156 CHAPTER 6. HAMILTON’S EQUATIONS
later subdivide these into coordinates and velocities. We will take the space in which x takes values to be some general n-dimensional space we call , M which might be ordinary Euclidean space but might be something else, like the surface of a sphere1. Given a differentiable function f of n independent variables xi, the differential is Chapter 6 n ∂f df = dxi. (6.1) ∂xi Xi=1 Hamilton’s Equations What does that mean? As an approximate statement, this can be regarded as saying
n ∂f df ∆f f(xi +∆xi) f(xi)= ∆xi + (∆xi∆xj), ≈ ≡ − ∂xi O We discussed the generalized momenta Xi=1
∂L(q, q,˙ t) with some statement about the ∆xi being small, followed by the dropping of pi = , the “order (∆x)2” terms. Notice that df is a function not only of the point ∂q˙i x , but also of the small displacements ∆x . A very useful mathematical ∈M i and how the canonical variables q ,p describe phase space. One can use language emerges if we formalize the definition of df , extending its definition { i j} phase space rather than q , q˙ to describe the state of a system at any to arbitrary ∆x , even when the ∆x are not small. Of course, for large { i j} i i moment. In this chapter we will explore the tools which stem from this ∆xi they can no longer be thought of as the difference of two positions in phase space approach to dynamics. and df no longer has the meaning of the difference of values of f at M two different points. Our formal df is now defined as a linear function of these ∆xi variables, which we therefore consider to be a vector ~v lying in 6.1 Legendre transforms an n-dimensional vector space Rn.Thusdf : Rn R is a real-valued M× → function with two arguments, one in and one in a vector space. The dx M i The important object for determining the motion of a system using the La- which appear in (6.1) can be thought of as operators acting on this vector grangian approach is not the Lagrangian itself but its variation, under arbi- space argument to extract the i0th component, and the action of df on the trary changes in the variables q andq ˙, treated as independent variables. It argument (x,~v)isdf (x,~v)= i(∂f/∂xi)vi. is the vanishing of the variation of the action under such variations which This differential is a special case of a 1-form, as is each of the operators determines the dynamical equations. In the phase space approach, we want P dxi. All n of these dxi form a basis of 1-forms, which are more generally to change variablesq ˙ p, where the p are components of the gradient of → i the Lagrangian with respect to the velocities. This is an example of a general ω = ωi(x)dxi, procedure called the Legendre transformation. We will discuss it in terms of Xi the mathematical concept of a differential form. where the ω (x) are n functions on the manifold . If there exists an ordi- Because it is the variation of L which is important, we need to focus i M our attention on the differential dL rather than on L itself. We first want nary function f(x) such that ω = df , then ω is said to be an exact 1-form. to give a formal definition of the differential, which we will do first for a 1Mathematically, is a manifold, but we will not carefully define that here. The M function f(x1, ..., xn)ofn variables, although for the Lagrangian we will precise definition is available in Ref. [16].
155 6.1. LEGENDRE TRANSFORMS 157 158 CHAPTER 6. HAMILTON’S EQUATIONS
Consider L(qi,vj,t), where vi =˙qi. At a given time we consider q and v where d¯Q is not an exact differential, and the heat Q is not a well defined as independant variables. The differential of L on the space of coordinates system variable. Though Q is not a well defined state function, the differential and velocities, at a fixed time, is d¯Q is a well defined 1-form on the manifold of possible states of the system. ∂L ∂L ∂L It is not, however, an exact 1-form, which is why Q is not a function on that dL = dqi + dvi = dqi + pidvi. manifold. We can express d¯Q by defining the entropy and temperature, in ∂qi ∂vi ∂qi Xi Xi Xi Xi terms of which d¯Q = TdS, and the entropy S and temperature T are well If we wish to describe physics in phase space (qi,pi), we are making a change defined state functions. Thus the state of the gas can be described by the of variables from vi to the gradient with respect to these variables, pi = two variables S and V , and changes involve an energy change ∂L/∂vi, where we focus now on the variables being transformed and ignore the fixed q variables. So dL = p dv , and the p are functions of the v dE = TdS pdV. i i i i i j − determined by the function L(vi). Is there a function g(pi) which reverses P We see that the temperature is T = ∂E/∂S . If we wish to find quantities the roles of v and p, for which dg = i vidpi? If we can invert the functions |V appropriate for describing the gas as a function of T rather than S, we define p(v), we can define g(pi)= i pivi(pj) L(vi(pj)), which has a differential P− the free energy F by F = TS E so dF = SdT pdV , and we treat dg = p dv + vPdp dL = p dv + v dp p dv − − − − i i i i − i i i i − i i F as a function F (T,V ). Alternatively, to use the pressure p rather than V , Xi Xi Xi Xi Xi we define the enthalpy X(p, S)=Vp+E, dX = Vdp+TdS. To make both = vidpi changes, and use (T,p) to describe the state of the gas, we use the Gibbs Xi free energy G(T,p)=X TS = E + Vp TS, dG = Vdp SdT. Each of − − − as requested, and which also determines the relationship between v and p, these involves a Legendre transformation starting with E(S, V ). ∂g Unlike Q, E is a well defined property of the gas when it is in a volume vi = = vi(pj), V if its entropy is S,soE = E(S, V ), and ∂pi giving the inverse relation to pk(v`). This particular form of changing vari- ∂E ∂E T = ,p= . ables is called a Legendre transformation. In the case of interest here, ∂S ∂V V S the function g is called H(qi,pj,t), the Hamiltonian,
∂2E ∂2E ∂T ∂p H(qi,pj,t)= pkq˙k(qi,pj,t) L(qi, q˙j(q`,pm,t),t). (6.2) As = we can conclude that = . We may also k − ∂S∂V ∂V ∂S ∂V ∂S X S V consider the state of the gas to be described by T and V ,so Then for fixed time,
dH = (dp q˙ + p dq˙ ) dL = (˙q dp p dq ) , ∂E ∂E k k k k − k k − k k dE = dT + dV k k ∂T ∂V X X V T
∂H ∂H =˙qk, = pk. (6.3) 1 p 1 ∂E 1 ∂E ∂p ∂q − k q,t k p,t dS = dE + dV = dT + p + dV, T T T ∂T "T ∂V !# V T Other examples of Legendre transformations occur in thermodynamics.
The energy change of a gas in a variable container with heat flow is sometimes from which we can conclude written ∂ 1 ∂E ∂ 1 ∂E = p + , dE =d¯Q pdV, ∂V T ∂T ! ∂T "T ∂V !# − V T T V
6.1. LEGENDRE TRANSFORMS 159 160 CHAPTER 6. HAMILTON’S EQUATIONS and therefore In Section 2.1 we discussed how the Lagrangian is unchanged by a change ∂p ∂E T p = . of generalized coordinates used to describe the physical situation. More pre- ∂T − ∂V V T cisely, the Lagrangian transforms as a scalar under such point transforma-
tions, taking on the same value at the same physical point, described in the This is a useful relation in thermodynamics. Let us get back to mechanics. Most Lagrangians we encounter have the new coordinates. There is no unique set of generalized coordinates which describes the physics. But in transforming to the Hamiltonian language, decomposition L = L2+L1+L0 into terms quadratic, linear, and independent of velocities, as considered in 2.4.2. Then the momenta are linear in velocities, different generalized coordinates may give different momenta and different Hamiltonians. An nice example is given in Goldstein, a mass on a spring at- pi = j Mijq˙j + ai, or in matrix form p = M q˙ + a, which has the inverse 1 · 1 1 tached to a “fixed point” which is on a truck moving at uniform velocity v , relationq ˙ = M − (p a). As H = L L , H = (p a) M − (p a) L . T P · − 2 − 0 2 − · · − − 0 As a simple example, with a = 0 and a diagonal matrix M, consider spherical relative to the Earth. If we use the Earth coordinate x to describe the mass, coordinates, in which the kinetic energy is the equilibrium position of the spring is moving in time, xeq = vT t, ignoring a 1 2 1 2 negligible initial position. Thus U = 2 k(x vT t) , while T = 2 mx˙ as usual, 2 2 1 2 1 2 − 2 1 2 m 1 p p and L = mx˙ k(x vT t) , p = mx˙, H = p /2m + k(x vT t) . The T = r˙2 + r2θ˙2 + r2 sin2 θφ˙2 = p2 + θ + φ . 2 − 2 − 2 − r 2 2 2 equation of motion isp ˙ = mx¨ = ∂H/∂x = k(x v t), of course. This 2 2m r r sin θ ! − − − T shows that H is not conserved, dH/dt =(p/m)dp/dt + k(˙x v )(x v t)= − T − T Note that the generalized momenta are not normalized components of the (kp/m)(x v t)+(kp/m kv )(x v t)= kv (x v t) = 0. Alterna- − − T − T − T − T − T 6 ordinary momentum, as pθ = p~ eˆθ, in fact it doesn’t even have the same tively, dH/dt = ∂L/∂t = kv (x v t) = 0. This is not surprising; the 6 · − − T − T 6 units. spring exerts a force on the truck and the truck is doing work to keep the The equations of motion in Hamiltonian form (6.3), fixed point moving at constant velocity. On the other hand, if we use the truck coordinate x0 = x vT t,we 1 2 1 − 2 ∂H ∂H may describe the motion in this frame with T 0 = mx˙ 0 , U 0 = kx0 , L0 = q˙k = , p˙k = , 2 2 ∂p − ∂q 1 2 1 2 k q,t k p,t mx˙ 0 kx0 , giving the correct equations of motion p0 = mx˙ 0,˙p0 = mx¨0 = 2 − 2 ∂L0/∂x0 = kx0. With this set of coordinates, the Hamiltonian is H0 = are almost symmetric in their treatment of q and p. If we define a 2N − −2 1 2 x˙ 0p0 L0 = p0 /2m + kx0 , which is conserved. From the correspondence dimensional coordinate η for phase space, − 2 between the two sets of variables, x0 = x v t, and p0 = p mv , we see that − T − T the Hamiltonians at corresponding points in phase space differ, H(x, p) ηi = qi 2 2 1 2 − for 1 i N, H0(x0,p0)=(p p0 )/2m =2mvT p mvT =0. ηN+i = pi ≤ ≤ − − 2 6 Thus the Hamiltonian is not invariant, or a scalar, under change of gen- we can write Hamilton’s equation in terms of a particular matrix J, eralized coordinates, or point transformations. We shall see, however, that it is invariant under time independent transformations that are more general 2N ∂H 01IN N than point transformations, mixing coordinates and momenta. η˙j = Jjk , where J = × . ∂ηk 1I N N 0 kX=1 − ×
