THERE ARE BASICALLY TWO approaches to conservative mechanics, due principally

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1 CHAPTER 2 LAGRANGE S EQUATIONS THERE ARE BASICALLY TWO approaches to conservative mechanics, ue principally to Lagrange an Hamilton respectively Each has avantages over the other, although the Lagrangian approach is almost always important as a first step in fining the Hamiltonian framework Newton s escription of mechanics really requires one to be working in Eucliean (Cartesian) coorinates, an Lagrange 1 foun a very elegant approach which allows one to use ifferent coorinate systems In this chapter we escribe this approach; this approach has the ae avantage of making simple constraints easy to hanle 21 LAGRANGE S EQUATIONS OF MOTION To stuy the motion of a system of N particles escribe in Chapter 1, one starts with Newton s secon law which states: ṙi = 1 m i F tot i, (21) where F tot is the total force acting on particle i (i = 1,, N ), an r i i is the position vector of particle i This is a system of 3N secon orer ifferential equations in 3N variables (x 1, y 1, z 1,, x N, y N, z N ) The space of all possible configurations is therefore R 3N an is calle the configuration space of the system One coul alternatively be consiering N particles moving in the plane R 2, an the configuration space woul then be R 2N or N particles moving on a line, in which case it woul be R N Recall that the total number of coorinates require is calle the number of egrees of freeom of the system In this book, we eal only with conservative forces, so that we assume there is a function on configuration space V (r 1,, r N ) calle the potential (energy), such that F tot = i i V, where as usual i V = V / r i = ( V / x i, V / y i, V / z i ) T 1 Joseph-Louis Comte e Lagrange live , an publishe the first eition of his great treatise Mécanique Analytique (Analytical Mechanics) in 1788, just 100 years after Newton s Principia (1687) 23

2 24 LAGRANGE S EQUATIONS EUCLIDEAN COORDINATES Although one of the main avantages of Lagrange s approach over Newton s is the flexibility in the coorinate system use, we begin by introucing Lagrange s approach in Eucliean coorinates Lagrange foun he coul formulate Newton s laws in a manner that i not epen on using Eucliean coorinates, an he was le to introuce a function equal to the ifference between the kinetic an potential energies, now calle the Lagrangian In Eucliean coorinates this is, L(q, v)=t (v) V (q) (22) where q = (r 1,,r N ) R 3N v = (v 1,,v N ) R 3N T (v) = 1 m 2 i v i 2 i an V is the potential energy The set {(q, v)} of all possible configurations 2 an velocities is calle the Lagrangian phase space an the Lagrangian is therefore a function on this phase space L : R 3N R 3N R DEFINITION 21 Given any smooth function L(q, v) on the phase space, Lagrange s equations are t ( ) L v = L q (23) This is a vector expression with 3N components, so consists of 3N ifferential equations, one for each component of q, each of secon orer in q, so t ( L v i ) = L q i i = 1,2,,3N (or i = 1,,2N if the motion is in a plane, etc) In practice, when solving a problem, one writes the 3N equations out one by one Since v= q, Lagrange s equations are often written t ( L q ) = L q (24) or one by one, t ( L q i ) = L q i i = 1,2,,3N 2 Note that the symbol q for configuration is not in bol face we o not think of configurations as vectors, but just as coorinates Aition of configurations oes not (usually) have any meaning On the other han velocity q is a vector CM JM February 14, 2015

3 21 Lagrange s equations of motion 25 REMARK 22 Lagrange s equations are also known as the Euler-Lagrange equations This is because there is an important interpretation of Lagrange s equations as a variational problem, the so-calle principle of least action, see Section 27 EXAMPLE 23 (Lagrangian for a simple spring) Consier a particle of mass m on the en of a spring of spring constant k, an let x R be the extension of the spring Since the configuration of the system is etermine by a single variable x, it is a system with 1 egree of freeom The Lagrangian phase space is just R R in this example, so q = x an v= v = ẋ Then Consequently, the Lagrangian is simply an Lagrange s equation requires: an T (v)= 1 2 mv 2, V (x)= 1 2 k x2 L(x, v)=t V = 1 2 mv k x2, t so that Lagrange s equation gives us ( ) L = v t (mv)=m v = m x, L x = k x m x= k x, which is (of course!) the same as the equation one obtains irectly from Newton s secon law (as F = k x for a spring) This example is inicative of the general equivalence: PROPOSITION 24 In Cartesian coorinates, Lagrange s equations (23) are equivalent to Newton s equations (21) PROOF: This is an easy calculation: for L/ v i = m i v i = p i, an L/ r i = gra i V Lagrange s equation is then just ṗ i = gra i V as require REMARK 25 Notice that L/ v=p, the momentum, an L/ q = grav which is the force We will return to this later EXAMPLE 26 Fin the equations of motion of a particle of mass m moving in the plane with potential energy V (x, y)= 1 2 (x2 + y 2 ) February 14, 2015 CM JM

4 26 LAGRANGE S EQUATIONS Solution: We nee to fin the Lagrangian The kinetic energy is 1 2 m(ẋ2 + ẏ 2 ), so that Lagrange s equations give us: for x: L(x, y, ẋ, ẏ)= 1 2 m(ẋ2 + ẏ 2 ) 1 2 (x2 + y 2 ) t ( L ẋ ) = (mẋ) = m x, t an L x = x Lagrange s equation for x is then m x = x for y: this is analogous t ( L ẏ ) (m ẏ) = = m y, t L y = y The equations of motion are therefore m x= x, m y = y (25) EXAMPLE 27 (Two boy problem) Consier 2 particles of masses m 1,m 2, moving in space uner a mutual gravitational interaction (a system with 6 egrees of freeom) Let r 1, r 2 be the position vectors of the two particles The gravitational potential energy is etermine by their mutual istance: V (r 1, r 2 )= Gm 1m 2 r 1 r 2, where G is Newton s universal gravitational constant The kinetic energy is so that the Lagrangian is given by Then for j = 1,2 we have T (v 1, v 2 )= 1 2 m 1 v m 2 v 2 2, L(q, v)= 1 2 m 1 v m 2 v Gm 1m 2 r 1 r 2 (26) ( ) L = m j vj = t v ṗ j j L = Gm 1m 2 r j r 1 r 2 3 (r j r k ), CM JM February 14, 2015

