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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

For an affine parameter $\lambda$, the geodesic Lagrangian isVarying $S=\int \mathcal{L}d\lambda$ gives the geodesic equations

The rotational Killing vector fields of the spherical symmetry conserve the angular-momentum vector. Its direction is fixed, so every nonradial orbit lies in the plane through the symmetry centre orthogonal to that vector. A spatial rotation can make this the equatorial plane. Equivalently, the initial conditions $\theta =\pi /2$ and $\dot{\theta }=0$ solve the $\theta$ equation uniquely. This is the planarity of geodesics in spherical symmetry; a radial geodesic has zero angular momentum and may be assigned any plane.

On $\theta =\pi /2$, the cyclic coordinates $t$ and $\varphi$ produce geodesic conserved quantities from Killing vectors,For a timelike geodesic parametrized by proper time, the proper-time normalization is $g_{ab}{\dot{x}}^a{\dot{x}}^b=-1$. Substitution of $E$ and $L$ gives the first-order radial equationThis is the timelike geodesic effective potential for static Anti-de Sitter spacetime with curvature radius one.

A geodesic passing through $r=0$ must have $L=0$. If it moves away from the origin, $E>1$, andTaking proper time $s=0$ at departure givesIt turns around at $s=\pi /2$ and first returns to the origin atThis energy-independent refocusing is the characteristic radial timelike geodesic in anti-de Sitter spacetime result; for curvature radius $a$, the answer is $\pi a$.

For a constant electromagnetic field tensor, the relativistic Lorentz force is a linear system with constant coefficients, so

PutThe initial four-velocity is $(u^0,u^1,u^2,u^3)={\gamma }_0(c,0,v_0,0)$. For parallel fields along $x$, the field tensor has one hyperbolic block in the $(0,1)$ plane and one rotation block in the $(2,3)$ plane. With the orientation convention in the question,Taking the initial spacetime point to be the origin and integrating $u^{\mu }=d{x}^{\mu }/d\tau$ gives the relativistic motion in parallel electric and magnetic fieldsThe expressions have their continuous $a\to 0$ or $b\to 0$ limits when one field vanishes; reversing the sign of $q$ reverses the electric acceleration and magnetic rotation.

Eliminating proper time between the first two equations yieldsAt early coordinate time,which is uniformly accelerated motion with the transverse relativistic inertia factor ${\gamma }_0$. At late positive time,so the longitudinal speed approaches the speed of light.

The $y$-$z$ projection satisfiesIt is therefore a circle of radius and proper-time period

For the pitch magnitude assume $v_0>0$ and put $\alpha =|q|E/(mc)$, $\Omega =|q|B/m$, andNumber the $n$th proper-time period as $[(n-1)\Delta \tau ,n\Delta \tau ]$. Its longitudinal displacement isConsequently, as $n\to \infty$,where

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

For $A^{\mu }=(0,0,Bx,0)$, the only independent nonzero component of the electromagnetic field tensor isThe temporal and $z$ components of the Lorentz force show that $u^0$ and $u^3$ are constant. Taking $u^3=0$, the transverse equations arewhere dots now denote proper time derivatives. Witha solution isDirect differentiation verifies both equations. Hence the trajectory is a relativistic cyclotron orbit with proper angular frequency $omega=qB/m$; the sign fixes its orientation.

The radial equation for a circular geodesic, or equivalently $V^{\prime }(r)=0$, gives the Angular velocity of a circular Schwarzschild geodesicOne orbit therefore takes Schwarzschild coordinate time

Observer $O_3$ is static at infinity, where coordinate time equals proper time, so

For the static observer $O_2$ at radius $r$, $dr=d\theta =d\varphi =0$, and the Schwarzschild spacetime metric givesThus

Along $O_1$,where $d\varphi /dt=\Omega$ was used. Hence the Proper period of a circular Schwarzschild geodesic is

Let $x^{\mu }=(ct,\mathbf{x})$, let $\tau$ be proper time, and define the four-velocity and four-momentum byIf $F^{\mu \nu }$ is the electromagnetic field tensor, the covariant Lorentz force law isThe temporal component and three spatial components respectively giveHere $\mathcal{E}=\gamma m{c}^2$ is the relativistic energy and $\mathbf{p}=\gamma m\mathbf{v}$ is the relativistic momentum. In the nonrelativistic limit, $\gamma \to 1$, so the spatial equation becomesthe usual Lorentz-force law.