J is like a multidimensional version of the iσy which we meet in quantum- 6.2 Variations on phase curves mechanical descriptions of spin 1/2 particles. It is real, antisymmetric, and because J 2 = 1I, it is orthogonal. Mathematicians would say that J de- In applying Hamilton’s Principle to derive Lagrange’s Equations, we consid- − scribes the complex structure on phase space, also called the symplec- ered variations in which δqi(t) was arbitrary except at the initial and final tic structure. times, but the velocities were fixed in terms of these, δq˙i(t)=(d/dt)δqi(t). 6.3. CANONICAL TRANSFORMATIONS 161 162 CHAPTER 6. HAMILTON’S EQUATIONS
In discussing dynamics in terms of phase space, this is not the most natural became simpler. We switched from cartesian to center-of-mass spherical coor- variation, because this means that the momenta are not varied indepen- dinates to discuss planetary motion, for example, or from the Earth frame to dently. Here we will show that Hamilton’s equations follow from a modified the truck frame in the example in which we found how Hamiltonians depend Hamilton’s Principle, in which the momenta are freely varied. on coordinate choices. In all these cases we considered a change of coordinates We write the action in terms of the Hamiltonian, q Q, where each Q is a function of all the q and possibly time, but not of → i j the momenta or velocities. This is called a point transformation. But we tf have seen that we can work in phase space where coordinates and momenta I = piq˙i H(qj,pj,t) dt, ti " − # enter together in similar ways, and we might ask ourselves what happens if Z Xi we make a change of variables on phase space, to new variables Qi(q, p, t), and consider its variation under arbitrary variation of the path in phase P (q, p, t), which also cover phase space, so there is a relation (q, p) (Q, P ) i ↔ space, (qi(t),pi(t)). Theq ˙i(t) is still dqi/dt, but the momentum is varied free given by the two descriptions of the same point in phase space. We should of any connection toq ˙i. Then not expect the Hamiltonian to be the same either in form or in value, as we
tf saw even for point transformations, but there must be a new Hamiltonian tf ∂H ∂H δI = δpi q˙i δqi p˙i + dt + piδqi , K(Q, P, t) from which we can derive the correct equations of motion, ti " − ∂pi ! − ∂qi !# Z i i i ti X X X ∂K ∂K Q˙ i = , P˙i = . where we have integrated the pidδqi/dt term by parts. Note that in order ∂Pi −∂Qi to relate stationarity of the actionR P to Hamilton Equations of Motion, it is necessary only to constrain the q (t) at the initial and final times, without Q i The analog of η for our new variables will be ζ = , and the relation imposing any limitations on the variation of pi(t), either at the endpoints, as P we did for qi(t), or in the interior (ti,tf ), where we had previously related pi η ζ means each can be viewed as a function of the other. Hamilton’s ↔ ∂K andq ˙j. The relation betweenq ˙i and pj emerges instead among the equations equations for ζ are ζ˙ = J . If this exists, we say the new variables of motion. · ∂ζ (Q, P ) are canonical variables and the transformation (q, p) (Q, P )isa Theq ˙i seems a bit out of place in a variational principle over phase space, → and indeed we can rewrite the action integral as an integral of a 1-form over canonical transformation. Note that the functions Qi and Pi may depend a path in extended phase space, on time as well as on q and p. These new Hamiltonian equations are related to the old ones,η ˙ = J · ∂H/∂η, by the function which gives the new coordinates and momenta in I = pidqi H(q, p, t)dt . − ! terms of the old, ζ = ζ(η, t). Then Z Xi We will see, in section 6.6, that the first term of the integrand leads to a very dζi ∂ζi ∂ζi ζ˙i = = η˙j + . important form on phase space, and that the whole integrand is an important dt ∂ηj ∂t Xj 1-form on extended phase space. Let us write the Jacobian matrix Mij := ∂ζi/∂ηj. In general, M will not be a constant but a function on phase space. The above relation for the velocities 6.3 Canonical transformations now reads
We have seen that it is often useful to switch from the original set of coor- ∂ζ ζ˙ = M η˙ + . dinates in which a problem appeared to a different set in which the problem · ∂t η
6.3. CANONICAL TRANSFORMATIONS 163 164 CHAPTER 6. HAMILTON’S EQUATIONS
The gradients in phase space are also related, also be continuous, or depending on a parameter. For example, with rotating coordinates we considered x = r cos(θ + φ),y = r sin(θ + φ), which can be ∂ ∂ζj ∂ T viewed as a set of possible discrete transformations (r, θ) (x, y) depending = , or η = M ζ . ∂η ∂η ∂ζ ∇ ·∇ ↔ i t,η j i t,η j t,ζ on a parameter φ, which might even be a function of time, φ(t)=Ωt. For each X fixed φ this transformation φ :(x, y) (r, θ) is an acceptible point trans- P 7→ Thus we have formation, which can be used to describe phase space at any time t.Thuswe 1 may also consider a composition, the point tranformation φ(t+∆t) − from ˙ ∂ζ ∂ζ T ∂ζ P Pφ(t) ζ = M η˙ + = M J ηH + = M J M ζ H + the polar coordinate system at time t to that at t +∆t. This is an infinites- · ∂t · ·∇ ∂t · · ·∇ ∂t imal point transformation, and is what we used in discussing the angular = J ζ K. ·∇ velocity in rigid body motion. Let us first consider a canonical transformation which does not depend Just as for orthogonal transformations, symplectic transformations can be on time, so ∂ζ/∂t = 0. We see that we can choose the new Hamiltonian to divided into those which can be generated by infinitesimal transformations |η be the same as the old, K = H (i.e. K(ζ,t):=H(η(ζ),t)), and get correct (which are connected to the identity) and those which can not. Consider a mechanics, provided transformation M which is almost the identity, Mij = δij + Gij,orM = 1I + G, where is considered some infinitesimal parameter while G is a finite M J M T = J. (6.4) matrix. As M is symplectic, (1 + G) J (1 + GT )=J, which tells us that · · · · to lowest order in , GJ + JGT = 0. Comparing this to the condition for We will require this condition even when ζ does depend on t, but then we the generator of an infinitesimal rotation, Ω = ΩT , we see that it is similar − need to revisit the question of finding K. except for the appearence of J on opposite sides, changing orthogonality to The condition (6.4) on M is similar to, and a generalization of, the condi- symplecticity. The new variables under such a canonical transformation are T tion for orthogonality of a matrix, = 1I,which is of the same form with ζ = η + G η. OO · J replaced by 1I. Another example of this kind of relation in physics occurs The condition (6.4) for a transformation η ζ to be canonical does not → in special relativity, where a Lorentz transformation Lµν gives the relation involve time — each canonical transformation is a fixed map of phase-space between two coordinates, xµ0 = ν Lµνxν, with xν a four dimensional vector onto itself, and could be used at any t. We might consider a set of such with x4 = ct. Then the conditionP which makes L a Lorentz transformation maps, one for each time, giving a time dependant map g(t):η ζ. Each → is such map could be used to transform the trajectory of the system at any 100 0 time. In particular, consider the set of maps g(t, t0) which maps each point 010 0 η at which a system can be at time t0 into the point to which it will evolve L g LT = g, with g = . · · 001 0 at time t. That is, g(t, t0):η(t0) η(t). If we consider t = t0 +∆t for 7→ 000 1 infinitesimal ∆t, this is an infinitesimal transformation. As ζi = ηi +∆tη˙i = 2 − ηi +∆t k Jik∂H/∂ηk,wehaveMij = ∂ζi/∂ηj = δij +∆t k Jik∂ H/∂ηj∂ηk, 2 The matrix g in relativity is known as the indefinite metric, and the condition so Gij =P k Jik∂ H/∂ηj∂ηk, P on L is known as pseudo-orthogonality. In our current discussion, however, P 2 2 J is not a metric, as it is antisymmetric rather than symmetric, and the word T ∂ H ∂ H (GJ + JG )ij = Jik J`j + Ji`Jjk which describes M is symplectic. ∂η`∂ηk ∂η`∂ηk ! Xk` So far the only examples of canonical transformation which we have dis- ∂2H cussed have been point transformations. Before we continue, let us recall that = (JikJ`j + Ji`Jjk) . ∂η`∂ηk such transformations can be discrete, e.g. (x, y, z) (r, θ, φ), but they can Xk` → 6.4. POISSON BRACKETS 165 166 CHAPTER 6. HAMILTON’S EQUATIONS
