Chapter 5 Applications of the Lagrangian formalism

5.1 The simple pendulum

Consider a point particle of mass \(m\) that is attached by means of a rigid bar of negligible mass to a fixed point \(O\). It is submitted to the gravitational force. The reaction force of the bar on the point particle is perpendicular to the motion and thus produces no work.

We assume that the motion takes place in a vertical plane. For instance, this plane is determined by two straight lines, the first from \(O\) to the point particle when the pendulum is at equilibrium, and the second from \(O\) to the point where the particle is released with zero velocity.

Let us choose coordinates on \(\mathbb R^3\) where this plane is the plane \((x,z)\) with \(O\) sitting at the origin, so that the motion is constrained to take place on the circle \(S\) of radius \(l\) centered at the origin.

The equation for the constraint is

\begin{equation} x^2+z^2=l^2\label {eq:190}. \end{equation}

Going to polar coordinates, with angle \(\theta \) chosen as the deviation from the negative \(z\) axis,

\begin{equation} \label {eq:191} x=r\sin \theta ,\quad z=-r\cos \theta , \end{equation}

the constraint becomes

\begin{equation} \label {eq:192} r-l=0. \end{equation}

The Lagrange coordinate in this case is \(\theta \), and velocities are given by

\begin{equation} \label {eq:195} \dot x=l\cos \theta \dot \theta ,\quad \dot z=l\sin \theta \dot \theta . \end{equation}

In terms of the Lagrange coordinate, the kinetic energy is

\begin{equation} \label {eq:194} T=\frac 12 m (\dot x^2+\dot z^2)=\frac 12 m (\dot x^2+\dot z^2)=\frac 12 m l^2 \dot \theta ^2 \end{equation}

while the potential energy is

\begin{equation} \label {eq:193} V=mgz=-mgl\cos \theta . \end{equation}

The Lagrangian is

\begin{equation} \label {eq:196} L=T-V=m(\frac 12 l^2 \dot \theta ^2+gl\cos \theta ). \end{equation}

The Euler-Lagrange derivative is

\begin{equation} \label {eq:197} \frac {\partial L}{\partial \theta }-\frac {d}{dt}\frac {\partial L}{\partial \dot \theta }=-mgl\sin \theta -m l^2\ddot \theta , \end{equation}

and the equation of motion is equivalent to

\begin{equation} \label {eq:198} \boxed {\ddot \theta +\omega ^2 \sin \theta =0,\quad \omega ^2=\frac {g}{l}}. \end{equation}

5.1.1 Small oscillations

For small oscillations, \(\theta \ll 1\), \(\sin \theta \sim \theta \), the equation describes “harmonic” oscillations, with general solution given by

\begin{equation} \label {eq:199} \theta (t)=A\sin (\omega t+\theta _0),\quad \omega =\sqrt {\frac {g}{l}}. \end{equation}

5.1.2 Conservation of energy

In the general case, like in many systems, the energy which is given by \(E=T+V\), or explicitly

\begin{equation} \label {eq:201} E=m(\frac 12 l^2 \dot \theta ^2-gl\cos \theta ), \end{equation}

in the current case, is a conserved quantity. By this one means that the energy is constant on all solutions to the equations of motion. Indeed,

\begin{equation} \label {eq:200} \frac {d}{dt}E=ml^2\dot \theta (\ddot \theta +\frac {g}{l}\sin \theta ) \end{equation}

vanishes for all solutions of (5.9).

Instead of solving the second order equation, whose general solution contains two integration constants, one may also solve the first order equation provided by (5.11), which is equivalent to

\begin{equation} \label {eq:202} \dot \theta ^2=2\omega ^2[\frac {E}{mgl}+\cos \theta ], \end{equation}

in terms of \(E\) and an additional integration constant. Since the left hand side of this equation is always positive, it follows in particular that \(E\) must be such that

\begin{equation} \label {eq:203} - 1\leq \frac {E}{mgl}. \end{equation}

5.1.3 Points of equilibrium

The relation (5.13) provides a relation between velocities and Lagrange coordinates as needed in the discussion of points of equilibrium. The velocity vanishes if

\begin{equation} \label {eq:204} \cos \theta =-\frac {E}{mgl} \end{equation}

which is possible only if

\begin{equation} \label {eq:205} -1\leq \frac {E}{mgl} \leq 1. \end{equation}

Setting

\begin{equation} \label {eq:206} \cos \alpha =-\frac {E}{mgl},\quad 0\leq \alpha \leq \pi , \end{equation}

we may write

\begin{equation} \label {eq:207} \dot \theta ^2=2\omega ^2(\cos \theta -\cos \alpha ). \end{equation}

