Derivation of hyperbolic motion in one spatial dimension
Derivation
In this article we derive the formulas for hyperbolic motion in one dimension. Let’s start by considering a point-like particle and choose an inertial frame \(\mathcal{R}\) such that the trajectory of the particle passes through the origin. We here assume that both the velocity and the acceleration of the particle are directed along the \(x\) axis. It is reasonable to expect that the trajectory of such a particle can written as \((x(t),\,0,\,0,\,t)\).
When we say that the particle is uniformly accelerated, we mean that the particle accelerates uniformly from one inertial frame to the next. This means that, if we consider an arbitrary point \(\mathrm{Q}\in\insR^4\) of the particle trajectory and we express the trajectory in an intertial frame tangent to the trajectory in \(\mathrm{Q}\) we expect to always get the same expression, independently from the choice of \(\mathrm{Q}\).
So let us return to the reference frame \(\mathcal{R}\) and let us choose a time \(t_1\). The particle will be at \(x_1 \equiv x(t=t_1)\) on the \(x\) axis and have a velocity \(u_1 \equiv u(t=t_1)\) also directed along the \(x\) direction. At a subsequent time \(t_2 = t_1 + \Delta t\) the particle will be at position \(x_2\) with velocity \(u_2\). Let us consider the inertial reference frame, \(\mathcal{R}'\), where \(x_1'\) and \(u_1'\) are both zero. This frame moves with velocity \(u_1\) with respect to \(\mathcal{R}\). In this frame the particle is seen accelerating from standstill between time \(t_1'\) and time \(t_2'\). If \(t_2'\) is sufficiently close to \(t_1'\), then the classical limit must hold and the velocity must be well approximated by \(u' = a \, \Delta t' = a \, (t_2' - t_1')\), where \(a\) is the proper acceleration. We can now compute the velocity in the reference frame \(\mathcal{R}\) using the relativistic law of composition of velocities:
\[u = \frac{u_1 + a \, \Delta t'}{1 + (u_1 a\, \Delta t')/c^2},\]and
\[\Delta u = u - u_1 = \frac{1 - u_1^2/c^2}{1 + (u_1 a\, \Delta t')/c^2} \, a \, \Delta t'.\]The limit \(\Delta t' \rightarrow 0\) gives:
\[\Delta u = u - u_1 = \left(1 - u_1^2/c^2\right) \, a \, \Delta t'.\]\(\Delta t'\) is the proper time, i.e. an interval of time measured from the reference frame which moves with the particle. This is related to the time seen from \(\mathcal{R}\) as follows: \(\Delta t = \gamma_1 \Delta t'\), where \(\gamma_1 = 1/\sqrt{1 - u_1^2/c^2}\). Substituting this into the previous equation we get:
\[\deriv{u}{t} = a \, \gamma^{-3} = \left(1 - u^2/c^2 \right)^{3/2},\]where the index 1 was omitted as this derivation can be done for any value of the time \(t_1\).
This equation can be solved doing the substitutions \(\sin \theta = u/c\) and \(T = at/c\):
\[\deriv{\theta}{T} = \cos^2 \theta.\]which is easily integrated to:
\[\int_{\thetai}^{\theta} \frac{\mathrm{d}\Theta}{\cos^2 \Theta} = \tan \theta - \tan \thetai = T.\]Substituting \(\sin \theta = u/c\), we have:
\[\begin{equation} \sin \theta = \frac{u}{c} = \frac{T + \Ti}{\sqrt{1 + (T + \Ti)^2}}, \label{eq:sin_theta} \end{equation}\]where we introduced \(\Ti \equiv \tan \thetai\). Note that, taking \(T = 0\) (i.e., \(t = 0\)) in the equation above, we get to:
\[\frac{\ui}{c} = \frac{\Ti}{\sqrt{1 + \Ti^2}} \Rightarrow \Ti = \frac{\gammai \ui}{c}.\]From \eqref{eq:sin_theta} we can compute \(\gamma = (1 - (u/c)^2)^{-1/2}\):
\[\gamma = \left[1 - \frac{(T + \Ti)^2}{1 + (T + \Ti)^2} \right]^{-1/2} = \sqrt{1 + (T + \Ti)^2}\]We can integrate the velocity formula \eqref{eq:sin_theta} to obtain the position:
\[x = \int_0^{\tf} u \, \mathrm{d}t = \frac{c^2}{a} \int_0^{\Tf} \frac{u}{c} \, \mathrm{d}T\]Changing variable of integration to \(\Theta\) as above:
\[x = \frac{c^2}{a} \int_{\thetai}^{\theta} \frac{\sin \Theta}{\cos^2 \Theta} \, \mathrm{d}\Theta = -\frac{c^2}{a} \int_{\cos \thetai}^{\cos \theta} \frac{\mathrm{d}\cos \Theta}{\cos^2 \Theta}\]which gives:
\[x = \frac{c^2}{a} \left( \frac{1}{\cos \theta} - \frac{1}{\cos \thetai} \right).\]From Eq. \eqref{eq:sin_theta} we deduce that \(\cos^{-1} \theta = \sqrt{1 + (T + \Ti)^2} = \gamma\) and \(\cos^{-1} \thetai = \sqrt{1 + \Ti^2} = \gammai\). We thus have:
\[\begin{equation*} x = \frac{c^2}{a} \left( \gamma - \gammai \right) \end{equation*}\]and,
\[\bbox[lightyellow, 10px, border: 2px solid orange]{ \begin{equation*} x(t) = \frac{c^2}{a} \left( \sqrt{1 + \left(\frac{at}{c} + \frac{\gammai \ui}{c}\right)^2} - \gammai \right). \end{equation*} }\]This expression has a removable singularity in \(a = 0\), which isn’t well suited for numerical computation. We can solve the issue by multiplying the right hand side by \((\gamma + \gammai)/(\gamma + \gammai)\):
\[\begin{eqnarray*} x & = & \frac{c^2}{a} \frac{\gamma^2 - \gammai^2}{\gamma + \gammai} \\ & = & \frac{c^2}{a(\gamma + \gammai)} \left[1 + \left(\frac{at}{c} + \frac{\gammai \ui}{c}\right)^2 - \gammai^2\right]\\ & = & \frac{a t^2 + 2 \gammai \ui \, t}{\gamma + \gammai}. \end{eqnarray*}\]Above, we used the identity: \(1 + \frac{\gammai^2 \ui^2}{c^2} - \gammai^2 = 0\).
