Just Enough Classical Mechanics

© R. J. Wilkes, 1/20/98

Introduction

Quantum mechanics was derived from the Lagrangian and Hamiltonian formulations of classical mechanics, and uses much of their terminology and conventions. These formulations use energy and momentum considerations, rather than vector analysis of forces, to arrive at the equations of motion. Since many students do not get a chance to study the non-Newtonian formulations of mechanics, it is useful to briefly introduce them. The aim here is to present just enough classical mechanics to motivate and explain the corresponding quantum mechanical forms.

Consider a spring-mass system. Newton’s 2^nd Law, F=ma, gives us the equation of motion:

m(d²x/dt² ) = mx’’ = -kx

(using the notation ‘=d/dt) with solutions of the form

x(t)=Asin(w t+B)

where w ² = k/m, and A and B are arbitrary amplitude and phase constants respectively, which arise from the fact that our solution comes from integrating the equation of motion twice, each resulting in an integration constant. The presence of these two arbitrary constants means there is a twofold infinity of particular motions possible, and the selection of one particular pair of constants selects one of these solutions. The constants A and B can be related to the initial conditions (x₀, v₀) desired, for example.

Another approach to finding the motion of the oscillator is via energy considerations. The spring force -kx can be related to the potential energy stored in the spring in the usual way:

V(x) = (1/2)kx²; F=-dV/dx.

Note that any constant could be added to V without affecting F. The total energy (assuming no additive constant, which amounts to setting V=0 at x=0) is given by

E = KE + PE = (1/2)m(dx/dt)² + (1/2)kx².

Taking the time derivative:

dE/dt = mv(dv/dt) + kx(dx/dt), or

(dx/dt)(md²x/dt² + kx) = 0.

Conservation of energy requires dE/dt=0, since there is no external force doing work on the mass; we say that E is a constant of the motion.

Going back to the expression for E and solving for dx/dt:

dx/dt = {2(E-(1/2)kx²)/m}^(1/2), or

dt = dx {2(E-(1/2)kx²)/m}^-(1/2) , which has only t on one side and only x on the other, and so can be integrated to give

t= (m/k)(1/2)arcsin(x(k/2E) ^(1/2)) + t ,

where now the two constants of the motion (integration constants) are the system energy E and t , the time at which the mass passes through x=0 with positive velocity.

Solving for x:

x=(2E/K)(1/2)sin(w [t-t ])

which is identical in form to the solution obtained from Newton’s Laws.

The advantage of the energy approach is that it can handle anharmonic oscillators, where the force is a nonlinear function of x and therefore the potential energy is not a simple parabola. We still have

dt = dx {2(E-V(x))/m}^-(1/2) + t

and the problem can always be solved (if necessary, by numerical integration). Notice that if we plot V(x) and draw a horizontal line at V=E, the ordinates at which the V curve cuts E are the turning points of the motion. Even if V is monotonic, so that the motion is non-oscillatory, the energy method still applies, and the single turning point is readily identified. This general method of finding the motion of particles under the action of forces representable by potentials was developed by Lagrange in the 18^th century, and further developed by Hamilton in the 19^th century. It provided the basis for the development of quantum mechanics.

Generalized coordinates, velocities and momenta

We consider systems of N particles which have no fixed spatial relationships (independent particles). Particles are defined as pieces of matter so small that their internal structure and motions, if any, are of no consequence to their gross motion, so they have only one property of external importance: their mass. In short, classical particles are point masses. At any time t, the 3N specific coordinates of the particles making up the system under consideration are called the configuration of the system. It is convenient to use Cartesian (x,y,z) coordinates, since these can always be easily related to an inertial coordinate frame (ie, one moving with constant velocity in 3-space). We can regard the system as a single point in a 3N-dimensional configuration space rather than as a constellation of N points in 3-space; the motion of the particles comprising the system is then represented by motion of this single point in configuration space.

In many cases, it is useful to define generalized coordinates which are not simple Cartesian coordinates, or even necessarily inertial coordinates. For example, spherical coordinates are convenient for representing motion under forces whose potentials are spherically symmetrical. All that is required is that the functional relations which connect the generalized coordinates as a function of time to inertial Cartesian coordinates,

q_i(t) = q_i(x₁,x₂...x_3N; t)

exist and be differentiable as required. (In classical mechanics, time is a parameter, not a fully-equal coordinate as in special relativity). For each generalized coordinate q_i we can define a corresponding generalized velocity dq_i/dt = q’_i . Just as the generalized coordinates need not have dimensions of length, the generalized velocities need not have dimensions of [length/time].

