Lagrangian & Hamiltonian Proof of the Beltrami Identity ’s Second Law for a single

1 2 2 2 Rather than starting from the traditional beginnings dL L  L  L L m x  y  z  V(,,) x y z  xx  Chain Rule 2   of Conservation of , Conservation of ii dt ti   xii  x zˆ rx xˆ  yy ˆ  z zˆ or Newton’s Second Law, the Lagrangian (and related d L L  Euler-Lagrange Equation Hamiltonian) formulations of offer a more d L L d L dx L dx L dt dt x x  L x,, x t    ii systematic solution mechanism for mechanical dt x x dt x dt  x dt  t dt ii d V V yˆ systems, especially those with multiple variables. mx    mx   dL L d  L  L dt xi x   xxii  ˆ xyzˆ ˆ  ˆ 1 In , the Lagrangian L is defined as: dt t dt  x  x d V V x i ii my    my   dt y y LTV dL L d L i     xi Product rule d V V dt ti dt xi mz    mz   Kinetic Potential dt zi z energy energy d L L Lx  i  mVr   dti  xi t d L L LL ˆ ˆ ˆ The Euler-Lagrange equation i.e. using rx x  y y  z z   0 Lx i  constant yields the equations of  Cartesian dt xii x txi i VVV   basis vectors V xyzˆ  ˆ  ˆ x  y  z xi or angular coordinate 1707-1783 dx Conservation of Energy for a single particle i.e. we associate a with a gradient x  i Rate of change of position or of , which is entirely i angle with t 1 2 dt L2 mv V(,,) x y z consistent with any conservative field such as or electromagnetism. L1 m x2  y 2  z 2  V(,,) x y z The idea is that the Euer-Lagrange equation 2   extremizes* a quantity called the Action L Hamiltonians & total energy Beltrami Identity L xi k i x t2 i From the penultimate line of the proof S Lxx 1, 2 ,... xxxNN , 1 , 2 ,... xtdt ,  t of the Beltrami Identity: 1 LLL   Joseph-Louis k  L  x  y  z x  y  z d L L Lagrange 1736-1813 In many problems the Lagrangian is independent Lxi 2 2 2 2  1 dti  x t of time, so the Euler-Lagrange equation reduces k2 m x  y  z  V  xmx  ymy  zmz i to the Beltrami Identity. This is often the best starting 2 point for problem solving in . k 1 m x2  y 2  z 2  V Define the Hamiltonian 2   ‘Canonical 2 H p x L 1 2 2 2  ii monentum’ L E 2 m x  y  z  V() E   k i If  0 this means L t LHpi  i.e. we can associate energy with the (negative) x L pxii   i Lxconstant of the constant in the Beltrami Identity xpii  i x i i dH L So we can associate the    Hamiltonian with the total dt t energy. H is constant if L is Sir William Rowan time invariant. Hamilton 1805–1865

*i.e. yields either a maximum or minimum value Mathematics topic handout – Mechanics – Lagrangian and Hamiltonian dynamics Dr Andrew French. www.eclecticon.info PAGE 1 Example #1: dynamics of a Hookean From above: An ellipse defined in plane 2 polar coordinates ˆ L11 mx22 kx 22JJJ θ 22 rr       rˆ x is the spring extension m mr2 m 2 r 3 rd d L L k is the spring constant 1 2  2 GM dA 2 r d dt x x rr r2 Semi- mx   kx i.e. Hooke’s Law, SHM... minor 2 2 is the eccentricity r ba 1- J GM  axis r 2 3  2 of the ellipse This is perhaps a bit contrived to start with the m r r 2b  quadratic Elastic Potential Energy expression! b2 .. But at least this shows consistency with the  1 If orbits are elliptical (Kepler’s First Law) 2 a(1 ) Newtonian method. a aa1122   r  cos  1  Example #2: Kepler’s Laws for a 1  cos r planet orbiting a massive star (i.e. Semi-major axis 2 2a no movement of star due to gravitational a 1  sin  r2 sin  r    attraction to planet, only vice versa) 1  cos 2 a 1  2  dA122 1 J 1 2 2 2 2 2r  2  2 G1   a vrr  rθˆ  v  r  r Plane polar J sin dt m coordinates r   2 GMm ma 1   1 2 2 2 So equal areas are swept out in equal , which is L2 m r  r    Note sign! r 2 GPE is negative JJJcos a 1   Kepler’s Second Law. so –V becomes  d L L r  22  1  2  positive ma11 r ma   mr dt r r  Since equal areas are swept out in equal times, the orbital period is the area of the ellipse divided by the rate GMm 2 of area sweep. This proves Kepler’s Third Law. mr  mr  2 2 r J22 GMa 1   J   1  GM Confirm Newton II, with 2 3 2 2 2 2 rr  2   m r r r m a1  r 22 2 correct formula for   ab  a 1  Ellipse area is r centripetal PP   J2 GM J 2 J 2 dA 1 GM1  2 a ab d L L     2    m2 r 3 r 2 m 2 r 3 m2ar1  22 dt dt     2 234 d 2 22 Pa mr  0 Conservation of J  GMm a 1    GM dt   mr2 J So elliptical orbits satisfy the laws of Kepler’s Third Law: The square of the orbital motion, and yield an expression for total period of a planet is directly proportional to the The last result is the first glimpse of the utility of angular momentum J, which is constant. cube of the semi-major axis of its orbit. the Lagrangian method. The conservation of angular momentum comes straight from the fact that the Lagrangian has no explicit dependence on angle, since KE doesn’t, and Newton’s Law of Gravitation acts purely radially.

