Lectures on Theoretical Mechanics. Dietrich Belitz Department of Physics University of Oregon

Transcription

1 Lectures on Theoretical Mechanics Dietrich Belitz Department of Physics University of Oregon

2 Contents Acknowledgments, and Disclaimer 0 1 Mathematical principles of mechanics 1 1 Philosophical comments The role of theoretical physics The structure of a physical theory Review of basic mathematical concepts and notation Number sets and functions Differentiation Stationary points Ordinary differential equations Integration Paths and path functionals Surfaces Line and surface integrals The implicit function theorem Calculus of variations Three classic problems Path neighborhoods A fundamental lemma The Euler-Lagrange equations General discussion of the Euler-Lagrange equations First integrals of the Euler-Lagrange equations Example: The brachistochrone problem Covariance of the Euler-Lagrange equations Gauge invariance Nöther s theorem The principle of least action The axioms of classical mechanics Conservation laws Problems for Chapter Mechanics of point masses 39 1 Point masses in potentials A single free particle i

3 ii Contents 1.2 Galileo s principle of relativity Potentials The equations of motion Systems of point masses Simple examples for the motion of point masses Galileo s law of falling bodies Particle on an inclined plane Particle on a rotating pole The harmonic oscillator One-dimensional conservative systems Definition of the model Solution to the equations of motion Unbounded motion Bounded motion Equilibrium positions Motion in a central field Reduction to a one-dimensional problem Kepler s second law Radial motion Equation of the orbit Classification of orbits Kepler s problem Kepler s laws of planetary motion Introduction to perturbation theory General concept Digression - Fourier series Oscillations and Fourier series Summary thus far, and the anharmonic oscillator Perturbation theory for the central field problem Perturbations of Kepler s problem Digression - introduction to potential theory Mercury s perihelion advance due to an oblate sun Scattering theory Scattering experiments Classical theory for the scattering cross section Scattering by a central potential Rutherford scattering Scattering by a hard sphere N-particle systems Closed N-particle systems The two-body problem Problems for Chapter

4 Contents iii 3 The rigid body Mathematical preliminaries - vector spaces and tensor spaces Metric vector spaces Coordinate transformations Proper coordinate systems Proper coordinate transformations Tensor spaces Orthogonal transformations Three-dimensional Euclidean space Special orthogonal transformations - the group SO (3) Rotations about a fixed axis Infinitesimal rotations - the generators of SO (3) Euler angles The rigid body model parameterization of a rigid body Angular velocity The inertia tensor Classification of rigid bodies Angular momentum of a rigid body The continuum model of rigid bodies The Euler equations Simple examples of rigid-body motion The physical pendulum Cylinder on an inclined plane The force-free symmetric top The force-free asymmetric top Problems for Chapter

5 Acknowledgments These lectures are based in part on notes taken in lectures delivered by Prof. Wolfgang Götze at the Technical University Munich between 1976 and In their current form they have been taught as a 15-week course for first-year graduate students at the University of Oregon. The original handwritten notes were typeset by Spencer Alexander during the academic year 2013/14. Julian Smith helped with the first round of proofreading. Thanks are due to everyone who has alerted me to typos since. All remaining mistakes and misconceptions are, of course, the author s responsibility. Disclaimer These notes are still a work in progress. If you notice any mistakes, whether it s trivial typos or conceptual problems, please send to dbelitz a uoregon.edu. 0

6 Chapter 1 Mathematical principles of mechanics 1 Philosophical comments 1.1 The role of theoretical physics The diagram below is meant to give a rough idea of how theoretical physics ( Theory ) is interconnected with some related subjects. Figure 1.1: The role of theory. Examples: Natural phenomena: Planetary motion, atomic spectra, galaxy formation, etc. Experiment: Measure speed of light, drop objects from leaning tower in Pisa, etc. Mathematics: ODEs, Calculus of Variations, Group Theory, etc. Paradigms: Balls and springs, fields, strings, etc. 1

7 2 Chapter 1. Mathematical principles of mechanics 1.2 The structure of a physical theory The flowchart below is meant to describe the logical process behind the construction of physical theories. Figure 1.2: The structure of a physical theory. Remark 1: Occasionally, physical problems spawn the invention of new mathematics (e.g., calculus, differential geometry). Remark 2: Most theorists deal with steps 4-6 in Figure 1.2. Some very good theorists deal with step 3 (e.g., Dirac, Landau). Very few exceptional theorists deal with steps 1 and 2 (e.g., Bohr, Einstein). Occasional theorists deal with setp 4 in an innovative way (e.g., Newton, Witten). Remark 3: This course explains steps 2 through 5, using the specific example of classical mechanics. Remark 4: Guidelines for building axioms and models often include intuition, beauty, simplicity etc.; they may or may not include direct experimental input. (Famous example: GR did not!) 2 Review of basic mathematical concepts and notation 2.1 Number sets and functions We will often use the following familiar number sets: The natural numbers, The integers, N = {1, 2, 3,...}. (1.1) Z = {..., 2, 1, 0, 1, 2,...}. (1.2)

8 2. Review of basic mathematical concepts and notation 3 The rational numbers, Q = {..., 1, 12, 15,..., 0,..., 13, 12, 1,... }. (1.3) Remark 1: Care must be taken to eliminate equivalent fractions (1/2 = 2/4, etc.). See Math books for details. The real numbers, R = Q {a suitable completion} (1.4) Remark 2: A rigorous definition of R is quite involved, see a suitable math book. We will also use the Cartesian product set defined as the set of all ordered real n-tuplets. R n R R = {(x }{{} 1,..., x n ) x i R}, (1.5) n times Remark 3: Elements of R n are denoted by x = (x 1,..., x n ), x i R, and are called n-vectors, or simply vectors. 1-vectors are called scalars. This is actually a special case of a much more general concept; see any book on vector spaces. Consider functions f : R n R m. We say f is an (m-vector-valued) function of n real variables and write f (x) f (x 1,..., x n ) = y, x R n, y R m. (1.6) Example 1: f : R 3 R, f (x) = x ( x x x 2 3) 1/2 is a real-valued function of x R 3 called the norm of x. Remark 4: For m = 1, we write f instead of f. 2.2 Differentiation Definition 1: a) Let n = m = 1. We define the derivative of f, as provided the limit exists. f df : R R, (1.7) dx f 1 (x) lim [f (x + ε) f (x)]. (1.8) ε 0 ε b) For n > 1, m = 1 we define partial derivatives of f, f x i i f : R n R, (1.9)

