The Foucault pendulum - the physics (and maths) involved

It's easy enough to understand the motion of a pendulum suspended at one of the Earth's poles, because in this case the point of suspension is not accelerating. At any other latitude, the point of suspension accelerates and this introduces some more complicated time-dependent forces acting on the pendulum. At the poles, the (apparent) precession period is 23.9 hours, whereas elsewhere it is longer by a factor of the reciprocal sine of the latitude. But why? And is the precession uniform? On the way, we'll derive expressions for the fictitious forces mentioned in the introduction.

This page is a physical analysis of the motion. I have kept it as simple as I could (though see footnote), but it does use vector calculus – without this tool, the analysis would be very long and awkward. Before you start on the analysis below, see the Introduction to the Foucault pendulum.

We treat the earth as rotating about its axis with angular velocity Ω relative to an inertial frame. At a point O on the surface with latitude λ, we choose a Cartesian coordinate system with the x axis horizontal and South, the y axis horizontal and East and the z axis vertical and up. In the diagram, the i,k plane (the North-South, vertical plane at the position of our pendulum) is shaded. P is the point on the earth's axis closest to O and ro is the vector from P to O. If the position vector of a moving particle is r relative to O, then its position vector relative to P is rp = r + ro. Further let r be sufficiently small that the gravitational field does not vary significantly between O and r. Now P is in an inertial frame, so the equation of motion for the particle is:        mr"p = mG + F          (1) where m is the mass of the particle, G is the gravitational field (i.e. force per unit mass) at O, F is the sum of all forces other than gravity, and plain text dashes signify derivatives with respect to time.

(1) is an equation of motion in the inertial frame, and it is quite simple. r_p and r_o are vectors in an inertial frame. However, r is a vector in a non-inertial frame. We shall now write the equation of motion in terms of r, so it will refer to motion in the rotating frame. The extra terms that will appear in the right hand side when we make the substitution will be the fictitious forces mentioned in the introduction.

For the motion in the frame of the earth at O the acceleration is r" = r"_p − r"_o. Point O rotates with the angular velocity of the earth Ω, so

       r"_o = − Ω² r_o = - Ω²r_o(sin λ i − cos λ k)           (2)

so    mr" = mG + F + mΩ² r_o(sin λ i − cos λ k)           (3).

Consider now a pendulum consisting of a particle at the end of a string length a attached to a point with coordinates in the O frame (0,0,L). Suppose initially that the pendulum were stationary and vertical in that frame. The tension To is the only external force and:        To = mgk           (4) where g is the (local) apparent gravitational acceleration at O. (Note that g and G are neither parallel nor equal in magnitude because of the rotation of the earth.)

Since r' = r" = 0 here, (3) gives:

       G + Ω²r_o(sin λ i − cos λ k) = −gk           (6)

Now consider a moving pendulum with tension T, and substitute from (6) to remove G (whose direction and magnitude are difficult to measure directly).

       mr"= T − mgk           (7).

This equation can be resolved into the i,j,k coordinates considering that the O frame rotates with Ω = − Ω cos λ i − Ω sin λ k.

       r = r_xi + r_yj + r_zk

so    r' = (r_x)'i + (r_y)'j + (r_z)'k + r_x i' + r_y j'+ r_zk'

Now i' is the motion of a point at i relative to O due to the rotation Ω of the frame so i' = Ω X i and similarly j' = Ω X j and k' = Ω X k, so

r' = {(r_x)' i + (r_y)' j + (r_z)' k} + Ω X r

which, by definition,

       = r' + Ω X r

       and r" = r" + 2 Ω X r' + Ω X (Ω X r)           (8).

where bold dashes indicate partial differentials with respect to time.

For r_o, at point at rest with respect to the surface of the Earth, we have already included the effects of Ω in the local (apparent) acceleration (in g and k ). For a pendulum small compared with the earth, r is not far from r_o and Ω is small, so the last term in (8) disappears and (7) and (8) together the give the equation of motion of the pendulum in the frame of the earth's surface:

       m r" = T − mgk − 2m Ω X r'           (9).

Note the significance of the terms: the first is the string tension, the second is the apparent weight in the rotating frame and the third term, which depends on the velocity of the pendulum and on the earth's rotation, is the Coriolis "force".

Resolving in coordinates fixed with respect to the earth gives:

       mx" = T_x + 2mΩ sin λ y'                 (9.1),

       my" = T_y − 2mΩ (sin λ x' + cos λ z')       (9.2)

and    mz" = T_z − mg + 2mΩ cos λ y'           (9.3).

For z << a, motion in the z (vertical) direction is small and so z' and z" are approximately zero. Equation (9.3) therefore becomes:

       T_z = mg − 2mΩ cos λ y'           (9.3*),

whose last term is very small.

For small oscillations (x, y, z all << a), the components of the tension T in the string are

       T_x = − T x/a , . . . T_y = − T y/a, . . . T_z = T(L − z)/a           (10).

Let us define the constant ω² = g/a and substitute (10) and (9.3*) in (9.1) and (9.2), taking advantage of z' approx = 0 :

       x" − 2ω sin λ y' + ω²x = 0           (9.1*)

       y" + 2ω sin λ x' + ω²y = 0           (9.2*).