5 22 More general coorinates 27 where k j Lagrange s equations are therefore p 1 = Gm 1m 2 r 1 r 2 3 (r 1 r 2 ) p 2 = Gm 1m 2 r 1 r 2 3 (r 2 r 1 ) which are of course the same as Newton s equations 22 MORE GENERAL COORDINATES It may seem that we have gaine little in rewriting Newton s equation in Lagrange s form However the power of Lagrange s insight is in the following Suppose we no longer use Eucliean (Cartesian) coorinates, but some others such as polar coorinates, or spherical polar coorinates (which might escribe particles moving on the surface of a sphere) classically such non-eucliean coorinates were calle generalize coorinates then Newton s laws no longer hol, at least not in the form m q= grav Lagrange wante to write the equations of motion in a manner that i not epen upon using Eucliean coorinates, an we now procee to show how Lagrange s equations o this There is another important avantage to Lagrange s equations, which is that it can eal with many constraine systems, as we will see in the next section First we go through the argument with an example, an then procee to aapt the example to eal with the general setting EXAMPLE 28 Let us reo example 26 but in polar coorinates We have V (x, y)= 1 2 (x2 + y 2 ), so in polar coorinates, V (r,θ)= 1 2 r 2 (it happens to be inepenent of θ) The kinetic energy is (see Problem 18), T = 1 2 (ṙ 2 + r 2 θ 2 ) Thus, Lagrange s equations give us: L(r,θ,ṙ, θ)= 1 2 (ṙ 2 + r 2 θ 2 ) 1 2 r 2 for r : t ( L ṙ ) = (mṙ ) = m r, t an (on t forget the r term in the kinetic energy) Thus m r = r θ 2 r L r = mr θ 2 r for θ: t ( L θ ) = t (mr 2 θ)=2mr ṙ θ+ mr 2 θ, February 14, 2015 CM JM

6 28 LAGRANGE S EQUATIONS (using the prouct rule) On the other han L 2 θ = 0, giving mr θ+2mr ṙ θ= 0 Thus in polar coorinates the equations of motion are, { m r = mr θ 2 r r θ = 2ṙ θ (27) The question is, are these the equations of motion arising from Newton s laws (as in Example 26)? The answer is, of course, affermative See Problem 22 Recall that kinetic energy is efine using Cartesian coorinates, an for this reason in all cases, we must start with Cartesian coorinates Let us write x=(x 1, x 2,, x 3N ) to escribe the Cartesian coorinates of all the N particles in the system Next write q = (q 1,, q n ) for the new coorinates (polar, spherical, whatever, ) for a system with n egrees of freeom Then write ψ for the transformation converting non-cartesian coorinates to Cartesian ones: Thus ψ is a smooth 3 map ψ : R n R 3N x=ψ(q) For example, using polar coorinates for a single particle moving in the plane, we have x= r cosθ an y = r sinθ an z= 0, so that Here 3N = 3 an n= 2 x r cosθ x= y =ψ(r, θ) = r sinθ (28) z 0 To use Lagrange s equations in the new coorinates q requires us to erive the expression for kinetic energy in the new coorinates, for example using ṙ, θ in polar coorinates (as in Problem 18) Now using the coorinates q to escribe the motion means that we will use q=( q 1, q 2,, q n ) T as velocity vector An we want to express the kinetic energy in terms of q rather than the v=ẋ use in its efinition If x=ψ(q) an q epens on time t, then x(t )=ψ(q(t )) an the velocity is relate by ifferentiating with respect to t : x=dψ(q) q, (29) by the chain rule, where Dψ(q) is the Jacobian matrix of ψ at q It is the 3N n matrix, efine by (Dψ) i j = ψ i q j 3 smooth means ψ can be ifferentiate as many times as we want; in practice 2 or 3 times ifferentiable is sufficient CM JM February 14, 2015

7 23 Constraine systems 29 (That s the i th row an j th column) For example, for polar coorinates we have from (28) that cosθ r sinθ Dψ= sinθ r cosθ (210) 0 0 The relation between q an ẋ is then x=dψ(q) q= cosθ r sinθ ( ) r cosθ r θ sinθ sinθ r cosθ ṙθ = r sinθ+ r θ cos θ ; that is, ẋ= ṙ cosθ r θ sinθ an ẏ = ṙ sinθ+ r θ cosθ (an ż= 0), as we saw above REMARK 29 Notice that q is not consiere to be a vector, although it oes have several components it is meaningless to a polar coorinates, or multiply them by a scalar On the other han, the velocity q is a vector: aition an scalar multiplication of velocities is meaningful That sai, we o write x (Cartesian coorinates) as a vector The next step is to write the kinetic energy in terms of q, similar to that for polar coorinates in the example above However, since the approach is the same for constraine systems, we postpone the general case until after escribing constraine systems 23 CONSTRAINED SYSTEMS Many systems involve constraints of some kin If there are no constraints, then each particle is free to move in 3-imensions, so with N particles there are 3N egrees of freeom Constraints ecrease the number of egrees of freeom Here we list some of the principal type of constraint: 1 A particle on a plane, for example a smooth table Here the constraint is lying on the table, an in Cartesian coorinates imposes the conition z = 0, so reucing the number of egrees of freeom from 3 to 2 The system coul be parametrize by (x, y), or using polar coorinates, by (r,θ) 2 In a planar penulum, firstly the bob is constraine to move in a vertical plane, so reucing the number of egrees of freeom from 3 to 2, but then the inelastic ro imposes another (inepenent) conition that the istance of the bob from the fixe point of the ro is constant, equal to l, in other wors one has x 2 + y 2 = l 2 (in appropriate coorinates) The en result is that it is sensible to use just the angle θ as configuration variable, an there is just the one egree of freeom 3 A bea moving on a smooth wire: this has just 1 egree of freeom, an the bea is constraine to the wire by appropriate equations The system coul be parametrize by the variable parametrizing the curve 4 A cyliner rolling own an incline plane 5 A ball rolling on a table top February 14, 2015 CM JM