9 4 Chapter 1. Mathematical principles of mechanics as 1 i f lim ε 0 ε [f (x 1,..., x i + ε,..., x n ) f (x 1,..., x i,..., x n )]. (1.10) We define the gradient of f, f x f : Rn R n (1.11) as f ( 1 f,..., n f). (1.12) c) For n = 1, m > 1, we define the derivative of f, as d) For n = m, we define the divergence of f, as df dx : R Rm, (1.13) ( df dx df1 dx,... df ) m. dx (1.14) div f f : R n R, (1.15) f n i f i where the last line uses the common summation convention. e) For n = m = 3 we define the curl of f, as where i=1 i f i, (1.16) curl f f : R 3 R 3, (1.17) ( f) i ε ijk j f k, (1.18) +1 if (ijk) is an even permutation of (1, 2, 3), ε ijk = 1 if (ijk) is an odd permutation of (1, 2, 3), 0 if (ijk) is not a permutation of (1, 2, 3), is the Levi-Civita tensor, or the completely antisymmetric tensor of rank 3. Remark 1: We will define tensors in general, and deal with them in detail, in Chapter 3. (1.19) Definition 2: Let I = [t 0, t 1 ] R be a real interval, and let x : I R n be a function of t I. Let f : R n I R be a real-valued function of x and t. Then the total derivative of f with respect to t is the function df : I R, (1.20) dt defined by df dt (t ) t f (x (t ), t ) + i f (x (t ), t ) dx i dt (t ). (1.21)

10 2. Review of basic mathematical concepts and notation 5 Example 1: Let f : R 2 R, f (x) = x = x x2 2 be the norm of the vector x. Then, f x i = x i x, f = x x, (1.22) so f is the normalized vector. Let x : [0, 2π] R, x (t) = (a cos t, b sin t), with a, b R. Then, df dt = i f dx i dt = x 1 x a sin t + x 2 x b cos t = 1 ( b 2 a 2) sin t cos t = 1 2 x ( b 2 a 2) sin 2t a2 cos 2 t + b 2 sin 2 t. (1.23) Problem 1: Total derivative Consider Example 1 above: Let x : R R 2 be a function defined by and let f : R 2 R be a function defined by x (t) = (a cos t, b sin t), a, b R, f (x) = x x2 2. Discuss the behavior of the total derivative, df/dt, and give a geometric interpretation of the result. Hint: First determine the geometric figure in R 2 that x provides a parametric representation of, then consider the geometric meaning of x (t) = f (x (t)). 2.3 Stationary points Definition 1: Let f : R n R be a function. f is called stationary at x 0 R n if i f (x 0 ) = 0, (i = 1,..., n). (1.24) Theorem 1: A necessary condition for f to have an extremum in the point x 0 is that f is stationary at x 0. Proof: MATH 281 or equivalent. Remark 1: While necessary, f being stationary at x 0 is not a sufficient condition for f to have an extremum at x 0, due to the possibility of saddle points.

11 6 Chapter 1. Mathematical principles of mechanics Corollary 1: Let the arguments x = (x 1,..., x n ) of f be constrained to a set S R n defined by N constraints g j (x 1,... x n ) c j, j = 1,..., N, c j R, (1.25) with N functions g j : R n R. Then a necessary condition for f to have an extremum at x 0 S is i f ( x 0 ) = 0, (1.26) where f (x) = f (x) + N λ j g j (x), (1.27) and the n+n unknowns x 0 1,..., x 0 n; λ 1,..., λ N are to be determined from the n+n conditions given in Eqs. (1.25) and (1.26). Proof: MATH 281 or equivalent. Remark 2: The constants λ i are called Lagrange multipliers. Example 1: Let f (x, y, z) = x y + z be defined on the 2-sphere j=1 S 2 = { (x, y, z) x 2 + y 2 + z 2 = 1 }. (1.28) Then, f (x, yz) = x y + z + λ ( x 2 + y 2 + z 2). (1.29) A necessary condition for extrema is and 1 + 2λx = 0, 1 + 2λy = 0, 1 + 2λz = 0, (1.30) x 2 + y 2 + z 2 = 1. (1.31) Solving this system of four equations, we find that candidates for extrema are (x, y, z) = ± 1 3 (1, 1, 1). (1.32) Problem 2: Extrema subject to constraints Consider Example 1 above: Let f : R 3 R be a function defined by f (x, y, z) = x y + z. Let S 2 = { (x, y, z) x 2 + y 2 + z 2 = 1 } be the 2-sphere. a) Show that the extremal points for f on S 2 are (1, 1, 1) / 3, and ( 1, 1, 1) / 3, as claimed in the lecture. b) Determine which of these extremal points, if any, are a maximum or a minimum, and determine the corresponding extremal values of f on S 2.

12 2. Review of basic mathematical concepts and notation 7 Problem 3: Minimal distance on the 2-sphere Consider the 2-sphere, S 2 = { (x, y, z) x 2 + y 2 + z 2 = 1 } embedded in R 3. Find the point on S 2 that is closest to the point (1, 1, 1) R 3, and determine the distance between the two points. Proposition 1: (Taylor expansion) A function f : R n R that is n-times differentiable at x can, in a neighborhood of x, be represented by a power series f (x 1 + ε, x 2,..., x n ) = f (x 1,..., x n ) + ε f (x 1,..., x n ) +... x 1 M 1 = m! εm m f x m (x 1,..., x n ) + R M, (1.33) 1 m=0 with R M a remainder, and analogously for the other variables x 2,..., x n. Proof: MATH or equivalent. Remark 3: Taylor s theorem gives an explicit bound for the remainder R M ; see your favorite calculus book for details. 2.4 Ordinary differential equations Definition 1: a) Let I = [t, t + ] R be a real interval, and let y : I R n be a function of t. Let f : R n I R n be a function of y and t. Then dy = f (y, t), (1.34) dt or, in components, dy i dt = f i (y, t), (i = 1,..., n) (1.35) is called a system of n ordinary differential equations, or ODEs, of first order. b) Let t 0 I, y 0 R n. A function y : I R n that obeys equation (1.34) and has the property that y (t 0 ) = y 0 (1.36) is said to solve equation (1.34) under the initial condition given by equation (1.36). Theorem 1: Let f (y, t) and its partial derivatives with respect to each y i be bounded and continuous in a neighborhood of N (y 0, t 0 ) of (y 0, t 0 ) R n+1 I. Then there exists a unique solution y (t) to equation (1.34) under the initial condition given by equation (1.36). Proof: MATH 256