Now let us define the complex variable ζ = x + iy = |ζ|exp(i φ) (which represents the position of the pendulum in the horizontal plane). Multiplying (9.2*) by i and adding it to (9.1*) gives the differential equation for motion in the horizontal plane:

       ζ" + 2iω sin λ . ζ' + ω²ζ = 0           (11).

This is a version of the equation of the harmonic oscillator, but note that the coefficient of the the first derivative of ζ is purely imaginary and so it does not contribute damping.

It may be verified by substitution that the solution to (11) is:

       ζ = {Aexp(iωt) + Bexp(−iωt)} exp(−iωt sin λ)           (12)

where ω² has been neglected in comparison with Ω² and where the complex constants A and B depend on the initial conditions of position and velocity. (Two simple cases are A = +/-B, both real, corresponding to motion passing through the equilibrium point. Note, however, that if the pendulum is released from a point of maximum amplitude, it never passes through ζ = 0.)

Equation (12) is readily interpreted: if the earth were not rotating (Ω = 0), the solution is just that for a simple pendulum with period t = 2π/ω whose elliptical path is given by

ζ_s = {Aexp(iωt) + Bexp(−iωt)}.

Subject to the approximations already made, the axis of this ellipse does not vary with time. The same situation arises at the equator (where λ = 0). With Ω not equal to 0 and at latitude λ not equal to 0, the complex vector ζ_s rotates (in the frame of the earth) with angular frequency − Ω sin λ, i.e. with a period 2π/(Ω sin λ) = T_earth/(sin λ) where T_earth is the period of rotation of the earth*. Before we check the direction of precession, the author congratulates readers who have persevered and offers in felicitation this puzzle: why does the longest day of the year have neither the earliest sunrise nor the latest sunset? A clue is to be found early in the introductory page. Now back to the precession: Ω is positive (the earth rotates anti-clockwise as seen from the North Pole); λ is negative in the southern hemisphere, so − Ω sin λ is positive: the pendulum should precess anti-clockwise (from above) in the frame of the earth when viewed from the Southern hemisphere.

Complications

As usual, the real situation is more complicated: a finite pendulum which does not swing in a plane traces an ellipse whose axis precesses (in the direction of the orbit) even in a Newtonian frame. If the orbit of a pendulum passes through the origin, we see from symmetry that it should not precess in a Newtonian frame. To demonstrate the precession due to the Earth's rotation, the pendulum should move slowly over a small angle. Any asymmetries in the support, launching or drag of the pendulum which would give rise to elliptical orbits must be removed. This is one reason why Foucault pendula are usually long, and firmly and symmetrically mounted.

Finally, a complication that applies in principle to all pendulums, including those of clocks. Taking x and y components of equation (8) shows an asymmetry in the x and y directions:

       x" − 2Ω sin (λ) y'+((g/a)−Ω² sin (λ)²) x = 0, and

       y" + 2Ω sin (λ) x'+((g/a)−Ω²) y = 0

Thus, at positions other than the poles, there is a small difference in the pendulum frequencies for the NS and EW directions. Ω/2π is of course about once per day, and for Foucault pendula, (g/a)^1/2/2π is of order once every several seconds, so this variation is small. For a grandfather clock, (g/a)^1/2/2π is faster and so the variation with orientation is smaller still. It is occasionally said that one might detect the difference in the timekeeping of a good pendulum clock when it is rotated between a NS and an EW swing. However, other effects, such as the angle of the floor, may be larger.

* Special cases. The special cases when the pendulum swings NS or EW are of course simpler. If one assumes that the precession rate is constant (we have just seen that this is not quite true), one can estimate the precession rate by considering one of these special cases. This is done in Richard Feynman's Lectures on Physics (vol 1).

Consider a time at which the pendulum swings North-South. At the Northern end of the swing, the pendulum bob is further from the axis (Southern hemisphere) by a distance Z.sin λ, where Z is the amplitude of the swing, so the support point overtakes the pendulum bob with a relative horizontal speed ZΩ = 2πA.sin λ/T_earth. Divide the circumference 2πA of the envelope of the pendulum's path relative to the earth by this speed and one has an estimate of the period of precession T_earth/sin λ.

Thanks to Norman Phillips of Merrimack USA and Jacques Gilbert of Canada for suggesting improvements to an earlier version of this discussion.

Links to other sites with information on the Foucault pendulum:

• A Foucault pendulum at the South Pole
  
• Our site gives an introduction to the Foucault pendulum; inertial frames in history and cosmology.
• A page about Coriolis forces.
• For more on relative motion (both Galileo's and Einstein's versions, see Einstein Light.
• For more on basic mechanics, see Physclips.

A review of Umberto Eco's novel "Foucault's Pendulum". In my opinion, the central joke is indeed funny and had me laughing out loud several times towards the end. It is also filled with interesting scholarly references. However I am not convinced it was funny enough nor sufficiently well researched to justify the tedium regularly encountered on the way.

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© 1996. Modified 9/8/04 J.Wolfe@unsw.edu.au, phone 61- 2-9385 4954 (UT + 10, +11 Oct-Mar). School of Physics, University of New South Wales, Sydney, Australia. Joe's scientific home page Joe's educational pages Joe's music page