# Time dilation

### Gamma factor

From now on, I will often use the notation $$\gamma = \gamma(v) = \frac{1}{\sqrt{1 - v^2}}$$

For reference, we note that $$\gamma \gt 1 \text{ for all } v \in (-1, 1)$$ and $$\lim_{v \rightarrow 1} \gamma = \infty$$

If we set $c \neq 1$, meaning we replace $$v \rightarrow \frac{v}{c}$$ then $$\gamma = \gamma(v) = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

### Proper time

In figure 4.5 we consider an event $E$ in the time-like region of a coordinate frame $O$.

Since the event is time-like, we can draw in the world-line of a clock ($O'$) travelling at velocity $v$ say, which intersects the origin $O$, at time $t' = 0$ and $E$ at time $t' = \tau$ as measured by the clock so that $\tau$ is the proper time between $O$ and $E$.

Nb. Remember that space-time is being stretched and squeezed, hence lengths and times on space-time diagrams are scaled according to which frame you are using to measure them. To help calibrate the scale of frames, we include relevant invariant hyperbolae, which allow us to compare lengths and times in all frames.

Figure 4.5 - Time dilation

We would like to calculate the coordinate time of $E$ in the $O$ frame.

We could use Lorentz transformations, but we could also notice that the $O$ components of $E$ satisfy the two equalities \begin{align*}x - vt &= 0\\t^2 - x^2 &= \tau^2\end{align*}

Thus, \begin{align*}\tau &= \sqrt{t^2 - x^2}\\&= t \sqrt{1 - v^2}\end{align*} or $$t = \gamma \tau \gt \tau$$ since $\gamma \gt 1$.

This means that the coordinate time $t$ is greater than the proper time $\tau$. This phenomenon is called time dilation.

In the limit, where $v \rightarrow 0$, $t \rightarrow \tau$ and we get back Galilean universal time. If we put $c \neq 1$ back in, $$t = \frac{\tau}{\sqrt{1 - \frac{v^2}{c^2}}} \rightarrow \infty \text{ as } v \rightarrow c$$

Notice that we cannot say the following about the event $E$. We cannot say that since we measured the event to occur at time $t$ in the coordinate (stationary) frame, then the time measured in the moving frame would be dilated - in a relativistic fashion. The problem with this is that $t$ isn't the proper time.

Figure 4.6 - Time dilation again

If we now consider that $E$ occurs a proper time $t = \tau$ in the stationary frame and compare it with the time as measured by the clock (moving at speed $v$) in figure 4.5, we would need to draw in the clock's line of simultaneity that intersects $t = \tau$, which is $$t - vx = \tau$$

But we also know, from Lorentz, that $$t' = \gamma\left(t - vx\right) = \gamma \tau$$

Hence, as before, coordinate time, $t'$ is greater than proper time, $t = \tau$.

Twins are born at a certain location, which we'll call the origin. One is born at rest and one is born with a certain speed, $v$. This is not a realistic experiment of course, just a thought experiment.

After a certain coordinate time $T$ say, the moving twin instantaneously reverses direction, $E_1$, comes back at the same speed as it went out with and arrives at the origin at the event $E_2$ at time $2T$, which the time the stationary twin experiences
The moving twin experiences twice the proper time $\tau$ between the events $O$ and $E_1$ (by symmetry the second leg of the journey will take the same amount of time as the first, for both observers), which is related to the coordinate time of $E_1$ by the
expression for time dilation, $$T = \gamma \tau$$
So, $$2T = 2 \gamma \tau \gt 2 \tau$$ and the stationary twin experiences more time than the moving twin. That is, the stationary twin will be older than the moving twin.
The supposed paradox is that surely, since everything is relative, we can reverse the roles of the twins - the stationary one is in fact moving at velocity $-v$ relative to the previously moving twin. Then, surely the previously stationary twin would age less than the previously moving twin.