The factor in parentheses in the last line is ( J J + J J ) which is anti- The Poisson bracket is a fundamental property of the phase space. In − ik j` i` jk symmetric under k `, and as it is contracted into the second derivative, symplectic language, ↔ T which is symmetric under k `, we see that (GJ + JG )ij = 0 and we ↔ ∂u ∂v T have an infinitesimal canonical transformation. Thus the infinitesimal flow [u, v]= Jij =( ηu) J ηv. (6.8) ∂ηi ∂ηj ∇ · ·∇ of phase space points by the velocity function is canonical. As compositions Xij of canonical transformations are also canonical2, the map g(t, t ) which takes 0 If we describe the system in terms of a different set of canonical variables η(t )intoη(t), the point it will evolve into after a finite time increment t t , 0 − 0 ζ, we should still find the function f(t) changing at the same rate. We may is also a canonical transformation. think of u and v as functions of ζ as easily as of η. Really we are thinking Notice that the relationship ensuring Hamilton’s equations exist, of u and v as functions of points in phase space, represented by u(η)=˜u(ζ) T T ∂ζ and we may ask whether [˜u, v˜]ζ is the same as [u, v]η. Using η = M ζ , M J M ζ H + = J ζ K, ∇ ·∇ · · ·∇ ∂t ·∇ we have T T with the symplectic condition M J M = J, implies ζ (K H)= J ∂ζ/∂t, T T T T · · ∇ − − · [u, v]η = M ζ u˜ J M ζ v˜ =( ζ u˜) M J M ζ v˜ so K differs from H here. This discussion holds as long as M is symplectic, ·∇ · · ∇ ∇ · · · ∇ =( u˜)T J v˜ =[˜u, v˜] , even if it is not an infinitesimal transformation. ∇ζ · ∇ζ ζ so we see that the Poisson bracket is independent of the coordinatization 6.4 Poisson Brackets used to describe phase space, as long as it is canonical. The Poisson bracket plays such an important role in classical mechanics, Suppose I have some function f(q, p, t) on phase space and I want to ask how and an even more important role in quantum mechanics, that it is worthwhile f, evaluated on a dynamical system, changes as the system evolves through to discuss some of its abstract properties. First of all, from the definition it phase space with time. Then is obvious that it is antisymmetric:
df ∂f ∂f ∂f [u, v]= [v, u]. (6.9) = q˙i + p˙i + − dt ∂qi ∂pi ∂t Xi Xi ∂f ∂H ∂f ∂H ∂f It is a linear operator on each function over constant linear combinations, = + . (6.5) but is satisfies a Leibnitz rule for non-constant multiples, ∂qi ∂pi − ∂pi ∂qi ∂t Xi Xi The structure of the first two terms is that of a Poisson bracket, a bilinear [uv, w]=[u, w]v + u[v, w], (6.10) operation of functions on phase space defined by which follows immediately from the definition, using Leibnitz’ rule on the ∂u ∂v ∂u ∂v partial derivatives. A very special relation is the Jacobi identity, [u, v]:= . (6.6) ∂qi ∂pi − ∂pi ∂qi Xi Xi [u, [v, w]]+[v, [w, u]]+[w, [u, v]] = 0. (6.11) Thus Eq. 6.5 may be rewritten as We need to prove that this is true. To simplify the presentation, we introduce df ∂f =[f,H]+ . (6.7) some abbreviated notation. We use a subscript ,i to indicate partial derivative dt ∂t with respect to ηi,sou,i means ∂u/∂ηi, and u,i,j means ∂(∂u/∂ηi)/∂ηj.We 2 T T T will assume all our functions on phase space are suitably differentiable, so If M = M1 M2 and M1 J M1 = J, M2 J M2 = J, then M J M = (M M ) J( M T · M T )=M (·M · J M T ) M T =· M· J M T = J,soM is· canonical.· u = u . We will also use the summation convention, that any index 1 · 2 · · 2 · 1 1 · 2 · · 2 · 1 1 · · 1 ,i,j ,j,i 6.4. POISSON BRACKETS 167 168 CHAPTER 6. HAMILTON’S EQUATIONS which appears twice in a term is assumed to be summed over3. Then [v, w]= and we see that this combination of Poisson brackets involves the commutator v,iJijw,j, and of differential operators. But such a commutator is always a linear differential operator itself, [u, [v, w]] = [u, v,iJijw,j] 2 ∂ ∂ ∂fj ∂ ∂ =[u, v,i]Jijw,j + v,iJij[u, w,j] DhDf = hi fj = hi + hifj ij ∂ηi ∂ηj ij ∂ηi ∂ηj ij ∂ηi∂ηj = u,kJk`v,i,`Jijw,j + v,iJiju,kJk`w,j,`. X X X 2 ∂ ∂ ∂hi ∂ ∂ In the Jacobi identity, there are two other terms like this, one with the Df Dh = fj hi = fj + hifj ∂ηj ∂ηi ∂ηj ∂ηi ∂ηi∂ηj substitution u v w u and the other with u w v u, giving Xij Xij Xij → → → → → → a sum of six terms. The only ones involving second derivatives of v are the first term above and the one found from applying u w v u to the so in the commutator, the second derivative terms cancel, and → → → second, u,iJijw,kJk`v,j,`. The indices are all dummy indices, summed over, so ∂fj ∂ ∂hi ∂ their names can be changed, by i k j ` i, converting this second DhDf Df Dh = hi fj → → → → − ij ∂ηi ∂ηj − ij ∂ηj ∂ηi term to u,kJk`w,jJjiv,`,i. Adding the original term u,kJk`v,i,`Jijw,j, and using X X v = v , gives u J w (J +J )v = 0 because J is antisymmetric. Thus ∂fj ∂hj ∂ ,`,i ,i,` ,k k` ,j ji ij ,`,i = hi fi . the terms in the Jacobi identity involving second derivatives of v vanish, but ∂ηi − ∂ηi ! ∂ηj Xij the same argument applies in pairs to the other terms, involving second derivatives of u or of w, so they all vanish, and the Jacobi identity is proven. This is just another first order differential operator, so there are no second This argument can be made more elegantly if we recognize that for each derivatives of g left in (6.12). In fact, the identity tells us that this combina- function f on phase space, we may view [f, ] as a differential operator on tion is · functions g on phase space, mapping g [f,g]. Calling this operator D , → f DhDf Df Dh = D[h,f] (6.13) we see that − An antisymmetric product which obeys the Jacobi identity is what makes ∂f ∂ Df = Jij , a Lie algebra. Lie algebras are the infinitesimal generators of Lie groups, ∂ηi ! ∂ηj Xj Xi or continuous groups, one example of which is the group of rotations SO(3) which we have already considered. Notice that the “product” here is not as- which is of the general form that a differential operator has, sociative,[u, [v, w]] =[[u, v],w]. In fact, the difference [u, [v, w]] [[u, v],w]= 6 − ∂ [u, [v, w]] + [w, [u, v]] = [v, [w, u]] by the Jacobi identity, so the Jacobi iden- D = f , − f j tity replaces the law of associativity in a Lie algebra. Lie groups play a j ∂ηj X major role in quantum mechanics and quantum field theory, and their more where fj are an arbitrary set of functions on phase space. For the Poisson extensive study is highly recommended for any physicist. Here we will only bracket, the functions fj are linear combinations of the f,j, but fj = f,j. mention that infinitesimal rotations, represented either by the ω∆t or Ω∆t 6 With this interpretation, [f,g]=Df g, and [h, [f,g]] = DhDf g.Thus of Chapter 4, constitute the three dimensional Lie algebra of the rotation group (in three dimensions). [h, [f,g]]+[f,[g, h]] = [h, [f,g]] [f,[h, g]] = D D g D D g − h f − f h Recall that the rate at which a function on phase space, evaluated on the =(DhDf Df Dh)g, (6.12) system as it evolves, changes with time is − 3This convention of understood summation was invented by Einstein, who called it the df ∂f = [H, f]+ , (6.14) “greatest contribution of my life”. dt − ∂t 6.4. POISSON BRACKETS 169 170 CHAPTER 6. HAMILTON’S EQUATIONS where H is the Hamiltonian. The function [f,g] on phase space also evolves 6.5 Higher Differential Forms that way, of course, so In section 6.1 we discussed a reinterpretation of the differential df as an example of a more general differential 1-form, a map ω : Rn R.We d[f,g] ∂[f,g] M× → = [H, [f,g]] + saw that the dx provide a basis for these forms, so the general 1-form can dt − ∂t { i} ∂f ∂g be written as ω = i ωi(x) dxi. The differential df gave an example. We =[f,[g, H]]+[g, [H, f]] + ,g + f, defined an exact 1-form as one which is a differential of some well-defined " ∂t # " ∂t # P function f. What is the condition for a 1-form to be exact? If ω = ωidxi ∂g ∂f = f, [H, g]+ + g, [H, f] is df , then ωi = ∂f/∂xi = f,i, and P " − ∂t !# " − ∂t !# 2 2 dg df ∂ωi ∂ f ∂ f = f, g, . ωi,j = = = = ωj,i. " dt # − " dt # ∂xj ∂xi∂xj ∂xj∂xi
Thus one necessary condition for ω to be exact is that the combination ωj,i If f and g are conserved quantities, df/dt = dg/dt = 0, and we have the − ωi,j = 0. We will define a 2-form to be the set of these objects which must important consequence that d[f,g]/dt = 0. This proves Poisson’s theorem: vanish. In fact, we define a differential k-form to be a map The Poisson bracket of two conserved quantities is a conserved quantity. We will now show an important theorem, known as Liouville’s theo- ω(k) : Rn Rn R M× ×···× → rem, that the volume of a region of phase space is invariant under canonical k times transformations. This is not a volume in ordinary space, but a 2n dimen- | {z } 2n which is linear in its action on each of the Rn and totally antisymmetric in sional volume, given by integrating the volume element i=1 dηi in the old its action on the k copies, and is a smooth function of x . At a given coordinates, and by Q ∈M point, a basis of the k-forms is4 2n 2n 2n ∂ζi P dζi = det dηi = det M dηi dxi1 dxi2 dxik := ( 1) dxiP 1 dxiP 2 dxiPk. ∂η | | ∧ ∧···∧ − ⊗ ⊗···⊗ i=1 j i=1 i=1 PXSk Y Y Y ∈
