k}^{\sigma }){ϵ}^{\mu \nu }\cos \Psi =0$. The gauge condition gives $-k_{\sigma }{ϵ}^{\nu \sigma }\sin \Psi =0$. Thus a propagating mode satisfiesThe wavevector is null, so the gravitational wave propagates at the vacuum speed of light. The amplitude is transverse as a four-dimensional tensor in Lorenz gauge in linearized gravity. This condition alone does not set all time components or the trace to zero: residual gauge symmetry of linearized gravity remains. Of the ten symmetric components, four Lorenz constraints and four independent residual coordinate choices leave two physical gravitational wave polarizations.

For a grazing null geodesic, take the impact parameter as $b\simeq R=6.0\times 10^6\mathrm{m}$. The leading Schwarzschild light deflection isWith the gravitational constant $G\simeq 6.6743\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and speed of light $c\simeq 2.998\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$,Here $GM/(R{c}^2)\simeq 2.5\times 10^{-4}\ll 1$, so the weak-field approximation is appropriate. A ray passing at a larger impact parameter has a correspondingly smaller deflection.

For $N\ge 2$, each adjacent pair of walkways has relative rapidity ${\varphi }_N=\mathrm{artanh}(1/N)$. There are $N$ such steps between $P_0$ and $P_N$, so the equal-rapidity chain of collinear Lorentz boosts has total rapidity $N{\varphi }_N$. Thus the actual velocity isIt is strictly below the speed of light for every finite $N\ge 2$. Adding the adjacent velocities as ordinary numbers ignores the relativistic velocity-addition formula.

For large $N$, the Taylor series of the inverse hyperbolic tangent givesConsequentlyMore precisely, $v_N/c=\tanh 1+{\mathrm{sech}}^2(1)/(3N^2)+O(N^{-4})$, so the limit is approached from above. The case $N=1$ would prescribe an adjacent relative velocity equal to the speed of light; it does not define an admissible walkway inertial frame.

When both speeds are much smaller than $c$, expand the denominator of the velocity-addition formula:This is the Galilean transformation limit. For $0\le \beta =v/c<1$ and $0\le {\beta }^{\prime }=v^{\prime }/c<1$,If both $v,v^{\prime }\to c$ from below, the right side tends to zero, soNo sum of these subluminal speeds exceeds the speed of light.