Finding t and x in terms of proper time?

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The discussion focuses on finding time and position variables for a particle undergoing uniform acceleration, represented by a constant acceleration vector a=g. The user derived an expression for the four-vector inner product but is unsure how to proceed from there. They clarify that in the instantaneous rest frame, the particle experiences a constant acceleration of 1 g. The conversation emphasizes the importance of using invariant properties of acceleration to relate components in different frames. The goal is to derive the time and position equations based on the established relationships.
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Homework Statement
A particle is moving along x, uniformly accelerated at a=g=constant.
(a) find x and t as a function of proper time (provided at t=0, x=0 and v=0)
Hint: (now a and u are 4-vectors) consider u and a. What are a.a, u.u and a.u? Use these to find the particle's 4-velocity and integrate to find position.
Relevant Equations
Still 4-velocities
u.u=1
a.u=0
u=(γ,γv)
a=(γ[SUP]4[/SUP]v.a, γ[SUP]2[/SUP]a + γ[SUP]4[/SUP](v.a)v)
On the right hand side, v and a are three vectors
I tried finding a.a (four vector inner product) and I got to γ4{(v.a)2(1-γ4v.v - 2γ2) - a.a}, where again a and v are three vectors on the rhs (sorry to be confusing). a.a = g2 since it's a constant.

I have no idea where to go from here to find the time and position. Please help!
 
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Physicsphysics said:
Homework Statement:: A particle is moving along x, uniformly accelerated at a=g=constant.
By"uniformly accelerated", I suspect that they mean that ##a^1 = g## in the instantaneous rest frame of the particle. So, if you were to move with the particle you would always experience ##1 g## of acceleration.

Relevant Equations:: Still 4-velocities
u.u=1
a.u=0
u=(γ,γv)
a=(γ4v.a, γ2a + γ4(v.a)v)
Instead of the last equation, you can use the fact that a.a is an invariant. Thus, a.a in the fixed inertial frame must equal a.a in the instantaneous rest frame of the particle. This allows you to relate ##a^0## and ##a^1## in the fixed frame.
 

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