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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?

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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.

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There are a multitude of ways to explain this --

  1. Think about this in analogy to electromagnetism: In this context, think about a distribution of charges at rest with respect to each other. There is one special Lorentz frame where there is no magnetic field, namely, where you are at rest with respect to the charge distribution. Now, take a test charge, and hold it stationary with respect to the charge distribution. Obviously, this will require a force $F$. Now, look at this situation from a boosted frame. Here, there will be a magnetic field, and the charge distribution will be Lorentz contracted. It turns out that the magnetic field will arise in precisely the way necessary to make the force transform correctly between the two frames. In general relativity, you obviously can have the same situation -- a line of mass can exert a force $F$ on a stationary object in one frame, and then you can Lorentz boost the frame. To avoid a contradiction, a "magnetic" gravitational force must arise. It turns out that this force is precisely what happens when you frame drag.
  2. Boost schwarzschild: If you want a more concrete model, there is a coordinate system where the Schwarzschild metric can be written as: $$g_{ab} = \eta_{ab} + C \ell_{a}\ell_{b}$$ where $\ell_{a} = (-1,1,0,0)$, $\eta_{ab} = {\rm diag}(-1,1,r^{2},r^{2}\sin^{2}\theta)$ and $C = 2M/r$ This is convenient, because you have a background Minkowski metric, and you have the bit of the metric that is dependent on the Schwarzschild metric. Therefore, when you Lorentz boost the metric, you don't transform the Minkowski metric. Thus, applying a Lorentz matrix to this metric gives you the Schwarzchild metric as observed by a relativistically moving observer. Even this simple example will then show that the subsequent motion will involve forces that pull test particles in the direction of the matter distribution -- $g_{ti}$ will pick up r-dependent terms, which will tell you that an observer close to the black hole has a rest frame that is in motion with respect to an observer at infinity
  3. directly analyze the Kerr metric: Finally, we can just naïvely look at the Kerr metric. Here, we will have, in Boyer-Lindquist coordinates, $g_{t\phi}$ terms. These terms tell us that a timelike observer near the black hole will have their time coordinate shifted with respect to a timelike observer far from the black hole -- my notion of being "stationary with respect to the black hole" will depend on how far away from the hole I am. In fact, there is a surface called the erogosphere beyond which it is impossible for me to be stationary with respect to infinity -- in order for my path to stay timelike, I will HAVE to pick up motion relative to infinity.

But no matter how this is explained, note that whether or not this is a "force" depends on what you're trying to do. If you're just drifting along a geodesic, you never feel a force. If I'm trying to do something like stay at rest relative to infinity, though, yes, I'm going to have to exert a force (or for the case of an observer beyond the erogsphere but outside the horizon, I will be unable to do so, even with an infinite force)

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  • $\begingroup$ "It turns out that the magnetic field will arise in precisely the way necessary to make the force transform correctly between the two frames." Does this mean elecromagnetism can be formalized in a way so there is only electric field and magnetic field is just a relativistic effect? That's interesting... $\endgroup$ – Calmarius Jan 4 '15 at 12:51
  • $\begingroup$ @Calmarius: that's close to it, but not quite. It'd probably be more proper to say that it would be inconsistent to have JUST an electric or JUST a magnetic field -- they are tied to each other. And if you have one, and special relativity, consistency demands that you have the other. $\endgroup$ – Jerry Schirmer Jan 5 '15 at 2:40
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Frame dragging has several different but of course closely related meanings in the literature.

However there are a couple of other canonical meanings that are worth mentioning. In Kerr space-time one has a time-like Killing field $\xi^{\alpha}$ and an axial Killing field $\psi^{\alpha}$. The congruence of observers $u^{\alpha} = \xi^{\alpha}/|\xi|$ following orbits of $\xi^{\alpha}$ are of course the observers who are at rest with respect to the central mass and hence are at rest with respect to the asymptotic Lorentz frame at spatial infinity in which the central mass is at rest.

