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e Fig. 7.15 Solution The constraint imposed on the point A is such that the line of nodes is fixed. Choosing the fixed and moving reference frames as in Fig. 7.15, we obtain that the vector ω is the sum of ω = − ˙ϕ e 3 (7.120)
and of ω e = − ˙θN = − ˙θ(cos ϕ e 1 + sin ϕ e 2 ), (7.121) where N is the unit vector of the node lines (we assume the line coincides with the ξ
1 -axis). The signs in equations (7.120), (7.121) are due to the fact that θ and ϕ increase in the negative direction of the rotations. Recall that J 1
J 2 = 1 2 J 3 = 1 4 M R
2 and note that (A − O) × φ = e 3 × φ (cos ϕe 1 + sin ϕe 2 ). Then Euler’s equations can be written as − 1
M R 2 d dt ( ˙
θ cos ϕ) = − 1 4 M R
2 ˙ θ ˙ ϕ sin ϕ − φ sin ϕ, − 1
M R 2 d dt ( ˙
θ sin ϕ) = 1 4 M R 2 ˙ θ ˙ ϕ cos ϕ + φ cos ϕ, 7.15 The mechanics of rigid bodies: dynamics 271 the third one giving ¨ ϕ = 0. We can choose the reference system in such a way that θ(0) = ϕ(0) = 0. In addition, ˙ θ(0) = ω 0 e , ˙ ϕ(0) = ω
0 3 . Clearly ˙ ϕ(t) = ω 0 3 . Multiplying the first equation by cos ϕ, and the second by sin ϕ and adding them, we find ¨ θ = 0 ⇒ ˙θ = ω 0 e . From the latter we can easily deduce the expression for the reaction φ: φ = −
2 2 ω 0 3 ω 0 e N. Note that the constraint contributes the torque M = 1 2 M R 2 ω 0 3 ω 0 e N × e 3 , which can be expressed as M = J 3 ω 0 3 de 3 /dt, in agreement with what was discussed regarding gyroscopic effects. Problem 3 Consider the precession by inertia of a rigid body around a point O with respect to which the ellipsoid of inertia is not of revolution ( J 1
J 2
J 3
suitable Lyapunov function prove that the rotations around the middle axis of the ellipsoid of inertia are unstable. Solution We write the system (7.54) in the form ˙ ω = f(ω), with f 1 = J 1 − J 3 J 1 ω 2 ω 3 , f 2 = J 3 − J 1 J 2 ω 3 ω 1 , f 3 = J 1 − J
2 J 3 ω 1 ω 2 . Consider the equilibrium point (0, ω 0 2 , 0), ω 0 2 = / 0, and prove that the function Λ ∗ ( ω) = J
1 J 2 J 3 ω 1 (ω 2 − ω 0 2 )ω 3 satisfies the hypotheses of the ˇ Cetaev instability theorem (Theorem 4.10). Examine first of all ˙ Λ ∗
∇ ω Λ ∗ ·f(ω):
˙ Λ ∗ = J 2 J 3 ( J 2 − J 3 )ω 2 ω 2 3 (ω 2 − ω 0 2 ) + J 1 J 3 ( J 3 − J
1 )ω 2 1 ω 2 3 + J 1 J 2 ( J 1 − J 2 )ω 2 1 ω 2 (ω 2 − ω 0 2 ). The first and the last term have sign opposite to the sign of ω 2 (ω
− ω 0 2 ), while the second term is positive if ω 1 ω
= / 0. To fix ideas, assume that ω 0 2
the inequality ω 2 (ω 2 − ω
0 2 ) < 0 is satisfied for 0 < ω 2 < ω 0 2 . In this strip we have ˙ Λ ∗ > 0. Hence if we take Ω 1
1 > 0,
ω 3 > 0, all the hypotheses of ˇ Cetaev’s theorem are satisfied. 272 The mechanics of rigid bodies: dynamics 7.15
Fig. 7.16 Problem 4 In a horizontal plane a system of three equal rods AB, BC, CD hinged to one another at B and C, is constrained at the ends A, D by hinges at two fixed points located at a mutual distance a. Let l be the length of each rod. Then we assume that 0 < a < 3l. The points B and D are mutually attracting by an elastic force. Study the equilibrium of the system. Solution Set a/l = γ ∈ (0, 3). Choose ϕ ∈ (−π, π), the angle the rod AB makes with the line AD (Fig. 7.16). This angle may be subject to further limitations, as we request that BD ≤ 2l.
