Hamiltonian Mechanics

Transcription

1 Alain J. Brizard Saint Michael's College Hamiltonian Mechanics 1 Hamiltonian The k second-order Euler-Lagrange equations on con guration space q =(q 1 ; :::; q k ) _q j j ; (1) can be written as k rst-order di erential equations, known as Hamilton's equations, on ak-dimensional phase space z =(q 1 ;:::; q k ; p 1 ;:::; p k )as dq j and dp j ; () where the Hamiltonian function H(q; p; t) is de ned fromthe Lagrangian function L(q; _q; t) by the Legendre transformation H(q; p; t) = p _q(q; p;t) L[q; _q(q; p;t);t]; (3) where p j (q; _q; t) = (q; _q; t) _q j de nes the j th -component of the canonical momentum. 1D Hamiltonian Dynamics The one degree-of-freedom Hamiltonian dynamics of a particle of mass m is based onthe Hamiltonian H(x; p) = p + U (x); (5) m where p = m _x is the particle's momentum and U (x)isthepotential energy. The Hamilton's equations ()forthishamiltonian are dx = p m and dp = du (x) dx : (6) 1

2 Since the Hamiltonian (and Lagrangian) is time independent, the energy conservation law states that H(x; p) = E. Inturn,thisconservation law implies that the particle's velocity _x can be expressed as s _x(x; E) = [E U(x)]; (7) m where the sign of _x is determined from the initial conditions. It is immediately clear that physical motion is possible only if E U (x); points where E = U(x) areknownasturning points. Furthermore, motion is bounded for a xed value of energy E if two roots exist (labeled x 1 <x )..1 Dynamical Solution x(t; E) and Oscillation Period T (E) The dynamical solution x(t; E) of the Hamilton's equations (6) is rst expressed as t(x; E) = r m Z x x ds q E U(s) ; (8) where x is the particle's initial position (x 1 <x <x )andweassume that _x() >. Next, inversion of the relation (8) yields the solution x(t; E). Forbounded motion in one dimension, the particle bounces back and forth between the two turning points x 1 and x >x 1.Theperiod of oscillation T(E) isafunction of energy alone T (E) = Z x x1 dx _x(x; E) = p m Z x x1 dx q E U(x) : (9) As a rst example, we consider the case of a particle of mass m attached to aspringof constant k, forwhichthe potential energy is U(x) = 1 kx.themotionof a particle with total energy E is always bounded, with turning points x 1; = q E=k = a: We start with the solution t(x; E) forthecase of x(; E) =+a, sothat _x(t; E) < for t>, and r m Z a r ds m x t(x; E) = p k x a s = k arccos : (1) a Inversion q of this relation yields the well-known solution x(t; E) =a cos(t), where = k=m. UsingEq.(9),we nd the period of oscillation r m T(E) = 4 k Z a which turns out to be independent of energy E. dx p a x = ¼ ;

3 Our second example involves the case of the pendulum of length ` and mass m in a gravitational eld g. The Hamiltonian in this case is H = 1 m` _ + mg` (1 cos ): The total energy of the pendulum is determined from its initial conditions; assuming that the pendulum starts from rest at an angular deviation (E) fromthevertical, the total energy is E = mg` (1 cos ) = arccos 1 E ; mg` and the angular velocity _ is solved as s q _(; E) = (cos cos ) = sin sin ; (11) where = q g=` denotes the characteristic angular frequency and, thus, are the turning points for this problem. By making the substitution sin = = k sin', where k(e) = sin[ (E)=] = s E mg` < 1 and ' = ¼=when =,thesolutionof the pendulum problemis t(; E) = Z ¼= (;E) d' q1 k sin ' ; (1) where (; E) =arcsin(k 1 sin =). The inversion of this relation yields(t; E) expressed in terms of elliptic functions, while the period of oscillation is de ned as T(E) = 4 = ¼ Z ¼= d' q1 k sin ' = 4 Z ¼= d' 1 + k sin ' k 4 + : (13) We note here that if k 1(or 1) the period of a pendulum is independent of energy, T ' ¼=. However, we also note that as E mg` (or ¼), the period of the pendulum becomes in nitely large, i.e., T 1as k 1. In this limit ( ¼), the pendulum equation (11) yields the separatrix equation _' = cos ', where' = =. The solution to the separatrix equation is expressed in terms of the transcendental equation sec '(t) = cosh( t + ); where cosh =sec' represents the initial condition. We again note that ' ¼= (or ¼) onlyast 1. 3 ;