13 8 Chapter 1. Mathematical principles of mechanics Example 1: The system of ODEs with is uniquely solved by dx 1 dt = 2x 1 + 4x 2 + 2, dx 2 dt = x 1 x 2 + 4, (1.37) x 1 (t = 0) = 8, x 2 (t = 0) = 1 (1.38) x 1 (t) = e 2t 4e 3t + 3, x 2 (t) = e 2t 4e 3t + 1. (1.39) Remark 1: The initial condition uniquely determines the solution of a first-order ODE. Problem 4: System of ODEs Solve the system of first order ODEs considered in Example 1 above: ẋ = 2x + 4y + 2, ẏ = x y + 4. Corollary 1: Let y : I R n be a function of t, and let ẏ = dy/dt be its derivative. Let f (y, ẏ, t) : R n R n I R n be bounded and continuously differentiable in a neigborhood of (y 0, ẏ 0, t 0 ) R n R n I. Then there exists a unique function y such that Proof: MATH 256. y (t 0 ) = y 0, ẏ (t 0 ) = ẏ 0, ÿ (t) = f (y, ẏ, t). (1.40) Remark 2: ÿ = f (y, ẏ, t) is a system of second-order ODEs. Remark 3: The initial point y 0 and the initial tangent vector ẏ 0 uniquely determine the solution of a second-order ODE. Example 2: ÿ (t) = y has the general solution y (t) = c 1 cos t + c 2 sin t. The initial conditions y (t = 0) = 1 and ẏ (t = 0) = 0 determine c 1 = 1, c 2 = Integration Definition 1: Let f : I R be a scalar function of t. Then the Riemann integral is defined as the limit of a sum, F = ˆ t+ t dt f (t) N 1 lim N i=1 Remark 1: F exists if f is bounded and continuous on I. (t i+1 t i ) f (t i ). (1.41) Remark 2: The generalization for f : I J R, F = t + t dt u + u du f (t, u) is straightforward.

14 2. Review of basic mathematical concepts and notation 9 Remark 3: Let B = {f : I R f is bounded and continuous}. Then, F : B R maps functions onto numbers. Such mappings are called functionals. Proposition 1 (Integration by parts): An integral of the form t + t as ˆ t+ t dt f (t) d dt g (t) = ˆ t+ t ˆ t+ t dt f (t) d dtg (t) can be rewritten ( ) d dt dt f (t) g (t) + f (t + ) g (t + ) f (t ) g (t ) ( ) t d dt dt f (t) + g (t) + f (t) g (t). (1.42) t Proof: MATH Paths and path functionals Definition 1: a) Let I = [t, t + ] R and let q : I R n be continuously differentiable. Then the set C {q (t) t I} R n is called a path, or curve, in R n, and q (t) is called a parameterization of C with parameter t. b) C inherits an order from the order defined on I as follows: q (t 1 ) < q (t 2 ) is defined to be true if and only if t 1 < t 2. c) q ± q (t ± ) are the start and end points of C. d) The tangent vector τ (t) to C in the point q (t) is defined as τ (t) d q (t) q (t). (1.43) dt d) If q (t) 0 t I, then the curve is called smooth.

15 10 Chapter 1. Mathematical principles of mechanics Figure 1.3: A parameterized path. Definition 2: Let L : R n R n I R be a function of q, q, and t. Let L be twice continuously differentiable with respect to all arguments. Let C be a path with parameterization q (t). Then we define a functional S L (C) as S L (C) := ˆ t+ Remark 1: For a given L, S L (C) is characteristic of the path C. t dt L (q (t), q (t), t). (1.44) Remark 2: If C changes slightly, C C + δc, then S L (C) changes slightly as well, S L S L + δs L. Example 1: Let n = 2 and let C be a path with parameterization q (t). The length l C of C is given by l C = lim N N 1 i=1 N 1 = lim N i=1 [q (t i+1 ) q (t i )] 2 (t i+1 t i ) N 1 = lim t ( q/ t) 2 N = ˆ t+ i=1 [q (t i+1 ) q (t i )] 2 That is, the choice L (q, q, t) = q 2 = q yields S L (C) = l C. [t i+1 t i ] 2 t dt q 2 (t). (1.45)

16 2. Review of basic mathematical concepts and notation 11 Example 2: Let n = 3. With t representing time, let a point mass m move on a trajectory x (t) in a potential U (x). The kinetic energy is given by E kin = ẋ 2 /2m. Consider L (x, ẋ, t) = E kin (ẋ) U (x). Then, ˆ t+ S = dt L (x, ẋ) (1.46) t is called the action for the trajectory x (t). Problem 5: Passage time Consider a path C in R 2 with a parameterization q (t), and a point mass moving along C with speed v (q). Let T (C) be the passage time of the particle from q to q +. Find the function L (q, q, t) such that the functional S L (C) is equal to T (C). Problem 6: Get by with a little help from your Friend TM Spring-loaded camming devices, also known as Friends TM or Camalots TM, depending on the manufacturer, are used by rock climbers to protect the climber in case of a fall. The devices consist of four metal wedges that pairwise rotate against one another so that the outside edge of each pair of wedges moves along a curve. The cam is placed in a crack with parallel walls, where the springs hold it in place. The camming angle α is defined as the angle between the line from the center of rotation to the contact point with the rock and the tangent to the curve in the contact point. The figure below shows the camming angle for an almost-extended cam in a wide crack, and a largely retracted cam in a narrow crack. Figure 1.4 If you fall, you want to get as much help from you Friend TM as the laws of physics let you. To ensure that, you want the camming angle to be the same irrespective of the width of the crack (see the figure). Determine the shape of the curve the cam surfaces must form to ensure that is the case. a) Parametrize the curve, using polar coordinates, as (r (t), ϕ (t)). Find the tangent vector in the point P = (r, ϕ) in Cartesian coordinates. b) Define the angle β between the tangent in P and the line through P that is perpendicular to the radius vector from the origin to P. How is β related to α? Show that tan β = (dr/dϕ) /r.

