Spacetime Interval - Classical Physics
Spacetime interval can be introduced also in classical Physics.
It answers to the question: "Is it possible reach C from A using only the velocities +-v?".
But spacetime intervals ARE NOT INVARIANT in classical Physics: the value of v change in different inertial frames. In RR there is a single velocity that is invariant in any inertial system, the velocity of the LIGHT, so if we calculate the spacetime interval using the "c" velocity instead of "v", we can obtain an invariant.
Here we are using a frame (x,t) with specific units: the light speed is 1 light-second/second, so from O, the yellow line can represent the motion of a light beam.
As you can see, in the initial configuration we can reach C from A moving 1.5 seconds with velocity v and returning with velocity -v for 0.5 seconds.
If d = (x_C- x_A)² - v² (t_C - t_A)² = 0 we can reach C from A using only the velocity +v or -v: in effect
(x_C- x_A)² - v² (t_C - t_A)² = 0 -->
(x_C- x_A)²= v² (t_C - t_A)² -->
(x_C- x_A)/ (t_C - t_A) = +-v
if d < 0 we can reach C from A using a mixed path with +v or -v velocities.
Why? if C can be reached from A with a velocity v' < v it is always possible to go beyond the x_C position and return back.
if d > O it is'nt possible.
Also a moving frame has the same d value: this is not explained in the same image, anyway - for example - if the frame is moving at velocity v, the line from A to the vertex of the hyperbola mark a velocity -v, and from vertex to C is 2v in original, but only +v for this frame.
In galilean relativity intervals of time are invariant for frames that move each other with constant velocity: not so for RR. If we use v=c, space and time are measured as different for different frames, but if we divide the values, we must obtain the same result for light speed.
In RR "d" is invariant only if v=c : the speed of light is invariant, so we need to place the cursor only in +1 or -1 light-second/second positions.
Look also to https://www.geogebra.org/m/wqf5bEum