Hamiltonian mechanics and its applications
Dr. Sukanta Deb and Dr. Subhash Kumar Department of Physics, Acharya Narendra Dev College (University of Delhi) Govindpuri, Kalkaji, New Delhi 110019
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Learning Objectives: • Statement of Hamilton’s principle, the foundational principle of classical mechanics. • Defintion of Hamiltonian and conservation of energy. • Derivation of Hamilton’s canonical equations of motion from the defintion of Hamiltonian. • Derivation of Hamilton’s canonical equations of motion from the Hamilton’s principle. • Application of Hamilton’s equations to derive the equations of motion of various physical systems such as simple pendulum, compound pendulum, projectile motion, etc.
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Hamilton’s Principle Hamilton’s principle is the generalization and reformulation of the principle of least action. It is one of the foundational principles of classical mechanics. The principle is named after Sir William Rowan Hamilton (1805 − 1865), an eminent Irish physicist, astronomer, and mathematician. The formalism of this principle has been outlined in his two papers published in 1834 and 1835 [1]. Hamilton’s principle allows to determine the equation of motion for a system without requiring to specify the total force acting on the system. Hence it offers tremendous advantages over Newton’s laws of equation of motion which requires specification of the total force and sometimes it becomes very difficult to know all the forces acting on the system. The principle may be stated as follows [1]: “Out of all the possible paths along which a dynamical system may move from one point to another within a specified interval of time (consistent with constraints, if any), the actual path followed is the one which minimizes the time integral of the Lagrangian function for the system." In terms of the calculus of variations, Hamilton’s principle becomes Z t2 L(qi , q˙i , t)dt = 0 δ t1
Hamilton’s principle is applicable to systems even when the energy of the particle or system of particles is not conserved but still can be represented in terms of a potential function. Such cases arise when the forces represented by constraints do work on the system or when the potential energy depends on t explicitly. In fact, the Hamilton’s principle can be applied to all systems called the Hamiltonian systems whose dynamics is fully governed by a potential energy function and constraints [2]. Sir William Rowan Hamilton (1805 − 1865) was an Irish physicist, astronomer, and mathematician. He made important contributions to classical mechanics, algebra, and optics. He is best known for his contribution to mathematical physics in the reformulation of Newtonian mechanics, now called Hamiltonian mechanics. This work has led to the substantial development in classical field theories such as electromagnetism, and plays a pivotal role in the development of quantum mechanics. He developed the mathematical theory of quaternions. Hamilton was a genius at a very early age. At the age of 13 he learned 15 languages and his ability to find mistakes in famous work of mathematics impressed many [3] (Photo credit: Enterprise Ireland).
Conservation of energy (Definition of Hamiltonian): If the Lagrangian function L of a system does not explicitly depend on time, i.e., time is homogeneous in an inertial reference frame, then the translation of the system along time t leaves it unchanged. Let us consider the Lagrangian L of a system as L = L(q1 , q2 , . . . , qi , . . . , qn , q˙1 , q˙2 , . . . , qi , . . . , q˙n , t) ∂L ∂L ∂L ∂L ∂L ∂L ∂L dq1 + dq2 + · · · + dqn + dq˙1 + dq˙2 + · · · + dq˙n + dt ∂q1 ∂q2 ∂qn ∂ q˙1 ∂ q˙2 ∂ q˙n ∂t ∂L dq1 ∂L dq2 ∂L dqn ∂L dq˙1 ∂L dq˙2 ∂L dq˙n ∂L = + + ···+ + + + ···+ + ∂q1 dt ∂q2 dt ∂qn dt ∂ q˙1 dt ∂ q˙2 dt ∂ q˙n dt ∂t n n X X ∂L dqi ∂L dq˙i ∂L = + + ∂q dt ∂ q ˙ dt ∂t i i i=1 i=1
