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Why does Lense-Thirring precession look like a dipole?

+ 3 like - 0 dislike
162 views

1. Background:

Lense-Thirring precession is the rotation undergone by the spin of a particle located in the gravitational field of a massive spinning body. In terms of asymptotically inertial coordinates $(t,\vec x)$ in a four-dimensional space-time, and if we denote by $\vec J$ the angular momentum of the source, the angular velocity of precession of a particle at position $\vec x$ is

$$\vec\Omega = \frac{1}{r^3}\left(-\vec J+3\frac{(\vec J\cdot\vec x)\vec x}{r^2}\right)$$

where $r\equiv\sqrt{\vec x\cdot\vec x}$ and the dot denotes the scalar product of spatial vectors. We use units such that $G=c=1$. (For the derivation of this formula, see e.g. Misner-Thorne-Wheeler, section 40.7.)

2. My question:

The dependence of the angular velocity $\vec\Omega$ on the source's angular momentum $\vec J$ and on the spatial position $\vec x$ is exactly the same as that of the electric field generated by an electric dipole. Under such an identification, $\vec\Omega$ is identified with the electric field while $\vec J$ is identified with the dipole moment. My question is the following: is there an intuitive explanation for why the precessional angular velocity has to be of the same form as the field sourced by a dipole?

Just to make things clear: I'm not looking for a mathematical proof that the above formula for $\vec\Omega$ is correct. Instead, I'd like to find an intuitive (but nevertheless rigorous) argument that makes the above result obvious. Indeed, the standard derivation of the formula for $\vec\Omega$ relies on some relatively advanced mathematical tools, but the result is so simple and pretty that I suspect there's a deeper reason for the apparent coincidence with the formula from electrostatics. (Of course, my expectation may be wrong.)

asked Nov 15, 2016 in Theoretical Physics by anonymous [ no revision ]

I can explain this, but not succinctly. It would take me several thousand words. I'd have to explain the electron and its electromagnetic field, and how the thing we call an electric field maps out the linear force that results from the interaction of two (or more) electromagnetic fields. Then I'd have to show how gravitomagnetism relates to electromagnetism, what a spinor is, why an electron moves rotationally in a magnetic field, and so on. It's just too much, and I can't think of a shortcut.

@JohnDuffield: The answer field allows several thousand words. You can make shortcuts by assuming that readers know the field that an electron generates, what a spinor is, etc. Just start with the main points of the explanation and add details later if needed....

I'm sorry Arnold, it's just too much.

2 Answers

+ 1 like - 0 dislike

The reason is that linearized gravity is a set of partial differential equations which is very similar in nature to electromagnetism. Namely in vacuum, weak fields, and under Lorentz gauge, the trace-reversed perturbation  follows

$$\Box \bar{h}_{\mu\nu} =0$$

Which is very much like the equation for the electromagnetic potential in Lorentz gauge

$$\Box A_\mu = 0$$

This analogy runs so deep that, under appropriate post-Newtonian assumptions, it is even possible to rewrite the weak-field perturbation $\bar h_{\mu\nu}$ through a "gravitoelectromagnetic vector" $A^\mathrm{(g)}_\mu$ such that the dynamical equations are written as
$$\Box A^\mathrm{(g)}_\mu = 4 \pi G j^\mathrm{(m)}_\mu$$

where $j^\mathrm{(m)}_\mu$ is the local matter current. There are some post-Newtonian corrections to the naive $j^\mathrm{(m)}_\mu \sim \rho u_\mu$ (see Braginsky, Caves and Thorne 1977) but our general intuitions about what this current means are correct. The time-like component corresponds to the "static" matter density, whereas the space-like corresponds to moving masses, very much like in the electric current $j_\mu$ with static and moving charges.

I.e., the gravitomagnetic field of a small circulating loop of matter will be qualitatively the same as the magnetic field of a small electric-current loop, at least in the far field. Amongst other things, the far field will exhibit the same multipolar structure. There are some naughty details to this (such as what's up with dipolar gravitational waves in this approximation), but this is the general idea - dragging effects exhibit analogous mathematical structures as magnetic effects because weak-field, post-Newtonian gravity is essentially like electromagnetism only with masses playing the roles of charges.

answered May 28 by Void (1,505 points) [ no revision ]

@Void Thanks a lot! This is exactly the kind of answer I was hoping to get. Now that I've read it, it does seem kind of obvious... Though it does seem hard, from that perspective, to justify the absence of gravitational dipole radiation.

The intuitive argument can be actually constructed the same way as in the usual linearized gravity. The second derivative of the "gravitational dipole" of an isolated system vanishes because of momentum conservation. So it is not that the dipole radiation is not a solution, it is just an unphysically sourced one.

@Void Marvellous -- thanks for the illuminating explanations!

+ 0 like - 2 dislike

I believe there is no frame-dragging effect of gravity on mass, only on electromagnetic energy!
Curiously, the Lense-Thirring effect in Gravity Probe B has the same value than the geodetic effect of the Earth around the Sun.
Kinetic energy of a rotating body like Earth will increase gravity but more in the sense of classical force than classical aether!
An interesting experiment!
Understanding Gravity Probe-B experiment without math
http://www.molwick.com/en/gravitation/082-gravity-probe-b.html

answered May 28 by anonymous [ no revision ]

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