Reducing General Relativity to Special Relativity in limiting case I understand that general relativity is applicable to gravitational fields and special relativity is applicable to case when there is no gravity. But is there a derivation on how to reduce General Relativity to Special Relativity in limiting case, much like how General Relativity is reduced to Newtonian gravity in weak-gravity case?
Edit: By reducing I mean, how can we derive the Lorentz transformation from General Relativity under appropriate limits?
 A: If you take a spacetime whose metric is a solution of Einstein's equations, then pick a point in that spacetime and introduce a locally inertial coordinate system there, the metric at that point can be brought into the Minkowski form.  Effectively by working in a locally inertial frame, you are seeing what freely falling observers would see - you are removing the effects of gravity by falling along with it.  To make this work fully, you also have to restrict yourself to an infinitesimally small region, otherwise, the tidal effects of gravity will become noticeable, and again the metric will deviate from the Minkowski values.
Of course the set of locally inertial coordinates is not unique, but the transformation from one set to another would have to preserve the Minkowski metric, i.e. they would be Lorentz transformations.  It is in this limiting sense (inertial frame, infinitesimally small region) that GR "reduces" to SR, and the general coordinate transformations allowed by GR become restricted to the Lorentz transformations of SR.
A: Here is a mathematically rigorous elaboration of twistor59’s answer, which makes no mention at all of Einstein’s field equations.

Let $ (M,g) $ be an arbitrary space-time manifold, and let $ p \in M $. Suppose that $ (U,\phi: U \to \mathbf{R}^{4}) $ and $ (V,\psi: V \to \mathbf{R}^{4}) $ are two local coordinate charts of $ M $ that satisfy the following two properties:


*

*$ p \in U \cap V $.

*Letting $ \phi = (x^{0},x^{1},x^{2},x^{3}) $ and
$ \psi = (y^{0},y^{1},y^{2},y^{3}) $, we have
$$
  \forall a,b \in \{ 0,1,2,3 \}: \qquad
    g \!
    \left( \!
    \left. \frac{\partial}{\partial x^{a}} \right|_{p},
    \left. \frac{\partial}{\partial x^{b}} \right|_{p}
    \right)
  = \eta_{a b}
  = g \!
    \left( \!
    \left. \frac{\partial}{\partial y^{a}} \right|_{p},
    \left. \frac{\partial}{\partial y^{b}} \right|_{p}
    \right),
  $$
where $ \eta_{a b} $ denotes the Minkowski inner product
$ \eta: \mathbf{R}^{4} \times \mathbf{R}^{4} \to \mathbf{R} $ in component form.


Let $ X \in T_{p} M $, i.e., $ X $ lies in the tangent space to $ M $ at $ p $. It is a basic result of differential geometry that $ \left( \! \left. \dfrac{\partial}{\partial x^{a}} \right|_{p} \right)_{a = 0}^{3} $ and $ \left( \! \left. \dfrac{\partial}{\partial y^{b}} \right|_{p} \right)_{b = 0}^{3} $ are ordered bases for $ T_{p} M $, so there exist scalars $ \lambda^{0},\lambda^{1},\lambda^{2},\lambda^{3} $ and $ \mu^{0},\mu^{1},\mu^{2},\mu^{3} $ such that
$$
  \lambda^{a} \cdot \left. \frac{\partial}{\partial x^{a}} \right|_{p}
= X
= \mu^{b} \cdot \left. \frac{\partial}{\partial y^{b}} \right|_{p}.
$$
Furthermore, $ \mu^{b} = \lambda^{a} \left. \dfrac{\partial y^{b}}{\partial x^{a}} \right|_{p} $ for each $ b \in \{ 0,1,2,3 \} $. Now,
\begin{align}
    g(X,X)
& = g \!
    \left(
    \lambda^{a} \cdot \left. \frac{\partial}{\partial x^{a}} \right|_{p},
    \lambda^{b} \cdot \left. \frac{\partial}{\partial x^{b}} \right|_{p}
    \right) \\
& = \lambda^{a} \lambda^{b}
    g \!
    \left( \!
    \left. \frac{\partial}{\partial x^{a}} \right|_{p},
    \left. \frac{\partial}{\partial x^{b}} \right|_{p}
    \right) \\
& = \eta_{a b} \lambda^{a} \lambda^{b}; \\
    g(X,X)
& = g \!
    \left(
    \mu^{a} \cdot \left. \frac{\partial}{\partial y^{a}} \right|_{p},
    \mu^{b} \cdot \left. \frac{\partial}{\partial y^{b}} \right|_{p}
    \right) \\
& = \mu^{a} \mu^{b}
    g \!
    \left( \!
    \left. \frac{\partial}{\partial y^{a}} \right|_{p},
    \left. \frac{\partial}{\partial y^{b}} \right|_{p}
    \right) \\
& = \eta_{a b} \mu^{a} \mu^{b}.
\end{align}
If $ T: \mathbf{R}^{4} \to \mathbf{R}^{4} $ denotes the linear map whose matrix representation is $ \left[ \! \left. \dfrac{\partial y^{b}}{\partial x^{a}} \right|_{p} \right]_{a,b = 0}^{3} $, then
$$
(\spadesuit) \qquad
  \eta(\mathbf{v},\mathbf{v})
= g(X,X)
= \eta(T(\mathbf{v}),T(\mathbf{v})),
$$
where $ \mathbf{v} = (\lambda^{0},\lambda^{1},\lambda^{2},\lambda^{3}) $ and $ T(\mathbf{v}) = (\mu^{0},\mu^{1},\mu^{2},\mu^{3}) $. However, $ X $ is an arbitrary tangent vector to $ M $ at $ p $, so $ (\spadesuit) $ actually holds for all $ \mathbf{v} \in \mathbf{R}^{4} $. Hence, $ T: \mathbf{R}^{4} \to \mathbf{R}^{4} $ fixes the origin and preserves space-time intervals, which automatically implies that $ T $ is a Lorentz transformation on $ \mathbf{R}^{4} $.