⇒ dL = ⇒
dL dt
⇒
dL dt
(1)
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We know that the Lagrange equation for a conservative system is given by ∂L d ∂L =0 − ∂qi dt ∂ q˙i ∂L d ∂L ⇒ = ∂qi dt ∂ q˙i Therefore, equation (1) becomes
⇒
d dt
X n n ∂L dq˙i ∂L dL X dqi d ∂L + = + dt dt dt ∂ q˙i ∂ q˙i dt ∂t i=1 i=1 n n X ∂L dL X d ∂L ∂L ⇒ q¨i q˙i + + = dt dt ∂ q ˙ ∂ q ˙ ∂t i i i=1 i=1 n ∂L ∂L dL X d q˙i + = ⇒ dt dt ∂ q ˙ ∂t i i=1 n dL X d ∂L ∂L ⇒ ∵ pi = = [q˙i pi ] + dt dt ∂t ∂ q˙i i=1 ! n X ∂L pi q˙i − L = − ∂t i=1
(2)
The quantity in the parentheses is called the Hamiltonian function H, i.e., n X
H=
pi q˙i − L
i=1
Equation (2) becomes dH ∂L =− dt ∂t If the Lagrangian function does not depend on t explicitly, ∂L =0 ∂t which implies that dH = 0 ⇒ H = constant. dt Now pi =
∂ ∂T ∂V ∂L = (T − V ) = (∵ = 0 for a conservative system.) ∂ q˙i ∂ q˙i ∂ q˙i ∂ q˙i
Therefore, the Hamiltonian function H becomes H= H=
n X
i=1 n X i=1
q˙i pi − L q˙i
∂T −L ∂ q˙i
Euler’s theorem states that if f (yi ) is a homogeneous function of yi of degree n, then n X i=1
yi
∂f = nf ∂yi
(3)
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Applying Euler’s theorem, equation (3) becomes H = 2T − L ⇒ H = 2T − (T − V ) ⇒H=T +V ⇒ H = E (constant) Thus we find that the Hamiltonian H represents the total energy of the system E and is conserved under the condition that 1. the system is a conservative one, i.e., the potential energy V is independent of q˙i , 2. kinetic energy T of the system is a homogeneous quadratic function of the q˙i and 3. the transformation equation does not involve time, i.e. rj = rj (qi ).
Hamilton’s Canonical Equations of Motions From the defintion of the Hamiltonian (H), we know that it is related with the Lagrangian (L) as: H(qi , pi , t) =
n X
q˙i pi − L(qi , q˙i , t)
(4)
i=1
Since H = H(q1 , q2 , . . . , qi , . . . , qn , p1 , p2 , . . . , pi , . . . , pn , t) dH =
n X ∂H i=1
∂qi
dqi +
n X ∂H i=1
∂pi
dpi +
∂H dt ∂t
(5)
From equation (4), we have n n X X ∂L ∂L ∂L dqi − dq˙i − dt ∂q ∂ q ˙ ∂t i i i=1 i=1 i=1 i=1 n n n n X X X X ∂L ∂L ∂L ⇒ dH = dqi − dt ∵ pi = q˙i dpi + pi dq˙i − pi dq˙i − ∂qi ∂t ∂ q˙i i=1 i=1 i=1 i=1
dH =
⇒ dH =
n X
q˙i dpi +
n X
n X
q˙i dpi −
n X ∂L ∂L dqi − dt ∂qi ∂t i=1
i=1
pi dq˙i −
(6)
From the Lagrange’s equation we know that d ∂L − ∂qi dt
Substituting
∂L ∂ q˙i
∂L =0 ∂ q˙i ∂L d ∂L ⇒ = ∂qi dt ∂ q˙i ∂L dpi ⇒ = ∂qi dt ∂L = p˙i ⇒ ∂qi
= p˙i in equation (6), we get dH =
n X i=1
q˙i dpi −
n X i=1
p˙i dqi −
∂L dt ∂t
(7)
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Comparing equations (5) and (7), we get ∂H ∂pi ∂H p˙i = − (8) ∂qi ∂H ∂L =− ∂t ∂t These first two equations are known as the Hamilton’s canonical equations of motion. In the case of the Hamilton’s equations, the coordinates qi ’s and momenta pi ’s form 2n canonical equations and replace n Lagrange equations (where n denotes the number of degrees of freedom of the system) and these equations are first order differential equations. On the other hand, the Lagrange equations are of second order. The steps to use canonical equations in solving a problem are outlined below: q˙i =
1. Construct the Hamiltonian (H) as a function of the generalised coordinates (qi ) and momenta (pi ). In some cases, it may be possible to do this directly. Otherwise, for complicates cases 2. the Lagrangian (L) is to be set up and the generalised momenta are calculated using the equation pi =
∂L . ∂ q˙i
The equations of motion can then be obtained from the Hamitlton’s canonical equations. From the Hamilton’s equation, we have ∂L ∂H =− ∂t ∂t If H does not contain qi , ∂H = 0 ⇒ p˙i = 0 ⇒ pi = constant. ∂qi Such a coordinate (qi ) is called a cyclic or ignorable coordinate.
