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For a frame moving at velocity $V\hat{\mathbf{x}}$, put $\Gamma =(1-V^2/c^2{)}^{-1/2}$. The Lorentz transformation of the energy-momentum four-vector isExpansion givesThe transverse components are unchanged, hence $\boxed{E^{\prime 2}-c^2|{\mathbf{p}}^{\prime }{|}^2=E^2-c^2|\mathbf{p}{|}^2}$. For a particle of rest mass $m$ this invariant is $m^2{c}^4$; the same algebra also applies to the total four-momentum of a system.

Use the inverse Lorentz transformation, with ${\gamma }_u=(1-u^2/c^2{)}^{-1/2}$:Taking differentials along the object's trajectory givesThis is the collinear relativistic velocity addition law. It concerns the velocity of one object measured in two frames. By contrast the rate of change of the distance between two objects in one laboratory frame can be as large asattained by oppositely directed light signals; for massive objects it is approached but not attained. This is not an object's speed in the other's rest frame and cannot transmit a signal faster than $c$.

Let the source's proper emission period be $T_0$, and let it approach a stationary observer directly with speed $v=\beta c>0$. In the observer's inertial frame, consecutive emissions are separated by $\gamma T_0$ through time dilation. The second crest starts $v\gamma T_0$ nearer the observer, reducing the arrival interval toSince light travels at $c$ in each inertial frame, the longitudinal relativistic Doppler effect isRecession reverses the sign of $\beta$. More generally, if $\Theta$ is the angle between the source velocity and the photon propagation direction in the observer's frame, a Lorentz transformation of the photon four-momentum gives ${\nu }_0=\gamma {\nu }_{\mathrm{obs}}(1-\beta \cos \Theta )$.

The common Doppler scaling of spectral-line separations follows because every line receives the same wavelength factor $D$. The supplied wavelengths have rest separation $50\mathrm{nm}$; the measured separation gives $D=51/50=1.02$. Defining ${\beta }_{\mathrm{rec}}>0$ for recession, inversion givesSource contradiction: a strictly approaching source has $D<1$ and must reduce the wavelength separation below $50\mathrm{nm}$, so no positive approach speed satisfies the stated $51\mathrm{nm}$. For example, approach at $0.02c$ gives a separation of $49.0098\mathrm{nm}$. The speed inferred from the numerical data is therefore about $0.0198c$ receding; retaining the word “approaching” would require a negative signed approach velocity. The two observed wavelengths consistent with the given separation are $357$ and $408\mathrm{nm}$.

Let $S^{\prime }$ be the runner's inertial frame, with origin at the entry event. The Lorentz transformation isThe emission event $E=(L,0)$ has coordinates $x_E^{\prime }=\gamma L$, $t_E^{\prime }=-\gamma vL/c^2$. It is earlier than entry in $S^{\prime }$: entry and far-end emission are simultaneous only in the tunnel frame. This is relativity of simultaneity, not an extra delay of the light source.

The photon propagates along $-x$ in the tunnel frame and has energy $E_{\gamma }=hc/{\lambda }_0$ and momentum $p_x=-E_{\gamma }/c$. A Lorentz transformation into the approaching runner's frame givesSince $E_{\gamma }^{\prime }=hc/{\lambda }^{\prime }$, the approaching-source relativistic Doppler effect isIt is a blueshift. With $q={\lambda }^{\prime }/{\lambda }_0=500.0/589.1$, solving $q^2=(1-\beta )/(1+\beta )$ givesThis shift depends on the relative velocity along the light ray, not on the tunnel length or the location of detection.

Align both frames' $x$ axes with the motion and make their origins coincide at $t=t^{\prime }=0$. With $S^{\prime }$ moving along $+x$ at speed $v_2$ relative to $S$, the Lorentz transformation and its inverse arewhere ${\gamma }_2=(1-v_2^2/c^2{)}^{-1/2}$; the transverse coordinates are unchanged. Along the object's path, $d{x}^{\prime }/d{t}^{\prime }=v_1$. Taking differentials of the inverse transformation givesThis is relativistic velocity addition. For $v_1,v_2\ll c$ it reduces to the Galilean sum; if $v_1=c$, the result is still $c$. Two subluminal positive speeds give a subluminal result.

