Mechanics – A Brief Introduction to Alternative Formulations

Concepts of primary interest: Newtonian Mechanics Lagrangian Mechanics

Generalized Coordinates qi L Generalized Momenta pi  qi Hamiltonian Mechanics Poisson Brackets Sample calculations: Box on an online Object rolling down an incline 1D particle motion Tools of the Trade Total Derivatives Newton’s ‘overdot’ notation

Alternative Formulations of Mechanics – a very quick look The goal of this handout is to introduce two alternative formulations of mechanics that apply well to problems with conservative forces and normal constraint forces. Energy methods proved so fruitful in introductory mechanics that extensions must exist that essentially allow an energy viewpoint to replace the Newtonian force viewpoint. This energy viewpoint is a prerequisite for quantum mechanics so we will just bite the bullet, push the ‘I believe’ button and proceed. In some sense, the energy-view alternatives, the Lagrangian and the Hamiltonian formulations, can be derived or deduced from Newton’s Laws. No such logical development is to be presented here. In fact, the prescriptions in this handout have restricted validity. The whole truth is to found in your junior year mechanics course. Our point of departure is Newton’s Second Law:   ma Fi V() r Fj. [MechAlt.1] i,, forceson j forces do the particle nowork The forces have been separated into those derived from potentials and those do to normal constraints, the classes of forces handled most easily in the Lagrangian formalism. The normal force on a box sliding down an incline and the tension force exerted by a string on a particle tethered to move around a circular path are examples of normal constraint forces. As they are directed at right

Contact: [email protected] angles to the displacement, normal constraint forces due no work.

Our example problems are to be a box sliding down a frictionless incline and an object rolling down an incline without slipping.

The sliding box could be treated using the Cartesian coordinates x and y, but, as the box moves along a line, using two coordinates is over-kill. Instead a coordinate q is chosen – basically the distance from the top of the incline to the box as measured along a parallel to the inclined surface. The restriction to motion along a line means that the motion has a single degree of freedom. In the developments that follow care is to be taken to restrict the number of coordinates chosen to be equal to the degrees of freedom for the motion. A bead might be constrained to move along a curved wire requiring a generalized or unusual coordinate to describe its single-degree-of-freedom position. All coordinates are to be called generalized coordinates are to be represented by qi where the index i assumes j values, the number of degrees of freedom. The total time derivatives of the coordinates dq are called coordinate velocities: q  i . i dt

q y

x 

Example I: A problem with a conservative force, the gravitational attraction due to the earth, and a normal constraint force that does no work and serves to prevent the box from moving into the space occupied by the incline. As the mass of the earth is huge compared to that of the box and the problem is restricted to altitudes near the earth’s surface, the potential energy can be represented as mgy. As illustrated, q is the distance that the box has moved parallel to the incline. It is the generalized coordinate.

The kinetic energy for uniform translation can always be computed from its Cartesian form:

1122ˆˆ1 ˆˆˆˆ 22 T(xyz 22,mvm,x2iy )jzk()xiy jzk) m()xyz (

9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 2 Given the choice of coordinates, it follows that: (, the incline angle is a constant.)

x = xo + q cos; y = yo - q sin xq cosy  q  sin Note that zand z are not needed here. Substituting,

1 2 Tqq(,) 2 mq and Vqq( , ) mgyo  mgq sin

122212 Exercise: Show that 2mx() y z 2 mq  for the example above. The Lagrangian process: The description that follows is specialized to simple cases. The term usually means that it is true for our applications, but may need modification for more general applications.

I.) Choose a set of generalized coordinates (q1, …, qj) where j is the number of degrees of freedom.

II.) Find expressions for the total kinetic energy T and potential energy V in terms of the generalized coordinates and there corresponding coordinate velocities, the qi . The natural variables for a Lagrangian description are the q’s, the qs ' and the time t.

jj LLL  dL dqik dq  dt qqi k t ik11 That is: The natural variables show up nicely in the differential of the function. This introduction does not include cases with time dependent forces so all functions are to be written as forms depending on the q’s and qs ' . When partial derivatives are indicated, each

qi is to be considered to be an independent variable  even independent of its conjugate qi.

