I read this page¹ for reference. Without it, I would never figure out how to approximate. My notes focus on the parts that I find difficult to grasp. What I find easy are omitted.

Let V be a 3 dimensional vector space over \mathbb{R} equipped with the standard inner product. Let \mathbf{O} denote the origin. Let \mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3} be the standard basis.

The symbol t will denote the time, which I have no idea how to define. We assume that the functions in t below are smooth.

Angular velocity

Consider rotation about the origin in the space V. The motion can be described by g(t) : V \rightarrow V for some g(t)\in\mathop{\mathrm{\mathrm{SO}}}(V) and a particle which is at the point \mathbf{P}(0)\in V at time 0 is at \mathbf{P}(t) := g(t)\mathbf{P}(0) at time t. The velocity of the particle is \mathbf{P}'(t) = g'(t)\mathbf{P}(0). The linear map g'(t)g(t)^{-1} : V\rightarrow V is skew-symmetric, i.e., g'g^{-1} + (g'g^{-1})^{*} = 0, which is easy to see from the Lie algebra point of view. Thus it has an eigenvalue 0. The other two eigenvalues are ib(t) and -ib(t) for some b(t)\in\mathbb{R}. The eigenspace for 0 is of dimension 1 or 3. When the dimension is 1, on the complement, each point has angular velocity perpendicular to the position vector with amplitude |b(t)|. In other words, the motion is rotation about the axis given by the eigenspace for 0 with angular speed |b(t)|. We choose a smooth section \mathbf{a}: \mathbb{R}\rightarrow V such that \mathbf{a}(t) is a unit vector in the 0-eigenspace of g'(t)g(t)^{-1}. Each \mathbf{a}(t) gives the axis of rotation and an orientation of the complementary plane and g'(t)g(t)^{-1} determines the signed amplitude b(t) where CCW is considered as the positive direction. More explicitly, choose a right-handed orthonormal basis \mathbf{v}_{1}(t),\mathbf{v}_{2}(t),\mathbf{a}(t) of V and if g'(t)g(t)^{-1} is represented by the matrix \begin{aligned} \begin{pmatrix} 0 & -b(t) & 0 \\ b(t) & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \end{aligned} then b(t) is the signed amplitude. Let \bm{\Omega}(t) = b(t) \mathbf{a}(t), which is the angular velocity of the particles at time t. We note that \bm{\Omega}(t) \times \mathbf{v}(t) = g'(t) g^{-1}(t) \mathbf{v} for all \mathbf{v}\in V, so we get the following.

Coordinates with respect to frames

Set \mathcal{I}= \{ \mathbf{e}_{1},\mathbf{e}_{2}, \mathbf{e}_{3} \}. Let \mathbf{r}(t)\in V. Write r_{\mathbf{e}}(t) for the \mathbf{e}-coordinate of \mathbf{r}(t), for \mathbf{e}\in \mathcal{I}. Then \mathbf{r}(t) = \sum_{\mathbf{e}\in \mathcal{I}} r_{\mathbf{e}}(t) \mathbf{e}.

Let \mathcal{R}= \{ \mathbf{f}_{1}, \mathbf{f}_{2},\mathbf{f}_{3} \} be a rotating frame in V based at \mathbf{P}, where, for \mathbf{f}\in\mathcal{R}, \mathbf{f}: \mathbb{R}\rightarrow V and \mathbf{P}: \mathbb{R}\rightarrow V are functions depending on the time. Write r_{\mathbf{f}}(t) for the \mathbf{f}(t)-coordinate of \mathbf{r}(t), for \mathbf{f}\in \mathcal{R}. Then \mathbf{r}(t) = \mathbf{P}(t) + \sum_{\mathbf{f}\in \mathcal{R}} r_{\mathbf{f}}(t) \mathbf{f}(t).

