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Transcript

Lagrangian and Hamiltonian Dynamics Chapter 7 Claude Pruneau Physics and Astronomy Minimal Principles in Physics • Hero of Alexandria 2nd century BC. – Law governing light reflection minimizes the path length. • Fermat’s Principle – Refraction can be understood as the path that minimizes the time - and Snell’s law. • Maupertui’s (1747) – Principle of least action. • Hamilton (1834, 1835) Hamilton’s Principle Of all possible paths along which a dynamical system may move from one point to another within a specified time interval (consistent with any constraints), the actual path followed is that which minimizes the time integral of the difference between the kinetic and potential energy. Hamilton’s Principle • In terms of calculus of variations: t2 T U dt 0 t1 • The is a shorthand notation which represents a variation as discussed in Chap 6. • The kinetic energy of a particle in fixed, rectangular coordinates is of function of 1st order time derivatives of the position T T ( x&i ) • The potential energy may in general be a function of both positions and velocities. However if the particle moves in a conservative force field, it is a function of the xi only. U U(xi ) Hamilton’s Principle (cont’d) • Define the difference of T and U as the Lagrange function or Lagrangian of the particle. L T U L(xi , x&i ) • The minimization principle (Hamilton’s) may thus be written: t2 L(xi , x&i )dt 0 t1 Derivation of Euler-Lagrange Equations • Establish by transformation… x2 f {y, y'; x}dx 0 x1 t2 L(xi , x&i )dt 0 t1 xt yi (x) xi (t) yi '(x) x&i (t) f {yi (x), yi '(x); x} L(xi (t), x&i (t)) f d f 0 yi dx yi ' L d L 0, xi dt x&i i 1, 2, 3 Lagrange Equations of Motion L d L 0, xi dt x&i i 1, 2, 3 • L is called Lagrange function or Lagrangian for the particle. • L is a function of xi and dxi/dt but not t explicitly (at this point…) Example 1: Harmonic Oscillator Problem: Obtain the Lagrange Equation of motion for the one-dimensional harmonic oscillator. Solution: • Write the usual expression for T and U to determine L. L T U 12 m& x 2 12 kx 2 • Calculate derivatives. L kx x L m& x & x d L m& x& dt & x Example 1: Harmonic Oscillator (cont’d) • Combine in Lagrange Eq. L d L 0, xi dt x&i m& x& kx 0 i 1, 2, 3 Example 2: Plane Pendulum Problem: Obtain the Lagrange Equation of motion for the plane pendulum of mass “m”. l Solution: • Write the expressions for T and U to determine L. T 12 m( x&2 y&2 ) 12 I 2 12 ml 2&2 U mgl(1 cos ) L T U 12 ml 2&2 mgl(1 cos ) Example 2: Plane Pendulum (cont’d) • Calculate derivatives of L by treating as if it were a rectangular coordinate. L 1 2 &2 2 ml mgl(1 cos ) mgl sin L 1 2 &2 2 ml mgl(1 cos ) ml 2& & & d L ml 2&& dt & • Combine... mglsin ml 2&& 0 g sin 0 l Remarks • Example 2 was solved by assuming that could be treated as a rectangular coordinate and we obtain the same result as one obtains through Newton’s equations. • The problem was solved by involving kinetic energy, and potential energy. We did not use the concept of force explicitly. Generalized Coordinates • Seek generalization of coordinates. • Consider mechanical systems consisting of a collection of n discrete point particles. • Rigid bodies will be discussed later… • We need n position vectors, I.e. 3n quantities. • If there are m constraint equations that limit the motion of particle by for instance relating some of coordinates, then the number of independent coordinates is limited to 3n-m. • One then describes the system as having 3n-m degrees of freedom. Generalized Coordinates (cont’d) • Important note: if s=3n-m coordinates are required to describe a system, it is