2
$\begingroup$

I am wondering whether the region $H:=\{(t,x):x^2−t^2<1\}$ of $(1+1)$-dimensional Minkowski spacetime, equipped with the restriction $g_H$ of the standard Minkowski metric $g=−\mathrm{d} t\otimes \mathrm{d}t + \mathrm{d}x \otimes \mathrm{d} x$, is globally conformally equivalent to the vertical strip $S:=\{(t,x):|x|<1\}$, again equipped with the restriction $g_S$ of the Minkowski metric. To further clarify: by "globally conformally equivalent" I mean that there should be a diffeomorphism $\phi : H \to S$ with $\phi_* (g_H)=\Omega^2 g_S$ for some smooth, strictly positive function $\Omega$.

I posted this question on Mathematics Stack Exchange some time ago as I had the feeling there might be a straightforward argument involving null geodesics, but to this day I still haven't found the answer. Thanks for your help.

$\endgroup$
3
$\begingroup$

In $H$ you have two lightlike geodesics such that every lightlike geodesic intersects one of these two, namely the geodesics $\{x=y\}$ and $\{x=-y\}$. In $S$ you do not have such two geodesics. Hence, the regions are not conformally equivalent

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.