Lecture Notes for PHY 405 Classical Mechanics

Transcription

1 Lecture Notes for PHY 45 Classical Mechanics From Thorton & Marion s Classical Mechanics Prepared by Dr. Joseph M. Hahn Saint Mary s University Department of Astronomy & Physics October 11, 25 Chapter 6: Calculus of Variations A mathematical method that will be used in Chapter 7 to obtain the Lagrange equations and Hamilton s principle, which are very useful reformulations of Newtonian mechanics. Functions and functionals The problem: determine the path y(x) such that J = f [y(x), y (x); x] dx is an extremum (ie, a min or max) where x = independent variable (might be time, distance, angle, etc) y(x) = dependant function y = dy/dx J = integral of the function of y(x), ie, a functional and x 2 are fixed integration endpoints (could be times, distances, etc) 1

2 Fig Solution: Euler s equation consider y(α, x) = y(, x) + αη(x) where y(x) = y(, x) = desired path that minimizes J(α) so y(α, x) = alternate paths that result in larger J when α > and η(x) = an (almost) arbitrary function obeying η( ) = η(x 2 ) = J(α) = f [y(α, x), y (α, x); x] dx We want α to be such that J is an extremum what does this tell us about J(α)? J α = for J to have an extremum α= 2

3 by Chain Rule, where and so y α + f y ) dx (since x is indep of α) y α J x2 ( f α = y y α = η y α = dy α dx = ( dy α dx + α dη ) α= dx J x2 ( f α = y η + f ) dη dx = y dx = dη dx Do the 2 nd integral by parts: udv = uv x 2 vdu u = f dv = dη dx = dη y dx du = d ( ) f dx v = η dx y so 2 nd integral = f x 2 ( ) y η d f η(x)dx dx y and recall that η( ) = η(x 2 ) = by definition so J α = [ f y d ( )] f η(x)dx = dx y for any η(x) that is (almost) arbitrary. What does this tell us about the integrand? f y d ( ) f dx y = This is Euler s equation, whose solution yields the path y(x) that minimizes/maximizes J. 3

4 Note that many physics problems (optics, mechanics, etc) also seek the path y(x) that minimize J. Ex. 6.1, the brachistochrone problem Brachistochrone=Greek for path of shortest delay. This classic physics problem was solved by Bernoulli in A bead slides along a frictionless wire due to gravity, from rest at point (, y 1 ) = (, ) to point (x 2, y 2 ). What is the shape of the wire y(x) that minimizes the travel time? Fig where independent coordinate x = bead s vertical position and y(x) = its horizontal position. 4

5 Since dt = ds/v = time for bead to traverse small distance ds, the total travel time is t = (x2,y 2 ) (,) where ds = dx 2 + dy 2 = (1 + y 2 ) 1/2 dx = small path segment ds v particle s energy E = 1 2 mv2 mgx = (recall zeropoint is arbitrary) so velocity v(x) = 2gx x2 1 + y and t = 2 2gx dx What is our functional J? What is f? The total travel time t is thus minimized when the path y(x) satisfies Euler s Eqn.: f y d ( ) f = dx y where f(y, y 1 + y ; x) = 2 2gx f since y =, f y = constant C = (1 + y 2 ) 1/2 2y 2 2gx so y 2 = 2gC 2 x(1 + y 2 ) y 2 (1 2gC 2 x) = 2gC 2 x so y = dy dx = 2gC 2 x 1 2gC 2 x = y 2gx(1 + y 2 ) 5

6 Now what? dy = y(x) = x 2gC 2 x 1 2gC 2 x dx Recall that C = some unknown constant, so set 2gC 2 = 1/2a where a is some other constant: x2 x/2a x2 y(x 2 ) = 1 x/2a = xdx 2ax x 2 after multiplying by 2ax upstairs & downstairs. To solve this integral, change variables: note denominator x = a(1 cos θ) so dx = a sin θdθ θ a 2 (1 cos θ) sin θdθ and y = 2a2 (1 cos θ) a 2 (1 2 cos θ + cos 2 θ) so y = = a 2 (1 cos 2 θ) = a sin θ θ a(1 cos θ)dθ or y(θ) = a(θ sin θ) and x(θ) = a(1 cos θ) 6

7 This is the equation for a cycloid= a curve traced by a point on a circle of radius a rolling along the x = plane: x(θ) = x c + x(θ) y(θ) = y c + y(θ) where x + c = a & y c = aθ is the motion of the cycloid s center and x(θ) = a cos θ y(θ) = a sin θ are the bead s displacements from the center The radius a is chosen so that the bead passes thru endpoint (x 2, y 2 ). Fig Note also that a straight wire does not minimize travel time. 7

8 Euler s Eqn. for N dimensional problem write where suppose f = f(y 1, y 2, y 3,..., y 1, y 2, y 3,... ; x) and f(y i, y i; x) y i (α, x) = y i (, x) + αη i (x) where i = 1, 2, 3,..., N y i (, x) = y i (x) = trajectory along i axis that minimizes J η i ( ) = η i (x 2 ) = at trajectory endpoints recall J = so dj dα = y i and note that α = η i Integrate 2 nd term by parts: u = f du = so the 2 nd term = and dj dα = y i f(y i, y i ; x)dx N i=1 d ( ) f dx y i dx [ N f x 2 η i y i=1 i N i=1 ( f y i y i α + f y ) i y i dx α [ f y i d dx dv = y i α dx = η i x dx = dη i v = η i ( ) ] d f dx y i η i dx ( )] f η i (x)dx y i = when J is an extremum Since the η i (x) are (almost) arbitrary functions of x, each individual i th integrand in the [] must also be zero: f d ( ) f y i dx y i = which is Euler s eqn. in N dimensions 8