Gravitomagnetic arguments also predict that a flexible or fluid toroidal mass undergoing minor axis rotational acceleration (accelerating "smoke ring" rotation) will tend to pull matter through the throat (a case of rotational frame dragging, acting through the throat). In theory, this configuration might be used for accelerating objects (through the throat) without such objects experiencing any g-forces. —Wikipedia

Assume I have a 30g doughnut (made of a flexible material that cannot be broken or torn apart). The major radius of my doughnut is 5cm, and the minor radius is 3cm. What should be its "minor axis rotational acceleration" in order to make the gravitomagnetic acceleration in the center exactly 10 m/s²?

As I don't know how to do general relativity, I tried to use the simpler GEM equations but the maths is still too advanced for me. For example I don't know how to compute the mass flux.


Forward's donut

The formula in Forward's classic paper is $$G=-\frac{d}{dt}\left(\frac{\eta N T r^2}{4\pi R^2}\right )$$ where $NT$ is the total mass current ($N$ windings of pipes carrying a heavy liquid in a spiral around the torus - here we will use the donut mass) and $\eta=3.73\cdot 10^{-26}$ m/kg. So plugging in your numbers, $r=0.03$ m, $R=0.05$ m and we assume $v(t)=0.03 a t$ kg m/s for some acceleration $a$ I get $3.2057\cdot 10^{-29}a$ , so to get "antigravity" we need $a=3.1194\cdot10^{29}$ m/s$^2$. That donut better be pretty indestructible.

(The acceleration is actually in principle physically possible, just a few orders of magnitude above electrons in wakefield accelerators, way below the Planck acceleration).

Tajmar's donut

One somewhat similar calculation can be found in Tajmar, M. (2010). Homopolar artificial gravity generator based on frame-dragging. Acta Astronautica, 66(9-10), 1297-1301. For a pair of rotating disks he states the field at the center as $B_g=(4G/c^2)mr\omega$ where $m$ is the disk mass, $r$ their radius and $\omega$ their angular velocity. The prefactor is $4G/c^2\approx 3\cdot 10^{-27}$.

Note that the gravitomagnetic field acts on a particle with mass only if it is moving, just as a magnetic field will only affect moving charges. The force is at right angles to the field and velocity, and proportional to the speed. This is why there has to be an accelerating flow around the torus in Forward's paper: had it been constant there would have been a constant gravitomagnetic field, and there would not have been any acceleration of particles inside the torus.

Tajmar suggests having a cabin moving at constant velocity along a hallway with the spinning disks to provide a velocity. However, the final model in the paper has a ring-shaped cabin surrounded by two rings of spinning disks that themselves spin around the centre. This way one can enjoy artificial gravity without having to move. While this model is a bit donut-shaped it doesn't correspond to a plausible motion of donut dough.


The math is way beyond my abilities, but since no one has answered yet I thought I would take a theoretical stab at this one for you. My guess would be that you could not achieve acceleration of 10m/s^2 since Frame Dragging and GR only deal with warping of spacetime. As the Wikipedia article states there would be no G-forces, which as I understand it would mean no acceleration at all since acceleration and gravity are equal under GR.

  • $\begingroup$ No, it is not correct that acceleration and gravity are equal in GR. Accelerating something with an electric field does not make any gravity. $\endgroup$ Nov 17 '18 at 21:46
  • $\begingroup$ But the force felt by that something would be indiscernible between acceleration and gravity. No? $\endgroup$ Nov 17 '18 at 21:50
  • $\begingroup$ In GR there is proper acceleration (an acceleration compared to an inertially free-falling observer) en.wikipedia.org/wiki/Proper_acceleration In the case near the donut there would be no proper acceleration (it is all due to spacetime) but free-falling observers would get an increasing velocity relative to the donut. Had it been charged then charged observers would have experienced a proper acceleration too. $\endgroup$ Nov 17 '18 at 23:19

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.