Fermi–Walker Transport and Nonrotating Frames

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An accelerating observer’s frame is nonrotating when its spatial axes undergo only the succession of boosts required to keep the time axis equal to the observer’s four-velocity. Mathematically, those axes are Fermi–Walker transported. A torque-free gyroscope supplies the operational standard.

Required background. Special Relativity provides Minkowski causal geometry, four-vectors, and mass-shell kinematics; Relativistic Observers and Measurement provides observer rest spaces and invariant decompositions of measured quantities. Helpful background. Calculus and Linear Algebra provides multivariable differentiation, linear maps, inner products, and basis changes; Ordinary Differential Equations provides initial-value problems, linearization, and asymptotic solution methods.

Fermi–Walker transport and nonrotation

Along a worldline with four-velocity $u^\mu$ and four-acceleration $a^\mu=u^\nu\nabla_\nu u^\mu$ , define the Fermi–Walker derivative of a vector $X^\mu$ by

\frac{D_{\rm FW}X^\mu}{d\tau} =u^\nu\nabla_\nu X^\mu -(u^\mu a_\nu-a^\mu u_\nu)X^\nu.

In Cartesian inertial components, Fermi–Walker transport is therefore

\boxed{\frac{dX^\mu}{d\tau} =(u^\mu a_\nu-a^\mu u_\nu)X^\nu.}

The sign can be checked by setting $X^\mu=u^\mu$ : the right-hand side becomes $a^\mu$ because $u^2=-1$ and $u\mathbin{\cdot}a=0$ . Thus the time axis transports into the next four-velocity, as required.

The generator

K_{\mu\nu}=u_\mu a_\nu-a_\mu u_\nu=-K_{\nu\mu}

is antisymmetric. If $X$ and $Y$ are both Fermi–Walker transported, metric compatibility gives

\begin{aligned} \frac{D}{d\tau}(X\mathbin{\cdot}Y) &=(KX)\mathbin{\cdot}Y+X\mathbin{\cdot}(KY)\\ &=(a\mathbin{\cdot}X)(u\mathbin{\cdot}Y) -(u\mathbin{\cdot}X)(a\mathbin{\cdot}Y)\\ &\quad +(u\mathbin{\cdot}X)(a\mathbin{\cdot}Y) -(a\mathbin{\cdot}X)(u\mathbin{\cdot}Y)=0. \end{aligned}

This proves, rather than assumes, that lengths and mutual angles are preserved. It also preserves spatiality: differentiating $u\mathbin{\cdot}X=0$ gives $a\mathbin{\cdot}X+u\mathbin{\cdot}KX=0$ .

Choose a carried tetrad with $e_{\hat0}=u$ and Fermi–Walker transport each spatial vector $e_{\hat i}$ . Since $u\mathbin{\cdot}e_{\hat i}=0$ ,

\frac{D e_{\hat i}^{\mu}}{d\tau} =u^\mu(a\mathbin{\cdot}e_{\hat i}).

Its change has no component along any other spatial axis. There is a boost-like change along $u^\mu$ , but no instantaneous spatial rotation. If $a^\mu=0$ , the equation becomes parallel transport and reduces to constant components in an inertial Cartesian basis.

A gyroscope’s spin four-vector $S^\mu$ obeys $S\mathbin{\cdot}u=0$ . In the torque-free idealization its spin is Fermi–Walker transported, so a frame fixed to such gyroscopes is nonrotating. The separation between the Fermi–Walker and freely chosen spatial-rotation parts is displayed in Gourgoulhon 2010, PDF pp. 26–30, printed slide 18.

The definition is local along one worldline. It does not automatically construct a spatially extended rigid frame. Different neighboring worldlines can have different accelerations, and their simultaneity slices can eventually intersect.

Thomas precession

Two noncollinear Lorentz boosts do not compose to a pure boost. Their rotational remainder makes a Fermi–Walker gyroscope precess relative to axes kept parallel in a chosen inertial laboratory. Let $B(\mathbf v)$ be the standard rotation-free boost from laboratory components to the momentarily comoving frame. Its components are

\begin{aligned} B^0{}_0&=\gamma,& B^0{}_j&=-\gamma v_j,\\ B^i{}_0&=-\gamma v^i,& B^i{}_j&=\delta^i{}_j +\frac{\gamma^2}{\gamma+1}v^iv_j, \end{aligned} \qquad \gamma=(1-v^2)^{-1/2}.

