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

Question

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 $R'$.

Ng Chung Tak · Answer 1 · 2018-02-12T18:27:53.363

0

Rindler spacetime

\begin{align} F' &= m_0 a_0 \\[5pt] \begin{pmatrix} t \\ x \end{pmatrix} &= \begin{pmatrix} \left( \frac{c}{a_0}+\frac{\xi}{c} \right) \sinh \frac{a_0\tau}{c} \\ \left( \frac{c^2}{a_0}+\xi \right) \cosh \frac{a_0\tau}{c}-\frac{c^2}{a_0} \end{pmatrix} \\[5pt] \begin{pmatrix} \tau \\ \xi \end{pmatrix} &= \begin{pmatrix} \frac{c}{a_0} \tanh^{-1} \frac{ct}{\frac{c^2}{a_0}+x} \\ -\frac{c^2}{a_0}+\sqrt{\left( \frac{c^2}{a_0}+x \right)^2-(ct)^2} \end{pmatrix} \\[5pt] ds^2 &= (c^2+a_0\xi)^2 d\tau^2-d\xi^2 \\[5pt] v &= \frac{\partial x}{\partial t} \\[5pt] &= c\tanh \frac{a_0\tau}{c} \\[5pt] &= \frac{a_0 t}{\sqrt{1+\left( \frac{a_0 t}{c} \right)^2}} \\[5pt] \gamma &= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \\[5pt] &= \cosh \frac{a_0\tau}{c} \\[5pt] &= \sqrt{1+\left( \frac{a_0 t}{c} \right)^2} \\[5pt] \frac{dx}{dt} &= \frac{c\frac{d\xi}{d\tau}+\left( 1+\frac{a_0\xi}{c^2} \right) v} {\left( 1+\frac{a_0\xi}{c^2} \right)+\frac{v}{c^2} \frac{d\xi}{d\tau}} \tag{velocity addition} \end{align}

See also another situation here.

edited Feb 12 '18 at 18:27

answered Feb 12 '18 at 18:11

Ng Chung Tak

1,377

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 $x'(t')$ ? – François Ritter Feb 12 '18 at 18:23
Actually, $(\xi,\tau)=(x',t')$. The metric is governed by proper acceleration $a$, proper time $\tau$. – Ng Chung Tak Feb 12 '18 at 18:25
In my problem, $a$ is the coordinate acceleration, not the proper acceleration. So your $a$ is not my $a$, correct ? – François Ritter Feb 12 '18 at 18:27
Notation revised – Ng Chung Tak Feb 12 '18 at 18:30
Knowing that $a_{0}=a\gamma^3$, do you agree that your answer is:
$x'(t) = -\frac{c^2}{a\gamma^3}+\sqrt{\left( \frac{c^2}{a\gamma^3}+x \right)^2-(ct)^2}$
– François Ritter Feb 12 '18 at 18:36
I am sorry Ng Chung Tak, but I believe Rindler coordinates are working only for a constant proper acceleration. In my problem, the coordinate acceleration is constant, but the proper acceleration varies, so one cannot use the Rindler coordinates ! – François Ritter Feb 12 '18 at 18:46
You never have constant acceleration forever, otherwise the velocity will exceed $c$. – Ng Chung Tak Feb 12 '18 at 19:02

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

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