Order of magnitude of gravitational vector potential & its gradient on Earth

Question

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:

• Poisson, Will: Gravity: Newtonian, Post-Newtonian, Relativistic (CUP 2014)
• Petit, Luzum: IERS Conventions, International Earth Rotation and Reference Systems Service, Technical Note 36.

This post imported from StackExchange Physics at 2025-01-21 21:50 (UTC), posted by SE-user pglpm

Comments

"Gravity" is 0.1% stronger at the pole than at the equator because of centrifugal force of the Earth's rotation. Is that part of what you want?

This post imported from StackExchange Physics at 2025-01-21 21:50 (UTC), posted by SE-user mmesser314

---

@mmesser314 Not quite. That effect lies within Newtonian theory. The gravitational vector potential is outside of Newtonian theory. So to speak, it is connected to the extra force that a mass exerts because it's moving (like an electric charge in movement generates a magnetic field and a magnetic force). This is also clear from the approximate expression of the vector potential from the mass distribution creating it: $V_j \approx \int \frac{\rho\,\pmb{v}_j}{r}\,\mathrm{d}^3r$. You can do a search for "gravito-magnetism".

This post imported from StackExchange Physics at 2025-01-21 21:50 (UTC), posted by SE-user pglpm

---

Then you are also not looking for gradients because the Earth isn't perfectly spherical? That too is explained by Newtonian gravity.

This post imported from StackExchange Physics at 2025-01-21 21:50 (UTC), posted by SE-user mmesser314

---

@mmesser314 Correct, not that either. Here's an example. Take an inertial reference frame, and a perfectly spherical distribution of mass there. Take a small test mass at rest in the inertial frame, outside the spherical distribution. According to Newtonian gravity, the gravitational potential and force at the test mass don't depend on whether the spherical mass distribution is rotating (around an axis at its centre) or not. According to general relativity (and experiment) they do, instead. This effect is what the gravitational vector potential describes.

This post imported from StackExchange Physics at 2025-01-21 21:50 (UTC), posted by SE-user pglpm

Answer 1

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.

This post imported from StackExchange Physics at 2025-01-21 21:50 (UTC), posted by SE-user pglpm