# Can a drop of water be set in rotational motion by rotating mass around it?

We are in empty space and see a spherical drop of water. Around the drop we have placed a massive shell with uniform density. The drop is positioned at the center. Then we set the shell in rotational motion (by small rockets on the side). Will the drop start rotating (slowly)? Will frame drag cause a torque?

The Newtonian idea of gravity predicts a zero gravity field inside the sphere. General relativity predicts frame dragging. The mass-energy-momentum tensor includes momentum and that's what we see in this case.

So, will it rotate? Will the shell and the droplet be eventually rotating in tandem? Of course we must stop the acceleration before a black hole develops...

Can we say the rotating sphere induces torsion?

• Note that in the electromagnetic analogue, a rotating charged shell would not induce a torque on a charged object inside. So I wouldn't expect a rotation at the linearized gravity level either (using gravitomagnetism as an analogy.) There might be non-linear effects, though. May 31 at 19:50
• @MichaelSeifert Is frame-dragging non-linear? May 31 at 21:14

Apparently Thirring computed this in 1918: Phys. Z. 19, 33 (1918) (in German) the central (corrected result) for the acceleration of a test particle inside a slowly rotating mass shell (of mass $$M$$, radius $$R$$, and angular momentum $$\vec{\omega}$$) is given by $$\vec{a}=−2d_1(\vec{\omega} \times\vec{v} )−d_2[\vec{\omega} \times(\vec{\omega} \times\vec{r})+2(\vec{\omega} \cdot\vec{r})\vec{\omega} ],$$ with the constants $$d_1 = 4MG/3Rc^2$$ and $$d_2 = 4MG/15Rc^2$$ for the Coriolis- and centrifugal contributions respectively, according to H. Pfister (2005) On the history of the so-called Lense-Thirring effect. This expression is valid only close to the center of the sphere: $$|\vec{r}|\ll R$$.
• If I'm not mistaken, the first term wouldn't apply a torque to a mass at rest ($\vec{v}=0$). I'm not immediately sure whether the second one would, though. Jun 1 at 12:20
• Going through the vector identities, the second term applies a specific torque of $3 (\vec{\omega}\cdot \vec{r}) (\vec{r} \times \vec{\omega})$. If you set up an integral of this quantity over a solid body, you can show that it must vanish for any body with a rotational symmetry axis parallel to $\vec{\omega}$. So at this level of approximation, a small spherical fluid drop at rest at the center of the sphere would not start rotating. Jun 1 at 13:32