I'm studying Susskind's GR TTM book, in which he gives a nice explanation of why differential geometry is needed for GR. But there is one gap that I want to fill.
The argument is: through a thought experiment, it seems that a uniform gravitation field can be seen as an artifact of going from an inertial frame $R$ to another one $R'$ that's uniformly accelerating w.r.t. $R$. More precisely, I can do a (curvilinear) coordinate transformation to $R'$ coordinate system. So the general question is: if there is a gravitational field $F$, how do we know if it's real or just the result of some weird coordinate transformation (i.e. a fake gravitational field)?
So I'll finally try to frame the above in a more concrete way: given a force field $F(x,y,z,t)$, does there exist a coordinate transformation $(x,y,z,t)\to (x',y',z',t')$ that removes the effects of the field, i.e. $F(x',y',z',t')=0$ [or is it supposed to be $F'(x',y',z',t')=0$? I'm not sure]?
Fine so far (except the $F$ vs $F'$ confusion). The book then mentions that this question is similar to the question of deciding whether a geometry is flat. It goes into a long discussion, finally concluding that a manifold is flat if it's locally flat everywhere. And it's locally flat at a point $p$ if the metric tensor at $p$, via a coordinate transformation $(x,y,z,t)\to (x',y',z',t')$, gets transformed into the identity matrix (actually should be the Minkowski metric tensor but for simplicity the book is discussing Riemannian geometry at this point).
Individually I understand the logic on both sides, but I'm not sure about the connection (see the bold italic statement). I see the following similarity:
- The effects of a "fake" gravitational field (as seen in $R$) can be eliminated globally by switching to a reference frame $R'$ with a specific acceleration $\vec{a}(t)$ w.r.t. $R$ (via a global coordinate transformation). But the effects of any arbitrary gravitational field can be eliminated at any point locally via a similar procedure of switching to some other local reference frame (via a local coordinate transformation)
- This is similar to how a manifold is globally flat if there exists a coordinate transformation that diagonalizes the metric tensor everywhere, vs. how for a general smooth manifold, we can still diagonalize the metric tensor at a point via a local coordinate transformation
But this is still just a vague similarity to me. Consider an extended region of spacetime in which, w.r.t. some coordinate system (i.e. reference frame) $(x,y,z,t)$, we notice a gravitational field $F$. We also notice that the transformation to a different coordinate system $(x',y',z',t')$ "gets rid of" $F$.
If the above is true, is there some mathematical result that guarantees existence of a coordinate transformation diagonalizing the metric tensor in that particular region of spacetime? Or conversely, is there a result that if a diagonalizing coordinate transformation exists in a spacetime region, then there also exists a transformation that "gets rid of" $F$?
If these kind of results don't exist, how can we say that the above two scenarios (points 1 and 2) are mathematically equivalent?