Doesn't spin-spin interactions in gravity disobey Galilean relativity?

Question

In gravity (GR) apparently there are forces which occur which are related to the spin of two masses.

For example if we had a rotating gravitational source and dropped, say, some particles into it with instrinsic spin, they would fall differently under gravity.

From what we learned at school, every object no matter what it's made of must fall the same under gravity. It only depends on the mass of the source. (i.e. drop a feather and a rock they must fall the same on the moon).

So was Galileo wrong?

What's more, using this spin-spin interaction seems to suggest that one might use this interaction to create an anti-gravity device. (In the sense that one could create an object that fell every so slightly slower towards a spinning object such as the sun or Earth).

In fact I've seen papers which suggest that two spinnning black holes could even repel each other.

Since gravity is equivalent to acceleration, is there an equivalent accelerated frame that causes different spinning particles to behaving differently?

I'm trying to reconcile this spin-spin interaction with what I was previously taught about how things must fall the same under gravity.

Answer 1

Absolutely, Galileo was wrong. You were taught according to the Newtonian model, which is incorrect, though a (very) good approximation at low speeds. In general relativity, a spinning object has quite different properties than a stationary one, because gravity is not a force dependent on mass, but rather the result of the curvature of spacetime, which is different when the object is spinning.

Consider the prototypical examples: the Schwarzschild and Kerr black holes, respectively: $$\mathrm{d}s^2=-\left(1-\frac{r_s}{r}\right)\mathrm{d}t^2+\left(1-\frac{r_s}{r}\right)^{-1}\mathrm{d}r^2+r^2\mathrm{d}\theta^2+r^2\sin^2\theta\,\mathrm{d}\phi^2$$ $$\mathrm{d}s^2=\left(1-\frac{r_sr}{\Sigma}\right)\mathrm{d}t^2+\frac{\Sigma}{\Delta}\mathrm{d}r^2+\Sigma\mathrm{d}\theta^2+\left(r^2+a^2+\frac{r_sra^2}{\Sigma}\sin^2\theta\right)\sin^2\theta\mathrm{d}\phi-\frac{2r_sra\sin^2\theta}{\Sigma}\mathrm{d}t\mathrm{d}\phi$$

Where $r_s$ is the Schwarzschild radius, $a=\frac{J}{M}$, $\Sigma=r^2+a^2\cos^2\theta$, and $\Delta=r^2-r_sr+a^2$.

As you can see, they are quite different. There are cross terms between $t$ and $\phi$, as you can see, which imparts angular momentum to things which approach it. In fact, if you get too close, you will be forced to move along with its motion, the effect called frame dragging. You can think of the object as dragging spacetime along with it as it spins, which causes these effects (this is a bit inaccurate, but easy to explain; for a more technical explanation, go here).

Answer 2

The formulation of the equivalence principle requires a bit more care in the context of general relativity. This becomes obvious from the fact that if you have a large object in general relativity it becomes unclear what exactly is meant by "the" trajectory of an the object. By extension it is unclear what one would mean by the trajectories of two different objects being the same.

For this reason the (relevant part of) the strong equivalence is formulate as follows: (Wikipedia)

The gravitational motion of a small test body depends only on its initial position in spacetime and velocity, and not on its constitution.

Small here means in relation to the curvature of the gravitational field through which the object is moving.

So isn't this principle violated for spinning objects in general relativity? The answer is no, because the spin of a small test body is necessarily also small. One can show that (assuming the dominant energy condition) the spin of a small body is at most $O(m^2)$ where $m$ is the test body's mass. This means that the effects of the body's spin coupling to the background curvature are at worst of the same order as corrections to its motion due to the generation of gravitational waves (i.e. the gravitational self-force).