Since λ = BD/l = (1 + γ 2 − 2γ cos ϕ) 1/2 , it must be that λ 2 ≤ 4, and hence γ 2 − 2γ cos ϕ − 3 ≤ 0, yielding 0 < γ < cos ϕ + 3 + cos 2 ϕ. The function cos ϕ + 3 + cos 2
at ϕ = ±π. It follows that for γ ∈ (1, 3) the above inequality defines the range ( −ϕ
, ϕ 0 ) for ϕ, where ϕ 0 ∈ (0, π) (ϕ 0 → 0 for γ → 3 and ϕ 0 → π for γ → 1). When γ ∈ (0, 1] there are no prohibited values for ϕ. If ϕ is an admissible, positive value, we find two configurations for the rods BC, CD, corresponding to the two isosceles triangles of side l that can be constructed on the segment BD. Let α be the angle ˆ BDA and β be the angle ˆ BDC. Then it is easy to deduce that sin α = sin ϕ
λ , cos β = 1 2 λ, and hence also that cos α =
(λ 2 − sin 2 ϕ) 1/2 λ = γ − cos ϕ λ , sin β = 1 2 (3 − γ
2 + 2γ cos ϕ) 1/2 .
ϕ → −ϕ. For γ ∈ (0, 1] at equilibrium we have the two configurations that min- imise λ 2
(ϕ = π). Obviously, in the first case the equilibrium is stable, and in the second 7.15 The mechanics of rigid bodies: dynamics 273
0
Fig. 7.17 case it is unstable. When γ ∈ (1, 3) there still exists the solution ϕ = 0, with the corresponding configuration of stable equilibrium, and the solutions ϕ = ±ϕ 0
which λ 2 = 4, i.e. the segment BD by the rods BC, CD being made collinear. The corresponding equilibrium configurations are unstable. Problem 5 Outline the dynamic analysis of the following plane system (Fig. 7.17): a disc (mass M , radius R) rotating around the fixed centre O; a rod (mass m, length l), with the end A hinged at a point of the disc at a distance a < l from the centre and with the end B sliding along a fixed line (x-axis) through O. Suppose that the constraints are frictionless and that the only force applied is an elastic force which pulls B towards the point B 0 on the x-axis at a distance a + l from O. Solution The system has one degree of freedom. Choose as the Lagrangian coordinate the angle ϕ formed by the radius OA with the x-axis, and determine the angle ψ =
ˆ ABO. By the sine theorem we find sin ψ = a
sin ϕ, cos ψ =
1 − a 2 l 2 sin 2 ϕ (7.122) (note that ψ cannot reach π/2). The coordinates of the centre of mass G of the rod are G =
a cos ϕ + 1 2 l cos ψ, 1 2 l sin ψ ,
274 The mechanics of rigid bodies: dynamics 7.15
1
2
1
2
1
2
Fig. 7.18 from which we obtain ˙ G 2 = a
2 sin
2 ϕ ˙
ϕ 2 + 1 4 l 2 ˙ ψ 2 − al sin ϕ sin ψ ˙ϕ ˙ψ. It follows that the kinetic energy is given by T =
1 4 M R 2 ˙ ϕ 2 + 1 2 m a 2 sin
2 ϕ ˙
ϕ 2 + 1 4 l 2 ˙ φ 2 + al sin ϕ sin ψ ˙ ϕ ˙ ψ +
1 12 l 2 ˙ ψ 2 . In addition to equation (7.122) we must use ˙ ψ =
a l cos ϕ ˙ ϕ 1 − a 2 l 2 sin
2 ϕ to obtain T = 1 2 ˙ ϕ 2 ⎧ ⎨ ⎩ 1 2 M R 2 + ma
2 ⎛ ⎝sin 2 ϕ +
1 3 cos 2 ϕ 1 − a 2 l 2 sin 2 ϕ + sin 2 ϕ a l cos ϕ
1 − a 2 l 2 sin 2 ϕ ⎞ ⎠ ⎫ ⎬ ⎭ . The potential energy is V =
1 2 k(B − B 0 ) 2 = 1 2 k[a(1
− cos ϕ) + l(1 − cos ψ)] 2 . We now have all the elements necessary to write down Lagrange’s equation, which is very complicated. However it is easy to answer a question such as the following: if ˙ ϕ 0 is the angular velocity of the disc corresponding to ϕ = 0, what is the velocity corresponding to ϕ = π? To answer, it is enough to use T + V = E = 1 2 ˙ ϕ 2 0 1 2 M R 2 + 1 3 ma 2 .
