Editing Geometric phase (section)

=== Foucault pendulum ===

One of the easiest examples is the [[Foucault pendulum]]. An easy explanation in terms of geometric phases is given by Wilczek and Shapere:<ref>{{cite book |editor1-last=Wilczek |editor1-first=F. |editor2-last=Shapere |editor2-first=A. |date=1989 |title=Geometric Phases in Physics |url=https://archive.org/details/geometricphasesp00shap |url-access=limited |location=Singapore |publisher=World Scientific |page=[https://archive.org/details/geometricphasesp00shap/page/n18 4] }}</ref>

{{blockquote|How does the pendulum precess when it is taken around a general path ''C''? For transport along the [[equator]], the pendulum will not precess. [...] Now if ''C'' is made up of [[Earth's geodesic|geodesic]] segments, the [[precession]] will all come from the angles where the segments of the geodesics meet; the total precession is equal to the net [[spherical excess|deficit angle]] which in turn equals the [[solid angle]] enclosed by ''C'' modulo 2''π''. Finally, we can approximate any loop by a sequence of geodesic segments, so the most general result (on or off the surface of the sphere) is that the net precession is equal to the enclosed solid angle.}}

To put it in different words, there are no inertial forces that could make the pendulum precess, so the precession (relative to the direction of motion of the path along which the pendulum is carried) is entirely due to the turning of this path. Thus the orientation of the pendulum undergoes [[parallel transport]]. For the original Foucault pendulum, the path is a circle of [[latitude]], and by the [[Gauss–Bonnet theorem]], the phase shift is given by the enclosed solid angle.<ref>{{cite journal |title=Foucault pendulum through basic geometry |author1=Jens von Bergmann |author2=HsingChi von Bergmann |journal=Am. J. Phys. |volume=75 |year=2007 |issue=10 |pages=888–892 |doi=10.1119/1.2757623 |bibcode=2007AmJPh..75..888V }}</ref>

==== Derivation ====
{{cleanup merge|Foucault pendulum|21=section|date=July 2023}}
[[File:Parallel Transport.svg|thumb|Parallel transport of a vector around a closed loop on the sphere: The angle by which it twists, {{mvar|α}}, is proportional to the area inside the loop.]]

In a near-inertial frame moving in tandem with the Earth, but not sharing the rotation of the Earth about its own axis, the suspension point of the pendulum traces out a circular path during one [[sidereal time|sidereal]] day.

At the latitude of Paris, 48 degrees 51 minutes north, a full precession cycle takes just under 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 by just over 270 degrees. If the plane of swing was north–south at the outset, it is east–west one sidereal day later.

This also 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 an exchange must have occurred.

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 can be demonstrated, by composing the infinitesimal rotations, that the precession rate is proportional to the [[Orthogonal projection|projection]] of the [[angular velocity]] of Earth onto the [[Normal (geometry)|normal]] direction to Earth, which implies that the trace of the plane of oscillation will undergo parallel transport. After 24 hours, the difference between initial and final orientations of the trace in the Earth frame is {{math|1=''α'' = −2''π'' sin ''φ''}}, which corresponds to the value given by the [[Gauss–Bonnet theorem]]. {{mvar|α}} is also called the [[holonomy]] or geometric phase of the pendulum. When analyzing earthbound motions, the Earth frame is not an [[inertial frame]], but rotates about the local vertical at an effective rate of {{nowrap|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.<ref>{{cite journal |bibcode=1972QJRAS..13...40S |title=The Description of Foucault's Pendulum |journal=Quarterly Journal of the Royal Astronomical Society |volume=13 |pages=40 |last1=Somerville |first1=W. B. |year=1972}}</ref><ref>{{cite journal |doi=10.1119/1.14972 |title=A simple geometric model for visualizing the motion of a Foucault pendulum |journal=American Journal of Physics |volume=55 |issue=1 |pages=67–70 |year=1987 |last1=Hart |first1=John B. |last2=Miller |first2=Raymond E. |last3=Mills |first3=Robert L. |bibcode=1987AmJPh..55...67H}}</ref>

