# Question about spacetime interval in acceleration in special relativity

In special relativity the spacetime interval between two events can be represented by the equation $${\Delta_s}^2={\Delta_x}+{\Delta_y}^2+{\Delta_z}^2-c^2{\Delta_t}^2$$ with ${\Delta_s}^2$ being the space time interval between the two events, $\Delta_x$ being the distance between the two events in the x dimension, $\Delta_y$ being the distance between the events in the y dimension, ${\Delta_z}$ being the distance between the two events in the z dimension $c$ being the speed of light, and $\Delta_t$ being the distance between the points in the time dimension.

In a 4d euclidean space the distance formula is $${\Delta_d}^2={\Delta_x}^2+{\Delta_y}^2+{\Delta_z}^2+{\Delta_w}^2$$ and if you put in real numbers for $\Delta_x$, $\Delta_y$, and $\Delta_z$, and either $0$ or a pure imaginary number for $\Delta_w$, then ${\Delta_x}^2$, ${\Delta_y}^2$, and ${\Delta_z}^2$ are all positive, while ${\Delta_w}^2$ either $0$ or negative, similar to how ${\Delta_x}^2$, ${\Delta_y}^2$, and ${\Delta_z}^2$ are all positive while $c^2{\Delta_t}^2$ is either $0$ or negative in special relativity.

This means that four dimensional spacetime can also be described as three real dimensions representing space and one imaginary dimension of time so we can substitute in ${\Delta_w}^2$ for $-c^2{\Delta_t}^2$ and have $\Delta_w$ always be either $0$, or a pure imaginary number to make ${\Delta_w}^2$ negative. So the space time interval between two events can now be represented as $${\Delta_s}^2={\Delta_x}^2+{\Delta_y}^2+{\Delta_z}^2+{\Delta_w}^2$$ and get the same results as we would using the previous equation for the space time interval.

Everything, moving slower than the speed of light, can be said to have the same rate of change in spacetime with the total rate of change in space time, for all particles, being represented by the equation $${\iota_s}^2={\iota_x}^2+{\iota_y}^2+{\iota_z}^2+{\iota_w}^2$$ with $\iota_s$ being the total rate of change in space time, $\iota_x$ being the rate of change in the x direction, $\iota_y$ being the rate of change in the y direction, $\iota_z$ being the rate of change in the z direction, and $\iota_w$ being the rate of change in the w direction. $\iota_x$, $\iota_y$, and $\iota_z$ are all real numbers, while $\iota_w$ is an imaginary number and $|\iota_w|\ge\sqrt{{\iota_x}^2+{\iota_y}^2+{\iota_y}^2}$.

This also means that different reference frames can be described as pure imaginary rotations in spacetime with the spacetime angle between two world lines a and b, being represented by the equation $$arctan\left(\frac{i\sqrt{{\Delta_v}^2}}{c}\right) \;=\theta$$ with $\Delta_v$ being the speed that the world lines a and b have relative to each other, and $\theta$ being the spacetime angle between the two world lines. So the spacetime angle between two world lines is always imaginary and the spacetime angle between the world line of a massive particle and the world line of a massless particle is $i{\infty}$.

If there are two world lines g and f, and g is in an inertial reference frame and f is in a non inertial reference frame, and f is accelerating at a constant rate relative to g, then the pure imaginary spacetime angle between g and f will change at a constant rate in f's reference frame.

In euclidian space, with every dimension being real, if the angle between something moving at a constant rate, through space, and a straight line, that is not moving, changes at a constant rate, and in a constant direction, then this something that is moving moves in a circle and there is a point in space, in which the distance between this point, and every point on the circle is the same.

In special relativity an object accelerating at a constant rate traces out a hyperbola in spacetime and just as the parametric equation of a circle, or any type of ellipse, uses sine and cosine, the parametric equation of a hyperbola uses hyperbolic sine and hyperbolic cosine, and $cosh(x)=cos(ix)$, while $isinh(x)=sin(ix)$. Just as $cos^2(x)+sin^2(x)=1$, $(isinh(x))^2+(cosh(x))^2=1$

So in special relativity if an object accelerates at a constant rate forever, is there a point in spacetime, in which the spacetime interval between that point and every point on the objects world line is the same?

First, let's note that we got hyperbola only in $(r,t)$ plane where $r$ is any spatial coordinate. When you examine $(x,y)$, $(x,z)$ and $(y,z)$ planes, you won't get a hyperbola there, but rather a straight line (assuming that acceleration is collinear to the velocity). Second, we know that canonical equation of hyperbola (in $(x,y)$ plane) is as follows:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$
$\Delta s^2 = \Delta r^2 - c^2 \Delta t^2$