4 3 Legendre Transformation We now investigate the Legendre transformation (3), which allows the transformation from alagrangiandescription of a dynamical system in terms of a Lagrangian function L(r; _r;t) to a Hamiltonian description of thesamedynamical system in terms of a Hamiltonian function H(r; p;t), where the canonical momentum p is de ned as p i = =@ _x i.inparticular, we want to know the condition(s) under which the Legendre transformation can be used. ALagrangian function for which the Legendre transformation is applicable is said to be regular. It turns out that the condition under which the Legendre transformation can be used is associated with the condition under which the inversion of the relation p(r; _r;t) _r(r; p;t) is possible. To simplify our discussion, we focus on motion in two dimensions (labeled x and y). The general expression of the kinetic energy term of a Lagrangian with two degrees of freedom L(x; _x; y; _y) =K(x; _x; y; _y) U(x; y) is K(x; _x; y; _y) = _x + _x _y + _y = 1 _rt M _r; where _r T =(_x; _y) anhemass matrix M is M = Here, the coe±cients,, and may be function of x and/or y. Thecanonical momentum vector (4) is thus de ned as p = M _r 1 C A : 1 p x B C A p y 1 C B _y or 9 p x = _x + _y >= >; : (14) p y = _x + _y The Lagrangian is said to be regular if the matrix M is invertible, i.e., if itsdeterminant = 6= : In the case of a regular Lagrangian, we readily invert (14) to obtain _r(r; p;t) = M 1 p _x _y 1 C A = 1 4 _x 1 C A 1 C B 1 p x C A p y

5 or 9 _x = ( p x p y )= >= >; ; (15) _y = ( p y p x )= and the kinetic energy term becomes K(x; p x ;y;p y ) = 1 pt M 1 p: Lastly, under the Legendre transformation, we nd H = p T ³ 1 M 1 p pt M 1 p U = 1 pt M 1 p + U: Hence, we clearly see that the Legendre transformation is applicable only if the mass matrix M is invertible. 4 Particle MotioninanElectromagneticField 4.1 Euler-Lagrange Equations The equations of motion for a charged particle of mass m and charge e moving in an electromagnetic eld represented by the electric eld E and magnetic eld B are dx dv = v (16) = e m E + dx B c ; (17) where x denotes the position of the particle and v its velocity. By treating the coordinates (x; v)asgeneralized coordinates (i.e., ±v is independent of ±x), we now show that the equations of motion (16) and (17) can be obtained as Euler- Lagrange equations from the Lagrangian L(x; _x; v; _v; t) = m v + e c A(x;t) _x e (x;t)+ m jvj ; (18) where and A are the electromagnetic potentials in terms of which electric and magnetic elds are de ned E = r and B = r A: (19) c 5

6 Note that these expressions for E and B satisfy Faraday's law r E = c t B and Gauss' law r B =. First, we look at the Euler-Lagrange equation for _x = m v + e c A d = m _v + _x c which yields Eq. (17), since m _v = e r + @x = e ra _x e r ; c + e c _x r A = ee + e c + _x ra _x B; () where the de nitions (19) were used. Next, we look at the Euler-Lagrange equation for _v = d = = m _x m v; which yields Eq. (16). Because =@ _v =, wenotethatwecoulduseeq.(16) asa constraint which could be imposed apriori on the Lagrangian (18) to give L(x; _x; t) = m j _xj + e c A(x;t) _x e (x;t): (1) The Euler-Lagrange equation in this case is identical to Eq. () with _v = Äx. 4. Energy Conservation Law We now show that the second Euler equation (i.e., the energy conservation law), expressed as d L _x _v _v ; is satis ed exactly by thelagrangian (18) and the equations of motion (16) and (17). First, from the Lagrangian (18), we nd L _x = e _x _v = L m v + e c A = m jvj + e 6 : _x

7 Next, we nd d L _x _v = mv _v e Using Eq. (16), we readily nd mv _v = e E v and e E v e + _x r = + _x r _x ; which is shown to be satis ed exactly by substituting the de nition for E. : 4.3 Gauge Invariance The electric and magnetic eldsde ned in (19) are invariant under the gauge transformation c and A A + râ; () where Â(x;t)isanarbitrary scalar eld. Although the equations of motion (16) and (17) are manifestly gauge invariant, the Lagrangian (18) is not manifestly gauge invariant since the electromagnetic potentials and A appear explicitly. Under a gauge transformation, however, we nd L L + e c râ _x e 1 = L + e c As is generally known, since Lagrangians can only be de ned up to the exact time derivative of a time-dependent function on con guration space (i.e., equivalent Lagrangians yield the same Euler-Lagrange equations), we nd that a gauge transformation keeps the Lagrangian within the same equivalence class. dâ : 4.4 Canonical Hamilton's Equationss The canonical momentum p for a particle of mass m and charge e in an electromagnetic eld is de ned as p(x; v;t) _x = m v + e A(x;t): (3) c The canonical Hamiltonian function H(x; p; t)isnowconstructed through the Legendre transformation H(x; p;t) = p _x(x; p;t) L[x; _x(x; p;t);t] = e (x;t)+ 1 m 7 p e c A(x;t) ; (4)