Energy–momentum transformations and invariant. Let the primed inertial frame move at signed velocity $u$ along the unprimed $x$ axis. With $\Gamma =(1-u^2/c^2{)}^{-1/2}$, a Lorentz transformation of four-momentum givesUse $p^2=p_x^2+p_y^2+p_z^2$. The longitudinal terms satisfyThe transverse momenta are unchanged, soFor one electron, the relativistic energy-momentum relation identifies this Lorentz invariant as $E_0^2=m_e^2{c}^4$.

Symmetric electron scattering. Assume an isolated elastic collision of the two electrons, with no emitted radiation. The incident electron has total energy $E$ and momentum magnitude $p$, so $p^2{c}^2=E^2-E_0^2$; the target electron initially contributes rest energy $E_0$ and zero momentum. In the nondegenerate configuration with final track angles $+\theta$ and $-\theta$, transverse momentum conservation givesBecause both final particles have the same rest mass, they also have equal total energies. Conservation of energy givesLongitudinal momentum conservation is $p=2p_f\cos \theta$. Combining these expressions, with $E>E_0$, proves the symmetric relativistic equal-mass scattering angle relation:The low-kinetic energy limit has $\theta \to 45^{\circ }$, recovering the familiar right angle between the outgoing tracks in nonrelativistic equal-mass scattering. Relativistic incident energies give $\theta <45^{\circ }$, approaching zero at very high energy.

Accelerating potential. For $\theta =30^{\circ }$, ${\tan }^2\theta =1/3=2E_0/(E+E_0)$, so $E=5E_0$. The acceleration from rest supplies kinetic energy, not total energy. If $e>0$ is the elementary charge, the required accelerating electric potential difference has magnitudeusing electron rest energy $E_0\simeq 0.511\mathrm{MeV}$. The electric potential difference is oriented to accelerate the negative electron; the result above specifies its magnitude.

Write $m_e$ for the electron rest mass and $\epsilon =m_e{c}^2$. The incident positron has total energy $E$ and satisfies $p^2{c}^2=E^2-{\epsilon }^2$. Adding the target electron's four-momentum, the squared total centre-of-momentum energy isAt the fixed-target production threshold, all three final particles are at rest relative to one another, so $W=3M_0{c}^2$. ConsequentlyThis is the positron's total energy, not its kinetic energy. It assumes $3M_0\ge 2m_e$ and addresses kinematic allowance; other selection rules could still forbid a proposed reaction. For $M_0{c}^2=135\mathrm{MeV}$ and $m_e{c}^2\simeq 0.510999\mathrm{MeV}$, $E_{\min }\simeq 160.494\mathrm{GeV}$.

At threshold the three products share the centre-of-momentum frame's laboratory velocity. Their common Lorentz factor and speed areFor the subsequent two-photon decay, each photon has rest-frame energy $E_{\gamma }^{\ast }=M_0{c}^2/2$. A Lorentz transformation of a photon emitted at rest-frame angle ${\theta }^{\ast }$ to the boost gives $E_{\gamma }={\gamma }_{\mathrm{CM}}{E}_{\gamma }^{\ast }(1+{\beta }_{\mathrm{CM}}\cos {\theta }^{\ast })$. Therefore the kinematically allowed laboratory range isor numericallyThe two photon energies in a single event are correlated: their sum is the parent's laboratory energy ${\gamma }_{\mathrm{CM}}{M}_0{c}^2$. When one attains the upper endpoint, the other attains the lower endpoint. Angular distributions and detector acceptance can affect which part of this range is observed.

Let $\theta$ describe the direction towards the star, measured from the Earth's positive direction of motion. The arriving photon propagates in the opposite direction, so in the stellar frame its four-momentum hasA Lorentz transformation into the Earth's frame, moving at $\beta c$ along $x$, givesThe observed stellar bearing is opposite to the transformed photon momentum. Since a photon satisfies $|{\mathbf{p}}^{\prime }|=E_{\gamma }^{\prime }/c$,This is relativistic aberration. For small $\beta$, ${\theta }^{\prime }=\theta -\beta \sin \theta +O({\beta }^2)$: stellar directions shift towards the direction of the observer's motion. Keeping the distinction between stellar bearing and photon propagation direction explains the plus signs.

For a Lorentz boost along $x$, the Lorentz transformation isTo calculate a photograph, compare photons that arrive together, which were generally emitted at different times. Put the lamina in the $xy$ plane, moving along $+x$, and take the distant camera to lie along negative $y$. Let $(X,Y)$ be a material point's coordinates in the lamina's rest frame. At a laboratory time $t$, its position satisfies $x=X/\gamma +vt$, $y=Y$.

