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Notes for Classical Mechanics II course, CMI, Spring 2013
Govind S. Krishnaswami, May 16, 2013
Some books on classical mechanics are mentioned on the course web site http://www.cmi.ac.in/ ∼ govind/teaching/cm2-e13. These lecture
notes are very sketchy and are no substitute for attendance and taking notes at lectures. Please let me know (via [email protected]) of any
comments or corrections.
Contents
1
2
Review of Newtonian and Lagrangian mechanics of point particles
2
1.1
Newton’s force law, configuration space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.2
Energy, Angular momentum, conserved quantities, dynamical variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.3
Lagrangian formulation of mechanics and principle of extremal action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.4
From symmetries to conserved quantities: Noether’s theorem on invariant variational principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
Hamiltonian formalism of mechanics
8
2.1
Hamilton’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
2.2
Hamiltonian from Lagrangian via Legendre transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.3
Poisson brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.4
Variational principles for Hamilton’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.5
Lagrange’s and Hamilton’s equations take same form in all systems of coordinates on Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.6
Canonical transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.7
3
2.6.1
Form of Hamilton’s equations are preserved iff Poisson brackets are preserved . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.6.2
Canonical transformations for one degree of freedom: Area preserving maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.6.3
Generating function for infinitesimal canonical transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.6.4
Liouville’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.6.5
Generating functions for finite canonical transformations from variational principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Action-Angle variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.7.1
Action-angle variables for the harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.7.2
Generator of canonical transformation to action-angle variables: Hamilton-Jacobi equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.7.3
Generating function for transformation to action-angle variables for harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.7.4
Action-angle variables for systems with one degree of freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.7.5
Time dependent Hamilton-Jacobi equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.7.6
Example: Solving HJ by separation of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
Rigid body mechanics
Lab and co-rotating frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.2
Infinitesimal displacement and angular velocity of rigid body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.3
Kinetic energy and inertia tensor or matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
3.4
Angular momentum of a rigid body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.5
Equations of motion of a rigid body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.6
Force-free motion of rigid bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.6.1
Free motion of rigid rotator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.6.2
Free motion of symmetrical top . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.7
Euler angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.8
Euler equations for a rigid body in body-fixed frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.9
Ellipsoid of inertia and qualitative description of free motion of rigid body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.8.1
4
34
3.1
Euler equations for force-free motion of symmetric top . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Dynamics of continuous media
50
4.1
Lagrangian and Eulerian descriptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.2
Vibrations of a stretched string . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
4.2.1
Wave equation for transverse vibrations of a stretched string . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
4.2.2
Separation of variables and normal modes of a vibrating string . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
1
4.3
1
4.2.3
Right and left-moving waves and d’Alembert’s solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.2.4
Conserved energy of small oscillations of a stretched string . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Fluid dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.3.1
Material derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.3.2
Incompressibility of a fluid flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
4.3.3
Conservation of mass: continuity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
4.3.4
Euler equation for inviscid flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
4.3.5
Flow visualization: Streamlines, streak-lines and path-lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
Review of Newtonian and Lagrangian mechanics of point particles
1.1
Newton’s force law, configuration space
• A point particle moving along a wire in the shape of a line or circle has one degree of freedom, namely
its position (coordinate) along the wire. A point-like molecule in the air has three degrees of freedom,
e.g., its cartesian coordinates with respect to a chosen set of axes. N such molecules have 3N degrees
of freedom. The set of possible locations of the molecules forms the configuration space of the system.
For a particle on a plane, the configuration space is R2 , two-dimensional Euclidean space.
• The zeroth law of classical (Newtonian) mechanics can be regarded as saying that the trajectory x(t)
of a particle is a (twice) differentiable function of time. This is a law that applies to tennis balls, planets
etc. But it fails for Brownian motion (movement of pollen grains in water). It also fails for electrons in
an atom treated quantum mechanically.
• Newton’s force law says that the trajectory ~r(t) = (x, y, z) of a particle is determined by the force acting
on it. In cartesian coordinates,
m~r¨ = f~ or ṗ = f,
or
m ẍi = fi .
(1)
Here the momentum ~p = m~v = m~r˙ . We also denote x1 = x, x2 = y, x3 = z, so xi are the coordinates.
Sometimes the coordinates are denoted qi instead of xi .
• In many interesting cases, the force can be written as the negative gradient of a potential energy1
∂V
∂V
∂V
ŷ +
ẑ
f~ = −∇V = − x̂ +
∂x
∂y
∂z
(2)
E.g. V = mgz for the gravitational potential energy and so F~ = −mgẑ points downwards. In this case,
Newton’s second law is
∂V
ṗ = −∇V or m ẍi = − .
(3)
∂xi
Since Newton’s equation is second order in time, to solve it we need two sets of initial conditions, the
initial position vector and velocity vector (~r, ~r˙) or the initial position and momentum.
• We say that the state of the particle is specified by giving its instantaneous position and momentum.
The set of possible instantaneous states of the particle is called its phase space. For a particle moving
in three dimensional space, its configuration space is three dimensional R3 (consisting of all possible
instantaneous locations) and its phase space is six-dimensional R6 (locations and momenta). Draw all
the phase space trajectories for a free particle with one degree of freedom.
1
Force points in the direction of greatest decrease in potential energy
2
• The path of the particle ~x(t) (satisfying Newton’s equation and initial conditions) in configuration
space is called its trajectory. Also of interest is the trajectory in phase space (~x(t), ~p(t)).
1.2
Energy, Angular momentum, conserved quantities, dynamical variables
• The energy of a particle moving in three dimensional space (say under the gravitational force) is a sum
of kinetic and potential energies
1
1 2
E = T + V = m( ẋ2 + ẏ2 + ż2 ) + V(x, y, z) =
p x + p2y + p2z + V(x, y, z)
2
2m
(4)
ẋ = dx
dt is the x-component of velocity. We often think of energy as a function of coordinates and
momenta E(x, y, z, p x , py , pz ). So energy is a function on the phase space.
• Newton’s equation implies that the energy is a constant of motion if the forces are conservative (expressible as the gradient of a potential). We say the energy is conserved in time/is a conserved quantity.
Ė = m
X
ẋi ẍi +
i
X ∂V
ẋi = 0.
∂x
i
i
(5)
• Conserved quantities are useful. They help us solve/understand Newton’s equation for the trajectory.
E.g., for one degree of freedom, we may integrate once and get an (implicit) expression for x(t):
r
Z x
1 2
dx
2
dx0
(E − V(x)) ⇒ t − t0 =
E = m ẋ + V(x) ⇒
=
(6)
q
2
dt
m
2
x0
0
m (E − V(x ))
One may wonder how this formula for energy arose from Newton’s equation. Let us consider one degree
of freedom. We wish to integrate m ẍ = − dV
dx with respect to time in order to solve the equation of motion.
To do so we notice that ẋ is an integrating factor. For, multiplying the equation by ẋ, both sides become
total time derivatives:
!
dV dx
1 d ẋ2
dV
d 1 2
dE
m ẍ ẋ = −
or
m
=−
or
m ẋ + V =
= 0.
(7)
dx dt
2 dt
dt
dt 2
dt
The constant energy E can be regarded as the constant of integration arising from integrating Newton’s
equation once. The second constant of integration is x0 above. The same sort of integrating factor also
works for several degrees of freedom.
• If there is no force, then each of the components of momentum is conserved, since ~p˙ = F~ = 0 (this
is Newton’s first law). If the force only acts downwards, then the horizontal components of momentum
p x , py are conserved.
• The angular momentum of a particle about a fixed point (origin) is ~L = ~r × ~p , where ~r is the position
vector of the particle from the origin. In components
L x = ypz − zpy ,
Ly = zp x − xpz ,
Lz = xpy − yp x .
(8)
• Newton’s force law then implies that the rate of change of angular momentum is the torque:
L̇ = ṙ × p + r × ṗ =
1
~p × ~p + ~r × F~ = ~r × F~ ≡ ~τ ≡ k.
m
3
(9)
E.g. For a projectile moving under the vertical gravitational force, the torque must be in the horizontal
plane. So the vertical component of angular momentum Lz = xpy − yp x must be conserved. Since p x
and py are also conserved, we conclude that the trajectory (x, y, z)(t) must be such that its projection on