Dervation of Hamilton’s equations from Hamilton’s principle From the Hamilton’s principle, we know that δ ⇒
Z
Z
t2
L(qi , q˙i , t)dt = 0
t1 t2
δL(qi , q˙i , t)dt = 0 [∵ δ(dt) = 0]
t1
From the defintion of the Hamiltonian we know that it is related with the Lagrangian as X pi q˙i − L(qi , q˙i , t) H(qi , pi , t) = i
⇒ L(qi , q˙i , t) =
X
pi q˙i − H(qi , pi , t)
i
⇒ δL =
X
(pi δ q˙i + q˙i δpi ) − δH(qi , pi , t)
i
X ∂H ∂H pi δ q˙i + q˙i δpi − δqi − δpi ∂qi ∂pi i X ∂H ∂H [Adding and subtracting p˙i δqi ] pi δ q˙i + p˙i δqi − p˙i δqi + q˙i δpi − δqi − δpi ⇒ δL = ∂qi ∂pi i X d ∂H ∂H ⇒ δL = )δqi + (q˙i − )δpi (pi δqi ) − (p˙i + dt ∂qi ∂pi i ⇒ δL =
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Substituting the value of δL in the Hamilton’s principle, we have Z t2 X ∂H d ∂H )δqi + (q˙i − )δpi dt = 0 (pi δqi ) − (p˙i + dt ∂qi ∂pi t1 i Z t2 ∂H ∂H (p˙i + ⇒ pi δqi |tt21 − )δqi − (q˙i − )δpi dt = 0 ∂qi ∂pi t1 Z t2 ∂H ∂H (p˙i + ⇒− )δqi − (q˙i − )δpi dt = 0 ∂qi ∂pi t1 In this equation the first term vanishes since δqi = 0 at the terminal points 1 and 2. Also, since qi′ s and p′i s are all independent δqi and δpi are arbitrary at all points of the path. So to get zero in the right hand side, the coefficient of δqi and δpi must be separately zero. ∂H ∂H = 0 and q˙i − =0 ∂qi ∂pi ∂H ∂H ⇒ p˙i = − and q˙i = ∂qi ∂pi
p˙i +
These are the Hamilton’s canonical equations of motion.
Legendre Transformation The procedure for transition from Lagrangian to Hamiltonian formulation corresponds to changing the variables in the mechanical functions from (q, q) ˙ to (q, p). The procedure for switching variables in this manner is facilitated by the Legendre transformation. This kind of transformation is very useful for this type of change of variable [4]. Let us consider any arbitrary function defined by f = f (x, y). The total derivative of this function is given by df =
∂f ∂f dx + dy ∂x y ∂y x
⇒ df = udx + vdy, where
Let us define a function
∂f ∂f u= and v = ∂x y ∂y x g(x, y, u) = ux − f (x, y)
(9)
⇒ dg = d(ux) − df = udx + xdu − udx − vdy ⇒ dg = xdu − vdy Thus we find g to be a function of u and y, i.e., g = g(u, y). In order to have an explicit expression for g(u, y), the equation ∂f u(x, y) = ∂x y must be inverted to obtain
x = x(u, y).
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It is then inserted into equation (9) so that g(u, y) = ux(u, y) − f (x(u, y), y) This is called the Legendre transform. We find that the Legendre transform allows a means to transform one function f (x, y) to a different function g(u, y), where u = ∂f ∂x without any loss of information. Infact, the function f (x, y) can always be y
recovered from g(u, y) by observing the fact that ∂g ∂f ∂g = x(u, y) and = −v = − . ∂u y ∂y u ∂y
This assures that the inverse Lagrange transform
f =u takes us back to the orginal function.
∂g −g ∂u y
Adrien-Marie Legendre (1752−1833) was a famous French mathematician. He developed mathematical theories of Legendre transformations, Legendre polynomials and elliptical integrals. He made remerkable contributions in the field of statistics, number theory, abstract algebra, and mathematical analysis. The least squares method was developed by him which finds wide range of applications in linear regression, statistics, signal processing and curve fitting problems. In classical mechanics, the Legendre transformation is used to switch from Lagrangian forulation to Hamiltonian formulation. The Legendre transfomation also helps finding enthalpy, Helmholtz and Gibbs free energies from the internal energy in statistical mechanics [5].
