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In units $c=1$, a receiver on the hyperbolic motion worldline $t=a^{-1}\sinh (a\tau )$, $x=a^{-1}\cosh (a\tau )$ has retarded null coordinate $p=t-x=-a^{-1}{e}^{-a\tau }$. A right-moving light ray emitted at $(t_0,L)$ has $p=t_0-L$, so it meets the receiver only if $t_0<L$. The photon wave covector is proportional to $(1,-1)$, and contraction with the receiver's four-velocity gives its measured frequency as ${\nu }_0(\dot{t}-\dot{x})={\nu }_0{e}^{-a\tau }$. The intersection condition gives the second displayed equality. Emission at $t_0=L$ lies on the Rindler horizon and never reaches the receiver at finite proper time; $L$ is the supremum of receivable emission times, not an attained latest time.

Let an observer start at rest at $T=0$, $X=R=c^2/a$ and accelerate towards increasing $X$ with constant proper acceleration $a>0$. A signal emitted simultaneously from $X=R-d$ travels along $X=R-d+cT$. The observer follows hyperbolic motion, $X=R\cosh (a\tau /c)$ and $cT=R\sinh (a\tau /c)$. Their intersection obeysA finite future reception exists only for $0\le d<c^2/a$. At equality the ray lies on the limiting null horizon and never reaches the observer; for larger separation it remains outside the observer's receivable past. The proper reception time tends to $d/c$ as $a\to 0$. The simultaneous emission and initial-rest conditions are essential; changing either changes the timing problem.

Newtonian mechanics describes motion using forces, masses and acceleration in inertial frames with absolute time. It is the low-speed limit of special relativity; for constant proper acceleration with $|at|\ll c$, relativistic hyperbolic motion reduces to $x\simeq a{t}^2/2$.

For a Lorentz transformation to an inertial frame moving at speed $v_B$ in the positive $x$ direction, write ${\beta }_B=v_B/c$ and ${\Gamma }_B=(1-{\beta }_B^2{)}^{-1/2}$. Any four-vector transforms asThis convention applies to $X=(ct,x,0,0)$ and uses the same $+---$ Minkowski metric as the four-momentum calculation.

The definition of proper time gives $d\tau =dt\sqrt{1-u^2/c^2}=dt/\gamma$ for $|u|<c$. Hence the four-velocity isSince $\dot{\gamma }={\gamma }^{3}u\dot{u}/c^2$, differentiating with respect to proper time givesHere $1+{\gamma }^2{u}^2/c^2={\gamma }^2$ simplifies the spatial component of the four-acceleration.

At each event, boost with $v_B=u$ to the instantaneous rest inertial frame. Then $A^{\prime 0}=0$ and $A^{\prime 1}=\gamma {\gamma }^4\dot{u}(1-u^2/c^2)={\gamma }^3\dot{u}$. Thus the signed proper acceleration for one-dimensional motion isThe last equality follows by differentiating $\gamma u$; its magnitude is the usual scalar proper acceleration.

If this signed proper acceleration is constant and the particle starts from rest, integration gives $\gamma u=at$, soIntegrating once more and imposing $x(0)=0$ yields the hyperbolic motion worldlineFor $a=0$ the continuous limiting solution is $x=0$. Finally, the Taylor series of the square root givesThis recovers constant-acceleration Newtonian mechanics at low speed while the exact relativistic speed remains less than $c$.

The meetings occur where Bob's constant position $x=\alpha c{t}_0$ intersects Alice's positive-$x$ branch of the hyperbolic motion:Bob is at rest in this inertial frame, so his elapsed proper time is

Parametrize Alice's world line bySince $t$ increases with $\eta$, the proper time element isAt the meetings, $\cosh \eta =\alpha$, giving endpoints $\eta =\pm \mathrm{arcosh}\alpha$. Thus Alice's elapsed proper time isHere $\mathrm{arcosh}$ is the nonnegative real inverse hyperbolic cosine. As a check, if $\alpha =\cosh {\eta }_{\ast }$ with ${\eta }_{\ast }>0$, then Bob ages by $2t_0\sinh {\eta }_{\ast }$ and Alice by $2t_0{\eta }_{\ast }$; $\sinh {\eta }_{\ast }>{\eta }_{\ast }$ confirms that Bob again ages more.

Let $a>0$ be the constant proper acceleration away from the emitter and set $R=c^2/a$. Take the signal emission to be simultaneous, in the initial inertial frame, with the ship being at rest and its clock reading $\tau =0$. Place that initial ship event at $T=0$, $X=R$. Its hyperbolic motion isA photon emitted from $X=R-d$ at $T=0$ towards the departing ship follows $X_{\gamma }=R-d+cT$. At reception,Therefore light reception by a uniformly accelerated observer occurs afterThe same result follows in Rindler coordinates: the ship stays at $x=R$, the emitter starts at $x=R-d$, and the right-moving null ray has $x=(R-d)e^{\alpha t}$. The onboard lapse $d\tau =\alpha Rdt/c$ converts its coordinate catch-up time to the expression above.