From the perspective of an Earth-bound coordinate system (the measuring circle and spectator are Earth-bounded, also if terrain reaction to Coriolis force is not perceived by spectator when he moves), using a rectangular coordinate system with its {{mvar|x}} axis pointing east and its {{mvar|y}} axis pointing north, the precession of the pendulum is due to the [[Coriolis force]] (other [[fictitious forces]] as gravity and centrifugal force have not direct precession component, Euler's force is low because Earth's rotation speed is nearly constant). Consider a planar pendulum with constant natural frequency {{mvar|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 centrifugal force, opposed to the gravitational restoring force, can be neglected). The Coriolis force at latitude {{mvar|φ}} is horizontal in the small angle approximation and is given by
<math display="block">
\begin{align}
 F_{\text{c},x} &= 2m \Omega \dfrac{dy}{dt} \sin\varphi, \\
 F_{\text{c},y} &= -2m \Omega \dfrac{dx}{dt} \sin\varphi,
\end{align}
</math>
where {{math|Ω}} is the rotational frequency of Earth, {{math|''F''<sub>c,''x''</sub>}} is the component of the Coriolis force in the {{mvar|x}} direction, and {{math|''F''<sub>c,''y''</sub>}} is the component of the Coriolis force in the {{mvar|y}} direction.

The restoring force, in the [[small-angle approximation]] and neglecting centrifugal force, is given by
<math display="block">
\begin{align}
 F_{g,x} &= -m \omega^2 x, \\
 F_{g,y} &= -m \omega^2 y.
\end{align}
</math>

[[File:Foucault_pendulum_precession_vs_latitude.svg|thumb|upright=1.2|Graphs of precession period and precession per sidereal day vs latitude. The sign changes as a Foucault pendulum rotates anticlockwise in the Southern Hemisphere and clockwise in the Northern Hemisphere. The example shows that one in Paris precesses 271° each sidereal day, taking 31.8 hours per rotation.]]
Using [[Newton's laws of motion]], this leads to the system of equations
<math display="block">
\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}
</math>

Switching to complex coordinates {{math|1=''z'' = ''x'' + ''iy''}}, the equations read
<math display="block">
 \frac{d^2z}{dt^2} + 2i\Omega \frac{dz}{dt} \sin \varphi + \omega^2 z = 0.
</math>

To first order in {{math|{{sfrac|Ω|''ω''}}}}, this equation has the solution
<math display="block">
 z = e^{-i\Omega \sin \varphi t} \left(c_1 e^{i\omega t} + c_2 e^{-i\omega t}\right).
</math>

If time is measured in days, then {{math|1=Ω = 2''π''}} and the pendulum rotates by an angle of {{math|−2''π'' sin ''φ''}} during one day.
<!--Foucault's original paper (Comptes Rendus Vol. 32, 1851, p. 135) is wholly descriptive. He remarks in effect that the rate of precession of the plane of rotation can be obtained analytically or geometrically. From a geometric perspective much of the explanation given above comes down to the statement that the rate of change of azimuth of a star on the horizon depends only on the observer's latitude and is the same as the rate of precession of the pendulum (Journal of the Royal Astronomical Society of Canada, Vol. 63, 1970, pp. 227–228). Binet (Comptes Rendus Vol.32, 1851, p. 197) gives a simple analytical derivation of the rate. [[Arthur Cayley]] (Collected Works Vol. 4, 1891, pp. 534–537) says that[[Poisson]] examined the problem in 1838 and concluded that there would be no effect due to the rotation of the Earth. [[Kamerlingh Onnes]] based his doctoral dissertation on the accurate determination of the Earth's rotation by means of the pendulum (Schulz-DuBois, American Journal of Physics, Vo. 28, 1970, p. 173). -->