2. Hamilton’s Principle Lagrangian & Hamiltonian Dynamics Newton’s 2 nd Law: F = (dp/dt) –This is a 100% correct description of particle motion in an Inertial Frame.  It will give the correct differential Equations of Motion. When solved (for given initial conditions) we will get r(t) & v(t). –For relatively simple motion & in rectangular coordinates, these equations of motion are relatively simple. –For more complicated motion & if we use non-rectangular coordinates, the equations of motion can become complicated & difficult to deal with. e.g., Particle motion on a spherical surface: Recall the Ch. 1 homework on acceleration in spherical coordinates!

3. Introduction Newton’s 2 nd Law: F = (dp/dt) –Consider, for example, particle motion on a surface.  Forces exist (Forces of Constraint) which keep the particle in contact with the surface (“Normal Forces”). –For a smooth, horizontal surface these constraint forces are simple: F c = N = - mg –However, for something more complicated, say, a bead sliding down a wire: F c = something very complicated! Perhaps even impossible to calculate! 2 nd Law: F = (dp/dt) F  F total  To use Newton’s 2 nd Law directly, we must know all forces acting on the particle exactly.  Sometimes Newton’s 2 nd Law is impractical!

4. Philosophy & History To overcome such practical difficulties, several alternate procedures were developed Hamilton’s Principle  Lagrangian Dynamics & Hamiltonian Dynamics These are the basis of much of the modern theory of matter! All such procedures obtain equations of motion which are 100% equivalent to those from Newton’s 2 nd Law: F = (dp/dt)  These alternate procedures are NOT new theories! They are reformulations of Newtonian Mechanics in a different mathematical language that sometimes might be more convenient than a direct application of F = (dp/dt). Hamilton’s Principle: Is also applicable outside of particle mechanics. For example, to fields in E&M.

5. Hamilton’s Principle (HP): –Based on experiment! Philosophical Discussion (again!) –HP: Leads to no new physical theories, only new formulations of old theories! –HP can be used to unify several theories: Mechanics, E&M, Optics, … –HP: Very elegant & far reaching –HP: “More fundamental” than Newton’s Laws??!! –HP: Is given as a (single, simple) postulate. –In the following, we consider conservative systems only. HP & Lagrange’s Equations can be extended to non- conservative systems (see Graduate Texts!)

6. Hamilton’s Principle Sect. 7.2 Philosophy & History There are several “Minimal” Principles in Physics. –These assume that Nature always minimizes certain quantities when physical processes take place –These are very common in the history of physics Brief list of others: –Hero, 200 BC: Optics: Hero’s Principle of Least Distance: A l ight ray traveling from one point to another by reflection from a plane mirror always takes the shortest path. By geometric construction:  Law of Reflection: θ i = θ r Says nothing about the Law of Refraction!

7. “Minimal” Principles Continued –Fermat, 1657: Optics: Fermat’s Principle of Least Time: A light ray travels in a medium from one point to another by a path that takes the least time. Law of Reflection: θ i = θ r  Law of Refraction: “Snell’s Law” (Prob. 6.7) n 1 sinθ 1 = n 2 sinθ 2 –Maupertuis, 1747: Mechanics: Maupertuis’s Principle of Least Action: Dynamical motion takes place with minimum action Action  (Distance)  (Momentum) = (Energy)  (Time) Based on Theological Grounds! (???!!!!) Lagrange: Put this on a firm mathematical foundation. Principle of Least Action  Hamilton’s Principle

8. Hamilton’s Principle (1834-35) –Of all of the possible paths of a mechanical system, the path actually followed is the one which minimizes the time integral of the difference in the kinetic & potential energies. That is, the actual path is the one which makes the variation of the following integral vanish: δ∫[T - U] dt = 0 (limits t 1 < t < t 2 ) δ  the arbitrary variation, just discussed in Ch. 6! The δ of an integral is similar to the derivative of a function. See sections 6.3 & 6.7!