III) Form the Lagrangian (function) as:

LqqqqTqqqqVqqqq(11 , ;...;j ,jj ) (11 ,  ;...; ,j ) (11 ,  ;...;j ,j ) . [MechAlt.2]

IV.) For each coordinate, compute the conjugate generalized momentum. L p  [MechAlt.3]. qi qi

Note that the process calls for a partial derivative with respect to qi . The variables qi and qi are to be considered as independent variables as these partial derivatives are computed.

V.) Compute the equations of motion for each coordinate degree of freedom.

9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 3 dL L d jj    where qqi k  [MechAlt.4] dt qii q dtik11  q i  qk  t In this last expression, a total time derivative is to be computed. Note the mixing of the d d conventional /dt and the Newtonian dot to represent total time derivatives. The /dt form is used for emphasis. Example I: Box on the incline: I.) There is one degree of freedom and the single coordinate q, the distance moved down the incline is chosen to be the coordinate.

1 2 II.) It has been shown that: Tqq(,) 2 mq and Vqq( , ) mgyo  mgq sin .

1 2 III.) Lqq(,) = Tqq(,) Vqq (,) 2 mq mgyo mgqsin . L IV.) p mq q q

dL d L V.) mq mq m gsin q g sin . The box accelerates down dt q dt qi along the incline with a linear acceleration g sin as expected.

Example II: An symmetric object of radius R with moment if inertia I rolls down the same incline without slipping. As before there is one degree of freedom, and the net angle of rotation  is to be chosen as the one generalized coordinate. In terms of this coordinate, the center of mass coordinates of the rolling object are: X = Xo + q cos; Y = Yo - q sin and its components of velocity are: Xq   cos and Yq  sin where q = R. The kinetic energy decomposes into the translational energy of the center of mass and the kinetic energy of rotation about the center of mass.

1122  TMXYI2[]2CM . The potential energy depends on the center of mass position only. V = M g Y. Example II: Rotationally symmetric object rolling w/o slipping on the incline  the distance

that the center of mass has moved parallel to the incline is q = R  . I.) There is one degree of freedom; we choose the single coordinate , the net angle of rotation.

1122 2  II.) It follows that: TMRI(,) 22CM  andVVMgR(,) 0  sin  .

9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 4 1122 2 III.) LMRIVMgR(,) 220  sin  . L IV.) p MR2  I  .   

Note that the momentum conjugate to the angle  has the form of an angular momentum, but it is not just I  , the angular momentum due to the rotation about the center of mass. The contribution MR2  is the angular momentum associated with the center of mass motion relative to the instantaneous point of contact (or any other point of the surface of the incline). The angular momentum of a rigid rotating body decomposes into the angular momentum associated with the center of mass motion plus that due to the body’s rotation about the center of mass. Adopting a single coordinate to describe the rotation and translation requires that the momentum conjugate to  represents the combination of the coupled motions. 2 dL d 22L gMRsin  V.) MR I MR I MgRsin  .   2 dt dt R MR I 2  2 1 If the object is a uniform cylinder, I = ½ M R and   3 gR sin .

Center of Mass Theorems:

Exercise: Repeat the object rolling down the incline problem, but choose q, the displacement of the center of mass along a line parallel to the incline. Find expressions for the kinetic energy T(q, q ) and the generalized momentum pq. Discuss the result for pq given that it represents the total mass- motion and that the rotation and translation are coupled.