the horizontal plane is a straight line Lz = xpy − yp x . Again, knowledge of conserved quantities allowed
us to clarify the trajectory.
• Position, energy, momentum and angular momentum are examples of dynamical variables. Dynamical
variables are sometimes called observables, since they may be observed/measured. They are real-valued
physical quantities defined at each point along any trajectory. So they are functions of the instantaneous
location and momentum (or velocity). Most often, our dynamical variables will be smooth functions on
phase space. For e.g. the x-component of angular momentum is one such function L x = ypz − zpy .
In general, dynamical variables change along the trajectory. Conserved quantities are those dynamical
variables that are constant along all trajectories.
1.3
Lagrangian formulation of mechanics and principle of extremal action
• The Lagrangian formulation is an alternative way of obtaining and studying Newton’s equation of
motion m~q¨ = −∇V(q) for a system of particles interacting via forces derivable from a potential V .
• Let us define the Lagrange function (Lagrangian) L(q, q̇) = T − V(q) = 12 mq̇2 − V(q). Then we observe
that Lagrange’s equation
d ∂L ∂L
∂V
=
⇒ mq̈ = −
(10)
dt ∂q̇ ∂q
∂q
reproduces Newton’s equation. We denote coordinates by q rather than x to emphasize they need not be
Cartesian coordinates. Let us briefly describe how Lagrange’s equations arise.
• We consider the problem of determining the classical trajectory that a particle must take if it was at qi
at ti and q f at t f . Instead of specifying the initial velocity, we give the initial and final positions at these
times. Which among all the paths that connect these points, is the one that solves Newton’s equation?
We seek a variational principle, called the principle of extremal action (sometimes called least action)
that should determine the classical trajectory.
• The idea is to find an action functional S [q], which is a real-valued function of paths on configuration
space connecting q(ti ) and q(t f ) such that the classical trajectory is a local extremum of S [q]. In
other words, if we modify the classical trajectory qcl (t) → qcl (t) + δq(t), holding the end points fixed
(δq(ti ) = δq(t f ) = 0), then the action S [q] must remain stationary to 1st order in δq(t). This imposes
certain conditions on qcl (t) called Euler-Lagrange equations.
• The action that does this job is
S [q] =
tf
Z
L(q, q̇) dt
(11)
ti
where L is the Lagrangian. For a particle in a potential V ,
1 X i i
L(q, q̇) = T − V = m
q̇ q̇ − V(q)
2
i
(12)
• The Euler-Lagrange equations are got by computing the infinitesimal change in action δS under
qi (t) → qi (t) + δqi (t).
(
)
n Z tf
X
∂L i 0
0 ∂L
i 0
δS =
dt
δq (t ) + i δq̇ (t ) + O(δq)2
i
∂q
∂q̇
i=1 ti
4
=
tf
Z
ti
!
∂L
d ∂L
∂L
∂L
δq (t )
− 0 i dt0 + δqi (t f ) i
− δqi (ti ) i
+ O(δq)2
i
dt ∂q̇
∂q
∂q̇ (t f )
∂q (t f )
i 0
(13)
We integrated by parts to isolate the coefficient of δq. The last two ‘boundary terms’ are zero due to the
initial and final conditions and so the condition δS = 0 can be reduced to a condition that must hold at
each time, since δqi (t0 ) are arbitrary at each intermediate time. So choosing, roughly, δqi (t0 ) = 0 except
for a specific time t0 = t we get the Euler-Lagrange (EL) (or just Lagranges’s) equations
∂L
d ∂L
−
= 0,
i
dt
∂q (t)
∂q̇i (t)
i = 1, 2, . . . n.
(14)
• Moreover, the momentum conjugate to the coordinate qi is defined as
pi =
∂L
∂q̇i
(15)
Now suppose we have a particle moving in the potential V(xi ). For the above Lagrangian, we can check
that in cartesian coordinates, pi are the usual momenta and Lagrange’s equations are just Newton’s
equations
∂L
∂V
pi =
= mq̇,
and
ṗi = − .
(16)
∂q̇i
∂qi
The Lagrangian scheme has the advantage of generating all the equations of motion from a single function. Moreover, the Euler-Lagrange equations may be literally carried over to any coordinate system, so
qi need not be Cartesian coordinates and are often called generalized coordinates. There are as many
generalized coordinates qi as there are degrees of freedom. So for a pair of particles in a room, there
would be six generalized coordinates q1 , · · · , q6 .
∂L
= 0, is called a cyclic coordinate. The
• A coordinate qi that does not appear in the Lagrangian ∂q
i
d ∂L
∂L
corresponding momentum is conserved since ṗi = dt ∂q̇i = ∂qi = 0.
• It is important to bear in mind that the Lagrangian L(q, q̇) is a function of the coordinates q and
velocities q̇, and that the momentum p is a derived concept.
• The example of a harmonic oscillator (particle connected to a spring of force constant k ). Here the
restoring force −kx arises from a potential V(x) = 12 kx2 , where x is the extension of the spring. So,
1
1
L = m ẋ2 − kx2
2
2
and Lagrange’s equation
d(m ẋ)
dt
(17)
= −kx reproduces Newton’s equation.
• For a particle on a plane in a central potential, in polar coordinates
m
L = T − V = (ṙ2 + r2 θ̇2 ) − V(r)
2
(18)
The momenta conjugate to (r, θ) are
pr =
∂L
= mṙ,
∂ṙ
pθ =
∂L
= mr2 θ̇
∂θ̇
(19)
They coincide with the radial component of linear momentum and the angular momentum. Moreover,
the first of Lagrange’s equations is
ṗr = mr̈ =
∂L
= mrθ̇2 − V 0 (r).
∂r
5
(20)
This is the balance of radial acceleration, centripetal ‘force’ and central force. On the other hand,
ṗθ =
d
∂L
(mr2 θ̇) =
=0
dt
∂θ
⇒
mrθ̈ = −2mṙθ̇.
(21)
This states the conservation of angular momentum, and involves the so-called Coriolis term on the rhs
when written out. Note that Newton’s equations do not take the same form in all systems of coordinates.
There is no force in the θ̂ direction, yet the naive ‘angular acceleration’ mθ̈ is non-zero. On the other
∂L
hand, Lagrange’s equations dtd ∂∂Lq̇i = ∂q
are valid in all systems of coordinates. So q could be a Cartesian
i
or polar coordinate for instance.
• A Lagrangian for a given system of equations is not uniquely defined. For instance, we may add a
constant to L(q, q̇, t) without affecting the EL equations. We may also multiply the Lagrangian by a
constant. Another source of non-uniqueness arises from the freedom to add the total time derivative of
any (differentiable) function F(q, t) to the Lagrangian. The change in the action is
Z tf
dF
L → L + Ḟ ⇒ S → S +
dt = F(q(t f ), t f ) − F(q(ti ), ti )
(22)
dt
ti
But this quantity has zero variation since ti , t f , q(ti ), q(t f ) are all held fixed as the path is varied. So the
addition of Ḟ to L does not affect the EL equations. Notice that we could not allow F to depend on q̇
since δq̇(ti ), δq̇(t f ) , 0 in general and such an F would modify the EL equations.
1.4
From symmetries to conserved quantities: Noether’s theorem on invariant variational principles
• Newton/Lagrange
equations of classical mechanics have been formulated as conditions for the action
R
S = L dt to be extremal. Many concepts (such as symmetries) may be formulated more simply in
terms of the action/Lagrangian than in terms of the equations of motion.
• If a coordinate q j is absent in the Lagrangian (q j is a cyclic coordinate), then the corresponding
∂L
conjugate momentum p j = ∂∂L
is conserved in time. This follows from Lagrange’s equation ṗ j = ∂q
j .
q̇ j
If the Lagrangian is independent of a coordinate, then in particular, it is unchanged when this coordinate
is varied δL = 0 under q j → q j + δq j . We say that translations of q j are a symmetry of the Lagrangian.
This relation between symmetries and conserved quantities is deeper.
• A transformation of coordinates qi → q̃i is a symmetry of the equations of motion (eom) if it leaves
the equations of motion unchanged: i.e., the eom for q̃ is the same as that for q. Symmetries usually
allow us to produce new solutions from known ones. For example, the free particle equation mq̈ = 0 is
left unchanged by a translation of the coordinate q → q̃ = q + a for any constant length a. Now q = 0
is one static solution. We may use the symmetry under translations to produce other static solutions,
namely q = a for any a, i.e., the particle is at rest at location with coordinate a rather than at the origin.
Incidentally, the momentum of a free particle is conserved in time. We will see that such symmetries are
associated with conserved quantities. On the other hand, the equation of motion of a particle attached
to a spring mq̈ = −kq is non-trivially modified by a translation of the coordinate q → q̃ = q + a since
q̃ satisfies a different equation mq̃¨ = −kq̃ + ka. Moreover, p = mq̇ is not (in general) conserved for a
particle executing simple harmonic motion, the momentum is zero at the turning points and maximal at
the point of equilibrium.
• A symmetry
of the action is a transformation that leaves S unchanged. E.g. the free particle action
Rt
S [q] = t f 21 mq̇2 dt is unchanged under the shift q → q + a. Since the eom are the conditions for S to be
i
stationary, a symmetry of the action must also be a symmetry of Lagrange’s equations. Noether’s theorem
6
constructs a conserved quantity associated to each infinitesimal symmetry of the action. Let us see how
in the simplest case. Suppose the infinitesimal change qi → qi + δqi leaves the Lagrangian unchanged to
linear order. Then it is automatically an infinitesimal symmetry of the action. Let us explicitly calculate
the first variation of the action for paths between the times t1 and t2 , S [q + δq] = S [q] + δS [q]. Upto
terms of order (δq)2 we get
#
!
#
Z t2 "
Z t2 "
∂L
∂L
d
∂L
d ∂L
∂L
δS =
δqi i + δq̇i i dt =
δq
− δqi
dt
δqi i +
∂q̇ #
dt ∂q̇
∂q
∂q̇
∂q " dt
t1
t1 Z
t2
∂L
∂L
∂L
d ∂L
= δqi (t2 ) (t2 ) − δqi (t2 ) (t1 ) +
δqi
−
dt
(23)
i
∂q̇
∂q̇
dt
∂q̇
∂q
t1
So far, this is true for any path and for any infinitesimal change δq. Let us now specialize to infinitesimal
changes about a trajectory, so that q(t) satisfies Lagrange’s equations and the last term vanishes. Further
more, we assume that the transformation is an infinitesimal symmetry of the Lagrangian, so that δS = 0.
Thus we have
∂L
∂L
(24)
0 = δS = δqi (t2 ) i (t2 ) − δqi (t2 ) i (t1 ).
∂q̇
∂q̇
∂L
Since t1 , t2 are arbitrary, the quantity δq ∂L
∂q̇ must be constant along a trajectory. Notice that ∂q̇i is the
momentum pi In other words, an infinitesimal symmetry q → q + δq of the Lagrangian implies that the
quantity Q = pi (t)δqi (t) = ~p · δ~q is a constant of the motion, i.e. the dynamical variable Q has the same
value at all points along a trajectory.
• E.g. 1: We already saw that the free particle Lagrangian is translation invariant with δqi = ai where
ai are the components of an arbitrary infinitesimal vector. It follows that Q = ai pi = ~p · ~a is a conserved
