Time taken for two relativistic particles to meet

[phosgene]

Homework Statement

A particle P undergoes the hyperbolic motion

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

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

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

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 \frac{a}{c}ln(3), while according to a clock moving with Q, the elapsed time is \frac{2c}{\sqrt{3}a}

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.

t'=\gamma (t-\frac{Vx}{c^2})

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:

x_{P}=x_{Q}
\frac{c}{a}(c^2 + a^2 t^2)^\frac{1}{2} = \frac{1}{2}ct + \frac{c^2}{a}
Square both sides and move all terms to one side to get:
\frac{3}{4}t^2 - \frac{c}{a}t=0
t(\frac{3}{4}t - \frac{c}{a})=0
t=0 \ or\ t=\frac{4c}{3a}
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.

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

To find the relative velocities between frames, we take time derivatives of x_Q and x_P:

\frac{d}{dt}x_{Q}(t)=v_{Q}(t)=\frac{1}{2}c
\frac{d}{dt}x_{P}(t)=v_{P}(t)=\frac{act}{\sqrt{c^2+a^2 t^2}}
v_{P}(\frac{4c}{3a})=\frac{4}{5}c

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:

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

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

 

[Orodruin]

Why are you Lorentz transforming? It is a question of integrating the proper time over the respective world lines.

 

[phosgene]

Hi, thanks for the reply. The book doesn't cover the technique you mention and I don't know how to do it. The question references a previous one in which the particles aren't accelerating and the method is to use a Lorentz transform for velocity.

 

[PeroK]

Hi, thanks for the reply. The book doesn't cover the technique you mention and I don't know how to do it. The question references a previous one in which the particles aren't accelerating and the method is to use a Lorentz transform for velocity.

Where you've gone wrong is this:

You calculate the coordinate time in the reference frame when the particles meet. You calculate the velocity of each particle in that frame at this time. But, this is an instantaneous velocity. The Lorentz Transformation relates the elapsed time between two frames moving at a constant relativity velocity of $V$.

This is not going to be valid to calculate an elapsed time over a period when the relative velocity between the two particles has been changing.

 

[phosgene]

Thanks for that, I'll go back to the drawing board and see what I can come up with.

 

[PeroK]

Thanks for that, I'll go back to the drawing board and see what I can come up with.

Do you know how to calculate the proper time of a particle with variable velocity?