Mathematics topic handout – Mechanics – Lagrangian and Hamiltonian dynamics Dr Andrew French. www.eclecticon.info PAGE 2 d L L Applying the Euler-Lagrange equation Example #3: Moving plane*   for m dt x22 x h x g d 2 mx2  m x 1  x 2  tan  mg tan  dt d 2 mx2  m x 1  x 2  tan  mg tan

d Hence: Mx12 mx 0 Mx  mx   0 Which is a statement of conservation dt 12 M of linear momentum in the horizontal Consider a rectangular block of mass m xx21 direction i.e. where there is no m frictionlessly from the top net external force (there is gravity, of a planar wedge of mass M. This also M 2 but this acts downwards) Mx11  mx1  tan  mg tan slides frictionlessly on a horizontal surface. m 22 Assuming the block has not reached the Mx1  mx 1tan  Mx 1 tan   mg tan  horizontal surface, after t seconds the horizontal x M1  tan22  m tan   mg tan  separation is: 1     2 M sin sin 1 sin x  m  mg 2 x  x12  x 1 22 1 tan 2 , tan  cos cos  cos  cos cos h  x  x tan vertical drop of block  12 2 x1  M  msin  mg sin  cos  The Lagrangian for the system is: LTV mg sin cosg sin  cos  x1 22 2 2 2 M 1 1 1 Mmsinm sin T2 Mx12  2 mx  2 mh Mgsin cos  g sin  cos  x   2 22m V mgh Mmsin 1M sin Gravitational potential energy L 1 Mx2  1 mx 2  1 m x  x2 tan 2   mg x  x  tan i.e. constant acceleration motion for both 21 2 2 2 1 2 1 2 block and wedge. If system starts from rest: Limiting case when Mm

g sin cos g sin cos x1  0 x2  gt sin cos d L L Applying the Euler-Lagrange equation xt1  2 xt2  2 M m 1 2  m  sin  1 sin  x  0 for mass M 1 x2  2 gt sin cos dt x11 x M g sin cos g sin cos xt 1 2 xt 1 2 d 2 1 2 M 2 2 2 m 2 Mx1  m x 1  x 2  tan  mg tan m  sin  1 M sin  dt 2 .. which is consistent with no horizontal movement when the angle tends to 90o. Mx11  m x  x2  tan mg tan i.e. in this case the block simply falls vertically!

* Morin, Classical Mechanics problem 6.1 Mathematics topic handout – Mechanics – Lagrangian and Hamiltonian dynamics Dr Andrew French. www.eclecticon.info PAGE 3 Example #4: Hoop & Pulley* Now since the string is inextensible: sin  sin  eq    sin  sin  cos   cos  sin  h  R  h  h0  R eq eq V  MgR  MgRcos  mgh  mgR sin m  cos   g 0 M eq cos22 sin 1 m2 sin m  1  2 M M 2 cos  1  sin The Lagrangian for the system is therefore: Positive root O m 1 22 sin MM Mm  in range of  germane  h LTV m 1 to this system 1 22 sin MM  (M  m )( M  m ) L2  m  M R  MgRcos   MgR  mgh0  mgR 

Hence: Applying the Euler-Lagrange equation: Mass M is fixed to a circular hoop g m M sin d L L    of negligible mass, which can freely rotate  R m M about fixed origin O. Between M and m dt   is a light inextensible string, which runs over 1 22 m Mm  1 ( M  m )( M  m ) 2 M m R  MgRcos   mgR   MgR  mgh0 g MM  a light and frictionless pulley. When the   d R m M system is disturbed by a small amount from M  m R2   MgRsin  mgR  equilibrium it undergoes small dt   g m m  ( M  m )( M  m ) of f. 2  M  m R   MgRsin  mgR  R m M The of mass M (which is g m M sin   g M m tangential to the hoop) is: R m M    R M m vR  Equilibrium occurs when: Since the string is inextensible, This is the equation of simple harmonic motion  2  this must also be the speed of   0 where frequency: So as M becomes much larger  the vertically hanging mass m. f  mM sin  0 than m, mass M will tend to 2 the bottom of the hoop 1 m The total kinetic energy is therefore:  sin  M  i.e.  = 0 which makes sense! Hence frequency of small oscillations is: 112 2 2 2 T22 MR mR Let us consider small angle deviations 1 4 from this point: 1 g M m f  2  The total (gravitational) potential energy R M m of the system is:  eq ,   eq ,  1 sin  m V Mg R  Rcos   mgh eq M