quantity. In other words, the component of momentum in any direction is conserved, which we knew.
• E.g. 2: Now consider a particle in a central potential V(q2 ) so that the Lagrangian is
1
L(q, q̇) = mq̇ · q̇ − V(q · q)
2
(25)
Let us first show that L is invariant under rotations of three dimensional space ~q → R~q where R is any
S O(3) rotation matrix (Rt R = I, det R = 1). Recall that the dot product is defined as a · b = at b for any
column vectors a, b and that (Ra)t = at Rt for any matrix R and t denotes transposition. Thus
1
1
L(Rq, Rq̇) = mq̇Rt Rq̇ − V(qt Rt Rq) = mqt q − V(qt q) = L(q, q̇).
2
2
(26)
So the Lagrangian is invariant under rotations. Noether’s theorem, however, refers to infinitesimal transformations, rotations in this case. So let us find a formula for an infinitesimal rotation.
• Suppose we make an infinitesimal rotation of the vector q about the axis n̂ by a small angle θ counterclockwise. Then the vector q sweeps out a sector of a cone. Suppose q makes an angle φ with respect n̂,
so that the opening angle of the cone is φ. Then the rotated vector q̃ also makes an angle φ with respect
to the axis n̂. Let δq = q̃ − q be the infinitesimal change in q. By looking at the base of this cone, we
find that it is a sector of a circle with radius q sin φ and opening angle θ . So we find that |δq| = θq sin φ.
Moreover δq points in the direction of n̂ × q. Thus, under a counter-clockwise rotation about the axis n̂
by a small angle θ , the change in ~q is
δq = θ n̂ × q
and
δq̇ = θ n̂ × q̇
In particular, we see that δq and δq̇ are orthogonal to q and q̇ respectively.
7
(27)
• Now let us check that the Lagrangian is invariant under infinitesimal rotations:
1
1
1
L(q + δq, q̇ + δq̇) ≈ mq̇2i + mq̇ · δq̇ + mδq̇ · q − V(q2 + q · δq + δq · q) = L(q, q̇)
2
2
2
(28)
The last equality follows on account of the orthogonality properties just mentioned. Thus the Lagrangian
(and action) are invariant under infinitesimal rotations. The resulting conserved quantity from Noether’s
theorem is
Q = ~p · θ n̂ × ~q = θ n̂ · (~q × ~p) = θ ~L · n̂.
(29)
Since Q is conserved for any small angle θ and for any axis of rotation n̂, we conclude that the component of angular momentum in any direction is conserved. So the angular momentum vector is a constant
~
of motion ddtL = 0, a fact we are familiar with from the Kepler problem for the 1/r central potential. We
also knew this since the torque ~r × f~ on such a particle about the force centre vanishes: the moment arm
and force both point radially.
2
2.1
Hamiltonian formalism of mechanics
Hamilton’s equations
• Newton’s equation of motion F = ma is a second order equation for the coordinate of the particle. It
can be rewritten as pair of first order equations for the position and momentum using the Hamiltonian or
energy function
p2
H(x, p) = T + V =
+ V(x)
(30)
2m
First write m ẍ = dV
dx as the pair of equations: p = m ẋ, (which is merely the definition of momentum)
∂V
and ṗ = − ∂x . Then observe that these equations could also be written as
ẋ =
∂H
p
= ,
∂p
m
and
ṗ = −
∂H
dV
=−
∂x
dx
(31)
Note that here we regard the hamiltonian H(x, p) as a function on phase space, i.e., as a function of
positions and momenta rather than positions and velocities. For several degrees of freedom, Hamilton’s
equations take the form
∂H
∂H
and ṗi = −
(32)
ẋi =
∂pi
∂xi
Hamilton’s equations are 1st order in time. So there are twice as many of them compared to Newton’s
or Lagrange’s equations. Hamilton’s equations treat position and momentum on a more equal footing,
except for a sign.
• Moreover, it is easier to write down Hamilton’s equations, since we only need to know one function
(the hamiltonian). By contrast, to write Newton’s equations we need to know the force, which is a vector.
We will define the hamiltonian in general in the next section. Here we take a simple-minded approach
based on previous physical experience.
• One of the main advantages of the Hamiltonian formalism over the Newtonian one is that Hamilton’s
equations take the same form in all systems of coordinates on configuration space. By contrast, Newton’s
equations m ẍi = −∂i V look quite different in curvilinear coordinates. Let us illustrate this assertion with
the example of a particle free to move on a plane without any force acting on it. Newton’s equations are
8
then m ẍ = 0 and mÿ = 0 and the energy is E = m2 ( ẋ2 + ẏ2 ). The coordinates are x, y and the components
of momenta in the x̂, ŷ directions are p x = m ẋ and py = mẏ.
• Let us transform to polar coordinates x = r cos φ and y = r sin φ. Then the components of velocity are
v = ( ẋ, ẏ)
where
ẋ = ṙ cos φ − r sin φ φ̇,
ẏ = ṙ sin φ + r cos φ φ̇.
(33)
So ẋ2 + ẏ2 = ṙ2 + r2 φ̇2 . So the energy is
E=H=
m 2
ṙ + r2 φ̇2
2
(34)
Since we wish to write H in terms of momenta rather than velocities, let us introduce the radial momentum as the component of linear momentum in the r̂ = (cos φ, sin φ) direction.
pr =
~p · ~r
= r̂ · ~p = mr̂ · ~v = mṙ
r
(35)
We also introduce the angular momentum corresponding to a displacement of φ holding r fixed. The
moment arm is r and the velocity corresponding to a shift in φ is rφ̇ so pφ = rmrφ̇. Check that pφ is
just the angular momentum about the vertical axis Lz = xpy − yp x = m(xẏ − y ẋ) = mr2 φ̇.
• Note that even though we call pφ the momentum conjugate to φ, it does not have dimensions of
momentum, it actually has dimensions of angular momentum. In the same way we will refer to r, φ as
coordinates, though they don’t both have dimensions of length.
• The energy is now expressed in polar coordinates and momenta
E=H=
p2φ
p2r
+
2m 2mr2
(36)
• Now Newton’s equations ẍ = ÿ = 0 can be expressed in polar coordinates. One can check that
mr̈ = mrφ̇2
(37)
Notice that Newton’s equation in polar coordinates has a different form to that in cartesian coordinates.
If we had naively regarded mr̈ as an acceleration and observed that there is no force on the particle, we
would have got a wrong equation. Though there is no force on the particle, neither r̈ nor φ̈ is zero. Of
course, we sometimes say that mrφ̇2 is a centripetal force, but this is not really a force, it is just another
term in the acceleration due to the curvilinear coordinates.
• On the other hand, let us write down Hamilton’s equations, which we claimed take the same form in
all coordinate systems
pφ
∂H
φ̇ =
=
,
∂pφ mr2
∂H
pr
ṙ =
= ,
∂pr
m
∂H
ṗφ = −
= 0,
∂φ
p2φ
∂H
ṗr = −
=
∂r
mr3
(38)
• Let us check if these equations agree with Newton’s equations. The first two of Hamilton’s equations
just reproduce the definitions of pr and pφ . The third says that the angular momentum pφ = Lz is
conserved. The last one along with the first two is equivalent to Newton’s equation mr̈ = mrφ̇2 for the
balance of radial acceleration and centrifugal force.
• The hamiltonian formulation allows us to identify conserved quantities easily. A cyclic coordinate
q is one that does not appear in the hamiltonian so that ∂H
∂q = 0. Its conjugate momentum is then
9
automatically conserved ṗ = − ∂H
∂q = 0. In our case, φ does not appear in the hamiltonian, so the
conjugate momentum (angular momentum) is conserved. This was not quite obvious from Newton’s
equations, though we knew it.
∂H
• So we have checked that in this case, Hamilton’s equations ẋi = ∂p
and ṗi = − ∂H
∂xi take the same
i
form in Cartesian as well as polar coordinates. In other words we can choose xi = (x, y) or (r, φ) in (32).
• In the above example, we could motivate the formulae for H, pr , pφ based on simple physical arguments. More generally, there is a systematic procedure to find the conjugate momenta and hamiltonian
starting from the Lagrangian.
2.2
Hamiltonian from Lagrangian via Legendre transformation
• The Lagrangian L(q, q̇) that leads to the equations of motion, is a function of coordinates and velocities. On the other hand, the Hamiltonian H(q, p) is a function of coordinates and momenta. In the simple
case of a particle in a potential, L = T − V while H = T + V , so they are not the same function. In
general, H(q, p) is obtained from L(q, q̇) via a Legendre transformation. To pass from L to H we need
to eliminate the independent variable q̇ in favor of the independent variable p.
• To do so, notice that the definition of conjugate momentum p =
∂L
∂q̇
is in fact the condition for the
q̇)
function pq̇ − L(q, q̇) to be extremal with respect to q̇. Now we may invert the relation p = ∂L(q,
∂q̇
and solve for q̇ as a function of q and p. The hamiltonian is defined as the result of substituting this
expression for q̇ in pq̇ − L(q, q̇). In other words, H(q, p) is the extremal value of pq̇ − L(q, q̇). For
several degrees of freedom
h
i
H(q, p) = extq̇i pi q̇i − L(q, q̇)
(39)
where the extermination is carried out with respect to all the generalized velocities.
• Let us check that this reproduces the energy of a particle in a potential V
1
L = T − V = mq̇2 − V(q),
2
and
p=
∂L
= mq̇
∂q̇
⇒
q̇ =
p
.
m
(40)
For this q̇,
p 1 p2
p2
− m 2 + V(q) =
+ V.
(41)
m 2 m
2m
which we recognize as the energy. Do the same for a free particle moving on a plane, first in cartesian
H = pq̇ − L = p
and then in polar coordinates. Show that the hamiltonian is H(r, θ, pr , pθ ) =
p2r
2m
+
p2θ
2mr2
.
• Another way of viewing the Legendre transformation is to look at the differential of the Lagrangian,
i.e., the infinitesimal change in the Lagrangian under a small change in the independent variables q, q̇:
dL(q, q̇) =
∂L
∂L
dq +
dq̇ = ṗ dq + p dq̇
∂q
∂q̇
(42)
where we used the definition of momentum and Lagrange’s equation:
p=
∂L
,
∂q̇
and
ṗ =
∂L
.
∂q
(43)
Now we want the independent variable to be p instead of q. So consider H = pq̇ − L and compute its
differential
dH = p dq̇ + q̇ d p − ṗ dq − p dq̇ = q̇ d p − ṗ dq
(44)
10
This makes it clear that the independent variables in H are q and p alone, and moreover, that
q̇ =
∂H
,
∂p
and
ṗ = −
∂H
.
∂q
(45)
These we recognize as Hamilton’s equations which we introduced in an ad hoc fashion before. So the
hamiltonian is a function on phase space and the solution of hamilton’s equations (q(t), p(t)) gives the
trajectory on phase space. Hamilton’s equations being first order in time require the specification of
initial positions and initial momenta (qi (t = 0), pi (t = 0)).
• The Legendre transform H(q, p) = extq̇ pq̇ − L(q, q̇) is defined provided L(q, q̇) is a convex function
of q̇, which allows us to solve p = ∂L
∂q̇ for q̇ in terms of q, p. L is convex in q̇ provided the second
derivative
∂2 L
∂q̇2