in the new, where we have used the fact that the change of variables requires For example, in three dimensions there are three independent 2-forms at a T point, dx dx , dx dx , and dx dx , where dx dx = dx dx dx a Jacobian in the volume element. But because J = M J M , det J = 1 ∧ 2 1 ∧ 3 2 ∧ 3 1 ∧ 2 1 ⊗ 2 − 2 ⊗ T 2 · · dx , which means that, acting on ~u and ~v, dx dx (~u,~v)=u v u v . The det M det J det M = (det M) det J, and J is nonsingular, so det M = 1, 1 1 ∧ 2 1 2 − 2 1 ± product is called the wedge product or exterior product, and can be and the volume element is unchanged. ∧ extended to act between k - and k -forms so that it becomes an associative In statistical mechanics, we generally do not know the actual state of a 1 2 distributive product. Note that this definition of a k-form agrees, for k =1, system, but know something about the probability that the system is in a 4 particular region of phase space. As the transformation which maps possible Some explanation of the mathematical symbols might be in order here. Sk is the group of permutations on k objects, and ( 1)P is the sign of the permutation P , which is plus values of η(t1) to the values into which they will evolve at time t2 is a canon- − ical transformation, this means that the volume of a region in phase space or minus one if the permutation can be built from an even or an odd number, respectively, of transpositions of two of the elements. The tensor product of two linear operators into does not change with time, although the region itself changes. Thus the prob- a field is a linear operator which acts on the product space,⊗ or in other words a bilinear n n ability density, specifying the likelihood that the system is near a particular operator with two arguments. Here dx dx is an operator on R R which maps the i ⊗ j × point of phase space, is invariant as we move along with the system. pair of vectors (~u,~v)touivj. 6.5. HIGHER DIFFERENTIAL FORMS 171 172 CHAPTER 6. HAMILTON’S EQUATIONS with our previous definition, and for k = 0 tells us a 0-form is simply a k =3: There is only one basis 3-form available in three dimensions, dx 1 ∧ function on . The general expression for a k-form is dx dx . Any other 3-form is proportional to this one, though the M 2 ∧ 3 proportionality can be a function of xi . In particular dxi dxj dxk = (k) { } ∧ ∧ ω = ωi ...i (x)dxi dxi . dx dx dx . The most general 3-form C is simply specified by 1 k 1 ∧···∧ k ijk 1 ∧ 2 ∧ 3 i1<... B = Bijdxi dxj = Bijdxi dxj. C = A B = Ai Bjk dxi dxj dxk ∧ ⊗ ∧ i j The exterior derivative We can summarize the action of the exterior derivative in three dimensions in this diagram: We defined the differential of a function f, which we now call a 0-form, giving a 1-form df = f,idxi. Now we want to generalize the notion of differential dd dd fω- (1) Aω~ - (2) Bω~ - (3) - 0 so that d can actP on k-forms for arbitrary k. This generalized differential f ∼ A ∼ B ∇ ∇× ∇· d : k-forms (k + 1)-forms → Now that we have d operating on all k-forms, we can ask what happens is called the exterior derivative. It is defined to be linear and to act on if we apply it twice. Looking first in three dimenions, on a 0-form we get one term in the sum over basis elements by d2f = dA for A~ f, and dA A,sod2f f. But the curl of a ∼∇ ∼∇× ∼∇×∇ gradient is zero, so d2 = 0 in this case. On a one form d2A = dB, B~ A~ ∼∇× d (f (x)dx dx )=(df (x)) dx dx and dB B = ( A~). Now we have the divergence of a curl, i1...ik i1 ik i1...ik i1 ik ∼∇· ∇· ∇× ∧···∧ ∧ ∧···∧ which is also zero. For higher forms in three dimensions we can only get zero = fi1...ik,jdxj dxi1 dxik . ∧ ∧···∧ because the degree of the form would be greater than three. Thus we have Xj a strong hint that d2 might vanish in general. To verify this, we apply d2 to Clearly some examples are called for, so let us look again at three dimen- (k) ω = ωi1...ik dxi1 dxik . Then sional Euclidean space. ∧···∧ Pdω = (∂ ω ) dx dx dx j i1...ik j ∧ i1 ∧···∧ ik j i1 Lemma, is that if ω is a closed k-form on a coordinate neighborhood U of a so manifold M, and if U is contractible to a point, then ω is exact on U. We will ∂f˜ ∂f ∂xi ∂yk ignore the possibility of global obstructions and assume that we can write df˜ = dyk = dxj closed k-forms in terms of an exterior derivative acting on a (k 1)-form. k ∂yk ijk ∂xi ∂yk ∂xj − X X ∂f = δijdxj = f,idxi = df . Coordinate independence of k-forms ∂xi Xij Xi We have introduced forms in a way which makes them appear dependent We impose this transformation law in general on the coefficients in our k- on the coordinates xi used to describe the space . This is not what we forms, to make the k-form invariant, which means that the coefficients are 7 M want at all . We want to be able to describe physical quantities that have covariant, intrinsic meaning independent of a coordinate system. If we are presented with another set of coordinates yj describing the same physical space, the ∂xi ω˜j = ωi points in this space set up a mapping, ideally an isomorphism, from one ∂yj Xi coordinate system to the other, ~y = ~y(~x). If a function represents a physical k ∂xi` field independent of coordinates, the actual function f(x) used with the x ω˜j1...jk = ωi1...ik . ∂yj ! coordinates must be replaced by another function f˜(y) when using the y i1,iX2,...,ik `Y=1 l coordinates. That they both describe the physical value at a given physical point requires f(x)=f˜(y) when y = y(x), or more precisely8 f(x)=f˜(y(x)). Integration of k-forms This associated function and coordinate system is called a scalar field. Suppose we have a k-dimensional smooth “surface” S in , parameterized If we think of the differential df as the change in f corresponding to M by coordinates (u1, ,uk). We define the integral of a k-form an infinitesimal change dx, then clearly df˜ is the same thing in different ··· (k) coordinates, provided we understand the dyi to represent the same physical ω = ω dx dx i1...ik i1 ∧···∧ ik displacement as dx does. That means i1<... ∂yk over S by dyk = dxj. j ∂xj k X (k) ∂xi` ω = ωi1...ik (x(u)) du1du2 duk. S ∂u ··· ˜ ˜ i1,i2,...,i `=1 ` ! As f(x)=f(y(x)) and f(y)=f(x(y)), the chain rule gives Z Z X k Y We had better give some examples. For k = 1, the “surface” is actually ∂f ∂f˜ ∂y ∂f˜ ∂f ∂x = j , = i , a path Γ : u x(u), and ∂xi j ∂yj ∂xi ∂yj i ∂xi ∂yj 7→ X X u max ∂xi 7Indeed, most mathematical texts will first define an abstract notion of a vector in ωidxi = ωi(x(u)) du, Γ umin ∂u the tangent space as a directional derivative operator, specified by equivalence classes of Z X Z X parameterized paths on . Then 1-forms are defined as duals to these vectors. In the ~ first step any coordinatizationM of is tied to the corresponding basis of the vector space which seems obvious. In vector notation this is Γ A d~r, the path integral of n M ~ · R . While this provides an elegant coordinate-independent way of defining the forms, the the vector A. R abstract nature of this definition of vectors can be unsettling to a physicist. For k =2, 8More elegantly, giving the map x y the name φ,soy = φ(x), we can state the ∂x ∂x → (2) i j relation as f = f˜ φ. ω = Bij dudv. ◦ ZS Z ∂u ∂v 6.5. HIGHER DIFFERENTIAL FORMS 177 178 CHAPTER 6. HAMILTON’S EQUATIONS In three dimensions, the parallelogram of ,with∂C its boundary. Let ω bea(k 1)-form. Then Stokes’ theorem M − which is the image of the rectangle [u, u + says du] [v, v + dv] has edges (∂~x/∂u)du and × v (∂~x/∂v)dv, which has an area equal to the dω = ω. (6.15) magnitude of u ZC Z∂C This elegant jewel is actually familiar in several contexts in three dimen- ∂~x ∂~x “dS~”= dudv sions. If k =2,C is a surface, usually called S, bounded by a closed path ∂u × ∂v! Γ=∂S.Ifω is a 1-form associated with A~, then ω = A~ d~`.Nowdω is Γ Γ · ~ the 2-form ~ A~, and dω = ~ A~ dS~, so we see that this Stokes’ and a normal in the direction of “dS”. Writing Bij in terms of the corre- ∼ ∇× S S ∇× · R R theorem includes the one we first learned by that name. But it also includes sponding vector B~ , Bij = ijkBk,so R R other possibilities. We can try k = 3, where C = V is a volume with surface S = ∂V . Then if ω B~ is a two form, ω = B~ dS~, while dω ~ B~ , (2) ∂~x ∂~x S S ω = ijkBk dudv ~ ~ ∼ · ∼ ∇· S S ∂u ∂v so V dω = BdV , so here Stokes’ generalR R theorem gives Gauss’s theo- Z Z !i !j ∇· rem. Finally, we could consider k =1,C = Γ, which has a boundary ∂C ∂~x ∂~x R R = Bk dudv = B~ dS,~ consisting of two points, say A and B. Our 0-form ω = f is a function, and S ∂u ∂v S 10 Z × !k Z · Stokes’ theorem gives df = f(B) f(A), the “fundamental theorem of Γ − calculus”. so ω(2) gives the flux of B~ through the surface. R RSimilarly for k = 3 in three dimensions, 6.6 The natural symplectic 2-form ∂~x ∂~x ∂~x ijk dudvdw ∂u! ∂v! ∂w! We now turn our attention back to phase space, with a set of canonical X i j k coordinates (qi,pi). Using these coordinates we can define a particular 1- is the volume of the parallelopiped