We have the equality \begin{aligned} \sum_{\mathbf{e}\in \mathcal{I}} r_{\mathbf{e}}(t) \mathbf{e}= \mathbf{P}(t) + \sum_{\mathbf{f}\in \mathcal{R}} r_{\mathbf{f}}(t) \mathbf{f}(t). \end{aligned} Taking derivatives up to second order, we get \begin{aligned} \sum_{\mathbf{e}\in \mathcal{I}} r_{\mathbf{e}}'(t) \mathbf{e}&= \mathbf{P}'(t) + \sum_{\mathbf{f}\in \mathcal{R}} (r_{\mathbf{f}}'(t) \mathbf{f}(t) + r_{\mathbf{f}}(t) \mathbf{f}'(t))\\ \sum_{\mathbf{e}\in \mathcal{I}} r_{\mathbf{e}}''(t) \mathbf{e}&= \mathbf{P}''(t) + \sum_{\mathbf{f}\in \mathcal{R}} (r_{\mathbf{f}}''(t) \mathbf{f}(t) + 2 r_{\mathbf{f}}'(t) \mathbf{f}'(t) + r_{\mathbf{f}}(t) \mathbf{f}''(t)). \end{aligned}

We consider the case where the rotation is given by g:\mathbb{R}\rightarrow \mathop{\mathrm{\mathrm{SO}}}(V) or equivalently by \bm{\Omega}: \mathbb{R}\rightarrow V. Thus we have \mathbf{P}'(t) = \bm{\Omega}(t) \times \mathbf{P}(t) and \mathbf{f}'(t) = \bm{\Omega}(t) \times \mathbf{f}(t). Substituting, we find \begin{aligned} \sum_{\mathbf{e}\in \mathcal{I}} r_{\mathbf{e}}'(t) \mathbf{e}&= \bm{\Omega}(t) \times \mathbf{P}(t) + \sum_{\mathbf{f}\in \mathcal{R}} (r_{\mathbf{f}}'(t) \mathbf{f}(t) + r_{\mathbf{f}}(t) \bm{\Omega}(t) \times \mathbf{f}(t))\\ \sum_{\mathbf{e}\in \mathcal{I}} r_{\mathbf{e}}''(t) \mathbf{e}&= \bm{\Omega}'(t) \times \mathbf{P}(t) + \bm{\Omega}(t) \times (\bm{\Omega}(t) \times \mathbf{P}(t))\\ & + \sum_{\mathbf{f}\in \mathcal{R}} (r_{\mathbf{f}}''(t) \mathbf{f}(t) + 2 r_{\mathbf{f}}'(t) \bm{\Omega}(t) \times \mathbf{f}(t) + r_{\mathbf{f}}(t) (\bm{\Omega}'(t) \times \mathbf{f}(t) + \bm{\Omega}(t) \times (\bm{\Omega}(t) \times \mathbf{f}(t)) )). \end{aligned}

Assume the motion of a particle is given by \mathbf{r}(t). According to Newton’s second law of motion, \begin{aligned} \sum_{\mathbf{e}\in \mathcal{I}} r_{\mathbf{e}}''(t) \mathbf{e}= \frac{1}{m(t)} \mathbf{F}(t), \end{aligned} where m(t) is the mass of the particle and \mathbf{F}(t) is the total force acting on the particle at time t. We get differential equations in r_{\mathbf{f}}’s.

Next we make some simplifying assumptions and approximations. Assume \bm{\Omega} is constant. Assume |\mathbf{P}(t)| \gg |r_{\mathbf{f}}(t)| for all \mathbf{f}\in\mathcal{R}. Assume m(t) = m is constant. We get an approximate differential equation for the motion in the rotating frame \begin{aligned} \frac{1}{m} \mathbf{F}(t) = \bm{\Omega}(t) \times (\bm{\Omega}(t) \times \mathbf{P}(t)) + \sum_{\mathbf{f}\in \mathcal{R}} (r_{\mathbf{f}}''(t) \mathbf{f}(t) + 2 r_{\mathbf{f}}'(t) \bm{\Omega}(t) \times \mathbf{f}(t) ). \end{aligned} Rearranging, we get \begin{aligned} \label{eq:motion}\tag{eq:motion} \sum_{\mathbf{f}\in \mathcal{R}} r_{\mathbf{f}}''(t) \mathbf{f}(t) = \frac{1}{m} \mathbf{F}(t) - \bm{\Omega}(t) \times (\bm{\Omega}(t) \times \mathbf{P}(t)) - \sum_{\mathbf{f}\in \mathcal{R}} 2 r_{\mathbf{f}}'(t) \bm{\Omega}(t) \times \mathbf{f}(t) , \end{aligned} where -m \bm{\Omega}(t) \times (\bm{\Omega}(t) \times \mathbf{P}(t)) is called the centrifugal force and -m \sum_{\mathbf{f}\in \mathcal{R}} 2 r_{\mathbf{f}}'(t) \bm{\Omega}(t) \times \mathbf{f}(t) is called the Coriolis force.

