Einstein-Infeld-Hoffmann (EIH) Lagrangian for a Test-Particle as Limit of Schwarzschild-Geodesic Consider a test particle of mass $m$ which is in orbit around a spherical-symmetric body with mass $M$. It therefore has a position as described by the coordinates $r,\phi$, and its motion can be described by the Lagrangian $L$ of the Einstein-Infeld-Hoffmann-Equations:
$$L = \frac{mv^2}{2}+ \frac{GmM}{r}+\frac{mv^4}{8c^2} + \frac{3GmMv^2}{2c^2r}-\frac{kmM\left(m+M\right)}{2c^2r^2},\tag{1}$$
where $v$ is the particles velocity.
But the orbit of the test-particle can also be described by the Schwarzschild-Metric and the corresponding Lagrangian $\mathcal{L}$
$$\mathcal{L} = -\frac{1}{2}\left[-\left(1-\frac{2 G M}{c^2 r}\right) c^2 \dot{t}^2 + \left(1-\frac{2 G M}{c^2 r}\right)^{-1}\dot{r}^2 + r^2 \dot{\varphi^2}\right].\tag{2}$$
Where the dot denotes the derivative with respect to the proper time of the particle $\tau$ along the world line.
I know that the Newtonian Lagrangian for a testparticle can be derived by requiring $\frac{v}{c}\rightarrow 0$.
Since $L$ simply adds some extra terms to the Lagrangian, it should be possible to do something similar here.
But what kind of expansion is needed to arrive at $L$ from $\mathcal{L}$?
 A: TL;DR: OP's Lagrangians (1) & (2) are not directly related beyond the 0PN approximation.

*

*OP's eq. (2) is (up to normalization) a gauge-fixed version of the square root action
$$\begin{align} L_0~=~~~~&-mc\sqrt{-g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}\cr
~\stackrel{\text{static gauge}}{=}&-mc\sqrt{g_{00}-v^2 +{\cal O}(1PN)}\cr
~=~~~~&-mc^2+\frac{1}{2}mv^2+ \frac{GMm}{r} +{\cal O}(1PN)
\end{align} \tag{2'}$$
for a massive point particle, cf. e.g. this related Phys.SE post. To 0PN order eq. (2') agrees with OP's Lagrangian (1) up to the rest energy $E_0=mc^2$, which can be ignored.


*Let us mention that OP's version of the EIH Lagrangian (1) is under the extra assumption $\frac{m}{M}\ll 1$, so that $\frac{V}{v}\simeq \frac{m}{M}\ll 1$. The EIH Lagrangian is derived in the 1PN approximation.


*The EIH Lagrangian consists not just of the matter action of the point particle(s), but it also contains a term from the metric tensor field. The latter is some gauge-fixed, linearized, truncated remnant of the Einstein-Hilbert action. $\leftarrow$ This in principle answers OP's main question in the negative.


*For a derivation of the EIH Lagrangian, see e.g. Refs. 1 & 2.
References:

*

*M. Maggiore, Gravitational Waves: Volume 1: Theory and Experiments, 2008; eq. (5.54).


*W.D. Goldberger & I.Z. Rothstein, An Effective Field Theory of Gravity for Extended Objects, Phys. Rev. D 73 (2006) 104029, arXiv:hep-th/0409156; eq. (40).
