
CHAPTER VII
THE CANONICAL EQUATIONS
§40. Hamilton's equations
THE formulation of the laws of mechanics in terms of the Lagrangian, and
of Lagrange's equations derived from it, presupposes that the mechanical
state of a system is described by specifying its generalised co-ordinates and
velocities. This is not the only possible mode of description, however. A
number of advantages, especially in the study of certain general problems of
mechanics, attach to a description in terms of the generalised co-ordinates
and momenta of the system. The question therefore arises of the form of
the equations of motion corresponding to that formulation of mechanics.
The passage from one set of independent variables to another can be
effected by means of what is called in mathematics Legendre's transformation.
In the present case this transformation is as follows. The total differential
of the Lagrangian as a function of co-ordinates and velocities is
dL =
This expression may be written
(40.1)
since the derivatives aL/dqi are, by definition, the generalised momenta, and
aL/dqi = pi by Lagrange's equations. Writing the second term in (40.1) as
= - Eqi dpi, taking the differential d(piqi) to the left-hand
side, and reversing the signs, we obtain from (40.1)
The argument of the differential is the energy of the system (cf. §6);
expressed in terms of co-ordinates and momenta, it is called the Hamilton's
function or Hamiltonian of the system:
(40.2)
t The reader may find useful the following table showing certain differences between the
nomenclature used in this book and that which is generally used in the English literature.
Here
Elsewhere
Principle of least action
Hamilton's principle
Maupertuis' principle
Principle of least action
Maupertuis' principle
Action
Hamilton's principal function
Abbreviated action
Action
- -Translators.
131
132
The Canonical Equations
§40
From the equation in differentials
dH =
(40.3)
in which the independent variables are the co-ordinates and momenta, we
have the equations
=
(40.4)
These are the required equations of motion in the variables P and q, and
are called Hamilton's equations. They form a set of 2s first-order differential
equations for the 2s unknown functions Pi(t) and qi(t), replacing the S second-
order equations in the Lagrangian treatment. Because of their simplicity and
symmetry of form, they are also called canonical equations.
The total time derivative of the Hamiltonian is
Substitution of qi and pi from equations (40.4) shows that the last two terms
cancel, and so
dH/dt==Hoo.
(40.5)
In particular, if the Hamiltonian does not depend explicitly on time, then
dH/dt = 0, and we have the law of conservation of energy.
As well as the dynamical variables q, q or q, P, the Lagrangian and the
Hamiltonian involve various parameters which relate to the properties of the
mechanical system itself, or to the external forces on it. Let A be one such
parameter. Regarding it as a variable, we have instead of (40.1)
dL
and (40.3) becomes
dH =
Hence
(40.6)
which relates the derivatives of the Lagrangian and the Hamiltonian with
respect to the parameter A. The suffixes to the derivatives show the quantities
which are to be kept constant in the differentiation.
This result can be put in another way. Let the Lagrangian be of the form
L = Lo + L', where L' is a small correction to the function Lo. Then the
corresponding addition H' in the Hamiltonian H = H + H' is related to L'
by
(H')p,a - (L')
(40.7)
It may be noticed that, in transforming (40.1) into (40.3), we did not
include a term in dt to take account of a possible explicit time-dependence
§41
The Routhian
133
of the Lagrangian, since the time would there be only a parameter which
would not be involved in the transformation. Analogously to formula (40.6),
the partial time derivatives of L and H are related by
(40.8)
PROBLEMS
PROBLEM 1. Find the Hamiltonian for a single particle in Cartesian, cylindrical and
spherical co-ordinates.
SOLUTION. In Cartesian co-ordinates x, y, 2,
in cylindrical co-ordinates r, , z,
in spherical co-ordinates r, 0, ,
PROBLEM 2. Find the Hamiltonian for a particle in a uniformly rotating frame of reference.
SOLUTION. Expressing the velocity V in the energy (39.11) in terms of the momentum p
by (39.10), we have H = p2/2m-S rxp+U.
PROBLEM 3. Find the Hamiltonian for a system comprising one particle of mass M and n
particles each of mass m, excluding the motion of the centre of mass (see §13, Problem).
SOLUTION. The energy E is obtained from the Lagrangian found in §13, Problem, by
changing the sign of U. The generalised momenta are
Pa = OL/OV
Hence
-
= (mM/14)
=
=
Substitution in E gives
41. The Routhian
In some cases it is convenient, in changing to new variables, to replace
only some, and not all, of the generalised velocities by momenta. The trans-
formation is entirely similar to that given in 40.
