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For one particle at a fixed Lorentz factor, the Liénard radiated-power formula gives $P_{\parallel }=C_q{\gamma }^6{a}_{\parallel }^2$ and $P_{\perp }=C_q{\gamma }^4{a}_{\perp }^2$, where $C_q=q^2/(6\pi {ϵ}_0{c}^3)$. Consequently equal coordinate acceleration magnitudes make transverse motion radiate less by ${\gamma }^2$. Equal applied laboratory force magnitudes instead give $a_{\parallel }=F/(m{\gamma }^3)$ and $a_{\perp }=F/(m\gamma )$, so $P_{\perp }/P_{\parallel }={\gamma }^2$. These opposite orderings concern distinct constraints and are consistent. An energy budget without a time, length, force, or acceleration constraint cannot by itself select a unique minimum-loss accelerator design.

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 a uniform magnetic field, a nonrelativistic particle of mass $m$ and nonzero charge $q$ completes one circular transverse orbit in time $T_c=2\pi /{\omega }_c$, where ${\omega }_c=|q|B/m$. Parallel velocity changes the helix pitch but not this period. In relativistic motion the constant Lorentz factor multiplies the period: $T_c=2\pi \gamma m/(|q|B)$. The direction of rotation depends on the charge sign; the period is positive.

An isotropic two-photon decay of a massive parent with laboratory Lorentz factor $\gamma >1$ has the displayed normalized opening-angle probability density, with $\beta =\sqrt{1-{\gamma }^{-2}}$ and $2\arcsin (1/\gamma )<A<\pi$. To derive it in units $c=1$, the laboratory photon energies are $E_{1,2}=\gamma M(1\pm \beta z)/2$, where $z$ is uniform on $[-1,1]$. The diphoton invariant mass gives ${\sin }^2(A/2)=1/[{\gamma }^2(1-{\beta }^2{z}^2)]$. Apply the change of variables for probability densities to both signs of $z$. Its minimum-angle peak is an inverse-square-root singularity. At $\gamma =1$, the probability distribution is concentrated at $A=\pi$.

For an incoming timelike Schwarzschild geodesic with asymptotic Lorentz factor $k\ge 1$, the critical unstable circular orbit obeysAt $k=1$, $r_c=4\mu$ and $|h_c|=4\mu c$. For $k>1$, the critical radius lies between $3\mu$ and $4\mu$ and $|h_c|>4\mu c$. The marginally bound angular-momentum bound remains sufficient at larger energies but is no longer the exact threshold. Origin at infinity and an inward initial branch are essential; an orbit already inside a barrier is a different initial value problem.

For an unbound equatorial Schwarzschild spacetime geodesic, let $E_{\infty }$ and $L_{\infty }$ be its conserved energy and angular momentum measured relative to infinity. The ratio $J=c^2{L}_{\infty }/E_{\infty }$ has dimensions of area divided by time. For a massive test particle, $E_{\infty }=m{c}^2{\gamma }_{\infty }$ and $L_{\infty }=m\ell$, so $J=\ell /{\gamma }_{\infty }=b{v}_{\infty }$, where $\ell$ is specific angular momentum and $b$ the impact parameter. For a photon, $E_{\infty }=c{p}_{\infty }$ and $L_{\infty }=b{p}_{\infty }$, giving $J=bc$ without assigning a finite Lorentz factor or a proper time to the photon. This ratio removes the arbitrary normalization of a null affine parameter.

For two inertial frames with equal speed in a laboratory, a photon is measured at equal angular frequency by their comoving observers exactly when its laboratory propagation unit vector $\mathbf{n}$ obeys the displayed condition. Indeed, measured angular frequency is $-k\cdot U=\gamma {\omega }_{\mathrm{lab}}(1-\mathbf{n}\cdot \mathbf{v}/c)$, and equal laboratory Lorentz factors cancel. Thus the permitted directions form a great circle on the unit sphere. In the plane of the two laboratory velocities, only the internal angle bisector line remains. The velocity information alone fixes neither direction along that line nor coplanarity of the ray.