NOT necessary these s coordinates be rectangular or curvilinear coordinates. • One can choose any combination of independent parameters as long as they completely specify the system. • Note further that these coordinates (parameters) need not even have the dimension of length (e.g. in our previous example). • We use the term generalized coordinates to describe any set of coordinates that completely specify the state of a system. • Generalized coordinates will be noted: q1, q2, …, qn. Generalized Coordinates (cont’d) • A set of generalized coordinates whose number equals the number s of degrees of freedom of the system, and not restricted by the constraints is called a proper set of generalized coordinates. • In some cases, it may be useful/convenient to use generalized coordinates whose number exceeds the number of degrees of freedom, and to explicitly use constraints through Lagrange multipliers. – Useful e.g. if one wishes to calculate forces due to constraints. • The choice of a set of generalized coordinates is obviously not unique. – They are in general (infinitely) many possibilities. • In addition to generalized coordinates, we shall also consider time derivatives of the generalized coordinates called generalized velocities. Generalized Coordinates (cont’d) Notation: q1,q2 ,L ,qs q&1, q&2 ,L , q&s or or {qi } {q&i } i 1,..., s i 1,..., s Transformation • Transformation: The “normal” coordinates can be expressed as functions of the generalized coordinates - and vice-versa. x ,i x ,i (q1 ,q2 ,L ,qs ,t), x ,i (q j ,t), =1,2,...,n i 1,2, 3 j 1,2,..., s Transformation (cont’d) • Rectangular components of the velocties depend on the generalized coordinates, the generalized velocities, and the time. x ,i x& ,i (q j ,q&j ,t) • Inverse transformations are noted: q j q j (x ,i ,t) q&j q&j (x ,i , x& ,i ,t) • There are m=3n-s equations of constraint… fk (x ,i ,t) 0, k 1,2,..., m Example: Generalized coordinates • Question: Find a suitable set of generalized coordinates for a point particle moving on the surface of a hemisphere of radius R whose center is at the origin. • Solution: Motion on a spherical surface implies: x 2 y2 z 2 R2 0, z0 • Choose cosines as generalized coordinates. x q1 , R y q2 , R q12 q22 q23 1 z q3 R Example: Generalized coordinates (cont’d) • q1, q2, q3 do not constitute a proper set of generalized of coordinates because they are not independent. • One may however choose e.g. q1, q2, and the constraint equation x 2 y 2 z 2 R2 Lagrange Eqs in Gen’d Coordinates • Of all possible paths along which a dynamical system may move from one point to another in configuration space within a specified time interval, the actual path followed is that which minimizes the time integral of the Lagrangian for the system. Remarks • Lagrangian defined as the difference between kinetic and potential energies. • Energy is a scalar quantity (at least in Galilean relativity). • Lagrangian is a scalar function. • Implies the lagrangian must be invariant with respect to coordinate transformations. • Certain transformations that change the Lagrangian but leave the Eqs of motion unchanged are allowed. • E.G. if L is replaced by L+d/dt f(qi,t), for a function with continuous 2nd partial derivatives. (Fixed end points) • The choice of reference for U is also irrelevant, one can add a constant to L. Lagrange’s Eqs • The choice of specific coordinates is therefore immaterial L T ( x& ,i ) U(x ,i ) T (q j , q&j ,t) U(q j ,t) L(q j , q&j ,t) • Hamilton’s principle becomes 2 L(q j , q&j ,t) 0 1 Lagrange’s Eqs xt yi (x) qi (t) yi '(x) q&i (t) f {yi (x), yi '(x); x} L(qi (t), q&i (t)) L d L 0, qi dt q&i i 1, 2,..., s “s” equations “m” constraint equations Applicability: 1. Force derivable from one/many potential 2. Constraint Eqs connect coordinates, may be fct(t) Lagrange Eqs (cont’d) • Holonomic constraints fk (x ,i ,t) 0, k 1,2,..., m • Scleronomic constraints – Independent of time • Rheonomic – Dependent on time Example: Projectile in 2D • Question: Consider the motion of a projectile under gravity in two dimensions. Find equations of motion in Cartesian and polar coordinates. • Solution in Cartesian coordinates: 1 2 mv 2 U mgy T with L T U 1 2 mv mgy 2 U 0 at y 0. L d L 0 x dt & x d 0 m& x0 dt & x& 0 L d L 0 y dt & y d mg m& y0 dt & y& g Example: Projectile in 2D (cont’d) • In polar coordinates… 1 m r&2 r 2&2 2 U mgr sin with U 0 at 0. T L d L 0 r dt & r d mr&2 mgsin m& r 0 dt r&2 gsin & r& 0 L T U 1 m r&2 r 2&2 mgr sin 2 L d L 0 dt & d mgr cos mr 2& 0 dt gr cos 2r& r& r 2&& 0 Example: Motion in a cone • Question: A particle of mass “m” is constrained to move on the inside surface of a smooth cone of hal-angle a. The particle is subject to a gravitational force. Determine a set of generalized coordinates and determine the constraints. Find Lagrange’s Eqs of motion. z Solution: Constraint: z r cot 0 r 2 degrees of freedom only! 2 generalized coordinates. y x Example: Motion in a cone (cont’d) • Choose to eliminate “z”. v 2 r&2 r 2&2 z&2 r&2 r 2&2 r&2 cot 2 z r cot U mgz mgr cot r&2 csc 2 r 2&2 L is 1 2 1 mv m r&2 csc 2 r 2&2 mgr cot 2 2 L d L independent of . 0 dt & L constant=mr 2& & L T U mr 2& mr 2 is the angular momentum relative to the axis of the cone. Example: Motion in a cone (cont’d) • For r: L d L 0 r dt & r r r&2 sin2 gsin cos 0 Lagrange’s Eqs with underdetermined multipliers • Constraints that can be expressed as algebraic equations among the coordinates are holonomic constraints. • If a system is subject to such equations, one can always find a set of generalized coordinates in terms of which Eqs of motion are independent of these constraints. • Constraints which depend on the velocities have the form f x ,i , x& ,i ,t 0 Non holonomic constraints unless eqs can be integrated to yield constrains among the coordinates. • Consider A x& B 0 i i i 1, 2, 3 i • Generally non-integrable, unless Ai f , xi Bi f 0, t f f (xi ,t) • One thus has: f f & x x i t 0 i i • Or… df 0 dt • Which yields… f (xi ,t) constant 0 • So the constraints are actually holonomic… Constraints… • We therefore conclude that if constraints can be expressed fk f q dqi t dt 0 i i • Constraints Eqs given in differential form can be integrated in Lagrange Eqs using undetermined multipliers. • For: fk q dqi 0 i i • One gets: L d L fk k (t) 0 q j dt q&j q j k Forces of Constraint • The underdetermined multipliers are the forces of constraint: fk Q j k q j k Example: Disk rolling incline plane Example: Motion on a sphere 7.6 Equivalence of Lagrange’s and Newton’s Equations • Lagrange and Newton formulations of mechanic are equivalent • Different view point, same eqs of motion. • Explicit demonstration… L d L 0, xi dt & xi i=1,2,3 T U d T U 0, xi dt & xi T 0 xi and U d T , xi dt & xi U Fi , xi U 0, & xi i=1,2,3 i=1,2,3 i=1,2,3 i=1,2,3 d T d 3 1 2 d x j m& xi p&i m& dt & xi dt & xi j 1 2 dt Fi p&i , i=1,2,3 xi xi (qi ,t) xi j xi x q&j i q j t & xi xi q&j q j Generalized momentum T pj q&j Generalized force defined through virtual work W W Fi xi i xi W Fi qj q j i, j W Q j q j j xi Q j Fi q j i For a conservative system: U Qj q j pj T 1 2 xi m& q&j q&j i 2 p j m& xi i remember p j m& xi i & xi q&j & xi xi q&j q j xi q j xi d xi p j m&x&i m& xi q dt q i j j d xi 2 xi 2 xi q&k dt q j k qk q j q j t xi 2 xi 2 xi p j m&x&i mx&i q&k mx&i q j i,k qk q j q j t i