To simplify the formulae, let us at first suppose that there are only two
co-ordinates q and E, say, and transform from the variables q, $, q, $ to
q, $, p, & where P is the generalised momentum corresponding to the co-
ordinate q.
134
The Canonical Equations
§42
The differential of the Lagrangian L(q, $, q, §) is
dL = dq + (al/dg) ds (0L/as) d
d,
whence
= (0L/d) d.
If we define the Routhian as
= pq-L,
(41.1)
in which the velocity q is expressed in terms of the momentum P by means
of the equation P = 0L/dq, then its differential is
dR = - ds - (aL/a)
(41.2)
Hence
DRIP, p = OR/dq,
(41.3)
(41.4)
Substituting these equations in the Lagrangian for the co-ordinate $, we have
(41.5)
Thus the Routhian is a Hamiltonian with respect to the co-ordinate q
(equations (41.3)) and a Lagrangian with respect to the co-ordinate $ (equation
(41.5)).
According to the general definition the energy of the system is
E -p-L =
In terms of the Routhian it is
E=R-R,
(41.6)
as we find by substituting (41.1) and (41.4).
The generalisation of the above formulae to the case of several co-ordinates
q and & is evident.
The use of the Routhian may be convenient, in particular, when some of
the co-ordinates are cyclic. If the co-ordinates q are cyclic, they do not appear
in the Lagrangian, nor therefore in the Routhian, so that the latter is a func-
tion of P, $ and $. The momenta P corresponding to cyclic co-ordinates are
constant, as follows also from the second equation (41.3), which in this sense
contains no new information. When the momenta P are replaced by their
given constant values, equations (41.5) (d/dt) JR(p, $, 5)108 = JR(P, &, §) 128
become equations containing only the co-ordinates $, so that the cyclic co-
ordinates are entirely eliminated. If these equations are solved for the func-
tions (t), substitution of the latter on the right-hand sides of the equations
q = JR(p, $, E) gives the functions q(t) by direct integration.
PROBLEM
Find the Routhian for a symmetrical top in an external field U(, 0), eliminating the cyclic
co-ordinate 4 (where 4, , 0 are Eulerian angles).
§42
Poisson brackets
135
SOLUTION. The Lagrangian is = see
§35, Problem 1. The Routhian is
R = cos 0);
the first term is a constant and may be omitted.
42. Poisson brackets
Let f (p, q, t) be some function of co-ordinates, momenta and time. Its
total time derivative is
df
Substitution of the values of and Pk given by Hamilton's equations (40.4)
leads to the expression
(42.1)
where
(42.2)
dqk
This expression is called the Poisson bracket of the quantities H and f.
Those functions of the dynamical variables which remain constant during
the motion of the system are, as we know, called integrals of the motion.
We see from (42.1) that the condition for the quantity f to be an integral of
the motion (df/dt = 0) can be written
af(dt+[H,f]=0
(42.3)
If the integral of the motion is not explicitly dependent on the time, then
[H,f] = 0,
(42.4)
i.e. the Poisson bracket of the integral and the Hamiltonian must be zero.
For any two quantities f and g, the Poisson bracket is defined analogously
to (42.2):
(42.5)
The Poisson bracket has the following properties, which are easily derived
from its definition.
If the two functions are interchanged, the bracket changes sign; if one of
the functions is a constant c, the bracket is zero:
(42.6)
[f,c]=0.
(42.7)
Also
[f1+f2,g]=[f1,g)+[f2,g]
(42.8)
[f1f2,g] ]=fi[fa,8]+f2[f1,8] =
(42.9)
Taking the partial derivative of (42.5) with respect to time, we obtain
(42.10)
136
The Canonical Equations
§42
If one of the functions f and g is one of the momenta or co-ordinates, the
Poisson bracket reduces to a partial derivative:
(42.11)
(42.12)
Formula (42.11), for example, may be obtained by putting g = qk in (42.5);
the sum reduces to a single term, since dqk/dqi = 8kl and dqk/dpi = 0. Put-
ting in (42.11) and (42.12) the function f equal to qi and Pi we have, in parti-
cular,
[qi,qk] = [Pi, Pk] =0, [Pi, 9k] = Sik.
