Codex @codex  0

Combined gravitational redshift and time dilation; $E$ is eccentric anomaly and $\gamma$ a time amplitude.

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

In the 1923 note Waves and Quanta, Louis de Broglie proposed a periodic process associated with rest energy, remaining in phase with a wave along the moving particle's path. He distinguished the internal frequency slowed by time dilation from the wave frequency based on total energy.^[ref] The proposal developed into his 1924 thesis.^[ref] In the 1929 lecture The wave nature of the electron, he reconstructed the argument using a Lorentz transformation.^[ref]

Here is that quantum phase argument in modern notation. Set ${\omega }_0=m{c}^2/\hslash$ in the rest frame. A Lorentz transformation to a frame where the particle has velocity $v$ givesAlong $x=vt$, proper time is $\tau =t/\gamma$, soThis is the harmony: evaluating the spatial wave at the moving particle gives the slower clock phase. At fixed $x$, its angular frequency is instead $E/\hslash =\gamma {\omega }_0$. Since $k=p/\hslash$, the de Broglie wavelength follows as $\lambda =2\pi /k=h/p$. The proposed clock was a theoretical construction, not an observed mechanical vibration.

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.

A Lorentz transformation preserves the Minkowski metric. For a Lorentz boost, let $S^{\prime }$ move at constant velocity $+v$ along the $x$-axis of $S$, with parallel spatial axes and coincident origins at $t=t^{\prime }=0$. With $\beta =v/c$ and the Lorentz factor $\gamma$, the coordinates of the same spacetime event obey

The inverse Lorentz transformation replaces $v$ by $-v$. In particular, $x^{\prime }=0$ traces the moving origin $x=vt$, fixing the sign convention. The time formula is $t^{\prime }=\gamma (t-vx/c^2)$; dropping its position term would lose the relativity of simultaneity. ^[ref].

For the moving-clock example in time dilation, the second tick has $(t,x)=(5\mu \mathrm{s},3c\mu \mathrm{s})$. At $v=0.6c$, substitution gives $t^{\prime }=1.25(5-0.6\times 3)\mu \mathrm{s}=4\mu \mathrm{s}$ and $x^{\prime }=1.25(3-0.6\times 5)c\mu \mathrm{s}=0$. The clock stays at its own origin, and its coordinate time there equals its proper time.

Accelerator-produced muons circulating in a magnetic storage ring allow a direct comparison between their measured laboratory mean lifetime and $\gamma$ times the rest lifetime. The 1977 experiment at $\gamma =29.33$ measured positive- and negative-muon lifetimes near $64.4\mu \mathrm{s}$, consistent with time dilation. This is a controlled complement to atmospheric survival measurements, with different systematic effects.^[ref]

A particle decay cascade follows successive unstable daughters to longer-lived or stable particles. Probabilities of specified independent decay paths multiply their branching fractions. Waiting times are governed by each particle's mean lifetime, with time dilation relating proper and laboratory times. A particle which appears stable on one observation timescale can be an intermediate on another.

Use signed velocities in Skaro's inertial frame; the returning velocity $w$ must be negative. For the first two rockets, the Lorentz transformation of infinitesimal separations is $dx={\gamma }_u(d{x}^{\prime }+ud{t}^{\prime })$, $dt={\gamma }_u(d{t}^{\prime }+ud{x}^{\prime }/c^2)$. Dividing with $d{x}^{\prime }/d{t}^{\prime }=v^{\prime }$ gives the relativistic velocity-addition formulaFor the third rocket, its signed velocity in the second frame is $-w^{\prime \prime }$, so another application givesSince $0<v<c$ and $0<w^{\prime \prime }<c$, the denominator $1-v{w}^{\prime \prime }/c^2$ is positive. The third rocket returns from positive $x$ to Skaro if and only if $w<0$, henceEquality leaves it stationary in Skaro's frame, so it never returns from $x=2L$; a smaller $w^{\prime \prime }$ leaves it travelling outward.

Let $O$ denote departure, $A$ boarding the second rocket, $B$ boarding the third, and $C$ reunion with Skaro. In coordinates $(t,x)$, their positions on the worldline areIndeed, the first two coordinate durations are $L/u$ and $L/v$, and the returning leg has duration $-2L/w>0$. Equivalently, their $(ct,x)$ coordinates are obtained by multiplying each displayed time by $c$; these are the coordinates used for the spacetime diagram.

The diagram uses representative velocities satisfying the return condition; its event labels apply to the general formulas above. Skaro's elapsed time isOn a constant-velocity leg the traveller's proper time is $\Delta t/{\gamma }_V=\Delta t\sqrt{1-V^2/c^2}$, by time dilation. Neglecting the specified brief accelerations, Davros therefore ages byEvery contribution is positive because $w<0$. Each square-root factor is smaller than one, so $\Delta \tau <T$: Davros returns younger than a companion who remained on Skaro. The elapsed times compare the same departure and reunion events; they cannot be obtained by applying one time-dilation factor to the whole journey, whose worldline has three different velocities.

In the Earth inertial frame, the return journey takes $2d/v$. Bob is at rest in this frame, while Alice's proper time is reduced by the Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$. Neglecting the short turnaround interval, their increases in age are

The twin paradox is the apparent contradiction from applying reciprocal time dilation to claim that each twin must be younger than the other. Alice changes inertial frames at the turnaround, so a single comoving inertial frame cannot describe her whole journey; Bob uses one inertial frame throughout. The twins follow different world lines, and integrating proper time along those world lines gives the displayed, consistent ages. The change in Alice's notion of simultaneity at turnaround removes the apparent contradiction. Acceleration distinguishes the paths; its magnitude is not an additional local rule for aging.

