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In the original rest frame each leg takes $d/c$, so the round trip takes $2d/c$. For the moving frame, use the coordinate differences between the same three events. The Lorentz transformation givesThe outbound leg is shorter because the mirror approaches the outgoing ray; the inbound leg is longer because the source retreats from the returning ray. Their sum is larger than the original round-trip proper time by the time dilation factor $\gamma$.

These are coordinate times for return to the original source, not a claim that the pulse returns to the second observer's body at event $C$. If that observer starts at $A$, it intercepts the returning light at a different event, earlier than $C$, at its own time $2\gamma d(1-\beta )/c$. Distinguishing those events is necessary when interpreting “seen” or “return”.

Use the inertial frame in which the departure point and destination are stationary, with separation $d$. The astronauts' elapsed proper time is $\tau =t/\gamma$, whereas $d=vt$. HenceThe Earth-frame travel time is $t=\sqrt{41}\mathrm{yr}\simeq 6.40\mathrm{yr}$ and the speed is approximately $0.625c$. The spacecraft's relativistic kinetic energy is thereforeThe five years are not the time to use in $v=d/t$ in the departure frame. This is the cruising kinetic energy, with departure and arrival acceleration intervals omitted, rather than a complete propulsion energy budget.

Belinda remains in one inertial frame with $v=5c/13$, so the time dilation factor is $\sqrt{1-v^2/c^2}=12/13$. Her elapsed proper time isThe reunion event is shared by both twins, making their clock readings directly comparable without any convention for synchronizing separated moving clocks.

Alison's outbound proper time is $(13/5)(12/13)=12/5$ years. Her return Earth-frame duration is $T-13/5=26/7$ years, with time dilation factor $\sqrt{1-(12/13{)}^2}=5/13$. Adding the two segments,Alison is two years younger than Belinda at reunion. Her turnaround changes inertial frames, and the faster return segment accumulates less proper time per unit Earth-frame time. The twin paradox is resolved by comparing the complete worldlines, not by applying one constant time dilation factor to Alison's whole journey.

Einstein's postulates state that the laws of physics have the same form in every inertial frame, and that light in vacuum has the same speed of light $c$ in every inertial frame, independently of the motion of its source. For a primed frame moving with velocity $v{\mathbf{e}}_x$ relative to an unprimed frame, with coincident origins at $t=t^{\prime }=0$, the Lorentz transformations areThe following worldlines use Earth-frame time in years and distance in light years, so numerically $c=1$. The instantaneous turnaround is the idealized limit of a short acceleration stage; the elapsed proper time is calculated along each inertial segment.

Interpret the quoted muon energy as total relativistic energy, and neglect energy loss in the atmosphere. The Lorentz factor, speed and Earth-frame travel time areThe muon decay law is exponential in proper time. Equivalently, time dilation makes the Earth-frame mean lifetime $\gamma {\tau }_0$. The exponential survival probability over the flight isWith $h=15000\mathrm{m}$ and ${\tau }_0=2.2\times 10^{-6}\mathrm{s}$, the elapsed proper time is $8.841\times 10^{-7}\mathrm{s}$. ThereforeThis uses the stated single incident energy and vertical trajectories. If the number were a kinetic energy, one would instead use $\gamma =1+T/(m{c}^2)$; the distinction is not silently absorbed into the rest energy.

With metric convention $(+---)$, the invariant interval is $c^2\Delta {\tau }^2=c^2\Delta t^2-|\Delta \mathbf{r}{|}^2$ along an inertially moving clock. For constant speed $v$ in the ground frame, $|\Delta \mathbf{r}|=v\Delta t$, soThe aircraft records proper time $\Delta \tau =1000\mathrm{s}$. The ground duration is measured by a network of clocks at rest in the ideal ground frame, with Einstein synchronisation. Its excess isThus the ground-frame duration is longer by about $0.50\mathrm{ns}$. This is the special-relativistic time dilation contribution under the stated constant-speed assumptions; gravitational potential differences and Earth's rotation are outside this model.

First work in the receiver's rest frame. If a source recedes at speed $\beta c$, two successive emissions separated by source proper time $T_0$ occur $\gamma T_0$ apart in this frame. During that interval the source retreats by $\beta c\gamma T_0$, so the arrival interval is $T_R=\gamma T_0(1+\beta )$. The longitudinal relativistic Doppler effect is consequentlyA negative $\beta$ describes approach and gives a blueshift.