The Hamiltonian Process: The Hamiltonian picture is built on the Lagrangian picture by a process L involving the conjugate momenta computed as: p  . The Hamiltonian (function) is defined to qi qi

j jj LLL be Hqpt(,ii ,) pq kk Lqqt (,,) ii . As dL dqik dq  dt ,  qqi k t k 1 ik11

jjjj   LLL dH(, qii p ,) t d pkk q dL(,,) q ii q t pkk dq q k dp k  qqt dqi dq k dt . kki111i k1k L As ppq i , i qi

jjj LL dH(, qii p ,) t pkdqkqk dp k dqi pm dqm dt qti ki11m1 jj  LL dH(, qii p ,) t q k dp k qt dqi dt k11i i

As they appear as the differentials, qi, pi and t are the natural variables for the hamiltonian. If pk is

9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 5 varied and all other variables are held constant, we find: H dH(, qii p ,) t q k dp k qk others held constant pk

If qk is varied and all other variables are held constant, we find:

LHLdL dH(, qii p ,) t dqi    p k qqqqi kkdt   k Note that the Euler Lagrange relations were used to complete the last step.

The natural variable set for the Hamiltonian is the coordinates, their conjugate momenta

and time { … qi … ; … pi … ; t}rather than the coordinates, their conjugate momenta and

time q’s, the qs ' and time t { … qi … ; … qi … ; t}. This introduction does not include examples with time dependence so all functions are to be written as forms depending on the

q’s and p’s. When partial derivatives are indicated, each qj and pi is to be considered to be

an independent variable  even independent of its conjugate pj or qj.

I.) Execute the Lagrangian process to the point that you have chosen a set of generalized coordinates, computed the set of conjugate momenta and expressed the Lagrangian as

Lq(11 , q ;...; qj , qj ) .

j   II.) Compute the Hamiltonian as: Hqpt(ii , ,) pqkk Lqqt( ii , ,). [MechAlt.5] k 1

III) Express the result as a function of the natural variable set for the Hamiltonian form of mechanics. Each appearance of a qi by must be replaced by expressing it in terms of pi and the q’s.

IV.) For each coordinate qi, evaluate the two Hamilton’s equations: HH qpiiand  [MechAlt.6] piiq The process yields 2j first order differential equations describing the motion rather than the j second order differential equations that the Lagrangian process yields. Newton’s Second Law and the Lagrangian process generate a set of second order differential equations describing the motion of the system. The Hamilton’s process yields pairs of first order equations that must be combined to recover the Newton-like second order equations.

9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 6

Example 3: Consider a particle constrained to move along the x axis in a potential V(x). Specialize each result to the special case V(x) = ½ k x2 as an exercise..

I. There is one degree of freedom and the generalized coordinate chosen is x. It follows that

1 2 the Lagrangian is: Lxx(,) Txx (,) Vx ()2 mx  Vx () and that the momentum

conjugate to x is p L mx . x x II. The Hamiltonian becomes:

11221 212 HxppxLxxmxxmxVx(,xx ) (,)  22 ()  mxVx   ()  2 mx 2kx which is just the energy. Expect the Hamiltonian to be the energy for our time- independent applications. III. Transform the Hamiltonian by eliminating each coordinate velocity in favor of

expression involving just the coordinates and momenta. As px = m x ,

2 px Hxp(,x )2m Vx (). IV. Hamilton’s equations become:

HHVpx  xand pFxx kx. pmx x x

Exercise: Apply the Hamiltonian procedure to each of the incline examples. Combine the Hamilton’s equations in each case to recover the familiar Newton’s Law form.

NOTE: The hamiltonian is the total energy whenever it is time independent. H Constants of the Motion: The relation pi  shows that pi is a constant of the motion of the qi

Hamiltonian is independent of qi. (Recall that we are restricting our attention to Hamiltonians that do not depend explicitly on time.)

Exercise: The total time derivative of the hamiltonian can be expressed as: dHjj H dq H dp H jj H H H ik    qpik dtik11 qik dt p dt t ik  11 q ik p t

9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 7 dH H Show that /dt = 0 if the hamiltonian has not explicit time dependence ( /t = 0).