8 30 LAGRANGE S EQUATIONS The constraint in each of the last two examples involves velocity: the constraint of rolling is expresse by saying that the point of contact at any instant has zero velocity This type of constraint is harer to eal with than the constraints which only involve position in their efinition, like those numbere 1,2,3 in the list above These latter constraints which arise as restrictions on the possible configurations an are calle holonomic constraints The harer ones involving the velocities as well as the configurations are therefore calle non-holonomic constraints We will just be ealing with holonomic constraints A simple example of a constraine system is the plane penulum as escribe above: while the bob moves in the plane, it is constraine by the inextensible ro to lie on a circle Similarly in the Atwoo machine (see Problem sheet 1), the two masses are constraine to lie at a fixe istance apart (istance as measure up over the pulley an own again equal to the length of the inextensible string) In fact almost all the examples we are intereste in have some constraints, an in orer to procee further we must therefore inclue these constraints in the Lagrangian formalism As in the previous section, we will choose coorinates aapte to the problem, an which moreover respect the constraints For example, for the plane penulum we only nee the one coorinate θ, an the expression in Cartesian coorinates is then x = lsinθ an y = lcosθ (see the iagram on p 33) For a general (constraine) system, we introuce coorinates q = (q 1, q 2,, q n ) The number n is calle the number of egrees of freeom of the system So for example, the penulum with q = θ has just 1 egree of freeom As in the previous section, we will parametrize the (allowe) configurations by a map ψ, so ψ : R n R 3N, with x=ψ(q) The (allowe) velocities are then given by v=dψ(q) q, (211) with q=( q 1, q 2,, q n ) T (a column vector) Again, Dψ is the Jacobian matrix of ψ, but now it is a 3N n matrix (3N rows an n columns) ( ) lsinθ For example, for a simple penulum the parametrization is x=ψ(θ)=, so lcosθ the velocity is v=dψ(θ) θ= ( ) lcosθ θ, lsinθ an geometrically one can see that this velocity is tangent to the circle at the point ψ(θ) Let us write Q R 3N for the image of ψ; that is Q is the set of configurations that are allowe by the constraints it is calle the configuration space We are going to assume that Q is a regular subset 4 of R 3N, which happens when the Jacobian matrix Dψ always has rank equal to n (so the linear map u Dψ u is always injective) The space of allowe velocities at x=ψ(q) is the space T x Q = {Dψ(q) q q R n } Vectors in T x Q are calle tangent vectors to Q at x (an in geometry, the space T x Q is calle the tangent space to Q at x = ψ(q)) The allowe velocities are calle virtual velocities in some texts 4 also calle a submanifol CM JM February 14, 2015

9 23 Constraine systems 31 In other wors, T x Q is the linear subspace of R 3N equal to the image of the Jacobian matrix Dψ(q), where x=ψ(q) It is the space of allowe velocities, by equation (211) DEFINITION 210 A constraint force F on a system is one which is orthogonal to the allowe velocities of the system: N F j ẋ j = 0 j=1 Here the vector F j is the effect of the force on particle j In terms of q this conition is, F Dψ q ( q)=0, for all q R n, where now F is written as a vector in R 3N Or again this is equivalent to F T Dψ(q)= 0, (prouct of row vector an matrix) The reson for this efinition of constraint force is the following property PROPOSITION 211 A constraint force oes no work on the system, so in particular oes not alter the kinetic energy PROOF: The kinetic energy of the system is T = N 1 j=1 2 m j v j 2 Then ue to the force F we have, T N t = m j v j v j = v j F j j=1 (F j = m v j by Newton s secon law) So it follows from the efinition of constraint force that T /t = 0 This fact is behin why Lagrange s metho works for constraine systems: the constraints o not effect the kinetic energy, an Lagrange s metho is base on kinetic energy EXAMPLE 212 Let us show irectly that the tension in the string in the Atwoo machine (see Problem sheet 1) is a constraint force Now, two masses are moving with the same spee in opposite irections, write v m = v M Both are vertical vectors so write v M = ve 3 The tension in the string is the same for each of the masses, say T e 3 (T > 0 as both act upwars) Then (T e 3 ) v M + (T e 3 ) v m = (T e 3 ) (v v)e 3 = 0 Therefore the tension is a constraint force Note that the tension in the (inextensible) string is what is forcing the spees of the masses to be equal: if the string were elastic, then the tension woul not be a constraint force, an inee kinetic energy woul be converte into elastic energy February 14, 2015 CM JM

10 32 LAGRANGE S EQUATIONS The fact that the velocities are equal an opposite in this example is what is meant by allowe velocities in the efinition Similarly in the plane penulum, any allowe velocity is tangent to the circle of motion (which is perpenicular to the tension in the ro) More generally, if a system of N particles moves in a way so that all the inter-particle istances are constant (for example they are all connecte by light inextensible ros) then the forces maintaining their constant separations (tensions in sai ros) are constraint forces 24 LAGRANGIANS IN GENERAL COORDINATES Whether it is just that the problem suggests using non-cartesian coorinates, or that the system is constraine is some way forcing a use of other coorinates, in orer to escribe the state of the system one often nees to introuce some ifferent (non-cartesian) coorinates We will enote the new coorinates by q = (q 1, q 2,, q n ) For example, for the plane penulum one only nees a single variable θ, so in that case q = θ If we have two particles in the plane, an want to use polar coorinates, then we woul have coorinates q = (r 1, θ 1, r 2, θ 2 ) The Cartesian coorinates x will epen on the chosen coorinates q through some (ifferentiable) map ψ : R n R 3N, with x=ψ(q) The number n is the number of egrees of freeom of the system In these new coorinates, the new Lagrangian is as before equal to kinetic - potential The new potential is simply obtaine by writing V in terms of q instea of x, an the new kinetic energy requires writing v (or ẋ) in terms of q as in (211) In other wors we substitute (x, ẋ)=(ψ(q), Dψ(q) q) That is, L(q, q)=t (ψ(q), Dψ(q) q) V (ψ(q)) We call this the Lagrangian on the configuration space THEOREM 213 Lagrange s equations on the configuration space are equivalent to Newton s laws with the constraint There are several ways of proving this theorem, but the most beautiful is the one using Hamilton s principle of least action an is given in 27 The basic example of a constraint force is any set of forces on the particles that maintain the istances between them, so that the system behaves as a rigi boy EXAMPLE 214 (The plane penulum: Lagrangian approach) CM JM February 14, 2015