is positive for all q̇. This condition is satisfied by L(q̇) = 12 mq̇2 as long as m > 0. On
the other hand, let us attempt to compute the Legendre transform of L = 41 q̇4 − 12 q̇2 . We expect to run
into trouble. Indeed, there is often more than one solution q̇ for a given p when we try to solve for q̇ in
3
p = ∂L
∂q̇ = q̇ − q̇. In this case, the Legendre transform H is not single-valued.
• When the Legendre transform is defined, the Lagrangian can be re-obtained from H(q, p) by an (inverse) Legendre transform
L(q, q̇) = ext p pq̇ − H(q, p) .
(46)
2.3
Poisson brackets
• Consider a particle (or system of particles) with configuration space Rn with generalized coordinates
qi and generalized momenta pi = ∂∂Lq̇i . To motivate the idea of Poisson brackets, let us use Hamilton’s
∂H
and ṗi = − ∂H
equations ( q̇i = ∂p
∂qi ) to find the time evolution of any dynamical variable f (q, p; t). f is
i
in general a function on phase space, which could depend explicitly on time. However, we will mostly
be interested in observables that aren’t explicitly time-dependent, but which depend on time only via q(t)
and p(t).
!
!
n
n
df
∂ f X ∂ f dqi ∂ f d pi
∂ f X ∂ f ∂H ∂ f ∂H
∂f
=
+
+
+
−
+ { f, H}.
=
=
dt
∂t i=1 ∂qi dt
∂pi dt
∂t i=1 ∂qi ∂pi ∂pi ∂qi
∂t
(47)
Here we introduced Poisson’s bracket of f with the hamiltonian. More generally, the p.b. of two dynamical variables gives another dynamical variable defined as2
!
n
X
∂ f ∂g
∂ f ∂g
−
.
{ f, g} =
∂qi ∂pi ∂pi ∂qi
i=1
(48)
• So the time derivative of any observable is given by its Poisson bracket with the hamiltonian (aside
from any explicit time-dependence). From here on, we will restrict to observables that are not explicitly
time-dependent, unless otherwise stated. Hamilton’s equations for time evolution may now be written
q̇ = {q, H} and
ṗ = {p, H}.
(49)
If H isn’t explicitly dependent on time, then time does not appear explicitly on the RHS of hamilton’s
equations. We say that the ODEs for q and p are an autonomous system.
2
Some authors define the p.b. with an overall minus sign relative to our definition.
11
• One advantage of Poisson brackets is that the time evolution of any observable f (q, p) is given by an
equation of the same sort f˙ = { f, H}. We say that the hamiltonian generates infinitesimal time evolution
via the Poisson bracket (more on this later).
• If { f, g} = 0 we say that f Poisson commutes with g. In particular, f is a constant of motion iff
it ‘Poisson commutes’ with the hamiltonian, f˙ = 0 ⇔ { f, H} = 0. We begin to see the utility of the
Poisson bracket in the study of conserved quantities.
• The Poisson bracket has some notable properties. The P.b. of any dynamical variable with a constant
is zero. The Poisson bracket is anti-symmetric in the dynamical variables { f, g} = −{g, f }. In particular,
the p.b. of any observable with itself vanishes {h, h} = 0. A special case of this encodes the conservation
of energy. Assuming H isn’t explicitly dependent on time,
dH
= {H, H} = 0.
dt
(50)
• Since the above formula for the p.b. involves only first order derivatives of f , the p.b. satisfies the
Leibnitz/product rule of differential calculus. Check that
{ f g, h} = f {g, h} + { f, h}g and
{ f, gh} = { f, g}h + g{ f, h}.
(51)
Anti-symmetry ensures that the Leibnitz rule applies to the second entry as well.
• In the Poisson bracket { f, g} we refer to f as the function in the first slot or entry and g as occupying
the second. A simple consequence of the definition is that the Poisson bracket is linear in each entry.
Verify that { f, cg} = c{ f, g} and { f, g + h} = { f, g} + { f, h} etc. where c is a real constant.
• The fundamental Poisson brackets are between the basic dynamical variables, namely coordinates and
momenta. The above formulae give for one degree of freedom
{q, p} = 1
or {p, q} = −1,
while {q, q} = 0
and
{p, p} = 0.
(52)
The last two equations are in fact trivial consequences of the anti-symmetry of the p.b. For n-degrees of
freedom we have the fundamental p.b. among the coordinates and momenta
{qi , p j } = δi j ,
and
{qi , q j } = {pi , p j } = 0
for 1 ≤ i, j ≤ n.
(53)
These are sometimes called the canonical (‘standard’) Poisson bracket relations between coordinates and
conjugate momenta.
• Poisson’s theorem: Perhaps the most remarkable feature of the Poisson bracket is that it can be used
to produce new conserved quantities from a pair of existing ones. Poisson’s theorem states that if f
and g are conserved, then so is { f, g}. Let us first illustrate this with a couple of examples. For a
free particle moving on a plane we know that p x and py are both conserved. Their Poisson bracket is
{p x , py } = 0, which is of course a trivially conserved quantity. As a second example, consider a particle
moving in three dimensions under the influence of a central potential. We know that L x = ypz − zpy and
Ly = zp x − xpz are both conserved. We compute {L x , Ly } by using bi-linearity, the Leibnitz rule and other
properties of the p.b. and find {L x , Ly } = Lz . And indeed, we know that Lz is also a conserved quantity.
Similarly we check that
{L x , Ly } = Lz ,
{Ly , Lz } = L x
12
and {Lz , Lz } = Ly .
(54)
• Jacobi identity: More generally, Poisson’s theorem is a consequence of the Jacobi identity. For any
three dynamical variables f, g and h, the following cyclic sum of ‘double’ Poisson brackets vanishes:
{ f, {g, h}} + {h, { f, g}} + {g, {h, f }} = 0.
(55)
Using anti-symmetry we could write the Jacobi identity also as
{{ f, g}, h} + {{g, h}, f } + {{h, f }, g} = 0.
(56)
Before we prove the Jacobi identity, let us use it to establish Poisson’s theorem. Suppose f, g are conserved so that each of them Poisson commutes with the hamiltonian h = H , i.e., { f, h} = {g, h} = 0.
Then the Jacobi identity implies that
{{ f, g}, H} = 0
⇒
d
{ f, g} = 0.
dt
(57)
So the p.b. of two conserved quantities is again a conserved quantity.
• Poisson tensor: To prove the Jacobi identity, it is convenient to introduce a compact notation. Let us
combine the coordinates and momenta into a 2n-component ‘grand’ coordinate on phase space
~ξ = (ξ1 , ξ2 · · · , ξn , · · · , ξ2n ) = (~q, ~p) = (q1 , · · · , qn , p1 , · · · , pn )
Then check the fundamental Poisson bracket relations may be expressed in terms of ξi
!
0 I
{ξi , ξ j } = ri j where r =
.
−I 0
(58)
(59)
Here r is a 2n×2n block matrix with n×n blocks consisting of the identity and zero matrices as indicated.
The constant matrix ri j is sometimes called the Poisson ‘tensor’ of the canonical p.b. relations.
• The p.b. of any pair of observables may now be written in terms of the ‘fundamental’ p.b. between
coordinates and momenta. Show that
! X
n
2n
2n
X
X
∂ f ∂g
∂ f ∂g
∂ f ∂g
{ f, g} =
−
=
{ξ
,
ξ
}
=
ri j ∂i f ∂ j g.
(60)
i
j
∂qi ∂pi ∂pi ∂qi
∂ξi ∂ξ j
i=1
i, j=1
i, j=1
All the properties of the canonical Poisson brackets are encoded in the Poisson tensor. Of particular
importance to us is the anti-symmetry of the Poisson tensor (equivalent to the antisymmetry of the p.b.)
and the constancy of the components ri j .
• Let us now prove the Jacobi identity. We wish to evaluate the cyclic sum
J = {{ f, g}, h} + {{g, h}, f } + {{h, f }, g}.
(61)
∂f
≡ fi . We use the Poisson
Below, we will use subscripts on f, g, h to denote partial differentiation ∂ξ
i
tensor and the Leibnitz rule to write the first term of J as
h
i
(62)
{{ f, g}, h} = fi g j ri j hl rkl = fik g j hl + fi g jk hl ri j rkl
k
Adding its cyclic permutations,
h
i
J = fik g j hl + fi g jk hl + gik h j fl + gi h jk fl + hik f j gl + hi f jk gl ri j rkl .
13
(63)
If J has to vanish for any smooth functions f, g, h on phase space, then the terms involving 2nd derivatives of f must mutually cancel as must those involving 2nd derivatives of g or h. So let us consider
Jf
=
=
=
fik g j hl ri j rkl + f jk gl hi ri j rkl = fik g j hl ri j rkl + fik gl h j r ji rkl
fik g j hl ri j rkl + fik g j hl rli rk j = fik g j hl ri j rkl + fki g j hl rlk ri j
fik g j hl ri j rkl − fik g j hl ri j rkl = 0.
(64)
We relabeled indices of summation i ↔ j, j ↔ l and i ↔ k in the three successive equalities and
2
2
finally used the equality of mixed partial derivatives ∂ξ∂ i ξfk = ∂ξ∂ k ξf i and antisymmetry of the Poisson tensor
rkl = −rlk . Thus we have shown that J f = 0 and by cyclic symmetry, Jg = Jh = 0. Thus J = 0 and the
Jacobi identity has been established. As a corollary we obtain Poisson’s theorem on conservation of p.b.
of conserved quantities.
2.4
Variational principles for Hamilton’s equations
• WeR seek an extremum principle for Hamilton’s equations, just as we had one for Lagrange’s equations:
S = Ldt and δS = 0. Hamilton’s variational principle for his equations is given by the functional of a
path on phase space (qi (t), p j (t))
Z tf
S [q, p] =
pi q̇i − H(q, p) dt.
(65)
ti
Recall that L(q, q̇) = ext p (pq̇ − H(q, p)), which motivates the formula for S [q, p]. We ask that this functional be stationary with respect to small variations in the phase path (q(t), p(t)) while holding δq(ti ) = 0
and δq(t f ) = 0. Note that we do not constrain δp(ti ) or δp(t f ). That would be an over specification3 .
We find upon integrating by parts in the second term,
#
Z tf "
∂H
∂H
δqi −
δpi dt + . . . .
(66)
S [q + δq, p + δp] = S [q, p] +
q̇i δpi − ṗi δqi −
∂qi
∂pi
ti
For δS = 0 to first order, the coefficients of δp and δq must vanish, as these variations are arbitrary.
Thus we recover Hamilton’s equations at all times ti < t < t f :
q̇i =
∂H
∂pi
and
ṗi = −
∂H
.
∂qi
(67)
Hamilton’s equations treat position and momentum on an equal footing except for a sign. But the above
boundary conditions treat them asymmetrically. This is a clue that there is another variational principle
for Hamilton’s equations. Consider the functional of a path on phase space
Z tf
S̃ [q, p] =
−q ṗ − H(q, p) dt
(68)
ti
which we extremize with respect to small variations δq, δp while holding δp(ti ) = δp(t f ) = 0. Then
integrating by parts and denoting partial derivatives by subscripts,
Z tf h
Z tf h
i
i
δS̃ =
− ṗδq − qδ ṗ − Hq δq − H p δp dt =
( ṗ + Hq )δq + (q̇ − H p )δp dt
(69)
ti
ti
So δS̃ = 0 also implies Hamilton’s equations. We will exploit both these variational principles later.
3
There would in general not be any trajectory joining specified values of q and p at both ti and t f . Demonstrate this in the
case of a free particle.
14
2.5
Lagrange’s and Hamilton’s equations take same form in all systems of coordinates on Q
• One of the advantages of having a variational principle is that it allows us to show that the equations