8 where v(x; p; t)was obtained by inverting p(x; v; t)from Eq.(3). Using the canonical Hamiltonian function (4), we immediately nd = 1 p e m c A ; = e r e c ra _x; from which we recover the equations of motion (16) and (17) once we use the de nition (3) for the canonical momentum. 5 Charged Spherical Pendulum in a Magnetic Field Asphericalpendulum of length ` and mass m carries a positive charge e and moves under the action of a constant gravitational eld (with acceleration g) and a constant magnetic eld B (see Figure). The position vector of the pendulum is and thus its velocity v =_x is x = ` [sin (cos Á bx +siná by) cos bz ] ; v = ` _ [cos (cosá bx +siná by) +sin bz ]+` sin _ Á ( sin Á bx +cosá by); and the kinetic energyofthependulum is m` K = ³ _ +sin _ Á : 8

9 5.1 Lagrangian Because the charged pendulum moves in a magnetic eld B = B bz, wemustinclude the magnetic term v ea=c in the Lagrangian [see Eq. (18)]. Here, the vector potential A must be evaluated at the position of the pendulum and is thus expressed as A = B` sin ( siná bx +cosá by) ; and, hence, we nd e c A v = B` sin Á: _ Lastly, the charged pendulum is under the in uence of two potential energy terms: gravitational potential energy ( mg` cos) andmagnetic potential energy ( ¹ B = ¹ z B), where ¹ = e (x v) mc denotes the magnetic moment of a charge e moving about a magnetic eld line. Here, it is easy to nd ¹ z B = eb ` sin Á c _ and by combining the various terms, the Lagrangian for the system is " # _ L(; ; Á Á) _ = m` +sin c Á _ + mg` cos ; (5) where the cyclotron frequency c is de ned as c = eb mc : 5. Euler-Lagrange equations The Euler-Lagrange equation for _ = m` _ _ = m` = mg` sin + m` ³ Á _ c Á _ sin cos or Ä + g` sin = ³ Á _ c Á _ sin cos The Euler-Lagrange equation for Á immediately leads to aconstant of the motion for the system since the Lagrangian (5) is independent of the azimuthal angle Á and hence p Á Á _ = m` sin ³ Á _ c 9

10 is a constant of the motion, i.e., the Euler-Lagrange equation for Á states that _p Á =. Since p Á is a constant of the motion, we can use it the rewrite Á _ in the Euler-Lagrange equation for as p Á _Á = c + m` sin ; and thus _Á c Á _ p Á = m` sin c so that Ä + g` " # sin = sin cos p Á m` sin c : Not surprisingly the integration of the second-order di erential equation for is outrageously complex. It turns out, however, that the Hamiltonian formalism allows us glimpses into the gobal structure of general solutions of this equation. 5.3 Hamiltonian The Hamiltonian for the system is obtained through the Legendre transformation H = _ + Á Á _ L p = + 1 ³ m` m` sin pá + m` c sin mg` cos: (6) The Hamilton's equations for (;p )are = p m` = mg` sin + m` sin cos while the Hamilton's equations for (Á; p Á )are _Á = fá; Á = " # p Á m` sin c ; p Á m` sin + c _p Á = fp Á = : It is readily checked that these Hamilton equations lead to the same equations as the Euler-Lagrange equations for and Á. So what have we gained? It turns out that a most useful application of the Hamiltonian formalism resides in the use of the constants of the motion to plot Hamiltonian orbits in 1

11 phase space. Indeed, for the problem considered here, a Hamiltonian orbit is expressed in the form p (; E;p Á ), i.e., each orbit is labeled by values of the two constants of motion E (the total energy) and p Á the azimuthal canonical momentum (actually an angular momentum): p = s m` (E + mg` cos ) 1 ³ sin p Á + m` sin c : Hence, for charged pendulum of given mass m and charge e with a given cyclotron frequency c (and g), we can completely determine the motion of the system once initial conditions are known (from which E and p Á can be calculated). 11