7.15 The mechanics of rigid bodies: dynamics 275 Since V (π) = 2ka 2 , the new kinetic energy is T = 1
˙ ϕ 2 1 2 M R 2 + 1 3 ma 2 = 1 2 ˙ ϕ 2 0 1 2 M R 2 + 1 3 ma 2 − 2ka
2 (we must of course have ˙ ϕ
0 > 4ka 2 1 2 M R 2 + 1 3 ma 2 for ϕ = π to be reached). The analysis of the equilibrium of the system is trivial, given that to ϕ = 0 and ϕ = π there correspond the minimum and maximum of the potential energy. Since the potential energy has only variations of O(ϕ 4
near ϕ = 0, the small oscillations are not harmonic. Problem 6 To illustrate the use of the cardinal equations in the systems made of rigid parts mutually constrained, we examine the simplest case: two rods AB, BC, hinged at B and constrained by hinges at A, C (Fig. 7.18). Apply to the two rods two generic loads F 1 , F
2 at the two internal points P 1 , P
2 , respectively. If l 1 , l
2 are
the lengths of the rods, determine all the constraint reactions. Solution
The cardinal equations F 1 + F 2 + φ A + φ
C = 0,
(P 1 − A) × F 1 + (P
2 − A) × F
2 + (C
− A) × φ C = 0 (7.123) are three scalar equations in the unknowns φ Ax , φ
Ay , φ
Cx , φ
Cy . Hence the system is underdetermined. This is due to the fact that the system is not a rigid system. (However, if the two rods are welded at B, yielding a rigid system, why are the equations still insufficient? How must the constraints be modified to make them sufficient?) To solve the problem, it is necessary to write the cardinal equations for every rigid component. In the present case, this is equivalent to writing six equations for the four unknowns mentioned plus the two components of the reaction in B. Let ξ 1 = |P 1 − A|, ξ 2 = |P 2 − C|. Denote by φ B the force that the hinge transmits to the rod AB (the force transmitted to the rod BC is −φ B ); we find for the equilibrium of the rod AB the equations F 1x
Ax + φ
Bx = 0,
F 1y + φ Ay + φ
By = 0,
ξ 1 cos ϕ 1 F 1y − ξ 1 sin ϕ 1 F 1x + l 1 cos ϕ 1 φ By − l 1 sin ϕ 1 φ Bx = 0, 276 The mechanics of rigid bodies: dynamics 7.15 and for BC the equations F 2x + φ Cx − φ
Bx = 0,
F 2y + φ Cy − φ
By = 0,
−ξ 2 cos ϕ 2 F 2y − ξ 2 sin ϕ 2 F 2x + l 2 cos ϕ 2 φ By + l 2 sin ϕ 2 φ Bx = 0. The two torque balance equations can be solved independently to obtain φ Bx , φ
By . Note that the determinant of the coefficients is l 1 l 2 sin(ϕ 1 + ϕ 2 ) and that the solvability condition is 0 < ϕ 1 + ϕ 2 < π, i.e. the two rods cannot be collinear. Once φ
Bx , φ
By are known, the remaining equations are trivial. This example includes the case that gravity is the only force acting on the system. It is also interesting to analyse the case that the rods are not carrying any load, and that the force is applied at the hinge point. This is the simplest case of a truss (more generally, we can have systems with more than two rods concurring at the same node). When a rod carries no load, the cardinal equations imply that the forces at the extreme points must constitute a balanced pair. Hence the rod is under pure tension or pure compression. In the elementary case that the two rods have a force F applied at B the problem is solved immediately by decomposing F in the direction of the two rods. Problem 7 The effect of tides caused by the Moon is to constantly increase the length of the day. Compute what is the eventual duration of the day when it coincides with the lunar month (the period of revolution of the Moon around the Earth). The radius of the Earth is r = 6.4 × 10 3
T /m L = 81, the Earth–Moon distance is R = 3.8 × 10 5
velocity ω T of the Earth and the angular velocity ω L of revolution of the Moon around the Earth is ω T /ω L = 28.