The Lorentz transformations may therefore be thought of as coordinate transformations at an event between local inertial reference frames.
A: A well-known technique is post-Minkowskian approximation. It applies to the weak-field limit of general relativity, and gravity shows up as correction terms to the Minkowski metric in powers of Newton’s gravitational constant $ G $.
If, in addition to powers of $ G $, you expand the metric in powers of $ \left( \dfrac{v}{c} \right)^{2} $ (i.e., in addition to a weak gravitational field, you consider slow motion), you will arrive at the post-Newtonian regime.
A: When you say "how can we derive the Lorentz transformation from General Relativity" this is really asking "how is the Minkowski metric a solution of the vacuum Einstein equation", because Special Relativity is just the geometry defined by the Minkowski metric.
If you take the Einstein equation and turn off gravity by setting $ G = 0 $, you get the vacuum Einstein equation $ G_{a b} = 0 $. The Minkowski metric is a solution of this equation, but of course there are lots of others. From your question I'd guess you're hoping that the Einstein equation will simplify in the absence of gravity, and this will make it obvious how Special Relativity emerges. Sadly this isn't the case, because even in the absence of mass, or $ G $ set to zero, gravity waves are still allowed.
I don't think there is any way to simplify the Einstein equation to make the Minkowski metric the only solution. You can require that the first derivatives of the metric vanish, but this is really getting the flat space solution by requiring that space not be curved, which is a bit of a tautology. The problem is that SR the Minkowski metric is an assumption i.e. it's where you start from. In GR the Minkowski metric is just one among many solutions so there's nothing fundamental about it.
Have a look at http://en.wikipedia.org/wiki/Einstein_tensor if you want to play around with the Einstein tensor to try and extract the Minkowksi metric.
A: Other answers have already addressed the relationship between General Relativity and the Minkowski metric, but it seems you are most interested in getting from the Minkowski metric to the Lorentz transformation.  So let's do that.
Given a set of coordinates in which the metric takes the standard Minkowski form
$ds^2 = dt^2 - dx^2 - dy^2 - dz^2$
you want to find another set of coordinates in which the metric also takes the same form
$ds^2  = d \bar{t}^2 - d \bar{x}^2 - d \bar{y}^2 - d \bar{z}^2$
Consider a linear boost along the x axis.  We want to choose coordinates such that $\bar{x} = 0$ where $x - vt = 0$; the general solution is:
$\bar{x} = \gamma(x - vt)$
$\bar{t} = at - bx$
where $\gamma$, $a$, and $b$ are unknowns.
Then
$dt^2 - dx^2 = d\bar{t}^2 - d\bar{x}^2$
$ = (a dt - b dx)^2 - \gamma^2 (dx - v dt)^2$
$ = (a^2 - \gamma^2 v^2) dt^2 - (\gamma^2 - b^2) dx^2 + (\gamma^2 v - ab) dx dt$
so
$a^2 - \gamma^2 v^2 = 1$
$\gamma^2 - b^2 = 1$
$\gamma^2 v - ab = 0$
so
$a = \sqrt{1 + \gamma^2 v^2}$
$b = \sqrt{\gamma^2 - 1}$
so
$\gamma^2 v = \sqrt{\gamma^2 v^2 + 1} \sqrt{\gamma^2 - 1}$
$\implies \gamma^4 v^2 = (\gamma^2 v^2 + 1)(\gamma^2 - 1)
= \gamma^4 v^2 + \gamma^2 - \gamma^2 v^2 - 1$
$\implies \gamma^2(1 - v^2) = 1$
$\implies \gamma = \sqrt{\frac{1}{1-v^2}}$
Calculating a and b (exercise left to the reader) gives $a = \gamma$ and $b = \gamma v$, so that
$\bar{t} = \gamma(t - vx)$
completing the Lorentz transformation as expected.