Interpret $633\mathrm{nm}$ as the laser's rest-frame wavelength and the stated detector velocity as measured in the initial laboratory frame. Choose $+x$ from source to sensor, with detector velocity $-0.2c$. The required relative recession speed follows fromBy relativistic velocity addition, the source velocity ${\beta }_{S}c$ in the laboratory obeys $({\beta }_S+0.2)/(1+0.2{\beta }_S)=-{\beta }_{\mathrm{rec}}$. HenceThe laser must move away from the sensor at about $0.5784c$ in that laboratory frame. Although the sensor moves toward the source's initial location, the retreating source produces the required net redshift. The frame qualification matters: a velocity permanently fixed relative to the moving source would specify the Doppler factor already, rather than leave it adjustable.

The laboratory is the parent's rest frame. In this relativistic two-body decay, conservation of momentum gives equal and opposite daughter momenta, of magnitude $p$. Their total relativistic energies satisfySubtracting the relativistic energy-momentum relations eliminates $p$:Solving the two linear equations givesSince three-momentum obeys $p{c}^2=Ev$, the daughter speeds areTheir directions are opposite; the initial state does not specify their common axis. The kinetic energies are $K_1=5m{c}^2/8$ and $K_2=3m{c}^2/8$. Their sum $m{c}^2$ is the loss of rest energy when the total daughter rest mass is $3m$ instead of $4m$.

The proper mean lifetime ${\tau }_0$ is the mean lifetime measured in the unstable particle's own rest frame, or equivalently the mean elapsed proper time before decay. In the exponential decay model, its survival probability to proper time $\tau$ is $e^{-\tau /{\tau }_0}$. For the daughter of rest mass $m$, the Lorentz factor is ${\gamma }_1=E_1/(m{c}^2)=13/8$, so time dilation gives a laboratory mean lifetime of ${\gamma }_1{\tau }_0=3.25\mu \mathrm{s}$.

At constant speed $v_1$, travelling a distance $L$ requires laboratory time $L/v_1$ and proper time $L/({\gamma }_1{v}_1)$. Its mean decay length isfor ${\tau }_0=2\times 10^{-6}\mathrm{s}$. HenceThe mean lifetime sets the scale of an exponential distribution; it is not a fixed deadline, so travelling beyond one mean decay length has a nonzero probability.

Take inertial axes parallel, with frame $S^{\prime }$ moving at $+v\hat{\mathbf{x}}$ relative to $S$ and origins coincident at zero time. The Lorentz factor is ${\gamma }_v=(1-v^2/c^2{)}^{-1/2}$; $\Delta x,\Delta y,\Delta t$ are separations of the same pair of events in $S$, and primed quantities are their separations in $S^{\prime }$. The transverse coordinates are unchanged. For two events on an inertial clock's worldline,The proper time $\Delta \tau$ is that clock's elapsed time; uniform speed $w$ gives $\Delta t={\gamma }_w\Delta \tau$. Coordinate intervals between spatially separated events must instead be Lorentz-transformed with their spatial separation included.

For the round trip in the station frame, the rendezvous distance is $u{t}_{\mathrm{out}}$, and the return has the same speed and distance. Thus $t_{\mathrm{round}}=2t_{\mathrm{out}}=10a/(\sqrt{7}c)$. Its departure and arrival both occur at the station, so $\Delta x=0$ for these endpoints:This uses the return speed as measured by the station. The elapsed time aboard A, apart from an ideal instantaneous turn, would be $\Delta t_{S,\mathrm{round}}/{\gamma }_u$.

For the crash, A's station-frame velocity components are $(3c/5,\sqrt{7}c/5)$, while B's are $(3c/5,0)$. Here ${\gamma }_u=5/3$ and ${\gamma }_v=5/4$. Four-momentum conservation, with no separately escaping radiation, gives the wreckage totals in the station frame:In B's original inertial frame the same totals areThe invariant wreckage mass is $m\sqrt{14/3}$, exceeding $2m$ because internal energy remains in the wreckage. Its momentum cannot be specified without naming a frame; both natural frames have been given.