Solution For simplicity suppose that the axis of rotation is orthogonal to the plane of the Moon’s orbit, that the Earth is a homogeneous sphere and that the Moon is a point mass. The total angular momentum |L| with respect to the centre of mass of the two bodies is equal to the sum of the contributions due to the rotation of the Earth and of the Moon around the centre of mass and of the rotation of the Earth around its axis: |L| = m L
T R m L + m
T 2 ω L + m
T m L R m L + m T 2 ω L + 2 5 m T r 2 ω T = I L ω L + I T ω T , (7.124) where I T = 2 5 m T r 2 , I L = [m L m T /m L + m T ]R 2 . When the day is equal to the lunar month, the angular velocity of the Earth is equal to that of the revolution of the Moon around the Earth; we denote them by ω (they both have to
7.15 The mechanics of rigid bodies: dynamics 277 change to keep |L| constant). Under this condition, by Kepler’s third law, the Moon–Earth distance is equal to d = R(ω L /ω)
2/3 . The angular momentum is equal to |L| = (I
L + I
T )ω,
(7.125) where I
L = [m
L m T /m L + m T ]d 2 . Comparing (7.124) with (7.125), and setting x = ω/ω
T , since I L /I
= x 4/3
(ω T /ω L ) 4/3 , we find x + a
ω L ω T x 1/3 = 1 + a, (7.126)
where a = I L ω L /I T ω T . From equation (7.126) we obtain the fourth-degree equation x(1 + a
− x) 3 − a 3 ω L ω T = 0. (7.127) Substituting the approximate numerical values a 3.8, ω L
T 1/28, we find that (7.127) has two real roots, x f 1/55 and x p 4. The solution x f cor-
responds to the future: the day has a duration of 55 present days, and the Earth–Moon distance is about 6 × 10 5
p corresponds to a past state when the day’s duration was 6 present hours and the Earth–Moon distance was only 2 × 10 4
considering the rotation axes of the Moon and of the Earth as orthogonal to the plane of the orbit. In reality, the respective inclinations are 23.5 ◦ and 5
◦ , and
hence the inclination of the lunar orbit with respect to the Earth’s equator varies between 18.5 ◦ and 28.5
◦ . On the other hand, the angular momentum due to the Moon’s rotation around its own axis is very small, with an approximate value of 2 5 m L r 2 L ω L , with r
L = 1.7
× 10 3 km. Comparing this with the angular momentum of the Earth 2 5 m T r 2 ω T we see that the ratio is of order 10 −3 . 8 ANALYTICAL MECHANICS: HAMILTONIAN FORMALISM 8.1 Legendre transformations Within the Lagrangian formalism, the phase space for the equations of motion makes use of the coordinates (q, ˙q). We start this chapter by studying the coordinate transformations in this space. We shall see that this study has wide and significant developments. The first objective is the transformation of the equations of motion into a form, the so-called canonical form, whose particular structure highlights many important properties of the motion. This objective is realised by an application of a transformation due to Legendre. In this section, we study the most important properties of this transformation. For simplicity, consider a real function f of a real variable w, defined in an interval (a, b) (not necessarily bounded), with continuous, positive second derivative, f (w) > 0. (8.1)
Because of (8.1) the equation f (w) = p (8.2) uniquely defines a function w = w(p), with p variable in an open interval (c, d) where w (p) exists and is continuous. Geometrically, we can interpret w(p) as the abscissa of the point where the graph of f (w) is tangent to the line of slope p (Fig. 8.1). D efinition 8.1 The Legendre transform of f(w) is the function g(p) = pw(p) − f[w(p)]. (8.3) A significant property of this transform is that it is involutive. The meaning of this is expressed in the following. T heorem 8.1 The function (8.3) in turn has a Legendre transform, which coincides with the initial function f (w). Proof
To verify that g(p) admits a Legendre transform it is sufficient to check that g (p) > 0. Differentiating (8.3), and using (8.2), we find g (p) = w(p) + pw (p) − f [w(p)]w (p) = w(p), (8.4)
280 Analytical mechanics: Hamiltonian formalism 8.1