The velocity vanishes if \(\theta =\pm \alpha \). The gradient of the potential

\begin{equation} \label {eq:208} \frac {\partial V}{\partial \theta }=mgl \sin \theta \end{equation}

vanishes as well if and only if

\begin{equation} \label {eq:209} \theta =\alpha =0\quad {\rm or}\quad \theta =\alpha = \pi . \end{equation}

These two possibilities define the points of equilibrium where the pendulum is at rest, \(\theta (t)=0\) or \(\theta (t)=\pi \). (One can show that the former is a stable point of equilibrium whereas the latter is unstable. )

5.1.4 Motion if \(\alpha =\pi \)

In the case \(\alpha =\pi \), the relation becomes

\begin{equation} \label {eq:210} \dot \theta ^2=\frac {2g}{l}(\cos \theta +1). \end{equation}

Since

\begin{equation} \label {eq:211} \cos ^2\frac {\theta }{2}=(\frac {e^{i\frac {\theta }{2}}+e^{-i\frac {\theta }{2}}}{2})^2=\frac {e^{i\theta }+e^{-i\theta }+2}{4}=\frac {1}{2}(1+\cos \theta ), \end{equation}

this becomes

\begin{equation} \label {eq:212} \dot \theta ^2=\frac {4g}{l}\cos ^2\frac {\theta }{2}, \end{equation}

so that

\begin{equation} \label {eq:213} \dot \theta =\pm 2\omega \cos \frac {\theta }{2}\iff \pm \omega dt=\frac {d\frac {\theta }{2}}{\cos \frac {\theta }{2}}. \end{equation}

By integration, it follows that

\begin{equation} \label {eq:214} \pm \omega (t-t_0)=\big [\ln \tan (\frac {x}{4}+\frac {\pi }{4})\big ]^\theta _{\theta _0}. \end{equation}

Indeed,

\begin{equation} \frac {d}{dx}\ln \tan (\frac {x}{4}+\frac {\pi }{4})= \frac {\cos (\frac {x}{4}+\frac {\pi }{4})}{4\sin (\frac {x}{4}+\frac {\pi }{4})\cos ^2(\frac {x}{4}+\frac {\pi }{4})} =\frac {1}{2\sin (\frac {x}{2}+\frac {\pi }{2})}=\frac {1}{2\cos \frac x2}. \end{equation}

Let us assume that \(t_0=0\), \(\theta _0=0\), and \(\dot \theta (t_0)>0\). This means that one has to choose the plus sign in equations (5.24) and (5.25) and that the pendulum starts at the stable point of equilibrium with an initial velocity such that one may reach the unstable point of equilibrium. Equation (5.25) simplifies to

\begin{equation} \label {eq:216} \omega t=\ln \tan (\frac {\theta }{4}+\frac {\pi }{4}), \end{equation}

and the time to reach \(\theta =\pi \) diverges,

\begin{equation} \label {eq:217} \lim _{\theta \to \pi }\ln \tan {(\frac {\theta }{4}+\frac {\pi }{4})}=\infty . \end{equation}

5.1.5 Oscillations

Consider now the case where \(0<\alpha <\pi \), or equivalently, on account of (5.17), the case where \(-1<\frac {E}{mgl}<1\).

From \(\ddot \theta =-\frac {g}{l}\sin \theta \) and the fact that \(\sin \theta >0\) for \(\theta =\alpha > 0\), it follows that the acceleration is negative while we have already seen that the velocity vanishes. Hence, if \(\theta \) was increasing towards \(\alpha \), it will start decreasing after reaching \(\alpha \). For \(\theta =-\alpha <0\), \(\sin \theta <0\), the acceleration is positive and if \(\theta \) was decreasing towards \(-\alpha \), it will start increasing after having reaching \(-\alpha \). We thus have an oscillation between \(-\alpha \) and \(\alpha \).

