The expression for the acceleration of a near-earth satellite as presented in the IERS Technical note is given by \begin{equation} \label{eq:problemeq} \tag{1} \frac{d^2\mathbf{r}}{dt^2} = \frac{GM_E}{c^2r^3} \left\{\left[2(\beta+\gamma)\frac{GM_E}{r} - \gamma \dot{\mathbf{r}} \cdot \dot{\mathbf{r}} \right] \mathbf{r} + 2(1+\gamma)(\mathbf{r}\cdot\dot{\mathbf{r}})\dot{\mathbf{r}} \right\}. \end{equation}

We are working in the Parametrised Post-Newtonian formalism, hence the dimensionless constants $\beta,\gamma$.

From what I can gather from here and here this expression can be derived from the "Schwarzschild isotropic one-body point mass metric".

Now, I know what the Schwarzschild metric looks like in isotropic coordinates but I can't see where Eq. (\ref{eq:problemeq}) comes from.

Another point to make is that in GR the dimensionless parameters $\beta,\gamma$ are equal to unity and when substituted above gives a well known formula for Schwarzschild precession which is given by \begin{equation} \label{eq:problemeq2} \tag{1} \frac{d^2\mathbf{r}}{dt^2} = \frac{GM_E}{c^2r^3} \left\{\left[4\frac{GM_E}{r} - \dot{\mathbf{r}} \cdot \dot{\mathbf{r}} \right] \mathbf{r} + 4(\mathbf{r}\cdot\dot{\mathbf{r}})\dot{\mathbf{r}} \right\}. \end{equation}

Again, I don't recall this either.

Any suggestions?

  • 2
    $\begingroup$ The derivation of that expression seems to be rather involved. I looked through the various references in IERS Technical Note No. 36 and AU Resolutions B1.3 and B1.4 (2000) and the best starting point I could find is [Klioner 2001; arxiv.org/abs/astro-ph/0107457] eq. (3) and the references for it [Will 1993] and [Klioner & Soffel 2000; arxiv.org/abs/gr-qc/9906123]. $\endgroup$
    – N0va
    Feb 20, 2017 at 9:54
  • $\begingroup$ It's definitely quite involved. I don't expect it to be done in one sitting :) On the hunt for some resources that can be understood. I find the old school PN guys quite difficult to follow! $\endgroup$ Feb 20, 2017 at 11:41
  • $\begingroup$ The eq of motion for your case is $\frac{d^2x^\sigma}{ds^2} = \Gamma^\sigma_{\mu\nu} \frac{dx^\mu}{ds}\frac{dx^\nu}{ds} $ . Just plug in the Schwarzschild metric tensor $g_{\mu\nu}$ into $ \Gamma^{\sigma}_{\mu\nu} = 1/2 g^{\sigma\alpha}( \frac{\partial g_{\mu\alpha}}{\partial x_\nu} + \frac{\partial g_{\nu\alpha}}{\partial x_\mu} -\frac{\partial g_{\mu\nu}}{\partial x_\alpha}) $ and do the math on the eq of motion. This is, i suppose, the simplest approach. $\endgroup$
    – Mihai B.
    May 18, 2017 at 11:03
  • $\begingroup$ @MihaiB. Unfortunately this wouldn't work. That what you would do in a completely relativistic framework. However, this is expressed in terms of the PPN formalism. $\endgroup$ May 18, 2017 at 11:05
  • 2
    $\begingroup$ That's very similar to the form used in Yeomans, D. K., et al. "Cometary orbit determination and nongravitational forces." Comets II (2004): 137-151, who reference Anderson J. D., et al. "Experimental test of general relativity using time-delay data from Mariner 6 and Mariner 7." Astrophys. J., 200, 221–233. The only difference is the use of the Gaussian gravitational constant $k$ instead of $GM_E$ $\endgroup$ May 22, 2017 at 11:07

1 Answer 1


I see that you are referencing my old paper. Well in my paper I was referencing to the other of your two references, the book by JPL. There is an expression numbered 4-26 on page 4-19 in that book that reduces to the expression you wrote above in the case of one static large mass and one small "test-body". The derivation of expression 4-26 is provided on page 4-22 to page 4-24 in the same book by the Jet Propulsion Laboratory. The derivation is a bit beyond me I must say but still.


Edit I just wanted to add, in case someone arrives here looking for an expression, that the expression above only works well in the weak field limit. As is seen it approximates GR by introducing two velocity-dependent terms and one repulsive inverse cube term. This does not work well in the strong field regime. As the repulsive term gets stronger as you close in on the black hole, "bouncing" will occur. In the simulations below the green circle represents the Schwarzschild radius and the red circle the radius of the "innermost stable circular radius". As seen in this post, the post-Newtonian expansion is available also at the 3PN level, including more terms. Maybe that expression will work better in stronger fields.

enter image description here

  • $\begingroup$ Appreciate the response and information. I figured it out quite some time ago ;) $\endgroup$ Mar 25, 2019 at 7:16

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