Examples Example 1. Obtain Hamiltonian from Lagrangian using the Legendre transformation. Let us consider a particle of mass m with generalized coordinate q and corresponding generalized velocity q. ˙ Let T and V represent the kinetic and potential energies of the particle, respectively. The Lagrangian of the particle is defined as L(q, q) ˙ = T − V. Now, let us try to transform L(q, q) ˙ to a new function H(p, q), where p is the momentum. Let us now apply the Legendre transformation as discussed above in order to carry out the transformation L(q, q) ˙ → H(p, q). Let us put f ≡L x ≡ q˙ (Switching variable) y≡q ∂L ∂L = = p (Canonical momentum) u= ∂x y ∂ q˙ q g≡H Kineteic energy of the particle (T ) = 12 mq˙2 and the potential energy of the particle is a function of q only, i.e., V = V (q). Now ∂T ∂L = = mq˙ p= ∂ q˙ q ∂ q˙ q
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From the Legendre transformation we know that g(u, y) = ux(u, y) − f (x(u, y), y) Using the equivalences as listed above, we get H(p, q) = pq˙ − L(q, ˙ q) We now calculate 1 1 H = pq˙ − L(q, ˙ q) = (mq) ˙ q˙ − mq˙2 + V (q) = mq˙2 + V (q) = T + V = E 2 2 Therefore we find that the Hamiltonian (H) denotes the total energy (E) of the system in this case. Example 2. Compute the Legendre transform of f (x) = ex . Also show that the inverse Legendre transform takes back to the original function. We know that the Legendre transform of f is given by g = ux − f ⇒ g = ux − ex with ∂f = ex ∂x ⇒ x = log u
u=
Substituting the value of x back, we get g = ux − f = u log u − f (log u) = u log u − u. This gives the required transform of the function f (x) = ex . The inverse Legendre transform can be obtained using ∂g −g f =u ∂u y
with
∂g ∂g ∂f = x(u, y) and = −v = − . ∂u y ∂y u ∂y
Now
x= Therefore, f =u
∂g 1 = u + log u − 1 = log u ⇒ u = ex ∂u u
∂g − g = ex x − u log u + u = ex x − ex x + ex = ex ∂u
which is the original function. Example 3. Write down the Hamiltonian for a simple pendulum and obtain its equations of motion. For a simple pendumlum, 1 1 ˙ 2 = 1 ml2 θ˙2 Kinetic Energy(T ) = mv 2 = m(lθ) 2 2 2 Potential Energy(V ) = mgl(1 − cos θ)
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Therefore, the Lagrangian (L) is L=T −V =
1 2 ˙2 ml θ − mgl(1 − cos θ) 2
∂L = ml2 θ˙ ∂θ pθ ⇒ θ˙ = ml2
⇒ pθ =
Hence,
1 2 pθ 2 pθ 2 + mgl(1 − cos θ) ml 2 4 + mgl(1 − cos θ) = 2 m l 2ml2 Therefore, the Hamilton’s equations of motion are Hamiltonian(H) = T + V =
pθ ∂H ∂H = and p˙θ = − = mgl sin θ θ˙ = ∂pθ ml2 ∂θ Now, ∂H p˙θ = − = mgl sin θ ∂θ d ml2 θ˙ = −mgl sin θ ⇒ dt ⇒ ml2 θ¨ = −mgl sin θ ⇒ lθ¨ + g sin θ = 0 ⇒ lθ¨ + gθ = 0 (For small values of θ, sin θ ≈ θ) g ⇒ θ¨ + θ = 0 l This gives the required equation of motion for a simple pendulum. Example 4. Using Hamilton’s equations obtain the equation of motion for an ideal spring-mass system. The Lagrangian of the system is given by 1 1 mx˙ 2 − kx2 , k = Spring constant of the system 2 2 L = mx˙ ⇒ px = ∂ x˙ px ⇒ x˙ = m
L=T −V =
The Hamiltonian (H) of the system is given by 1 mx˙ 2 − 2 px 2 + ⇒H= 2m
H=T +V =
1 2 kx 2 1 2 kx 2
Hamilto’s equations are x˙ =
∂H px ∂H = and p˙x = − = −kx ∂px m ∂x
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Now p˙x = −kx ⇒
d (mx) ˙ = −kx dt ⇒ m¨ x = −kx
⇒x ¨+
k x=0 m