11 24 Lagrangians in general coorinates 33 Traitionally, the mathematical penulum consists of a particle of mass m the bob attache to one en of a light inextensible ro of length l, the other en of which is attache to a fixe point The penulum is free to swing in a plane We can escribe the position of the penulum by the angle θ the ro makes with the vertical Figure 21 [Here light means the ro has negligible mass an inextensible means that any elastic effects are ignore] θ l m Figure 21: The plane penulum As we saw in Chapter 1 (Example 111) gravitational force is conservative an the potential epens on the height of the particle above some fixe level The height above the lowest point of the circle is here l(1 cos θ), so V = mgl(1 cos θ) Let us fin the Lagrangian in terms of θ an θ The potential energy is V, an for the kinetic energy we nee that the Eucliean coorinates corresponing to θ are ( ) ( ) x lsinθ x= = y lcosθ The velocity is then, so the kinetic energy is The Lagrangian is therefore Now we write Lagrange s equations (23) ( ) ( ) ẋẏ lcosθ θ x= = lsinθ θ, T ( θ)= 1 2 m ẋ 2 = 1 2 m(ẋ2 + ẏ 2 )= 1 2 ml2 θ 2 L(θ, θ)= 1 2 ml2 θ 2 mgl(1 cos θ) t ( ) L θ t (ml2 θ) = L θ Simplifying (since m an l are constant) gives = mglsinθ θ= g l sinθ This is inee the equation of motion one erives using Newton s law using both the gravitational force an the force ue to the tension in the ro (the constraint force: it is always perpenicular to the velocity) Thus, Lagrange s equations continue to hol true even for this constraine system February 14, 2015 CM JM

12 34 LAGRANGE S EQUATIONS Notice that L/ θ = ml 2 θ, which is the angular momentum of the penulum about the origin, an L/ θ= mglsinθ, which is the torque of the gravitational force about the centre of the circle, so in this case Lagrange s equation is saying that torque is equal to rate of change of angular momentum KINETIC ENERGY AND THE MASS-INERTIA MATRIX Since the kinetic energy is always quaratic in the velocity vector q (whatever generalize coorinates we are using), it can be written T (q, q)= 1 q T K(q) q, where q is a column vector an K(q) is a symmetric matrix epening on q, an calle the mass-inertia matrix (also the kinetic 2 matrix) For example, in polar coorinates the kinetic energy of a particle in the plane is T = 1 2 m(ẋ2 + ẏ 2 )= 1 2 m(ṙ 2 + r 2 θ 2 ) The mass-inertia matrix is then ( ) m 0 K= 0 mr 2, so that with q=(ṙ, θ) T one has inee 1 q T K q= m(ṙ 2 + r 2 θ 2 ) θ l m (x, y) FIGURE 22: The elliptic penulum EXAMPLE 215 (The elliptic penulum) This consists of a light ro of length l with a bob of mass m at one en, hanging in a vertical plane, with its other en not fixe buts free to slie on a horizontal axis, see Figure 22 Determine the Lagrangian an euce the equations of motion Solution: We will use x, θ as coorinates, because specifying the coorinates (x, y) of the bob oes not etermine the configuration of the penulum The relation is y = lcosθ, so ẏ = lsinθ θ First we fin the kinetic energy The only part with mass is the bob, which has Cartesian coorinates (x, y), so the kinetic energy is T = 1 2 m(ẋ2 + ẏ 2 ) = 1 2 m (ẋ 2 + l 2 sin 2 θ θ 2) (212) CM JM February 14, 2015

13 24 Lagrangians in general coorinates 35 The mass-inertia matrix is therefore, K(x,θ)= ( ) m 0 0 ml 2 sin 2 θ Notice that when sinθ = 0 this matrix is egenerate (putting (ẋ, θ) = (0,1) gives T = 0: this correspons to keeping the mass fixe an just moving the slier infinitesimally, an of course there is no kinetic energy for such a motion) This will have repercussions for the equations of motion The potential energy is V = mg y = mg l cos θ, so the Lagrangian is given by L(x, ẋ,θ, θ)=t V = 1 2 m (ẋ 2 + l 2 sin 2 θ θ 2) + mglcosθ (213) The erivatives require for Lagrange s equation are: t t ( ) L ẋ L x ( ) L θ = t = 0 (mẋ) = m x, = ml 2 ( sin 2 θ ) θ t = ml 2( 2sinθ cos θ θ 2 + sin 2 θ θ ) L θ = ml 2 sinθ cosθ θ 2 mglsinθ so Lagrange s equations give us: m x = 0 ml 2( 2sinθ cosθ θ 2 + sin 2 θ θ ) = ml 2 sinθ cosθ θ 2 mglsinθ or, simplifying an rearranging, we get { x = 0 sin 2 θ θ+ sinθ cosθ θ 2 = g l sinθ You will notice the first equation is conservation of the x-component of momentum; physically, this is to be expecte because both the constraining force an gravity act vertically so have no component in the x-irection The secon equation is singular when θ = 0 or π, as the coefficient of θ vanishes there This is closely relate to the fact that the matrix K is not invertible at those points We will not attempt to say anything about the solutions here The general formula for K is foun by substituting v=dψ(q) q into the expression for the kinetic energy The expression in Eucliean (Cartesian) coorinates can be written February 14, 2015 CM JM