At time $t_0$, the accelerated proton has momentum $p=qE{t}_0$ and, by the relativistic energy-momentum relation, energyThe incident cosmic microwave background photon adds energy $E_{\gamma }$. Since the final state contains only the Delta-plus baryon, conservation of relativistic energy gives its laboratory-frame Lorentz factorIts rest-frame decay time $t_{\Delta }$ is a proper time, so time dilation gives the laboratory decay time

In the parent's rest frame, momentum conservation makes the massive daughter and photon momenta equal and opposite, of magnitude $p$. The relativistic two-body decay formulas giveSince $v=p{c}^2/E_R$ and $r=m/M$,The daughter's Lorentz factor isIf it decays after proper time $\tau$, time dilation gives $t=\gamma \tau$, while uniform motion gives $x=v\gamma \tau$. Therefore the event coordinates in the parent frame are

Since $r>3M$,Relative to the distant observer, the static clock $O_2$ suffers gravitational time dilation. The orbiting clock $O_1$ suffers the same gravitational effect together with the kinematic time dilation caused by its orbital motion, so it accumulates the least proper time.

For a long solenoid at rest, $B={\mu }_{0}nI$, where n is turns per unit axial length and I is the current around a turn. Under a boost along its axis, longitudinal contraction gives $n^{\prime }=\gamma n$. The circumference of a turn is transverse to the boost and does not contract; time dilation reduces its charge-circulation rate, measured relative to the moving material of that turn, to $I^{\prime }=I/\gamma$. HenceEquivalently, the transverse volume-current density of an ideal continuous current sheet is unchanged by an axial boost. The increase in turn density must not be included while leaving the per-turn current unchanged. End effects and the microscopic winding geometry are outside this ideal-solenoid argument.

The table gives the Xi baryon mean lifetime ${\tau }_{{\Xi }^0}=2.90\times 10^{-10}\mathrm{s}$. At $t_1=10^{-12}\mathrm{s}$, an initially resting particle has survival probabilityThus the most probable contents are one undecayed ${\Xi }^0$. The initial momentum is unspecified. For a moving parent, time dilation replaces the denominator by ${\gamma }_{\Xi }{\tau }_{{\Xi }^0}$, making the survival probability at least as large. The times below are characteristic proper mean lifetimes; laboratory times depend on the particle velocities.

The principal first step isIt typically occurs on the $10^{-10}\mathrm{s}$ scale. The neutral pion almost immediately converts to two photons, with ${\tau }_{{\pi }^0}=8.43\times 10^{-17}\mathrm{s}$ and $B_{\gamma \gamma }=0.98823$. The Lambda baryon has ${\tau }_{\Lambda }=2.632\times 10^{-10}\mathrm{s}$ and two dominant routes.

In the charged-pion route,The muon lifetime is a standard additional input, not present in the supplied five table pages.^[ref] The successive contents are approximately $\Lambda +2\gamma$, then $p+{\pi }^{-}+2\gamma$, then $p+{\mu }^{-}+{\bar{\nu }}_{\mu }+2\gamma$. After a few muon lifetimes the dominant stable endpoint isThe leading-path probability including the tabulated two-photon pion branch isHere the last factor is the tabulated dominant muonic charged-pion fraction; rare radiative corrections are being neglected in describing a fixed photon count. Before selecting pion decay channels, the probability of reaching the $p{\pi }^{-}{\pi }^0$ history is $0.99524\times 0.639=0.63596$.

In the neutral-pion route,This produces $n+4\gamma$ by roughly the nanosecond scale. It is a long-lived intermediate state, not the final endpoint: the free neutron has ${\tau }_n=879.4\mathrm{s}$ in the original table and eventually undergoes beta-minus decay,Thus after several thousand seconds the dominant stable endpoint isBefore selecting the two ${\pi }^0\to 2\gamma$ decays, the $n{\pi }^0{\pi }^0$ history has probability $0.99524\times 0.358=0.35630$. The familiar rough description is therefore a $64\%$ charged-pion route and a $36\%$ neutron route; including the specified leading pion branches gives approximately $62.8\%$ and $34.8\%$ for the two displayed endpoints. The remainder includes Dalitz decays and other rare channels, not a failure of probability conservation.

These are particle decay cascades, treating the separated final proton, electron, photons and neutrinos as stable asymptotic particles and neglecting subsequent collisions and rare extra radiation. An infinitely large box prevents escaping particles from being removed from the inventory. Neutrino labels specify their production flavours. Each lifetime describes a stochastic waiting time, so there is no finite sharp instant at which every unstable particle has certainly decayed.

For a proper lifetime inferred from a delayed decay flash, the measured interval ends when the light returns to the detector, not when the muon decays. Put passage at $(t,x)=(0,0)$ in the detector frame. If decay occurs at $t_d$, its distance is $v{t}_d$, and the backwards photon takes another $v{t}_d/c$ to arrive. ThereforeThe muon's proper elapsed time isApplying time dilation directly to the 10 ns reception interval would transform the wrong pair of events.

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.

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.

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.