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Physics 319 Classical Mechanics. G. A. Krafft Old Dominion University Jefferson Lab Lecture 11. Calculus of Variations. General method for finding functions that extremize integrals Solution by Euler-Lagrange equation Newton’s Law of motion is an example
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Physics 319Classical Mechanics G. A. Krafft Old Dominion University Jefferson Lab Lecture 11
Calculus of Variations • General method for finding functions that extremize integrals • Solution by Euler-Lagrange equation • Newton’s Law of motion is an example • Equivalent to Hamilton’s “Principle of Least Action” • Equations of motion can be written and solved in Euler-Lagrange form • Euler-Lagrange equations apply in arbitrary coordinate systems, including accelerating or rotating coordinate systems, in contrast to F=mawhich only applies in inertial systems. • Easy identification of constants of motion
Foundational Result Finding extreme values of functions is easily accomplished by standard methods in calculus. Suppose instead we wish to find the function y(x)which makes the integral take on extreme values, either maximum or minimum. Our goal, find a necessary (but not sufficient!) condition that y(x) must satisfy when the integral extremal. Suppose g(x) is continuous and for all h(x) differentiable with h(a) = h(b) =0, vanishes. What can we say about g(x)?
Suppose g(x)≠0 somewhere, call the location x0. By continuity, there is a small band [x1,x2] containing x0 where g(x)>g(x0)/2 if g(x0) is positive or g(x)<g(x0)/2 if g(x0) is negative. Consider the function h(x) = sign[g(x0)](x-x1)2(x-x2)2 for x inside [x1,x2] and vanishing outside [x1,x2] . It is clearly strictly non-negative or non-positive, differentiable, and has a non-zero integral. Call its integral over the closed interval I[x1,x2]. But then contradicting the hypothesis. Therefore, g(x)must vanish throughout [a,b] .
Euler-Lagrange Equation y(x)+δy(x)(wrong) a b
Example: Shortest Distance • “The shortest distance between two points is a straight line” • Argument entirely symmetrical under the interchange x ↔ y. Euler-Lagrange then yields
Classical Brachistochrone Problem • What is the curve x(y) between two points that provides the shortest time to descend, assuming the velocity is zero at the highest point? • Euler-Lagrange equation gives (0,0) (x1,y1)
Cycloid Solution • Trigonometric Substitution • Solution for x(θ) • Cycloid solution
Following the problem hint • Remarkable fact: does not depend on θ0!
Two Dimensional Solution • Can also do two dimensions separately as a function of time • Euler-Lagrange
Using the first Euler-Lagrange Equation • Or using the second • Form of solution does not depend on coordinates used!
Fermat’s Principle of Least Time • The travel time of a light ray, moving from point A to point B, is minimum along the actual path • In general, velocity of light depends on index of refraction n = c/v (>1) which can be a function of position n(x,y,z) • If path is in a plane (will happen if the index of refraction does not vary in some direction), the problem is to minimize the quantity • Strictly speaking, the path is stationary, e.g. reflected rays are not on distance minima
Snell’s Law 1: Problem 6.4 • Actual light path has least time
Snell’s Law 2 • Suppose used θ1 as independent variable instead