I hate this "Awww... Snap!" thing with SE on chrome.

Basically, it's because a Lorentz Transformation maps between spatial coordinates and temporal ones.

I just realised that instead of writing this old (and "new") answer rubbish (in quotes now), I could have just done the Lorentz transform . So, in units where $\hbar=c_0=G=k_e= \ell_s=1 $ (who cares if other than $c_0$, any of them are actually involved),

$$\gamma \left[ {\begin{array}{*{20}{c}} 1&{ - v} \\ { - v}&1 \end{array}} \right]\gamma \left[ {\begin{array}{*{20}{c}} 1&{ - v} \\ { - v}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_0} - 2\pi R} \\ {{x_1} + 2\pi R} \end{array}} \right] = \left[ \begin{gathered} {x_0} \\ {x_1} + 2\pi R \\ \end{gathered} \right]$$

That's what we want .

$$ \begin{gathered} \gamma \left[ \begin{gathered} {x_0} - 2\pi R - v{x_1} - 2\pi Rv \\ {x_1} + 2\pi R - v{x_0} + 2\pi Rv \\ \end{gathered} \right] = \left[ \begin{gathered} {x_0} \\ {x_1} + 2\pi R \\ \end{gathered} \right] \\ \left. \begin{gathered} \frac{{1 - \sqrt {1 - {v^2}} }}{{\sqrt {1 - {v^2}} }}{x_0} - 2\pi R\left( {1 + v} \right) - \frac{v}{{\sqrt {1 - {v^2}} }}{x_1} = 0 \\ \frac{{1 - \sqrt {1 - {v^2}} }}{{\sqrt {1 - {v^2}} }}{x_1} + 2\pi Rv - \frac{v}{{\sqrt {1 - {v^2}} }}{x_0} = 0 \\ \end{gathered} \right\} \\ \end{gathered} $$

And those are the equations you're asking for (right?) .

3 choices:

- Spend the rest of your life solving them .
- Use Wolfram Alpha .
- Consider it "Solvable in Principle".

Choice (2) is recommended.