14 36 LAGRANGE S EQUATIONS as T = 1 2 vt M v m 0 0 where M is the mass matrix If there is just one particle, then M = 0 m 0 In general, 0 0 m M is the iagonal matrix with iagonal entries (m 1,m 1,m 1,m 2,m 2,m 2,,m N,m N,m N ); that is, in 3 3 blocks, m 1 I m 2 I M = 0 0 0, m N I 3 where I 3 is the 3 3 ientity matrix Thus substituting v=dψ q q gives T = 1 2 vt M v = 1 2 (Dψ q q) T M (Dψ q q) = 1 2 q T (Dψ q ) T M (Dψ q q) = 1 ( q T Dψ T q 2 q q) M Dψ q T K(q) q = 1 2 so that K(q)=Dψ T q M Dψ q (214) If the system has n egrees of freeom, then K(q) is an n n symmetric matrix REMARK 216 Notice that the kinetic energy T was originally just a function of the velocity v=ẋ, but is now a function of both position (configuration) an velocity; this is because when using non-eucliean coorinates the Jacobian matrix Dψ q often brings q explicitly into the expression (214) for the kinetic energy From now on, we will therefore always write T (q, q) DEFINITION 217 A LagrangianL= 1 2 q T K q V (q) is regular or non-egenerate if K(q) is an invertible matrix Since one assumes all the masses of the particles in a system are non-zero, the mass matrix M is an invertible matrix It then follows from (214) that K is egenerate iff Dψ q fails to be injective In other wors, the cause 5 of a Lagrangian to be egenerate is that the coorinate system is not regular, such as occurs with polar coorinates when r = 0 an etdψ (0,θ) = 0 (see equation (210)) In many cases, this can be resolve by choosing a better system of coorinates, but sometime the egeneracy is inherent to the system (as in the elliptic penulum above) The following efinition extens the observation mae in Remark 25 DEFINITION 218 Consier a Lagrangian system with (generalize) coorinates (q 1,, q n ) 5 One can also imagine a system where one of the particles has zero mass in which case M is itself egenerate, an then K may also be egenerate even with regular coorinates CM JM February 14, 2015

15 24 Lagrangians in general coorinates 37 For each coorinate q j, the quantity p j := L q j, is calle the generalize momentum corresponing to q j, or conjugate to q j The quantity L/ q j is calle the generalize force on the system, associate to q j (or, conjugate to q j ) This quantity is name momentum because, in stanar Eucliean coorinates for a particle,l= 1 2 m q 2 V (q), so that L q = m q which is precisely the usual momentum An in Example 219 below, L/ θ is the angular momentum about the z-axis, so L/ q encompasses both types of momentum we know For example, in the plane penulum, p = p θ = L θ = ml2 θ, an this is equal to the angular momentum about the origin In the elliptic penulum above, we have from (213) p x = L = mẋ, ẋ p θ = L θ = ml2 sin 2 θ θ If, as in the efinition, the quantity L/ q is the generalize force, then Lagrange s equation reas, generalize force equals rate of change of generalize momentum which shoul soun familiar if you remove the wor generalise! For general coorinates, if the system is a simple mechanical one withl(q, q) = 1 2 q T K(q) q V (q), then p=k(q) q In particular, the Lagrangian being regular is equivalent to the linear map q p = K q being invertible EXAMPLE 219 (The spherical penulum) This is a penulum that is not confine to a plane, where the bob moves on a circle, but now the bob can move on a sphere For coorinates we use spherical polar coorinates q = (θ,ϕ), with θ [0,π] measuring the angle with the ownwar vertical, an ϕ [0, 2π) the angle between the arm of the penulum an a fixe vertical plane (so θ= π/2 latitue, an ϕ is the longitue) The potential energy is given by V = mgl(1 cos θ) (taking the ownwar configuration as zero) To fin the kinetic energy, one nees to express the velocity in Eucliean terms: we have r=ψ(θ, ϕ)=(lsinθ cosϕ, lsinθ sinϕ, lcosθ) February 14, 2015 CM JM

16 38 LAGRANGE S EQUATIONS an so r=l θ(cos θ cosϕ,cosθ sinϕ,sinθ)+l ϕ( sin θ sinϕ,sinθ cosϕ,0) One then fins that the kinetic energy, which in Eucliean terms is T = 1 2 m ṙ 2, becomes in spherical polar coorinates, T = 1 2 ml2( θ 2 + sin 2 θ ϕ 2), (215) as the reaer shoul check The mass-inertia matrix is then given by ( ) ml 2 0 K(θ,ϕ)= 0 ml 2 sin 2 θ The Lagrangian fails to be regular when sinθ= 0, so at θ= 0, π which correspon to the penulum being either upwar or ownwar This is because spherical polar coorinates are not regular at these points The Lagrangian is thus L(q, q)= 1 2 ml2( θ 2 + sin 2 θ ϕ 2) mgl(1 cos θ) Lagrange s equations for this system are then, after simplifying, { θ = sinθ cosθ ϕ 2 g l sinθ t (ml2 sin 2 θ ϕ) = 0 The final 0 arises because L is inepenent of ϕ so L/ ϕ = 0 This correspons to there being no force in the ϕ-irection In general, iflis inepenent of a particular coorinate, that coorinate is sai to be cyclic, so in this example the coorinate ϕ is cyclic The quantity L/ ϕ=ml 2 sin 2 θ ϕ is the angular momentum of the penulum aroun the vertical axis φ R x α FIGURE 23: Cross-section of a cyliner rolling own an incline plane CM JM February 14, 2015

17 24 Lagrangians in general coorinates 39 EXAMPLE 220 (Cyliner rolling own incline plane) Consier a uniform cyliner of mass M, raius R an length l, rolling without slipping own a plane incline at an angle α to the horizontal Fin the equation of motion Solution: Referring to Figure 23 we can choose either x or φ as coorinate Here x is the istance move along the plane (eg by the centre of mass), an φ is the angle mae by a fixe raial line with the vertical Provie one starts with x = φ = 0, the two are relate by the rolling conition x= Rφ To fin the Lagrangian, we nee first to calculate the kinetic energy Recall from Chapter 1, the kinetic energy is the sum of the kinetic energy of the centre of mass, whose velocity here is ẋ, an the kinetic energy T 0 of the motion of the particles relative to the centre of mass: T = 1 2 M ẋ2 + T 0 Now T 0 is the energy ue to the rotation about the axis of the cyliner, which as we saw in Chapter 1 is T 0 = 1 2 I ω2, where I = 1 2 MR 2, the moment of inertia about the axis of the cyliner Thus, using φ as the coorinate, we have ω= φ an T = 1 2 M ẋ2 + T 0 = 1 2 MR 2 φ I φ 2 = 3 2 I φ 2 The potential energy is gravitational, so V = M g h = M g x sinα = M g Rφsinα (up to aing a constant), so that the Lagrangian is given by Lagrange s equations now rea L(φ, φ)= 3 2 I φ 2 + M g Rφsinα t ( 3I φ ) = M g R sinα so that 3I φ= M g R sinα Substituting for I = 1 2 MR 2, the acceleration of the cyliner is therefore Or, with x = Rφ, x = 2 3 g sinα φ= 2g sinα 3R REMARK 221 In that last example, one can consier x, φ as inepenent variables, so February 14, 2015 CM JM