which is the image of [u, u + du] [v, v + form ω = p dq . For a point transformation Q = Q (q ,...,q ,t)we × 1 i i i i i 1 n dv] [w, w + dw]. As ωijk = ω123ijk, this is exactly what appears: may use theP same Lagrangian, reexpressed in the new variables, of course. × 11 Here the Qi are independent of the velocitiesq ˙j, so on phase space dQi = ∂xi ∂xj ∂xk ω(3) = ω dudvdw = ω (x)dV. j(∂Qi/∂qj)dqj. The new velocities are given by ijk 123 ∂u ∂v ∂w 123 Z Z X Z P ∂Qi ∂Qi ∂Q˙ i ∂Qi Notice that we have only defined the integration of k-forms over subman- Q˙ i = q˙j + , so = . ∂qj ∂t ∂q˙j ∂qj ifolds of dimension k, not over other-dimensional submanifolds. These are Xj the only integrals which have coordinate invariant meanings. Also note that 10Note that there is a direction associated with the boundary, which is induced by a the integrals do not depend on how the surface is coordinatized. direction associated with C itself. This gives an ambiguity in what we have stated, for We state9 a marvelous theorem, special cases of which you have seen often example how the direction of an open surface induces a direction on the closed loop which before, known as Stokes’ Theorem. Let C be a k-dimensional submanifold bounds it. Changing this direction would clearly reverse the sign of A~ d~`.Wehavenot worried about this ambiguity, but we cannot avoid noticing the appearence· of the sign in R 9For a proof and for a more precise explanation of its meaning, we refer the reader to this last example. 11 ∂Qi the mathematical literature. In particular [14] and [3] are advanced calculus texts which We have not included a term ∂t dt which would be necessary if we were considering give elementary discussions in Euclidean 3-dimensional space. A more general treatment a form in the 2n + 1 dimensional extended phase space which includes time as one of its is (possibly???) given in [16]. coordinates. 6.6. THE NATURAL SYMPLECTIC 2-FORM 179 180 CHAPTER 6. HAMILTON’S EQUATIONS Thus the old canonical momenta, also know M is invertible and that J 2 = 1, so if we multiply this known 1 − equation from the left by J M − and from the right by J M, we learn ∂L(q, q,˙ t) ∂L(Q, Q,˙ t) ∂Q˙ ∂Q − · · p = = j = P j . that i ˙ j ∂q˙i q,t j ∂Qj q,t ∂q˙i q,t j ∂qi X X 1 T 1 J M − M J M J M = J M − J J M − · · · · · · − · · · · Thus the form ω1 may be written 1 = J M − M = J T ·T · ∂Qj = J J M J M = M J M, ω1 = Pj dqi = PjdQj, − · · · · · · i j ∂qi j X X X which is what we wanted to prove. Thus we have shown that the 2-form ω2 so the form of ω1 is invariant under point transformations. This is too lim- is form-invariant under canonical transformations, and deserves its name. ited, however, for our current goals of considering general canonical transfor- One important property of the 2-form ω2 on phase space is that it is mations on phase space, under which ω1 will not be invariant. However, its non-degenerate. A 2-form has two slots to insert vectors — inserting one exterior derivative leaves a 1-form. Non-degenerate means there is no non-zero vector ~v on phase ω := dω = dp dq space such that ω2( ,~v) = 0, that is, such that ω2(~u,~v) = 0, for all ~u on phase 2 1 i ∧ i · Xi space. This follows simply from the fact that the matrix Jij is non-singular. is invariant under all canonical transformations, as we shall show momen- tarily. This makes it special, the natural symplectic structure on phase Extended phase space space. We can reexpress ω2 in terms of our combined coordinate notation ηi, because One way of looking at the evolution of a system is in phase space, where a given system corresponds to a point moving with time, and the general Jijdηi dηj = dqi dpi = dpi dqi = ω2. equations of motion corresponds to a velocity field. Another way is to con- − ∧ − ∧ ∧ Xi One way to see this is to simply work out what dω3 is and apply it to the vector ξ, which is proportional to ~v =(˙qi, p˙i, 1). So we wish to show 6.6.1 Generating Functions dω3( ,~v) = 0. Evaluating · Consider a canonical transformation (q, p) (Q, P ), and the two 1-forms → dp dq ( ,~v)= dp dq (~v) dq dp (~v)= dp q˙ dq p˙ ω1 = i pidqi and ω10 = i PidQi. We have mentioned that the difference of i ∧ i · i i − i i i i − i i X dH dt( ,~v)=XdH dt(~v) dt dHX(~v) X X theseP will not vanish in general,P but the exterior derivative of this difference, 13 ∧ · − d(ω ω0 )=ω ω0 =0,soω ω0 is an closed 1-form. Thus it is exact , ∂H ∂H ∂H 1 − 1 2 − 2 1 − 1 = dqi + dpi + dt 1 and there must be a function F on phase space such that ω1 ω10 = dF . ∂qi ∂pi ∂t ! − 14 X X We call F the generating function of the canonical transformation ∂H ∂H ∂H If the transformation (q, p) (Q, P ) is such that the old q’s alone, without dt q˙i + p˙i + → − ∂qi ∂pi ∂t ! information about the old p’s, do not impose any restrictions on the new Q’s, X X ∂H ∂H ∂H ∂H then the dq and dQ are independent, and we can use q and Q to parameter- = dqi + dpi dt q˙i +˙pi ize phase space15. Then knowledge of the function F (q, Q) determines the ∂qi ∂pi − ∂qi ∂pi ! X X X transformation, as ∂H ∂H dω3( ,~v)= q˙i dpi p˙i + dqi · − ∂pi ! − ∂qi ! X ∂F ∂F ω1 ω10 = (pidqi PidQi)=dF = dqi + dQi ∂H ∂H − − ∂qi ∂Qi + q˙i +˙pi dt i i Q q X X ∂qi ∂pi ! X ∂F ∂F =0 = pi = , Pi = . ⇒ ∂q − ∂Q i Q i q where the vanishing is due to the Hamilton equations of motion. There is a more abstract way of understanding why dω ( ,~v) If the canonical transformation depends on time, the function F will 3 · vanishes, from the modified Hamilton’s principle, which states also depend on time. Now if we consider the motion in extended phase v dt that if the path taken were infinitesimally varied from the physi- δη space, we know the phase trajectory that the system takes through extended cal path, there would be no change in the action. But this change phase space is determined by Hamilton’s equations, which could be written in any set of canonical coordinates, so in particular there is some Hamiltonian is the integral of ω3 along a loop, forwards in time along the first trajectory and backwards along the second. From Stokes’ the- K(Q, P, t) such that the tangent to the phase trajectory, ~v, is annihilated by dω0 , where ω0 = PidQi K(Q, P, t)dt. Now in general knowing that two orem this means the integral of dω3 over a surface connecting 3 3 − these two paths vanishes. But this surface is a sum over infinitesimal parallel- 2-forms both annihilateP the same vector would not be sufficient to identify ograms one side of which is ~v ∆t and the other side of which12 is (δ~q(t),δ~p(t), 0). them, but in this case we also know that restricting dω3 and dω30 to their As this latter vector is an arbitrary function of t, each parallelogram must in- action on the dt = 0 subspace gives the same 2-form ω2. That is to say, if dependently give 0, so that its contribution to the integral, dω3((δ~q, δ~p, 0),~v)∆t = 13We are assuming phase space is simply connected, or else we are ignoring any compli- cations which might ensue from F not being globally well defined. 12It is slightly more elegant to consider the path parameterized independently of time, 14 This is not an infinitesimal generator in the sense we have in Lie algebras — this and consider arbitrary variations (δq,δp,δt), because the integral involved in the action, generates a finite canonical transformation for finite F . being the integral of a 1-form, is independent of the parameterization. With this approach 15Note that this is the opposite extreme from a point transformation, which is a canonical we find immediately that dω ( ,~v) vanishes on all vectors. transformation for which the Q’s depend only on the q’s, independent of the p’s. 3 · 6.6. THE NATURAL SYMPLECTIC 2-FORM 183 184 CHAPTER 6. HAMILTON’S EQUATIONS ~u and ~u 0 are two vectors with time components zero, we know that (dω in momenta to allow [Q ,P ]=δ to remain unchanged. This also doesn’t 3 − i j ij dω30 )(~u,~u 0) = 0. Any vector can be expressed as a multiple of ~v and some change H.Forλ = 1, this is the identity transformation, for which F =0, vector ~u with time component zero, and as both dω3 and dω30 annihilate ~v, of course. we see that dω dω0 vanishes on all pairs of vectors, and is therefore zero. Placing point transformations in this language provides another example. 