Determination of the vertical direction

We now turn to model the motion of a pendulum on the surface of the Earth. We assume that the rotation of the Earth is given by the constant angular velocity \bm{\Omega}. This defines the north pole and the south pole, by requiring that \bm{\Omega} points from the south pole to the north pole. We assume that the Earth is a sphere. Assume the hanging point of the pendulum is at \mathbf{P} (depending on the time t). Since the pendulum is small relative to the size of the Earth, we will ignore the variation of the gravitational field near \mathbf{P}(t). Let \mathbf{O} be the point on the axis of rotation of the Earth that is closest to \mathbf{P}. Let \mathbf{e}_{3} be the unit vector pointing in the direction of \bm{\Omega}. Choose unit vectors \mathbf{e}_{1},\mathbf{e}_{2} so that \mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3} form a righthanded system and \mathbf{e}_{1} has the same direction as \mathbf{P}(0). In reality, the vectors \mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3} also depend on the time, but we will ignore the variation. Set \Omega := ||\bm{\Omega}||. It is easy to see that \mathbf{P}(t) = \cos(\Omega t)\mathbf{e}_{1} + \sin(\Omega t)\mathbf{e}_{2}.

Let \mathbf{f}_{3}(t) be the unit upward vertical direction based at \mathbf{P}(t), that is, the opposite direction of the string that hangs a motionless (relative to \mathbf{P}) object. We will exclude the case that \mathbf{P} is at the poles of the Earth. Set \mathbf{f}_{2}(t) = \frac{\bm{\Omega}\times \mathbf{f}_{3}(t)}{||\bm{\Omega}\times \mathbf{f}_{3}(t)||} and \mathbf{f}_{1}(t) = \mathbf{f}_{2}(t) \times \mathbf{f}_{3}(t).

Now we determine \mathbf{f}_{3}. Assume the angle between \mathbf{P}(t) and \bm{\Omega} is \varphi. Apply [eq:motion] to the case where r_{\mathbf{f}}'=r_{\mathbf{f}}''=0. We get \begin{aligned} 0 = \mathbf{F}(t) - m \bm{\Omega}(t) \times (\bm{\Omega}(t) \times \mathbf{P}(t)) . \end{aligned} By our setup, \bm{\Omega}\perp \mathbf{P}, so that \bm{\Omega}\times (\bm{\Omega}\times \mathbf{P}(t)) = - \Omega^{2} \mathbf{P}(t). The hanging object has gravitational force \mathbf{G}(t) and tension from the string acting on it. The gravity points from \mathbf{P}(t) to the centre of the Earth, so it is in the direction of \begin{aligned} - ( \sin(\varphi)\cos(\Omega t) \mathbf{e}_{1} + \sin(\varphi)\sin(\Omega t) \mathbf{e}_{2} + \cos(\varphi) \mathbf{e}_{3}) \end{aligned} Write mk for the magnitude of the gravity for some constant k\in\mathbb{R}. (Here we use the approximation that locally the gravity is considered uniform.) The direction of \mathbf{f}_{3}(t) is in the direction of the tension, which is m times \begin{aligned} \label{eq:multiple-of-f3}\tag{eq:multiple-of-f3} (k\sin(\varphi) + \Omega^{2}) (\cos(\Omega t) \mathbf{e}_{1} + \sin(\Omega t) \mathbf{e}_{2}) + k\cos(\varphi) \mathbf{e}_{3}. \end{aligned}

