Proper time in an acceleration frame

[LCSphysicist]

Homework Statement

.

Relevant Equations

.

$$d\tau = \sqrt{-g_{\mu v}\frac{dx^{\mu}}{d\lambda} \frac{dx^{v}}{d \lambda}} d \lambda $$
Calling $d \lambda = d t$, t is coordinate time in an inertial frame.
$$\Delta \tau = \int \sqrt{-(-1+v^2/c^2)} dt$$
$$\Delta \tau \approx \Delta t - \frac{g^2*\Delta t^2}{6c^2}$$
where $$\Delta t = 2 \sqrt{L/g}, L = 30,000ly$$

So we just need to substitute it. Is this right?

 

[anuttarasammyak]

Your second equation of time integration gives $\tau(t)$. To integrate it we need to know $v(t)$. You can get it from the relation of v(t) and constant acceleration below
\frac{d}{dt}\frac{v}{\sqrt{1-v^2/c^2}}=g
By further integrating v(t) with time you get x(t) so you can get $\tau(x)$ to know proper time for one way the amount of which should be doubled for a round trip.

 

[PeroK]

Homework Statement:: .
Relevant Equations:: .

View attachment 278319
$$d\tau = \sqrt{-g_{\mu v}\frac{dx^{\mu}}{d\lambda} \frac{dx^{v}}{d \lambda}} d \lambda $$
Calling $d \lambda = d t$, t is coordinate time in an inertial frame.
$$\Delta \tau = \int \sqrt{-(-1+v^2/c^2)} dt$$
$$\Delta \tau \approx \Delta t - \frac{g^2*\Delta t^2}{6c^2}$$
where $$\Delta t = 2 \sqrt{L/g}, L = 30,000ly$$

So we just need to substitute it. Is this right?

No, it's not right. I don't think you've understood the problem at all. The proper acceleration must be $g$. You've given no explanation on how you calculated that integral.

Have you heard of the hyberbolic trig functions?

The answer you give can't be right. You're going to get $\Delta \tau$ to be negative.