3 − 3 Thus ω3 ω30 is a closed 1-form, which must be at least locally exact, and For a point transformation, Qi = fi(q1,...,qn,t), which is what one gets with − 16 indeed ω ω0 = dF , where F is the generating function we found above . a generating function 3 − 3 Thus dF = pdq PdQ+(K H)dt,or − − F2 = fi(q1,...,qn,t)Pi. P P i ∂F X K = H + . ∂t Note that ∂F2 ∂fj pi = = Pj The function F (q, Q, t) is what Goldstein calls F1. The existence of F ∂qi ∂qi Xj as a function on extended phase space holds even if the Q and q are not is at any point ~q a linear transformation of the momenta, required to preserve independent, but in this case F will need to be expressed as a function of the canonical Poisson bracket, but this transformation is ~q dependent, so other coordinates. Suppose the new P ’s and the old q’s are independent, so while Q~ is a function of ~q and t only, independent of p~, P~ (q, p, t) will in general we can write F (q, P, t). Then define F2 = QiPi + F . Then have a nontrivial dependence on coordinates as well as a linear dependence P on the old momenta. dF = Q dP + P dQ + p dq P dQ +(K H)dt 2 i i i i i i − i i − For a harmonic oscillator, a simple scaling gives X X X X = Qi dPi + pi dqi +(K H)dt, − p2 k 1 X X H = + q2 = k/m P 2 + Q2 , so 2m 2 2 ∂F ∂F ∂F q Q = 2 ,p= 2 ,K(Q, P, t)=H(q, p, t)+ 2 . i ∂P i ∂q ∂t where Q =(km)1/4q, P =(km) 1/4p. In this form, thinking P i i − θ The generating function can be a function of old momenta rather than the of phase space as just some two-dimensional space, we seem to old coordinates. Making one choice for the old coordinates and one for the be encouraged to consider a second canonical transformation Q new, there are four kinds of generating functions as described by Goldstein. Q, P θ, , generated by F1(Q, θ), to a new, polar, coordi- −→F1 P 1 Let us consider some examples. The function F1 = i qiQi generates an nate system with θ = tan− Q/P as the new coordinate, and interchange of p and q, P we might hope to have the radial coordinate related to the new momentum, = ∂F1/∂θ Q.AsP = ∂F1/∂Q θ is also Q cot θ, Qi = pi,Pi = qi, P −1 2 | | − we can take F1 = 2 Q cot θ,so 1 1 1 which leaves the Hamiltonian unchanged. We saw this clearly leaves the = Q2( csc2 θ)= Q2(1 + P 2/Q2)= (Q2 + P 2)=H/ω. form of Hamilton’s equations unchanged. An interesting generator of the P −2 − 2 2 1 second type is F2 = i λiqiPi, which gives Qi = λiqi, Pi = λi− pi, a simple Note as F1 is not time dependent, K = H and is independent of θ, which is change in scale of the coordinates with a corresponding inverse scale change therefore an ignorable coordinate, so its conjugate momentum is conserved. P P Of course differs from the conserved Hamiltonian H only by the factor 16 From its definition in that context, we found that in phase space, dF = ω ω0 , P 1 − 1 ω = k/m, so this is not unexpected. With H now linear in the new which is the part of ω ω0 not in the time direction. Thus if ω ω0 = dF 0 for some 3 − 3 3 − 3 other function F 0, we know dF 0 dF =(K0 K)dt for some new Hamiltonian function momentumq , the conjugate coordinate θ grows linearly with time at the − − P K0(Q, P, t), so this corresponds to an ambiguity in K. fixed rate θ˙ = ∂H/∂ = ω. P 6.6. THE NATURAL SYMPLECTIC 2-FORM 185 186 CHAPTER 6. HAMILTON’S EQUATIONS Infinitesimal generators, redux through phase space rather than as a change in coordinates. More generally, we may view a canonical transformation as a diffeomorphism18 of phase Let us return to the infinitesimal canonical transformation space onto itself, g : with g(q, p)=(Q, P ). M→M ζ = η + g (η ). For an infinitesimal canonical transformation, this active interpretation i i i j gives us a small displacement δη = [η, G] for every point η in phase space, so we can view G and its associated infinitesimal canonical transformation Mij = ∂ζi/∂ηj = δij + ∂gi/∂ηj needs to be symplectic, and so Gij = ∂gi/∂ηj satisfies the appropriate condition for the generator of a symplectic matrix, as producing a flow on phase space. G also builds a finite transformation by G J = J GT . For the generator of the canonical transformation, we repeated application, so that we get a sequence of canonical transformations · − · λ need a perturbation of the generator for the identity transformation, which g parameterized by λ = n∆λ. This sequence maps an initial η0 into a se- quence of points gλη , each generated from the previous one by the infinites- can’t be in F1 form (as (q, Q) are not independent), but is easily done in F2 0 λ+∆λ λ λ imal transformation ∆λG,sog η0 g η0 =∆λ[g η0,G]. In the limit form, F2(q, P)= i qiPi + G(q, P, t), with pi = ∂F2/∂qi = Pi + ∂G/∂qi, − ∆λ 0, with n allowed to grow so that we consider a finite range of λ,we Qi = ∂F2/∂Pi = qPi + ∂G/∂Pi,or → have a one (continuous) parameter family of transformations gλ : , M→M Q q 01I ∂G/∂q satisfying the differential equation ζ = i = i + i = η + J G, Pi pi 1I 0 ∂G/∂pi ·∇ − dgλ(η) = gλη, G . where we have ignored higher order terms in in inverting the q Q relation dλ → h i and in replacing ∂G/∂Pi with ∂G/∂pi. The change due to the infinitesimal transformation may be written in This differential equation defines a phase flow on phase space. If G is not terms of Poisson bracket with the coordinates themselves: a function of λ, this has the form of a differential equation solved by an exponential, λ λ[ ,G] δη = ζ η = J G = [η, G]. g (η)=e · η, − ·∇ which means In the case of an infinitesimal transformation due to time evolution, the small 1 gλ(η)=η + λ[η, G]+ λ2[[η, G],G]+.... parameter can be taken to be ∆t, and δη =∆t η˙ =∆t[H, η], so we see that 2 the Hamiltonian acts as the generator of time translations, in the sense that it maps the coordinate η of a system in phase space into the coordinates the In the case that the generating function is the Hamiltonian, G = H, this system will have, due to its equations of motion, at a slightly later time. phase flow gives the evolution through time, λ is t, and the velocity field on This last example encourages us to find another interpretation of canoni- phase space is given by [η, H]. If the Hamiltonian is time independent, the cal transformations. Up to now we have taken a passive view of the transfor- velocity field is fixed, and the solution is formally an exponential. mation, as a change of variables describing an unchanged physical situation, Let us review changes due to a generating function considered in the just as the passive view of a rotation is to view it as a change in the descrip- passive and alternately in the active view. In the passive picture, we view η tion of an unchanged physical point in terms of a rotated set of coordinates. and ζ = η+δη as alternative coordinatizations of the same physical point A in But rotations are also used to describe changes in the physical situation with phase space. For an infinitesimal generator F2 = qiPi +G, δη = J G = 17 i ∇ regards to a fixed coordinate system , and similarly in the case of motion [η, G]. A physical scalar defined by a function u(η) changes its functional through phase space, it is natural to think of the canonical transformation P generated by the Hamiltonian as describing the actual motion of a system 18An isomorphism g : is a 1-1 map with an image including all of (onto), M→N 1 N which is therefore invertible to form g− : . A diffeomorphism is an isomorphism 17 1 N→M We leave to Mach and others the question of whether this distinction is real. g for which both g and g− are differentiable. 6.6. THE NATURAL SYMPLECTIC 2-FORM 187 188 CHAPTER 6. HAMILTON’S EQUATIONS (A) form tou ˜, but not its value at a given physical point, sou ˜(ζA)=u(ηA). For ζ1 η (B) = ζ (A) the Hamiltonian, there is a change in value as well, for H˜ or K˜ may not be 1 1 η2 η the same as H, even at the corresponding point, (A) 2 δη A ζ2 (A) A η (A) η2 2 B η (B)= ζ (A) ∂F2 ∂G 2 K˜ (ζA)=H(ηA)+ = H(ηA)+ . 2 ∂t ∂t A η1 η 1 η (A) η (A) Now consider an active view. Here a canonical transformation is thought 1 1 of as moving the point in phase space, and at the same time changing the Passive view of the canonical trans- Active view of the transformation, functions u u˜, H K˜ , where we are focusing on the form of these func- formation. Point A is the same moving from point A in phase space → → tions, on how they depend on their arguments. We think of ζ as representing point, whether expressed in coor- to point B. So ηB = ζA. The func- the η coordinates of a different point B of phase space, although the coordi- dinates ηj or ζj, and scalar func- tionu ˜ then differs from u when eval- nates ηB = ζA. We ask howu ˜ and K˜ differ from u and H at B, evaluated tions take the same value there, so uated on the same coordinates η. at the same values of their arguments, not at what we considered the same u(ηA)=˜u(ζA). 19 physical point in the passive view. Let Note that if the generator of the transformation is a conserved quantity, the Hamiltonian is unchanged, in that it is the same function after the trans- formation as it was before. That is, the Hamiltonian is form invariant. ∂u So we see that conserved quantities are generators of symmetries of the ∆u =˜u(ηB) u(ηB)=˜u(ζA) u(ζA)=u(ηA) u(ζA)= δηi − − − − ∂ηi problem, transformations which can be made without changing the Hamil- ∂u tonian. We saw that the symmetry generators form a closed algebra under = [η ,G] = [u, G] i Poisson bracket, and that finite symmetry transformations result from expo- − i ∂ηi − X nentiating the generators. Let us discuss the more common conserved quan- tities in detail, showing how they generate symmetries. We have already seen that ignorable coordinates lead to conservation of the corresponding momen- tum. Now the reverse comes if we assume one of the momenta, say pI ,is conserved. Then from our discussion we know that the generator G = pI ∆H = K˜ (η ) H(η )=K˜ (ζ ) H(η ) B − B A − B will generate canonical transformations which are symmetries of the system. ∂G ∂G Those transformations are = H(ηA)+ H(ηA) δη ηH = [H, G] ∂t − − ·∇ ∂t − ! A δqj = [qj,pI ]=δjI,δpj = [pj,pI ]=0. dG = . dt Thus the transformation just changes the one coordinate qI and leaves all the other coordinates and all momenta unchanged. In other words, it is a translation of qI . As the Hamiltonian is unchanged, it must be independent 19 We differ by a sign from Goldstein in the definition of ∆u. of qI , and qI is an ignorable coordinate. 