Let us define

\begin{equation} \label {eq:218} k^2=\frac 12 (1+\frac {E}{mgl})\iff k^2=\frac 12 (1-\cos \alpha )\iff k=\sin \frac {\alpha }{2},\quad 0<k<1. \end{equation}

The equation of motion (5.13) becomes

\begin{equation} \label {eq:219} \dot \theta ^2=2\omega ^2[2k^2+\cos \theta -1]=4\omega ^2[k^2-\sin ^2\frac {\theta }{2}]. \end{equation}

The integral to be computed is

\begin{equation} \label {eq:220} \pm \omega (t-t_0)=\frac {1}{2}\int ^\theta _{\theta _0}\frac {d\xi }{\sqrt {k^2-\sin ^2\frac {\xi }{2}}}. \end{equation}

The change of variables \(\sin \frac {\theta }{2}=k\sin \varphi \), \(\sin \frac {\xi }{2}=k\sin \eta \), which implies that \(\frac 12 \cos \frac {\xi }{2}d\xi =k\cos \eta d\eta \), brings this integral to the form

\begin{equation} \label {eq:221} \pm \omega (t-t_0)=\frac 12 \int ^\varphi _{\varphi _0} \frac {2k\cos \eta d\eta }{\cos {\frac {\xi }{2}} \sqrt {k^2-k^2\sin ^2\eta }}=\int ^\varphi _{\varphi _0}\frac {d\eta }{\cos \frac {\xi }{2}} =\int ^\varphi _{\varphi _0}\frac {d\eta }{\sqrt {1-k^2\sin ^2\eta }}. \end{equation}

In terms of the elliptic integral of first kind defined by

\begin{equation} \label {eq:222} F(\varphi ,k)=\int ^\varphi _0\frac {d\eta }{\sqrt {1-k^2\sin ^2\eta }}=-F(-\varphi ,k), \end{equation}

we have

\begin{equation} \label {eq:223} \pm \omega (t-t_0)=F(\varphi ,k)-F(\varphi _0,k). \end{equation}

The period \(T\) of the motion is determined as follows. Because \(k=\sin \frac {\alpha }{2}\) and \(\sin \frac {\theta }{2}=k\sin \varphi \), if \(\theta =\pm \alpha \), \(\varphi =\pm \frac {\pi }{2}\), while \(\theta =0\) corresponds to \(\varphi =0\). \(T\) is twice the time needed to go from \(-\alpha \) to \(\alpha \). Indeed, for \(\theta \) going from \(-\alpha \) to \(+\alpha \), \(\dot \theta \geq 0\), and one uses the \(+\)sign in (5.34), so that the time needed is \(F(\frac {\pi }{2})-F(-\frac {\pi }{2})\). For \(\theta \) going from \(\alpha \) to \(-\alpha \), \(\dot \theta \leq 0\), and one uses the minus sign in (5.34), so that the time needed is \(-[F(-\frac {\pi }{2})-F(\frac {\pi }{2})]\).

\begin{equation} \label {eq:224} \omega T=2[ F(\frac {\pi }{2},k)-F((-\frac {\pi }{2},k)]=4F(\frac {\pi }{2},k)=4K(k), \end{equation}

where the complete elliptic integral is defined by

\begin{equation} \label {eq:225} K(k)=F(\frac {\pi }{2},k). \end{equation}

Suppose now that \(k^2=\sin ^2\frac {\alpha }{2}\ll 1\), which means that oscillations are small, and let us develop in \(\sin ^2\frac {\alpha }{2}\). To first order, the period becomes

\begin{multline} \label {eq:1b} T=\frac {4}{\omega }\int ^{\frac {\pi }{2}}_0\frac {d\eta }{\sqrt {1 -\sin ^2\frac {\alpha }{2}\sin ^2\eta }} =\frac {4}{\omega }\int ^{\frac {\pi }{2}}_0 d\eta (1+\frac 12\sin ^2\alpha \sin ^2\eta )+O(\sin ^4\frac {\alpha }{2})\\= \frac {4}{\omega }(\frac {\pi }{2}+\frac 14\sin ^2\frac {\alpha }{2} \int ^{\frac {\pi }{2}}_0d\eta (1-\cos 2\eta ))+O(\sin ^4\frac {\alpha }{2})= \frac {2\pi }{\omega }(1+\frac 14 \sin ^2\frac {\alpha }{2}) +O(\sin ^4\frac {\alpha }{2}). \end{multline} The leading term is \(\frac {2\pi }{\omega }\), the period of the harmonic oscillator.