This is the equation of motion for the ideal mass-spring system. Example 5. Let us consider a particle of mass m moving in one dimension under the influence of a force F (x, t) =
k −(t/τ ) e x2
where k and τ represent positive constants. Calculate the Hamiltonian function. Make a comparision between the Hamiltonian and the total energy, and discuss the conservation of energy of the system. Comparing k F (x, t) = 2 e−(t/τ ) x with ∂V , F (x, t) = − ∂x where V denotes the potential energy of the system, we find V =
k −(t/τ ) e x
Therefore, the Lagrangian is given by L=T −V =
k 1 mx˙ 2 − e−(t/τ ) 2 x ∂L ⇒ px = ∂ x˙ ⇒ px = mx˙
The Hamiltonian is given by H = px x˙ − L 1 k ⇒ H = mx˙ 2 − mx˙ 2 + e−(t/τ ) 2 x k 1 ⇒ H = mx˙ 2 + e−(t/τ ) 2 x ⇒H=T +V Thus we find that the Hamiltonian is equal to the total energy T + V . But the total energy of the system is not conserved because H contains t explicitly. Example 6. A particle having mass m moves under the influence of gravity along the helix given by z = kθ, r = constant, where k represents a constant and z-axis is directed towards the vertical. Obtain the Hamilitonian equations of motion. In cylindrical polar coordinates (r, φ, z), we have i 1 h Kinetic Energy(T ) = m r˙ 2 + r2 θ˙2 + z 2 2
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Figure 1: Helix described by equations z = kθ, r = constant.
Potential Energy(V ) = mgz Given, ˙ = constant ⇒ r˙ = 0 z = kθ ⇒ z˙ = k θr The Lagrangian is given by 2 1 r m 2 z˙ 2 + z˙ 2 − mgz 2 k 2 ∂L r ⇒ pz = + 1 z˙ =m ∂ z˙ k2
L=T −V =
⇒ z˙ =
pz m
r2 k2
+1
The Hamiltonian is given by H = pz z˙ − L = pz
pz m r2 k2
⇒ H = pz z˙ − L = pz
2 1 r 2 2 − m 2 z˙ + z˙ + mgz 2 k +1 pz m
r2 k2
+1
−
⇒H=
pz 2
2m 2m
r2 k2 + pz 2 r2 k2 +
1 1
+ mgz + mgz
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Figure 2: Simple pendnulum whose length is shortened by
dl dt
= −α = constant.
The Hamilitons’ equations of motion are given by z˙ =
∂H = ∂pz m
pz r2 k2 +
This is the required equation of motion of the particle.
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and p˙z = −
∂H = −mg ∂z
∂H = −mg Now, p˙z = − ∂z 2 r + 1 z¨ = −mg m k2 g ⇒ z¨ = r2 k2 + 1
Example 7. Consider a simple plane pendulum having a mass m attached to a string of length l. The length of the string is shortened at a constant rate dl = −α = constant. dt after the predulum is set into motion. Obtain the Hamiltonian and the total energy, and discuss the conservation of energy for the system. Kinetic energy (T ) is given by # " 2 i 2 1 h 1 dl 1 2 + lθ˙ = m α2 + l2 θ˙2 (T ) = mv = m 2 2 dt 2 Potential energy (V ) is given by V = mgl(1 − cos θ) Therefore, the Lagrangian (L) is L=T −V =
i 1 h 2 m α + l2 θ˙2 − mgl(1 − cos θ) 2 ∂L ⇒ pθ = = ml2 θ˙ ∂θ pθ ⇒ θ˙ = ml2
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Hence, 1 1 Hamiltonian(H) = pθ θ˙ − L = ml2 θ˙2 − mα2 + mgl(1 − cos θ) 6= T + V 2 2 Since H 6= T + V, the total energy is not conserved in this system. Example 8. A particle of mass m is attracted to a force center with the force of magnitude and find Hamilton’s equations of motion. Given k F (r) = − 2 r ∂V k ⇒− = 2 ∂r r k ⇒ V (r) = − r The Lagrangian of (L) the particle moving in the central force field is given by 1 L = T − V = m r˙ 2 + r2 θ˙2 − V (r) 2 The generalised momenta (pr , pθ ) are givn by
k r2 .