Fig. 8.1
and therefore g (p) = w (p) = {f [w(p)]} −1 > 0. (8.5) To construct the Legendre transform of g(p) we must firstly define the function p(w) through the equation g (p) = w. (8.6) Comparing this equation with (8.4), we find that p(w) is the inverse function of w(p). Through p(w) we obtain the expression for the Legendre transform h(w) of g(p), simply by an application of the definition: h(w) = wp(w) − g[p(w)]. (8.7) Finally, inserting (8.3) into (8.7) yields h(w) = wp(w) − {p(w)w − f(w)} = f(w). (8.8) We note that the use of Legendre transforms yields the Young inequality: pw ≤ f(w) + g(p), (8.9) where f and g are strictly convex functions which are the Legendre transform of each other. The inequality (8.9) is proved starting from the function F (w, p) = pw − f(w)
(8.10) (where f is any function which admits a Legendre transform). Indeed, note that ∂F ∂w
− f (w), ∂ 2 F ∂w 2 = −f (w) < 0, 8.1 Analytical mechanics: Hamiltonian formalism 281 and hence the maximum of F for every fixed p is taken when w = w(p), defined by (8.2). This maximum value thus coincides with g(p), i.e. F (w, p)
≤ g(p), which yields (8.9). This proves in particular that in equation (8.9) equality holds along the curve w = w(p). Example 8.1 Compute the Legendre transform of f (w) = aw n , a > 0, n > 1, and prove that the Cauchy inequality 2pw
≤ εw 2 + 1 ε p 2 , ∀ ε > 0 (8.11) can be deduced from the Young inequality (8.9), which for this choice of f has the form pw ≤ a w n + (n
− 1) p na n/ (n−1)
(8.12) (see also Problem 6.1); it suffices to choose n = 2, a = ε/2. The previous considerations extend without difficulty to the case of a real function f (w), with w ∈ R , and with continuous second partial derivatives, such that the quadratic form associated with the Hessian matrix ∂ 2
∂w h ∂w k (8.13)
is positive definite. In this case it is possible to invert the system ∂f ∂w k = p
k , k = 1, . . . , , (8.14) and thus to define the vectorial function w = w(p). It is now clear how it is possible to define the Legendre transform of f (w). D efinition 8.2 The Legendre transform of f(w) is g(p) = p · w(p) − f[w(p)]. (8.15) We can also prove that f (w) in turn represents the Legendre transform of g(p); it is enough to note that ∇g = w(p), and hence that the Hessian matrix of g(p) coincides with the Jacobian matrix of w = w(p), and therefore with the inverse of the Hessian matrix of f (w). The latter is also positive definite. We can then define the function p = p(w) by inverting the system ∂g ∂p k = w
k , k = 1, . . . , . (8.16) 282 Analytical mechanics: Hamiltonian formalism 8.2 We conclude that p(w) is the inverse of w(p). Finally, we proceed as for (8.7) and (8.8) to obtain the final result. Analogously we can extend the Young inequality: p · w ≤ f(w) + g(p), (8.17) where the equals sign holds for w = w(p). Remark 8.1 The Legendre transform is paired with the invertible variable transformation from w k
k , defined by (8.14). The inverse transform is defined by (8.16). Remark 8.2 According to Theorem 4.1, the Lagrangian function L(q, ˙q, t) of a system admits a Legendre transform with respect to the variables ˙ q k , for every fixed q and t. Remark 8.3 The Legendre transform can be defined by inverting the signs in the right-hand side of (8.3); this is equivalent to considering the transform of −f, a common trick in thermodynamics. 8.2 The Hamiltonian We are ready to pass from the Lagrangian formalism to a new representa- tion in phase space, by a Legendre transformation of the Lagrangian variables (Remark 8.2). Only the kinetic variables ˙ q k are transformed, and replaced by the corresponding variables p k , while the Lagrangian is replaced by its Legendre transform, called the Hamilton function or Hamiltonian. Therefore the transformation is obtained by expressing explicitly the vector ˙q = ˙q(q, p, t) from the system (linear in the variables ˙ q k , as clearly follows from (4.32))
∂L ∂ ˙
