We saw in the last section that the kinetic and potential energies in the continuum limit can be written as integrals over x of densities, and so we may also write the Lagrangian as the integral of a Lagrangian density L(x),
L = T − U = Z L
0 dxL(x), L(x) =
1
2ρ ˙y2(x, t)− 1
2τ ∂y(x, t)
∂x
!2
.
This Lagrangian, however, will not be of much use until we figure out what is meant by varying it with respect to each dynamical degree of freedom or its corresponding velocity. In the discrete case we have the canonical momenta Pi = ∂L/∂ ˙yi, where the derivative requires holding all ˙yj fixed, for j 6= i, as well as all yk fixed. This extracts one term from the sum 12ρPa ˙yi2, and this would appear to vanish in the limit a → 0. Instead, we define the canonical momentum as a density, Pi → aP (x = ia), so
P (x = ia) = lim1 a
∂
∂ ˙yi
X
i
aL(y(x), ˙y(x), x)|x=ai.
We may think of the last part of this limit, if we also define a limiting operation
lima→0
Here δ(x0− x) is the Dirac delta function, defined by its integral,
Z x2
We also need to evaluate δ which is again defined by its integral,
Z x2
where after integration by parts the surface term is dropped because δ(x− x0) = 0 for x6= x0, which it is for x0 = x1, x2 if x∈ (x1, x2). Thus
δ
δy(x)L =−Z `
0 dx0τ∂y
∂x(x0)δ0(x0− x) = τ∂2y
∂x2, and Lagrange’s equations give the wave equation
ρ¨y(x, t)− τ∂2y
∂x2 = 0.
Exercises
5.1 Three springs connect two masses to each other and to immobile walls, as shown. Find the normal modes and frequencies of oscillation, assuming the system remains along the line shown.
m k
a 2k
2a k
a
m
5.2 Consider the motion, in a vertical plane of a double pendulum consist-ing of two masses attached to each other and to a fixed point by inextensible strings of length L. The upper mass has mass m1 and the lower mass m2. This is all in a laboratory with the ordinary gravitational forces near the surface of the Earth.
a) Set up the Lagrangian for the motion, assuming the strings stay taut.
b) Simplify the system under the approximation that the motion involves only small deviations from equilibrium.
Put the problem in matrix form appropriate for the pro-cedure discussed in class.
c) Find the frequencies of the normal modes of oscilla-tion. [Hint: following exactly the steps given in class will be complex, but the analogous procedure reversing the order of U and T will work easily.]
L
L m1
m2
5.3 (a) Show that if three mutually gravitating point masses are at the vertices of an equilateral triangle which is rotating about an axis normal to the plane of the triangle and through the center of mass, at a suitable angular velocity ω, this motion satisfies the equations of motion. Thus this configuration is an equilibrium in the rotating coordinate system. Do not assume the masses are equal.
(b) Suppose that two stars of masses M1 and M2 are rotating in circular orbits about their common center of mass. Consider a small mass m which is approximately in the equilibrium position described above (which is known as the L5 point). The mass is small enough that you can ignore its effect on the two stars. Analyze the motion, considering specifically the stability of the equilibrium point as a function of the ratio of the masses of the stars.
Hamilton’s Equations
We discussed the generalized momenta pi= ∂L(q, ˙q, t)
∂ ˙qi ,
and how the canonical variables{qi, pj} describe phase space. One can use phase space rather than {qi, ˙qj} to describe the state of a system at any moment. In this chapter we will explore the tools which stem from this phase space approach to dynamics.
6.1 Legendre transforms
The important object for determining the motion of a system using the Lagrangian approach is not the Lagrangian itself but its variation, un-der arbitrary changes in the variables q and ˙q, treated as independent variables. It is the vanishing of the variation of the action under such variations which determines the dynamical equations. In the phase space approach, we want to change variables ˙q → p, where the pi are part of the gradient of the Lagrangian with respect to the velocities.
This is an example of a general procedure called the Legendre trans-formation. We will discuss it in terms of the mathematical concept of a differential form.
Because it is the variation of L which is important, we need to focus our attention on the differential dL rather than on L itself. We first
147
want to give a formal definition of the differential, which we will do first for a function f (x1, ..., xn) of n variables, although for the Lagrangian we will later subdivide these into coordinates and velocities. We will take the space in which x takes values to be some general space we call M, which might be ordinary Euclidean space but might be something else, like the surface of a sphere1. Given a function f of n independent variables xi, the differential is
df =
Xn i=1
∂f
∂xidxi. (6.1)
What does that mean? As an approximate statement, this can be regarded as saying
df ≈ ∆f ≡ f(xi+ ∆xi)− f(xi) =
Xn i=1
∂f
∂xi∆xi+O(∆xi∆xj),
with some statement about the ∆xi being small, followed by the drop-ping of the “order (∆x)2” terms. Notice that df is a function not only of the point x ∈ M, but also of the small displacements ∆xi. A very useful mathematical language emerges if we formalize the definition of df , extending its definition to arbitrary ∆xi, even when the ∆xi are not small. Of course, for large ∆xi they can no longer be thought of as the difference of two positions in M and df no longer has the meaning of the difference of two values of f . 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 an n-dimensional vector space Rn. Thus df :M × Rn → R is a real-valued function with two arguments, one in M and one in a vector space. The dxi which appear in (6.1) can be thought of as operators acting on this vector space argument to extract the i0th component, and the action of df on the argument (x, ~v) is df (x, ~v) =Pi(∂f /∂xi)vi.
