Another Solution To Brachistochrone Problem

Question

Recalling the statement of the problem :

Given two points A and B in a vertical plane, what is the curve traced out by a point acted on only by gravity, which starts at A and reaches B in the shortest time.

And as we can show with the method of variational calculus that the curve is turn out to be cycloid (Figure-A).

$$x=r(\phi-\sin\phi)$$ $$y=r(1-\cos\phi)$$

where $\phi$ is a real parameter, corresponding to the angle through which the rolling circle has rotated. For given $\phi$, the circle's center lies at $(x, y) = (r\phi, r)$

In the brachistochrone problem, the motion of the body is given by the time evolution of the parameter: $$\phi(t)=\omega t, \ \ \omega =\sqrt{\frac{g}{r}}$$

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Now, look at the angle, call $\theta$, that the velocity vector makes with the vertical (Figure-B).

$$\tan(\theta)=\frac{\dot{x}}{\dot{y}}=\tan\frac{\phi}{2}$$ or $$\theta \propto t$$

In other words, In the $\theta-t$ plane, the curve is just a straight line. So What's the intuitive reason that the curve which minimizes the time should be a straight line curve in the $\theta-t$ plane?

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Answer 1

A good question! This is best answered by making use of intrinsic coordinates. The reason for this is twofold: firstly, as you have very nearly found, the equation of a cycloid is particularly simple when written in intrinsic coordinates; and secondly, the variational equation for the curve also takes on a a particularly simple form in these coordinates. I will explain both of these points now.

Intrinsic coordinates

In any curvilinear coordinate system , a curve is described by specifying the values of the coordinates as functions of each other e.g. $y=f(x)$, or as a functions of some parameter, e.g. $y=y(t), x=x(t)$. Here, each coordinate is understood to be defined relative to some 'grid'. An alternative, and somewhat esoteric, way to describe a curve is to specify its arc length, $s$, and its angle of inclination (relative to some chosen direction in space), $\psi$. This does not require reference to any fixed grid of coordinates, and instead uses the 'intrinsic' properties of the curve itself.

(N.B. this image shows the y axis pointing up - we will take it to be pointing down instead.)

Cartesian coordinates and intrinsic coordinates are related by $$ \frac{\text{d}y}{\text{d}s} = \sin\psi, \qquad \frac{\text{d}x}{\text{d}s} = \cos\psi $$ These awkward equations are part of the reason that intrinsic coordinates are so rarely used.

(I haven't quite figured out Geogebra yet!)

As you have seen, for a cycloid described by parametric equations $$ x(\phi) = r(\phi-\sin\phi), \qquad y(\phi) = r(1-\cos\phi) $$ we get $$ x' = r(1-\cos\phi), \qquad y' = r\sin\phi $$ so that $$ \frac{\text{d}y}{\text{d}\phi} \frac{\text{d}\phi}{\text{d}x}=\frac{r\sin\phi}{r(1-\cos\phi)} = \cot\frac{\phi}{2}= \frac{\text{d}y}{\text{d}s}\frac{\text{d}s}{\text{d}x} = \tan\psi \implies \psi = \frac{\pi}{2}-\frac{\phi}{2}. $$ We also have $$ s(\phi) = \int_0^\phi \text{d}\phi'\sqrt{x'^2+y'^2} = r^2\int_0^\phi \text{d}\phi' \sqrt{2-2\cos\phi'}=4r-2r\sqrt{2}\sqrt{1-\cos\phi}\cot\frac{\phi}{2}. $$ so that, combining the two, we get the very simple $$ s(\psi) = 4r(1-\sin\psi). $$ This is the first part done - we have used intrinsic coordinates to show how such a simple expression arises. I will now demonstrate why it is so simple in these coordinates.

