Special Relativity, accelerating frames, proper time

[iamaelephant]

Homework Statement

A particle has a constant acceleration in a laboratory from 0 to 0.5c in 2 seconds. What time elapses for the particle (i.e. what is the proper time for the particle)?
Hint You will have to integrate the proper time of the particle over the two seconds as measured in the laboratory frame. You may need to look up an integral such as $$\int\sqrt{a^2 - x^2} dx$$

Homework Equations

You tell me

The Attempt at a Solution

I'm pretty sure this isn't a difficult problem, I think I may be going about it the wrong way.
I figure I need to integrate dt' over the interval of acceleration, but I'm confused about how I go about this. Any help would be greatly appreciated.

 

[Dick]

It's constant acceleration in the lab frame. That means v=a*t. dt'=dt*sqrt(1-v^2/c^2). You are right, this shouldn't be at least that hard to start. Please start and tell us where you are confused.

 

[iamaelephant]

Hi Dick, sorry for the late reply, busy weekend. Thanks for your help, I think I got it but I'd appreciate if you (or someone) could check to see if I did things right.

$$dt' = \frac{dt}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{dt}{\sqrt{1-\frac{(at)^2}{c^2}}}$$
$$= \frac{dt}{\sqrt{1 - t^2(\frac{a}{c})^2}}$$
$$= \frac{(\frac{c}{a})^2}{\sqrt{(\frac{c}{a})^2 - t^2}} dt$$
Therefore
$$t' = \frac{c}{a} \int \frac{dt}{\sqrt{(\frac{c}{a})^2 - t^2}}$$
$$= \frac{c}{a} \arcsin{\frac{t^2}{(\frac{c}{a})^2}$$
$$= 4 \arcsin{\frac{2}{16}}$$
$$= 0.501 seconds$$

I'm not sure how to put limits of integration into Latex but it's implied throughout to be t=0 and t=2s

The answer seems reasonable but I'd appreciate if someone could quickly go over my maths to see if I bungled it. Thanks! :D

 

[iamaelephant]

Also just so I'm clear, does "proper time" mean the amount of time that has passed in the frame of some object, regardless of whether it is accelerating or not? The words don't even appear in my textbook unfortunately.