- #1
LCSphysicist
- 644
- 163
- Homework Statement
- N
- Relevant Equations
- N
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?
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?