# How does an uniformly accelerated particle see the world in 1+1D?

Let's consider an observer $O$ associated with an inertial referential frame $R$ in a flat spacetime (no gravitational field). This observer is looking at an uniformly accelerated particle $P$ moving in a constant direction (no Thomas precession). The coordinates of the particle in $R$ are ($x$,$t$). The constant acceleration of the particle is called $a$, I define it here as a coordinate acceleration ($a = d^2x/d^2t$) and not a proper acceleration. Therefore in this problem the proper acceleration is not constant, one cannot use the Rindler coordinates.

The equation of motion of the particle $P$ view from $R$ is, $$x(t) = \frac{1}{2}at^2+v_{0}t+x_{0}$$

What is the equation of motion of the observer $O$ view from the particle $P$ in its non inertial referential $R'$ ?

You can use $(x',t')$ as the coordinates of $O$ in the referential $R'$, and $\tau$ the proper time of $R$, $\tau'$ the proper time of $R'$.

• Thank you for this answer. However I am sorry to not understand how it does answer my question, what is the equation of motion for $x'(t')$ ? Feb 12, 2018 at 18:23
• Actually, $(\xi,\tau)=(x',t')$. The metric is governed by proper acceleration $a$, proper time $\tau$. Feb 12, 2018 at 18:25
• In my problem, $a$ is the coordinate acceleration, not the proper acceleration. So your $a$ is not my $a$, correct ? Feb 12, 2018 at 18:27
• Knowing that $a_{0}=a\gamma^3$, do you agree that your answer is: $x'(t) = -\frac{c^2}{a\gamma^3}+\sqrt{\left( \frac{c^2}{a\gamma^3}+x \right)^2-(ct)^2}$ Feb 12, 2018 at 18:36