Taking spherical coordinates as an example, the generalized coordinates are (r, q , f ), which are connected to the Cartesian inertial coordinates by

r=(x²+y²+z²) ^(1/2), q =arccos(z/r); f =arctan(y/x),

with inverse relationships

x=r sin(q ) cos(f ); y=r sin(q ) sin(f ), z=r cos(q ).

Another connection between the Cartesian and generalized coordinates is the set of partial derivatives connecting the two coordinate sets. For spherical coordinates, these are

¶ x/¶ r=sin(q ) cos(f ) ; ¶ y/¶ r=sin(q ) sin(f ) ; ¶ z/¶ r=cos(q ), etc. (A full list is given below).

Notice the spherical coordinates q , f are dimensionless angles, and their corresponding generalized velocities have dimensions of [1/time]. Furthermore, dq /dt is not the q component of the particle’s vector velocity in 3-space (which is vq =r dq /dt).

We should not view q’=dq/dt as a quantity that contains no more information than is already contained in q, because q(t) has a particular form only in the context of one specific motion of the system. If we consider all the possible motions accessible to the system, we can see that specification of the 3N values of the q_i is not sufficient to determine the future dynamical behavior of the system, as required by the deterministic character of classical physics. We need to specify also the 3N values of dq_i/dt. In other words, we must treat all 6N quantities (3N q_i(t) and 3N dq_i(t)/dt) on an equal basis.

We express the velocities with respect to Cartesian coordinates in terms of the generalized coordinates and velocities:

dx_k/dt = x’_k = S ¶ x_k(t)/¶ t + (¶ x_k(t)/¶ q_j) q’_j.

We can then write the kinetic energy in terms of the generalized coordinates using this expression:

T=(1/2) å _k m_k [dx_k(t)/dt]² = (1/2) å _k m_k [¶ x_k(t)/¶ t + å _j (¶ x_k(t)/¶ q_j) q’_j]²

= (1/2) å _k m_k[x’_k + å _j (¶ x_k(t)/¶ q_j) q’_j]²

Here we use m_k to represent the mass of the particle associated with coordinate q_k. Thus T=T(q_j, q’_j, t). We define the generalized (or canonical) momentum associated with coordinate q_j by p_j=¶ T/¶ q’_j ; we say p_j is the momentum conjugate to q_j . For example, if q=x, then we recover the usual expression for momentum, p=¶ T/¶ x’ = mx’=mv, while for a single particle in spherical coordinates, we get (check this using the partial derivatives as given in the next section):

p₁=mq’₁ = m r’ = p_r

p₂=mq²₁ q’₂ = m r^2q ’ = pq

p₃=mq²₁ sin²(q₂) q’₃ = m r² sin² (q ) f ‘ = pf .

The generalized momenta can be used (instead of the generalized velocities) along with the generalized coordinates to form a 6N dimensional space in which the dynamical behavior of the system is represented by the motion of a single point with time. The 6N-dimensional space spanning the generalized coordinates and momenta is called phase space.

Lagrangian formulation of mechanics

We can express Newton’s 2^nd Law, written in inertial coordinates, in terms of generalized coordinates. This will lead to the Lagrangian formulation of mechanics. Note that Lagrangian mechanics does not replace Newton’s Laws, but provides (as we shall see) a method for writing down a set of second-order differential equations to describe the motion of even a very complicated system simply and surely. Moreover, it often simplifies the discovery of constants of the motion, allowing us to replace the corresponding second-order equations with first-order equations. But most importantly, the Lagrangian formulation leads us to the Hamiltonian formulation, which provides the basis for quantum mechanics.

To begin, we take the partial derivatives of T(q, q’;t) with respect to the q’:

¶ T/¶ q’_j = å _k m_k [¶ x_k/¶ t + å _m (¶ x_k/¶ q_m) q’_m] ¶ x_k/¶ q_j

The left-hand side of this equation is just p_j and the quantity in square brackets on the right is x’_k, so we have

p_j = å m_k x’_k ¶ x_k/¶ q_j .

Taking the time derivative of this, we get

p’_j = å m_k x’’_k ¶ x_k/¶ q_j + å m_k x’_k d(¶ x_k/¶ q_j)/dt

If we limit ourselves to the case where all forces are derivable from potentials, the only case of interest in quantum mechanics, we have

m_k x’’_k = -¶ V/¶ x_k .