q k = p k , k = 1, . . . , , (8.18) and the Hamiltonian is then defined by H(p, q, t) = p · ˙q(p, q, t) − L(q, ˙q(p, q, t), t). (8.19) D
k , q
k ) are called conjugated canonical variables. The p k
The reason for the latter terminology is that the variables p k are the Lagrangian components of the linear momentum in R 3n (see (4.33)). Remark 8.4 When there are no generalised potentials, we have ∂L/∂ ˙ q k
q k , and the transformation (8.18) depends only on the geometric structure of the holonomic system under consideration, and not on the system of applied forces. 8.2 Analytical mechanics: Hamiltonian formalism 283 Verify that if the Lagrangian L is replaced by the Lagrangian L = cL or L = L + d dt F (q) (Remark 4.8), we obtain the respective momenta p = cp, p = p + ∇ q
the corresponding Hamiltonians H = cH, H = H. Example 8.2 For an unconstrained point particle (P, m), of Cartesian coordinates q i , we have T = 1 2 m 3 i =1 ˙ q 2 i , and hence p i = m ˙
q i , ˙ q i = p i /m. It follows that p · ˙q = p 2 /m and if the particle is subject to a field of potential energy V (q) we easily obtain H(p, q) = p 2
+ V (q). (8.20)
In particular, the Hamiltonian of the harmonic oscillator of frequency ω is H(p, q) = p 2
+ 1 2 mω 2 q 2 . (8.21) In the previous example the Hamiltonian coincides with the total energy. This fundamental property is valid in more general situations. T heorem 8.2 In a holonomic system with fixed constraints, of Lagrangian L(q, ˙q), and without any generalised potential, the Hamilton function H(p, q) represents the total mechanical energy of the system. Proof From equation (8.2) it follows immediately that H does not depend explicitly on t. The kinetic energy T is a homogeneous quadratic form in ˙ q k , and consequently p · ˙q = k =1 ∂T ∂ ˙ q k ˙ q k = 2T, (8.22)
from which it follows that H(p, q) = 2T − L(q, p) = T (p, q) + V (q). (8.23)
Remark 8.5 When there are generalised potentials, Theorem 8.18 is no longer valid. In this respect see Problem 8.1. Remark 8.6 Since T (hence L) is a quadratic function of ˙ q k , equations (8.18) are an invert- ible linear system. It follows that, in the Hamiltonian formalism, T becomes a quadratic function of p k (homogeneous if the constraints are fixed). 284 Analytical mechanics: Hamiltonian formalism 8.3 8.3
Hamilton’s equations The main advantage in using the Legendre transformation in the phase space is that the equations of motion then take the form ˙ p k = − ∂H ∂q k , k = 1, . . . , , ˙ q
= ∂H ∂p k , k = 1, . . . , . (8.24) Equations (8.24) are called Hamilton’s canonical equations and can be easily verified. The second group coincides with equations (8.16), and describes the transformation ˙q = ˙q(p, q, t). This can also be obtained by directly differentiating the two sides of (8.19) with respect to p k and using (8.18). Differentiating (8.19) with respect to q k we obtain ∂H ∂q k = p · ∂ ˙q ∂q k − ∂L ∂q k − ∇ ˙ q L · ∂ ˙q ∂q k = − ∂L ∂q k , where we have once again used (8.18). Finally, recall that thanks to (8.18), Lagrange’s equations (4.75) can be written in the form ˙ p k = ∂L ∂q k , k = 1, . . . , . (8.25)
This yields immediately the first group of (8.24). Remark 8.7 The Hamiltonian obviously has the same regularity as the Lagrangian. The exist- ence and uniqueness of the solution of the initial value problem for equations (8.24) is thus guaranteed. P roposition 8.1 If q = q(t), p = p(t) are solutions of the system (8.24), we have d dt H(p(t), q(t), t) = ∂ ∂t H(p(t), q(t), t). (8.26)
Proof The proof follows directly from (8.24). Remark 8.8 From equations (8.26) we again find that if ∂H/∂t = 0 the Hamilton function H is a constant of the motion, called the generalised energy integral. In what follows, it is convenient to use a vector notation for equations (8.24), introducing the vector x = p q and the 2 × 2 matrix I = 0
1 0 , (8.27) 8.4 Analytical mechanics: Hamiltonian formalism 285 where 1 and 0 are the identity and the null × matrix, respectively. Equations (8.29) can then be written as ˙x = I∇H.