Time Dependence and Poisson Brackets: In the Hamiltonian formulation, every dynamical quantity can be expressed as a function of the coordinates, their conjugate momenta and time { … qi … ; … pj … ; t}. That is: every quantity can be represented as: Q(… qi … ; … pj … ; t). The time derivative of the quantity is therefore: dQjj Q dq Q dp Q j Q H j  Q  H Q ik      dtik11 qik dt p dt t i  1 qiikk p k  1  p  q t This expression is somewhat unwieldy so we introduce a notation, the Poisson Bracket.

jj n FG   F  G Def : {,}FG   ik11qpii   p kk  q Using this notation, the time derivative of a general dynamic quantity Q is: dQ Q {,QH } . dt t The most fundamental bracket is that for a coordinate and one of the conjugate momenta. It vanishes except for the case of a coordinate and its conjugate momentum.

jjjj qpmn  qp mn {,}qpm n  mi ni 00mn iki111qpii  pq kk k1

Samples: {x, px} = 1; {y, px} = 0.

dQ Q Exercise: Use the relation {,QH } to show that the hamiltonian is a constant of the dt t motion as long as it has no explicit time dependence.

Skim the next paragraph. Review it carefully as each topic arises in your study of quantum mechanics.

The Relation to Quantum Mechanics: In the coordinate representation of Schrödinger quantum mechanics, one forms the operator for a dynamical quantity Q(… qi … ; … pj … ; t) by replacing each momentum pk by -i times a partial derivative with respect to the t conjugate coordinate qk.

9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 8 ˆ  QQ(... qi ,..;..., i ,... ; t ) q j ˆ ˆ  Following this prescription, x  x and pix  x . The z component of angular momentum can be ˆ ˆˆ  chosen as a more complex example: Lz = x py – y px; so Lppixiyz xyyxyx  . The

2 px operator that is the granddaddy example of them all is the 1D hamiltonian: Hxp(,x )2m Vx ().

2 2 Hxp(, )px Vx () Hˆ 1 i  Vx ()2 2 Vx () x 2m 2m  x  2m x2

The classical Poisson brackets map to commutators in quantum mechanics. The commutator of two dynamical variables is i times the Poisson bracket of the corresponding Hamiltonian dynamical variables. ˆˆ {,}qpm n mn  qpm , n   i mn A particular example is: ˆ {,x pxpxx } 1 ,  i

Conservation or constants of the motion:

L dL  L dL p pi    i   qi dt qii q dt qi

The momentum pi is a constant of the motion if the Lagrangian does not depend on qi, the coordinate conjugate to that momentum. For the planar orbit problem : Lr(, ,, r ) ½ mr [222 r  ] V () r ,

2  there is no dependence of the angular coordinate  so the conjugate momentum pmr   is a constant of the motion. 

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Question: Two masses m1 and m2 on a horizontal straight frictionless track are connected by a spring of spring constant k, as shown in the figure to the left. The spring is

 initially at its equilibrium length. If x1 and x2 are the displacements of the masses from their equilibrium positions, the lagrangian L for the system is given by which of the following? The dot denotes differentiation with respect to time.

Note that the generalized coordinates chosen (see the figure) are the displacements of each mass from equilibrium. This choice is common. The Lagrangian is to be the kinetic energy minus the potential energy. The kinetic energy is the sum of the kinetic energies of the two particles, T (or K) =

22 2 ½½mx11 mx 2 2, and the potential energy is ½ times the spring constant times (x2 – x1) , the stretch

2 of the spring squared.  U = ½ k (x2 – x1) . Note that it helps to roll the words around in your mind as you attempt to decipher a problem.

22 2 Conclusion: Lx(,1212 x ,, x x ) ½ mx 11  ½ mx 22  ½( k x 2 x 1 ) B

22 Note that four of the candidates identify the kinetic energy as: ½½mx11 mx 2 2, and three identify 2 the potential as ½ k (x2 – x1) . Only two have the negative sign for L = T – U. Possibility (E) includes the awe-inspiring reduced mass; it has the wrong sign for the potential. In addition, (E) depends only on the relative position and relative velocity. That is: the center of mass motion has

9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 10 2 2 been suppressed. For example, if q = x2 – x1, then (E) is ½  q + ½ k q . Forgetting that it should be

½  q 2 - ½ k q2, there is the issue of the kinetic energy.