* Morin, Classical Mechanics problem 6.10 Mathematics topic handout – Mechanics – Lagrangian and Hamiltonian dynamics Dr Andrew French. www.eclecticon.info PAGE 4 Now consider small angle approximations: Example #5: with free support* The total kinetic energy of the system is therefore: 1 2 sin  , cos   1  2  112 2 2 M T22 Mx  m xmm  y  2 X g 112 2 2 2 2 2 2 2 M m x  mlsin   ml  cos   0 T22 Mx  m x 2 xl cos  l cos   l sin   M  m x  ml22   ml 10 1   T11 Mx2  m x 2 2 xl cos  l 2  2    2  l 22   221 M  m x  ml    ml    2 ml    0 m The potential energy of the system (taking zero to be M  m x  ml  0 x at the block height) is: V mgl cos l xcos   g sin   0 Hence: A mass m swings from the central underside X of rectangular block of mass l  x  g  0 Hence the Lagrangian for the system is: M m  g  l   ml   0 M. The mass m is connected to X via a x   g  l light inextensible string of length l. LTV l  M  m  m  g M  m 112 2 2 2 g L22 Mx  m x 2 xl cos  l   mgl cos  m Block M is free to slide horizontally without   1  M  . l

Applying the Euler-Lagrange equation: 2  SHM:    f   cos t  It is assumed oscillations of the system are 2 0 small and the is such that M d L L does not collide with the side walls.  g 1 g 1 dt x x f 1 1  m  2  1 m  2 2 l M l M d Let x be the of block M Mx mx  ml cos   0 from the centre point between the side dt x  g  l General solution is CF+ PI walls. The horizontal position of the 2 M m x  mlsin   ml  cos   0 m CF is solution to: pendulum is therefore: x   g  g 1  M  x0  x  At  B x m gcos  t   CF xm  x l sin M 0 d L L x  x  l cos  m  PI is of the dt   xPI  Ccos t  form: 2 m The vertical displacement of the d 2 Ccos  t     g cos  t   mxlcos ml    mxl  sin   mgl sin  M 0 pendulum is: dt m M 00gmm l1 ml 2 Cg  2  MM0 1    ylm  cos mxlcos mxl sin   ml    mxl  sin   mgl sin   g m M l  xcos   g sin   0 ylm   sin Sensible initial ml x( t )  At  B 0 cos t   conditions mM  0,B 0

(tt )  0 cos    xA(0)

* Morin, Classical Mechanics problem 6.14 Mathematics topic handout – Mechanics – Lagrangian and Hamiltonian dynamics Dr Andrew French. www.eclecticon.info PAGE 5 Example #6: Two , one swinging* Applying the Euler-Lagrange equation: Hence:

d L L  cos t   g  dt r r   sin  t    d 2mr  mr2  mg  mg cos h dt 1122 112 r24 r g r 22 r  g cos  1 112 2 2 2 2 r 24 r sin  t    g  cos   t    g r 11 r2sin 2  t    g  2 cos 2  t   Two masses are connected via d L L 24     r light and inextensible string which  dt   112 2 2 run over light and frictionless pulleys. r 22 gsin  t    cos   t    d 2 mr   mgr sin r 11 g21  cos 2  t    cos 2  t   Since in general the left mass may move, dt 22     we must assume the most general g r 1 g221 3 cos  t   expression ( in polar coordinates r,  )    sin 22   for the speed of the right mass. r 2 1 The average value of the harmonic term is: cos t  2 Since the string is inextensible, the Let us assume a small angle rate of increase of r is also the upward 1223 1 1 approximation ** : Hence: r2 g1  2 2  8 g of the left mass. g hr    r So we might expect the amplitude of the pendulum swing to slowly increase and therefore the left mass to slowly rise. h() t  h0  r 12 1 1 2 1 2 1 2 r2 r  2 g11  2    2   4 g  Hence total kinetic energy is:

112 2 2 2 Let us also assume r changes slowly T22 mr  m r  r   compared to the pendulum speed. Total potential energy is (taking The angular equation is therefore SHM zero to be at the level of the pulley) (i.e. r is essentially time invariant compared to  ). V  mgh  mgr cos  is the angular V  mg h0  r  r cos   cos t  amplitude g of the SHM Hence the Lagrangian for the system is:   r LTV

