1
$\begingroup$

Frame-dragging is a general relativistic effect that can be imagined as the effect that the momentum of mass-energy has (the stress elements in the stress-energy tensor) on mass-energy in the surroundings of the moving mass-energy.
For example, a large moving sheet of mass will cause particles to be dragged along in the direction of motion of the sheet (on top of gravitating towards the sheet).

So I was wondering. Does the sun have a noticeable frame-dragging effect on the planets?

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
2
$\begingroup$

An effect, but not a noticeable effect.

Wikipedia gives a formula for the frame-dragging in a a Kerr metric: $$\Omega = \frac{r_{s} \alpha c}{r^{3} + \alpha^{2} r + r_{s} \alpha^{2}}$$

where $\Omega$ is the angular speed that the frame of reference rotates at, $r$ is the orbital radius (70 billion metres for Mercury), $r_s$ is the Schwartzchild radius of the sun (3000 m), $c$ is the speed of light and $\alpha = \frac{J}{Mc}$ is the angular momentum of the sun divided its mass and the speed of light. For the sun $\alpha\approx 320$ metres.

Plugging these values in gives $\Omega \approx 10^{-18}$ radians per second.

This isn't noticeable (compared to other relativistic effects and Newtonian perturbations) Frame dragging can be detected in carefully designed experiments on Earth satellites since the effect is greater at smaller orbit radius.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I'm not sure I get the meaning of $\Omega$. $\endgroup$
    – Deschele Schilder
    Jun 5, 2021 at 15:28
  • 1
    $\begingroup$ It’s “the angular speed that the frame of reference rotates at,” in other words, the effect imprinted on the planet. The quoted value of $ 10^{-18} $ radians per second is about $ \frac {6}{10000} $ arcsecond per century; indeed, not noticeable. $\endgroup$
    – Pierre Paquette
    Jun 5, 2021 at 18:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .