Deriving the relativistic rocket equations. 
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*I am trying to derive the relativistic rocket equations found here [(4),(5),(6),(7),(8)] but I do not understand proper time, proper velocity and proper acceleration.


Define a point $P$ with spacetime coordinates $(t,x,y,z)$ in reference frame $S$ and $(t',x',y',z')$ in frame $S'$ which is moving at velocity $v$ relative to frame $S$ (parallel to the x-axis). Using the Lorentz transformation, the coordinates are related in the following way:
$$t'=\gamma(t-vx/c^2)$$
$$x'=\gamma(x-vt)$$
$$y'=y$$
$$z'=z$$
where
$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}.$$


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*From my understanding, if $S$ is the Earth reference frame and $S'$ is the reference frame of a rocket with velocity $v$ moving in the x-axis direction relative to Earth then; a clock on Earth would measure an event E at time $t$ and a clock on the rocket would measure event E at time $t'$. Is this correct? Is $t'$ proper time?


The inverse Lorentz transformation is given by:
$$t=\gamma(t'+vx'/c^2)$$
$$x=\gamma(x'+vt')$$
$$y=y'$$
$$z=z'$$
Taking the differentials of $x$ and $t$,
$$dx=\gamma(dx'+vdt')=\gamma(v'+v)dt'$$
$$dt=\gamma(dt'+dx'v/c^2)=\gamma(1+v'v/c^2)dt'$$
Dividing $dx$ by $dt$,
$$\frac{dx}{dt}=v=\frac{v'+v}{1+v'v/c^2}$$
Differentiating gives,
$$\frac{dv}{dt}=a=\frac{dv'}{\gamma^2(1+v'v/c^2)^2dt}$$
Substituting $dt = \gamma(1+v'v/c^2)dt'$,
$$a=\frac{a'}{\gamma^3(1+v'v/c^2)^3}$$


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*How does constant acceleration work with special relativity and the Lorentz transformation?

*Is $v' = \frac{dx'}{dt'}$ proper velocity?

*From my understanding, $a$ is the acceleration of point $P$ measured from reference frame $S$ but what is $a'$? Is it the acceleration of point $P$ measured from reference frame $S'$? If $S'$ was the reference frame of a rocket with constant acceleration, is $a'$ the acceleration measured inside the rocket?  

*Is $a'$ proper acceleration? Wikipedia states that proper acceleration is $a' = \gamma^3a$. How did they get that?
Using $a' = \gamma^3 a = \gamma^3\frac{dv}{dt}$ and integrating with respect to $t$,
$$\frac{v}{\sqrt{1-v^2/c^2}} = a't$$
Rearranging for $v$ gives,
$$v = \frac{dx}{dt} = \frac{a't}{1+(a't/c)^2}$$ 
Integrating again with respect to $t$, 
$$x=\frac{c^2}{a'}(\sqrt{1+(a't/c)^2}-1)$$
Both of these equations can be found here.