9. δ∫[T - U] dt = 0 (limits t 1 < t < t 2 ) (1) Here: T = T(x i ) & U = U(x i ), (functions of all x i & (dx i /dt)!) –N particles in 3d, i = 1,2,..(3N) Define: The Lagrangian L L  T(x i ) - U(x i ) = L(x i,x i ) (1)  δ∫Ldt = 0 (2) This is identical to the abstract calculus of variations problem of Ch. 6 with the replacements: δJ  δ∫Ldt, x  t, y i (x)  x i (t) y i (x)  (dx i (t)/dt) = x i (t), f[y i (x),y i (x);x]  L(x i,x i ;t)  The Lagrangian satisfies Euler’s eqtns with these replacements! Hamilton’s Principle

10. δ∫Ldt = 0 (limits t 1 < t < t 2 ) (2)  The Lagrangian L = L(x i,x i ) satisfies Euler’s equations! Or, with the changes noted: (  L/  x i ) - (d/dt)[(  L/  x i )] = 0 (3) N particles in 3d: i = 1,2,..(3N)  Lagrange’s Equations of Motion (3) enables us to get differential equations of motion without (explicitly) using Newton’s 2 nd Law & without (explicitly) needing to calculate forces! It is a recipe for getting equations of motion from ENERGY!

11. Lagrange’s Equations Lagrange Equations are most useful in situations where direct application of Newton’s 2 nd Law is difficult or impossible! Recipe for solution using the Lagrangian formalism: –Step 1: Compute the PE in terms of the x i ; U = U(x i ). Compute the KE in terms of the x i = (dx i /dt). T=T(x i ). Form the Lagrangian: L  T(x i ) - U(x i ) = L(x i,x i ) –Step 2: For each x i, x i ; Obtain the Equation of Motion using: (  L/  x i ) - (d/dt)[(  L/  x i )] = 0 N particles in 3d: i = 1,2,..(3N) First, we’ll do examples using Lagrange’s Equations in problems where Newton’s 2 nd Law is relatively easy to apply & where we (may) already know the solution. Later, we’ll do many examples where it would be difficult or impossible to apply Newton’s 2 nd Law.

12. Simple Example #1 The 1d simple harmonic oscillator: –We already know the solution! We’re doing it just to illustrate the Lagrangian formalism! –Step 1: Compute U & T: L = T - U U = (½)kx 2, T = (½)mx 2, L = T - U = (½)mx 2 - (½)kx 2 –Step 2: Obtain the Lagrange Equation of motion using: (  L/  x) - (d/dt)[(  L/  x)] = 0 (  L/  x) = -kx; (  L/  x) = mx, (d/dt)[(  L/  x)] = mx  mx + kx = 0 Or x + (k/m)x = 0 Or x + (ω 0 ) 2 x = 0 Nothing new! Just used to illustrate the method!

13. Simple Example #2 The Plane Pendulum: –We already know the solution! We do it just to illustrate the Lagrangian formalism. –Step 1: Compute U, T; L = T-U U = mg (1-cosθ), T = (½)m 2 θ 2, L = T - U = (½)m 2 θ 2 - mg (1-cosθ) –Step 2: Obtain the Lagrange Equation of motion. Treat θ as if it were a rectangular coordinate! Justification: Next section!  In the formalism, make the replacement x  θ  the Lagrange Equation is (  L/  θ) - (d/dt)[(  L/  θ)] = 0 (  L/  θ) = - mg sinθ; (  L/  θ) = m 2 θ; (d/dt)[(  L/  θ)] = m 2 θ  m 2 θ + mg sinθ = 0 Or: θ + (g/ )sinθ = 0 Nothing new! Just used to illustrate the method!

14. Remarks on These 2 Simple examples Plane Pendulum: –We treated θ AS IF it were a rectangular coordinate x! Justification: Next section! –We obtained the known equation of motion!  Lagrange’s Equations are more general than the derivation in rectangular coordinates indicates! Both Examples: –Obtained known results using the Lagrange formalism. –No statements were made regarding forces & there was no (explicit) calculation of forces! –Instead of forces (properties of the interaction between the particle & its environment) we used ENERGY (properties of the particle itself).  Can get eqtns of motion for a particle using only energy!

15. Generalized Coordinates Section 7.3 The plane pendulum example: –Showed that we can use the Lagrangian formalism on coordinates which are not rectangular!  This suggests that we can choose any (complete set) of coordinates we want to do the problem & then use Lagrange’s Equations. –This section is the justification of this.

16. Consider a general mechanical system with n discrete point particles (these might be connected to form rigid bodies, Ch11). Arbitrary origin. – The state of the system at time t is specified by the n position vectors r n (t). 3d  We need 3n coordinates to describe the system. –Possibly, constraints exist which make the number of independent coordinates < 3n. Suppose there are m equations of constraint which connect some of the 3n coordinates to some others.  The number of degrees of freedom is s = 3n - m

17. For a general mechanical system with s = 3n - m degrees of freedom We needn’t necessarily choose s rectangular coordinates to describe the system. We are free to choose any set of s coordinates which completely describes the state of motion of the system. Depending on the problem: –We could choose s curvilinear (spherical or cylindrical) coordinates. Or: We could choose a mixture of rectangular coordinates (k = # rectangular coordinates) & curvilinear coordinates (s - k = # curvilinear coords) –The s coordinates needn’t have units of length! They could be dimensionless or they could have (almost) any units.

