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An object undergoing constant proper acceleration has the 4-vectors describing their location obey $$\frac{d\vec r }{d\tau }\cdot \frac{d\vec r}{d\tau }=-1$$, $$\frac{d^2\vec r}{d\tau ^2}\cdot \frac{d^2\vec r}{d\tau \:^2}=a^2=constant$$

How do I find the solution $\vec r\left(\tau \right)$. (For example for an object accelerating in the x-direction starting at $x=L$ at $t=0$). Also what would the final velocity be in the lab frame after a proper time $\tau _f $ has elapsed.

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If you pick a series of frames so that the acceleration is only in the $x$ direction, you will get the following system of differential equations:

$$-\left(\frac{dt}{d\tau}\right)^2 + \left(\frac{dx}{d\tau}\right)^2 = -1$$

$$-\left(\frac{d^2t}{d\tau^2}\right)^2 + \left(\frac{d^2x}{d\tau^2}\right)^2 = a^2$$

From the first equation one could guess that $\dot{t} = \cosh(k(\tau))$ and $\dot{x} = \sinh(k(\tau))$ for some function $k$. Then plugging in to the second equation,

$$\dot{k}^2 = a^2 \implies k(\tau) = \pm a\tau$$

This means that

$$t(\tau) = \frac{1}{a}\sinh(a\tau) + C_1 \hspace{20 pt} x(\tau) = \frac{1}{a}\cosh(a\tau) + C_2$$

Notice that the $\pm$ doesn't affect the parametrization of $t$, but does affect which direction $x$ accelerates in. To have an acceleration in the positive $x$ direction, pick $+a$. Then plugging in our initial conditions, we have

$$t(\tau) = \frac{1}{a}\sinh(a\tau) \hspace{20 pt} x(\tau) = \frac{1}{a}(\cosh(a\tau)-1) + L$$

where we choose the the object to start off in the same (instantaneous) rest frame as the laboratory. If the object was already at some speed, that changes "initial" value of $\tau$, but will only shift $\tau$ by a constant in our parametrizations (since proper time is the arc length parametrization of the curve, all unit speed parametrizations will be equivalent to $\tau$ mod adding a constant).

To find the speed in the lab frame after some proper time $\tau_f$, use the chain rule for parametric curves:

$$\frac{dx}{dt} = \frac{\frac{dx}{d\tau}}{\frac{dt}{d\tau}} = \tanh(a\tau_f)$$

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