In standard configuration the frames are inertial, their corresponding axes are parallel, S${}^{\prime }$ moves at $v$ along the positive $x$ axis of S, and their origins coincide at $t=t^{\prime }=0$. Here $(t,x,y,z)$ and $(t^{\prime },x^{\prime },y^{\prime },z^{\prime })$ label the same event, $c$ is the speed of light and $\gamma =(1-v^2/c^2{)}^{-1/2}$.

A clock fixed in S${}^{\prime }$ has $\Delta x^{\prime }=0$. The inverse time transformation givesso the moving clock accumulates less proper time between its two ticks than the elapsed S-coordinate time.

Without time dilation the flight time from the given altitude would be $66.7\mu \mathrm{s}$, about $30.3$ rest-frame mean lifetimes, giving survival fraction only $e^{-30.3}\simeq 6.8\times 10^{-14}$ even at speed $c$. Relativistic muons can instead reach the ground with appreciable survival. Momentum-selected decay measurements test this clock-rate dependence experimentally.^[ref]

A mean-lifetime survival energy estimate requires a mean flight time no greater than the dilated mean lifetime:This corresponds to survival probability at least about $e^{-1}$. There is no sharp minimum energy for an individual random decay. For a required probability $p$, use $\exp [-H/(\gamma v{\tau }_0)]\ge p$. Energy loss in the atmosphere is neglected in this estimate. In the muon's frame the atmospheric path is length-contracted to $H/\gamma \simeq 660\mathrm{m}$; the ground reaches it in about ${\tau }_0$, giving the same survival probability.

For the two-body decay, use the rest frame of A. Momentum conservation makes B and the photon oppositely directed with equal momentum magnitude $p$. Energy conservation and the energy–momentum relation giveSubstitution and cancellation of $E_B^2$ giveA nonzero emitted photon requires $m_A>m_B$.

Use the PDF's notation $f_0$ for the source-frame pulse frequency, and set $\beta =v/c$, $\gamma =(1-{\beta }^2{)}^{-1/2}$. Successive emissions occur at the same train-frame position and are separated by proper time $1/f_0$. The Lorentz transformation gives a station-frame emission interval $\Delta t_e=\gamma /f_0$ and displacement $v\Delta t_e$. For a receding beacon, the second pulse travels an extra distance $v\Delta t_e$, soThis derivation includes both time dilation and the change in light-travel time. Expanding for $\beta \ll 1$ gives $f/f_0=1-\beta +O({\beta }^2)$, recovering the stated nonrelativistic approximation.

Let ${\lambda }_0$ be the common wavelength of the identical lights at rest. The approaching and receding relativistic Doppler effect factors giveTheir ratio and product therefore determine the speed and rest wavelength separately:No assumption about the rest color was needed.

The last observation requires a transverse-geometry convention. The general observer-frame ray formula iswhere $\theta$ is the angle between source velocity and photon propagation in the platform frame at emission. Under the usual transverse-ray interpretation, $\theta =\pi /2$, so for $\beta =0.6$ and $\gamma =1.25$,This is the standard transverse redshift from time dilation.

There is a genuine distinction between emission and reception at the instant of passing. A side-mounted source emitting perpendicularly in its own frame instead has $\cos \theta =\beta$ by aberration of light, giving $\lambda ={\lambda }_3^{\prime }/\gamma =320\mathrm{nm}$. The same blue result occurs if “at the exact moment” means reception when the source's current position is at closest approach, with a small finite transverse offset. For offset $d$, reception and emission satisfy $t_r=t_e+\sqrt{v^2{t}_e^2+d^2}/c$. Setting $t_r=0$ gives $t_e=-\gamma d/c$ and $\cos \theta =\beta$; setting $t_e=0$ instead gives the transverse ray received at $t_r=d/c$. These limits remain different as $d\to 0$.

Thus 500 nm requires the transverse ray in the platform frame; 320 nm applies to reception at closest approach or transverse emission in the train frame. The wording does not distinguish these event/frame choices, so merely neglecting the source-observer separation is insufficient to select between them. The reusable derivation is transverse Doppler shift and observation time.