If $d=c^2/a$, the ray lies on the limiting null horizon and never intersects it at finite proper time; if $d>c^2/a$, it never reaches the ship. The reception time diverges as the emitter approaches this Rindler horizon. For $ad/c^2\ll 1$, ${\tau }_{\mathrm{rec}}=d/c+a{d}^2/(2c^3)+\cdots$, recovering the inertial light-travel time. A different emission time or initial relative velocity would require specifying that extra datum; the formula uses the initial-rest simultaneous-emission convention stated above.

For the stated outward-accelerating hyperbola, take positive signed proper acceleration and an initially stationary spaceship. Then $\eta =a\tau$, so $\dot{p}=e^{-a\tau }$ and $\dot{q}=e^{a\tau }$. The initial values are $p(0)=-1/a$, $q(0)=1/a$. Integration givesTaking their half-sum and half-difference yields the hyperbolic motionThe printed position data alone do not specify the initial velocity. If the initial rapidity is an arbitrary ${\eta }_0$, the equally valid positive-acceleration solution isThus the requested special expression implicitly selects ${\eta }_0=0$.

Let $f=|\mathbf{f}|$ and $\hat{\mathbf{f}}=\mathbf{f}/f$. A constant laboratory three-force means $d\mathbf{p}/dt=\mathbf{f}$, hence $\mathbf{p}=\mathbf{f}t$ from rest. Using $E^2=m^2{c}^4+p^2{c}^2$ givesThe dimensionless time $s=ft/(mc)$ makes the shapes transparent: $\gamma =\sqrt{1+s^2}$ starts at one with zero slope and becomes asymptotically linear; $\beta =s/\sqrt{1+s^2}$ starts linearly and approaches one without reaching it at finite time. The acceleration decreases as $f/(m{\gamma }^3)$.

Integrating the velocity gives the distance along the force,Thus a constant three-force produces hyperbolic motion with constant proper acceleration $f/m$, although its velocity is never Newtonian at large time. The following original curves supply the requested sketches.

For the electric-dominated crossed-field orbit, choose a boost $V=c^{2}B/E$ along $x$. Since $|E|>c|B|$, it is subluminal. The Lorentz transformation of electromagnetic fields gives $B_z^{\prime }=0$ and $E_y^{\prime }=E/\Gamma$, where $\Gamma =(1-V^2/c^2{)}^{-1/2}$. The initially stationary electron has $U^{\prime }(0)=(\Gamma c,-\Gamma V,0,0)$, not zero velocity in this frame.

Let $q=-e$, $a=q{E}_y^{\prime }/m_e$ and $s=a\tau /c$. Neglect radiation reaction, as in the prescribed external-field model. Integrating the relativistic Lorentz force in this purely electric frame gives the exact trajectoryIndeed $U^{\prime 0}=\Gamma c\cosh s$, $U^{\prime x}=-\Gamma V$, and $U^{\prime y}=\Gamma c\sinh s$ satisfy both the force equation and $U^{\prime 2}=c^2$. The constant transverse momentum must be retained.

The inverse boost gives a useful exact laboratory parametrization as well:For $E,B>0$, the lab electron starts accelerating toward negative $y$ and bends toward positive $x$, the direction of $\mathbf{E}\times \mathbf{B}$. It never acquires a cyclotron orbit in this electric-dominated case. At late times, $v_x\to V$ and $v_y\to \mathrm{sgn}(q{E}^{\prime })c/\Gamma$, so its speed approaches $c$ along an oblique asymptote. Reversing field signs reverses the appropriate directions. The zero-$B$ limit is the ordinary electric-field hyperbolic motion.

Let $m$ denote the constant rest mass, set $x(0)=0$, and put $a=f_x/m$. The relativistic three-force is $f_x=d{p}_x/dt$, soIt follows thatIntegrating the velocity with the initial position fixed givesThese formulas allow either sign of $a$; their $a\to 0$ limit is rest at the origin. For $a\ne 0$, the trajectory obeysThis is hyperbolic motion. Its coordinate acceleration decreases while its proper acceleration has constant magnitude $|a|$. In particular, a constant three-force does not give constant coordinate acceleration at relativistic speed.

For motion on a fixed axis, the signed spatial four-acceleration in the instantaneous rest inertial frame is $a={\gamma }^3\dot{u}$. Its magnitude is the scalar proper acceleration. Constant signed $a$ from rest gives $\gamma u=at$ and the hyperbolic motion worldline $x=(c^2/a)(\sqrt{1+a^2{t}^2/c^2}-1)$.

A particle released from rest in a constant electric field parallel to its motion follows a spacetime hyperbola and has rapidity proportional to proper time.

This force law realizes hyperbolic motion with constant proper acceleration.