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*I don't know how to derive these two formulas:
$$t=\frac{c}{a'}\sinh{(\frac{a't'}{c})}$$
$$T=\frac{c}{a'}\sinh^{-1}{\frac{a't}{c}}$$
 A: It's easy to get lost in all the derivatives of primed and unprimed coordinates.
But your working seems to be (at least mostly) consistent with the following setup.
We suppose the rocket is launched from the origin of an inertial frame of reference $S$ and travels along the $x$ axis in that frame.
At some point in spacetime there is an event $E_1$ in which the rocket passes some point.
We can suppose that this event occurs at time $t_1$ and distance $x_1$ from the origin in frame $S.$ This same event occurs at time $t'_1$ according to a clock on board the rocket.
In event $E_1$ the rocket is traveling at some velocity $v_1$ measured in frame $S.$
There is an inertial frame $S'$ in which the rocket is at rest at that moment.
It would be desirable if the event $E_1$ also occurs at the origin of the frame $S'$, which I do not think your Lorentz transformations allow. (You would need to add a constant term.)
But this also does not make any difference in computing acceleration,
since the constant term gets wiped out by differentiation.
Within a very small interval of time $\Delta t'$ (measured in frame $S'$), the rocket will accelerate to a new velocity.
Let event $E_2$ be the event that occurs at the same location as $E_1$ in frame $S'$
after this interval.
Then in event $E_2$ the rocket will be at rest within a new inertial frame of reference $S''.$
Let $t'_2 = t'_1 + \Delta t'.$
If $a'$ is the proper acceleration of the rocket at time $t_1$--that is, the acceleration felt by an observer in the rocket when it is at rest in frame $S'$--then the velocity of frame $S''$ relative to frame $S'$ is $a' \Delta t'$; that is, speed is acceleration times time.
Note that if $v_1'$ is the velocity of the rocket in frame $S'$ in the event $E_1,$ then $v_1'=0,$ since that is how $S'$ is defined: the rocket is at rest in that frame in that event.
Now let $t_2$ be the time of $E_2$ in frame $S$ and let $\Delta t = t_2 - t_1.$ Then
$\Delta t = \gamma \Delta t',$
where $\gamma = 1/\sqrt{1 - (v_1/c)^2}$ 
is the Lorentz factor between frames $S$ and $S'.$
(We only need one Lorentz factor, so I see no need for a subscript on $\gamma$.)
Then as measured in frame $S,$ at time $t_1$ the rocket's velocity is $v_1,$
and at time $t_1 + \Delta t$ the velocity is
$$ v_2 = \frac{v_1 + a' \Delta t'}{1 + v_1(a'\Delta t')/c^2}. $$
The change in velocity is
\begin{align}
\Delta v = v_2 - v_1 
&= \frac{(v_1 + a' \Delta t') - (v_1 + (a'\Delta t')v_1^2/c^2)}
        {1 + v_1(a'\Delta t')/c^2}\\
&= \frac{(a' \Delta t')(1 - v_1^2/c^2)}{1 + v_1(a'\Delta t')/c^2}\\
&= \frac{a' \Delta t'}{\gamma^2(1 + v_1(a'\Delta t')/c^2)}.
\end{align}
The acceleration of the rocket in frame $S$ at event $E_1$ is approximately
$$
\frac{\Delta v}{\Delta t}
= \frac{a' \Delta t'}{\gamma^2(1 + v_1(a'\Delta t')/c^2)} \cdot
  \frac{1}{\gamma \Delta t'}
= \frac{a'}{\gamma^3(1 + v_1 a'(\Delta t'/\gamma)/c^2)},
$$
which matches what you calculated if $v_1$ is what you called $v$
and $a'\Delta t'$ is what you called $v'.$
Note that if we reduce $\Delta t,$ as $\Delta t$ goes to zero $\Delta t'$ also goes to zero (and vice versa). So the acceleration of the rocket in frame $S$ comes out to
$$ 
a = \frac{dv}{dt}
= \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t}
= \lim_{\Delta t \to 0} \frac{a'}{\gamma^3(1 + v_1(a'\Delta t')/c^2)}
= \frac{a'}{\gamma^3}.
$$
That is, the way we get $a = a'/\gamma^3$ is fundamentally the same way we get any instantaneous acceleration from a relation between velocity and time, namely how any derivative is defined, as a limit of the ratio of two differences.

As you have already shown, from $a = a'/\gamma^3$ we can derive
$$
v = \frac{a't}{\sqrt{1 + (a't/c)^2}}.
$$
Therefore
$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \sqrt{\left(\frac{a't}{c}\right)^2 + 1}.
$$
If $\tau$ represents the time measured by a clock aboard the rocket throughout its flight
(which is the proper time aboard the rocket),
then from in the event $E_1$ described earlier we have $d\tau/dt = dt'/dt = 1/\gamma$.
We again have $d\tau/dt = 1/\gamma$ in any other event on board the rocket
(taking $\gamma$ as the Lorentz factor for the rocket relative to frame $S$ in that event),
so under the assumption that $\tau = 0$
when the rocket was launched we can find by integration that
$$
\tau = \int \frac1{\sqrt{\left(\frac{a't}{c}\right)^2 + 1}}\, dt = \frac c{a'} \sinh^{-1} \frac{a't}{c}.
$$
Solve that equation for $t$ and you have
$$
t = \frac c{a'} \sinh \frac{a'\tau}{c}.
$$

Be aware that I have used the
clock postulate
throughout this derivation, which is how I get to say that an observer aboard the rocket
experiences the passage of $\Delta t'$ time between events $E_1$ and $E_2$
(the same as in the inertial frame $S'$) and that $d\tau/dt = dt'/dt.$
This is an extra assumption we required beyond the Lorentz transformations.