5.1.6 Revolutions

The second generic case is when \(\frac {E}{mgl}>1\), or equivalently \(k>1\). The motion becomes a revolution since \(\dot \theta \) never vanishes, so it is either always strictly positive or strictly negative. \(\theta \) will thus become bigger than \(\pi \) or smaller than \(-\pi \). When setting \(k_1=\frac 1k\), \(k_1<1\) and equation (5.30) becomes

\begin{equation} \label {eq:r1} \dot \theta ^2=\frac {4\omega ^2}{k_1^2}(1-k_1^2\sin ^2\frac {\theta }{2}). \end{equation}

Integrating, it now follows directly from the definition (5.33) that

\begin{equation} \label {eq:r2} \pm \frac {\omega }{k_1}(t-t_0)=\frac 12 \int ^{\theta }_{\theta _0} \frac {d\eta }{\sqrt {1-k_1\sin ^2\frac {\eta }{2}}}=F(\frac {\theta }{2},k_1) -F(\frac {\theta _0}{2},k_1). \end{equation}

In this case, a period of one revolution corresponds to the time neded to go from \(\theta =-\pi \) to \(\theta =\pi \),

\begin{equation} \label {eq:pr} T=2\frac {k_1}{\omega }F(\frac {\pi }{2})=2\frac {k_1}{\omega }K(k_1). \end{equation}

5.2 The brachystochrone

The word means either “shortest time” or “the curve of quickest descent” and concerns a famous prize problem formulated by Bernoulli in the 17th century, and solved by himself, Newton and Leibniz. In the way it was solved by Bernoulli, it is famous in the history of mathematics because it has lead to the calculus of variations, which is at heart of the Lagrangian method.

5.2.1 Statement of the problem

Consider \(2\) points \(P\) and \(Q\) in a vertical plane, find the ideal constraint relating these two points that minimizes the time needed for a massive point particle sliding with no friction under the influence of gravity to reach \(Q\) when being released with zero velocity at \(P\).

5.2.2 Main equations

Without loss of generality, we may choose a coordinate system such that the vertical plane is \((x,z)\) with the arrival point \(Q\) at the origin, \(Q:(0,0)\), and the initial point somewhere above, to the left, \(P:(x^*,z^*)\) with \(x^*<0\) and \(z^*>0\).

The time needed to fall by a distance

\begin{equation} \label {eq:215} ds=\sqrt {dx^2+dz^2} \end{equation}

is

\begin{equation} \label {eq:226} |dt|=\frac {|ds|}{v}, \end{equation}

where \(v\) is the magnitude of the velocity along the curve.

One may describe the curve that one is looking for by the graph of the function \(x(z)\), with \(x(0)=0\) and \(x(z^*)=x^*\), so that

\begin{equation} \label {eq:227} dt=-\frac {\sqrt {1+(\frac {dx}{dz})^2}}{v}dz, \end{equation}

where the minus sign is due to the fact that if \(t\) increases, \(z\) decreases.

Conservation of energy gives

\begin{equation} \label {eq:228} \frac 12 m v_1^2+mgz_1=\frac 12 mv_2^2+mgz_2. \end{equation}

When applied to the current case with \(P\): \((x_2,z_2)=(x^*,z^*)\) and \(v_2=0\), while \((x_1,z_1)\) is a generic point \((x(z),z)\) along the curve where the velocity is \(v\), it follows that

\begin{equation} \label {eq:229} \frac 12 m v^2=mg(z^*-z)\Longrightarrow v=\sqrt {2g(z^*-z)}, \end{equation}

so that (5.43) becomes

\begin{equation} \label {eq:230} dt=-\sqrt {\frac {1+(\frac {dx}{dz})^2}{2g(z^*-z)}}dz. \end{equation}

Integrating gives the time needed for the fall as

\begin{equation} \label {eq:231} T=t_0-t_{z^*}=-\int _{z^*}^0\sqrt {\frac {1+(\frac {dx}{dz})^2}{2g(z^*-z)}}dz. \end{equation}

This time of fall is a functional of the function \(x(z)\), \(T=T[x(z)]\). But we know how to find extremals of such functionals. By the analogy \(T\leftrightarrow S\), \(x(z)\leftrightarrow q(\tau )\), the extremal with fixed end points is given functions for which the Euler-Lagrange derivative

\begin{equation} \label {eq:232} \frac {\delta L}{\delta z}=\frac {\partial L}{\partial x} -\frac {d}{dz}\frac {\partial L}{\partial x'},\quad x'=\frac {dx}{dz}, \end{equation}

vanishes with

\begin{equation} \label {eq:233} L(x,x',z)=\sqrt {\frac {1+(\frac {dx}{dz})^2}{2g(z^*-z)}}. \end{equation}

