- #1

phosgene

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## Homework Statement

A particle P undergoes the hyperbolic motion

[tex]x_{P}(t) = \frac{c}{a}(c^2 + a^2 t^2)^\frac{1}{2} [/tex]

along the x-axis of frame S, where c is the speed of light and a is a constant. A second particle, Q, undergoes the motion

[tex]x_{Q}(t) = \frac{1}{2}ct + \frac{c^2}{a}[/tex]

and so just passes P at time t=0. Show that according to a clock moving with P, the two particles meet again after a time [tex]\frac{a}{c}ln(3)[/tex], while according to a clock moving with Q, the elapsed time is [tex]\frac{2c}{\sqrt{3}a}[/tex]

## Homework Equations

The clock hypothesis is assumed, so the clocks are assumed to depend on instantaneous velocity, but not acceleration.

Lorentz transform for time between inertial frames with relative velocity V along the x-axis.

[tex]t'=\gamma (t-\frac{Vx}{c^2})[/tex]

## The Attempt at a Solution

Seems like a simple problem, solve for t when x

_{P}= x

_{Q}and then use the Lorentz transforms to get the time taken in the two frames corresponding to each particle being at rest, but my answers are not correct. Here's what I did:

[tex]x_{P}=x_{Q} [/tex]

[tex]\frac{c}{a}(c^2 + a^2 t^2)^\frac{1}{2} = \frac{1}{2}ct + \frac{c^2}{a}[/tex]

Square both sides and move all terms to one side to get:

[tex]\frac{3}{4}t^2 - \frac{c}{a}t=0[/tex]

[tex]t(\frac{3}{4}t - \frac{c}{a})=0[/tex]

[tex]t=0 \ or\ t=\frac{4c}{3a}[/tex]

t=0 corresponds to their initial meeting, so they must meet again at the latter. We plug that value into the equations for x

_{Q}or x

_{P}to get the spatial coordinate to use in the Lorentz time transformation.

[tex]x_{Q}(\frac{4c}{3a})=\frac{1}{2}c(\frac{4c}{3a}) + \frac{c^2}{a}=\frac{5c^2}{3a}[/tex]

To find the relative velocities between frames, we take time derivatives of x

_{Q}and x

_{P}:

[tex]\frac{d}{dt}x_{Q}(t)=v_{Q}(t)=\frac{1}{2}c[/tex]

[tex]\frac{d}{dt}x_{P}(t)=v_{P}(t)=\frac{act}{\sqrt{c^2+a^2 t^2}}[/tex]

[tex]v_{P}(\frac{4c}{3a})=\frac{4}{5}c[/tex]

Transforming to the frame S', which is the frame where x

_{P}is at rest when the particles meet in S, I get a time of zero:

[tex]t' =\gamma (t-\frac{Vx}{c^2}) = \frac{\frac{4c}{3a} - \frac{\frac{4c}{5} \frac{5c^2}{3a}}{c^2}}{\sqrt{1-\frac{16}{25}}} = \frac{ \frac{4c}{3a} - \frac{4c}{3a} } { \sqrt{\frac{9}{25} } }=0[/tex]

That's clearly not correct. What am I doing wrong?