(42.13)
The relation
[f,[g,h]]+[g,[h,f]]+[h,[f,g]] = 0,
(42.14)
known as Jacobi's identity, holds between the Poisson brackets formed from
three functions f, g and h. To prove it, we first note the following result.
According to the definition (42.5), the Poisson bracket [f,g] is a bilinear
homogeneous function of the first derivatives of f and g. Hence the bracket
[h,[f,g]], for example, is a linear homogeneous function of the second
derivatives of f and g. The left-hand side of equation (42.14) is therefore a
linear homogeneous function of the second derivatives of all three functions
f, g and h. Let us collect the terms involving the second derivatives of f.
The first bracket contains no such terms, since it involves only the first
derivatives of f. The sum of the second and third brackets may be symboli-
cally written in terms of the linear differential operators D1 and D2, defined by
D1() = [g, ], D2(b) = [h, ]. Then
3,[h,f]]+[h,[f,g]] = [g, [h,f]]-[h,[g,f]
= D1[D2(f)]-D2[D1(f)]
= (D1D2-D2D1)f.
It is easy to see that this combination of linear differential operators cannot
involve the second derivatives of f. The general form of the linear differential
operators is
where & and Nk are arbitrary functions of the variables .... Then
and the difference of these,
§42
Poisson brackets
137
is again an operator involving only single differentiations. Thus the terms in
the second derivatives of f on the left-hand side of equation (42.14) cancel
and, since the same is of course true of g and h, the whole expression is identi-
cally zero.
An important property of the Poisson bracket is that, if f and g are two
integrals of the motion, their Poisson bracket is likewise an integral of the
motion:
[f,g] = constant. =
(42.15)
This is Poisson's theorem. The proof is very simple if f and g do not depend
explicitly on the time. Putting h = H in Jacobi's identity, we obtain
[H,[f,g]]+[f,[g,H]]+[g,[H,fl]=0.
Hence, if [H, g] =0 and [H,f] = 0, then [H,[f,g]] = 0, which is the
required result.
If the integrals f and g of the motion are explicitly time-dependent, we
put, from (42.1),
Using formula (42.10) and expressing the bracket [H, [f,g]] in terms of two
others by means of Jacobi's identity, we find
d
[
(42.16)
which evidently proves Poisson's theorem.
Of course, Poisson's theorem does not always supply further integrals of
the motion, since there are only 2s-1 - of these (s being the number of degrees
of freedom). In some cases the result is trivial, the Poisson bracket being a
constant. In other cases the integral obtained is simply a function of the ori-
ginal integrals f and g. If neither of these two possibilities occurs, however,
then the Poisson bracket is a further integral of the motion.
PROBLEMS
PROBLEM 1. Determine the Poisson brackets formed from the Cartesian components of
the momentum p and the angular momentum M = rxp of a particle.
SOLUTION. Formula (42.12) gives [Mx, Py] = -MM/Dy = -d(yp:-2py)/dy
=
-Pz,
and similarly [Mx, Px] = 0, [Mx, P2] = Py. The remaining brackets are obtained by cyclically
permuting the suffixes x, y, Z.
6
138
The Canonical Equations
§43
PROBLEM 2. Determine the Poisson brackets formed from the components of M.
SOLUTION. A direct calculation from formula (42.5) gives [Mx, My] = -M2, [My, M]
= -Mx, [Mz, Mx] = -My.
Since the momenta and co-ordinates of different particles are mutually independent variables,
it is easy to see that the formulae derived in Problems 1 and 2 are valid also for the total
momentum and angular momentum of any system of particles.
PROBLEM 3. Show that [, M2] = 0, where is any function, spherically symmetrical
about the origin, of the co-ordinates and momentum of a particle.
SOLUTION. Such a function can depend on the components of the vectors r and p only
through the combinations r2, p2, r. p. Hence
and similarly for The required relation may be verified by direct calculation from
formula (42.5), using these formulae for the partial derivatives.
PROBLEM 4. Show that [f, M] = n xf, where f is a vector function of the co-ordinates
and momentum of a particle, and n is a unit vector parallel to the z-axis.
SOLUTION. An arbitrary vector f(r,p) may be written as f = where
01, O2, 03 are scalar functions. The required relation may be verified by direct calculation
from formulae (42.9), (42.11), (42.12) and the formula of Problem 3.