1 2 2 2 L2 m2 r  r  mg h0  r  r cos  2o r001m, g  9.81ms ,  5 , (0)  0,  t  0.01s Verlet solver ** sin  , cos   1  1  2 i.e. constant acceleration 2 motion between fixed time steps

* Morin, Classical Mechanics problem 6.4 Mathematics topic handout – Mechanics – Lagrangian and Hamiltonian dynamics Dr Andrew French. www.eclecticon.info PAGE 6 g Example #7: Bead on a rotating hoop* Applying the Euler-Lagrange equation: 1  Let us also consider small oscillations about: 0  cos 2 R g d L L    ,  1, cos    002 g dt   R d mR2  mR 2  2 sin  cos   mgR sin  dt sin sin 0    sin  0 cos   cos  0 sin   sin  0   cos  0

g 2 cos cos     cos  cos   sin  sin   cos    sin      cos  sin  0 0 0 0 0 R m 

sin cos   sin 0   cos  0 cos  0   sin  0  So equilibrium when: 2 2 2 sin cos  sin 0 cos 0  cos 0  sin 0  sin 0 cos 0 A bead of mass m is free to slide g 2 2 cos0 0 sin cos sin  cos    1  2sin  frictionlessly on a circular hoop R 0 0 0  of radius R. The hoop itself rotates at constant angular speed 1 g 0 cos 2  about a vertical axis. R g 2    cos  sin  sin0  0 R

The kinetic energy of the bead is: 0 0, g 2 2 2 (noting motion is constrained to the circle   sin 0  cos  0    sin  0 cos  0    1  2sin  0  R in the plane, and also circular into the plane)  The ‘bead at the top of the hoop’ g g2 g 2 2 2 is clearly an unstable equilibrium. 11 2   sin 0 22    sin  0    1  2sin  0  T22 m R m  Rsin   RRR Using a small angle perturbation from 1 2 2 2 2 g 2 T2 mR  sin   the bottom of the hoop: 2 2 2 2 2   22      sin 00    sin  R  Taking zero GPE to be the lowest point on the g 2      g 2 hoop, the potential energy of the bead is: R 2 2 2 2 2 2    22    cos 00  sin  R  V mgR(1 cos ) 22 which is SHM of frequency gg 2 2 2   2 2    2 4   sin 0 Hence the Lagrangian for the system is: RR 1 g 2 f 2  22 LTV R 2 2 2gg   2      sin 0    1 2 4        2 2   RR 1 2 2 2 2     L2 mR  sin   mgR cos   mgR which is SHM of frequency:

g 2 f 1 2 (t )  cos 2  ft  2 R22 0

* Morin, Classical Mechanics problem 6.11 Mathematics topic handout – Mechanics – Lagrangian and Hamiltonian dynamics Dr Andrew French. www.eclecticon.info PAGE 7 Example #8: Small oscillations of whirly Applying the Euler-Lagrange equation: Consider small perturbation in r about this value: bungs on a table* J 2 d L L r3   0 g dt   mMg

d rr0 ,1 mr2 0 dt   mr2 J i.e. conservation of J 2 angular momentum J m M r   Mg mr3 J 2 Mass m can slide without friction m  M 3  Mg on a horizontal table. It is connected d L L mr3 1   0  r0  via a light an inextensible string of length dt r r J2 mMg l to mass M through a hole in the table. 112 2 2 3 22m M r  mr  Mgl  Mgr m  M 2 1 r   Mg It is assumed that M only moves vertically. mJ 0 d 2 Mg Mg m  M r  mr  Mg 3  1 r   The kinetic energy of the system is: dt m M0 m M m  M r  mr 2  Mg 112 2 2 2   3Mg T22 m r  r   Mr    m M r0 J Note since the string is inextensible, the   2 which is SHM of frequency: downward velocity of M is r mr J 2 m  M r  mr  Mg 3Mg   24 f  1 Taking zero GPE to be the level of the table mr 2 r( t ) r00  cos 2  ft   m M r0 J 2 V  Mg() l  r m  M r   Mg mr3

Hence the Lagrangian for the system is: Equilibrium when: LTV J 2 112 2 2 2 3 Mg 0 L22 m r  r   Mr  Mgl  Mgr mr 1 112 2 2 2 3 L22 m  M r  mr  Mgl  Mgr J r0  mMg

i.e. of mass m

* Morin, Classical Mechanics pp 240-241 Mathematics topic handout – Mechanics – Lagrangian and Hamiltonian dynamics Dr Andrew French. www.eclecticon.info PAGE 8