Geodetic Effect - Formulae

Formulae

To derive the precession, assume the system is in a rotating Schwarzschild metric. The nonrotating metric is

$ds^2 = dt^2 \left(1-\frac{2m}{r}\right) - dr^2 \left(1 - \frac{2m}{r}\right)^{-1} - r^2 (d\theta^2 + \sin^2 \theta \, d\phi'^2) ,$

where c = G = 1.

We introduce a rotating coordinate system, with an angular velocity, such that a satellite in a circular orbit in the θ = π/2 plane remains at rest. This gives us

In this coordinate system, an observer at radial position r sees a vector positioned at r as rotating with angular frequency ω. This observer, however, sees a vector positioned at some other value of r as rotating at a different rate, due to relativistic time dilation. Transforming the Schwarzschild metric into the rotating frame, and assuming that is a constant, we find

$ds^2 = \left(1-\frac{2m}{r}-r^2 \beta\omega^2 \right)\left(dt-\frac{r^2 \beta\omega}{1-2m/r-r^2 \beta\omega^2} \, d\phi\right)^2 - dr^2 \left(1-\frac{2m}{r}\right)^{-1} - \frac{r^2 \beta - 2mr\beta}{1-2m/r - r^2 \beta\omega^2} \, d\phi^2$

with . For a body orbiting in the θ = π/2 plane, we will have β = 1, and the body's world-line will maintain constant spatial coordinates for all time. Now, the metric is in the canonical form

From this canonical form, we can easily determine the rotational rate of a gyroscope in proper time

$\Omega = \frac{\sqrt{2}}{4} e^\Phi ^{1/2} = \frac{ \sqrt{\beta} \omega (r -3 m) }{ r- 2 m - \beta \omega^2 r^3 } =\sqrt{\beta}\omega.$

where the last equality is true only for free falling observers for which there is no acceleration, and thus . This leads to

$\Phi,_i = \frac{2m/r^2 - 2r\beta\omega^2}{2(1-2m/r-r^2 \beta\omega^2)} = 0.$

Solving this equation for ω yields

$\omega^2 = \frac{m}{r^3 \beta}.$

This is essentially Kepler's law of periods, which happens to be relativistically exact when expressed in terms of the time coordinate t of this particular rotating coordinate system. In the rotating frame, the satellite remains at rest, but an observer aboard the satellite sees the gyroscope's angular momentum vector precessing at the rate ω. This observer also sees the distant stars as rotating, but they rotate at a slightly different rate due to time dilation. Let τ be the gyroscope's proper time. Then

$\Delta \tau = \left(1-\frac{2m}{r} - r^2 \beta\omega^2 \right)^{1/2} \, dt = \left(1-\frac{3m}{r}\right)^{1/2} \, dt.$

The −2m/r term is interpreted as the gravitational time dilation, while the additional −m/r is due to the rotation of this frame of reference. Let α' be the accumulated precession in the rotating frame. Since, the precession over the course of one orbit, relative to the distant stars, is given by:

$\alpha = \alpha' + 2\pi = -2 \pi \sqrt{\beta}\Bigg( \left(1-\frac{3m}{r} \right)^{1/2} - 1 \Bigg).$

With a first-order Taylor series we find

$\alpha \approx \frac{3\pi m}{r}\sqrt{\beta} = \frac{3\pi m}{r}\sin(\theta).$

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“I don’t believe in providence and fate, as a technologist I am used to reckoning with the formulae of probability.”
—Max Frisch (1911–1991)