# Special Relativity: Time Dilation

I am having difficulty solving the following problem:

How fast must a rocket travel relative to the earth so that time in the rocket "slows down" to half its rate as measured by earth-based observers?

My approach to the question:

$\Delta t$ is the time measured by the earth-based "rest frame" observers. $\Delta t_0$ is the time measured by the rocket in the "moving frame" relative to earth.

If the rocket "slows down" to half its rate as measured by earth-based observers, does that mean $\Delta t= \frac{\Delta t_0}{2}$ ?

If so, then $\Delta t= \frac{\Delta t_0}{2} = \gamma \Delta t_0$ , Where $\gamma$= $\frac{1}{\sqrt[]{1-u^2/c^2}}$

Rearranging becomes $\frac{1}{2} = \frac{1}{\sqrt[]{1-u^2/c^2}}$

so $1- \frac{u^2}{c^2} =2^2$

then $c^2 (1-4)=u^2$

then $u^2=-3c^2$

However, this cannot be correct, since I get a negative square root by solving for u.

All help is greatly appreciated!

## 1 Answer

The rocket's clock appears to be ticking slowly. This means that a time which I measure as, say, ten seconds, is only measured as, say, five seconds by the slow ticks of the rocket clock.

Hence, $\Delta t = 2 \Delta t_0$, not $1/2$.