18. Generalized Coordinates  Any set of s quantities which completely specifies the state of the motion of the system (for a system with s degrees of freedom). Standard Notation: q 1, q 2, q 3, … Or: q j (j = 1,2,… s) Proper Set of Generalized Coordinates  Any set of coordinates whose number = s (# degrees of freedom) and are not restricted by constraints. –Sometimes, it might be useful to use generalized coordinates whose number is > s & to explicitly take into account constraints using Lagrange multipliers (as in Ch. 6) –We’ll see: We need to do this to calculate forces of constraint.

19. Note: The choice of the set of s generalized coordinates to solve a problem is not unique. It depends on the problem & on personal taste! Example: Consider a disk rolling down an inclined plane. We could use z = height of the disk center of mass above the reference level and y = linear distance rolled down the plane or z and the angle θ (radians) = the angular distance rotated. For a disk of radius R, the equation of constraint is y = Rθ.  y & θ aren’t independent! zz

20. In addition to the generalized coordinates q j, in the Lagrangian formalism, we also have Generalized Velocities  (dq j /dt)  q j Suppose we start in rectangular coordinates & transform to a set generalized coordinates. In general, this involves equations of the form: x α,i = x α,i (q 1,q 2,q 3,..t) or x α,i = x α,i (q j,t) α, = 1,2,3 (rectangular indices) i = 1,2,3,…, n; j = 1,2, 3,…,s Also, there are velocity transformations: x α,i = x α,i (q j,q j,t) –The velocities in rectangular coordinates can depend on both the generalized coordinates & the generalized velocities There also might be equations of constraint: f k (x α,i,t) = 0, k = 1,2,.. m

21. For example, if we choose spherical coordinates as the generalized coordinates: x 1 = x = r sinθ cosφ, x 2 = y = r sinθ sinφ x 3 = z = r cosθ q 1 = r, q 2 = θ, q 3 = φ, q 1 = r, q 2 = θ, q 3 = φ x 1 = (dx/dt) = r sinθ cosφ + rθ cosθ cosφ - rφ sinθ sinφ x 2 = (dy/dt) = r sinθ sinφ + rθ cosθ sinφ + rφ sinθ cosφ x 3 = (dz/dt) = r cosθ - rθ sinθ x α,i = x α,i (q j,q j )

22. Example 7.1 Find a proper set of generalized coordinates for a point particle moving on the surface of a hemisphere of radius R whose center is at the origin. Constraint eqtn (motion on the surface of a hemisphere) : x 2 + y 2 + z 2 - R 2 = 0, z  0 (1) Following the text, we choose the cosines of the angles between the x, y, & z axes & a line connecting the particle with the origin: q 1 = (x/R), q 2 = (y/R), q 3 = (z/R)  (1) becomes: (q 1 ) 2 +(q 2 ) 2 + (q 3 ) 2 = 1 (2) Because of (1), q 1, q 2, q 3 aren’t independent  They aren’t a proper set of generalized coordinates! ((2) is a constraint eqtn.) For a proper set, we need only choose 2 of the 3 q j : All that’s needed on a sphere’s surface! In general, we can always use the constraint equations to reduce the number of generalized coordinates needed in a problem!

23. Use the (x,y) coordinate system in the figure to find the KE T, the PE U, & the Lagrangian L for a simple pendulum (length, bob mass m), moving in the xy plane. Determine the transformation equations from the (x,y) system to the coordinate θ. Find the equation of motion. Worked on the board! Example 7.2

24. The state of a system of n particles & subject to m constraints connecting some of the 3n rectangular coordinates is completely specified by s = 3n – m generalized coordinates.  Sometimes its convenient to represent the state of such a system by a point in an abstract s-dimensional space called CONFIGURATION SPACE. Each dimension in this space corresponds to one of the coordinates q j. This point specifies the CONFIGURATION of the system at a particular time. As the q j change in time (governed by the eqtns of motion) this point traces out a curve in configuration space. The exact curve depends on the initial conditions. Often we speak of “the path” of the system as it “moves” in configuration space. Obviously, this is NOT the same as the particle path as it moves in ordinary 3d space! Configuration Space