- #1

Shirish

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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).

**. 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).**

*The book then mentions that this question is similar to the question of deciding whether a geometry is flat*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?