Equations of motion in intrinsic coordinates

We now consider the velocity of a particle with position $\mathbf{r}(t)$, and rewrite it in terms of the arc length: $$ \mathbf{v}(t) = \frac{\text{d}\mathbf{r}}{\text{d}t} = \frac{\text{d}\mathbf{r}}{\text{d}s}\frac{\text{d}s}{\text{d}t} = \hat{\boldsymbol{\tau}}\dot{s}, $$ where $\hat{\boldsymbol{\tau}} = \text{d}\mathbf{r}/\text{d}s$ is a unit vector tangent to the path of the particle, and $\text{d}s/\text{d}t=v$ is its speed. Now, using the Frenet-Serret formulas, the time derivative of $\hat{\boldsymbol{\tau}}$ is $$ \frac{\text{d}\hat{\boldsymbol{\tau}}}{\text{d}t} =\frac{\text{d}\hat{\boldsymbol{\tau}}}{\text{d}s}\frac{\text{d}s}{\text{d}t}= \frac{1}{\rho} \hat{\mathbf{n}} \dot{s} $$ where $\hat{\mathbf{n}}$ is a unit vector orthogonal to $\hat{\boldsymbol{\tau}}$. This means that the acceleration of the particle is $$ \mathbf{a} =\frac{\text{d}\mathbf{v}}{\text{d}t} = \frac{\dot{s}^2}{\rho}\hat{\mathbf{n}}+\ddot{s}\hat{\boldsymbol{\tau}}. $$ If the particle moves under gravity, then the equation of motion is therefore $$ m\frac{\dot{s}^2}{\rho}\hat{\mathbf{n}}+m\ddot{s}\hat{\boldsymbol{\tau}} = mg\hat{\mathbf{y}}, $$ where $\hat{\mathbf{y}}$ is a unit vector in the $y$ direction. Taking the dot product with $\hat{\boldsymbol{\tau}}$, and using $\hat{\boldsymbol{\tau}}\cdot\hat{\mathbf{y}} = \sin\phi$, we get the equation of motion for the speed of the particle: $$ \ddot{s} = g\sin\psi.\tag{2} $$ This equation is very important. Since for the Brachistochrone we have the initial conditions $s(0)=\dot{s}(0)=0$, the above equation means that the form of $s(t)$ is entirely determined by $\psi(t)$. In other words, the function $\psi(t)$ entirely determines the shape of the curve.

Functional minimisation

Finally, consider the functional to be minimised for the Brachistochrone problem: the total time $T$ to traverse a curve $C$ under the influence of gravity is $$ T = \int_C \text{d}t $$ As it stands, the integrand is not in a form that allows the functional to be varied: the path dependence of $T$ is not explicit. In order to apply the Euler-Lagrange equations, it is necessary to rewrite the integral to be over a fixed domain. Two common ways to do this are $$ T = \int_C \frac{\text{d}s}{\dot{s}} = \int \sqrt{\frac{1+y'(x)^2}{2gy(x)}}\text{d}x $$ and $$ T = \int \sqrt{\frac{\dot{x}(t)^2+\dot{y}(t)^2}{2gy(t)}}\text{d}t. $$ This works because $y(x)$, or $x(t)$ and $y(t)$ together are sufficient to specify the shape of the curve. This is true more generally: any form of the integrand (over a fixed domain) that will allow the Euler Lagrange equations to return a complete description of the curve is sufficient. But we have just found a single function that does just that, namely $\psi(t)$! Assuming $\psi(t)$ is invertible (which it is over the domain we are interested in) we are therefore allowed to rewrite the integrand as $$ T = \int_C \text{d}t = \int\frac{\text{d}t}{\text{d}\psi} \text{d}\psi. \tag{3} $$ Applying the E-L equations $$ \frac{\text{d}}{\text{d}\psi}\frac{\partial\mathcal{L}}{\partial t'} = \frac{\partial\mathcal{L}}{\partial t} $$ with $\mathcal{L} = \frac{\text{d}t}{\text{d}\psi} = t'$, we find $$ \frac{\text{d}}{\text{d}\psi} t' =0 \implies t' = \text{const} \implies t=A\psi+B \text{ i.e. } \psi(t) = at+b $$ This is exactly what you found before. As I will show below, the parameter $a$ is your $-\omega/2$. (I am stuck on one final detail here - clearly we need $b=\pi/2$, but I can't see a simple reason for this).

To reiterate, it is only possible to write equation (3) because $s(t)$ is entirely determined by $\psi(t)$ by equation (2). This is the reason we are not allowed to write e.g. $T= \int(\text{d}t/\text{d}x) \text{d}x$ - because the function $x(t)$ alone is not sufficient to describe the curve - $y(t)$ is also required.