Multiply the neighboring boosts $B(\mathbf v+d\mathbf v)B^{-1}(\mathbf v)$ and retain terms linear in $d\mathbf v$ . The antisymmetric spatial part—the part no pure boost can supply—rotates the gyroscope relative to the fixed laboratory axes by

d\boldsymbol\theta_{\rm T} =\frac{\gamma^2}{\gamma+1}\,d\mathbf v\times\mathbf v.

Since $d\mathbf v=\mathbf A\,dt$ , the angular velocity is

\boxed{\boldsymbol\Omega_{\rm T} =\frac{\gamma^2}{\gamma+1}\,\mathbf A\times\mathbf v.}

This sign convention describes the gyroscope relative to the laboratory axes. The same boost decomposition and coefficient are given in Gourgoulhon 2010, PDF pp. 45–47, printed slides 31–33. At low speed,

\boldsymbol\Omega_{\rm T} \simeq \frac12\mathbf A\times\mathbf v.

Restoring units inserts a factor $c^{-2}$ . For transverse turning, $A=v\,|d\hat{\mathbf v}/dt|$ , so the precession is suppressed by $v^2/(2c^2)$ relative to the rate at which the velocity direction turns. Without that relation between $\mathbf A$ and $\mathbf v$ , it is not meaningful to call the effect simply “second order in velocity.” A coordinate triad that continually points radially and tangentially around a circular orbit is therefore not Fermi–Walker transported.

More precisely, the rotating three-vector consists of the spin’s components after applying $B(\mathbf v)$ to its four-vector at each event. Each instantaneous rest frame is aligned with the laboratory by this standard boost. The unboosted laboratory spatial components of the spin four-vector need not have constant Euclidean length and do not obey a pure three-dimensional rotation law. Specifying the rest-frame comparison is essential when using the precession formula quantitatively.

Operational distinctions and failures

An accelerometer determines $a^{\hat i}$ ; gyroscopes determine whether the chosen spatial triad rotates. Neither measurement alone specifies the other.
“Fixed orientation in coordinates” is not an invariant nonrotation rule. Polar or rotating coordinate bases can rotate even in flat spacetime.
Replacing Fermi–Walker transport by ordinary constant components requires an inertial worldline and an inertial Cartesian, or otherwise parallel, basis.
The Thomas rotation is not caused by spacetime curvature. It follows from Lorentz-group kinematics in Minkowski spacetime.
A spacecraft may deliberately rotate its cabin relative to its gyroscopes. That adds an independent spatial angular velocity to the Fermi–Walker law.

Transfer task: one circular orbit

An observer moves counterclockwise at constant speed $v$ on a laboratory circle of radius $R$ , with orbital angular speed $\Omega=v/R$ . How far does a Fermi–Walker gyroscope turn relative to the laboratory after one orbit?

Solution

The laboratory acceleration is inward with magnitude $A=v^2/R$ . Hence $\mathbf A\times\mathbf v$ points opposite the orbital angular-velocity vector, and

|\Omega_{\rm T}| =\frac{\gamma^2}{\gamma+1}\frac{v^3}{R} =(\gamma-1)\Omega,

where $\gamma^2v^2=(\gamma-1)(\gamma+1)$ . During the laboratory orbital period $2\pi/\Omega$ , the gyroscope accumulates

\boxed{\Delta\theta_{\rm T}=-2\pi(\gamma-1)},

with the minus sign indicating precession opposite the orbital sense. For $v\ll1$ , $\Delta\theta_{\rm T}\simeq-\pi v^2$ , which vanishes in the Newtonian limit.

Next, Accelerated Coordinate Systems uses a Fermi–Walker tetrad to extend one observer’s measurements into a local coordinate chart.

References

Éric Gourgoulhon, Special relativity from an accelerated observer perspective (PDF), PDF pp. 26–30 and 45–47, printed slides 18 and 31–33, Observatoire de Paris, 2010.