18 40 LAGRANGE S EQUATIONS escribing a system with 2 egrees of freeom This woul in fact be necessary if the cyliner were allowe to slie own the slope However, the no-slip (or rolling) conition is a holonomic constraint imposing the relationship x = Rφ between the two variables, thus reucing it to the 1 egree of freeom system we consiere above All our Lagrangians have been of the form kinetic minus potential, but others o arise For example, a charge particle in an electromagnetic fiel is subject to the Lorentz force (see 16) F = q(e + v B) The Lagrangian for this is L(r, v)= 1 2 m v 2 qφ(r)+ qa(r) v, (216) where r is the position vector of the particle, v=ṙits velocity, φ(r) is the electric potential, so E= φ, an A is the magnetic vector potential: B= A Notice that the last term is linear in the velocity, so is neither kinetic nor potential energy EXERCISE Show Lagrange s equations for (216) o inee give the same equation of motion as Newton s laws with the Lorentz force (Problem 212) REMARK 222 If there is in aition a non-conservative force F acting on the system, then one can moify Lagrange s equation to take this into account If we write F i as this non-conservative force acting on particle i, an write r i = r i (q 1,, q K ) then the result is ( ) L t q = L +Q j j q j where Q j = i F i r i q j, so as a vector, Q=F T Dψ q The quantity L/ q j +Q j is calle a generalize force corresponing to the generalize coorinate q j 25 NOETHER S THEOREM Notice that in Example 219 above, the fact that φ is a cyclic coorinate leas to a conservation law: ( ml 2 sin 2 θ φ ) = 0, t or in other wors, ml 2 sin 2 θ φ is constant This is a particular case of a general phenomenon that we will be returning to an using many times: THEOREM 223 (Noether) Let L(q, v) be a Lagrangian for a given system, an suppose that the coorinate q 1 is cyclic (ielis inepenent of q 1 ) Then the quantity L/ q 1 is conserve PROOF: for This follows immeiately from Lagrange s equations (as in the example above), t ( ) L q 1 = L q 1, an this latter term is 0 aslis inepenent of q 1 CM JM February 14, 2015

19 25 Noether s theorem 41 Since p = L/ q is the momentum associate to q, Noether s theorem tells us that if a variable q is cyclic then the corresponing momentum is conserve This result of Emmy Noether 6 is of great importance as it helps restrict the types of motion possible in a given system In the spherical penulum example this result was easy to apply as it jumpe out from the equations of motion However, this is not always the case, an one may have to make a juicious change of coorinates to see a cyclic variable, as in the following example EXAMPLE 224 Recall Example 27 where we consiere two particles of m 1 an m 2 attracte by their gravitational interaction (see also Exercise 16) The Eucliean coorinates r 1 an r 2 are not cyclic, but a change of coorinates brings out the law of conservation of momentum Write M = m 1 + m 2 an µ j = m j /M (j = 1,2), so that µ 1 + µ 2 = 1 Define q=(q +, q ) by q + = µ 1 r 1 + µ 2 r 2, q = r 1 r 2 That is, q + is the position of the centre of mass an q the relative position of the particles This change of coorinates can easily be inverte: r 1 = q + + µ 2 q, r 2 = q + µ 1 q Write this as (r 1, r 2 ) = ψ(q +, q ) We want to express the Lagrangian in terms of these new coorinates, an for that we use the ifferential of ψ an replace the Lagrangian (26) byl =L Dψ: L = 1 2 M v m 1m 2 2M v 2 + Gm 1m 2 q, where v + = q + etc Now all three components of q + are cyclic as q + oes not appear in the expression for L Noether s theorem therefore implies that the three components of L v + = M v + are conserve But M v + is the total momentum p as efine in 13, so we have shown that the total momentum is conserve (a fact we alreay prove in Corollary 13 on p 8) Finally, notice that in this example an in these coorinates, the mass-inertia matrix (see Definition 217) is ( ) M 0 K=, 0 m 1 m 2 M 6 Emmy Noether: Born in Germany, she emigrate to the US to avoi Nazi persecution She mae many contributions to mathematics, mostly in algebra She prove a result that implies the theorem above in 1915, the same year Einstein s paper on General Relativity appeare Einstein is sai to have been impresse by the power an scope of Noether s theorem which goes far beyon the statement above February 14, 2015 CM JM

20 42 LAGRANGE S EQUATIONS which is clearly positive efinite, so the Lagrangian is regular (Recall that a symmetric matrix S (or quaratic form) is positive efinite if for all x 0, one has x T Sx> 0, or equivalently if all the eigenvalues of S are positive) 26 * ROUTH REDUCTION Ewar Routh in his famous prize-winning essay in 1877 evelope (among other things) a systematic metho for taking avantage of the conserve quantities arising in Noether s theorem; this is now calle Routh reuction We begin with an example where there is just one cyclic coorinate: the planar central force problem EXAMPLE 225 Consier a particle of mass m in the plane attracte by a gravitational force to the origin We will use polar coorinates to take avantage of the cyclic variable The potential energy is V = k/r for some constant k, an the kinetic energy is 1 2 m(ṙ 2 + r 2 θ 2 ) The Lagrangian is L(r,θ,ṙ, θ)= 1 2 m(ṙ 2 + r 2 θ 2 )+ k r Clearly (an as we have seen before) θ is cyclic, an p θ = mr 2 θ is conserve (it is the angular momentum about the origin see Chapter 1) Fix a value µ for p θ ( µ for momentum) an efine the moifie Lagrangian or Routhian for the value µ by R µ (r,ṙ )=L(r, θ,ṙ ) µ θ an substituting for θ in terms of µ= p θ to get R µ (r,ṙ ) = 1 2 m(ṙ 2 + r 2 θ 2 )+ k r µ θ = 1 2 mṙ mr 2( µ mr 2 ) 2+ k r µ ( µ mr 2 ) = 1 2 mṙ 2 + k r µ2 2mr 2 Notice that this is the Lagrangian of a 1-egree of freeom system, an there is one for each value of µ Moreover, we can interpret the first term as kinetic energy an the secon two together as minus potential energy: V µ (r )= k r + µ2 2mr 2 This is equal to the original potential energy plus an extra term arising from the momentum value µ (an in this example can be interprete as a centrifugal contribution) This is calle the amene potential for the reuce problem The general proceure introuce by Routh is as follows CM JM February 14, 2015

