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Hamilton’s Principle Lagrangian & Hamiltonian Dynamics

Hamilton’s Principle Lagrangian & Hamiltonian Dynamics. Newton’s 2 nd Law: F = (dp/dt)

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Hamilton’s Principle Lagrangian & Hamiltonian Dynamics

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  1. Hamilton’s PrincipleLagrangian & Hamiltonian Dynamics • Newton’s 2nd 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!

  2. Introduction • Newton’s 2nd 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: Fc = N = - mg • However, for something more complicated, say, a bead sliding down a wire: Fc = something very complicated! Perhaps even impossible to calculate! 2nd Law: F = (dp/dt) F  Ftotal  To use Newton’s 2nd Lawdirectly,we must knowall forces acting on the particle exactly.  Sometimes Newton’s2nd Law is impractical!

  3. 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 2nd 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.

  4. 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!)

  5. Hamilton’s PrincipleSect. 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 light ray traveling from one point to another by reflection from a plane mirror always takes the shortest path. By geometric construction:  Law of Reflection:θi = θr Says nothing about the Law of Refraction!

  6. “Minimal” PrinciplesContinued • 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:θi = θr  Law of Refraction:“Snell’s Law” (Prob. 6.7)n1sinθ1 = n2sinθ2 • 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

  7. 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 t1 < t < t2) δ thearbitrary 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!

  8. Hamilton’s Principle δ∫[T - U] dt = 0 (limits t1 < t < t2)(1) Here: T = T(xi) & U = U(xi), (functions of all xi & (dxi/dt)!) • N particles in 3d, i = 1,2, ..(3N) • Define: The LagrangianL L  T(xi) - U(xi) = L(xi,xi) (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 , yi(x)  xi(t) yi(x)  (dxi(t)/dt) = xi(t), f[yi(x),yi(x);x]  L(xi,xi;t)  The Lagrangian satisfies Euler’s eqtns with these replacements!

  9. δ∫Ldt = 0 (limits t1 < t < t2)(2)  The Lagrangian L = L(xi,xi) satisfies Euler’s equations! Or, with the changes noted: (L/xi) - (d/dt)[(L/xi)] = 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 2nd Law & without (explicitly) needing to calculate forces! It is a recipe for getting equations of motion from ENERGY!

  10. Lagrange’s Equations • Lagrange Equations aremost useful in situations where direct application of Newton’s 2nd Law is difficult or impossible! • Recipe for solution using the Lagrangian formalism: • Step 1:Compute the PE in terms ofthe xi; U = U(xi). Compute the KE in terms of the xi = (dxi/dt).T=T(xi). Form the Lagrangian: L  T(xi) - U(xi) = L(xi,xi) • Step 2:For each xi, xi; Obtain the Equation of Motion using: (L/xi) - (d/dt)[(L/xi)] = 0 N particles in 3d: i = 1,2, ..(3N) • First, we’ll do examples using Lagrange’s Equations in problems where Newton’s 2nd 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 2nd Law.

  11. 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)kx2, T = (½)mx2, L = T - U = (½)mx2 - (½)kx2 • 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 + (ω0)2x = 0 Nothing new! Just used to illustrate the method!

  12. 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2θ2, L = T - U = (½)m2θ2 - 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2θ; (d/dt)[(L/θ)] = m2θ  m2θ + mgsinθ = 0 Or:θ + (g/)sinθ = 0 Nothing new! Just used to illustrate the method!

  13. Remarks on These 2 Simple examples • Plane Pendulum: • We treated θAS 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!

  14. 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.

  15. 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 rn(t). 3d  We need 3n coordinates to describe the system. • Possibly, constraints exist which make the number of independent coordinates < 3n. Suppose there arem equations of constraint which connect some of the 3n coordinates to some others. The number of degrees of freedomis s = 3n - m

  16. For a general mechanical system with s = 3n - m degrees of freedom • Weneedn’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.

  17. 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:q1, q2, q3, …Or:qj (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.

  18. 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 

  19. In addition to the generalized coordinates qj, in the Lagrangian formalism, we also have Generalized Velocities(dqj/dt)  qj • Suppose we start in rectangular coordinates & transform to a set generalized coordinates. In general, this involves equations of the form: xα,i = xα,i (q1,q2,q3,..t) or xα,i = xα,i (qj,t) α, = 1,2,3 (rectangular indices) i = 1,2,3,…, n; j = 1,2, 3,…,s • Also, there are velocity transformations:xα,i = xα,i (qj,qj,t) • The velocities in rectangular coordinates can depend on both the generalized coordinates & the generalized velocities • There also might be equations of constraint: fk(xα,i ,t) = 0, k = 1,2, .. m

  20. For example, if we choose spherical coordinates as the generalized coordinates: x1 = x = r sinθ cosφ, x2 = y = r sinθ sinφ x3 = z = r cosθ q1 = r, q2 = θ, q3 = φ, q1 = r, q2 = θ, q3 = φ x1 = (dx/dt) = r sinθ cosφ + rθ cosθ cosφ - rφsinθ sinφ x2 = (dy/dt) = r sinθ sinφ + rθ cosθ sinφ + rφsinθ cosφ x3 = (dz/dt) = r cosθ - rθ sinθ xα,i = xα,i (qj,qj)

  21. 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): x2 + y2 + z2 - R2 = 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: q1 = (x/R) , q2 = (y/R) , q3 = (z/R)  (1) becomes: (q1)2 +(q2)2 + (q3)2 = 1 (2) • Because of (1), q1, q2, q3aren’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 qj: 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!

  22. Example 7.2 • 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!

  23. Configuration Space • 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 qj. This point specifies the CONFIGURATION of the system at a particular time. As the qj 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!

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