This differential is a special case of a 1-form, as is each of the oper-ators dxi. All n of these dxi form a basis of 1-forms, which are more generally
ω =X
i
ωi(x)dxi.
1Mathematically, M is a manifold, but we will not carefully define that here.
The precise definition is available in Ref. [11].
If there exists an ordinary function f (x) such that ω = df , then ω is said to be an exact 1-form.
Consider L(qi, vj, t), where vi = ˙qi. At a given time we consider q and v as independant variables. The differential of L on the space of coordinates and velocities, at a fixed time, is
dL =X a change of variables from vi to the gradient with respect to these variables, pi = ∂L/∂vi, where we focus now on the variables being transformed and ignore the fixed qi variables. So dL = Pipidvi, and the pi are functions of the vj determined by the function L(vi). Is there a function g(pi) which reverses the roles of v and p, for which dg = Pividpi? If we can invert the functions p(v), we can define
as requested, and which also determines the relationship between v and p,
vi = ∂g
∂pi = vi(pj),
giving the inverse relation to pk(v`). This particular form of changing variables is called a Legendre transformation. In the case of interest here, the function g is called H(qi, pj, t), the Hamiltonian,
H =X
i
˙
qipi− L. (6.2)
Other examples of Legendre transformations occur in thermody-namics. The energy change of a gas in a variable container with heat flow is sometimes written
dE = d¯Q− pdV,
where d¯Q is not an exact differential, and the heat Q is not a well defined system variable. Instead one defines the entropy and temperature d
¯Q = T dS, and the entropy S is a well defined property of the gas.
Thus the state of the gas can be described by the two variables S and V , and changes involve an energy change
dE = T dS− pdV.
We see that the temperature is T = ∂E/∂S|V. If we wish to find quantities appropriate for describing the gas as a function of T rather than S, we define the free energy F by−F = T S−E so dF = −SdT − pdV , and we treat F as a function F (T, V ). Alternatively, to use the pressure p rather than V , we define the enthalpy X(p, S) = V p + E, dX = V dp+T dS. To make both changes, and use (T, p) to describe the state of the gas, we use the Gibbs free energy G(T, p) = X − T S = E + V p− T S, dG = V dp − SdT
Most Lagrangians we encounter have the decomposition L = L2+ L1+ L0 into terms quadratic, linear, and independent of velocities, as considered in 2.1.5. Then the momenta are linear in velocities, pi =
PjMijq˙j + ai, or in matrix form p = M · ˙q + a, which has the inverse relation ˙q = M−1· (p − a). As H = L2− L0, H = 12(p− a) · M−1· (p − a)− L0. As an example, consider spherical coordinates, in which the kinetic energy is
T = m 2
˙r2+ r2θ˙2+ r2sin2θ ˙φ2= 1
2m p2r+p2θ
r2 + p2φ r2sin2θ
!
. Note that pθ 6= ~p · ˆeθ, in fact it doesn’t even have the same units.
The equations of motion in Hamiltonian form,
˙
qk = ∂H
∂pk
q,t
, p˙k =− ∂H
∂qk
p,t
,
are almost symmetric in their treatment of q and p. If we define a 2N dimensional coordinate η for phase space,
ηi = qi ηn+i = pi
for 1≤ i ≤ N,
we can write Hamilton’s equation in terms of a particular matrix J ,
˙
ηj = Jij∂H
∂ηk, where J =
0 1IN×N
−1IN×N 0
.
J is like a multidimensional version of the iσywhich we meet in quantum-mechanical descriptions of spin 1/2 particles. It is real, antisymmetric, and because J2 =−1I, it is orthogonal. Mathematicians would say that J describes the complex structure on phase space.
For a given physical problem there is no unique set of generalized coordinates which describe it. Then transforming to the Hamiltonian may give different objects. An nice example is given in Goldstein, a mass on a spring attached to a “fixed point” which is on a truck moving at uniform velocity vT, relative to the Earth. If we use the Earth coordinate x to describe the mass, the equilibrium position of the spring is moving in time, xeq = vTt, ignoring a negligible initial position. Thus U = 12k(x− vTt)2, while T = 12m ˙x2 as usual, and L = 12m ˙x2 − 12k(x− vTt)2, p = m ˙x, H = p2/2m + 12k(x− vTt)2. The equations of motion ˙p = m¨x =−∂H/∂x = −k(x−vTt), of course, show that H is not conserved, dH/dt = (p/m)dp/dt + k( ˙x− vT)(x− vTt) =
−(kp/m)(x − vTt) + (kp/m − kvT)(x− vTt) = −kvT(x− vTt) 6= 0.
Alternatively, dH/dt = −∂L/∂t = −kvT(x− vTt) 6= 0. This is not surprising; the spring exerts a force on the truck and the truck is doing work to keep the fixed point moving at constant velocity.
On the other hand, if we use the truck coordinate x0 = x− vTt, we may describe the motion in this frame with T0 = 12m ˙x0 2, U0 = 12kx02, L0 = 12m ˙x0 2− 12kx02, giving the correct equations of motion p0 = m ˙x0,
˙
p0 = m¨x0 = −∂L0/∂x0 = −kx0. With this set of coordinates, the Hamiltonian is H0 = ˙x0p0 − L0 = p02/2m + 12kx02, which is conserved.
From the correspondence between the two sets of variables, x0 = x−vTt, and p0 = p − mvT, we see that the Hamiltonians at corresponding points in phase space differ, H(x, p)− H0(x0, p0) = (p2 − p02)/2m = 2mvTp− 12mvT2 6= 0.