21 26 * Routh reuction 43 Consier a Lagrangian system given by L(q, q), where some of the coorinates are cyclic Let us write q = (θ 1,,θ k, x 1,, x r ), where the θ i are all cyclic, while the x i are not; so the total number of egrees of freeom is n = k+ r Then by Noether s theorem the momenta p θ1,, p θk are conserve quantities The aim is to treat the p θj as constants (so parameters in some sense) an the remaining variables x 1,, x r (an the corresponing velocities) as the variables in a smaller reuce Lagrangian system The reuction procees by eliminating fromlthe θ variables using p θj = L θ j (j = 1,,k) (217) So we nee to assume that one can solve (217) for each θ j as a function of the p θi ; this is always possible if L = T V with T = 1 2 q T K q an K is positive efinite Define the RouthianR µ (x, ẋ) by7 R µ (x, ẋ)=l(θ, x, θ, k ẋ) µ j θj, after eliminating the θ j in favour of the µ j = p θj using (217), as was one in the example above j=1 THEOREM 226 (Routh, 1877) Consier the LagrangianL(θ, x, θ, ẋ) an suppose the θ coorinates are cyclic an that L is R-regular Then when the conserve quantity p θ = µ the variables x evolve as if governe by the RouthianR µ efine above PROOF: Now x evolves accoring to Lagrange s equations forl Since θ epens on x an ẋ (an the constant µ) we have R µ x i = L x i + j = L x i + j = L x i ( )( ) L θj θ j x i j ( )( ) L θj θ µ j j x i µ j θ j x i where the partial erivatives are taken at constant p θ = µ An likewise (with ẋ i replacing x i ), R µ ẋ = L i ẋ i ( ) ( ) Thus, L t ẋ = Rµ i t ẋ an therefore because x satisfies Lagrange s equations for i L, it also satisfies Lagrange s equations forr µ, as require 7 Routh calle this the moifie Lagrangian February 14, 2015 CM JM

22 44 LAGRANGE S EQUATIONS One thing to note is that even if the original system escribes a simple mechanical system the reuce one may not be of the same form However, it can always be written R µ (x, ẋ)= 1 x T 2 Kẋ+A(µ)ẋ V µ(x) Here A(µ) epens linearly on µ (an epens on the original form of the kinetic energy) The quantity V µ (x), which involves all the terms inepenent of ẋ is calle the amene potential In the example above, A(µ)=0 an V µ (r )= V (r )+µ 2 /2mr 2 DEFINITION 227 A motion where all the x-coorinates are constant an only the cyclic coorinates vary is calle a relative equilibrium or steay motion In this case one can show that the cyclic variables always vary uniformly; that is, θ is constant Thus a relative equilibrium is an equilibrium point for the reuce system governe by the Routhian Routh s motivation in his work was to fin a metho for etermining whether a relative equilibrium is stable or not, an his reuction proceure reuces the problem to that of the stability of an equilibrium point, using the amene potential See the next chapter for how stability is etermine from the potential energy 27 * HAMILTON S PRINCIPLE This section is inclue for its own intrinsic interest, an not use elsewhere in the book, so can be safely skippe The principal motivation for incluing it is that it provies a beautiful argument justifying Lagrange s equations, so proviing a proof of Theorem 213 William Rowan Hamilton 8 was a very important mathematician an contribute to a number of areas His first major contribution to science was in the fiel of optics, when he publishe an important paper at the age of 21 In his work on optics he was particularly fascinate by Fermat s principle of least time, which states that for a light ray to go from a point A to a point B through any meium, it will choose the path of shortest time One can easily euce the laws of reflection an refraction from this principle (see eg Feynman s Lectures in Physics, Volume I) When Hamilton turne his min to mechanics he was probably struck by the fact that like light, particles travel in straight lines unless eflecte by some external force (analogous to eflecting a light ray by reflection or refraction), so he aske whether there coul be a principle similar to Fermat s that governs mechanics Of course, it coul not simply be a least time principle as a particle can travel from A to B in various times accoring to how fast it travels Hamilton s solution to his problem is as follows LetL(q, v) be the Lagrangian for the mechanical system Fix the points A an B in the configuration space, an fix a time interval T Consier a smooth path γ(t ) in the configuration space with γ(0) = A an 8 Sir William Rowan Hamilton, , Astronomer Royal of Irelan CM JM February 14, 2015

23 27 * Hamilton s principle 45 γ(t )=B, an to such a path associate a real number, the action, S(γ)= T This integral epens on the path chosen 0 L(γ(t ), γ(t )) t Hamilton s principle, also known as the principle of least action or more correctly, of stationary action, states that the path taken by the mechanical system to go from A to B in time T is precisely the one for which the action is stationary (that is, it is a critical point of the action, although not usually a minimum) This means that, if we write γ λ (t ) := γ(t )+λη(t ) with η(0) = η(t ) = 0, which for λ small is a nearby path satisfying the same conitions γ λ (0)= A an γ λ (T )=B, then we require λ S(γ λ) λ=0 = 0, an we require this for every possible perturbation η(t ) THEOREM 228 (Hamilton) Hamilton s principle is equivalent to Lagrange s equations PROOF: The action of the path γ λ is: S(γ λ )= T 0 L(γ(t )+λη(t ), γ(t )+λ η(t )) t Differentiating with respect to λ at λ=0 gives T ( ) L λ S(γ L λ) λ=0 = η(t )+ η(t ) t q v Integrating the secon term by parts gives T [ L L η(t ) t = v v η(v) 0 0 ] T 0 T 0 t ( L v ) η(t ) t Since η(0) = η(t ) = 0 the first term on the right han sie vanishes, so that T [ L λ S(γ λ) λ= 0 = q ( )] L η(t ) t t v 0 The terms L/ q an L/ v are of course evaluate at (γ(t ), γ(t )) Note that the term in square brackets is just the left han sie of Lagrange s equations It is therefore clear that if γ(t ) is a path that satisfies Lagrange s equations, then for any η, S(γ λ )/λ=0 Conversely, if λ S(γ λ) λ= 0 = 0 for all perturbations, so for all η, then γ(t ) satisfies Lagrange s equation This is because: February 14, 2015 CM JM

