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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: $$\phi(t)=\omega t, \ \ \omega =\sqrt{\frac{g}{r}}$$


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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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.

Intrinsic coordinates

(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. enter image description here

(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.


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
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  • $\begingroup$ ...I am stuck on one final detail here... Initially $\theta$ was with the horizontal as your $\psi$. Later on OP changed it with the vertical...see his comment addressed to me under the question. $\endgroup$
    – Frobenius
    Mar 23 at 23:50
  • $\begingroup$ I understand how to relate the two angles. What is not clear to me is how to determine the constant $b$ without the prior knowledge that the curve is a cycloid. $\endgroup$ Mar 24 at 0:05
  • $\begingroup$ At time $\:t=0^+\:$ the particle starts to move downwards to the positive $\:y\:$ so $\:\psi=\dfrac{\pi}{2}^{-}\:$ $\endgroup$
    – Frobenius
    Mar 24 at 0:15
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    $\begingroup$ @Arthur Morris. Imagine you had the solution for minimum time with a slope at the origin O which we assume is not vertical and A is a point very near the origin on the curve. OA is approx a straight line, but wouldn't another cycloid with a vertical start provide a quicker time for the particle to move from O to A, finishing with horizontal motion, contradicting the assumption. The particle would be at the same speed at A not matter what path was taken. The only slope we could choose, where a quicker time for the small segment wasn't possible, would be vertical. $\endgroup$ Mar 24 at 8:37
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    $\begingroup$ The usual phenomenon : downvoters that don't explain why. We must respect the answers even if incorrect (it doesn't mean that this one is incorrect). $\endgroup$
    – Frobenius
    Mar 26 at 11:05
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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.

enter image description here

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

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}

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  • $\begingroup$ That's what Bernoulli did back then, I guess. It's just a matter How you recognize that the property derived hold for cycloid. $\endgroup$ Mar 26 at 19:46
  • $\begingroup$ I want you to help me with some other matter, I have a picture drawn by hand, not a picture graph sort of, I want it to be digital like the one you have drawn above, I tried using Geogebra, but I'm not familiar with it too much, Can you please help me making the plot? drive.google.com/file/d/1orgSiZDWn0SK3bF_n85pK1qNZj95MsiU/… $\endgroup$ Mar 26 at 19:51
  • $\begingroup$ i.stack.imgur.com/zuJH5.png $\endgroup$
    – Frobenius
    Mar 27 at 1:44
  • $\begingroup$ Great! I find your addendum pretty satisfactory. And Thanks to you for making the plot, it's perfect. I was using the Geogebra Calculator suite. Are you using the same? Is there any tutorial or something that I can follow to learn to make the plot like you? $\endgroup$ Mar 28 at 21:02
  • $\begingroup$ @Young Kindaichi : Sign in GeoGebra and read the manual or the on-line documentation and Help. Learning is not succeded in one stroke. Using it day by day you will reach a satisfactory level to understand by intuition how to use this amazing software for your needs. In GeoGebra I like very much the 3D drawing, LaTeX, animation, trace on and many many other tools. For your information I am not a user of yesterday. I use GeoGebra since 2016 mainly in my LaTeX notes and in PSE answers. You will be expert in a few months so you must be patient. $\endgroup$
    – Frobenius
    Mar 28 at 21:56

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