Special relativity and acceleration

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Homework Statement
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Relevant Equations
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A particle is moving along the x-axis. It is uniformly accelerated in the sense
that the acceleration measured in its instantaneous rest frame is always g, a constant.
Find x and t as functions of the proper time τ assuming the particle passes through
x0 at time t = 0 with zero velocity.I

n particle frame, the acceleration is constant and given by g.

So we have $$dv/d\tau = g \implies x = x_o + v_ot + g\tau^2/2$$
Using the initial conditions,$$ x = x_o + g\tau^2/2 $$

So now we have to transform it to the rest frame coordinates/lab frame.

$$\begin{pmatrix}
ct'\\x'

\end{pmatrix} = \begin{pmatrix}
\gamma & \gamma \beta \\
\gamma \beta & \gamma
\end{pmatrix}

\begin{pmatrix}
ct \\ x = x_o + g\tau^2/2

\end{pmatrix}$$

I am using beta instead of minus beta, because i am changing from a frame in motion to a frame in rest.

Now, assuming that $$(dv/dt) dt/d\tau = g \implies v= gt/\gamma$$
And so, $$\gamma = \sqrt{1+g^2t^2}$$
$$\beta = -gt/(c\gamma)$$

that implies $$t` = \sqrt{1+g^2t^2}(c\tau + \beta*( x_o + g\tau^2/2))/c$$
and $$x' = \sqrt{1+g^2t^2}(c \tau \beta +(x_o + g\tau^2/2))$$

But i am not sure about these results i get, i have the impression i am doing something wrong. Is it right?
 
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robphy said:
It might be helpful to see the form of the equations when [itex]\beta[/itex] is small. (Correspondence principle.)
I think i could have had a mistake when i integrated the v, reading it again... But, i think what's matter more to me is if the approach is right. I mean, solve it using the proper time, and so changing to the rest frame using the inverse matrix transform, what do you think?