24 46 LAGRANGE S EQUATIONS LEMMA 229 If a continuous function f : [0,T ] R satisfies T 0 f (t )η(t ) t = 0 for all smooth functions η satisfying η(0)=η(t )=0 then f = 0 This is not har to prove For etails see any book on the calculus of variations LAGRANGE S EQUATIONS AND NEWTON S LAWS An easy consequence of Hamilton s principle is the important Theorem 213 Inee, since the action associate to a parametrize path oes not epen on which particular coorinate system is being use (it epens only on what units of time one uses, since the integral is over time), it follows that Hamilton s principle oes not epen on the choice of coorinate system Consequently, Lagrange s equations are invariant An we have alreay shown that these equations coincie with Newton s equations in Cartesian (Eucliean) coorinates There is another important principle in mechanics, calle Maupertuis principle, which is closely relate to, an preates, Hamilton s principle an can be foun in many books REMARK 230 The argument we have been using assumes that the Lagrangian is inepenent of time If not, then one nees to specify not only the length T of the time interval, but also the precise interval over which the motion takes place: [t 1, t 2 ] an then the action is efine in the same manner: S(γ)= t2 t 1 L(γ(t ), γ(t ), t ) t PROBLEMS 21 Let V (x, y) = 3x y be the potential energy function governing the motion of a particle of mass m moving in the plane Determine the Lagrangian for this system an euce the equations of motion (You are not expecte to try to solve them) 22 Show that the equations of motion in Equations (25) (Cartesian) an (27) (polar) are equivalent [Hint: Differentiate x = r cosθ an y = r sinθ twice each The first time shoul give you x = ṙ cosθ r θ sinθ, an something similar for ẏ Then x = r cosθ 2ṙ θ sin θ r θ sinθ r θ 2 cosθ an something similar for y By consiering x cosθ+ y sinθ an x sin θ y cosθ eliminate x an y from the equations (25)] 23 (i) Let V (x, y)= 1 4 (x2 + y 2 ) 2 be the potential energy function governing the motion of a particle of mass m moving in the plane Determine the Lagrangian for this system an euce the equations of motion (ii) Let V (r,θ) = 1 4 r 4 be the potential energy function governing the motion of a particle of mass m moving in the plane Using polar coorinates as generalize coorinates, etermine the Lagrangian for this system, an euce the equations of motion [Partial answer: One of the equations is m r = mr θ 2 r 3 ] CM JM February 14, 2015

25 Problems Consier a system of two ientical interacting particles of mass m with potential V (r ), where r is the inter-particle istance Let r G be the position vector of the centre of mass, an let r = r 1 r 2 be the position vector of particle 1 relative to particle 2 Fin the Lagrangian using as generalize coorinates r G an r [Hint: first write r 1 an r 2 in terms of r G an r] 25 Consier the system consisting of two ientical masses of mass m that can move horizontally, joine with three ientical springs with spring constant k as shown below Denote by x, y the horizontal isplacements of the two masses x y k m k Write own the Lagrangian for this system an euce its equations of motion [We will solve these equations in the next chapter] [Partial answer: for the y coorinate, m y = k x 2k y] 26 Determine the Lagrangian an euce the equations of motion for the following system: A particle of mass m moves on a smooth table, an is attache to a light inextensible string of length l passing through a smooth hole in the table, the other en of which is attache to a weight of mass M [Suggestion: for generalize coorinates, use the polar coorinates of the position of the particle on the table, with centre at the hole: why is this enough?] Fin the relation between the height of M an the angular velocity of m require in orer to maintain a motion in which M is stationary, an m rotates uniformly (Such a motion is often calle a relative equilibrium) [Partial answer: the relation aske for at the en is ω 2 = M g /m(l z), where z is the istance of M below the table] 27 Fin the Lagrangian for the Atwoo machine from Ex 17 an euce the equations of motion Do this both for a light pulley, an for a uniform pulley with moment of inertia I 0 28 A spring penulum is a penulum where the ro is no longer inextensible, but is moelle as a light spring Determine the Lagrangian of such a system (you will nee to give symbols to represent the physical quantities involve), an erive its equations of motion 29 LetL(q, v) with q = (q 1,, q n ), be the Lagrangian for a given system Define the energy function E by E(q, v)= L v i L(q, v) i v i Show using Lagrange s equation, that the energy function is a conserve quantity m k M m February 14, 2015 CM JM

26 48 LAGRANGE S EQUATIONS 210 Ientify a cyclic variable in problem 26 an apply Noether s theorem to the system to fin the corresponing conserve quantity Now apply the proceure of Routh reuction to euce a 1-egree of freeom problem for the non-cyclic variable Fin the amene potential an the value of r giving a relative equilibrium with momentum µ Determine whether it is stable or unstable 211 Non-uniqueness of the Lagrangian: letl = al+b with a, b constants an a 0 Show that Lagrange s equations arising fromlan froml are ientical 212 (i) Show that the Lagrangian for a charge particle in an electromagnetic fiel given in (216) gives the same equations of motion as Newton s law with the Lorentz force F = q(e + v B) [This is a lengthy exercise in vector calculus] (ii) Show that for the same Lagrangian, the energy E as efine in problem 29 is inepenent of the magnetic fiel 213 Consier a charge particle moving in the uniform magnetic fiel B=e 3 (the unit vector in the z-irection), with no electric fiel Show that A=(0, x,0) T is a vector potential for B an fin the Lagrangian for the particle Write own the generalize momenta for each of the coorinates Use Noether s theorem to show that the z-component of the usual 9 momentum is conserve 214 Consier the following compoun Atwoo machine A smooth light pulley is attache to a fixe support, an a light inextensible string is passe over it To one en of the string is attache a weight of mass m 1 = 8 kg an to the other a smooth light pulley Over this secon pulley is passe another light inextensible string with a weight attache at each en, of masses m 2 = 2 kg an m 3 = 3 kg respectively The masses are subject to gravity How many egrees of freeom oes the system have? Fin the Lagrangian of the system, an hence the accelerations of the three masses Deuce that if they start from rest, then one of the masses oes not move [Hint: Let x, y, z be vertical isplacements of the three weights respectively Show that 2x+y+ z is constant] 9 as against generalize momentum CM JM February 14, 2015