17 12 Chapter 1. Mathematical principles of mechanics c) Solve the differential equation that results from requiring that β = constant along the curve. Discuss the solution, which is the desired shape of the cam. 2.7 Surfaces Definition 1: a) Let I t = [t, t + ] R and let I u = [u, u + ] R be intervals. Let r : I t I u R 3 be a continuously differentiable function r (t, u). Then, is called a surface in R 3 with parameterization r (t, u). S {r (t, u) (t, u) I t I u } (1.47) b) The standard normal vector n (t, u) of S in r (t, u) is defined as n (t, u) t r (t, u) u r (t, u). (1.48) c) If n is nonzero everywhere, then the surface is called smooth. Example 1: Let I ϕ = [0, 2π], I θ = [0, 2π], and consider which is a parameterization of S 2 in R 3. Then r (ϕ, θ) = (sin θ cos ϕ, sin θ sin ϕ, cos θ) (1.49) ϕ r = ( sin θ sin ϕ, sin θ cos ϕ, 0), (1.50) Thus θ r = (cos θ cos ϕ, cos θ sin ϕ, sin θ), (1.51) n (ϕ, θ) = ( sin 2 θ cos ϕ, sin 2 θ sin ϕ, sin θ cos θ ). (1.52) Remark 1: n (ϕ, θ) = sin θ. Remark 2: S 2 is not smooth but S 2 minus its poles is smooth. 2.8 Line and surface integrals Definition 1: a) Let f : R n R n be a function, and let C be a smooth path in R n with parameterization q (t) and tangent vector τ (t) = q (t). Then the line integral of f over C is defined as ˆ ˆ t+ dl f := dt τ (t) f (q (t)), (1.53) C t with measure dl τ (t) dt. b) The length of C is defined as ˆ t+ l (C) := dt τ (t). (1.54) t

18 2. Review of basic mathematical concepts and notation 13 Remark 1: This is consistent with the intuitive concept of the length of a curve, see 2.6 Example 1. Remark 2: Integration over a closed curve C is denoted by dl. C Definition 2: a) Let f : R 3 R 3 be a function and let S be a smooth surface in R 3 with parameterization r (t, u) and standard normal vector n (t, u). The surface integral of f over S is defined as ˆ S dσ f := ˆ t+ t dt ˆ u+ u du n (t, u) f (r (t, u)), (1.55) with measure dσ n (t, u) dtdu. b) The area of S is defined as A (S) := ˆ t+ t dt ˆ u+ u du n (t, u). (1.56) Example 1: Let S be the 2-sphere S 2. From 2.7, n (θ, ϕ) = sin θ, so that A (S 2 ) = ˆ 2π = 2π dϕ 0 ˆ 1 1 ˆ π 0 d cos θ dθ sin θ = 4π. (1.57) Remark 3: A flat surface can be parameterized by the Cartesian coordinates of its points (see Figure 1.5): r (x, y, z) = (x, y, 0) = n (x, y) = 0 1 = ˆ = A (S) = dxdy. (1.58) S

19 14 Chapter 1. Mathematical principles of mechanics Figure 1.5: A parameterized flat surface. Problem 7: Enclosed area Consider a closed curve C in R 2 with parameterization q (t) = (x (t), y (t)). Show that the area A enclosed by C can be written A = 1 ˆ dt [x (t) ẏ (t) y (t) ẋ (t)]. 2 C Hint: Start from 1.6 Remark 5. Find a function f : R 3 R 3 with the property n ( f) 1, where n is the normal vector for the area enclosed by C. Then use Stokes s theorem. Theorem 1 (Gauss s theorem): Let V R 3 be a volume with smooth surface (V ), and let f : R 3 R 3 be a function. Then ˆ ˆ dv f = dσ f, (1.59) V (V ) with dv = dxdydz the measure of the volume integral. Proof: MATH 281,2 or equivalent Theorem 2 (Stokes s theorem): Let S be a surface in R 3 bounded by a smooth curve (S). Then, ˆ dσ ( f) = dl f. (1.60) Proof: MATH 281,2 or equivalent S (S)

20 2. Review of basic mathematical concepts and notation The implicit function theorem Suppose we want to define a function y (x) locally by an implicit relation F (x, y) = 0. Theorem 1: For i = 1,..., n, let each F i : R m+n R n be a function of m + n variables such that F i (x, y) := F i (x 1,..., x m, y 1,..., y n ) (1.61) a) F i is continuously differentiable in neighborhoods N m (x 0 ) R m, N n (y 0 ) R n, b) F i (x 0, y 0 ) = 0 i, and c) det (F ij ) x0,y 0 0 with F ij F i / y j an n n matrix. Then there exist n functions f i : R m R such that f i (x 1,..., x m ) f i (x) (1.62) a) f i is unique and continuously differentiable in some neighborhood Ñ m (x 0 ), b) f i (x 0 ) = y 0 i i, and c) F i (x 1,..., x m, f 1 (x 1,..., x m ),..., f n (x 1,..., x m )) = 0 x Ñ m (x 0 ). Proof: The idea of the proof is as follows: If for a given x there is a y such that F (x, y) = 0, then for x = x + ɛ one can find a y = y + δ such that F (x, y ) = 0 provided that everying is sufficiently well-behaved. See an analysis course for how to implement this idea. Example 1: Let F (x, y) = x y 2. a) If x 0 = y 0 = 1, then F/ y x0,y 0 = 2 0, so f (x) = x is unique and continuously differentiable for x > 0. b) If x 0 = y 0 = 0, then F/ y x0,y 0 = 0, so f (x) = ± x is neither unique nor continuously differentiable. Corollary 1: With i = 1,..., n, let each F i (q) be a function of n+m variables q = (q 1,..., q n+m ) such that a) F i is continuously differentiable for q N n+m (q 0 ), b) F i (q 0 ) = 0 i, and c) rank ( F i / q j ) q0 = n. Then it is possible to choose m variables and find n continuously differentiable functions f j (x) such that x 1 q i1,..., x m q im (1.63)

21 16 Chapter 1. Mathematical principles of mechanics 1. F (x, f (x)) = 0, and 2. f 1 (x 0 ) = q 0 j 1,..., f n (x 0 ) = q 0 j n, where (i 1,..., i m, j 1,... j n ) is a permutation of (1,..., m + n). Proof: Since rank ( F i / q j ) = n, x and y can be chosen such that Theorem 1 applies. Remark 1: consider n + m variables q 1,..., q n+m with n independent constraints. Then n variables can be eliminated using y 1 = q j1,..., y n = q jn to obtain m independent variables x 1 = q i1,..., x m = q jm. Example 2: Let n = 1, m = 2, F (q 1, q 2, q 3 ) = q q q (which is a 2-sphere, see 2.3 Example 2). Consider the point (0, 0, 1). We have Chose and a function F (0, 0, 1) = 0, (1.64) F q 1 = F (0,0,1) q 2 = 0, (0,0,1) (1.65) F q 3 = 2 0. (0,0,1) (1.66) x = q 1, y = q 2, (1.67) f (x, y) = 1 x 2 y 2. (1.68) f (x, y) is continuously differentiable for x 2 +y 2 < 1, so z = q 3 can be eliminated and expressed in terms of x and y. 3 Calculus of variations 3.1 Three classic problems Consider the following three classic problems: a) The brachistochrone problem (John Bernoulli, 1696) A massive particle moves under the force of gravity from A to B along a curve C. Which curve C yields the shortes passage time? b) The geodesic problem (John Bernoulli 1697) Two points A and B on a 2-sphere S 2 (or any manifold) are connected by a curve C S 2. Which curve C has the shortest length? c) The isoperimetric problem (Pappus of Alexandria, 2nd century CE, and Jacob Bernoulli) Consider a closed curve C R 2 with fixed length l. For which curve C will that curve enclose the largest area? Remark 1: Each classic problem above is an extremal value problem that ask for the extremum of a functional under the variation of a function.