The spacecraft clock measures proper time $\tau$, while a clock at rest with Earth measures the launch-to-emission interval $t$. Neglect Earth's subsequent motion and gravity, as in the intended special relativity model. The Lorentz factor and time dilation giveIn Earth's inertial frame, the spacecraft's displacement at emission is $d=vt$. ThereforeHere $\tau =43200\mathrm{s}$ and the speed of light is $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$. The distance is evaluated at the emission event, using simultaneous Earth-frame endpoints, as required by relativity of simultaneity; the return signal has not yet reached Earth. A distance assigned on the spacecraft's simultaneous time slice is different: Earth has moved away at $c/3$ for $12\mathrm{h}$ in that frame, giving $c\tau /3=4$ light-hours. Specifying the reference frame is essential to the numerical answer.

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

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

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

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

Equating the lander and return-signal positions givesThis is measured from launch, so about 25.18 months pass on Earth-frame synchronized clocks. D occurs slightly before the lander reaches Earth at $t=74/35\mathrm{yr}$.

The two events A and D lie on the inertial lander's own worldline, and their proper time difference isEquivalently, applying the Lorentz transformation with $\Delta x=u\Delta t$ gives $\Delta t^{\prime }={\gamma }_u\Delta t(1-u^2/c^2)=\Delta t/{\gamma }_u$.

The proper time satisfies $dt/d\tau =\gamma (u)$. The collinear transformation from the Earth frame into the instantaneous rest frame gives proper acceleration $f={\gamma }^{3}du/dt$, soStarting from rest, integration gives $\mathrm{artanh}(u/c)=f\tau /c$. HenceSince $dx/d\tau =u\gamma =c\sinh (f\tau /c)$, a further integration givesUsing $\tau =10$ Julian years and $f=1\mathrm{m}{\mathrm{s}}^{-2}$ gives $f\tau /c\simeq 1.053$, so the star is about $5.77$ light years away in the Earth frame. The Earth-frame travel time is about $11.95$ years and the final speed is about $0.783c$. The model accelerates toward the star throughout; adding a braking leg would be a different itinerary.

Three experimental tests probe different consequences of special relativity. The Michelson-Morley experiment compares light round trips along perpendicular interferometer arms while changing the apparatus orientation. Its small observed orientation-dependent fringe shift contradicts the simple stationary-aether prediction and supports orientation-independent round-trip light speed. It does not, alone, exclude every alternative theory.^[ref]

In the muon storage-ring time-dilation experiment, fast circulating muons live longer in laboratory time than muons at rest. Measuring decay counts versus elapsed laboratory time and independently determining the beam Lorentz factor tests ${\tau }_{\mathrm{lab}}=\gamma {\tau }_0$. Circular motion includes acceleration, so this also tests the applicability of accumulated proper time along the stored trajectory.^[ref]

Bertozzi's electron speed-energy experiment measures electron time of flight and deposited kinetic energy independently. An electrostatic accelerator and a linear accelerator increase the energy while the measured speed approaches the light-speed limit. Calorimetry checks that the additional beam energy is real, rather than inferring it only from an assumed voltage-to-speed relation. The result supports relativistic $K=(\gamma -1)m{c}^2$ rather than extending $K=m{v}^2/2$ to arbitrarily large speeds.^[ref]

For the two ships, set launch time to zero and measure Earth time in years. Their Lorentz factors are ${\gamma }_A=5/4$ and ${\gamma }_B=5/3$. Bob's thirtieth birthday occurs after one year of his proper time, so its Earth coordinates areIf Alice sends at Earth time $t_A$, her position is $x_A=3c{t}_A/5$. The left-going light signal obeys $x=x_A-c(t-t_A)$. Intersection with Bob givesAlice's elapsed proper time is thereforeShe sends two months after her twenty-ninth birthday. This compares the emission and reception events; the captains' clocks need only be synchronized at launch, not kept numerically synchronized afterward.

The relative recession speed isThe longitudinal relativistic Doppler effect then givesEquivalently, the left-going photon has Earth frequency $f_0/[{\gamma }_A(1+3/5)]=f_0/2$, and Bob's frame supplies another factor ${\gamma }_B(1-4/5)=1/3$. Both methods give the same redshift.