Use plane polar coordinates
∂L ∂L = mr˙ and pθ = = mr2 θ˙ ∂ r˙ ∂ θ˙ pr pθ ⇒ r˙ = and θ˙ = 2 m mr 2 2 1 pθ pr + V (r) Therefore, H = T + V = m + r2 2 m mr2 pθ 2 1 2 pr + 2 + V (r) ⇒H= 2m r pr =
The Hamiltons equations of motion (qi =
∂H ∂pi ,
pi =
∂H ∂qi )
are
∂H pr = ∂pr m 2 ∂V pθ ∂H − = p˙r = − ∂r mr3 ∂r p ∂H θ = θ˙ = ∂pθ mr2 ∂H p˙θ = − =0 ∂θ From the last two relations of equation (10), we find that r˙ =
(10)
pθ = constant ⇒ mr θ˙ = constant 2
But mr2 θ˙ is the angular momentum. Therefore in the case of central force problem we find that the angular momentum is conserved. From the second relation of equation (10) we have ∂V ∂H pθ 2 − = ∂r mr3 ∂r 2 4 ˙2 k m r θ − 2 ⇒ m¨ r= mr3 r
p˙r = −
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Example 9. Obtain the Hamilton’s equation of motion for a compound pendulum. We know that for a compound pundulum 1 ˙2 Iθ 2 Potential energy(V ) = M gl cos θ Kinetic energy(T ) =
Therefore, the Lagrangian is given by L=T −V = ⇒ pθ =
1 ˙2 I θ + M gl cos θ 2
∂L = I θ˙ ∂ θ˙
The Hamiltonian is given by H=
X i
H= The Hamilton’s equations of motion (q˙i =
∂H ∂pi
1 pi q˙i − L = pθ θ˙ − I θ˙2 − M gl cos θ 2
pθ 2 − M gl cos θ 2I
∂H and p˙i = − ∂q ) are: i
pθ ∂H ∂H = and p˙θ = − = −M gl sin θ θ˙ = ∂pθ I ∂θ Now, p˙θ = −M gl sin θ ˙ ⇒ I θ¨ = −M gl sin θ (∵ pθ = I θ) M gl sin θ = 0 ⇒ θ¨ + I M gl θ = 0 (∵ sin θ = θ for small values of θ) ⇒ θ¨ + I
The above equation represents simple harmonic motion (SHM) with time period given by s I T = 2π M gl x + Ky ˆ, where k and Example 10. Consider a particle of mass m moving in two dimensions, subjected to a force F = −kxˆ K represent positive constants. Write down the Hamiltonian and Hamilton’s equation of motion using x and y as generalised coordinates. The kinetic energy is 1 T = m(x˙ 2 + y˙ 2 ). 2 The potential energy of the system is Z r 1 V =− F.dr = kx2 + Ky. 2 0 The Hamiltonian of the system is given by H=
1 1 (px 2 + py 2 ) + kx2 + Ky. 2m 2
The Hamilto’s equations of motion for x are x˙ =
px ∂h ∂H = and p˙x = − = −kx ∂px m ∂y
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The above two equations can be combined to yield x ¨ = −(k/m)x Thus x oscialltes in simple harmonic motion (SHM). The solution of this equation is given by p x = A cos (ωt − δ) with ω = k/m.