22 3. Calculus of variations 17 Remark 2: Problem (c) involves a constraint. Remark 3: Extrema of functions under variation of a number was discussed eariler, in 2.3. Extrema of functionals under variation of a function requires a new mathematical apparatus, the calculus of variations. 3.2 Path neighborhoods Consider intervals I t = [t, t + ] R, I ε = [ε, ε + ] R, and a function q : I ε I t R n, with q ε (t) = (q ε 1 (t),..., q ε 2 (t)) such that i) q ε (t) parametrizes a path C ε for any fixed ε I ε. ii) q ε=0 (t) q (t) parametrizes a path C C ε=0. iii) q ε (t ± ) = q (t ± ) ε I ε. iv) q ε (t) is continuously differentiable with respect to ε for any fixed t I t. Definition 1: Under the above conditions, the set of paths {C ε ε I ε } forms an ε-neighborhood of the path C, and q ε (t) parametrizes that neighborhood. Figure 1.6: Path neighborhoods. Definition 2: a) Let L be a function and let S L be a functional according to 2.6 Definition 2. Then, S L (C) is called stationary, and C is called an extremal of S L, if 1 lim ε 0 ε [S L (C ε ) S L (C)] = 0. (1.69) b) S L is called a minimal (or maximal) under variations of the path C if there exists an ε > 0 such that S L (C ε ) > S L (C) (or S L (C ε ) < S L (C)) for all paths C ε in an ε-neighborhood of C. Proposition 1: Stationarity of S L is a necessary condition for S L to be either minimal or maximal. Proof: Once the neighborhood has been parameterized, S L (C) can be considered a function f : R R of ε. Equation (1.69) can then be written as df/dε = 0, and the assertion then follows from 2.3.

23 18 Chapter 1. Mathematical principles of mechanics 3.3 A fundamental lemma Lemma 1: Let I = [t, t + ] R, and let f : I R be a continuous function. If ˆ t+ t dt η (t) f (t) = 0 (1.70) for every function η that is continuously differentiable on I and vanishes at t = t, then f (t) = 0 t I. (1.71) Proof: Suppose that Eq. (1.71) did not hold, so that t I : f (t ) 0. Then, either f (t ) > 0 or f (t ) < 0. Case 1. f (t ) > 0. Continuity requires that [ t, t+] = I I with t I such that f (t > 0) t I. Define {( ) t t 2 ( η (t) = t + t ) 2, t I 0 t / I. (1.72) which fulfills the requirements stated in the Lemma. Then ˆ t+ t dt η (t) f (t) = ˆ t + dt η (t) f (t) t }{{} >0 > 0, (1.73) so Eq. (1.70) does not hold. The supposition was thus false, and hence Eq. (1.71) must hold. Case 2. f (t ) < 0. The set of arguments for this case mirror that of Case 1 exactly. Figure 1.7: Proving Lemma 1. Corollary 1: Let f : I R n be a continuous function, and let ˆ t+ t dt η (t) f (t) = ˆ t+ t dt η i (t) f i (t) = 0 (1.74)

24 3. Calculus of variations 19 for every continuously differentiable function η : I R n that vanishes at t = t. Then, Proof: Problem 8. f (t) = 0 t I. (1.75) Remark 1: The lemma and corollary above remain true if we require η (or η) to be n times continuously differentiable. See Problem The Euler-Lagrange equations Theorem 1: Let L (q, q, t) be a function and let q (t) be a parameterization of the path C. Then the functional S L (C) = ˆ t+ t dt L (q (t), q (t), t) (1.76) is stationary, i.e., C is an extremal path of S L, if and only if q (t) obeys the Euler-Lagrange equations, d L = L, i = (1,..., n). (1.77) dt q i q i Proof: Let {C ε } be a neighborhood of C, and let q ε (t) be a parameterization of C ε. Then, 1 ε [S L (C ε ) S L (C)] = = = = = = ˆ 1 t+ ε t ˆ t+ [ ] dt L (q ε (t), q ε (t), t) L (q (t), q (t), t) [ 1 n L ( dt L (q ε=0 (t), q ε=0 (t), t) + ε t q i=1 i q i ε (t) q i (t) ) ε=0 n L ( + q i q i ε (t) q i (t) ) + O ( ε 2) ] L (q (t), q (t), t) i=1 ε ε=0 ˆ t+ n [ L 1 ( dt q i t q i ε ε (t) q i (t) ) + L 1 ( q i }{{} q i ε ε (t) q i (t) ) ] + O (ε) }{{} ˆ t+ t ˆ t+ t ˆ t+ t dt dt dt i=1 n [ L i=1 ηε i (t)ε 0 η i(t) q i η i (t) + L q i η i (t) n [ ( L d L η i (t) q i dt q i n [ d L + L dt q i q i i=1 i=1 ] + O (ε) η ε i (t)ε 0 η i(t) ) ] n t L + η i (t) + η i (t) q i=1 i + O (ε) t ] η i (t) + O (ε). (1.78) where the second line follows from Taylor-expanding about ε = 0 (see 2.3). In the limit

25 20 Chapter 1. Mathematical principles of mechanics ε 0, which must be 0 since C ε approaches C, note that 1 lim ε 0 ε [S L (C ε ) S L (C)] = 0 ˆ t+ n [ dt d L + L ] η i (t) = 0. (1.79) dt q i q i t i=1 But the path variations q ε are arbitrary, so that the function η is arbitrary. Then, from the 3.3 corollary, we must have d L + L = 0 i, (1.80) dt q i q i as desired. Remark 1: Equation 1.77 gives a set of n coupled ODEs for the components of q. Remark 2: Using the chain rule to write equation 1.77 more explicitly gives n j=1 2 L q i q j q j + Defining a symmetric matrix M as n j=1 2 L q i q j q j + 2 L q i t = L q i. (1.81) M ij (q, q, t) 2 L q i q j = M ji, (1.82) and defining a function F i (q q, t) L q i Equation 1.77 can be rewritten as n j=1 2 L q j 2 L q i q j q i t, (1.83) n M ij q j = F i. (1.84) j=1 Written in this form, the equation has the form of Newton s second Law. Remark 3: In general, second-order ODEs are not integrable in closed form. It is worth-while to study special cases that are integrable, but most books on mechanics give the impression that there are only these special cases. Example 1: We can find the paths C R 2 between two points (x, y ) and (x +, y + ) that make the length of C stationary, i.e., find the geodesics in R 2. From 2.6 Example 1, the length of