We can express this in terms of the generalized coordinates (holding t constant when taking partials with respect to space coordinates):

¶ V/¶ q_j = å (¶ V/¶ x_k )(¶ x_k/¶ q_j ) = - å m_k x’’_k (¶ x_k/¶ q_j ) .

Notice that the right-hand side of the equation above is the first term in the expression for p’_j . Looking at the second term in that equation, we note ¶ x_k/¶ q_j is a function of the q and t, so

d(¶ x_k/¶ q_j)/dt = ¶ ²x_k/¶ q_j¶ t + {q’₁ (¶ ²x_k/¶ q_1¶ q_j ) + q’₂ (¶ ²x_k/¶ q_2¶ q_j ) + ... }

Looking back at eqn ()

x’_k = å ¶ x_k/¶ t + (¶ x_k/¶ q_j) q’_j

we can take ¶ /¶ q_j of both sides, and note that q’j is not affected by the operator ¶ /¶ q_j since we consider q and q’ independent functions. We get

¶ x’_k/¶ q_j = ¶ ²x_k/¶ q_j¶ t + {q’₁ (¶ ²x_k/¶ q_1¶ q_j ) + q’₂ (¶ ²x_k/¶ q_2¶ q_j ) + ... }

So we have discovered that

d(¶ x_k/¶ q_j)/dt = ¶ x’_k/¶ q_j .

Let us look again at eqn ():

p’_j = å m_k x’’_k ¶ x_k/¶ q_j + å m_k x’_k d(¶ x_k/¶ q_j)/dt

We have already identified the first term as -¶ V/¶ q_j . In the second term, we can use our latest discovery to write

å m_k x’_k d(¶ x_k/¶ q_j)/dt = å m_k x’_k ¶ x’_k/¶ q_j .

Putting this into eqn ():

p’_j = å m_k x’’_k ¶ x_k/¶ q_j + å m_k x’_k ¶ x’_k/¶ q_j

But we can identify

å m_k x’_k ¶ x’_k/¶ q_j = ¶ T/¶ q_j ,

so we get

p’_j = -¶ V/¶ q_j + ¶ T/¶ q_j = ¶ /¶ q_j (T-V)

This tells us that the time rate of change of generalized momentum is not equal to the derivative of -V with respect to the generalized coordinates, as it would be for simple inertial coordinates and momenta, but rather it is the derivative of a new quantity with dimensions of energy, the Lagrangian:

L(q,q’;t) = T(q, q’;t) -V(q;t).

Since V is not a function of the q’, we get

¶ L/¶ q’_j = ¶ T/¶ q’_j = p_j

so the equations of motion

p’_j = ¶ /¶ q_j (T-V)

can be expressed very compactly as

{d/dt} ¶ L/¶ q’_j = ¶ L /¶ q_j , or

{d/dt} ¶ L/¶ q’_j - ¶ L /¶ q_j = 0.

These 3N equations are Lagrange’s equations.

To show the value of the Lagrangian formulation, let us apply it to a familiar problem: the motion of a small planet about a massive Sun (so that we can use an inertial coordinate system in which the Sun is at rest). Then the configuration of the system consists of the position of the planet. The potential energy is given by

V = -GMm/r

where r is distance from the Sun, m is the planet’s mass, G is the universal gravitational constant, and M is the solar mass. Given the spherical symmetry of the problem, it is natural to use spherical coordinates. The partial derivatives we will need are conveniently listed in the form of a matrices (called Jacobians) as:

ì dr ü = ì sinq cosf sinq sinf cosq ü ì dx ü

ï dq ô = ê (1/r)cosq cosf (1/r)cosq sinf -1/(r sinq ) ú ê dyú

î df þ = î -sinf /(r sinq ) cosf /(r sinq ) 0 þ î dz þ

and

ì dx ü = ì sinq cosf r cosq cosf -r sinq sinf ü ì dr ü

ï dyô = ê sinq sinf r cosq sinf r sinq cosf ú ê dq ú

î dzþ = î cosq -r sinq 0 þ î df þ

The kinetic energy is given by

T=(1/2) å m [dx_k/dt]²

= (1/2) å _k m [x’_k + S _j (¶ x_k/¶ q_j) q’_j]²

where x_k ={x,y,z}and q_j ={r,q ,f }. Substituting the appropriate partial derivatives from the Jacobian, and after some horrendous algebra, we find

T = (1/2) m ( r’² + r² q ‘² + r² sin^2q f ‘²).