Take overdots in this question to mean differentiation with respect to the orbiting particle's proper time $\tau$. Write $\mu =GM/c^2$, $x=r/\mu$ and $\ell =hc/(GM)$, with $h$ the specific angular momentum. Then the dimensionless timelike effective potential isThe last term is the relativistic correction to the Newtonian centrifugal-plus-gravitational potential. At large $r$ the two agree. Newtonian angular momentum gives an infinite inner centrifugal barrier for nonzero $h$; the relativistic correction lowers that barrier and permits plunge. The exterior problem starts at $r=2\mu$, where $U=-c^2/2$; extrapolating the potential towards $r=0$ is not a static exterior description.

Circular orbits obeyThere are two extrema when $h^2>12(GM/c{)}^2$: an inner maximum representing an unstable circular orbit and an outer minimum representing a stable one. At equality they merge at $r=6\mu$; below it neither exists. More explicitly, evaluate the curvature at a circular orbit:Timelike circular orbits require $r>3\mu$. They are radially stable for $r>6\mu$, so the marginally stable inner limit is $r_{\mathrm{ISCO}}=6GM/c^2$. The Newtonian potential instead has one stable minimum at $r=h^2/(GM)$, without an innermost stable circular orbit.

For the Schwarzschild circular-orbit local speed in the question, $\gamma$ is the Lorentz factor relative to a local Schwarzschild static observer, not $dt/d\tau$. That observer's clock obeys $dT=\sqrt{1-2\mu /r}dt$. Since the particle passes with local speed $v$, $rd\varphi /d\tau =\gamma v$. Circular normalization givesThus $v=c/2$ at the innermost stable circular orbit. Using Schwarzschild time in place of the local clock would give a different speed.

For the conserved particle energy measured at infinity, substitute $h^2$ in the circular energy equation. Setting $y=\mu /r$ givesTake the positive root for a future-directed exterior orbit. It approaches $m{c}^2-GMm/(2r)$ at large radius. The marginally bound circular Schwarzschild orbit solves $k=1$, giving $y(4y-1)=0$ and hence $r_{\mathrm{mb}}=4GM/c^2$ at finite radius. It is unstable; strictly bound circular orbits have $r>4\mu$, with $E<m{c}^2$. The phrase smallest bound orbit here denotes this marginal circular limit, not a prohibition on a bound-energy trajectory plunging below it.

At the innermost stable circular orbit,A radiatively efficient thin disc with a zero-torque inner boundary condition can therefore release about 5.7% of the accreted rest energy before the plunge. energy advection, stress at the inner edge and photon capture can alter the escaping luminosity, so this is a model efficiency rather than a universal emitted fraction.

For the Kerr ISCO spin limits of an equatorial prograde disc around a rapidly rotating Kerr black hole, the innermost stable circular orbit moves inward and its specific energy decreases. In the ideal extremal limit its Boyer-Lindquist radius tends to $\mu$, $E/(m{c}^2)\to 1/\sqrt{3}$, and the corresponding binding-energy efficiency tends to 42.3%. Retrograde rotation has the opposite trend: the extremal limit is $r_{\mathrm{ISCO}}=9\mu$, $E/(m{c}^2)=5/(3\sqrt{3})$, giving about 3.8%. The prograde limiting coordinate radius does not identify a massive orbit with the null horizon generator.^[ref]

Finally $\dot{t}=k/(1-2\mu /r)=1/\sqrt{1-3\mu /r}$. Dividing $\dot{\varphi }=h/r^2$ by this givesIts algebraic form is exactly Newtonian, but $r$ is the Schwarzschild spacetime areal radius and $t$ Schwarzschild time, normalized at infinity. Local clock rates, orbit stability and binding energy remain relativistic. With $G{M}_{\odot }\simeq 1.3271\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$,The marginally bound circular-orbit frequency is about $0.404\mathrm{kHz}$. These Schwarzschild ISCO orbital frequencies and marginally bound frequencies are rotation frequencies per distant stationary observer's time, as intended by using $d\varphi /dt$. A static observer at finite radius $r_o$ instead divides both frequencies by $\sqrt{1-2\mu /r_o}$, leaving their ratio unchanged. Identifying an observed quasi-periodic oscillation with the innermost stable circular orbit orbital frequency is an additional astrophysical assumption.

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 $z\to 0$.

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.