general relativity - How does a rotating object cause frame dragging?

Frame dragging is a consequence of general relativity.

But I don't really understand it. Of course I can find metaphors like the "honey metaphor" where stirring a honey can move the specks even if the spoon doesn't touch them. But I'm not satisfied with such simplistic explanations.

So if I understood it right rotating object cause a force that push nearby objects in the direction of rotation. Is that right, or something else happens?

Is this a real force? So if I'm in the vicinity of a massive rotating object would I measure acceleration as I dragged around?

Does this effect caused by the mere fact of the rotation or does it caused by the gravitational anomalies? I mean that most rotating objects are not perfectly homogeneous rotating spheres.

Can the stress energy tensor deal with rotation? I seems it can encode energy density, linear momentum, pressure and shear stress but doesn't seem to be encode angular momentum, can it?

Answer

First I'll address your stress tensor question. Angular momentum is encoded in the following tensor: $$M^{\mu\nu\rho}=x^\nu T^{\mu\rho}-x^\rho T^{\mu\nu}$$ We then define the angular momentum tensor as $$J^{\mu\nu}=\int_\Sigma M^{0\mu\nu}$$ where the measure on the space-like hypersurface $\Sigma$ is understood. There are problems with this, however. This is because it is very difficult to distinguish between the energy-momentum of the gravitational field and the stuff in it. Take a perfect fluid, for instance. The stress tensor contains the metric explicitly! Thus the stress tensor for a perfect fluid also contains gravitational information. Because of this, calling $J^{\mu\nu}$ a tensor is not really right. (It should be noted that the energy-momentum of a gravitational field CAN be covariantly defined for an asymptotically flat manifold.)

Your other question is answered in chapter 7.5 of Einstein Gravity in a Nutshell by A. Zee (2013). I'll hit the key points. A Kerr spacetime is axisymmetric and thus has a Killing vector $\xi=\partial_\varphi$. There is a well-known theorem that states if $u$ is the tangent of a geodesic, then $\langle\xi,u\rangle$ is a conserved quantity. Define the momentum $p=mu$. Then $L=\langle\xi,p\rangle$ is a constant of motion, the angular momentum. The general axisymmetric metric is $$ds^2=g_{tt}dt^2+g_{rr}dr^2+g_{\theta\theta}d\theta^2+g_{\varphi\varphi}d\varphi^2+2g_{t\varphi}dtd\varphi$$ In coordinates, $L=p_\varphi=g_{\varphi t}p^t+g_{\varphi\varphi}p^\varphi$. In the $(-+++)$ convention I am using, $g_{\varphi\varphi}>0$. Positive angular momentum is when the $\phi$ coordinate increases with increasing proper time. Hence, angular momentum is $+p_\varphi$.

It is conventional to define $L=ml$. Then $l=g_{t\varphi}\dot{t}+g_{\varphi\varphi}\dot\varphi$. Go out to infinity and prepare a test particle with $l=0$. Drop it towards the black hole. The asymptotic nature of the field means that far away, $g_{t\varphi}\sim0$ and $g_{\varphi\varphi}\sim1$. Thus $l=0$ means $\dot\varphi\sim0$, which is to be expected.

Here's the kicker: angular momentum is conserved. Thus $l\equiv 0$ identically. But suppose we, in our lab at infinity, take a look at the test particle. We see an angular velocity $$\omega\equiv \frac{d\varphi}{dt}=\dot\varphi/\dot t=-\frac{g_{t\varphi}}{g_{\varphi\varphi}}$$ which can totally be nonzero!

There is no force, technically speaking. The particle is just following a geodesic, which happens to take it along a path with angular velocity!

EDIT: Found this shortly after posting. Suppose $(M,g)$ is an axisymmetric spacetime with angular Killing vector $\psi$. The angular momentum of spacetime is given by the Komar integral $$J=\frac{1}{16\pi}\int_{S^2_\infty}\star d\psi$$ over the 2-sphere at infinity. See page 466 of Straumann (2013) for a proof. Now choose a hypersurface $\Sigma$ such that $\psi$ is tangent. Then $$J=-\int_\Sigma T_{\mu\nu}\psi^\mu n^\nu$$ where $T_{\mu\nu}$ are the components of the stress tensor and $n$ is a unit normal to $\Sigma$. This is problem 11.6 in Wald (1984).

EDIT II: I'd like to say a few words about energy-momentum in GR. (See chapter 3.7 of Straumann (2013) for the very thorough [and highly technical] discussion.) From the nonlinearity of Einstein's equation we know that gravitons have a nasty tendency to couple to each other. This means that it is in general not possible to separate the stress tensor like $$T=T_\text{matter}+T_\text{gravity}$$ because gravitons are coupling to everything and making life difficult. (However, if the metric is sufficiently close to the Minkowski metric, we may define the stress tensor of gravity to be the second order expansion of the Einstein tensor. [Chap 10.3 in Weinberg (1972)])

The solution is the so-called ADM formalism. It allows for a covariant description of energy, momentum and angular momentum on an asymptotically flat spacetime. We use the notation $$\tau^\alpha=T^\alpha+t_\text{LL}^\alpha$$ where $T^\alpha$ are the stress 1-forms and $t_\text{LL}^\alpha$ are the Landau-Lifshitz 1-forms. We then define $$\star M^{\alpha\beta}=x^\alpha\star\tau^\beta-x^\beta\star\tau^\alpha$$ Then $$J^{\mu\nu}=\int_\Sigma\sqrt{-g}\star M^{\mu\nu}$$ is the ADM angular momentum and is conserved if the gravitational field falls off sufficiently fast at spacelike infinity. Note that $T^\alpha$ is constructed out of the stress tensor and $t_\text{LL}^\alpha$ out of the tetrad.