Geodetic and Frame-Dragging Precession

A torque-free gyroscope can precess even while its spin is parallel transported. In weak stationary gravity, two physically distinct terms appear relative to asymptotically inertial axes: geodetic precession from motion through the mass-monopole curvature, and Lense–Thirring precession from the source’s angular momentum. Their different dependence on the orbital velocity and source spin makes their limits, directions, and signs independently testable.

Required background. We use The Einstein Equation for the weak stationary metric, Events, Intervals, and Minkowski Geometry for spin measured against asymptotic axes, and The Newtonian Gravitational Field and Potential for the mass-monopole orbit.

Helpful background. Tensor Fields and Natural Operations supports the spin and angular-momentum vectors, and Affine and Variational Geodesics supplies the test-body trajectory and parallel transport.

Spin transport in a weak stationary field

Let $S^\mu$ be the spin four-vector of a small, torque-free gyroscope moving with four-velocity $u^\mu$. In the test-spin approximation,

$$u^\nu\nabla_\nu S^\mu=0, \qquad S^\mu u_\mu=0.$$

This parallel-transport law assumes geodesic motion. A supported or otherwise accelerated gyroscope instead obeys Fermi–Walker transport; its measured precession also depends on the specified accelerated reference frame.

The calculation assumes

$$\frac{GM}{r}\ll1, \qquad \lvert\mathbf v\rvert\ll1, \qquad \frac{G\lvert\mathbf J\rvert}{r^2}\ll1,$$

with a slowly rotating isolated source, a gyro much smaller than the curvature scale, negligible spin–curvature force on its orbit, and a stationary exterior. Source multipoles, torques, comparable-mass effects, and radiation reaction are omitted.

Using $\Phi=-GM/r$, the leading exterior metric in asymptotically Cartesian coordinates can be written

$$\begin{aligned} ds^2={}&-(1+2\Phi)dt^2 +(1-2\Phi)\delta_{ij}dx^i dx^j \\ &-\frac{4G}{r^3} (\mathbf J\times\mathbf r)\cdot d\mathbf x\,dt. \end{aligned}$$

Expanding parallel transport, eliminating the small time component through $S\cdot u=0$, and expressing the spatial spin relative to the orthonormalized asymptotically Cartesian triad boosted into the gyro rest space gives

$$\frac{d\mathbf S}{dt} =\left(\boldsymbol\Omega_{\rm geo} +\boldsymbol\Omega_{\rm LT}\right)\times\mathbf S.$$

This equation defines the sign convention used below: a positive $\boldsymbol\Omega$ rotates $\mathbf S$ by the right-hand rule relative to the chosen asymptotically inertial spatial axes.

From coordinate spin to measured spin

The rest-frame conversion is the step that fixes the geodetic factor. Let $\mathbf s$ denote the coordinate spatial components $S^i$, define $\mathbf g=\boldsymbol\nabla\Phi$, and write the metric’s mixed components as

$$A_i\equiv g_{0i}=-\frac{2G}{r^3}(\mathbf J\times\mathbf r)_i.$$

The cross term in $ds^2$ is $2A_i\,dt\,dx^i$, explaining its factor of four above. To the retained weak-field and slow-motion order, the orthogonality condition gives $S^0=\mathbf v\cdot\mathbf s$, and the spatial transport equation becomes

$$\begin{aligned} \frac{d\mathbf s}{dt}={}& \mathbf v(\mathbf g\cdot\mathbf s) +\mathbf s(\mathbf v\cdot\mathbf g) -2\mathbf g(\mathbf v\cdot\mathbf s)\\ &-\frac12(\boldsymbol\nabla\times\mathbf A)\times\mathbf s. \end{aligned}$$

Coordinate components are not yet the gyro’s Euclidean spin vector. Orthonormalizing the spatial axes and then applying the rotation-free boost into the gyro rest frame gives, at this order,

$$\mathbf S=(1-\Phi)\mathbf s -\frac12\mathbf v(\mathbf v\cdot\mathbf s).$$

The first correction comes from $g_{ij}=(1-2\Phi)\delta_{ij}$; the second is the low-velocity Lorentz-boost correction. Differentiate this relation using $d\mathbf v/dt=-\mathbf g$ at Newtonian order. The term $\mathbf s(\mathbf v\cdot\mathbf g)$ cancels, and the result is

$$\frac{d\mathbf S}{dt} =\frac32\left[\mathbf v(\mathbf g\cdot\mathbf S) -\mathbf g(\mathbf v\cdot\mathbf S)\right] -\frac12(\boldsymbol\nabla\times\mathbf A)\times\mathbf S.$$

Using $(\mathbf g\times\mathbf v)\times\mathbf S=\mathbf v(\mathbf g\cdot\mathbf S)-\mathbf g(\mathbf v\cdot\mathbf S)$ identifies $\boldsymbol\Omega_{\rm geo}=\tfrac32\mathbf g\times\mathbf v$ and $\boldsymbol\Omega_{\rm LT}=-\tfrac12\boldsymbol\nabla\times\mathbf A$. Terms with another power of $\Phi$, $v^2$, or the source-spin perturbation have been omitted consistently. The cross-product form also checks that the measured spin length is preserved: $d|\mathbf S|^2/dt=0$.

Geodetic precession from the mass monopole

For general relativity, the leading geodetic term is

$$\boldsymbol\Omega_{\rm geo} =\frac32\,\boldsymbol\nabla\Phi\times\mathbf v =\frac{3GM}{2r^3}\,\mathbf r\times\mathbf v.$$

Because $\boldsymbol\nabla\Phi=GM\mathbf r/r^3$ points outward while the acceleration $-\boldsymbol\nabla\Phi$ points inward, the cross-product order matters. For a prograde circular orbit, $\mathbf r\times\mathbf v$ is the orbital angular-momentum direction, so $\boldsymbol\Omega_{\rm geo}$ is parallel to the orbital normal. Reversing the orbit reverses this vector.

