In 1+1 Minkowski space the distance between two points is given by$$ (x_1 -x_2)^2 -(t_1 - t_2)^2.$$
This is different from the Euclidean distance. But is it possible to come up with a 2D surface embedded in 3D Euclidean space such that the geodesic distance between two points on the surface is like that in Minkowski space?
No, it is not possible because the induced metric on any submanifold $N$ of the Euclidean space $E^3$ is necessarily positively defined, whereas the metric on $1+1$ Minkowski space is indefinite.
The reason is trivial: The scalar product $\langle u,v\rangle $ of two vectors $u,v$ in $N$ is, by definition, the scalar product in $E^3$ of these vectors viewed as vectors in $E^3$, so that $\langle u,u\rangle \geq 0$ -- where $\langle u,u\rangle=0$ implies $u=0$ -- in any cases, contrarily to what may happen in Lorentzian manifolds.