i Qj T p j Qj q j T pj q&j T & xi m& xi q j i q j d T T U Qj dt q&j q j q j d T T U Qj dt q&j q j q j Because U does not depend on q&j , one has d T U T U 0 dt q&j q j And with L T -U, d L L 0 dt q&j q j 7.10 Canonical Equations of Motion – Hamilton Dynamics Whenever the potential energy is velocity independent: pj L x j Result extended to define the Generalized Momenta: pj L q j Given Euler-Lagrange Eqs: One also finds: L d L 0 q j dt q j L p j q j p j The Hamiltonian may then be considered a function of the generalized coordinates, qj, and momenta pj: H p j q j L j … whereas the Lagrangian is considered a function of the generalized coordinates, qj, and their time derivative. H (qk , pk , t ) p j q j L(qk , q k , t ) j To “convert” from the Lagrange formulation to the Hamiltonian formulation, we consider: H H H dH dqk dpk dt q p t j k k But given: H p j q j L j One can also write: dH pk dq k q k dpk dL j L L L pk dqk qk dpk dqk dqk dt qk qk j t p k pk L dH q&k dpk p&k dqk dt t j That must also equal: H H H dH dqk dpk dt q p t j k k We then conclude: H q k pk H p k qk H L t t Hamilton Equations Let’s now rewrite: H H H dH dqk dpk dt pk j qk t p k q k And calculate: dH H p k q k q k p k dt t j 0 Finally conclude: If : H 0 t dH H dt t H is a constant of motion If, additionally, H=U+T=E, then E is a conserved quantity.: Some remarks • The Hamiltonian formulation requires, in general, more work than the Lagrange formulation to derive the equations of motion. • The Hamiltonian formulation simplifies the solution of problems whenever cyclic variables are encountered. •Cyclic variables are generalized coordinates that do not appear explicitly in the Hamiltonian. • The Hamiltonian formulation forms the basis to powerful extensions of classical mechanics to other fields e.g. Beam physics, statistical mechanics, etc. • The generalized coordinates and momenta are said to be canonically conjugates – because of the symmetric nature of Hamilton’s equations. More remarks • If qk is cyclic, I.e. does not appear in the Hamiltonian, then p k L H 0 qk qk • And pk is then a constant of motion. p k k • A coordinate cyclic in H is also cyclic in L. • Note: if qk is cyclic, its time derivative “q-dot” appears explicitly in L. •No reduction of the number of degrees of freedom in the Lagrange formulation: still “s” 2nd order equations of motion. •Reduction by 2 of the number of equations to be solved in the Hamiltonian formulation – since 2 become trivial… q k H k k where k is possibly a function of t. One thus get the simple (trivial) solution: q k (t ) k dt The solution for a cyclic variable is thus reduced to a simple integral as above. The simplest solution to a system would occur if one could choose the generalized coordinates in a way they are ALL cyclic. One would then have “s” equations of the form : q k (t ) k dt Such a choice is possible by applying appropriate transformations – this is known as Hamilton-Jacobi Theory. Some remarks on the calculus of variation t2 Hamilton’s Principle: L(qk , q k , t )dt 0 t1 Evaluated: L L q qk q qk dt 0 k k t1 t2 Where qk and q k are not independent! dqk d q k dt dt q k The above integral becomes after integration by parts: L d L q dt q qk dt 0 k k t1 t2 Which gives rise to Euler-Lagrange equations: L d L 0 qk dt q k Alternatively, Hamilton’s Principle can be written: pk qk H dt 0 t1 k t2 Which evaluates to: H H pk q k q k pk q qk p pk dt 0 k k 1 j 2 2 2 1 j 1 j pk q k dt pk Consider: Integrate by parts: d qk dt 2 2 1 j 1 j pk q k dt p k qk dt The variation may then be written: H q k p k 1 k 2 H pk p k qk qk dt 0 H q k p k 1 k 2 H pk p k qk qk dt 0 H q k p k H p k q k Hamilton Equations