Finally, setting $b=\pi/2$ and $a=-\omega/2$ integrating the equation of motion (2), we find $$ \frac{\text{d}^2 s}{\text{d}t^2} = g\sin\left(\frac{\pi}{2}-\frac{\omega}{2} t\right) \implies s(t) = -\frac{4g}{\omega^2}\sin\left(\frac{\pi}{2}-\frac{\omega}{2} t\right) +ct+d. $$ With $s(0) = \dot{s}(0) = 0$, we find $c=0, d = 4g/\omega^2$ so that $$ s(\psi) = \frac{4g}{\omega^2}(1- \sin\psi) $$ as before.

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References:

1. https://en.wikipedia.org/wiki/Curvilinear_coordinates
2. https://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas
3. https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec06.pdf

Answer 2

If by intuition we could first prove that \begin{equation} \dot{\!\!\theta}=\dfrac{\mathrm d \theta}{\mathrm d t}=b \texttt{(constant)} \tag{01}\label{01} \end{equation} then it could be proved easily that the curve is the brachistochrone cycloid. But I don't find a proof either using the Euler-Lagrange formalism or not (as OP wants).

If we want a proof that time minimisation leads to the cycloid curve without using the Euler-Lagrange formalism then we must make use of the elementary proof of Snell's Law due to Feynman and given in my answer here : Why one should follow Snell's law for shortest time?. The proof doesn't use even differential calculus !!!

First note that the Euler-Lagrange formalism gives the result \begin{equation} y\left[1+\left(\dfrac{\mathrm d y}{\mathrm d x}\right)^{2}\right]=D=\texttt{constant}\:, \quad D>0 \tag{02}\label{02} \end{equation} and this leads to the cycloid curve (with $r=D/2$).

Exactly this constant of the motion is derived from the analogy of the refraction of light in a medium of variable refractive index (in other terms by connection with Snell's Law of refraction).

In the Figure below we see the path of least time from point $\:\mathrm{A}_{0}\:$ to point $\:\mathrm{A}_{4}\:$ through 4 regions of variable speed, increasing towards positive $\:y$. This would be the light path with decreasing refraction index. Under the assumption of least time path $\:\mathrm{A}_{0}\mathrm{A}_{4}\:$ every intermediate path $\:\mathrm{A}_{j}\mathrm{A}_{j+2}\,(j=0,1,2)\:$ is a path of least time between points $\:\mathrm{A}_{j}\:$ and $\:\mathrm{A}_{j+2}\:$. So, \begin{equation} \dfrac{v_{1}}{\sin\theta_1}=\dfrac{v_{2}}{\sin\theta_2}=\dfrac{v_{3}}{\sin\theta_3}=\dfrac{v_{4}}{\sin\theta_4}=\textrm{constant} \tag{03}\label{03} \end{equation} Now, if instead of the discrete regions we have a continuum with speed $\:v(y)\:$ being a continuous smooth increasing function of $\:y\:$, then in place of the piece-wise rectilinear path we would have a continuous smooth curve and \begin{equation} \dfrac{v(y)}{\sin\theta}=v(y)\sqrt{1+\tan^{2}\theta}=v(y)\sqrt{1+\left(\dfrac{\mathrm{d} y}{\mathrm{d} x}\right)^{2}}=v(y)\sqrt{1+y'^{\,2}}=\textrm{constant} \tag{04}\label{04} \end{equation} In the case of brachistochrone $\:v(y)=\sqrt{2g\,y}\:$ so above equation yields \begin{equation} \sqrt{y\left(1+y'^{\,2}\right)}=\textrm{constant} \tag{05}\label{05} \end{equation} and squaring equation \eqref{02}.

It seems that equation \eqref{01} is a result and not a starting point.

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$

ADDENDUM

This is a response to the OP comment :

That's what Bernoulli did back then, I guess. It's just a matter How you recognize that the property derived hold for cycloid.

[OP means the property equation \eqref{02}].