6.6. THE NATURAL SYMPLECTIC 2-FORM 189 190 CHAPTER 6. HAMILTON’S EQUATIONS Second, consider the angular momentum component ω~ L~ = ω r p The use of the Levi-Civita to write L as a vector is peculiar to three · ijk i j k ijk for a point particle with q = ~r. As a generator, ~ω L~ produces changes dimensions; in other dimensions d = 3 there is no -symbol to make a vector 6 · out of L, but the angular momentum can always be treated as an antisym- δr` = [r`,ijkωirjpk]=ijkωirj[r`,pk]=ijkωirjδ`k = ij`ωirj metric tensor, Lij = xipj xjpi. There are D(D 1)/2 components, and the = (ω~ ~r) , − − × ` Lie algebra again closes which is how the point moves under a rotation about the axis ω~ . The mo- [L ,L ]=δ L δ L δ L + δ L . mentum also changes, ij k` jk i` − ik j` − j` ik i` jk δp = [p , ω r p ]= ω p [p ,r ]= ω p ( δ )= ω p We have related conserved quantities to generators of infinitesimal canon- ` ` ijk i j k ijk i k ` j ijk i k − `j − i`k i k = (ω~ p~)`, ical transformation, but these infinitesimals can be integrated to produce fi- × nite transformations as well. As we mentioned earlier, from an infinitesimal so p~ also rotates in the same way. generator G we can exponentiate to form a one-parameter set of transforma- By Poisson’s theorem, the set of constants of the motion is closed un- tions der Poisson bracket, and given two such generators, the bracket is also a α[ ,G] symmetry, so the symmetries form a Lie algebra under Poisson bracket. ζα(η)=e · η For a free particle, p~ and L~ are both symmetries, and we have just seen 1 = 1+α[ ,G]+ α2[[ ,G],G]+... η that [p`,Li]=ik`pk, a linear combination of symmetries, while of course · 2 · [pi,pj] = 0 generates the identity transformation and is in the algebra. What 1 = η + α[η, G]+ α2[[η, G],G]+.... about [Li,Lj]? As Li = ik`rkp`, 2 [Li,Lj]=[ik`rkp`,Lj] In this fashion, any Lie algebra, and in particular the Lie algebra formed = ik`rk[p`,Lj]+ik`[rk,Lj]p` by the Poisson brackets of generators of symmetry transformations, can be = ik`rkj`mpm + ik`jmkrmp` exponentiated to form a continuous group of finite transformations, called a − Lie Group. In the case of angular momentum, the three components of L~ =(δijδkm δimδjk) rkpm (δijδm` δimδj`) rmp` − − − form a three-dimensional Lie algebra, and the exponentials of these form a =(δ δ δ δ ) r p ia jb − ib ja a b three-dimensional Lie group which is SO(3), the rotation group. = kijkabrapb = ijkLk. (6.16) We see that we get back the third component of L~ , so we do not get a new kind of conserved quantity, but instead we see that the algebra closes on the 6.7 Hamilton–Jacobi Theory space spanned by the momenta and angular momenta. We also note that We have mentioned the time dependent canonical transformation that maps ~ it is impossible to have two components of L conserved without the third the coordinates of a system at a given fixed time t into their values at ~ 0 component also being conserved. Note also that ω~ L does a rotation the a later time t. Now let us consider the reverse transformation, mapping ~ · same way on the three vectors ~r, p~, and L. Indeed it will do so on any vector (q(t),p(t)) (Q = q ,P = p ). But then Q˙ =0,P˙ = 0, and the Hamilto- 20 → 0 0 composed from ~r, and p~, rotating all of the physical system . nian which generates these trivial equations of motion is K = 0. We denote 20If there is some rotationally non-invariant property of a particle which is not built by S(q, P, t) the generating function of type 2 which generates this transfor- out of ~r and p~, it will not be suitably rotated by L~ = ~r p~, in which case L~ is not the mation. It satisfies full angular momentum but only the orbital angular momentum× . The generator of a rotation of all of the physics, the full angular momentum J~, is then the sum of L~ and ∂S ∂S another piece, called the intrinsic spin of the particle. K = H(q, p, t)+ =0, with pi = , ∂t ∂qi 6.7. HAMILTON–JACOBI THEORY 191 192 CHAPTER 6. HAMILTON’S EQUATIONS so S is determined by the differential equation which can be easily integrated ∂S ∂S q H q, ,t + =0, (6.17) W = 2mα kmq2 dq + f(α) ∂q ! ∂t Z0 − q α 1 which we can think of as a partial differential equation in n + 1 variables q, t, = f(α)+ θ + sin 2θ , (6.20) ω 2 thinking of P as fixed and understood. If H is independent of time, we can solve by separating the t from the q dependence, we may write S(q, P, t)= where we have made a substitution sin θ = q k/2α, used ω = k/m, and W (q, P) αt, where α is the separation constant independent of q and t, but − made explicit note that the constant (in q) ofq integration, f(α), mayq depend not necessarily of P . We get a time-independent equation on α. For other hamiltonians, we will still have the solution to the partial ∂W differential equation for S given by separation of variables S = W (q, P) αt, H q, = α. (6.18) − ∂q ! because H was assumed time-independent, but the integral for W may not be doable analytically. The function S is known as Hamilton’s principal function, while the As S is a type 2 generating function, function W is called Hamilton’s characteristic function, and the equa- tions (6.17) and (6.18) are both known as the Hamilton-Jacobi equation. ∂F ∂W p = 2 = . They are still partial differential equations in many variables, but under some ∂q ∂q circumstances further separation of variables may be possible. We consider first a system with one degree of freedom, with a conserved H, which we will For our harmonic oscillator, this gives sometimes specify even further to the particular case of a harmonic oscillator. ∂W ∂q α 1 + cos 2θ Then we we treat a separable system with two degrees of freedom. p = = = √2αm cos θ. We are looking for new coordinates (Q, P ) which are time independent, ∂θ , ∂θ ω 2α/k cos θ and have the differential equation for Hamilton’s principal function S(q, P, t): q Plugging into the Hamiltonian, we have ∂S ∂S H q, + =0. ∂q ! ∂t H = α(cos2 θ + sin2 θ)=α, 2 1 2 For a harmonic oscillator with H = p /2m + 2 kq , this equation is which will always be the case when (6.18) holds. 2 We have not spelled out what our new momentum P is, except that it is ∂S ∂S + kmq2 +2m =0. (6.19) conserved, and we can take it to be α. The new coordinate Q = ∂S/∂P = ∂q ∂t ! ∂W/∂α t. But Q is, by hypothesis, time independent, so |q − For any conserved Hamiltonian, we can certainly find a separated solution of the form ∂W = t + Q. S = W (q, P) α(P )t, ∂α q − and then the terms in (6.17) from the Hamiltonian are independent of t.For the harmonic oscillator, we have an ordinary differential equation, For the harmonic oscillator calculation (6.20), dW 2 ∂W 1 1 α ∂θ θ =2mα kmq2, = f 0(α)+ (θ + sin 2θ)+ (1 + cos 2θ)=f 0(α)+ ∂α ω 2 ω ∂α ω dq ! − q q 6.7. HAMILTON–JACOBI THEORY 193 194 CHAPTER 6. HAMILTON’S EQUATIONS where for the second equality we used sin θ = q k/2α to evaluate and 1 4c2 η2 ξ2 4c2 q y˙ = ξξ˙ − ηη˙ − . ∂θ q k 1 4c s ξ2 4c2 − s4c2 η2 ! = − = tan θ. − − ∂α 2α cos θ s2α −2α Squaring, adding in the x contribution, and simplifying then shows that q m ξ2 η2 ξ2 η2 Thus θ = ωt + δ, for δ some constant. T = − η˙2 + − ξ˙2 . 8 4c2 η2 ξ2 4c2 ! As an example of a nontrivial problem with two degrees of freedom which − − is nonetheless separable and therefore solvable using the Hamilton-Jacobi Note that there are no crossed terms ξ˙η˙, a manifestation of the orthogo- ∝ method, we consider the motion of a particle of mass m attracted by Newto- nality of the curvilinear coordinates ξ and η. The potential energy becomes nian gravity to two equal masses fixed in space. For simplicity we consider only motion in a plane containing the two masses, which we take to be at 1 1 2 2 4Kξ U = K + = K + = −2 2 . ( c, 0) in cartesian coordinates x, y.Ifr1 and r2 are the distances from the − r1 r2 − ξ + η ξ η ! ξ η ± − − particle to the two fixed masses respectively, the gravitational potential is In terms of the new coordinates ξ and η and their conjugate momenta, we 1 1 U = K(r− + r− ), while the kinetic energy is simple in terms of x and y, − 1 2 see that T = 1 m(˙x2 +˙y2). The relation between these is, of course, 2 2/m H = p2(ξ2 4c2)+p2(4c2 η2) 2mKξ . ξ2 η2 ξ − η − − 2 2 2 r1 =(x + c) + y − 2 2 2 y Then the Hamilton-Jacobi equation for Hamilton’s characteristic function is r2 =(x c) + y − 2 2 2/m 2 2 ∂W 2 2 ∂W r1 r (ξ 4c ) +(4c η ) 2mKξ = α, 2 ξ2 η2 − ∂ξ ! − ∂η ! − Considering both the kinetic and − or potential energies, the problem will 2 not separate either in c c x ∂W 1 (ξ2 4c2) 2mKξ mαξ2 terms of (x, y) or in terms of (r1,r2), but it does separate in terms of elliptical − ∂ξ ! − − 2 coordinates ∂W 2 1 +(4c2 η2) + αmη2 =0. − ∂η 2 ξ = r1 + r2 ! η = r1 r2 If W is to separate into a ξ dependent piece and an η dependent one, the − first line will depend only on ξ, and the second only on η, so they must each From r2 r2 =4cx = ξη we find a fairly simple expressionx ˙ =(ξη˙ + ξη˙ )/4c. 1 − 2 be constant, with W (ξ,η)=Wξ(ξ)+Wη(η), and The expression for y is more difficult, but can be found from observing that 2 1 2 2 2 2 2 2 2 (r + r )=x + y + c =(ξ + η )/4, so 2 2 dWξ(ξ) 1 2 2 1 2 (ξ 4c ) 2mKξ αmξ = β − dξ ! − − 2 2 2 2 2 2 2 2 2 2 ξ + η ξη 2 (ξ 4c )(4c η ) y = c = − − , 2 2 dWη(η) 1 2 4 − 4c ! − 16c2 (4c η ) + αmη = β. − dη ! 