Brachistochrone curve

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A brachistochrone curve (Gr. βράχιστος, brachistos - the shortest, χρόνος, chronos - time) or curve of fastest descent, is the path that will carry a point-like body from one place to another in the least amount of time. The body is released at rest from the starting point and is constrained to move without friction along the curve to the end point, while under the action of constant gravity.

The brachistochrone is the cycloid

Given two points A and B, with A not lower than B, only one upside down cycloid passes through both points, has a vertical tangent line at A, and has no maximum points between A and B: the brachistochrone curve. The curve does not depend on the body's mass or on the strength of the gravitational constant.

The problem can be solved with the tools from the calculus of variations and optimal control.[1]

If the body is given an initial velocity at A, or if friction is taken into account, then the curve that minimizes time will differ from the one described above.

Johann Bernoulli's solution

According to Fermat’s principle: The actual path between two points taken by a beam of light is the one which is traversed in the least time. Johann Bernoulli used this principle to derive the brachistochrone curve by considering the trajectory of a beam of light in a medium where the speed of light increases following a constant vertical acceleration (that of gravity g).[2]

The Conservation of energy can be used to express the speed of a body in a constant gravitational field as:

v={\sqrt  {2gy}},

where y represents the vertical distance the body has fallen. The speed of motion of the body along an arbitrary curve does not depend on the horizontal displacement.

Johann Bernoulli noted that the law of refraction gives a constant of the motion for a beam of light in a medium of variable density:

{\frac  {\sin {\theta }}{v}}={\frac  {1}{v}}{\frac  {dx}{ds}}={\frac  {1}{v_{m}}},

where vm is the constant and \theta represents the angle of the trajectory with respect to the vertical.

The equations above allow us to draw two conclusions:

  1. At the onset, the angle must be zero when the particle speed is zero. Hence, the brachistochrone curve is tangent to the vertical at the origin.
  2. The speed reaches a maximum value when the trajectory becomes horizontal and the angle θ = 90°.

Simplifyingly assuming that the particle (or the beam) with coordinates (x,y) departs from the point (0,0) and reaches maximum speed after a falling a vertical distance D:

v_{m}={\sqrt  {2gD}}.

Rearranging terms in the law of refraction and squaring gives:

v_{m}^{2}dx^{2}=v^{2}ds^{2}=v^{2}(dx^{2}+dy^{2})

which can be solved for dx in terms of dy:

dx={\frac  {vdy}{{\sqrt  {v_{m}^{2}-v^{2}}}}}.

Substituting from the expressions for v and vm above gives:

dx={\sqrt  {{\frac  {y}{D-y}}}}dy

which is the differential equation of an inverted cycloid generated by a circle of diameter D.

Johann's brother Jakob showed how 2nd differentials can be used to obtain the condition for least time. A modernized version of the proof is as follows. If we make a negligible deviation from the path of least time, then, for the differential triangle formed by the displacement along the path and the horizontal and vertical displacements,

ds^{2}=dx^{2}+dy^{2}.

On differentiation with dy fixed we get,

2ds\ d^{2}s=2dx\ d^{2}x.

And finally rearranging terms gives,

{\frac  {dx}{ds}}d^{2}x=d^{2}s=v\ d^{2}t

where the last part is the displacement for given change in time for 2nd differentials. Now consider the changes along the two neighboring paths in the figure below for which the horizontal separation between paths along the central line is d2x (the same for both the upper and lower differential triangles). Along the old and new paths, the parts that differ are,

d^{2}t_{1}={\frac  {1}{v_{1}}}{\frac  {dx_{1}}{ds_{1}}}d^{2}x
d^{2}t_{2}={\frac  {1}{v_{2}}}{\frac  {dx_{2}}{ds_{2}}}d^{2}x

