Another Solution To Brachistochrone Problem

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?

• Mar 20 at 22:28
• Isn't theta better defined as the angle the velocity vector makes with the vertical? Mar 22 at 18:47
• @Frobenius Go ahead! But Do not change the soul question. Please Mar 22 at 21:11
• Take $\theta$ from the vertical! Mar 22 at 21:13
• @Frobenius What have you used to make the figure? Just curious to know, as it's quite a good job. Mar 23 at 6:13

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.

References:

• ...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. Mar 23 at 23:50
• 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. Mar 24 at 0:05
• At time $\:t=0^+\:$ the particle starts to move downwards to the positive $\:y\:$ so $\:\psi=\dfrac{\pi}{2}^{-}\:$ Mar 24 at 0:15
• @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. Mar 24 at 8:37
• 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). Mar 26 at 11:05

If by intuition we could first prove that $$$$\dot{\!\!\theta}=\dfrac{\mathrm d \theta}{\mathrm d t}=b \texttt{(constant)} \tag{01}\label{01}$$$$ 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 $$$$y\left[1+\left(\dfrac{\mathrm d y}{\mathrm d x}\right)^{2}\right]=D=\texttt{constant}\:, \quad D>0 \tag{02}\label{02}$$$$ and this leads to the cycloid curve (with $$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, $$$$\dfrac{v_{1}}{\sin\theta_1}=\dfrac{v_{2}}{\sin\theta_2}=\dfrac{v_{3}}{\sin\theta_3}=\dfrac{v_{4}}{\sin\theta_4}=\textrm{constant} \tag{03}\label{03}$$$$ Now, if instead of the discrete regions we have a continuum with speed $$\:v(y)\:$$ being a continuous smooth increasing function of $$\:y\:$$, then in place of the piece-wise rectilinear path we would have a continuous smooth curve and $$$$\dfrac{v(y)}{\sin\theta}=v(y)\sqrt{1+\tan^{2}\theta}=v(y)\sqrt{1+\left(\dfrac{\mathrm{d} y}{\mathrm{d} x}\right)^{2}}=v(y)\sqrt{1+y'^{\,2}}=\textrm{constant} \tag{04}\label{04}$$$$ In the case of brachistochrone $$\:v(y)=\sqrt{2g\,y}\:$$ so above equation yields $$$$\sqrt{y\left(1+y'^{\,2}\right)}=\textrm{constant} \tag{05}\label{05}$$$$ and squaring equation \eqref{02}.

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

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

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
$$$$y \biggl[1+\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} \right)^{2}\biggr]=D \quad \Longrightarrow \quad \mathrm{d}x=\sqrt{\dfrac{y}{D-y}}\;\mathrm{d}y \tag{A-01}\label{A-01}$$$$

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 $$$$y =D\,\sin^{2}\theta \tag{A-02}\label{A-02}$$$$ so from \eqref{A-01} $$$$\mathrm{d}x=2D\sin^{2}\theta\,\mathrm{d}\theta \tag{A-03}\label{A-03}$$$$ or $$$$x=D \int \limits_{0}^{\theta}\left( 1-\cos2\theta\right)\mathrm{d}\theta \tag{A-04}\label{A-04}$$$$ that is $$$$x=\dfrac{D}{2}\left( 2\theta-\sin 2\theta\right) \tag{A-05}\label{A-05}$$$$ while equation \eqref{A-02} is written as follows $$$$y=\dfrac{D}{2}\left( 1-\cos 2\theta\right) \tag{A-06}\label{A-06}$$$$ Defining $$$$\phi\equiv 2\theta\,, \quad D=2r \tag{A-07}\label{A-07}$$$$ equations \eqref{A-05},\eqref{A-06} are expressed as \begin{align} x\left(\phi\right) & = r\left(\phi-\sin\phi \right) \tag{A-08a}\label{A-08a}\\ y\left(\phi\right) & = r\left( 1-\cos \phi\right) \tag{A-08b}\label{A-08b} \end{align} the parametric equation of a cycloid, say $$\: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 $$$$\Delta t_{12}=t_{2}-t_{1}=\dfrac{1}{\sqrt{2g}}\int\limits_{1}^{2}\sqrt{\dfrac{1+y'^{\,2}}{y}}\,\mathrm{d}x \tag{A-09}\label{A-09}$$$$ Taking advantage of the constant expression in equation \eqref{05}, the expression under the integral is $$$$\sqrt{\dfrac{1+y'^{\,2}}{y}}\,\mathrm{d}x=\dfrac{\;1\;}{y}\sqrt{y\left(1+y'^{\,2}\right)}\,\mathrm{d}x=\dfrac{\sqrt{D}}{y}\,\mathrm{d}x=\dfrac{\sqrt{2r}}{r\left( 1-\cos \phi\right)}\,r\left( 1-\cos \phi\right)\,\mathrm{d}\phi=\sqrt{2r}\,\mathrm{d}\phi \nonumber$$$$ so $$$$t_{2}-t_{1}=\sqrt{\dfrac{\,r\,}{g}}\, \int\limits_{\!\!\phi_{1}}^{\:\:\:\phi_{2}}\mathrm{d}\phi=\sqrt{\dfrac{\,r\,}{g}}\, \left(\phi_{2}-\phi_{1}\right) \tag{A-10}\label{A-10}$$$$

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 $$$$\phi=\omega \,t \qquad \texttt{where} \quad \omega\stackrel{\texttt{def}}{=\!\!=}\sqrt{\dfrac{g}{r}} \tag{A-11}\label{A-11}$$$$ From the $$\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}

• That's what Bernoulli did back then, I guess. It's just a matter How you recognize that the property derived hold for cycloid. Mar 26 at 19:46
• 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/… Mar 26 at 19:51
• i.stack.imgur.com/zuJH5.png Mar 27 at 1:44
• 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? Mar 28 at 21:02
• @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. Mar 28 at 21:56