Let's say that a spaceship moves with speed $v$ relative to an observer in earth. Let's make the calculations for the reference system of earth, so the observer is stationary and the spaceship is moving with a speed close to the speed of light.
The proper time $\tau$ is the time the stationary observer in earth counts. But the clock in the spaceship is moving slower, so for the spaceship clock, time $t$ will have passed. And this $t$ will be less than $τ$ since the clock is ticking slower inside the spaceship, from the perspective of the unmoving observer. And to find out how smaller it is we need the Lorentz factor: $$t = τ\sqrt{1 - \frac{v^2}{c^2}}$$ which tells us that indeed, $t<τ$.
That's my understanding, but I'm not sure if what I just typed is correct. If someone can tell me if I'm right or if not, or if not what's my mistake then I'd greatly appreciate it.
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