For the path of least times these times are equal so for their difference we get,

d^{2}t_{2}-d^{2}t_{1}=0={\bigg (}{\frac  {1}{v_{2}}}{\frac  {dx_{2}}{ds_{2}}}-{\frac  {1}{v_{1}}}{\frac  {dx_{1}}{ds_{1}}}{\bigg )}d^{2}x

And the condition for least time is,

{\frac  {1}{v_{2}}}{\frac  {dx_{2}}{ds_{2}}}={\frac  {1}{v_{1}}}{\frac  {dx_{1}}{ds_{1}}}

History

Johann Bernoulli posed the problem of the brachistochrone to the readers of Acta Eruditorum in June, 1696. He published his solution in the journal in May of the following year, and noted that the solution is the same curve as Huygens's tautochrone curve. After deriving the differential equation for the curve by the method given above, he went on to show that it does yield a cycloid.[3][4] But his proof is marred by the fact that he uses a single constant instead of the three constants, vm, 2g and D, above. Five mathematicians responded with solutions: Isaac Newton, Jakob Bernoulli (Johann's brother), Gottfried Leibniz, Ehrenfried Walther von Tschirnhaus and Guillaume de l'Hôpital. Four of the solutions (excluding l'Hôpital's) were published in the same edition of the journal as Johann Bernoulli's. In his paper Jakob Bernoulli gave a proof of the condition for least time similar to that above before showing that its solution is a cycloid.[3]According to Newtonian scholar Tom Whiteside, Newton found the problem in his mail when he arrived home from the mint at 4 p.m., and stayed up all night to solve it and mailed the solution by the next post. This story gives some idea of Newton's power, since Johann Bernoulli took two weeks to solve it.[5] Whiteside said that Newton would have solved it in a few minutes in his younger days.

In an attempt to outdo his brother, Jakob Bernoulli created a harder version of the brachistochrone problem. In solving it, he developed new methods that were refined by Leonhard Euler into what the latter called (in 1766) the calculus of variations. Joseph-Louis Lagrange did further work that resulted in modern infinitesimal calculus.

Earlier, in 1638, Galileo had tried to solve a similar problem for the path of the fastest descent from a point to a wall in his Two New Sciences. He draws the conclusion (Third Day, Theorem 22, Prop. 36) that the arc of a circle is faster than any number of its chords,[6]

"From the preceding it is possible to infer that the quickest path of all [lationem omnium velocissimam], from one point to another, is not the shortest path, namely, a straight line, but the arc of a circle.
...
Consequently the nearer the inscribed polygon approaches a circle the shorter is the time required for descent from A to C. What has been proven for the quadrant holds true also for smaller arcs; the reasoning is the same."

We are warned earlier in the Two New Sciences (just after Theorem 6) of possible fallacies and the need for a "higher science." In this dialogue Galileo reviews his own work. The actual solution to Galileo's problem is half a cycloid. Galileo studied the cycloid and gave it its name, but the connection between it and his problem had to wait for advances in mathematics.

See also

References

  1. Ross, I. M. The Brachistochrone Paridgm, in A Primer on Pontryagin's Principle in Optimal Control, Collegiate Publishers, 2009. ISBN 978-0-9843571-0-9.
  2. Babb, Jeff; Currie, James (July 2008), "The Brachistochrone Problem: Mathematics for a Broad Audience via a Large Context Problem", TMME 5 (2&3): 169–184 
  3. 3.0 3.1 Struik, J. D. (1969), A Source Book in Mathematics, 1200-1800, Harvard University Press, ISBN 0-691-02397-2 
  4. Herman Erlichson (1999), "Johann Bernoulli's brachistochrone solution using Fermat's principle of least time", Eur. J. Phys. 20: 299–304, doi:10.1088/0143-0807/20/5/301 
  5. D.T.Whiteside, Newton the mathematician, in Bechler, Contemporary Newtonian Research, p. 122.
  6. Galileo Galilei (1638), Discourses regarding two new sciences, p. 239  This conclusion had appeared six years earlier in Galileo's Dialogue Concerning the Two Chief World Systems (Day 4).

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