The distant-camera approximation makes the variable part of the light travel time $Y/c$. Equal arrival time therefore requires $t+Y/c=\text{constant}$. Choose that constant to be zero. The point's photographed transverse coordinate becomesA stationary lamina rotated in its plane by $\theta$ would instead have the projected coordinate $X\cos \theta -Y\sin \theta$. The two projections agree whenThis is Terrell rotation. The shape defined by simultaneous laboratory positions still has length contraction; the photographic rotation comes from combining that contraction with unequal emission times. Reversing the viewing or motion direction reverses the apparent rotation's sign.

Use emission, reflection and return-to-source events $A,B,C$. In the source-and-mirror rest frame their $(ct,x)$ coordinates are $A=(0,0)$, $B=(d,d)$ and $C=(2d,0)$. The laser and mirror have vertical worldlines; the two light rays have slopes $\pm 1$ on a diagram with equal spatial and $ct$ scales.

For a frame moving with velocity $+v$ along the outgoing beam, let $\beta =v/c$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$. A Lorentz transformation givesBoth laser and mirror now move towards negative $x^{\prime }$, with worldlines $x^{\prime }=-v{t}^{\prime }$ and $x^{\prime }=d/\gamma -v{t}^{\prime }$. Light still has slopes $\pm 1$. The moving observer's own worldline is $x^{\prime }=0$; it should not be confused with the moving source worldline.

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”.

For a massive free particle with rest mass $m$ and velocity $\mathbf{v}$,The relativistic energy-momentum relation follows directly:This is the squared Minkowski norm of the four-momentum multiplied by $c^2$. The Lorentz transformation preserves that norm, so it has the same value in every inertial frame. In the particle's rest frame, the momentum is zero and the total energy is its rest energy $m{c}^2$.

Let $S^{\prime }$ move at $+v$ along $x$ relative to $S$. The Lorentz transformation of energy and momentum, from primed to unprimed components, isThe inverse equations have $v$ replaced by $-v$ and the primes interchanged. A photon making angle ${\theta }^{\prime }$ with $+x$ in $S^{\prime }$ has $p_x^{\prime }=(E^{\prime }/c)\cos {\theta }^{\prime }$. Thus its general Doppler factor isThe requested blueshifted expression assumes propagation parallel to the boost, ${\theta }^{\prime }=0$:The propagation-direction assumption matters: for an oppositely travelling photon, $D=\gamma (1-\beta )$, the reciprocal of the displayed factor. Relative frame speed alone does not select the longitudinal relativistic Doppler effect.

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.

For a Lorentz transformation with $S^{\prime }$ moving at $+v$ along $x$ relative to $S$,Differentiate along the object's trajectory and divide the spatial differentials by $d{t}^{\prime }$. Using $u_x=dx/dt$ and $u_y=dy/dt$ derives the relativistic velocity-addition formula:In the rest frame of ship A, take $v=c/2$ and the Earth-frame velocity of B as $(0,-c/2)$. Here $\gamma =2/\sqrt{3}$ and the common denominator is one, soThe required relative speed is the magnitude of this two-component velocity, not either component separately:The inverse Lorentz factor check is ${\gamma }_{AB}={\gamma }_A{\gamma }_B(1-{\mathbf{u}}_A\cdot {\mathbf{u}}_B/c^2)=4/3$, which gives the same relative speed.

Let $L_0=100\mathrm{m}$ be the train's proper length, $D=90\mathrm{m}$ the tunnel's length in the track reference frame, and $v=0.99c$. The Lorentz factor is $\gamma =7.0888$. At one track-frame time, length contraction makes the trainYes, the whole train fits inside at a single track-frame time. Set $t=0$ when the rear enters the near portal. Until the front reaches the far portal, the rear and front are at $vt$ and $vt+L$, respectively. Complete containment lastsa track-frame interval of approximately $256\mathrm{ns}$, counting boundary contact.

In the train's reference frame, its length is $100\mathrm{m}$, while the moving tunnel has simultaneous length $D/\gamma =12.696\mathrm{m}$. No single train-frame time has the whole train between the portals. The two answers refer to different simultaneity slices.

For an explicit check, denote the rear-entry event by $A$ and the front-exit event by $B$. In track coordinates,The Lorentz transformation givesThus the front leaves before the rear enters in train coordinates. These spatially separated events have a frame-dependent order, as allowed by relativity of simultaneity; neither answer asserts a different local event at a portal. No collision with closed doors has been stipulated.