Write g for the magnitude of [eq:multiple-of-f3], which is the constant \begin{aligned} ((k\sin(\varphi) + \Omega^{2})^{2} + (k\cos(\varphi))^{2})^{\frac{1}{2}} = (k^{2} + 2k\Omega^{2}\sin(\varphi) + \Omega^{4})^{\frac{1}{2}} . \end{aligned} Some explicit computation gives \begin{aligned} \mathrm{coord}_{_{\mathcal{I}}}( \mathbf{f}_3 ) &= g^{-1} ( (k\sin(\varphi) + \Omega^{2})\cos(\Omega t), (k\sin(\varphi) + \Omega^{2})\sin(\Omega t), k\cos(\varphi) )_{\mathcal{I}}\\ \mathrm{coord}_{_{\mathcal{I}}}( \mathbf{f}_2 ) &= (-\sin(\Omega t), \cos(\Omega t), 0)_{\mathcal{I}}\\ \mathrm{coord}_{_{\mathcal{I}}}( \mathbf{f}_1 ) &= g^{-1} ( k\cos(\varphi)\cos(\Omega t), k\cos(\varphi)\sin(\Omega t), - (k\sin(\varphi) + \Omega^{2}) )_{\mathcal{I}} \end{aligned} and \label{eq:Omega-cross-f}\tag{eq:Omega-cross-f} \begin{aligned} \Omega\times \mathbf{f}_{3} &= \frac{\Omega }{g} (k\sin(\varphi) + \Omega^{2}) \mathbf{f}_{2}\\ \Omega\times \mathbf{f}_{2} &= -\Omega (\cos(\Omega t) , \sin(\Omega t), 0)_{\mathcal{I}} = - \frac{\Omega }{g} [ k\cos(\varphi) \mathbf{f}_{1} + (k\sin(\varphi) + \Omega^{2}) \mathbf{f}_{3}]\\ \Omega\times \mathbf{f}_{1} &= \frac{\Omega }{g} k\cos(\varphi) \mathbf{f}_{2} \end{aligned}

Le pendule de Foucault

We set the pendulum in motion. Let \begin{aligned} \mathbf{G}^{\star}(t) = \mathbf{G}(t) - m \bm{\Omega}(t) \times (\bm{\Omega}(t) \times \mathbf{P}(t)) = - m g \mathbf{f}_{3}(t) . \end{aligned} Write \mathbf{T}(t) for the tension. We get \begin{aligned} \sum_{\mathbf{f}\in \mathcal{R}} r_{\mathbf{f}}''(t) \mathbf{f}(t) &= \frac{1}{m} \mathbf{T}(t) + \frac{1}{m} \mathbf{G}^{\star}(t) - \sum_{\mathbf{f}\in \mathcal{R}} 2 r_{\mathbf{f}}'(t) \bm{\Omega}(t) \times \mathbf{f}(t)\\ &= \frac{1}{m} \mathbf{T}(t) - g \mathbf{f}_{3}(t) - \sum_{\mathbf{f}\in \mathcal{R}} 2 r_{\mathbf{f}}'(t) \bm{\Omega}(t) \times \mathbf{f}(t). \end{aligned} Since \mathbf{T} is parallel to \sum_{\mathbf{f}\in \mathcal{R}} r_{\mathbf{f}}(t) \mathbf{f}(t), we find that \begin{aligned} \sum_{\mathbf{f}\in \mathcal{R}} r_{\mathbf{f}}''(t) \mathbf{f}(t) + g \mathbf{f}_{3}(t) + \sum_{\mathbf{f}\in \mathcal{R}} 2 r_{\mathbf{f}}'(t) \bm{\Omega}(t) \times \mathbf{f}(t) \end{aligned} is parallel to \sum_{\mathbf{f}\in \mathcal{R}} r_{\mathbf{f}}(t) \mathbf{f}(t). The length of \sum_{\mathbf{f}\in \mathcal{R}} r_{\mathbf{f}}(t) \mathbf{f}(t) is the length of the string L. These give rise to 3 differential equations in r_{\mathbf{f}_{1}}, r_{\mathbf{f}_{2}}, r_{\mathbf{f}_{3}}. The solutions describe the motion of the pendulum in the rotating frame. Using [eq:Omega-cross-f] and separating the coordinates, we get differential equations \begin{aligned} r_{\mathbf{f}_{1}}'' &= \frac{1}{m} T_{\mathbf{f}_{1}} + 2 \Omega g^{-1} r_{\mathbf{f}_{2}}' k \cos(\varphi)\\ r_{\mathbf{f}_{2}}'' &= \frac{1}{m} T_{\mathbf{f}_{2}} - 2 \Omega g^{-1} ( r_{\mathbf{f}_{1}}' k \cos(\varphi) + r_{\mathbf{f}_{3}}' (k\sin(\varphi) + \Omega^{2}))\\ r_{\mathbf{f}_{3}}'' &= \frac{1}{m} T_{\mathbf{f}_{3}} - g + 2 \Omega g^{-1} r_{\mathbf{f}_{2}}' (k\sin(\varphi) + \Omega^{2}). \end{aligned} We do not attempt to solve the differential equations.

