# Relativity and distance/time

1. Oct 27, 2014

### Meanwhile

Is it correct that the time it takes to travel a distance from the point of view of the traveler (measured by their clock) is the same as it is in classical mechanics?

I.e. if you start accelerating at a constant rate, at some point you will go "faster than light" if you define speed as distance divided by onboard time (despite you'll always stay sub c for an external observer)?

2. Oct 27, 2014

### Staff: Mentor

Yes. If the distance traveled is $D$, the time taken on the journey is $T$, and the speed is $S$ and all are measured by the same inertial observer, then $D=ST$, $T=D/S$, $S=D/T$. But you have to remember that bit about "all are measured by the same inertial observer" - you cannot mix the values seen by different inertial observers or measured in different reference frames in these formulas.

Thus, it does not follow that:
We know what your onboard time is, but what value do we use for the distance? As the ship accelerates, its speed increases so the distance between origin and destination as measured by the onboard observer is being reduced by length contraction. There is no moment in the journey, including at the end when the traveller zooms past the destination at very close to $c$, that the distance as measured by the traveler divided by the travel time comes out greater than $c$.

3. Oct 27, 2014

### A.T.

4. Oct 28, 2014

### Meanwhile

Thank you. I was always wandering why such an obvious concept is hidden so deep in the theory. I think relativity should be taught "backwards": proper velocity first, THEN time dilation and so on.