Constant proper acceleration in special relativity

Considering special relativity, what is the world line of a moving particle with constant proper acceleration $a$ observed in the still world frame? The proper acceleration is defined as the acceleration measured by the inertial observer to which the particle looks rest during the measurement.

Lorentz transformation of acceleration

The problem is about the Lorentz transformation of acceleration. For this purpose, we start by the Lorentz transformations in special relativity between a still frame $(t, x)$ and a moving frame $(t', x')$ with constant speed $v$ relative to the still frame: \begin{eqnarray} x'&=&\frac{x-vt}{\sqrt{1-{v^2}/{c^2}}}\,,\tag{1}\\ t'&=& \frac{t-vx/c^2}{\sqrt{1-{v^2}/{c^2}}}\,.\tag{2}\end{eqnarray} A motion with speeds $u_x\equiv \frac{dx}{dt}$ in the still frame and $u'_x\equiv \frac{dx'}{dt'}$ in the moving frame has the transformation: \begin{eqnarray} u'_x = \frac{u_x - v}{1-{vu_x}/{c^2}}\,,\tag{3} \end{eqnarray} which can be derived by taking the derivatives on both sides of Eq. (1) and (2). Then by taking derivatives on both sides of Eq.(2) and (3), we can prove the transformations between the acceleration in the still frame $a_x\equiv \frac{du_x}{dt}$ and that in the moving frame $a'_x\equiv \frac{du'_x}{dt'}$ as \begin{eqnarray} a'_x = \frac{\left(1-v^2/c^2\right)^{3/2}}{\left(1-vu_x/c^2\right)^3}a_x\,.\tag{4}\end{eqnarray}

Hyperbolic motion

Now back to our problem proposed at the beginning. In this scenario, the coordinates of the still world frame is $(t, x)$ and the coordinates of the moving particle frame is $(t', x')$. Since we consider the motion of the moving particle frame itself, we have \begin{eqnarray} u'_x = 0\,,\quad u_x=v\,, \quad a'_x=a\,,\quad a_x = \frac{dv}{dt}\,.\tag{5}\end{eqnarray} As a result, Eq. (4) becomes a differential equation \begin{equation} \boxed{\frac{dv}{dt}=a\left(1-\frac{v^2}{c^2}\right)^{3/2}}\,.\tag{6}\end{equation}

To solve the function $v(t)$ from Eq. (6), we let \begin{equation}v=c\tanh\beta\,.\tag{7}\end{equation} Eq. (6) then becomes $dt = \frac{c}{a}\cosh\beta\, d\beta$, from which we can solve \begin{equation}t = \frac{c}{a}\sinh\beta + C_1\,,\tag{8}\end{equation} with a constant of integration $C_1$. Furthermore, we have $dx = v dt = c\tanh\beta\, \frac{c}{a}\cosh\beta\, d\beta=\frac{c^2}{a}\sinh\beta\, d\beta$, and thus \begin{equation}x = \frac{c^2}{a}\cosh\beta + C_2\,,\tag{9}\end{equation} where $C_2$ is another constant of integration. Eq. (8) and Eq. (9) are the parametrized form of the hyperbolic motion!

To determine the constants of integration $C_1$ and $C_2$, we assume the initial condition $x=x_0$ and $v=v_0$ at $t=t_0$. As a result, we can solve from Eq. (7)-(9) that \begin{equation}C_1=t_0-\frac{v_0}{a}\left(1-\frac{v^2_0}{c^2}\right)^{-1/2}\,,\quad C_2=x_0-\frac{c^2}{a}\left(1-\frac{v^2_0}{c^2}\right)^{-1/2}\,.\quad \end{equation}

Combining Eq. (8) and (9), we have the trajectory \begin{equation} \boxed{\left[x-x_0+\frac{c^2}{a}\left(1-\frac{v^2_0}{c^2}\right)^{-1/2}\right]^2 - c^2\left[t-t_0+\frac{v_0}{a}\left(1-\frac{v^2_0}{c^2}\right)^{-1/2}\right]^2 = \left(\frac{c^2}{a}\right)^2}\,. \tag{10}\end{equation}

Finally, it is worth pointing out that the parameter $\beta$ in Eq. (7)-(9) is actually related to the proper time of the moving particle  $\tau$. This is because $d\tau = dt\sqrt{1-v^2/c^2}=\frac{c}{a}\cosh\beta\, \sqrt{1-\tanh^2\beta}d\beta=\frac{c}{a}d\beta$. We can simply choose \begin{equation}\beta = \frac{c}{a}\tau\,,\end{equation} which implies that $\tau = \frac{a}{c}\text{arctanh}\,\frac{v_0}{c}$ when $t=t_0$.

Note:

• For the special initial condition $x=\frac{c^2}{a}, v=0, \tau=0$ at $t=0$, all the above results can be simplified to \begin{equation}\boxed{t = \frac{c}{a}\sinh\left(\frac{c}{a}\tau\right)\,,\quad x = \frac{c^2}{a}\cosh\left(\frac{c}{a}\tau\right)} \end{equation}and\begin{equation}\boxed{x^2 -(ct)^2 = \left(\frac{c^2}{a}\right)^2}\,. \end{equation} This is the most common form we see in the literature.
• For the special initial condition $x=0, v=0, \tau=0$ at $t=0$, all the above results can be simplified to \begin{equation}{t = \frac{c}{a}\sinh\left(\frac{c}{a}\tau\right)\,,\quad x = \frac{c^2}{a}\left[\cosh\left(\frac{c}{a}\tau\right)-1\right]} \end{equation}and\begin{equation}{\left(x+\frac{c^2}{a}\right)^2 -(ct)^2 = \left(\frac{c^2}{a}\right)^2}\,. \end{equation}