Since there is no explicit dependence on \(x\), the equation takes the form

\begin{equation} \label {eq:234} \frac {d}{dz}\frac {\partial L}{\partial x'}=0. \end{equation}

5.2.3 Finding the curve

This equation may be integrated once to get

\begin{equation} \label {eq:235} \frac {\partial L}{\partial x'}=c, \end{equation}

for some integration constant \(c\), or more explicitly,

\begin{equation} \label {eq:236} \frac 12 \sqrt {\frac {2g(z^*-z)}{1+(x')^2}}\frac {2x'}{2g(z^*-z)}=c\iff \frac {x'}{\sqrt {(1+(x')^2)2g(z^*-z)}}=c. \end{equation}

Taking the square in order to solve for \(x'\) gives

\begin{equation} \label {eq:237} (x')^2=c^2(1+(x')^2)(2g(z^*-z)) \iff (x')^2(1-2gc^2(z^*-z))=2gc^2(z^*-z), \end{equation}

and then

\begin{equation} \label {eq:238} x'=\pm \frac {c\sqrt {2g(z^*-z)}}{\sqrt {(1-2gc^2(z^*-z))}}=\pm \frac {\sqrt {z^*-z}}{\sqrt {2R-(z^*-z)}},\quad 2R=\frac {1}{2g c^2}. \end{equation}

By integration, one gets

\begin{equation} \label {eq:241} x-x^*=\pm \int _{z^*}^z \frac {\sqrt {z^*-z'}}{\sqrt {2R-(z^*-z')}} dz'. \end{equation}

Setting

\begin{equation} \label {eq:239} Z=z^*-z=2R\cos ^2\frac {\theta }{2}=R(1+\cos \theta ), \end{equation}

we have that \(z\) decreases from \(z^*\) to \(0\), \(z^*\geq z\geq 0\), while \(Z\) increases from \(0\) to \(z^*\), \(0\leq Z\leq z^*(\leq 2R)\), and then \(\theta \) decreases from \(\pi \) to some \(\theta _0\), \(\pi \geq \theta \geq \theta _0(\geq 0)\). The integral becomes

\begin{multline} \label {eq:240} x-x^*=\mp \int _0^{z^*-z} dZ\frac {\sqrt Z}{\sqrt {2R-Z}}=\pm \int ^\theta _\pi d\xi 2R 2 \cos \frac {\xi }{2}\sin \frac {\xi }{2}\frac 12 \frac {\sqrt {2R}\cos \frac {\xi }{2}}{\sqrt {2R}\sin \frac {\xi }{2}}=\pm 2R\int ^\theta _\pi d\xi \cos ^2\frac {\xi }{2}\\ =\pm R\int ^\theta _\pi d\xi (1+\cos \xi )=\pm R(\theta -\pi +\sin \theta ), \end{multline} In order to fix the sign, starting from \(\theta =\pi \), \(z=z^*\), \(Z=0\), \(x=x^*\), let us now take \(x=x^*<0\) and \(\theta =\pi -\alpha \) for small positive \(\alpha \). We then have

\begin{equation} \label {eq:242} z-z^*=-R(1+\cos {(\pi -\alpha )})=-R(1-\cos \alpha )=-R(1-(1-\frac {\alpha ^2}{2} +O(\alpha ^4)))\sim -R\frac {\alpha ^2}{2}<0, \end{equation}

so that \(z\) indeed decreases when \(\alpha \) increases. Furthermore,

\begin{equation} \label {eq:243} x-x^*=\pm R(-\alpha +\sin {(\pi -\alpha )})=\pm R(-\alpha +\sin \alpha )= \pm R(-\alpha +\alpha -\frac {\alpha ^3}{3!}+O(\alpha ^5))\sim \mp R\frac {\alpha ^3}{3!}, \end{equation}

Since we want \(x\) to increase when \(\alpha \) increases, we need the lower (+) sign in the final expression, and thus the lower (-) sign in the final expression for \(x-x^*\) in (5.57).

The parametric equations for the curve we are looking for are thus

\begin{equation} \label {eq:244} \boxed {z-z^*=-R(1+\cos \theta ),\quad x-x^*=R(\pi -\theta -\sin \theta )}. \end{equation}

These are the equations for a cycloid, which is the curve followed by a point on the circumference of a disc that rolls without sliding on a line. More precisely, in the current case they describe an inverted cycloid.

