# Could a universe with something like special relativity based on position instead of velocity be logically consistent?

In a parallel universe if two objects occupy different positions in space then in the reference frame of each object the other has length contraction and time dilation. If the two objects are very close to each other then the amount of length contraction and time dilation is negligible but if two objects are very far apart then the length contraction and time dilation becomes significant. The amount of length contraction and time dilation that two objects have relative to each other depends on how far apart they are from each other.

• Since you're trying to replace velocity with its integral, position, you may want to look at this question doing the same for acceleration and its integral, velocity. – Samuel Dec 5 '15 at 0:25
• special relativity says that speed of light in a vacuum is capped at c regardless of whoses the observer, time dilation and length contraction depends on where the observer is looking hence its position is very important and can you clarify your question again what do you mean by position? – user6760 Dec 5 '15 at 1:44

Both time dilation and length contraction depend on something called the Lorentz factor, commonly denoted by $\gamma$. It is defined in terms of the velocity, $v$, and speed of light, $c$, as $$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$$ Some like to simplify this a bit by writing $\beta=v/c$, and substituting accordingly.
However, if we were to substitute in $x$ for $v$, we have some issues:
• Units. $x$ has units of distance, while $c$ has units of speed. I suppose you can throw in a constant there, but that seems a bit ad hoc. Speaking of which, that brings us to a different problem.
• The derivation. A rather elegant derivation is given in this answer, as well as some applications to energy and momentum. Basically, if we start with the assumption that the line element $ds^2$ is given by $$ds^2=c^2dt^2-dx^2-dy^2-dx^2$$ then we can find $\gamma$ with only a small bit of work, including introducing a quantity called $d\tau$, the proper time. The expression for the line element, by the way, assumes we use the $(+,-,-,-)$ metric signature.