of motion take the same form in any system of coordinates on configuration space Q. Let us first discuss
this for Lagrange’s equations.
• Consider a free particle moving in one dimension along the half line q > 0, with Lagrangian L(q, q̇) =
1
2
2
2 mq̇ and eom q̈ = 0. Now let us switch to a new coordinate on configuration space, Q = q . Then the
Lagrangian may be written in terms of Q and Q̇, obtaining
mQ̇2
1
= L̃(Q, Q̇)
L(q, q̇) = mq̇2 =
2
8Q
(70)
So we have a new Lagrange function L̃ that depends on both Q and Q̇. If Lagrange’s equations have the
∂L̃
same form as before, we must have dtd ∂∂Q̇L̃ = ∂Q
. We check that these eom reduce to the familiar ones for
q by using Q = q2 , Q̇ = 2qq̇ and Q̈ = 2q̇2 + 2qq̈:
d ∂L̃
∂L̃
2Q̈ Q̇2
=
⇒
= 2 ⇒ qq̈ = 0 ⇒ q̈ = 0.
dt ∂Q̇ ∂Q
Q
Q
(71)
Thus we have verified that Lagrange’s equations take the same form in both systems of coordinates q
and Q on configuration space.
2
Q̇
• It may be noted that the equations q̈ = 0 and 2QQ̈ = Q
2 are not of the same form, though they are
equivalent. Rather, what we found is that the eom, when expressed in terms of the respective Lagrange
∂L
d ∂L̃
∂L̃
functions take the same form dtd ∂L
∂q̇ = ∂q and dt ∂Q̇ = ∂Q .
• Let us indicate why this is true in general, using the principle of extremal action. First we make
the simple observation that the condition for a function to be stationary is independent of choice of
coordinates, as long as we make a non-singular change of coordinates. Consider f (x), it is extremal
when f 0 (x) = 0. Now change to a new coordinate X(x) (e.g. X(x) = 2x). Then f (x) = f (x(X)) ≡ f˜(X).
dX
Moreover, by the chain rule, f 0 (x) = f˜0 (X) dX
dx . If the change of variables is non-singular, dx , 0. It
follows that f 0 (x) = 0 ⇔ f˜0 (X) = 0.
• Aside: A change of variables qi → Qi is said to be non-singular if it is a smooth and invertible change
of variables (with smooth inverse). The change is non-singular in some neighbourhood of the point ~q0
i
if the matrix of first partials (Jacobian matrix) Ji j = ∂Q
∂q j is a non-singular matrix (i.e. has non-zero
determinant) at ~q = ~q0 .
R
• Recall that Lagrange’s equations are the conditions for stationarity of the action S [q] = L(q, q̇) dt .
Now suppose we make a non-singular change of coordinates on configuration space q → Q(q). We
indicate the formulae for 1 degree of freedom though the same holds for several degrees of freedom
qi → Qi (~q). Then, as in the free particle example above,
Z
Z
S [q] =
L(q, q̇) dt =
L̃(Q, Q̇) dt ≡ S̃ [Q] where L(q, q̇) = L̃(Q(q), Q̇(q, q̇)).
(72)
As before the conditions for stationarity under infinitesimal variations are the same: δS [q] = 0 iff
δS̃ [Q] = 0. Thus, Lagrange’s equations must take the same form
d ∂L ∂L
=
dt ∂q̇ ∂q
⇔
15
d ∂L̃
∂L̃
=
.
dt ∂Q̇ ∂Q
(73)
• We may apply the same sort of reasoning to argue that hamilton’s equations take the same form in
all systems of coordinates on configuration space. Suppose we change from q → Q. Then we get
a new Lagrange function L̃(Q, Q̇). We also get a new conjugate momentum P = ∂∂Q̇L̃ while p = ∂L
∂q̇ .
In general P will be a function of both q and p and conversely
p = p(Q, P). Hamilton’s equations
R
are the conditions for the action functional S [q, p] = [pq̇ − H(q, p)] dt to be extremal with respect
to infinitesimal variations in the phase path δS = 0. We also have a new hamiltonian H̃(Q, P) =
H(q(Q), p(Q, P)) and new action functional
Z
Z
S̃ [Q, P] = [PQ̇ − H̃(Q, P)] dt with S̃ [Q(q), P(q, p)] = S [q, p] = [pq̇ − H̃(q, p)] dt.
(74)
As before, the conditions δS = 0 imply the conditions δS̃ = 0 so that Hamilton’s equations take the
same form in both systems of coordinates on configuration space.
• Notice that though we changed coordinates on configuration space, we did not change the momenta
in an arbitrary manner. Instead, the change in momenta was induced by the change in coordinates via
the formula P = ∂∂Q̇L̃ where L̃(Q, Q̇) is the Lagrangian expressed in terms of the new coordinates and
velocities. We did not have the freedom to redefine the momenta in an arbitrary manner.
• One may wonder if Hamilton’s equations take the same form in a larger class of coordinate systems on
phase space as opposed to configuration space alone. Remarkably, Hamilton’s equations take the same
form in a slightly larger class of coordinate systems on phase space.
2.6
Canonical transformations
• The noun canon and the adjective canonical refer to something that is standard or conventional.
• Recall that the space of generalised coordinates and momenta is called phase space. Hamilton’s equa∂H
tions q̇ = ∂H
∂p , ṗ = − ∂q may be easier to solve (or understand qualitatively) in some systems of coordinates and momenta compared to others. For instance, there may be more cyclic coordinates in one
particular coordinate system. E.g., consider a particle moving on the plane subject to a central potential
V(r). In this case, the equations of motion are simpler to handle in plane polar coordinates r, θ than in
Cartesian coordinates x, y. From the Lagrangian
!
q
1 2
1 2
2
2
L = m ẋ + ẏ − V
x + y = m ṙ2 + r2 θ̇2 − V(r)
(75)
2
2
θ is a cyclic coordinate and its conjugate momentum Lz = pθ = mr2 θ̇ is conserved. On the other hand,
neither p x nor py is conserved. Of course, here we are using the fact that Hamilton’s equations take the
same form in cartesian as well as plane polar coordinates (a feature we checked earlier):
∂H
∂H
∂H
∂H
, ẏ =
, ṗ x = −
, ṗy = −
∂p x
∂py
∂x
∂y
∂H
∂H
∂H
∂H
ṙ =
, θ̇ =
, ṗr = −
, ṗθ = −
∂pr
∂pθ
∂r
∂θ
ẋ =
∂L
and py =
∂ ẋ
∂L
where pr =
and pθ =
∂ṙ
where p x =
∂L
∂ẏ
∂L
.
∂θ̇
(76)
We say that the transformation from Cartesian coordinates and conjugate momenta (x, y, p x , py ) to polar
coordinates and conjugate momenta (r, θ, pr , pθ ) is a canonical transformation. One may also check
(HW) that the fundamental Poisson brackets among coordinates and momenta are preserved
{x, p x } = {y, py } = 1,
{x, py } = {y, p x } = {x, y} = {p x , py } = 0 · · ·
16
{r, pr } = {θ, pθ } = 1,
{r, pθ } = {θ, pr } = {r, θ} = {pr , pθ } = 0 · · · .
(77)
• Suppose we start with a system of coordinates qi and conjugate momenta pi , in which Hamilton’s
equations take the standard form given above. A canonical transformation (CT) of coordinates and momenta from old ones (qi , pi ) to new ones (Qi , Pi ) is one that preserves the form of Hamilton’s equations.
But not every choice of coordinates and momenta is canonical. For example, we notice that hamilton’s
equations treat coordinates and momenta on a nearly equal footing. So suppose we simply exchange
coordinates and momenta by defining Q = p and P = q. Then the hamiltonian may be written in terms
of the new variables H(q, p) = H(P, Q) ≡ H̃(Q, P). We find that
Q̇ = ṗ = −
∂H
∂H̃
=−
∂q
∂P
and
Ṗ = q̇ =
∂H ∂H̃
=
∂p
∂Q
(78)
So the eom in the new variables do not have the form of Hamilton’s equations, they are off by a sign. So
(q, p) 7→ (p, q) is not a canonical transformation. We may also check that the transformation does not
preserve the fundamental p.b.
{q, p} = 1
while
{Q, P} = {p, q} = −1
(79)
• We have seen in the last section that any change of coordinates alone (‘point transformation’) qi → Qi ,
is automatically canonical. An example of
with the associated ‘induced’ change in momenta Pi = ∂∂L
Q̇i
such a canonical transformation is the one from Cartesian to plane polar coordinates for a particle moving
on a plane. The interesting thing is that there are canonical transformations that are more general than
those resulting from changes of coordinates (point transformations) on configuration space. Perhaps the
simplest such example is Q = p, P = −q for one degree of freedom. Check that Hamilton’s equations
retain their form as do the fundamental Poisson brackets.
• In these two examples of canonical transformations, along with Hamilton’s equations, the fundamental
p.b. among coordinates and momenta were also preserved. This is true in general as we argue below.
2.6.1
Form of Hamilton’s equations are preserved iff Poisson brackets are preserved
• It is worth noting that a transformation is canonical irrespective of what the hamiltonian is. The form
of Hamilton’s equations must be unchanged for any smooth H(q, p). This suggests it should be possible
to state the condition of canonicity without reference to the hamiltonian. We will show now (for one
degree of freedom) that a transformation preserves the form of Hamilton’s equations iff it preserves the
fundamental Poisson brackets.
• Proof: Suppose we make a smooth invertible change of coordinates and momenta q 7→ Q = Q(q, p)
and p 7→ P = P(q, p). The inverse transformation expresses the old coordinates and momenta in terms
of the new ones q = q(Q, P) and p = p(Q, P). The old coordinates satisfy canonical Poisson brackets
{q, p} = 1, {q, q} = {p, p} = 0. Under this change, the old hamiltonian H(q, p) transforms into a new
hamiltonian H̃(Q, P) = H(q(Q, P), p(Q, P)).
• Suppose the above transformation preserves the form of Hamilton’s equations, i.e.
q̇ = H p
and
ṗ = −Hq
and
Q̇ = H̃P
and
Ṗ = −H̃Q
(80)
where subscripts denote partial derivatives. We will show that {Q, P} = 1. The other two p.b. {Q, Q} =
{P, P} vanish by anti-symmetry.
17
• Now by hamilton’s equations and the chain rule, Q̇ = H̃P = Hq qP + H p pP = − ṗqP + q̇pP . We also
know that Q̇ = Qq q̇ + Q p ṗ. We want to equate these two expressions and extract information about the
p.b. {Q, P} = Qq P p − Q p Qq .
• Since it is the partial derivatives of Q, P that appear in the p.b., it is of interest to express qP , pP etc in
terms of partial derivatives of Q, P. How do we do this? The answer is given in the following Lemma.
P
q
• Lemma: qP = −Q p {Q, P} , pP = Qq {Q, P} , pQ = − {Q,P}
, and qQ =
Pp
{Q,P}
.
• Proof: Recall that for an invertible map of one variable, x → X(x) = f (x) with inverse x = x(X) = f −1 (X)
−1
dx
. This arises from differentiating the identity ( f ◦ f −1 )(X) =
the derivatives are reciprocally related dX
dx = dX
−1
0
0
f ( f (X)) = X with respect to X . We get f (x)x (X) = 1 or X 0 (x)x0 (X) = 1 as advertised.
• What is the analogous formula for the map F : (q, p) 7→ (Q, P) ? Here
F : R2 → R2 ,
F(q, p) = (Q, P)
F −1 : R2 → R2 ,
and
F −1 (Q, P) = (q, p).
(81)
~ = (Q, P) and ~x = (q, p) then
Again, F ◦ F −1 is the identity map, i.e., (F ◦ F −1 )(Q, P) = (Q, P) . If we let X
−1 ~
F (X) = ~x or Fi (~x) = Xi . Then
~ =X
~
F(F −1 (X))
⇒
∂Fi ∂x j
= δik .
∂x j ∂Xk
~ = Xi and differentiating
Fi (~x(X))
(82)
Here the matrices of first partials are
 ∂Q
∂Fi  ∂q
=
∂x j  ∂P
∂q
∂Q 

