Sorry, there were some serious typos/latex issues in that last post. Gamma should have been 2, not 3. Now v=(sqrt(3)/2)*c, which gives v/c < 1. I edited the post, so the issues should now be corrected.
So \gamma \equiv 2 \implies v = \frac{\sqrt{3}}{2} c.
We have: t' = \gamma (t - \frac{x v}{c^2}) and t = \gamma (t' + \frac{x' v}{c^2}).
It follows that t' = 2 (5 - \frac{\frac{\sqrt{3}}{2} c 5 \frac{\sqrt{3}}{2} c}{c^2}) = 2.5. But then going in reverse, t = 2 (2.5 + \frac{-...
Suppose we have two frames of reference, frame A and frame B, which move past each other with a velocity such that \gamma \equiv 2. In frame A is clock A and in frame B is clock B.
In frame A, clock A is at rest and clock B is speeding past. As a result of time dilation, when an observer in...