Next we introduce some further approximations. We approximate \Omega^{2}, r_{\mathbf{f}_{3}}' and r_{\mathbf{f}_{3}}'' by 0. We approximate r_{\mathbf{f}_{3}} by L. Then g\approx k. Then we have approximate differential equations \begin{aligned} r_{\mathbf{f}_{1}}'' &= \frac{1}{m} T_{\mathbf{f}_{1}} + 2 \Omega r_{\mathbf{f}_{2}}' \cos(\varphi)\\ r_{\mathbf{f}_{2}}'' &= \frac{1}{m} T_{\mathbf{f}_{2}} - 2 \Omega r_{\mathbf{f}_{1}}' \cos(\varphi) \\ 0 &= \frac{1}{m} T_{\mathbf{f}_{3}} - g + 2 \Omega r_{\mathbf{f}_{2}}' \sin(\varphi). \end{aligned} We can find the components of the tension easily \begin{aligned} T_{\mathbf{f}_{3}} &= m (g - 2 \Omega r_{\mathbf{f}_{2}}' \sin(\varphi) ).\\ T_{\mathbf{f}_{2}} &= \frac{r_{\mathbf{f}_{2}}}{r_{\mathbf{f}_{3}}} T_{\mathbf{f}_{3}}\\ T_{\mathbf{f}_{1}} &= \frac{r_{\mathbf{f}_{1}}}{r_{\mathbf{f}_{3}}} T_{\mathbf{f}_{3}}. \end{aligned} The term 2 \Omega r_{\mathbf{f}_{2}}' \sin(\varphi) is negligible compared to g. After all the approximations, which I have no idea how well they approximate, we arrive at the nice differential equations \begin{aligned} r_{\mathbf{f}_{1}}'' &= \frac{g}{L} r_{\mathbf{f}_{1}} + 2 \Omega r_{\mathbf{f}_{2}}' \cos(\varphi)\\ r_{\mathbf{f}_{2}}'' &= \frac{g}{L} r_{\mathbf{f}_{2}}- 2 \Omega r_{\mathbf{f}_{1}}' \cos(\varphi) \end{aligned} where “nice” means that finally we know how to solve the differential equations analytically.

Set a=\frac{g}{L} and b = 2 \Omega \cos(\varphi). We focus on solving the differential equations \begin{aligned} x'' &= a x + b y'\\ y'' &= a y- b x'. \end{aligned} There is a standard way for solving this. One gets exponentials as solutions. When 4a > b^{2} (which is the case here), the periods of the solutions are easily seen to be \begin{aligned} \frac{2\pi}{\frac{|b|}{2}} = \frac{2\pi}{ \Omega |\cos(\varphi)|}. \end{aligned}