Thus the Lagrangian for this system is

L = (1/2) m ( r’² + r² q ‘² + r² sin^2q f ‘²) + GMm/r .

The generalized momenta are p_j = ¶ L/¶ q’_j :

p_r = mr’ ; pq = mr^2q ‘ ; pf = mr² sin^2q f ‘.

The partial derivatives of L with respect to the q_j are:

¶ L/¶ r = m r q ‘² + m r sin^2q f ‘² - GMm/r²

¶ L/¶ q = (1/2) m r² sin2q f ‘²

¶ L/¶ f = 0.

So Lagrange’s equations for this system, {d/dt} ¶ L/¶ q’_j - ¶ L /¶ q_j = 0, become

{d/dt}( mr’) - (m r q ‘² + m r sin^2q f ‘² - GMm/r² ) = 0

{d/dt}( mr^2q ‘) - ((1/2) m r² sin2q f ‘²) = 0

{d/dt}( mr² sin^2q f ‘) = 0

These are the equations of motion in the chosen coordinate system.

The last equation tells us that the quantity pf = mr² sin^2q f ‘ is a constant of the motion. This is of course L_z , the z component of the angular momentum of the planet. The second term is missing in the last equation because the generalized coordinate f is not present in the Lagrangian. So we can define a rule: the generalized momentum conjugate to any generalized coordinate which is not present in the Langragian is a constant of the motion.

The Lagrangian formulation not only allows us to write down the equations of motion using a straightforward recipe, but also makes it easy to identify constants of the motion—provided we choose the right set of generalized coordinates. We want to use coordinates that reflect the symmetries of the problem, and find a set that yields the simplest possible Lagrangian : one with the maximum number of missing coordinates. Notice that if we re-analyzed the planet problem in Cartesian coordinates, none of the coordinates would be missing and none of the canonical momenta would be constants of the motion.

We can choose an even more advantageous coordinate system: notice we are allowing the planet’s orbit to have any orientation. Instead, we could require q =p /2 (and q ‘=0). This amounts to defining the planets orbital plane to be the x-y plane. Then

L = (1/2) m ( r’² + r² f ‘²) + GMm/r ,

so we can immediately see that pq = 0 is a constant of the motion also. Now the conserved quantity pf = mr² f ‘ is L, the total angular momentum of the planet.

As an aside, we can mention that the generalized coordinates can include electrical charge, and we can apply the Lagrangian formulation to problems involving electromagnetic fields. In this case, we have to treat the electric field energy as potential energy, and the magnetic field energy as kinetic energy. For example, the Lagrangian for motion of charge q in an LC circuit would be

L = T-V = (1/2) L q’² + (1/2) q² / C

yielding the equation of motion

Lq’’ + q/C = 0. This of course is simply the electromagnetic version of the harmonic oscillator, the resonant circuit. Using the Lagrangian formulation, electromechanical systems can be analyzed in an integrated manner, defining space coordinates and momenta for the mechanical components, and charge and current for the electrical parts.

Hamilton’s Principle

As we have seen, we can regard the detailed motion as a function of time of a system of N particles as representable by the motion of a single point in a 3N-dimensional configuration space. The path followed by the system point in configuration space when the system follows one of the natural motions possible under the laws of physics is called a dynamic path. However, we are free to consider other paths. For every point along a path in configuration space, we can define a corresponding value of the Lagrangian, since L is a function of q, q’ and t. The time evolution of a system of N particles is represented by the motion of the system point from its initial to its final position in configuration space. If we integrate the Lagrangian with respect to t along a particular path joining the initial and final positions, we obtain the action for that path:

S = ò L dt.

Notice action thus has dimensions of energy*time, or momentum*distance. (This use of the term action has been conventional since the 19^th century, and should not be confused with the meaning used in Newton’s day, when action was a synonym for momentum.)

Sir William Hamilton proved that the natural, or dynamical path for a system point in configuration space is the one for which the action integral is minimized. This is called Hamilton’s principle, or the Principle of Least Action. (Purists point out it should be called the Principle of Stationary Action, since Hamilton showed it must be an extremum, but not necessarily a minimum). In fact, one may begin by postulating Hamilton’s Principle and from it derive Lagrange’s equations. In other words, the Principle of Least Action, instead of Newton’s Laws, could serve as the fundamental basis of classical mechanics. We have already seen how de Broglie used Hamilton’s Principle and its analog in optics, Fermat’s Principle (or the Principle of Least Time, which says that a light ray follows the path which brings it from origin to destination in the minimum time) to infer by analogy the properties of matter waves.