2 − or 1 These are now reduced to integrals for Wi, which can in fact be integrated y = ξ2 4c2 4c2 η2 to give an explicit expression in terms of elliptic integrals. 4c − − q q 6.8. ACTION-ANGLE VARIABLES 195 196 CHAPTER 6. HAMILTON’S EQUATIONS 6.8 Action-Angle Variables so J is just a constant 1/ k/m times the old canonical momentum α, and thus its conjugate φ = qk/m(t + β), so ω = k/m as we expect. The Consider again a general system with one degree of freedom and a conserved important thing here isq that ∆φ =2π, even ifq the problem itself is not Hamiltonian. Suppose the system undergoes periodic behavior, with p(t) solvable. andq ˙(t) periodic with period τ. We don’t require q itself to be periodic as it might be an angular variable which might not return to the same value when the system returns to the same physical point, as, for example, the Exercises angle which describes a rotation. If we define an integral over one full period, 6.1 In Exercise 2.7, we discussed the connection between two Lagrangians, L1 and L2, which differed by a total time derivative of a function on extended configuration 1 t+τ space, J(t)= pdq, d 2π t L ( q , q˙ ,t)=L ( q , q˙ ,t)+ Φ(q , ..., q ,t). Z 1 { i} { j} 2 { i} { j} dt 1 n it will be time independent. As p = ∂W/∂q = p(q, α), the integral can be You found that these gave the same equations of motion, but differing momenta defined without reference to time, just as the integral 2πJ = pdq over one (1) (2) pi and pi . Find the relationship between the two Hamiltonians, H1 and H2, orbit of q, holding α fixed. Then J becomes a function of α alone,R and if we and show that these lead to equivalent equations of motion. assume this function to be invertible, H = α = α(J). We can take J to be our canonical momentum P . Using Hamilton’s Principal Function S as the 6.2 A uniform static magnetic field can be described by a static vector potential generator, we find Q = ∂S/∂J = ∂W(q, J)/∂J (dα/dJ)t. Alternatively, we A~ = 1 B~ ~r. A particle of mass m and charge q moves under the influence of this − 2 × might use Hamilton’s Characteristic Function W by itself as the generator, field. to define the conjugate variable φ = ∂W(q, J)/∂J, which is simply related (a) Find the Hamiltonian, using inertial cartesian coordinates. to Q = φ (dα/dJ)t. Note that φ and Q are both canonically conjugate (b) Find the Hamiltonian, using coordinates of a rotating system with angular − velocity ω~ = qB/~ 2mc. to J, differing at any instant only by a function of J. As the Hamilton- − Jacobi Q is time independent, we see that φ˙ = dα/dJ = dH/dJ = ω(J), which is a constant, because while it is a function of J, J is a constant in 6.3 Consider a symmetric top with one point on the symmetry axis fixed in space, as we did at the end of chapter 4. Write the Hamiltonian for the top. time. We could also derive φ˙ from Hamilton’s equations considering W as a Noting the cyclic (ignorable) coordinates, explain how this becomes an effective genenerator, for W is time independent, the therefore the new Hamiltonian one-dimensional system. is unchanged, and the equation of motion for φ is simply φ˙ = ∂H/∂J. Either way, we have φ = ωt + δ. The coordinates (J, φ) are called action-angle 6.4 (a) Show that a particle under a central force with an attractive potential variables. Consider the change in φ during one cycle. inversely proportional to the distance squared has a conserved quantity D = 1~r 2 · p~ Ht. − ∂φ ∂ ∂W d d (b) Show that the infinitesimal transformation generated by G := 1~r p~ scales ∆φ = dq = dq = pdq = 2πJ =2π. 2 · I ∂q I ∂q ∂J ! dJ I dJ ~r and p~ by opposite infinitesimal amounts, Q~ =(1+ )~r, P~ =(1 )p~, or for a 2 − 2 finite transformation Q~ = λ~r, P~ = λ 1p~. Show that if we describe the motion in Thus we see that in one period τ,∆φ =2π = ωτ,soω =1/τ. − terms of a scaled time T = λ2t, the equations of motion are invariant under this For our harmonic oscillator, of course, combined transformation (~r,p, ~ t) (Q,~ P,T~ ). → 2α 2π 2απ 2πJ = pdq = √2αm cos2 θdθ = 6.5 We saw that the Poisson bracket associates with every differentiable function s k 0 k/m f on phase space a differential operator D := [f, ] which acts on functions g I Z f · q 6.8. ACTION-ANGLE VARIABLES 197 198 CHAPTER 6. HAMILTON’S EQUATIONS on phase space by Df g =[f,g]. We also saw that every differential operator is Find as many relations as you can, expressible without coordinates, among these associated with a vector, which in a particular coordinate system has components forms. Consider using the exterior derivative and the wedge product. fi, where ∂ Df = fi . 6.8 We have considered k-forms in 3-D Euclidean space and their relation to ∂ηi X vectors expressed in cartesian basis vectors. We have seen that k-forms are in- A 1-form acts on such a vector by variant under change of coordinatization of , so we can use them to examine M the forms of the gradient, curl, divergence and laplacian in general coordinates in dxj(Df )=fj. three dimensional space. We will restrict our treatment to orthogonal curvilinear Show that for the natural symplectic structure ω2, acting on the differential oper- coordinates (q ,q ,q ), for which we have, at each point p , a set of or- 1 2 3 ∈M ator coming from the Poisson bracket as its first argument, thonormal basis vectorse ˆi directed along the corresponding coordinate, so that dqi(ˆej)=0fori = j. We assume they are right handed, soe ˆi eˆj = δij and ω2(Df , )=df , 6 · · eˆi eˆj = k ijkeˆk. The dqi are not normalized measures of distance, so we define × h (p) so that dq (ˆe )=h 1δ (no sum). which indicates the connection between ω2 and the Poisson bracket. i P i j i− ij (a) For a function f(q ,q ,q ) and a vector ~v = v eˆ , we know that df (~v)=~v ~ f. 6.6 Give a complete discussion of the relation of forms in cartesian coordinates 1 2 3 i i ·∇ Use this to find the expression for ~ f in the basise ˆ . in four dimensions to functions, vector fields, and antisymmetric matrix (tensor) ∇ P i fields, and what wedge products and exterior derivatives of the forms correspond (b) Use this to get the general relation of a 1-form ωidqi to its associated vector ~v = v eˆ . to in each case. This discussion should parallel what is done in section 6.5 for i i P ˜ (a) (a) (b) (b) three dimensions. [Note that two different antisymmetric tensors, Bµν and Bµν = (c) IfP a 1-form ω is associated with ~v and 1-form ω is associated with ~v , 1 B , can be related to the same 2-form, in differing fashions. They we know the 2-form ω(a) ω(b) is associated with ~v(a) ~v(b). Use this to find the 2 ρσ µνρσ ρσ ∧ × are related to each other with the four dimensional , which you will need to general association of a 2-form with a vector. P jk`m define, and are called duals of each other. Using one fashion, the two different (d) We know that if a 1-form ω is associated with a vector ~v, then dω is associated 2-forms associated with these two matrices are also called duals. with ~ ~v. Use this to find the expression for ~ ~v in orthogonal curvilinear ∇× ∇× (b) Let F be a 4 4 matrix defined over a four dimensional space (x, y, z, ict), coordinates. µν × with matrix elements F = B , for j, k, ` each 1, 2, 3, and F = iE = F . (e) If the 1-form ω is associated with ~v and the 2-form Ω is associated with F~ ,we jk jk` ` 4j j − j4 Show that the statement that F corresponds, by one of the two fashions, to a know that ω Ω is associated with the scalar ~v F~ . Use this to find the general ∧ · closed 2-form F, constitutes two of Maxwell’s equations, and explain how this association of a 3-form with a scalar. implies that 2-form is the exterior derivative of a 1-form, and what that 1-form is (f) If the 2-form Ω is associated with ~v, we know that dΩ is associated with the in terms of electromagnetic theory described in 3-dimensional language. divergence of ~v. Use this to find the expression for ~ ~v in orthogonal curvilinear ∇· (c) Find the 3-form associated with the exterior derivative of the 2-form dual to F, coordinates. and show that it is associated with the 4-vector charge current density J =(~j, icρ), (g) Use (a) and (f) to find the expression for the laplacian of a scalar, 2f = ~ ~ f, ∇ ∇·∇ where ~j is the usual current density and ρ the usual charge density. in orthogonal curvilinear coordinates. 6.7 Consider the following differential forms: 6.9 Consider the unusual Hamiltonian for a one-dimensional problem A = ydx+ xdy+ dz 2 B = y2 dx + x2 dy + dz H = ω(x +1)p, C = xy(y x) dx dy + y(y 1) dx dz + x(x 1) dy dz − ∧ − ∧ − ∧ where ω is a constant. D =2(x y) dx dy dz − ∧ ∧ E =2(x y) dx dy (a) Find the equations of motion, and solve for x(t). − ∧