Accelerometers, Gyroscopes, and Transported Frames

An ideal accelerometer detects departure from geodesic motion; an ideal torque-free gyroscope supplies a local standard of nonrotation. These are independent readings. Proper acceleration can be nonzero in flat spacetime, while a freely falling instrument can read zero acceleration in a curved spacetime even though separated test bodies reveal tides.

Required background. Observers, Rest Spaces, and Measured Components provides observer projectors and measured decompositions; Pseudo-Riemannian Metrics and Signature provides Lorentzian norms and causal classification; Frames, Coframes, and Anholonomy provides dual frames and local Lorentz freedom.

Helpful background. Events, Intervals, and Minkowski Geometry supplies the flat-spacetime limit; Tensor Fields and Natural Operations clarifies tensors versus their components; Proper Acceleration and Accelerometer Readings gives operational intuition for carried instruments.

Proper acceleration and its reading

Along a timelike worldline with unit tangent $u^\mu$, define the four-acceleration

$$a^\mu=u^\nu\nabla_\nu u^\mu.$$

Differentiating $u^2=-1$ gives $u\cdot a=0$, so $a^\mu$ lies in the instantaneous rest space. An accelerometer aligned with an orthonormal spatial triad reads

$$a^{\hat\imath}=e^{\hat\imath}{}_{\mu}a^\mu, \qquad \alpha=\sqrt{a_\mu a^\mu}.$$

The invariant $\alpha$ is the accelerometer’s magnitude reading. An ideal freely falling test body follows a timelike geodesic and has $a^\mu=0$. Conversely, an observer supported at a fixed position in a gravitational field generally has nonzero proper acceleration even if all spatial coordinate components remain constant.

Coordinate acceleration $d^2x^i/dt^2$ is not a substitute: it can be nonzero for a geodesic in curvilinear coordinates or vanish for a forced stationary worldline. Curvature is not determined by one accelerometer either. A finite apparatus can show tidal corrections across its size, but that is a multi-point effect beyond the point-observer reading.

Fermi–Walker nonrotation

Let $X^\mu$ be a vector carried along the observer. With $a^\mu$ defined above, Fermi–Walker transport is

$$u^\nu\nabla_\nu X^\mu =\left(u^\mu a_\nu-a^\mu u_\nu\right)X^\nu.$$

This sign is fixed by testing $X^\mu=u^\mu$: the right side becomes $a^\mu$ because $u\cdot a=0$ and $u^2=-1$. For a spatial vector $X\cdot u=0$, the law simplifies to

$$\nabla_u X^\mu=u^\mu(a\cdot X).$$

Its change has no component along any spatial direction. It undergoes exactly the boost-like adjustment needed to remain orthogonal to the changing time axis, with no additional instantaneous spatial rotation.

A carried tetrad is nonrotating when $e_{\hat0}=u$ and all three spatial legs are Fermi–Walker transported. Fermi axes, their realization by gyroscopes, and their distinction from radar coordinates are discussed in Perlick 2008, § 4, pp. 8–12, preprint PDF. If $a^\mu=0$, Fermi–Walker transport reduces to parallel transport. A freely falling observer can therefore realize a parallel-transported orthonormal frame with ideal gyroscopes, as described in Reall 2013, Ch. 8, pp. 84–85, PDF.

For an explicit flat-space check, an observer with constant proper acceleration $\alpha$ along $x$ carries

$$u^\mu=(\cosh\alpha\tau,\sinh\alpha\tau,0,0), \qquad e_{\hat1}{}^\mu=(\sinh\alpha\tau,\cosh\alpha\tau,0,0).$$

These obey $a^\mu=\alpha e_{\hat1}{}^\mu$ and $\nabla_u e_{\hat1}{}^\mu=\alpha u^\mu$, exactly the Fermi–Walker law. The axis has changing Cartesian components but no rotation in the observer’s rest space. This is why “nonrotating” cannot mean “every coordinate component stays constant.”

Gyroscope spin and preservation checks

Represent ideal gyroscope spin by a spacelike vector $S^\mu$ satisfying $S\cdot u=0$. In the torque-free point-gyroscope approximation, $S^\mu$ is Fermi–Walker transported. The transport generator

$$A_{\mu\nu}=u_\mu a_\nu-a_\mu u_\nu$$

is antisymmetric. Consequently Fermi–Walker transport preserves inner products. Directly,

$$\frac{d}{d\tau}(S\cdot S)=0, \qquad \frac{d}{d\tau}(S\cdot u)=0.$$

For the second equality, the $a\cdot S$ from differentiating $u$ cancels the contribution $u\cdot[u(a\cdot S)]$ from differentiating $S$. These checks fail if the two terms in the transport law are given the same sign.

Relative to a Fermi–Walker tetrad, the three spin components of an ideal gyroscope are constant. A more general carried triad can be written

$$\nabla_u e_{\hat\imath}{}^\mu =u^\mu(a\cdot e_{\hat\imath}) +\Omega_{\hat\imath}{}^{\hat\jmath}e_{\hat\jmath}{}^\mu, \qquad \Omega_{\hat\imath\hat\jmath}=-\Omega_{\hat\jmath\hat\imath}.$$

The antisymmetric $\Omega_{\hat\imath\hat\jmath}$ is the triad’s deliberate rotation relative to gyroscopes. Keeping coordinate components fixed does not generally make $\Omega$ vanish.

Curvature, loops, and approximation limits

Fermi–Walker transport is defined along one worldline and needs no nearby congruence. It does not construct a globally rigid reference frame. Orthogonal hypersurfaces based at different proper times may intersect, and neighboring worldlines may require different accelerations to maintain a chosen separation.

Nor does a gyroscope precession automatically prove curvature. Thomas precession occurs for accelerated motion in Minkowski spacetime because successive noncollinear boosts do not commute. Curvature becomes detectable through path dependence: parallel transport around a sufficiently small closed loop produces a leading change proportional to the Riemann tensor and the loop area. Comparing directions after separated paths therefore requires the paths and transport prescriptions to be stated.

The ideal spin law neglects torque, finite size, higher multipole moments, spin-curvature forces on the center of mass, and backreaction. The point-gyroscope model is controlled when the apparatus size and spin scales are small relative to curvature and acceleration scales and when external torques are negligible.

Exercises

Show directly that Fermi–Walker transport preserves both the spin norm and the constraint $S\cdot u=0$, assuming the constraint holds initially.

Solution

For a spatial spin vector, $\nabla_u S=u(a\cdot S)$. Hence

$$\frac{d}{d\tau}(S^2)=2S\cdot\nabla_u S =2(S\cdot u)(a\cdot S)=0.$$

Also,

$$\frac{d}{d\tau}(S\cdot u) =(\nabla_u S)\cdot u+S\cdot a =(u\cdot u)(a\cdot S)+a\cdot S=0.$$

Thus the initial orthogonality is retained and the spin magnitude remains fixed. When $a=0$, the equation becomes $\nabla_u S=0$, the expected parallel-transport law for a freely falling gyroscope.

References

• Perlick, Volker. “On the Radar Method in General-Relativistic Spacetimes”, in Lasers, Clocks and Drag-Free Control, Astrophysics and Space Science Library, Springer, 2008, pp. 131–152; Open PDF.
• Reall, Harvey S. General Relativity, PDF, Part III lecture notes, University of Cambridge, 2013, Ch. 8, pp. 84–85.