Hamiltonian Formulation of mechanics

In the Lagrangian formulation, the fundamental quantities are the q_i and q’_i . In the Hamiltonian formulation, the fundamental quantities are the q_i and the p_i . Although generalized velocities q’_j may appear in the Hamiltonian formulation, they are regarded as functions of the fundamental variables:

q’ = f(q_i ; p_i ; t)

where again, as in all classical physics, time is an independent parameter.

This has a subtle but important effect on the partial derivatives to be used. The generalized velocity q’ and the generalized momentum p are distinct quantities, so in the Lagrangian formulation, ¶ L/¶ q_k means "hold constant all q (except q_k), all q’, and t", while in the Hamiltonian formulation it means "hold constant all q (except q_k), all p, and t". Similar definitions hold for (¶ /¶ q’_j), (¶ /¶ p_j), and (¶ /¶ t). Therefore in the Lagrangian view, ¶ q’_k/¶ q_j =0 for all j,k, but not necessarily so in the Hamiltonian; ¶ p_k/¶ q_j =0 for all j,k in the Hamiltonian view but not necessarily so in the Lagrangian.

We define a function called the Hamiltonian:

H(q,p,t) = [ å _k p_k q’_k ] - L(q, q’, t).

Since the kinetic energy is quadratic in the q’, it follows that the p are linear functions of the q’, or conversely, the q’ are linear functions of the p. Thus the Hamiltonian must be quadratic in the p.

If we take the partial derivative of the Hamiltonian with respect to p_j :

¶ H/¶ p_j = q’_j + å _k p_k (¶ q’_k /¶ p_j ) - å _k (¶ L /¶ q’_k) (¶ q’_k /¶ p_j )

but (¶ L /¶ q’_k)= p_k so the summations cancel, and we have

¶ H/¶ p_j = q’_j .

Similarly, if we take the partial derivative of the Hamiltonian with respect to q_j :

¶ H/¶ q_j =[ å _k p_k (¶ q’_k /¶ q_j )] - [å _k (¶ L /¶ q’_k) (¶ q’_k /¶ q_j )] - ¶ L /¶ q_j

but the summations cancel again for the same reason, and we have

¶ H/¶ q_j = - ¶ L /¶ q_j .

These two equations are called Hamilton’s equations. Notice that the latter equation tells us that if any coordinate is missing from the Lagrangian (so ¶ L /¶ q_j = 0), it is missing also from the Hamiltonian, so either situation can be used to identify a constant of the motion.

Since (¶ L /¶ q_k)= p’_k from Lagrange’s equations, we can summarize Hamilton’s equations as follows:

q’_j = ¶ H/¶ p_j

p’_k = - ¶ H/¶ q_j .

Finally, if we take the partial derivative of the Hamiltonian with respect to time:

¶ H/¶ t = [ å _k p_k (¶ q’_k /¶ t )] - [å _k (¶ L /¶ q’_k) (¶ q’_k /¶ t)] - ¶ L /¶ t

once again the summations cancel, and we have

¶ H/¶ t = - ¶ L /¶ t .

This tells us that in the Hamiltonian formulation, time acts just like a coordinate; if t is missing from L, is is also missing from H. Now look at the total derivative of H(q, p, t):

dH/dt = H’ = [ å _k (¶ H /¶ q_k )q’_k + (¶ H /¶ p_k) p’_k] + ¶ H /¶ t

Putting Hamilton’s equations into the square brackets gives zero, so

H’ =¶ H /¶ t

ie, the full time derivative is equal to the partial time derivative, which tells us that H is constant if it does not explicitly contain t. Thus if t is missing from the Hamiltonian (and hence also missing from the Lagrangian), H is a constant of the motion. Notice that in this respect H itself behaves like a generalized momentum, since it is a constant of the motion if its "conjugate coordinate" t is not in H and L, although there is a minus sign involved: -H is the "momentum" conjugate to the "coordinate" t.

Recall that Hamilton’s principle was obtained by considering paths in configuration space, the 3N-dimensional space in which all of the coordinates of all of the particles at any instant of time are represented by a single point. Knowledge of the configuration of the system at any time t is insufficient to predict its future configuration; we need to know the 3N generalized momenta to determine the path of the system as time passes. If instead we consider a 6N dimensional phase space, where again the system is represented by a single point, we can determine the motion of this point in phase space from the Hamiltonian, given its initial position:

q_k(t+dt) =q_k(t) +dq_k = q_k(t) + q’_k dt = q_k + (H /¶ p_k )dt

p_k(t+dt) =p_k(t) +dp_k = p_k(t) + p’_k dt = p_k + (H /¶ q_k )dt .