$43. The action as a function of the co-ordinates
In formulating the principle of least action, we have considered the integral
(43.1)
taken along a path between two given positions q(1) and q(2) which the system
occupies at given instants t1 and t2. In varying the action, we compared the
values of this integral for neighbouring paths with the same values of q(t1)
and q(t2). Only one of these paths corresponds to the actual motion, namely
the path for which the integral S has its minimum value.
Let us now consider another aspect of the concept of action, regarding S
as a quantity characterising the motion along the actual path, and compare
the values of S for paths having a common beginning at q(t1) = q(1), but
passing through different points at time t2. In other words, we consider the
action integral for the true path as a function of the co-ordinates at the upper
limit of integration.
The change in the action from one path to a neighbouring path is given
(if there is one degree of freedom) by the expression (2.5):
8S =
Since the paths of actual motion satisfy Lagrange's equations, the integral
in 8S is zero. In the first term we put Sq(t1) = 0, and denote the value of
§43
The action as a function of the co-ordinates
139
8q(t2) by 8q simply. Replacing 0L/dq by p, we have finally 8S = pdq or, in
the general case of any number of degrees of freedom,
ES==Pisqu-
(43.2)
From this relation it follows that the partial derivatives of the action with
respect to the co-ordinates are equal to the corresponding momenta:
=
(43.3)
The action may similarly be regarded as an explicit function of time, by
considering paths starting at a given instant t1 and at a given point q(1), and
ending at a given point q(2) at various times t2 = t. The partial derivative
asiat thus obtained may be found by an appropriate variation of the integral.
It is simpler, however, to use formula (43.3), proceeding as follows.
From the definition of the action, its total time derivative along the path is
dS/dt = L.
(43.4)
Next, regarding S as a function of co-ordinates and time, in the sense des-
cribed above, and using formula (43.3), we have
dS
A comparison gives asid = L- or
(43.5)
Formulae (43.3) and (43.5) may be represented by the expression
(43.6)
for the total differential of the action as a function of co-ordinates and time
at the upper limit of integration in (43.1). Let us now suppose that the co-
ordinates (and time) at the beginning of the motion, as well as at the end,
are variable. It is evident that the corresponding change in S will be given
by the difference of the expressions (43.6) for the beginning and end of the
path, i.e.
dsp
(43.7)
This relation shows that, whatever the external forces on the system during
its motion, its final state cannot be an arbitrary function of its initial state;
only those motions are possible for which the expression on the right-hand
side of equation (43.7) is a perfect differential. Thus the existence of the
principle of least action, quite apart from any particular form of the Lagran-
gian, imposes certain restrictions on the range of possible motions. In parti-
cular, it is possible to derive a number of general properties, independent
of the external fields, for beams of particles diverging from given points in
140
The Canonical Equations
§44
space. The study of these properties forms a part of the subject of geometrical
optics.+
It is of interest to note that Hamilton's equations can be formally derived
from the condition of minimum action in the form
(43.8)
which follows from (43.6), if the co-ordinates and momenta are varied inde-
pendently. Again assuming for simplicity that there is only one co-ordinate
and momentum, we write the variation of the action as
= dt - (OH/dp)8p dt].
An integration by parts in the second term gives
At the limits of integration we must put 8q = 0, so that the integrated term
is zero. The remaining expression can be zero only if the two integrands
vanish separately, since the variations Sp and 8q are independent and arbitrary
dq = (OH/OP) dt, dp = - (dH/dq) dt, which, after division by dt, are
Hamilton's equations.
$44. Maupertuis' principle
The motion of a mechanical system is entirely determined by the principle
of least action: by solving the equations of motion which follow from that
principle, we can find both the form of the path and the position on the path
as a function of time.
If the problem is the more restricted one of determining only the path,
without reference to time, a simplified form of the principle of least action
may be used. We assume that the Lagrangian, and therefore the Hamilton-
ian, do not involve the time explicitly, SO that the energy of the system is
conserved: H(p, q) = E = constant. According to the principle of least action,
the variation of the action, for given initial and final co-ordinates and times
(to and t, say), is zero. If, however, we allow a variation of the final time t,
the initial and final co-ordinates remaining fixed, we have (cf.(43.7))
8S = -Hot.
(44.1)
We now compare, not all virtual motions of the system, but only those
which satisfy the law of conservation of energy. For such paths we can
replace H in (44.1) by a constant E, which gives
SS+Est=0.
(44.2)
t See The Classical Theory of Fields, Chapter 7, Pergamon Press, Oxford 1962.