- #1

LCSphysicist

- 646

- 162

- 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?