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In post-Newtonian approximation of gravitation, the metric on and around the Earth is taken to have the expression $$ \begin{bmatrix} -1+\frac{2}{c^2}U+ \frac{2}{c^4}(\psi-U^2) + \mathrm{O}(c^{-5}) & -\frac{4}{c^3} V_x + \mathrm{O}(c^{-5}) & -\frac{4}{c^3} V_y + \mathrm{O}(c^{-5}) & -\frac{4}{c^3} V_z + \mathrm{O}(c^{-5}) \\\\ -\frac{4}{c^3} V_x + \mathrm{O}(c^{-5}) & 1+ \frac{2}{c^2} U + \mathrm{O}(c^{-4}) & 0 & 0 \\\\ -\frac{4}{c^3} V_y + \mathrm{O}(c^{-5}) & 0 & 1+ \frac{2}{c^2} U + \mathrm{O}(c^{-4}) & 0 \\\\ -\frac{4}{c^3} V_z + \mathrm{O}(c^{-5}) & 0 & 0 & 1+ \frac{2}{c^2} U + \mathrm{O}(c^{-4}) \end{bmatrix} $$ in a coordinate system $(ct, x, y,z)$, where $U$, $\psi$, $V_i$ depend on all coordinates. $U$ is related to the Newtonian gravitational potential, and $(V_j)$ is the so-called gravitational vector potential. See eg Poisson & Will 2014, eqns (8.2). This metric is used for example for GPS purposes, see eg Petit & Luzum 2010.

On Earth, say at the equator, $\lVert\partial U/\partial x^i\rVert \approx g \approx 9.8\,\mathrm{m/s^2}$ in a radial direction -- the gravitational acceleration. Can anyone provide an order of magnitude for the vector potential $(V_j)$ and its spatial gradient: $$ V_j \approx \mathord{?}\,\mathrm{m^3/s^3} \qquad \frac{\partial V_j}{\partial x^i} \approx \mathord{?}\,\mathrm{m^2/s^3} $$ say at the equator? I've been looking in the references above and in the references given in this question and its answer, but I don't manage to find it.

I could approximately calculate it using the integral formulae given eg in Poisson & Will, eqns (8.4). But both for lack of time and to double-check such a calculation I'd like to find some reference where this order of magnitude is given. Cheers!

References:

pglpm
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1 Answers1

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An approximate formula for the gravitational vector potential is given by Mashhoon 2008: $$ (V_j) \approx G\, \frac{\pmb{J} \times \pmb{r}}{r^3} $$ where $G$ is the gravitational constant, $\pmb{J}$ the rotational momentum of the Earth, and $\pmb{r}=(x,y,z)$ (there's an additional $c$ factor in Mashhoon's formula, but that's included in the $c^3$ factor in the original question).

Using the values $G\approx 6.67\cdot 10^{11}\,\mathrm{m^3/(kg\,s^2)}$, $c=299\,792\,458\,\mathrm{m/s}$, $J\approx 7\cdot 10^{33}\,\mathrm{kg\,m^2/s}\ (0,0,1)$, and taking a point roughly on the equator with $x=R,y=0,z=0$, where $R\approx 6.38\cdot 10^6\,\mathrm{m}$ is Earth's radius, we obtain $$ \frac{1}{c^3}\,(V_j) \approx (0,\ 4\cdot 10^{-16},\ 0) \qquad \frac{1}{c^3}\,\biggl(\frac{V_i}{x^j}\biggr)_{ij} \approx \begin{bmatrix} 0 & -7\cdot 10^{-23} & 0 \\\\ -1\cdot 10^{-22} & 0 & 0 \\\\ 0&0&0 \end{bmatrix} \ \mathrm{m^{-1}} $$

The formula given in Mashhoon is valid "far from the source", so not valid for a point at the equator. But the estimates above should give a rough initial idea.

pglpm
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