Write ${\dot{x}}^{\mu }=d{x}^{\mu }/ds$ and $L=\sqrt{-g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }}$. Varying $T=\int Lds$ with fixed endpoints and then choosing proper time as parameter, for which $L=1$, gives the Euler-Lagrange equationExpanding the derivative, raising the free index with the inverse metric, and using the symmetry of ${\dot{x}}^{\alpha }{\dot{x}}^{\beta }$ gives the geodesic equationThese are the Christoffel symbols of the Levi-Civita connection.

The proper time elapsed along the smooth timelike curve isThe expression is invariant under every orientation-preserving reparametrization of the curve.

With $\gamma =(1-v^2/c^2{)}^{-1/2}$,Differentiation with respect to proper time givesOrthogonality $U\cdot A=0$ implies that in the instantaneous rest frame $A^{\prime \mu }=(0,a^{\prime })$. For collinear motion a Lorentz boost givesFor constant positive proper acceleration $a^{\prime }$,This hyperbola has future null asymptote $x=ct-c^2/a^{\prime }$. A right-moving signal emitted from $x_0$ at $t=0$ follows $x=ct+x_0$; if $x_0<-c^2/a^{\prime }$, it lies beyond that acceleration horizon and never intersects the worldline.

A circular orbit at $r=r_0$ requires $V^{\prime }(r_0)=0$, henceThis can be satisfied for every $r_0>0$. Moreover,so every such circular orbit is stable. The radial energy equation givesand future direction selects $E=1+r_0^2/a^2=f(r_0)$. Thereforeso proper time and coordinate time differ only by an additive constant on the orbit. Also $\dot{\varphi }=\pm 1/a$.

Proper time satisfies $c^{2}d{\tau }^2=c^{2}d{t}^2-d{x}^2$ and is invariant, so differentiating by it defines a four-vector. The path is timelike iff $R|\omega |<c$. Its next return occurs after coordinate time $2\pi /|\omega |$, so Bob agesWith $\gamma =(1-R^2{\omega }^2/c^2{)}^{-1/2}$,
$U^{\mu }=\gamma (c,R\omega \cos \omega t,-R\omega \sin \omega t,0)$ and $U\cdot U=c^2$ for signature $(+---)$.

A radial timelike geodesic has conserved $E=f\dot{t}$ and normalization ${\dot{r}}^2=E^2-f$. Choose $E>1$ and the outward sign. Thenso the observer reaches $r=b$ in finite proper time even though static coordinate $t$ diverges.

With signature $(-,+,+,+)$ and units $c=1$, parametrizing a timelike worldline by proper time makes its four-velocity satisfy $U^a{U}_a=-1$.

Every maximal radial timelike geodesic in positive-mass Kruskal spacetime is incomplete in at least one direction. Its equations are ${\dot{r}}^2=E^2-1+2M/r$ and $\ddot{r}=-M/r^2$. A finite turning point is a maximum, and each trajectory reaches $r=0$ at one or both ends. The remaining proper time is finite because $d\tau \sim \sqrt{r/(2M)}dr$. Complete circular timelike geodesics exist, but they have nonzero angular momentum.

In a constant magnetic field $B\hat{z}$ with no electric field, a charged relativistic particle moves in a circle in the perpendicular plane. Its angular frequency with respect to proper time iswhereas its angular frequency in an inertial coordinate time is ${\omega }_t=qB/(\gamma m)$. The orbit radius at speed $v$ is

For three-momentum $p=\gamma mv$, force and acceleration satisfy

When uniform $\mathbf{E}$ and $\mathbf{B}$ are parallel, the relativistic Lorentz force splits into independent hyperbolic motion along the fields and circular motion in the perpendicular plane. The circle is periodic in proper time, while electric acceleration makes the helix's longitudinal pitch grow exponentially from one proper-time cyclotron period to the next.

ChooseA particle released from rest at the origin has the proper-time trajectoryEliminating proper time gives the semicubical parabolaThis signed coefficient assumes that the positive $x$ direction is chosen along $\mathbf{E}$; reversing that axis changes the signs of both $x^3$ and its coefficient.

Schwarzschild time $t$ is the exterior time coordinate normalized to the proper time of static observers at infinity. At fixed exterior radius, $d\tau =\sqrt{1-2M/r}dt$. A received signal and its emission occur at different values of this coordinate.