2
$\begingroup$

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 $\mathcal{L}$?

$\endgroup$
0
0
$\begingroup$

TL;DR: OP's Lagrangians (1) & (2) are not directly related beyond the 0PN approximation.

  1. 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.

  2. 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.

  3. 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.

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

References:

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

  2. 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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.