An incident particle of rest mass $m$ and energy $E$ striking a stationary identical target has total squared invariant energy $W^2=2m{c}^2(E+m{c}^2)$. Producing $N\ge 2$ final particles of the same rest mass requires $W\ge Nm{c}^2$, giving the threshold. Equality means no final kinetic energy in the centre-of-momentum frame. From the Lorentz factor, the corresponding incident speed is at least $cN\sqrt{N^2-4}/(N^2-2)$.

For uniform circular motion in an inertial laboratory, the Lorentz factor is constant. With metric signature $+---$, four-acceleration is $A^{\mu }=(0,{\gamma }^2\mathbf{a})$ and proper acceleration has magnitude ${\gamma }^2{v}^2/r$. The instantaneous spatial acceleration is perpendicular to velocity, even though the particle has nonzero proper acceleration.

A rod of rest length $L_0$ moving parallel to its length has simultaneous laboratory length $L=L_0/\gamma$, where $\gamma$ is the Lorentz factor. Here $L_0$ is the rod's proper length measured in its rest frame. Transverse lengths are unchanged by the corresponding Lorentz transformation.

For a rod at rest in $S^{\prime }$ with endpoints $x^{\prime }=0$ and $x^{\prime }=10\mathrm{m}$, let $S^{\prime }$ move at $v=0.6c$ relative to the laboratory $S$. Record both endpoint positions at laboratory time $t=0$. The Lorentz transformation then gives $x^{\prime }=1.25x$, so their recorded positions are $0$ and $8\mathrm{m}$: the laboratory length is $10/1.25=8\mathrm{m}$. This is an operational definition: synchronized laboratory detectors register the endpoints at the same laboratory time. A photograph collects light emitted at different times and need not show this length. ^[ref].

For those same two endpoint spacetime events, $\Delta t^{\prime }=-1.25(0.6c)(8\mathrm{m})/c^2=-6\mathrm{m}/c$. Thus the front-end measurement occurs earlier in $S^{\prime }$ by about $20\mathrm{ns}$. Length contraction compares simultaneous endpoint measurements in each frame, which select different pairs of spacetime events; it does not mean that the rod has been compressed in its own rest frame.

The light cone of a spacetime event $O$ consists of $O$ together with the spacetime events having null separation from it. In the coordinates of an inertial frame centered on $O$, it obeysFor $t>0$ it traces an expanding spherical flash of light, with radius $ct$ at time $t$; for $t<0$ it describes possible incoming light rays. A spacetime diagram with one spatial coordinate shows the two lines $x=\pm ct$. The apparent $45^{\circ }$ angle uses equal scales for $x$ and $ct$ and is not a separate physical angle.

The light cone is the boundary, while its interior has timelike separation from $O$. An outgoing light ray follows the boundary; a future-directed massive observer's world line remains inside it. The exterior has spacelike separation: no signal travelling at or below the speed of light can connect its spacetime events directly to $O$. This separates possible causal influence from merely being assigned a later coordinate time.

In the diagram, $A=(4\mathrm{s},3\mathrm{ls})$ is inside the future boundary, $N=(3\mathrm{s},3\mathrm{ls})$ is on it, and $B=(2\mathrm{s},3\mathrm{ls})$ is outside. Their squared Minkowski intervals are respectively $-7,0,5$ in ${\mathrm{ls}}^2$. Thus $B$ occurs later than $O$ in the displayed inertial frame while still being unable to receive a light-speed-or-slower signal sent from $O$.

The usefulness is immediate for an experiment: a detector's reading can depend on a source event only if a causal influence can reach it. A Lorentz boost changes coordinates but preserves this classification. For a boost with $\beta =0.8$ and Lorentz factor $\gamma =5/3$, applying the Lorentz transformation gives, in light-second length units,The corresponding squared Minkowski intervals remain $-7,0,5$. In particular $B$ is now earlier than $O$, which is allowed for spacelike separation; $A$ remains later.