26 4. General discussion of the Euler-Lagrange equations 21 the path is given by l C = = = ˆ t+ t ˆ x+ x ˆ x+ dt ẋ 2 (t) + ẏ 2 (t) dx 1 + (dy/dx) 2 x dx 1 + y 2 ˆ x+ x dx L (y ), (1.85) where L(y, y, x) = L (y ) = 1 + y 2. From equation 1.77, d L dx y = 0 = L y = constant = y 1 + y 2 = y = constant = y (x) = c 1 x + c 2, (1.86) i.e., the extremal paths being sought are straight lines. Given boundary condtions the values of the constants c 1 and c 2 can be determined as y (x ) = y, (1.87) c 1 = y + y x + x, c 2 = x +y x y + x + x. (1.88) Remark 4: In Example 1 above, it was tacitly assumed that the extremals can be represented with y a single-valued function of x. For a more general proof, see problem General discussion of the Euler-Lagrange equations 4.1 First integrals of the Euler-Lagrange equations Proposition 1: Let L (q, q, t) = L (q 1,..., q n, q 1,..., q n, t) be independent of q i for a given 1 i n. Then, p i (q, q, t) L q i (q, q, t) (1.89) is conserved along any extremal path, i.e., if C is extremal, with c i (C) a constant that is characteristic of C. p i (q (t), q (t), t) c i (C) (1.90)

27 22 Chapter 1. Mathematical principles of mechanics Proof: where the second line follows from the Euler-Lagrange equations. Remark 1: A variable q i on which L does not depend is called cyclic. Remark 2: p i = L/ q i is called [the momentum] conjugate to q i. Proposition 2: Let L (q, q, t) = L (q, q) be independent of t. Then, is constant along any extremal path, i.e., d dt p i = d L dt q i = L q i = 0, (1.91) H (q, q, t) q i p i (q, q, t) L (q, q, t) = q i p i (q, q) L (q, q) = H (q, q) (1.92) H (q (t), q (t)) E (C) (1.93) if C is extremal, with E (C) a constant that is characteristic of C. Proof: dh dt d L = q i p i + q i L q i L q i dt q i q i q ( i d L = q i L ) dt q i q i ( L = q i L ) q i q i = 0, (1.94) where the third line follows from the Euler-Lagrange equations. Remark 3: If L is independent of t, then the variational problem is called autonomous. Remark 4: H, as defined in the first line of equation 1.92, is called Jacobi s integral. 4.2 Example: The brachistochrone problem Consider the following statement of the brachistochrone problem: A point mass slides withouth friction on an inclined plane, with inclination angle α, from point P 1 to point P 2. Which path yields the shortest passage time?

28 4. General discussion of the Euler-Lagrange equations 23 The problem can be solved as follows: Recall from 2.6 Example 1 that the length l C of a path C parameterized by q (t) = (x (t), y (t)) is given by l C = = ˆ t+ t ˆ t+ t dt q 2 (t) dt ẋ (t) 2 + ẏ (t) 2. (1.95) Then, the passage time T is given by T = ˆ t+ t dt ẋ (t) 2 + ẏ (t) 2 From the fact that the point mass is on an inclined plane, with the coordinates as in Figure 1.8. v (x (t), y (t)). (1.96) z = y sin α, (1.97) Figure 1.8: The inclined plane for the brachistochrone problem. Energy conservation can be written as (choosing y 1 = z 1 = 0) so that m 2 v2 = mgz = mgy sin α, (1.98) v (x, y) = v (y) = 2g sin α y = ay, (1.99) where a 2g sin α > 0. (1.100)

29 24 Chapter 1. Mathematical principles of mechanics Using x as the parameter, with y = dy/dx, where T = = ˆ x2 x ˆ 1 x2 dx 1 + y 2 v (y) The problem is autonomous, so that by 2.5 Proposition 2, x 1 dx L (y, y ), (1.101) 1 + y L (y, y 2 ) =. (1.102) v (y) H (y, y ) = y L y L 1 + y 2 = y 2 v (y) 1 + y 2 v (y) = constant, (1.103) so that Substituting gives 1 v (y) = constant 1 + y 2 = ay 1 + y 2 = constant = y ( 1 + y 2) = constant c 1 < 0. (1.104) y = cot t, t = cot 1 (y ) (1.105) y = = dx = dy y c cot 2 t = c 1 sin 2 t = 1 2 c 1 (1 cos 2t) = 2c 1 sin t cos tdt cot t = 2c 1 sin 2 tdt = c 1 (1 cos 2t) dt = x = c 2 + c 1 t 1 2 c 1 sin 2t, (1.106) or x = 1 2 c 1 (2t sin 2t) + c 2. (1.107) With initial conditions x 1 = x (t = 0) = 0, y 1 = y (t = 0) = 0, (1.108) equations and give x (t) = c (2t sin 2t), y (t) = c (1 cos 2t), (1.109) where c = c 1 /2 < 0.

30 4. General discussion of the Euler-Lagrange equations 25 Remark 1: Equation is a parametric equation for a cycloid with γ = 1. To see this more clearly, a reparameterization can be performed as τ = 2t γ, x = x + γc, to give x (τ) = ( c) (τ + π sin (τ + π)) + πc = ( c) (τ + sin τ), (1.110) y (τ) = c (1 cos (τ + π)) = c (1 + cos τ) = ( c) ( 1 cos τ). (1.111) Remark 2: The constant c is a scalar factor that can be chosen such that the cycloid contains P 2. Remark 3: If x 2 > 0, then we can think of t as being less than 0. Remark 4: For further discussion, see Problem Covariance of the Euler-Lagrange equations Consider a coordinate transformation of the form (the inverse of ϕ i below is assumed to exist) Q i (t) = ϕ i (q (t), t), q i (t) = ϕ 1 i (Q (t), t). (1.112) For instance, a transformation satisfying equation in R 2 is x = r sin ϕ, y = r cos ϕ r = x 2 + y 2, ϕ = tan 1 (x/y). (1.113) Question. Are the Euler-Lagrange equations invariant under a coordinate transformation of the form given in equation 1.112? Consider a path parameterization with tangent vector q (t). The componets of the tangent vector are q i (t) = q (t) = ϕ 1 (Q (t), t), (1.114) ϕ 1 i (Q, t) Q Q j + t ϕ 1 i (Q, t) j ( ) Q, Q, t. (1.115) ϕ 1 i Defining gives ( ) ( Λ Q, Q, ( ) ) t L ϕ 1 (Q, t), ϕ 1 Q, Q, t, t (1.116) S (C) = = ˆ t+ t dt L (q, q, t) ˆ t+ t dt Λ ( ) Q, Q, t. (1.117)

