# Time travel (Velocity and Mass)

I learned that if I move on a high velocity, if my watch shows 12:00 and a my home clock shows 12:00 before the trip, when I come back, my watch will be like 12:05 while the clock will be 12:10. I also hear about the same thing when we are near a massive object. Is there a formula about this? For example if I travel 1h at 100km/h, how many times will I be younger than the rest of the world?

There most certainly are formulas for these things! I don't know precisely what your background is but I'll try to explain it in very elementary terms. I'll discuss the case where there is no effect of gravity (or when this is negligible), since if gravity is incorporated the story becomes a bit more complicated.

So say we neglect gravity, and we consider two observers $\mathcal O_1$ and $\mathcal O_2$, where observer $\mathcal O_1$ is at rest, but the other observer may move as she likes. Both observers have clocks, and there is a relatively simple formula that relates the elapsed times $\Delta t$ and $\Delta \tau$ on the two clocks. We consider the situation in which the two observers are initially at the same location. Their clocks both show an elapsed time of zero at that point. Let $v(t)$ be the function that gives the velocity $v$ of observer $\mathcal O_2$ at each time $t$ on the clock of the stationary observer $\mathcal O_1$. (For simplicity we consider only motion in the $x$-direction.)

If observer $\mathcal O_1$ now measures a time $\Delta t$ on her clock, then observer $\mathcal O_2$ will measure a time \begin{align} \Delta \tau = \int_{0}^{\Delta t} \sqrt{1-\frac{v(t)^2}{c^2}}\, \text{d}t, \tag{1} \end{align} where $c$ denotes the speed of light. First of all let's check if this makes sense. Suppose observer $\mathcal O_2$ does not move at al! Then $v(t)=0$ and hence \begin{align} \Delta\tau = \int_0^{\Delta t}1\,\text{d}t = \Delta t \end{align} and so both stationary observers measure the same elapsed time, as expected. So far the formula seems to work!

Now let's consider a different example: observer $\mathcal O_2$ moves with constant velocity $v$ for a time $\Delta t/2$, and then back with velocity $-v$, again for a time $\Delta t/2$, making for a total time $\Delta t$ (on the clock of observer $\mathcal O_1$, as indicated by the use of $t$ instead of $\tau$), after which the observers meet again. Then, since $v^2=(-v)^2$, the integral $(1)$ becomes

\begin{align} \Delta \tau = \int_{0}^{\Delta t} \sqrt{1-\frac{v(t)^2}{c^2}}\, \text{d}t = \int_{0}^{\Delta t} \sqrt{1-\frac{v^2}{c^2}}\, \text{d}t = \sqrt{1-\frac{v^2}{c^2}}\int_{0}^{\Delta t} \, \text{d}t = \sqrt{1-\frac{v^2}{c^2}}\Delta t. \end{align} Let's see what we can learn from this. Since $v$ must always be smaller than the speed of light, $v^2/c^2$ is smaller then 1. Also, because of the squares, $v^2/c^2$ is always positive. Hence $v^2/c^2$ lies between $0$ and $1$, which means that $1-v^2/c^2$ also lies between $0$ and $1$. Finally then, the multiplication factor $\sqrt{1-v^2/c^2}$ lies between $0$ and $1$, and therefore the formula above tells us that $\Delta\tau$ is always smaller than or equal to $\Delta t$. In other words:

The moving observer experiences less time then the observer which is at rest.

To get a feeling for the actual numbers, suppose observer $\mathcal O_2$ moves at a velocity of $7000$ km/h, which, according to a little searching on the internet, seems to be about the top speed that is reached by the fastest fighter jets today. Then doing the calculation shows that

\begin{align} \Delta \tau \approx 0.99999999997 \Delta t, \end{align}

which is an extremely small difference in elapsed times. (It is a difference of about 1 second on a total time of 2000 years.) For normal everyday speeds the difference is even a lot smaller than this, so definitely not noticeable. On the other hand, for a hypothetical spaceship that moves at a velocity of $90\%$ of the speed of light, i.e., $v=0.90\,c$, we get

\begin{align} \Delta \tau \approx 0.44 \Delta t, \end{align}

so at that speed the stationary observer ages approximately twice as fast as the moving observer.