For a radial timelike geodesic of specific Killing energy $k=fdt/d\tau$, the static observer measures $v_{\mathrm{loc}}=(dr/dt)/f=\dot{r}/k$. Its relative Lorentz factor is $k/\sqrt{f}$, giving $v_{\mathrm{loc}}^2=c^2(1-f/k^2)$.

A Lorentz boost relates inertial frames moving at a constant relative velocity without an additional spatial rotation. In units with light speed one, the displayed transformation preserves $t^2-x^2$, with Lorentz factor $\gamma =(1-v^2{)}^{-1/2}$. Applying it to a static classical field-theory soliton produces a uniformly moving solution in a Lorentz-invariant theory.

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

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

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.

The proper mean lifetime of a free muon, approximately $2.2\mu \mathrm{s}$. Independent particle decays give survival probability $\exp (-\Delta \tau /{\tau }_{\mu })$. In a frame where the Lorentz factor is constant, $\Delta \tau =\Delta t/\gamma$; interactions of a stopped negative muon in matter can add capture channels.

The average of future-directed normalized four-velocities is generally timelike but is not normalized. For an isotropic ensemble with common Lorentz factor $\gamma$, the mean is $(\gamma c,0,0,0)$ and has squared norm $-{\gamma }^2{c}^2$. A normalized bulk velocity instead equals $c⟨v⟩/\sqrt{-g(⟨v⟩,⟨v⟩)}$. In a perfect fluid in general relativity its rest-frame energy eigenvector is the natural bulk choice; the rest-frame energy density includes random kinetic energy.

The parametrized post-Newtonian formalism expresses a weak-field metric tensor using phenomenological parameters so that different gravitational theories can be compared with the same observations. Its spatial-curvature parameter ${\gamma }_{\mathrm{PPN}}$ is one in general relativity; it is distinct from the kinematic Lorentz factor.

Inverse Compton scattering transfers energy from fast electrons to lower-energy photons. In the Thomson scattering limit the scattering is approximately elastic in the electron's instantaneous rest frame, but the transformations into and out of that frame can greatly increase the photon energy in the original frame.

In a blazar, relativistic jet electrons can scatter their own synchrotron radiation, producing synchrotron self-Compton radiation, or scatter photons from an accretion disk and its surrounding reprocessing material. The two spectral humps constrain electron Lorentz factors, the electron distribution and, with assumptions about geometry and relativistic beaming, the magnetic field. For the same electron population in the Thomson regime, the ratio of inverse-Compton to synchrotron power scales as $u_{\mathrm{ph}}/u_B$ in the plasma rest frame. It does not determine the field without information about the photon energy density and beaming.

In a galaxy cluster, hot thermal electrons scatter cosmic microwave background photons and slightly raise their mean energy, producing the thermal Sunyaev-Zeldovich effect. The spectral distortion measures the line-of-sight integral of electron pressure, approximately $y={\sigma }_T\int n_e{k}_B{T}_{e}d\ell /(m_e{c}^2)$ for nonrelativistic thermal electrons. Combined with X-ray emission, which weights density approximately quadratically, it constrains the gas density, temperature and mass distribution. These are distinct source populations and seed photon fields from the blazar example.

For the power derivation, use an isotropic incident radiation field of total photon energy density $u_{\mathrm{ph}}$ in the laboratory frame. Its stress-energy tensor has $T^{00}=u_{\mathrm{ph}}$ and $T^{zz}=u_{\mathrm{ph}}/3$, with zero mean flux. Boosting to an electron moving at $\beta c$ along $z$ givesThe radiation energy flux in this rest frame is $S_z^{\prime }=c{T}^{\prime 0z}$. Elastic Thomson scattering exerts the drag $F_z^{\prime }={\sigma }_T{S}_z^{\prime }/c$. The parallel three-force has the same value in the laboratory frame, so the electron's positive energy loss rate isThis isotropic inverse-Compton drag assumes photon energies small compared with $m_e{c}^2$ in the electron rest frame, negligible recoil and an isotropic seed field in the specified frame. An anisotropic photon field requires its directional moments; the recoil-dependent differential scattering cross-section of Compton scattering becomes necessary at larger rest-frame energies.

