Is it possible to derive the time dilation expression directly from the space-time interval? Is there any way of deriving the time dilation expression:
$$\Delta t'=\gamma\Delta t$$
from the space-time interval:
$$s^2=c^2t^2-x^2-y^2-z^2$$
I know how to derive the time dilation expression by using geometry and simplifying, but I was wondering if there is a way to do it directly from the invariant as it encodes a lot of special relativity within it. Thanks!
 A: First, imagine a stopwatch at rest that measures a time interval $\Delta t$. This concretely means there are two events: event 1 is that we start the stop watch at time $t_1$ and position $x_1$, and event 2 is that we stop the watch at time $t_2=t_1+\Delta t$ while the watch is still at position $x_2$. The spacetime interval in the stopwatch frame is then
\begin{equation}
\Delta s^2 = - c^2 (\Delta t)^2
\end{equation}
Now imagine a passing observer on a rocket ship, moving at velocity $v$, and where we denote coordinates in this frame with primes. From the observer's perspective, the stopwatch starts at time $t_1'$ and position $x_1'$, and is stopped at time $t_2'$ and position $x_2'$. Note that we do not know how $t_1'$ and $t_2'$ are related in this frame. But, we do now how $x_1'$ and $x_2'$ are related; in particular, $x_2' = x_1' - v (t_2'-t_1')$. Or, defining $\Delta x' \equiv x_2' - x_1'$ and $\Delta t'=t_2'-t_1'$, we have that $v=-\frac{\Delta x'}{\Delta t'}$.
We can also compute the spacetime interval between the same two events (stopwatch start and stop) in this frame. We get
\begin{eqnarray}
\Delta s^2 &=& -c^2 (\Delta t')^2 + (\Delta x')^2 \\
&=& -c^2 (\Delta t')^2 \left[1 - \frac{(\Delta x')^2}{c^2 (\Delta t')^2}\right] \\
&=& -c^2 (\Delta t')^2 \left[1-\frac{v^2}{c^2}\right] \\
&=& - \frac{c^2 (\Delta t')^2}{\gamma^2}
\end{eqnarray}
where in the last line I've defined the usual factor $\gamma \equiv (1-v^2/c^2)^{-1/2}$.
Since the spacetime interval is the same in any frame, we can equate these two expressions for the interval. We find
\begin{equation}
-c^2 (\Delta t)^2 = - \frac{c^2 (\Delta t')^2}{\gamma^2}
\end{equation}
or, rearranging,
\begin{equation}
(\Delta t') = \gamma (\Delta t)
\end{equation}
Since $\gamma > 1$, the time interval in the moving frame is larger than in the stationary frame, which is time dilation ("moving clocks run slow.")
In general, all of the phenomena of special relativity can be derived from the invariance of the spacetime interval.
A: For simplicity, work in one spatial dimension.
Consider an inertial timelike segment from $O$ to $Q$.
Write $\Delta s_{OQ}^2=\Delta t_{OQ}^2-\Delta x_{OQ}^2$,
where we interpret the spacetime-displacement $\Delta s_{OQ}$
and $\Delta t_{OQ}$ is the adjacent side and $\Delta x_{OQ}$ is the opposite side.
Let $\theta$ be the Minkowski-angle (called the "rapidity")
between the timelike directions $\Delta t_{OQ}$ and $\Delta s_{OQ}$. ​
Trigonometrically, time dilation factor $\gamma (=\cosh\theta)$ is
the ratio of the adjacent side to the hypotenuse $OQ$.
So, we compute
\begin{align}
\gamma
&=\frac{ADJ}{HYP}\\
&=\frac{\Delta t_{OQ}}{\Delta s_{OQ}}\\
&=\frac{\Delta t_{OQ}}{\sqrt{ \Delta t^2_{OQ} - \Delta x^2_{OQ}  }}\\
&=\frac{\Delta t_{OQ}}{\Delta t_{OQ}\sqrt{ 1 - \frac{\Delta x^2_{OQ}}{\Delta t^2_{OQ}}  }}
&=\frac{1}{\sqrt{ 1 -  \left(\frac{\Delta x_{OQ}}{\Delta t_{OQ}}\right)^2  }}
\end{align}
A: Setting c = 1: $\tau = \sqrt{t^2 - x^2 - y^2 - z^2}$
$$\frac{d\tau}{dt} = \frac{\partial \tau}{\partial t} + \frac{\partial \tau}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial \tau}{\partial y}\frac{\partial y}{\partial t} + \frac{\partial \tau}{\partial z}\frac{\partial z}{\partial t}$$
For constant velocity, $\frac{\partial x_i}{\partial t}=\frac{x_i}{t}$
$$\frac{\partial \tau}{\partial t} = \frac{1}{\sqrt{1 - v_x^2 - v_y^2 - v_z^2}} = \frac{1}{\sqrt{1 - v^2}}$$
$$\frac{\partial \tau}{\partial x}\frac{\partial x}{\partial t} = \frac{-x}{\sqrt{1 - x^2 - y^2 - z^2}}v_x = \frac{-v_x^2}{\sqrt{1 - v_x^2 - v_y^2 - v_z^2}} = \frac{-v_x^2}{\sqrt{1 - v^2}}$$
Similar for y and z.
$$\frac{d\tau}{dt} = \frac{1 - v_x^2 - v_y^2 - v_z^2 }{\sqrt{1 - v^2}} = \frac{1 - v^2}{\sqrt{1 - v^2}} = \sqrt{1 - v^2} = \frac{1}{\gamma} \implies \frac{dt}{d\tau} = \gamma$$
This type of derivation assumes that the expression for the Minkowski space is a postulate.