∂p 

∂P 
∂p

and
 ∂q
∂x j  ∂Q
=  ∂p
∂Xk  ∂Q
∂q 
 .
∂P 
∂p 

∂P

(83)
And the above condition says that the matrix product must be the identity
 ∂Q ∂Q   ∂q ∂q 
!

 

1 0
∂Q
∂P 
∂q
∂p 



=
.
 ∂P



∂p 
∂P   ∂p
0 1
∂q
∂p
∂Q
∂P
(84)
Thus we get a system of four equations for the four ‘unknowns’ qQ , qP , pQ , pP
Qq qP = −Q p pP ,
Pq qP + P p pP = 1,
Pq qQ = −P p pQ ,
Qq qQ + Q p pQ = 1.
(85)
These are in fact two pairs of linear equations in two unknowns. They are easily solved for the partials of the old
variables (e.g. by matrix inversion). The solution involves the p.b. {Q, P} = Qq P p − Q p Pq :
qP = −
Qp
,
{Q, P}
pP =
Qq
,
{Q, P}
pQ = −
Pq
,
{Q, P}
qQ =
Pp
.
{Q, P}
(86)
• Armed with this lemma, we return to the fact that the eom retains its form in the new variables:
h
i
ṗQ p + q̇Qq
Q̇
Q̇ = H̃P =
=
.
(87)
{Q, P}
{Q, P}
For this equation to be satisfied we must have {Q, P} = 1. Similarly,
Ṗ = −H̃Q = −Hq qQ − H p pQ = ṗ
i
Pq
Pp
1 h
Ṗ
+ q̇
=
Pq q̇ + P p ṗ =
.
{Q, P}
{Q, P} {Q, P}
{Q, P}
(88)
For this to hold, again we need {Q, P} = 1. Thus we have shown that if Hamilton’s equations retain their
form in the new coordinates, then the transformation also preserves the form of the fundamental p.b.
18
• Let us now prove the converse, namely, if the transformation (q, p) 7→ (Q, P) preserves the form of the
p.b., i.e., if {Q, P} = 1, then it must preserve the form of Hamilton’s equations. So we must show that
∂H̃
∂H̃
∂P = Q̇ and − ∂Q = Ṗ. Let us compute
∂H̃ ∂H
∂H
=
qP +
pP = − ṗqP + q̇pP = ṗQ p + q̇Qq = Q̇.
∂P
∂q
∂p
(89)
In the penultimate equality we used a result from the lemma and the assumption that {Q, P} = 1. So the
first of Hamilton’s equations holds in the new variables! Similarly, we show that the second of Hamilton’s
equations also holds in the new variables
∂H̃ ∂H ∂q ∂H ∂p
=
+
= − ṗqQ + q̇pQ = − ṗP p − q̇Pq = −Ṗ.
∂Q
∂q ∂Q ∂p ∂Q
(90)
So we showed that if the transformation preserves the p.b., then it is canonical.
• A similar result also holds for several degrees of freedom, though we do not discuss it here. So the
new coordinates and momenta are said to be canonical provided
Q̇i =
∂H̃
∂H̃
and Ṗi = −
or equivalently {Qi , P j } = δi j and {Qi , Q j } = {Pi , P j } = 0.
∂Pi
∂Qi
(91)
• Time evolution by any hamiltonian gives us examples of canonical transformations. Recall that
{qi (t), p j (t)} = δi j
and
{qi (t), q j (t)} = {pi (t), p j (t)} = 0
at all times t .
(92)
So the map from (qi (t1 ), pi (t1 )) to (qi (t2 ), pi (t2 )) which is a map from phase space to itself, is canonical
for any times t1 , t2 . So hamiltonian dynamics is in a sense, a tale about canonical transformations.
2.6.2
Canonical transformations for one degree of freedom: Area preserving maps
• To better understand the concept, let us focus on canonical transformations for systems with one
degree of freedom. So we have just one coordinate q and one canonically conjugate momentum p
which together parametrize phase space and satisfy canonical p.b. relations {q, p} = 1. Suppose we
transform to a new pair of coordinates and momenta Q(q, p) and P(q, p). Then what does it mean for
the transformation to be canonical? It means the new variables satisfy the same p.b., i.e.,
1 = {Q, P} =
∂Q ∂P ∂Q ∂P
−
∂q ∂p ∂p ∂q
The quantity that appears above is in fact the determinant of the Jacobian matrix of 1st partials
 ∂Q ∂Q 