They have a vanishing angular velocity, $\omega = \frac{d\phi}{dt} = 0$, relative to infinity; here $t$ and $\phi$ are the time and angular coordinates defined by $\xi^{\alpha}$ and $\psi^{\alpha}$ respectively. But they have a non-vanishing angular momentum $L = u_{\alpha}\psi^{\alpha} \neq 0$. This can be interpreted in terms of the Sagnac effect i.e. if the observer were to send out light beams in opposite directions around the same circuit then they would return at different times on the observer's clock (Ashtekar and Magnon 1975). This is a sense in which the observer is not rotating relative to infinity but rotating relative to the local space-time geometry because the observer finds the $\phi$ and $-\phi$ directions nonequivalent. Since this is due solely because of the rotation of the space-time, one can consider this frame dragging.

Furthermore if any one of these observers carries a set of spatial axes $e_i$ that are fixed with respect to infinity, so that the Lie derivative $\mathcal{L}_{\xi}e_i = 0$, then it can be shown that $$F_{u}e^{\alpha}_i = -(\xi_{\delta}\xi^{\delta})^{-1}\epsilon^{\alpha\beta\gamma\sigma}\xi_{\beta}\nabla_{\gamma}\xi_{\sigma} \neq 0$$ where $F_u$ is the Fermi-Walker derivative along $u$; see for example Lightman et al problem 11.10. What this means is even though the spatial axes are fixed with respect to the asymptotic Lorentz frame, they nonetheless precess relative to a set of inertial guidance gyroscopes comoving with the observer.

Since $\xi_{[\beta}\nabla_{\gamma}\xi_{\sigma]} \neq 0$ if and only if the space-time is stationary but not static, which intuitively means the space-time is rotating, this is another manifestation of frame dragging: the rotation of the space-time makes the local standard of non-rotation different from the global standard of non-rotation relative to infinity. This effect can be interpreted as a gravitomagnetic effect as pointed out by Jerry Schirmer: the stationary gyroscope couples to the gravitomagnetic field generated by the rotation of the central mass and precesses akin to how a charged dipole moment in a magnetic field precesses.

As an aside, it might be worth noting that this notion of non-rotation is in general not the same as that prescribed by the Sagnac effect. To illustrate this, consider now the congruence of zero angular momentum observers (ZAMOs) $u^{\alpha} = \nabla^{\alpha}t/|\nabla t|$. They are termed this of course because $L = 0$ identically for the congruence as is easy to check. One can write $\nabla^{\alpha}t = \xi^{\alpha} + \omega(r,\theta)\psi^{\alpha}$ where $\omega(r,\theta) = -g_{t\phi}/g_{\phi\phi}$. In other words the ZAMOs orbit the central mass with exactly the angular velocity necessary in order to be at rest with respect to the local space-time geometry, in the sense described above.

Consider now the time-like Killing field $\eta^{\alpha} = \xi^{\alpha} + \omega(r_0,\theta_0)\psi^{\alpha}$ for some fixed $r_0,\theta_0$. This describes a congruence of observers which coincides with the ZAMO congruence at $r_0,\theta_0$. Consider also a ZAMO at $r_0,\theta_0$ carrying a set of spatial axes $e_i$ adapted to the space-time symmetries. The result above for $F_u e_i$ can be easily extended to any time-like Killing field from which it follows after a calculation that $$F_{u}e^{\alpha}_i|_{(r_0,\theta_0)} = -(\eta_{\delta}\eta^{\delta})^{-1}\epsilon^{\alpha\beta\gamma\sigma}\eta_{\beta}\nabla_{\gamma}\eta_{\sigma}|_{(r_0,\theta_0)} \neq 0$$ See for example MTW problem 33.4.

Hence even though an individual ZAMO has no angular momentum, the observer's spatial axes still precess relative to local gyroscopes. Thus when one finds statements in the literature about how the ZAMO eliminates frame dragging it is really in the sense of just having zero angular momentum.

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