31 26 Chapter 1. Mathematical principles of mechanics Recall from 3.4 that S is stationary d L = L dt q i q i d Λ dt Q = Λ. (1.118) i Q i Theorem 1: The Euler-Lagrange equations are covariant, i.e., are invariable under coordinate transformations. ( ) Remark 1: We don t need to distinguish between L and Λ, so we can write L Q, Q, t instead of ( ) Λ Q, Q, t. 4.4 Gauge invariance Definition 1: Let C be a path with parameterization q (t), and let G (q, t) R be a continuously differentiable function. Then the transformation L 1 (q, q, t) L 2 (q, q, t) = L 1 (q, q, t) + d G (q, t) (1.119) dt is called a gauge transformation, with gauge function G. Theorem 1: If L 1 and L 2 are related by a gauge transformation, then any extremal of L 1 is also an extremal of L 2. Proof: S 2 = = ˆ t+ dt L 2 (q, q, t) t dt [L 1 (q, q, t) + ddt ] G (q, t) ˆ t+ t ˆ t+ ˆ t+ = dt L 1 (q, q, t) + dt d G (q, t) t t dt = S 1 + G (q (t + ), t + ) G (q (t ), t ) S 1 + G. (1.120) Under a variation of the path, t ± does not change, nor does q (t ± ). Then, G does not change under a variation of the path. Then, S 1 + G is stationary exactly when S 1 is stationary, i.e., S 2 is stationary if and only if S 1 is stationary. Remark 1: The converse of the theorem is not true. Remark 2: If equation holds, then L 1 and L 2 are called equivalent.

32 4. General discussion of the Euler-Lagrange equations Nöther s theorem Consider a family of coordinate transformations as were described in 4.3, with α R a continuous parameter. Let Q i = ϕ i α (q, t), (1.121) Q i = q i + αf i (q, t) + O ( α 2), for α 0. (1.122) Remark 1: We know that ϕ α (q, t) is differentiable with respect to α at α = 0 - see Taylor s theorem in 2.3. ( ) Theorem 1 (Nöther s theorem): Let L Q, Q, t and L (q, q, t) be equivalent except for terms of O ( α 2), i.e., let there be a gauge function G (q, t) such that Then, the quantity ( ) L Q, Q, t = L (q, q, t) + α d dt G (q, t) + O ( α 2). (1.123) j (q, q, t) p i (t) f i (q, t) G (q, t), (1.124) with p i = L/ q i conjugate to q i according to 4.1, is constant for all extremal paths, i.e., if C is extremal. j (q, q, t) j (C) = constant (1.125) Proof: ( ) L Q, Q, t ( = L q + αf, q + αf, ) t + O ( α 2) ( L = L (q, q, t) + α f i + L ) f i + O ( α 2) q i q i = L (q, q, t) + α d dt G (q, t) + O ( α 2), (1.126) where the second line follows from Taylor-expanding the first about α = 0, and the third line follows from equation Comparing like powers of α (specifically, the α 1 terms), d L G (q, t) = dt ( d = = d dt f i + L f i q i q i L dt q i ( L f i q i ) f i + L ) d q i dt f i = d dt (p if i ), (1.127)

33 28 Chapter 1. Mathematical principles of mechanics where the second line follows from the Euler-Lagrange equations and the fact that f i df i /dt. Subtracting the LHS from each side, as desired. 0 = d dt (p if i G) = j p i f i G = constant, (1.128) Remark 2: If equation holds, we say that the coordinate transformation Q = ϕ α (q, t) represents a continuous symmetry of L. Remark 3: A continuous symmetry implies the existence of an invariant quantity j. Example 1: Let q i0 be cyclic and consider the transformation Then, ( ) L Q, Q, t Q i = q i + αδ ii0. (1.129) = L (q 1,..., q i0 + α,..., q n, q, t) = L (q, q, t) + α L q i0 = L (q, q, t), (1.130) so that equation applies, with Then, the quantity G (q, t) = 0, f i (q, t) = δ ii0. (1.131) j = p i0 (1.132) is conserved along any extremal path. Note that this result was derived directly from the Euler-Lagrange equations in 4.1 as well. Example 2: Let q R 3, and consider a rotation about the q 3 -axis, so that the coordinate transformation is Q 1 = q 1 cos α + q 2 sin α = q 1 + αq 2 + O ( α 2), (1.133) Q 2 = q 1 sin α + q 2 cos α = q 2 αq 2 + O ( α 2). (1.134) Let L be invariant under such rotations, ( ) L Q, Q, t = L (q, q, t). (1.135)

34 4. General discussion of the Euler-Lagrange equations 29 Then, equation is satisfied with Then, the quantity is conserved along any extremal path. G (q, t) = 0, f (q, t) = (q 2, q 1, 0). (1.136) j = p 1 q 2 p 2 q 1 = (p q) 3 (1.137) Remark 4: In Example 1, the continuous symmetry is translational invariance in the i 0 -direction, as given in equation Remark 5: In Example 2, it was found that (p q) 3 is conserved if the Lagrangian is invariant under rotations about the q 3 -axis. Similarly, (p q) i is conserved if the Lagrangian is invariant under rotations about the q 1 -axis. Then, if the Lagrangian is invariant under rotations about all three axes, then the quantity is constant. l p q (1.138) Remark 6: Remark 2 also follows directly from the Euler-Lagrange equations. Let L (q, q, t) = L ( q, q, t). (1.139) Then, L = L q i q i q q = L q, (1.140) q Then, L = L q i q i q q = L q. (1.141) q d dt l = q p + q ṗ = q L q + q d dt = q L q + q L q L q = 0, (1.142) where the third line follows from the Euler-Lagrange equations, and the last line follows from the fact that both terms are the cross products of parallel vectors, which can be seen from equations and

