Just for fun, here's how the formula for the spacetime interval is derived. Start with the Lorentz transformations:
$x'=\gamma(x-vt)$
$t'=\gamma(t-\frac{vx}{c^2})$
Now consider the quantity $(ct')^2 - x'^2$:
$(ct')^2 - x'^2 = \gamma^2(ct - \frac{vx}{c})^2 - \gamma^2(x-vt)^2$
Now
$\gamma^2 = \frac{1}{1-(v/c)^2} = \frac{c^2}{c^2-v^2}$
So
$(ct')^2 - x'^2 = \frac{c^2}{c^2-v^2}[(ct - \frac{vx}{c})^2 - (x-vt)^2]$
$=\frac{c^2}{c^2-v^2}[(ct)^2 - 2vxt + (\frac{vx}{c})^2 - x^2 + 2vxt - (vt)^2]$
$=\frac{c^2}{c^2-v^2}[(ct)^2 - x^2 + (\frac{vx}{c})^2 - (vt)^2]$
$=\frac{c^2}{c^2-v^2}[(c^2-v^2) t^2 - x^2 (1 - (\frac{v}{c})^2)]$
$=\frac{c^2}{(c-v)(c+v)}[(c-v)(c+v) t^2 - x^2 (\frac{(c^2 - v^2)}{c^2})$
$=\frac{c^2}{(c-v)(c+v)}[(c-v)(c+v) t^2 - x^2 (\frac{(c-v)(c+v)}{c^2})$
$=\frac{c^2}{(c+v)}[(c+v) t^2 - x^2 (\frac{(c+v)}{c^2})$
$= (ct)^2 - x^2$.
We see that the quantity 's', defined by
$s^2 \equiv (ct')^2 - x'^2 = (ct)^2 - x^2$
has the same numerical value in both frames of reference. In other words, s is a spacetime invariant quantity, which is really useful.
We can extend this by considering not just coordinates in spacetime, but actual intervals (i.e. the differences between coordinate measurements for two spacetime events). It is easy to show that:
$(\Delta s)^2 \equiv (c\Delta t')^2 - (\Delta x')^2 = (c\Delta t)^2 - (\Delta x)^2$
which tells us that the "spacetime interval", $\Delta s$ between two events in different frames is a constant for any two given events, and it can be measured in any inertial frame.
It turns out that the sign of $\Delta s$ is also really useful for determining whether it is possible for two given spacetime events to be causally connected to one another. A special case is that any two events that have $\Delta s=0$ can communicate by exchanging a light signal.
$x'=\gamma(x-vt)$
$t'=\gamma(t-\frac{vx}{c^2})$
Now consider the quantity $(ct')^2 - x'^2$:
$(ct')^2 - x'^2 = \gamma^2(ct - \frac{vx}{c})^2 - \gamma^2(x-vt)^2$
Now
$\gamma^2 = \frac{1}{1-(v/c)^2} = \frac{c^2}{c^2-v^2}$
So
$(ct')^2 - x'^2 = \frac{c^2}{c^2-v^2}[(ct - \frac{vx}{c})^2 - (x-vt)^2]$
$=\frac{c^2}{c^2-v^2}[(ct)^2 - 2vxt + (\frac{vx}{c})^2 - x^2 + 2vxt - (vt)^2]$
$=\frac{c^2}{c^2-v^2}[(ct)^2 - x^2 + (\frac{vx}{c})^2 - (vt)^2]$
$=\frac{c^2}{c^2-v^2}[(c^2-v^2) t^2 - x^2 (1 - (\frac{v}{c})^2)]$
$=\frac{c^2}{(c-v)(c+v)}[(c-v)(c+v) t^2 - x^2 (\frac{(c^2 - v^2)}{c^2})$
$=\frac{c^2}{(c-v)(c+v)}[(c-v)(c+v) t^2 - x^2 (\frac{(c-v)(c+v)}{c^2})$
$=\frac{c^2}{(c+v)}[(c+v) t^2 - x^2 (\frac{(c+v)}{c^2})$
$= (ct)^2 - x^2$.
We see that the quantity 's', defined by
$s^2 \equiv (ct')^2 - x'^2 = (ct)^2 - x^2$
has the same numerical value in both frames of reference. In other words, s is a spacetime invariant quantity, which is really useful.
We can extend this by considering not just coordinates in spacetime, but actual intervals (i.e. the differences between coordinate measurements for two spacetime events). It is easy to show that:
$(\Delta s)^2 \equiv (c\Delta t')^2 - (\Delta x')^2 = (c\Delta t)^2 - (\Delta x)^2$
which tells us that the "spacetime interval", $\Delta s$ between two events in different frames is a constant for any two given events, and it can be measured in any inertial frame.
It turns out that the sign of $\Delta s$ is also really useful for determining whether it is possible for two given spacetime events to be causally connected to one another. A special case is that any two events that have $\Delta s=0$ can communicate by exchanging a light signal.