In summary:
\[\begin{equation} \bbox[lightyellow, 10px, border: 2px solid orange]{ \begin{aligned} \gamma(t) & = \sqrt{1 + \left( \frac{at}{c} + \frac{\gammai \ui}{c} \right)^2}, \\ u(t) & = \frac{at + \gammai \ui}{\gamma(t)}, \\ x(t) & = \frac{a t^2 + 2 \gammai \ui \, t}{\gamma(t) + \gammai}. \end{aligned} } \label{eq:gamma_u_x_of_t} \end{equation}\]Note that \(\gamma(t = 0) = \gammai\). Also, the formula for \(x(t)\) reduces to \(x(t) = \ui t\) when \(a = 0\) and to \(x(t) = a t^2 /2 + \ui t\) when \(\gamma(t) \approx 1\).
Proper time parametrization
If we call \(\tau(t)\) the proper time, we have:
\[\deriv{\tau}{t} = \frac{1}{\gamma} = \frac{1}{\sqrt{1 + (T + \Ti)^2}},\]which can be integrated as follows:
\[\tau = \int_0^{t} \frac{\mathrm{d}t}{\sqrt{1 + (T + \Ti)^2}} = \frac{c}{a} \, \int_{\Ti}^{T + \Ti} \frac{\mathrm{d}W}{\sqrt{1 + W^2}},\]We now change variable of integration to \(W = \sinh \phi\):
\[\frac{a \tau}{c} = \int_{\asinh \Ti}^{\asinh (T + \Ti)} \mathrm{d} w = \asinh (T + \Ti) - \asinh \Ti,\]Which gives:
\[T + \Ti = \sinh \left( \frac{a \tau}{c} + \asinh \Ti \right).\]We now introduce the quantity \(w\) defined as:
\[w = \frac{a \tau}{c} + \asinh \Ti = \frac{a \tau}{c} + \atanh \frac{\ui}{c},\]where we have used the identity \(\asinh \frac{\gammai \ui}{c} = \atanh \frac{\ui}{c}\). We have \(T + \Ti = \sinh w\) and \(\sqrt{1 + (T + \Ti)^2} = \cosh w\). We can thus write:
\[\begin{equation} \bbox[lightyellow, 10px, border: 2px solid orange]{ \begin{aligned} w(\tau) & = \frac{a \tau}{c} + \atanh \frac{\ui}{c}, \\ \gamma(\tau) & = \cosh w(\tau), \\ \frac{u(\tau)}{c} & = \tanh w(\tau) \\ x(\tau) & = \frac{c^2}{a} \left[ \cosh w(\tau) - \gammai \right], \\ t(\tau) & = \frac{1}{a} \left[c \, \sinh w(\tau) - \gammai \ui\right]. \end{aligned} } \label{eq:gamma_u_x_of_tau} \end{equation}\]The formulas for \(x(t)\) and \(u(t)\) have a removable singularity at \(a = 0\). I doubt this singularity can be removed as easily as done in the previous section. Take the case \(\ui = 0\), for example. We get \(t(\tau) = c \, \sinh (a \tau / c) / a\). This can be rewritten as: \(t(\tau) = \tau \, \sinhc (a \tau/c)\), where \(\sinhc\) is the hyperbolic sinc function. The problem of removing the singularity from \(t(\tau)\) is thus transformed to the problem of removing the singularity from \(\sinhc\).
We thus have to be a bit careful when numerically computing \(x(\tau)\) and \(t(\tau)\). For example, we could use a Taylor expansion in the vicinity of \(a = 0\).
Final velocity as a boundary condition
So far, we have written down the equations for hyperbolic motion assuming the position, velocity, and acceleration are known at an initial time \(t = 0\) (or \(\tau = 0\).) In certain cases, it is desirable to use a different set of boundary conditions.
In this section we rewrite the formulas for hyperbolic motion in terms of a desired final velocity. In other words, we assume we are given the position and velocity at time \(t = 0\) and the velocity at a later time \(t = T\).
We start from the velocity as in Eqs. \eqref{eq:gamma_u_x_of_t} and rewrite it as follows:
\[\gamma(t) u(t) = \gammai \ui + at.\]This equation shows very clearly that the spatial component of the four-velocity, \(\gamma u\), increases linearly with time. This allows to express the acceleration, \(a\), in terms of the initial and final velocities:
\[a = \frac{\gammaf \uf - \gammai \ui}{T}.\]In some cases, it may be a good idea to go a step further and introduce a new variable as follows:
\[\omega(t) = \frac{\gamma(t) u(t)}{c}.\]The formulas for hyperbolic motion can now be reparametrised on \(\omega\):
\[\begin{equation*} \begin{aligned} \gamma(\omega) & = \sqrt{1 + \omega^2}, \\ u(\omega) & = \frac{c\omega}{\gamma(\omega)}, \\ x(\omega) & = cT \, \frac{\gamma(\omega) - \gammai}{\omegaf - \omegai}. \end{aligned} \end{equation*}\]This parametrisation can be useful to join two pieces of hyperbolic trajectory, for example.