35 30 Chapter 1. Mathematical principles of mechanics 5 The principle of least action 5.1 The axioms of classical mechanics Definition 1: A system of point masses whose positions are completely determined by specify f real numbers is called a mechanical system with f degrees of freedom. Example 1: One free point mass in R 3. For this system, f = 3. Example 2: A mathematical pendulum that can swing in a plane. For this system, f = 1. Axiom 1. A mechanical system with f degrees of freedom is described by a path in R f, writeable with parameterization q (t) = (q 1 (t),..., q f (t)). (1.143) Each q i is called the i th generalized coordinate. Each q i is called the i th generalized velocity. The path parameter, t, is called time. Remark 1: Having the path be in R f is actually too restrictive, in general. For instance, the appropriate space for a planar pendulum (Example 2) is the 1-sphere S 1, not R. We will ignore such topological coordinates. Remark 2: Note that the mathematical structures postulated are paths (as opposed to, for example, fields). Remark 3: This is an axiom - questions like, what is time? are out of order. Axiom 2. Every mechanical system is characterized by a function L (q (t), q (t), t), called the Lagrangian. The physical paths minimize the functional where S is called the action. S = ˆ t+ t dt L (q (t), q (t), t), (1.144) Remark 4: Axiom 2 is known as the principle of least action, or Hamilton s principle. Remark 5: From 3.4 and Axiom 2, a necessary condition for a physical path is that it satisfies the f ODEs i.e., that it satisfies the Euler-Lagrange equations. d L (q (t), q (t), t) = L (q (t), q (t), t), (1.145) dt q i q i

36 5. The principle of least action 31 Remark 6: The [generalized] momentum and [generalized] force, respectively, are given by and Then, equation can be rewritten as which is Newton s second law. p (t) = (p 1,..., p f ) L q, (1.146) F (t) = d p (t). (1.147) dt F (t) = d p (t), (1.148) dt Remark 7: From 2.4 Remark 3, the initial coordinates q (t ) and the initial velocities q (t ) completely determine the path. 5.2 Conservation laws Definition 1: A quantity that is independent of time along the physical path is called a constant of motion, or a conserved quantity. Proposition 1: If the Lagrangian is independent of a coordinate q i, then that coordinate s conjugate momentum p i is conserved. Proof: From Newton s second law, F i = L q i = 0 = d dt p i = 0. (1.149) Remark 1: From 4.4, This is an example of Nöther s theorem. q i is cyclic = translational invariance = p i is conserved. (1.150) Remark 2: The fact that q i is cyclic is sufficient, but not necessary, for p i to be constant. Nöther s theorem is more general, and deeper, than the first integral version of the statement above and in 4.1. Remark 3: In the above proof, Newton s first law, or the law of inertia, was effectively found. It is writeable as F i = 0 = p i = constant. (1.151)

37 32 Chapter 1. Mathematical principles of mechanics Definition 2: a) A system whose Lagrangian has no explicit time dependence, so that is called conservative. b) In mechanics, Jacobi s integral is called energy. Proposition 2: For any conservative system, energy is conserved. Proof: The proof is given in 4.1 Proposition 2. L (q, q, t) = L (q, q), (1.152) Remark 4: Time translational invariance implies energy conservation. This is another example of Nöther s theorem, or rather, a generalization of Nöther s theorem (see problem 14). Definition 3: Consider f = 3 and Cartesian coordinates, q x = (x 1, x 2, x 3 ). (1.153) Then, l (t) x (t) p (t) (1.154) is called the angular momentum. Proposition 3: For a system with f = 3 that is invariant under rotations of x, angular momentum is conserved. Proof: The proof is given in 4.4 Example 2. 6 Problems for Chapter 1 1. Total derivative Consider 2.2 Example 1: Let x : R R 2 be a function defined by and let f : R 2 R be a function defined by x (t) = (a cos t, b sin t), a, b R, (1.155) f (x) = x x2 2. (1.156) Discuss the behavior of the total derivative, df/dt, and give a geometric interpretation of the result. Hint: First determine the geometric figure in R 2 that x provides a parametric representation of, then consider the geometric meaning of x (t) = f (x (t)).

38 6. Problems for Chapter Extrema subject to constraints Consider 2.3 Example 1: Let f : R 3 R be a function defined by Let S 2 = { (x, y, z) x 2 + y 2 + z 2 = 1 } be the 2-sphere. f (x, y, z) = x y + z. (1.157) a) Show that the extremal points for f on S 2 are (1, 1, 1) / 3, and ( 1, 1, 1) / 3, as claimed in the lecture. b) Prove which of these extremal points, if any, are a maximum or a minimum, and determine the respective extremal values of f on S System of ODEs Solve the system of first order ODEs considered in 2.4 Example 1: ẋ = 2x + 4y + 2, (1.158) ẏ = x y + 4. (1.159) 4. Minimal distance Consider the 2-sphere, S 2 = { (x, y, z) x 2 + y 2 + z 2 = 1 } embedded in R 3. Find the point on S 2 that is closest to the point (1, 1, 1) R 3, and determine the distance between the two points. 5. Passage time Consider a path C in R 2 with a parameterization q (t), and a point mass moving along C with speed v (q). Let T (C) be the passage time of the particle from q to q +. Find the function L (q, q, t) such that the functional S L (C) is equal to T (C). 6. Get by with a little help from your Friend TM Spring-loaded camming devices, also known as Friends TM or Camalots TM, depending on the manufacturer, are used by rock climbers to protect the climber in case of a fall. The devices consist of four metal wedges that pairwise rotate against one another so that the outside edge of each pair of wedges moves along a curve. The cam is placed in a crack with parallel walls, where the springs hold it in place. The camming angle α is defined as the angle between the line from the center of rotation to the contact point with the rock and the tangent to the curve in the contact point. The figure below shows the camming angle for an almost-extended cam in a wide crack, and a largely retracted cam in a narrow crack.

39 34 Chapter 1. Mathematical principles of mechanics Figure 1.9 If you fall, you want to get as much help from you Friend TM as the laws of physics let you. To ensure that, you want the camming angle to be the same irrespective of the width of the crack (see the figure). Determine the shape of the curve the cam surfaces must form to ensure that is the case. a) Parametrize the curve, using polar coordinates, as (r (t), ϕ (t)). Find the tangent vector in the point P = (r, ϕ) in Cartesian coordinates. b) Define the angle β between the tangent in P and the line through P that is perpendicular to the radius vector from the origin to P. How is β related to α? Show that tan β = (dr/dϕ) /r. (1.160) c) Solve the differential equation that results from requiring that β = constant along the curve. Discuss the solution, which is the desired shape of the cam. 7. Enclosed area Consider a closed curve C in R 2 with parameterization q (t) = (x (t), y (t)). Show that the area A enclosed by C can be written A = 1 ˆ dt [x (t) ẏ (t) y (t) ẋ (t)]. (1.161) 2 C Hint: Start from 1.6 Remark 5. Find a function f : R 3 R 3 with the property n ( f) 1, where n is the normal vector for the area enclosed by C. Then use Stokes s theorem. 8. Corollary to the Basic Lemma Prove 3.3 Corollary 1: Let f : I R n be a continuously differentiable function, and ˆ t+ t dt η (t) f (t) = 0 (1.162) for every continuously differentiable function η : I R n that obeys η (t ) = η (t + ) = 0. Then f (t) = 0 t I. (1.163)