 ∂Q ∂P ∂Q ∂P
∂q
∂p 
 =
det J = det  ∂P
−
.
∂P 
∂q ∂p ∂p ∂q
∂q
∂p
(93)
(94)
Both (q, p) and (Q, P) provide coordinate systems on phase space. We recall from calculus that the
Jacobian determinant is the factor relating area elements on phase space
dQ dP = (det J) dq d p
19
(95)
So a canonical transformation for a system with one degree of freedom is simply a transformation that
preserves the signed area element on phase space. Such a transformation is called area preserving.
• Pictorially, what is an area preserving map? The map F : R2 → R2 specified by q → Q(q, p) and
p → P(q, p) maps points on the plane to points on the plane. E.g. it could be the translation map
q → q + 1, p → p + 1 or a rotation. Under such a transformation, any domain D ⊂ R2 will be mapped
0
0
to a new
R R region D
R R = F(D). The map is area preserving if for any D with finite area, Ar( D) = Ar( D )
i.e., D dqd p = D0 dqd p. The above condition guarantees this, since
Ar( D) =
ZZ
dqd p =
D
Z
F(D)
dQdP
=
det J
ZZ
dQdP =
ZZ
F(D)
D0
dqd p = Ar( D0 ).
(96)
In the first equality we changed variables of integration and in the second we used det J = 1 and relabeled
the dummy variables of integration Q → q and P → p.
2.6.3
Generating function for infinitesimal canonical transformations
• The condition for a transformation from canonical coordinates and momenta (qi , pi ) to new ones
(Qi , Pi ) to be canonical is that the Poisson brackets must be preserved. It would be nice to find a more
explicit & convenient characterization of canonical transformations. Let us address this question for
infinitesimal canonical transformations, namely those that depart from the identity transformation by a
small amount. It turns out that any such canonical transformation can be expressed in terms of a single
‘generating’ function on phase space. In other words, we consider transformations of the form
Qi = qi + δqi (q, p)
and
Pi = pi + δpi (q, p)
where δqi , δpi
are small.
(97)
Now we impose the conditions that the new coordinates and momenta satisfy canonical p.b. up to terms
quadratic in δqi and δpi .
0 = {Qi , Q j } = {qi + δqi , q j + δq j } ≈ {qi , δq j } + {δqi , q j } ⇒
0 = {Pi , P j } ≈ {pi , δp j } + {δpi , p j } ⇒
δi j = {Qi , P j } ≈ {qi , p j } + {qi , δp j } + {δqi , p j } ⇒
∂δq j ∂δqi
−
= 0.
∂pi
∂p j
∂δp j ∂δpi
−
= 0.
∂qi
∂q j
∂δp j ∂δqi
+
= 0.
∂pi
∂q j
(98)
Thus we have re-written the conditions for an infinitesimal transformation to be canonical as a system
of homogeneous linear partial differential equations for the small changes in coordinates and momenta
δqi (q, p) and δpi (q, p). These equations look a bit like the condition for the curl of a vector field on the
plane to vanish. If E = (E x , Ey ) is an electric field on a plane, then ∇ × E = (∂ x Ey − ∂y E x )ẑ. Recall
that if the curl of a vector field on the plane vanishes ∇ × E = 0, then we can express the vector field as
the gradient of a scalar function E = −∇φ = (−∂ x φ, −∂y φ). This motivates us to express δqi and δp j as
derivatives of a scalar function f . Check that the above equations are identically satisfied if we put
δqi =
∂f
∂pi
and δpi = −
∂f
∂qi
(99)
for an arbitrary differentiable function f (q, p) on phase space. We say that f is an infinitesimal generator for the above infinitesimal canonical transformation. f is uniquely determined upto an additive
20
constant. f is to be regarded as a small quantity (infinitesimal) here. In the case of R2n phase space, all
infinitesimal canonical transformations may be obtained through appropriate choices of generators f . It
is also possible to build up finite canonical transformations by composing a succession of infinitesimal
ones. We will say more about finite canonical transformations later.
• We already noted that hamiltonian time evolution over any time interval is a canonical transformation
since it preserves the p.b. among the q’s and p’s. The above result allows us to interpret infinitesimal
time evolution as an infinitesimal canonical transformation and identify the corresponding generating
function. Hamilton’s equations for evolution over a small time δt give, to leading order in δt ,
Qi ≡ qi (t + δt) = qi (t) + δt
∂H
∂pi
and
Pi ≡ pi (t + δt) − δt
∂H
.
∂qi
(100)
In this case the small change in coordinates and momenta are
δqi =
∂(H δt)
∂pi
and
δpi = −
∂(H δt)
.
∂qi
(101)
From these equations we read off the infinitesimal generator of time evolution over a small time δt as
f = H(q, p) δt . Notice that f is indeed a small quantity. We often say that the hamiltonian generates
infinitesimal time evolution.
2.6.4
Liouville’s theorem
• We will apply the idea of infinitesimal generator for a canonical transformation to establish an interesting theorem of Liouville on the geometric nature of canonical transformations. Previously, we saw
that for one degree of freedom, canonical transformations preserve areas in phase space. This is a special
case of Liouville’s theorem. For n degrees of freedom, it says that canonical transformations preserve
2n-dimensional ‘volumes’ in phase space. In other words, suppose we are given a region D ⊂ R2n
in phase space and apply a canonical transformation which maps it to a new region D0 ⊂ R2n . Then
Vol(D) = Vol(D0 ). Alternatively, it says that the volume element in phase space is invariant under a
canonical transformation
n
n
Y
Y
dQi dPi =
dqi d pi
(102)
i=1
i=1
For a general transformation, the determinant of the Jacobian matrix of first partials appears as a prefactor on the rhs
 ∂Q ∂Q 
i
 i

∂q j
∂p j 
, where each sub-matix is an n × n block with 1 ≤ i, j ≤ n.
(103)
J =  ∂P

∂Pi 
i
∂q j
∂p j 2n×2n
So Liouville’s theorem says that det J = 1 for a canonical transformation.
• Let us establish Liouville’s theorem for infinitesimal canonical transformations by using our expressions for Qi , P j in terms of an infinitesimal generator4 f
Qi ≈ qi + 4
∂f
∂pi
and
Pi ≈ pi − ∂f
∂qi
is a small parameter which will help us keep track of infinitesimals, we will ignore quantities of order 2 .
21
(104)
Let us first look at the simple case of n = 2 degrees of freedom, where
Q1 ≈ q1 + f p1 , Q2 ≈ q2 + f p2 , P1 ≈ p1 − fq1 and P2 ≈ p2 − fq2
(105)
and sub-scripts denote partial derivatives. In this case the Jacobian matrix

 fq1 p1
 f
J ≈ I +  q1 p2
− fq1 q1
− fq1 q2
fq2 p1
fq2 p2
− fq2 q1
− fq2 q2
f p1 p1
f p1 p2
− f p1 q1
− f p1 q2

f p1 p2 

f p2 p2 
 = I + F
− f p2 q1 

− f p2 q2
(106)
departs from the identity by an infinitesimal matrix of second partials of f . Using the general formula5
det J = det[I + F] = 1 + tr F + O( 2 ) = 1 + O( 2 )
(107)
since F is traceless. So for two degrees of freedom we have shown that an infinitesimal canonical
transformation preserves the (4-dimensional) volume element in phase space.
• The case of n-degrees of freedom is analogous. The 2n × 2n Jacobian matrix is made of n × n blocks