For a circular orbit with angular frequency $\Omega_{\rm orb}$,

$$\lvert\boldsymbol\Omega_{\rm geo}\rvert =\frac{3GM}{2r}\,\Omega_{\rm orb}.$$

For a weak eccentric Kepler orbit, averaging over one radial period gives

$$\left\langle\boldsymbol\Omega_{\rm geo}\right\rangle =\frac{3GM}{2a(1-e^2)}\,n\,\widehat{\mathbf L}, \qquad n=\sqrt{\frac{GM}{a^3}}.$$

The $3/2$ factor combines the special-relativistic kinematics of successive boosts with spatial curvature; it is not obtained from Newtonian torque. Will states the more general coefficient $(\gamma+1/2)$ and its GR value in Will 2014, § 4.4.2, printed p. 51, Eq. (70), PDF.

Lense–Thirring precession from source rotation

Let $\mathbf n=\mathbf r/r$ and orient $\mathbf J$ by the source’s right-hand rotation axis. The leading frame-dragging term in GR is

$$\boxed{ \boldsymbol\Omega_{\rm LT} =\frac{G}{r^3} \left[3\mathbf n(\mathbf J\cdot\mathbf n)-\mathbf J\right] }.$$

For $r\gt0$, the vector identity $\boldsymbol\nabla\times(\mathbf J\times\mathbf r/r^3)=[3\mathbf n(\mathbf J\cdot\mathbf n)-\mathbf J]/r^3$ follows by differentiating the three Cartesian components for constant $\mathbf J$. Together with $\boldsymbol\Omega_{\rm LT}=-\tfrac12\boldsymbol\nabla\times\mathbf A$, it verifies both the sign and coefficient. On the positive rotation axis, $\mathbf n\parallel\mathbf J$, so $\boldsymbol\Omega_{\rm LT}=2G\mathbf J/r^3$. In the equatorial plane, $\mathbf J\cdot\mathbf n=0$, so $\boldsymbol\Omega_{\rm LT}=-G\mathbf J/r^3$. The sign change with latitude is the angular structure of a dipole. Will gives the PPN form and its GR reduction in Will 2014, § 4.4.1, printed pp. 49–51, Eqs. (68)–(69), PDF.

The source-spin reversal $\mathbf J\mapsto-\mathbf J$ reverses $\boldsymbol\Omega_{\rm LT}$ but leaves $\boldsymbol\Omega_{\rm geo}$ unchanged for the same orbit. Setting $\mathbf J=0$ removes frame dragging. Setting $\mathbf v=0$ at one event removes the instantaneous geodetic term while leaving $\boldsymbol\Omega_{\rm LT}$ for an instantaneously stationary freely falling gyro. A supported gyro is accelerated, however, so its evolution requires Fermi–Walker transport and a specified reference frame rather than an unqualified use of the two-term geodesic formula. Taking $M\to0$ removes the monopole contribution, and taking $G\to0$ removes both. Dimensions also agree: $GM\,\mathbf r\times\mathbf v/r^3$ and $GJ/r^3$ each have dimension inverse time in units with $c=1$.

What the split means

The separate connection coefficients used during the derivation are gauge dependent, but the predicted rotation of a transported spin relative to a specified asymptotic reference frame is not. The monopole–dipole split is meaningful at this leading order for an isolated stationary source because $M$ and $\mathbf J$ are its asymptotic mass and angular momentum. In a generic time-dependent or multipolar spacetime, a two-term formula is insufficient.

The result describes spin direction, not necessarily the orbital plane. Frame dragging also produces nodal precession of an orbit, but its coefficient and averaging protocol are a different observable. Nor do the formulas supply a dated experimental bound; real gyroscope and satellite analyses must model classical torques, multipoles, reference-star motion, and instrument systematics.

Worked orientation check

Compare two idealized configurations at the same radius around a source with $\mathbf J=J\mathbf e_z$: (A) an instantaneously stationary, freely falling gyro on the positive rotation axis; and (B) a gyro on a prograde equatorial circular geodesic, whose orbital angular momentum points along $+\mathbf e_z$. Find the directions of the leading terms in each configuration.

Solution

In configuration A, $\mathbf v=0$ at the chosen event, so $\boldsymbol\Omega_{\rm geo}=0$. Because $\mathbf n=\mathbf e_z$ and $\mathbf J\cdot\mathbf n=J$,

$$\boldsymbol\Omega_{\rm LT} =\frac{2GJ}{r^3}\mathbf e_z.$$

This statement is instantaneous: the freely falling gyro does not remain at rest on the axis. In configuration B, $\mathbf r\times\mathbf v$ follows the prograde orbital angular momentum, so $\boldsymbol\Omega_{\rm geo}$ points along $+\mathbf e_z$. In the equatorial plane $\mathbf J\cdot\mathbf n=0$, and therefore

$$\boldsymbol\Omega_{\rm LT} =-\frac{GJ}{r^3}\mathbf e_z.$$

Thus the geodetic and frame-dragging contributions oppose one another in configuration B. The comparison does not assign an impossible orbital angular momentum to a trajectory located on the rotation axis; it uses two separately realizable local geometries.

References

• Will, Clifford M. “The Confrontation between General Relativity and Experiment.” Living Reviews in Relativity 17, article 4 (2014), § 4.4, printed pp. 49–51. DOI; arXiv.