Separating the variables $\:x,y\:$ in equation \eqref{02} we have
\begin{equation} y \biggl[1+\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} \right)^{2}\biggr]=D \quad \Longrightarrow \quad \mathrm{d}x=\sqrt{\dfrac{y}{D-y}}\;\mathrm{d}y \tag{A-01}\label{A-01} \end{equation}

Since the $\:y$-axis is vertical downwards with the motion starting at $\:y=0\:$, then on one hand $\:y\ge 0\:$ and on the other hand $\:y\le D\:$ because of this same equation \eqref{02}, we can set \begin{equation} y =D\,\sin^{2}\theta \tag{A-02}\label{A-02} \end{equation} so from \eqref{A-01} \begin{equation} \mathrm{d}x=2D\sin^{2}\theta\,\mathrm{d}\theta \tag{A-03}\label{A-03} \end{equation} or \begin{equation} x=D \int \limits_{0}^{\theta}\left( 1-\cos2\theta\right)\mathrm{d}\theta \tag{A-04}\label{A-04} \end{equation} that is \begin{equation} x=\dfrac{D}{2}\left( 2\theta-\sin 2\theta\right) \tag{A-05}\label{A-05} \end{equation} while equation \eqref{A-02} is written as follows \begin{equation} y=\dfrac{D}{2}\left( 1-\cos 2\theta\right) \tag{A-06}\label{A-06} \end{equation} Defining \begin{equation} \phi\equiv 2\theta\,, \quad D=2r \tag{A-07}\label{A-07} \end{equation} equations \eqref{A-05},\eqref{A-06} are expressed as \begin{align} x\left(\phi\right) & = r\left(\phi-\sin\phi \right) \tag{A-08a}\label{A-08a}\\ y\left(\phi\right) & = r\left( 1-\cos \phi\right) \tag{A-08b}\label{A-08b} \end{align} the parametric equation of a cycloid, say $\:C\:$, generated by a circle of radius $\:r\:$ rolling without slipping on the $\:x\:$axis, see Figure-A and Figure-B in the question.

The time for the particle to move on the cycloid from a point 1 to a point 2 is \begin{equation} \Delta t_{12}=t_{2}-t_{1}=\dfrac{1}{\sqrt{2g}}\int\limits_{1}^{2}\sqrt{\dfrac{1+y'^{\,2}}{y}}\,\mathrm{d}x \tag{A-09}\label{A-09} \end{equation} Taking advantage of the constant expression in equation \eqref{05}, the expression under the integral is \begin{equation} \sqrt{\dfrac{1+y'^{\,2}}{y}}\,\mathrm{d}x=\dfrac{\;1\;}{y}\sqrt{y\left(1+y'^{\,2}\right)}\,\mathrm{d}x=\dfrac{\sqrt{D}}{y}\,\mathrm{d}x=\dfrac{\sqrt{2r}}{r\left( 1-\cos \phi\right)}\,r\left( 1-\cos \phi\right)\,\mathrm{d}\phi=\sqrt{2r}\,\mathrm{d}\phi \nonumber \end{equation} so \begin{equation} t_{2}-t_{1}=\sqrt{\dfrac{\,r\,}{g}}\, \int\limits_{\!\!\phi_{1}}^{\:\:\:\phi_{2}}\mathrm{d}\phi=\sqrt{\dfrac{\,r\,}{g}}\, \left(\phi_{2}-\phi_{1}\right) \tag{A-10}\label{A-10} \end{equation}

If the particle starts moving from rest ($t_1=0,\phi_1=0$) then at any moment $t_2=t$ for the angle $\phi_2=\phi$ we have \begin{equation} \phi=\omega \,t \qquad \texttt{where} \quad \omega\stackrel{\texttt{def}}{=\!\!=}\sqrt{\dfrac{g}{r}} \tag{A-11}\label{A-11} \end{equation} From the $\phi-$parametric equations of the cycloid, \eqref{A-08a} and \eqref{A-08b}, we have the $t-$parametric equations of this curve \begin{align} x\left(t\right) & = r\left(\omega \,t-\sin\omega \,t \right) \tag{A-12a}\label{A-12a}\\ y\left(t\right) & = r\left(\:\:1\,-\cos\omega \,t\right) \tag{A-12b}\label{A-12b} \end{align}