There are two conventions to qualify in the printed expression. First, the density multiplying $4/3$ must be $u_{\mathrm{ph}}$, not literally only the magnetic part of a propagating electromagnetic field. That field has equal electric and magnetic energy densities, so $u_{\mathrm{ph}}=2u_{\mathrm{mag}}$ and the corresponding formula is $P_{\mathrm{IC}}=(8/3){\sigma }_{T}c{u}_{\mathrm{mag}}{\beta }^2{\gamma }^2$. Second, this is net electron energy loss, rather than all incident plus scattered photon energy. At $\beta =0$ an electron still scatters power ${\sigma }_{T}c{u}_{\mathrm{ph}}$, but transfers no kinetic energy to the photons, so the inverse-Compton gain is zero.

For an individual incident photon of energy $\epsilon$, the electron-frame energy is ${\epsilon }^{\prime }=\gamma \epsilon (1-\beta \cos \theta )$. For a head-on photon followed by forward emission in the laboratory,Averaging an isotropic seed field gives $⟨{\epsilon }_{\mathrm{out}}⟩\simeq (4/3){\gamma }^2\epsilon$ for ultrarelativistic electrons. The original plot has an inverse-Compton maximum of order $10^8\mathrm{eV}$, a synchrotron maximum around $10^{-1}\mathrm{eV}$ and a disk component around $1\mathrm{eV}$. ThusThe head-on maximum gives the same order of magnitude, about $5\times 10^3$ to $1.6\times 10^4$. These are approximate readings of a broad spectrum, not precise fitted parameters. Disk and synchrotron seed photons need not be scattered by the same part of the electron distribution, and bulk jet Doppler factors also enter the observed peak relation. The indicated rest-frame seed energies remain well below the electron rest energy in this estimate.

Let $\mu =GM/c^2$, $f=1-2\mu /r$, $q=\sqrt{1-3\mu /r}=1/u^0$, and choose $\Omega >0$. A torque-free test gyroscope moving on a geodesic has a spin four-vector obeying parallel transport,The orthogonality condition says the temporal spin component vanishes in the body's rest frame, not necessarily in Schwarzschild coordinates. Here it givesIn addition to the radial Christoffel symbols in part (a), use ${\Gamma }_{tr}^t=\mu /(r^{2}f)$ and ${\Gamma }_{r\varphi }^{\varphi }=1/r$. On the equator the parallel transport equations becomeThe radial equation follows by substituting $s^t$ into $d{s}^r/dt+{\Gamma }_{tt}^r{s}^t+\Omega {\Gamma }_{\varphi \varphi }^r{s}^{\varphi }=0$. Eliminating $s^{\varphi }$ gives $d^2{s}^r/d{t}^2+{\Omega }^2{q}^2{s}^r=0$.

Let $s_0$ be the invariant spin magnitude. Initially radial spin means $s^r(0)=s_0\sqrt{f}$ and $s^t(0)=s^{\varphi }(0)=s^{\theta }(0)=0$. The complete solution isTo interpret the orientation, convert coordinate components into the orbiting orthonormal frame. The Lorentz factor relative to a Schwarzschild static observer is $\gamma =\sqrt{f}/q$, so the body's measured radial and tangential spin components areTheir squared sum is $s_0^2$, and their angle relative to the outward radial direction decreases uniformly at rateThe differing coordinate amplitudes in $s^r,s^{\varphi }$ must not be interpreted as a nonuniform physical spin rotation. After one orbital period the radial basis has turned by $2\pi$ while the spin has turned by $-2\pi q$ relative to it. The residual geodetic precession is $\Delta \psi =2\pi (1-q)$ in the orbital direction.