22  2 2  ½½mx11 mx 2 2 = ½(mmX12 ) + ½  q where X is the center of mass velocity.

Two particle systems are often best treated using center-of-mass and relative coordinates. In part, this is because the kinetic energy decomposes into the kinetic energy of the center of mass motion plus the kinetic energy associated with motion relative to the center of mass.

  Let V be the velocity of the center of mass of a system of particles, vi be the velocity of particle i   and ui be the velocity of particle i relative to the center of mass.

22 2 KKKtotal CM relative or ½ mvMVmi i ½ ½ iu i ii For a system with only two particles, the reduced mass can be used to get the special form:

22 2  222 ½½½½||½½m11 v m 2 v 2 MV v 2 v1  MV vrel

mm12   where M = m1 + m2,   and vvrel  21 v mm12 Note that for rigid bodies the motion decomposes into the center of mass motion plus 22 2 rotation about the CM. KKKtotal CM relative or ½ mvMVIi i ½  ½ CM  i

Tools of the Trade: Review - - Total Derivative: Given a function of several variables such as f(x,y,t), the total rate of change of the with respect to one variable is needed when the other arguments have an implicit dependence on that variable. That is: for f(x(t),y(t), t), df f dx f  dy f   [MechAlt.7] dt x dt y dt t Note that this expression includes all the variation in the value of f(x,y,t) when t varies.

Dot Notation: Dot  total derivative w.r.t. time Time derivatives are ubiquitous leading physicists to develop a shorthand notation to represent the total derivative with respect to time, the dot notation.

9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 11 df(, x y ,) t f f f ffxy [MechAlt.8] dt x y t

An "overdot" is a raised dot appearing above a symbol most commonly used in mathematics to indicate a (total) derivative taken with respect to time (e.g., dx x  dt ). The expression is voiced " dot," and was Newton’s notation for derivatives (which he called “fluxions”) Eric W. Weisstein. "Overdot." From MathWorld – Wolfram Web Resource. http://mathworld.com/Overdot.html

Problems 1.)

2.)

3.)

4.) An exercise in blindly computing partial and total derivatives: Mechanics can be cast into the Lagrangian alternative to the Newton form. For our purposes, the Lagrangian function is defined as a function of the coordinates, their times derivatives (the coordinate velocities) and time. The function often assumes the form of T, the kinetic energy, minus V, the potential energy, but, in some cases, its form is more general. For central force problems, ones in which the potential depends on r only, conservation of angular momentum restricts the motion to a plane so one can adopt polar coordinates {r, } leading to the associated coordinates velocitiesr,. Without explanation, the

1 222 Lagrangian is Lr(, ,, r )2 m r r V() r = T - V.

9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 12 dL  L In this formalism, the equations of motion are:  where q represents first r and then  dt q  q and q represents first r and then .

LLL  LdL  d  L d  Compute: ,,, , and. What is  ? Express it in ‘shorthand’. rrdtrdt     dt

What does r represent? Contrast r and ar. dL L dL L  V     2 Partial Answers:  20rr ; mr m r. dt  dt r r  r Note that the rate of change with respect to distance moved in a direction of a potential function is the negative of the component of the force in that direction associated with that potential. The rate of change with respect to an angular coordinate change of a potential function is the negative of the torque associated with that potential about that angular axis. A torque is the generalized force expected when the generalized coordinate is an angle. In polar coordinates,  2 ˆˆ  ˆˆ ar rrrr  2  arar  .

5.)

References:

1. K. F. Riley, M. P. Hobson and S. J. Bence, Mathematical Methods for Physics and Engineering, 2nd Ed., Cambridge, Cambridge UK (2002).

2. The Wolfram web site: mathworld.wolfram.com/

3. Donald A. McQuarrie, Mathematical Methods for Scientists and Engineers, University Science Books, Sausalito, CA (2003).

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