Thus the Hamiltonian contains all of the information needed to determine the future state {q,p} of the system.

Finally, we can re-analyze the planetary motion example using Hamilton’s equations. Recall that for this problem,

T = (1/2) m ( r’² + r² q ‘² + r² sin^2q f ‘²) ; V= - GMm/r ;

and the Lagrangian was therefore

L = T - V = (1/2) m ( r’² + r² q ‘² + r² sin^2q f ‘²) + GMm/r .

The generalized momenta are given by p_j = ¶ L/¶ q’_j:

p_r = mr’ ; pq = mr^2q ‘ ; pf = mr² sin^2q f ‘;

so we can easily find the connection between generalized velocities and momenta:

p_r /m = r’ ; pq / mr² = q ‘ ; pf / mr² sin^2q = f ‘.

The Hamiltonian is thus

H(q,p,t) = [ å _k p_k q’_k ] - L(q, q’, t)

= [(p_r² /m) + (pq ² / mr² ) + (pf ² / mr² sin^2q )]

-(1/2) [(p_r² /m) + (pq ² / mr² ) + (pf ² / mr² sin^2q )] - GMm/r

= (1/2) [(p_r² /m) + (pq ² / mr² ) + (pf ² / mr² sin^2q )] - GMm/r

= T + V = E

So evaluation of the Hamiltonian gives the total energy of the system. The equations of motion are obtained from the second of Hamilton’s equations, p’_k = - ¶ H/¶ q_j :

p’_r = (pq ² / mr³ ) + (pf ² / mr³ sin^2q ) - GMm/r²

p’q = pf ² cos q / (mr² sin^3q )

p’f = 0.

As before, we find pf = constant, and if we fix q =p /2, we get pf = mr² f ‘ = L, the magnitude of the angular momentum of the planet. Only one second-order equation is left,

p’_r = (pq ² / mr³ ) + (pf ² / mr³ sin^2q ) - GMm/r² = L² / (mr³ ) - GMm/r² , or

mr’’ = L² / (mr³ ) - GMm/r² .

However, if (in the q =p /2 coordinate system) simply set H=E we get a first-order differential equation which can be simply integrated:

(1/2) [mr’² + L² /(mr²)] - GMm/r = E, or

dr/dt = {(2/m)[E- L² /(mr²) + GMm/r }^1/2 .

Poisson Brackets

Consider a function of the generalized coordinates and momenta, F(q, p, t). We define

[F,G] = å _k { (¶ F/¶ q_k)(¶ G/¶ p_k ) - (¶ F/¶ p_k )(¶ G/¶ q_k) }

as the Poisson Bracket (PB) of F and G. From this definition, we see the PB is antisymmetrical:

[F,G] = - [G,F].

Possion brackets with the Hamiltonian are especially interesting:

[H,F] = å _k { (¶ H /¶ q_k)(¶ F/¶ p_k ) - (¶ H /¶ p_k )(¶ F/¶ q_k) }

= å _k { p’_k(¶ F/¶ p_k ) - q’_k(¶ F/¶ q_k) }

= -( dF/dt - ¶ F/¶ t )

So evidently

dF/dt = ¶ F/¶ t - [H,F].

Notice the above equation implies Hamilton’s equations (let F=q_k and F=p_k). One application of PBs is of special importance:

[p_m , q_j ] = å _k { (¶ p_m /¶ q_k)(¶ q_j/¶ p_k ) - (¶ p_m /¶ p_k )(¶ q_j /¶ q_k) }

= - å _k d _mk d _jk = - d _mj

where the Kronecker delta is defined as d _mj = 1 (j=k), =0 (otherwise). Similarly, we find

[p_m , p_j ] = [q_m , q_j ] = 0.

If we transform from one set of generalized coordinates and momenta to another, the form of Hamilton’s equations is preserved (and the coordinate transformation is said to be canonical) if the three PB equations given above are satisfied. So the PBs provide a test for canonical transformations.

Bibliography:

The standard reference on classical mechanics is Herbert Goldstein, Classical Mechanics, Addison-Wesley, 1980.

A more elementary treatment may be found in Keith Symon, Mechanics, Addison-Wesley, 1971.

For an intellectual experience, see Cornelius Lanczos, The Variational Principles of Mechanics, Dover, 1986