(8.28) Remark 8.9 The right-hand side of (8.28) is a vector field X(x, t), prescribed in phase space. Equations (8.28) are the equations for the flow lines of this field. In the autonomous case, when ∂H/∂t = 0, we find that the possible trajectories of the system in the phase space are the flow lines of the field X(x), which belong to the level sets H(p, q) = constant (and coincide with them for l = 1). P roposition 8.2 If H has two continuous second derivatives with respect to q k and p
k , then
div X = 0, (8.29)
where X(x, t) = I∇H. (8.30)
Proof It is of immediate verification. Equation (8.29) has significant consequences, which we will consider in the following section, with reference more generally to systems of the form ˙x = X(x, t), div X = 0, x ∈ R
n . (8.31) Remark 8.10 The Hamilton system (8.24) is less generic than (8.31), since it has the peculiarity that the projections X k of X on each of the planes (q k , p
k ), have zero divergence in the geometry of the respective planes, where X k is the vector of components (∂H/∂p k , −∂H/∂q k ). 8.4 Liouville’s theorem Hamilton’s equations (8.24) induce a transformation S t in the phase space into itself, depending on time. With every point x 0 ∈ R 2 and for all t > 0 we can associate the point x(t) obtained by integrating the Hamilton system with the initial condition x(0) = x 0 . The transformation is invertible because of the reversibility of the equations of motion. An important property of the transformation S t , which we call the flow in phase space associated with the Hamiltonian H, is the following. T heorem 8.3 (Liouville) In phase space, the Hamiltonian flow preserves volumes. This property is true in general for any system of the type (8.31). 286 Analytical mechanics: Hamiltonian formalism 8.4 Proof
We must show that in the flow we are considering, for every t > 0 the image Ω (t) of any domain Ω ⊂ R
2 with a regular boundary has the same measure as Ω . Consider the flow associated with any differential system of the kind ˙x = X(x, t). (8.32)
Let us ignore for the moment that div X = 0, and let us prove that d dt | Ω (t) | = Ω (t) div X(x, t) dx, (8.33)
where | Ω | denotes the measure of Ω . From (8.33) it clearly follows that | Ω (t) | is constant for any system of the type (8.31). Equation (8.33) expresses the balance of volumes depicted in Fig. 8.2. The variation of the volume in time dt can be expressed as ∂ Ω
X · N dσ dt, where N is the outgoing normal. Therefore d/dt | Ω (t) | is simply the outgoing flux of the field X(x, t) through ∂ Ω (t) and equation (8.33) immediately follows. Equation (8.33) highlights the physical significance of the divergence of a velocity field v: div v is the dilation rate of the unit volume. We shall see that this theorem has important applications in statistical mech- anics. It also gives information on the nature of singular points (i.e. the constant solutions) of (8.31). C orollary 8.1 A singular point of a system of the type (8.31) cannot be asymptotically stable. Proof
If x 0 were asymptotically stable, then there would exist a sphere of centre x 0 such that all trajectories starting inside the sphere would tend asymptotically to x 0
t → ∞, contradicting Theorem 8.3. Download 10.87 Mb. Do'stlaringiz bilan baham: 
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