For a particle of rest mass $m_0$ and velocity $\mathbf{u}$, the relativistic energy and momentum areFor a photon of wavelength $\lambda$ travelling along $\hat{\mathbf{n}}$,Here $h$ is Planck constant and $c$ is the speed of light.

Let $\epsilon$ and ${\epsilon }^{\prime }$ be the incident and scattered photon energies. The stationary electron initially has energy $m_e{c}^2$. By four-momentum conservation, its final energy and momentum areUse the electron's relativistic energy-momentum relation, rather than assuming its recoil is nonrelativistic:Cancelling common terms givesSince $\lambda =hc/\epsilon$, the Compton wavelength shift isThe word elastic here means that the final electron has the same rest mass as initially and the complete photon-electron system conserves total energy; the photon alone loses energy to recoil.

For backscattering, $1-\cos \theta =2$, and henceChoose the incident direction as $+x$, so that the scattered photon travels in the $-x$ direction. An observer moving with velocity $+\beta c\hat{\mathbf{x}}$ meets that photon head-on. The Lorentz transformation of four-momentum gives the observed energyThis is the approaching-observer relativistic Doppler effect. Set $D=\epsilon /{\epsilon }^{\prime }=2.511/0.511$ and solve:The observer must move opposite to the scattered photon, along the original incident direction, at approximately $0.9205c$. Moving with the scattered photon would redshift it instead.

Orient the $x$ axis along the displacement from the earlier event to the later one, so $\Delta x=d=1000\mathrm{m}$ and $\Delta t=10^{-6}\mathrm{s}$. For a parallel Lorentz transformation, simultaneity in the second inertial frame requiresThe second inertial frame moves towards the later event's location. The positive value $d^2-c^2\Delta t^2$ establishes spacelike separation, which permits this change of simultaneity. For a general velocity $\mathbf{V}$, the condition is $\mathbf{V}\cdot \Delta \mathbf{r}=c^2\Delta t$: only its component along the separation is fixed at the displayed value, and transverse components are allowed provided $|\mathbf{V}|<c$. Thus the parallel velocity has the smallest possible speed.

A physical inertial frame must have $|v|<c$, givingThere is no attained maximum: $d/c$ is the supremum. For the concrete boundary case $\Delta t=1000\mathrm{m}/c$, the events have null separation and the formula would require $v=c$; no inertial frame can make them simultaneous.

Invariant and collision. For a particle of rest mass $m$, special relativity gives relativistic energy and momentumTherefore the relativistic energy-momentum relation isIt is the squared Minkowski norm of the four-momentum, multiplied by $c^2$, and hence is preserved by every Lorentz transformation. Explicitly a boost with speed $U$ along $x$ gives $E^{\prime }=\Gamma (E-U{p}_x)$, $p_x^{\prime }=\Gamma (p_x-UE/c^2)$, with $p_y^{\prime }=p_y$, $p_z^{\prime }=p_z$. Expanding $E^{\prime 2}-c^2{p}_x^{\prime 2}$ gives $E^2-c^2{p}_x^2$ because ${\Gamma }^2(1-U^2/c^2)=1$. Spatial rotations preserve $p^2$, completing the invariance argument for all inertial frames.

For the incoming moving particle, $v=3c/5$ and $\gamma =5/4$. The total laboratory energy and momentum before the perfectly inelastic collision areFour-momentum conservation identifies these with the final particle's energy and momentum. Its rest mass is thereforeThe result exceeds the sum $3m_0$ of the incoming rest masses; some initial kinetic energy has become internal rest energy. Since $p{c}^2/E=v$ for a massive particle,

Two-body decay. Work first in the parent's rest frame, denoted by a star. Four-momentum conservation requires equal and opposite daughter momenta. Their equal rest masses give equal energies:Thus their rest-frame velocities are $v_{\ast \pm }=\pm 3c/\sqrt{10}$. Applying the relativistic velocity-addition formula with boost $v^{\prime }=3c/13$ gives the two laboratory velocities:Their laboratory energies follow directly from the Lorentz transformation of four-momentum:The corresponding signed laboratory momenta areTheir sums recover $E_i$ and $p_i$, and each satisfies $E_{\pm }^2-p_{\pm }^2{c}^2=(m_0{c}^2/2{)}^2$. One daughter moves forwards and the other backwards in the laboratory; both are collinear with the original flight axis. Numerically, $v_{+}\simeq 0.967615c$, $v_{-}\simeq -0.919138c$, $E_{+}\simeq 1.980756m_0{c}^2$ and $E_{-}\simeq 1.269244m_0{c}^2$.