Foucault Pendulum - Precession As A Form of Parallel Transport

Precession As A Form of Parallel Transport

From the perspective of an inertial frame moving in tandem with Earth, but not sharing its rotation, the suspension point of the pendulum traces out a circular path during one sidereal day. At the latitude of Paris a full precession cycle takes 32 hours, so after one sidereal day, when the Earth is back in the same orientation as one sidereal day before, the oscillation plane has turned 90 degrees. If the plane of swing was north-south at the outset, it is east-west one sidereal day later. This implies that there has been exchange of momentum; the Earth and the pendulum bob have exchanged momentum. (The Earth is so much more massive than the pendulum bob that the Earth's change of momentum is unnoticeable. Nonetheless, since the pendulum bob's plane of swing has shifted the conservation laws imply that there must have been exchange.)

Rather than tracking the change of momentum the precession of the oscillation plane can efficiently be described as a case of parallel transport. For that it is assumed that the precession rate is proportional to the projection of the angular velocity of Earth onto the normal direction to Earth, which implies that the plane of oscillation will undergo parallel transport. The difference between initial and final orientations is α = −2 sin(φ), in which case the Gauss-Bonnet theorem applies. α is also called the holonomy or geometric phase of the pendulum. Thus, when analyzing earthbound motions, the Earth frame is not an inertial frame, but rather rotates about the local vertical at an effective rate of 2π sin(φ) radians per day. A simple method employing parallel transport within cones tangent to the Earth's surface can be used to describe the rotation angle of the swing plane of Foucault's pendulum.

From the perspective of an Earth-bound coordinate system with its x-axis pointing east and its y-axis pointing north, the precession of the pendulum is described by the Coriolis force. Consider a planar pendulum with natural frequency ω in the small angle approximation. There are two forces acting on the pendulum bob: the restoring force provided by gravity and the wire, and the Coriolis force. The Coriolis force at latitude φ is horizontal in the small angle approximation and is given by

\begin{align} F_{c,x} &= 2 m \Omega \dfrac{dy}{dt} \sin(\varphi)\\ F_{c,y} &= - 2 m \Omega \dfrac{dx}{dt} \sin(\varphi) \end{align}

where Ω is the rotational frequency of Earth, Fc,x is the component of the Coriolis force in the x-direction and Fc,y is the component of the Coriolis force in the y-direction.

The restoring force, in the small angle approximation, is given by

\begin{align} F_{g,x} &= - m \omega^2 x \\ F_{g,y} &= - m \omega^2 y. \end{align}

Using Newton's laws of motion this leads to the system of equations

\begin{align} \dfrac{d^2x}{dt^2} &= -\omega^2 x + 2 \Omega \dfrac{dy}{dt} \sin(\varphi)\\ \dfrac{d^2y}{dt^2} &= -\omega^2 y - 2 \Omega \dfrac{dx}{dt} \sin(\varphi) \,. \end{align}

Switching to complex coordinates z = x + iy, the equations read

To first order in Ω/ω this equation has the solution

If we measure time in days, then Ω = 2π and we see that the pendulum rotates by an angle of −2π sin(φ) during one day.

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