The Twins Paradox and the speed of light

The basic answer is that the twin that went in the rocket has to accelerate to turn around and head back to Earth. Even though velocity is relative, acceleration is absolute. That is everybody will agree which twin is accelerating and which one isn't. This acceleration introduces an absolutely agreed upon difference in their paths through spacetime which is responsible for one being younger than the other when they are reunited.

The spacetime of special relativity, formally called Minkowski space, has a different way of calculating the distance between points.In a bit more detail:
Take a space with 4 space dimensions, which I'll label with x, y, z, w.
Also take Minkowski spacetime with 3 space and 1 time dimension and coordintes x, y, z, t.
If I pick two points in either space, then ds denotes the distance between them. dx, dy, dz, dt or dw denote the difference in the values of that coordinate between the two points.Then distance in the purely spatial space is calculated by:
ds^2 = dw^2 + dx^2 + dy^2 + dz^2And distance in Minkowski space is calculated by:
ds^2 = -dt^2 + dx^2 + dy^2 + dz^2The "weird" geometry comes down to that minus sign.

If nobody minds I'll do the calculation for a twin that stays on Earth and one that goes to the moon and back. All calculations are done in the twin who remains on Earth's frame.
I'm also going to use ds^2 = dt^2 - dx^2 - dy^2 - dz^2, which is equivalent to ds^2 = -dt^2 + dx^2 + dy^2 + dz^2, but easier to use in this case.Basic set up:
First of all I have to get the problem of units out of the way.Since I'm computing a spacetime distance I'm going to have to use the same units for all quantities. Distances in space are measured using meters. So let's take a meter as our measurement. However humans use seconds for measuring distances in time, in order to compute the spacetime distance I'll have to convert a second into meters.Well basically a second is (roughly) 300,000,000 meters in the temporal direction.Now I'll ignore y and z. So I'll be using ds^2 = dt^2 - dx^2.First of all the starting location of both twins will be labelled as:
(0 ; 0). That is t=0, x=0.Stationary twin:
If one twin sits where they are for four seconds they end up with coordinates:
(1,200,000,000 ; 0) or t=1,200,000,000 and x=0.Now I'll compute the spacetime distance. Since dx=0 (no difference or change in spatial coordinate) we just have ds^2 = dt^2.dt = 1,200,000,000 - 0 = 1,200,000,000
dt^2 = 1,440,000,000,000,000,000
Hence ds^2 = 1,440,000,000,000,000,000 and taking the square root:
ds = 1,200,000,000 meters.Moving twin:
The moon is roughly 384,000,000 meters from Earth. The second twin starts at Earth and travels to the moon in two seconds.
So they start at (0 ; 0) and end up at (600,000,000 ; 384,000,000).
The spatial difference is dx = 384,000,000 - 0 = 384,000,000
Similarly, dt = 600,000,000.
dx^2 = 147,456,000,000,000,000
dt^2 = 360,000,000,000,000,000ds^2 = 360,000,000,000,000,000 - 147,456,000,000,000,000 = 212,544,000,000,000,000.Taking the square root, ds = 461,024,945 meters.Assuming the twin takes an exactly similar journey back to Earth, that is they return in two seconds, then the distance for the return journey is again ds = 461,024,945 meters.Hence the total spacetime distance of the moving twin is
ds = 2 x 461,024,945 meters = 922,049,890 meters.Which is significantly less than the 1,200,000,000 meters of the stationary twin. Hence spacetime distance is reduced by moving through space.

No, if the ds^2 is negative between two points in spacetime it indicates that they cannot be reached from one another or influence each other.