I am having some doubts on myself regarding the above concepts in General Relativity.
First, I want to point out how I understand them so far.
A male observer follows a timelike worldline ($\gamma$) in spacetime (because he must have a proper time). He has a frame for himself.
A coordinate is a sets of numbers the observer uses to describe the spacetime in his frame (which is another way to say the spacetime in his view).
The locally flat coordinate of an observer at a time ($s\in\gamma$) is the coordinate (of his frame, of course) in which he sees the metric tensor at a neighborhood of his position be the flat metric (Christoffel symbols vanish):
$$g_{\mu\nu}(s)=\eta_{\mu\nu}$$
$$\Gamma_{\mu\nu}^\rho(s)=0$$
This coordinate depends on and is used naturally by the observer.
Now a locally inertial frame is a frame of any freely falling observer, or any observer following a geodesic ($l$). He may use or may not use the locally flat coordinate of himself. But he has a very special coordinate which is locally flat at every point is his worldline:
$$\forall s\in l:$$ $$g_{\mu\nu}(s)=\eta_{\mu\nu}$$ $$\Gamma_{\mu\nu}^\rho(s)=0$$
Do I have any misunderstanding or wrong use of terminology?
Now there should be an freely falling observer $A$ (with his special coordinate) and his wordline crosses the wordline of another (not freely falling) observer $B$. And at the cross point can I believe that the two coordinate (of two frames) may be chosen to be locally identical (or equal) (that is, there exists a linear transformation locally transform one to other)?
Best Answer
In the most general case described by general relativity, it is not possible to find a $neighbourhood$ covered by coordinates $x^\mu$ such that $g^{\mu\nu} = \eta^{\mu\nu}$ in all U. If it were so, you have a zero Riemann tensor, hence the space-time would be flat in all U. You may have space-times with such flat pieces (I think there is no problem in gluing this piece with non-flat pieces, but I may be wrong), but is not the most general case and not what is meant when we say that space-time is locally flat.
What we mean is that the tangent space, in any point is the Minkowski space-time.
This mean that, for any point p, you can find a basis for the tangent space at p (and associated "exponential" coordinates) so that the metric is diag(-,+,+,+) in these coordinates at this point p and the connession coefficients vanish at this point (not in a neighbourhood!)
You can think of these coordinates as those of a inertial observer. Note that there exists several possible coordinates, which are related by a Lorentz transform at the tangent space, and are associated to different observers.
In what sense you can think of these coordinates as those of a inertial observer? In the sense that as long as you are covering a sufficiently small neighbourhood of p, whose dimension will be "smaller the larger the Riemann tensor is at p", you may describe everything happening here as if you were in special relativity. One above all, the geodesics are of the form $d/dt^2 x(\tau) = 0$ and do not accelerate with respect to each other. Of course, actually they do, but these effects are small if you consider small neighbourhood of p and small Riemann at p.
Analogously, Earth is flat at a point in the sense that you can "confuse" the flat tangent space with the actual neighbourhood because the differences are difficult to detect if you zoom enough.