Special relativity and acceleration

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The discussion focuses on the analysis of a uniformly accelerated particle along the x-axis in the context of special relativity. The particle's acceleration is constant in its instantaneous rest frame, leading to the equations for position and time as functions of proper time. Transformations to the lab frame are applied using Lorentz transformations, but there are concerns about the accuracy of the derived equations. Participants highlight the importance of correctly applying the principles of special relativity, especially when the particle's velocity is not zero. The need for a reevaluation of the approach is emphasized, particularly regarding the integration of velocity and the validity of the equations in different frames.
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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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It might be helpful to see the form of the equations when \beta is small. (Correspondence principle.)
 
robphy said:
It might be helpful to see the form of the equations when \beta 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?
 
You only have ##dx/d\tau = g## instantaneously in any frame when the particle is at rest in that frame. As soon as the particle deviates from zero velocity in that frame, it is no longer valid. Therefore you willhave to rethink your approach.
 
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