Indeed, the equations for a standard cycloid are

\begin{equation} \label {eq:245} x=a(t-\sin t),\quad y=a(1-\cos t), \end{equation}

The identification in our case relies on \(t\to \pi -\theta \), \(a\to R\), \(x\to x-x^*\), \(y\to -(z-z^*)\).

For a disc that rolls without sliding, one should have that the velocity of the center of the disc should be \(v=R\dot \theta =\sqrt {(\frac {dx_C}{dt})^2+(\frac {dz_C}{dt})^2}\), and thus, by multipliying by \(\frac {dt}{d\theta }\), \(R=\sqrt {(\frac {dx_C}{d\theta })^2+(\frac {dz_C}{d\theta })^2}\). Here we have that the motion of the center of the disc is

\begin{equation} \label {eq:246} x_C=R(\pi -\theta ),\quad z_C=-R, \end{equation}

so that \(\frac {dx_c}{d\theta }=-R\), \(\frac {dz_C}{d\theta }=0\), which indeed satisfies the required relation.

Each point of the circumference of the disc is animated by the compostion of this translation with a rotation around the center of the disc at \((-R,0)\),

\begin{equation} \label {eq:247} x_{\rm circ}=-R\sin \theta ,\quad z_{\rm circ}=-R\cos \theta . \end{equation}

5.2.4 Properties of the motion

We start at \(\theta =\pi \), which corresponds to \((x^*,z^*)\), then \(\theta \) decreases to \(\theta _0\) which corresponds to \((0,0)\). This angle is determined by

\begin{equation} \label {eq:248} z^*=R(1+\cos \theta _0)\iff \frac {z^*-R}{R}=\cos \theta _0,\quad x^*=R(\pi -\theta _0+\sin \theta _0). \end{equation}

It follows from (5.55) that \(z^*-z\geq 0\) and \(z^*-z\leq 2R\), so that, if \(z=0\), \(0\leq z^*\leq 2R\). From the first equation, it then follows that there exists \(\pi \geq \theta _0\geq 0\).

The extremum for \(z\) is reached when \(\frac {dz}{d\theta }=R\sin \theta =0\), which implies that \(\theta =0\). It is a minimum since \(\frac {d^2z}{(d\theta )^2}=R\cos \theta \), which is positive at \(\theta =0\).

Since in addition \(\frac {dx}{d\theta }=-R(1+\cos \theta )\), the velocity vanishes at \(\theta =\pi \) and \(\theta =-\pi \), and there is thus an oscillation between these two extreme values.

If \(z^*=2R\), \(z-2R=-R(1+\cos \theta )\), the extremum is at \((0,0)\) and the motion is along the whole branch of the cycloid.

Finally, let us study the temporal dependence starting from (5.46),

\begin{equation} \label {eq:249} dt=-\sqrt {\frac {1+(\frac {dx}{dz})^2}{2g(z^*-z)}}dz\Longrightarrow \frac {dt}{d\theta }=-\sqrt {\frac {(\frac {dz}{d\theta })^2 +(\frac {dx}{d\theta })^2}{2g(z^*-z)}}. \end{equation}

Injecting the values for the cycloid,

\begin{equation} \label {eq:250} \frac {dt}{d\theta }=-\sqrt {\frac {(R\sin \theta )^2 +(R(1+\cos \theta ))^2}{2gR(1+\cos \theta )}}=-\sqrt {\frac {R}{g}}. \end{equation}

The duration of the motion for a generic point on the cycloid is

\begin{equation} \label {eq:251} t-t^*=-\sqrt {\frac {R}{g}}(\theta -\pi )=\sqrt {\frac {R}{g}}(\pi -\theta ). \end{equation}

It follows that the duration of the fall from \((x^*,z^*)\) to \((0,0)\) is

\begin{equation} \label {eq:252} t(\theta _0)-t^*=\sqrt {\frac {R}{g}}(\pi -\theta _0). \end{equation}

More importantly, a half period corresponds to the motion from \(\pi \) to \(-\pi \),

\begin{equation} \label {eq:253} t(-\pi )-t^*=\sqrt {\frac {R}{g}}2\pi . \end{equation}

Since this is true for all \(z^*\), the period is independent of the amplitude, that is to say of the initial heigth, and the oscillation is isochronous.

A video of this state of affairs can be found here,
https://www.facebook.com/TrustMyScience/videos/982002581956755/ .