On the other hand, the Hamilton’s equations of motion for y are y˙ =
py ∂H ∂H = and p˙y = − = −K ∂py m ∂y
The above two equations can be combined to yield y¨ = −(K/m) Thus, we find that y accelerates in the negative direction, 1 y = − (K/m)t2 + vy0 t + y0 , with consant acceleration − K/m 2 Example 11. A smooth wire is bent to take the shape of a helix which is described in the cylindrical polar coordinates given by ρ = R and z = kφ. R and k represent constants. A bead of mass m is allowed to slide on it. The z-axis is taken vertically upwards. Write down the Hamiltonian of the system using z as the generalized coordinate. Find out the Hamilton’s equations and hence the bead’s vertical acceleration z¨. In the limit that R → 0, what is z¨? Does this make sense? In cylindrical polar coordinates (ρ, φ, z), we have x = ρ cos φ, y = ρ sin φ, z=z The components of velocity (v) in cylindrical polar coordinates are given by ˙ z˙ = 0, Rφ,˙ z˙ = z˙ 0, R , 1 v = ρ, ˙ ρφ, k The kinetic energy of the bead is
1 R2 1 2 2 T = mv = mz˙ 1 + 2 2 2 k
The potential energy of the bead is
V = mgz The Lagrangian of the system is given by R2 1 2 L = T − V = mz˙ 1 + 2 − mgz 2 k R2 ∂L = mz˙ 1 + 2 ⇒ pz = ∂ z˙ k The Hamiltonian of the system is given by 1 R2 R2 2 H = pz z˙ − L = mz˙ 1 + 2 − mz˙ 1 + 2 + mgz k 2 k 2 1 R ⇒ H = mz˙ 2 1 + 2 + mgz 2 k pz 2 ⇒H= 2 + mgz 2m 1 + R k2 2
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The Hamilton’s equations of motion (q˙i = z˙ =
∂H ∂pi , p˙i
∂H = − ∂q ) are: i
∂H pz = ∂pz m 1+
Now let us take
R2 k2
and p˙z = −
∂H = −mg ∂z
∂H = −mg p˙z = − ∂z R2 ⇒ z¨m 1 + 2 = −mg k g ⇒ z¨ = − 2 1+ R k2
In the limit R → 0 the above equation reduces to z¨ = −g. This is correct since the helix reduces to a vertical frictionless wire on which the acceleration is just g vertically downward.
Summary In this chapter we have discussed the Hamilton’s principle and derived Hamilton’s canonical equations of motion. The generalised coordinates (qi ’s) and corresponding momenta (pi ’s) are used as independent variables in Hamilton’s theory [6]. The Hamilton’s canonical equations have been applied to derive the equations of motion for some simple physical configurations. These equations provide a simple but powerful tool for finding the equations of motion for any physical system. The Hamilton’s principle is the generalised formulation of the principle of least action. Apart from its applications to classical mechanics, the principle also applies to classical fields such as the electromagnetic and gravitational fields. Further extension of the principle can be found in quantum mechanics and quantum field theory.
Exercises 1. A particle is moving in the direction of the earth’s gravitational field. Write down the Hamiltonian and equation of motion of the particle. 2. A particle of mass m is projected upwards. Find the equation of motion. 3. Consider the equation of motion of a particle having mass m which is dropped from rest in a uniform gravitational field. The field is along the vertical axis which is taken as the z-axis. Obtain the Hamiltonian and its equation of motion. 4. Consider the equation of motion of a particle having mass m constrained to move on the surface of a cylinder defined by x2 + y 2 = R2 . The particle is subjected to a force given by F = −kr, where r denotes the distance of the particle from the origin. The negative sign indicates that the force is directed towards the origin. 5. The potential for an anharmonic oscillator is V = kx2 /2 + bx4 /4, where k and b are constants. Find Hamilton’s equations of motion. 6. Consider a smooth wire bent into the form of an helix given by x = a cos φ, y = a sin φ, z = bφ, where φ is a real parameter. a and b are positive constants. The wire is kept fixed with the z-axis pointing vertically upwards. A particle of mass m is allowed to slide freely on the wire. Find the Hamiltonian and obtain Hamilton’s equations for the system taking φ as a generalised coordinate.
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References [1] S.T. Thornton and J.B. Marion. Classical Dynamics of Particles and Systems. Brooks/Cole, 2004. [2] D.S. Lemons. Perfect Form: Variational Principles, Methods, and Applications in Elementary Physics. Perfect Form: Variational Principles, Methods, and Applications in Elementary Physics. Princeton University Press, 1997. [3] http://en.wikipedia.org/wiki/William_Rowan_Hamilton, 2015 (Accessed April 13, 2015 at 16:24 Indian Standard Time). [4] H. Goldstein, C.P. Poole, and J.L. Safko. Classical Mechanics. Addison Wesley, 2002. [5] “adrien-marie legendre," the famous people website. http://www.thefamouspeople.com/profiles/ adrien-marie-legendre-591.php, 2015 (Accessed April 11, 2015 at 17:35 Indian Standard Time). [6] R.D. Gregory. Classical Mechanics. Cambridge University Press, 2006.
