# Curvature and length contraction paradox

I was thinking about the Flat-Earth model and I confronted with the following issue. Consider a sphere and an observer who is leaving the sphere. Using length contraction principle from special relativity, I supposed the observer would see the sphere has shrunk parallel to the movement direction and so it seems like a disc. The curvature that the observer would measure can make arbitrary small in some points by choosing the rapidity large enough. Although, according to GR, the curvature is an invariant within all observers but the inertia observer calculate identical curvature at all points on the sphere, however, the moving observer would seemingly measure different value for the curvature. I always thought GR would reduce to SR if we restrict in local neighborhoods and it seems we can actually perform this experiment in a local way. I am sure I have made some mistakes in my conclusion. I would be thankful if you could clarify this for me.

## 1 Answer

The Riemann curvature tensor is a tensor field defined on a $4$-dimensional manifold. However, the curvature I was talking about is the Riemann curvature tensor defined by the induced metric on a $3$-dimensional sub-manifold. The sphere is the $3$-dimensional sub-manifold that is described by $t=const.$ in their chart, while the disc in the second observer's coordinate is described by $t^\prime=const$ where in general are different. So, the $3$-dimensional curvature need not be invariant under Poincare transformations in $4$ dimension.