∂2 f
n
n

δi j + ∂2 f
X
X
∂2 f
∂2 f
∂p
∂q
∂p
∂p


i
j
i
j


⇒
det
J
≈
1
+
−
=1
(108)
J = 

2
2

∂pi ∂qi
∂qi ∂pi
− ∂q∂i ∂qf j
δi j − ∂q∂i ∂pf j 
i=1
i=1
Thus, an infinitesimal canonical transformation preserves the volume element in phase space. Synthesising a finite canonical transformation by composing a succession of infinitesimal ones, we argue that
finite canonical transformations also preserve the phase volume.
• Remarks on Liouville’s theorem. Consider the gas molecules in a room, modeled as a system of N
classical point particles. The phase space is 6N dimensional with coordinates ~q1 · · · ~qN , ~p1 · · · ~pN . Now
owing to the difficulty of determining the initial values of these variables, we may at best be able to say
that the initial conditions lie with in a certain region D of phase space. Each of the phase points in D will
evolve in time and trace out a phase trajectory. In this manner D itself will evolve in time to a new region
D0 which contains the possible phase points at a later time. We are often not interested in locations and
momenta of individual gas molecules but average properties of the gas (such as pressure). These may be
obtained by computing an average over the region of phase space D0 . Liouvilles’s theorem says that this
region of phase space evolves in time as an incompressible fluid. In general, the shape of the region will
get distorted with time while maintaining a constant volume. This is much like how an initially localized
drop of ink in water behaves as it diffuses through the volume of water (possibly under stirring). The
volume of ink remains fixed, but the region it occupies gets distorted in shape.
2.6.5
Generating functions for finite canonical transformations from variational principles
• Transformations between different sets of canonical coordinates and momenta are called canonical
transformations. Here we seek to express finite canonical transformations in terms of generating functions. We have already done this for infinitesimal canonical transformations. To do so, we will use
Suppose the eigenvalues of J = I + F are λ1 , · · · , λ2n . Then from the characteristic equation det(J − λI) = det(F − (λ −
1)I) = 0 we see that the eigenvalues of F are λ1 − 1, · · · λ2n − 1 . Hence the eigenvalues of F are f1 = λ1−1 , · · · λ2n−1 . Thus
det J = λ1 · · · λ2n = (1 + f1 ) · · · (1 + f2n ) = 1 + ( f1 + · · · + f2n ) + O( 2 ) = 1 + tr F + O( 2 ) .
5
22
Hamilton’s variational principle for his equations. Consider the (possibly explicitly time-dependent)
map from (qi , p j ) 7→ (Qi , P j ) with the equations of transformation given by the functions
Qi = Qi (q, p, t)
and
Pi = Pi (q, p, t)
(109)
Such a change is canonical provided there is a new Hamiltonian K(Q, P, t) (previously called H̃ ) such
that the eom in the new variables take the same form as those in the old variables, i.e.,
Q̇i =
∂K
∂Pi
and
Ṗi = −
∂K
∂Qi
while q̇i =
∂H
∂pi
and
ṗi = −
∂H
.
∂qi
(110)
When the transformation is not explicitly dependent on time, K(Q, P) was got by expressing q, p in
terms of Q, P in the old Hamiltonian H(q, p). We will see that essentially the same thing continues to
be true, but with a slight modification. Now both these sets of Hamilton equations should be equivalent
in the sense that if we express Q and P in terms of q and p in the second set, they should reduce to the
old Hamilton equations.
• Each set of Hamilton’s equations follows from a variational principle:
Z tf
Z tf h
i
δ
pi q̇i − H(q, p) dt = 0 and δ
Pi Q̇i − K(Q, P) dt = 0.
ti
(111)
ti
The extrema of these two functionals are the same equations (just in different coordinates). One way
for this to happen is for the integrands to be the same. A more general way for this to happen is for the
to differ by the total derivative of a function F1 (q, Q, t) with respect to time. This is because
Rintegrands
tf
Ḟ
dt
=
F1 (q(t f ), Q(t f ), t f )− F1 (q(ti ), Q(ti ), ti ), and q as well as Q are held fixed at ti and t f when the
1
ti
Rt
variations are performed. So δ t f Ḟ1 dt = 0. Note that F1 cannot be taken as a function of p or P since
i
δp(ti ), δp(t f ), δP(ti ), δP(t f ) are unconstrained in Hamilton’s variational principle and the total derivative
of such a term would non-trivially contribute to the equations of motion. In other words, a way in which
the equations in both old and new variables take the hamiltonian form is for the relation
pi q̇i − H = Pi Q̇i − K +
dF1
,
dt
(112)
to hold for some function F1 (q, Q, t). Multiplying through by dt we get
pdq − Hdt = PdQ − Kdt +
dF1
dt.
dt
(113)
That the independent variables in F1 are q, Q, t is also consistent with the fact that the independent differentials appearing in the rest of the terms above are dt, dq, dQ. So as an equation among the independent
differentials dq, dQ, dt we have
pdq − Hdt = PdQ − Kdt +
∂F1
∂F1
∂F1
dq +
dQ +
dt.
∂q
∂Q
∂t
(114)
Comparing coefficients, we read off the relations
p=
∂F1
,
∂q
P=−
∂F1
∂Q
and
K(Q, P) = H(q, p) +
∂F1 (q, Q, t)
.
∂t
(115)
F1 (q, Q) is called the generator of the canonical transformation. The first two equations determine the
equations of transformation. The first may be solved to find Q = Q(q, p, t) and using it, the second
23
may be solved to express P = P(q, p, t). The last relation fixes the new hamiltonian in terms of the
old one and the generator. If F1 does not depend explicitly on time, then it just says that K(Q, P) =
H(q(Q, P), p(Q, P)) = H̃(Q, P) as before. But in general, the new and old hamiltonians differ by the
partial time derivative of the generator.
• Not every function F1 (q, Q) is a legitimate generator. E.g., F1 (q, Q) = q + Q would imply p = 1 and
P = −1 which in general cannot be solved to express Q, P in terms of q, p.
• A choice that does produce an invertible canonical transformation is F1 (q, Q) = qQ. This leads to the
canonical transformation Q = p and P = −q that exchanges coordinates and momenta up to a sign.
• F1 (q, Q, t) generates a finite canonical transformation. It is distinct from the infinitesimal generator
f (q, p) encountered before. However, F1 (q, Q, t) does not generate all finite canonical transformations,
unlike f (q, p) which does generate all infinitesimal canonical transformations. In particular, the identity
transformation Q = q, P = p is not expressible via a generating function F1 (q, Q). The latter expresses
(q,Q)
(q,Q)
p = ∂F1∂q
= p(q, Q) and P = − ∂F1∂Q
= p(q, Q). But for the identity transformation, it is not possible
to express P as a function of Q and q.
• Roughly, F1 is good at generating canonical transformations that are in the vicinity of the one that
exchanges coordinates and momenta upto a sign Q = p, P = −q. It is not a good way of generating
canonical transformations in the vicinity of the identity transformation.
• To find a generator for other canonical transformations, we make use of the second variational principle
S̃ [Q, P] for Hamilton’s equations. Here the momenta are held fixed at the end points δP(ti ) = δP(t f ) = 0.
For the old hamilton equations, we use the first variational principle S [q, p] where δq(ti ) = δq(t f ) = 0:
Z tf
Z tf
δ
[pq̇ − H(q, p)] = 0 and δ
[−QṖ − K(Q, P)] = 0.
(116)
ti
ti
These two variational principles give the same equations even if the integrands differ by the total time
derivative of a function F2 (q, P, t) since δq, δP are held fixed at the end points. So we must have
pdq − Hdt = −QdP − Kdt +
∂F2
∂F2
∂F2
dq +
dP +
dt
∂q
∂P
∂t
(117)
Thus F2 (q, P) generates a canonical transformation, with the equations of transformation given by
p=
∂F2
,
∂q
Q=
∂F2
∂P
and
K+H+
∂F2
.
∂t
(118)
• It is easily seen that if F2 (q, P) = qP, then the resulting transformation is the identity Q = q, p = P.
• The difference between the generating functions F1 (q, Q) and F2 (q, P) lies in the independent variables that they depend on. As we have seen, F1 (q, Q) cannot be used to get the identity transformation and one checks that F2 (q, P) cannot be used to get the exchange transformation Q = p, P = −q.
But there are many CT that may be generated by both a generating function F1 (q, Q) and one of type
F2 (q, P). In these cases, one wonders whether F1 and F2 are related by a Legendre transform, as they
produce the same CT. From the above two relations among differentials,
− QdP + dF2 = PdQ + dF1 ⇒ dF2 (q, P) = d F1 (q, Q) + QP
where
P=−
In other words, up to an additive constant, F2 = QP + F1 with P given as above, or
F2 (q, P, t) = extQ QP + F1 (q, Q, t) .
24
∂F1
.
∂Q
(119)
(120)
• We may obtain two more types of generators F3 (p, Q, t) and F4 (p, P, t) for finite canonical transformations by suitable choices of variational principles for the old and new Hamilton equations.
S̃ [q, p]
& S [Q, P]
=⇒
F3 (p, Q)
while
S̃ [q, p]
& S̃ [Q, P]
=⇒
F4 (p, P)
(121)
As with F2 (q, P), F3 (p, Q) and F4 (p, P) may be obtained via Legendre transforms
F3 (p, Q) = extq [F1 (q, Q) − qp]
and
F4 (p, P) = extQ [QP + F3 (p, Q)].
(122)
• One may wonder whether there are generating functions F5 (q, p) and F6 (Q, P) for finite canonical
transformations. The above variational approach does not lead to such generating functions. In Hamilton’s action principle, both q and p cannot be held fixed at the end points, so the total time derivative
of F5 would non-trivially modify hamilton’s equations and not lead to a canonical transformation in
general. Similarly, a generating function of type F6 (Q, P) is also disallowed in general.
• E.g. if F2 (q, P) = f (q)P then Q = f (q) is a point transformation of coordinates while p = P ∂∂qf . It is
often convenient to work with a generating function of the second type, depending on the old coordinates
and new momenta. In the absence of explicit time dependence, F2 (q, P) is sometimes denoted W(q, P).
The above arguments show that F2 generates a canonical transformation and must therefore preserve
Poisson brackets.
2.7
Action-Angle variables
• Among canonical systems of coordinates and momenta, those which admit more cyclic coordinates
for a given hamiltonian are more suited to solving Hamilton’s equations. The momenta pi = Ii conjugate
to cyclic coordinates are constants of motion, determined by the initial conditions. The best possibility
is if all the coordinates qi = θi are cyclic. Then the hamiltonian is a function of the conjugate momenta
alone H = H(I1 , · · · In ) which are constants of motion I˙i = − ∂H
∂θi = 0. Furthermore, the coordinates are
then linear functions of time since we may easily integrate Hamilton’s equations.
θ̇i =
∂H(~I)
= ωi
∂Ii
which implies
θi (t) = θi (0) + ωi t
and
Ii (t) = Ii (0)
(123)
since ωi are constant in time. We call such a canonical system angle-action variables. The coordinates θi
are called angle variables and their conjugate constant momenta Ii are action variables. This terminology
will be explained below.
2
p
• E.g., the position and momentum of a free particle H = 2m
are angle and action variables since
{q, p} = 1, momentum is conserved p(t) = p(0) and position is linear in time q(t) = q(0) + p(0)t
m .
• Not every hamiltonian system with n degrees of freedom admits action-angle variables. A pre-requisite
is that the system possess n independent conserved quantities that Poisson commute with each-other so
that they can serve as action variables. When they exist, angle-action variables for a given hamiltonian are
not uniquely determined. E.g., we may add a constant to the action or angle variables without affecting
their Poisson brackets nor the conservation of the action variables. We may also rescale the action
variable Ii by constant λi and the angle variable θi by 1/λi while retaining their status as action-angle
variables.
25
2.7.1
Action-angle variables for the harmonic oscillator
q
p2
1 2
k
2m + 2 kq with ω =
m.
−ω2 q, whose solution is
• The harmonic oscillator hamiltonian is H =
and ṗ = −kq imply Newton’s equation q̈ =
q(t) = A sin(ω(t − t0 ))
Hamilton’s equations q̇ = p/m
p(t) = Amω cos(ω(t − t0 )) ≡ B cos(ω(t − t0 )).
and
(124)
q
A > 0 implies clockwise motion in the q- p phase plane along an ellipse with semi-axes A = 2E
k and
√
B = 2mE in the q and p directions. The phase point begins at the the northern-most point of the
ellipse at t = t0 . The constant A is fixed in terms of the energy E = 12 mω2 A2 . So
r
q(t) =
2E
sin θ(t)
k
and
p(t) =
√
2mE cos θ(t)
where
θ(t) = ω(t − t0 ).
(125)
Our aim is to identify canonically conjugate angle and action variables for this system. θ(t) varies
linearly with time, so it is a natural candidate for an angle variable. It is of course the angle subtended
by the phase point with respect to the p axis measured clockwise. This explains the name angle. Taking
a quotient we express the angle variable in terms of q and p:
 √ 
 q mk 
 .
θ = arctan 
(126)
p
• An action variable must be a constant of motion. Since there is only one degree of freedom, there can
be at most one independent constant of motion, which can be taken as energy E . All other constants of
motion must be functions of energy. So we suppose that I = I(E). We try to fix I(E) by requiring that it
be canonically conjugate to θ . Since I depends on q, p only via E , we have
#
"
∂θ ∂I ∂θ ∂I
∂θ 0
∂E ∂θ 0
∂E
p ∂θ
∂θ
0
1 = {θ, I} =
.
(127)
−
=
I (E)
−
I (E)
= I (E)
− kq
∂q ∂p ∂p ∂q ∂q
∂p ∂p
∂q
m ∂q
∂p
Taking the differential of the formula tan θ =
√
q km
p
for θ we get (using cos2 θ = p2 /2Em)
!
√
∂θ ωp
∂θ
ωq
dq q d p
ω
(p dq − q d p) ⇒
− 2 cos2 θ =
=
and
=− .
dθ = km
p
2E
∂q 2E
∂p
2E
p
Thus the condition that angle and action be canonically conjugate becomes an ODE for I(E)
!
ω p2 1 2
1
E
1 = I 0 (E)
+ kq
⇒ I 0 (E) =
⇒ I(E) = + c
E 2m 2
ω
ω
(128)
(129)
I is not uniquely determined, as mentioned earlier. For simplicity we pick c = 0 and arrive at a pair of
angle-action variables for the harmonic oscillator
 √ 
 q km 
E(q, p)
θ(q, p) = arctan 
(130)
 and I(q, p) =
p
ω
The formula E = Iω is a classical precursor to the quantum mechanical formula E = ~ω = hν where h
is Planck’s unit of action. By construction we know that θ(t) = ω(t − t0 ) is linear in time, I is constant
in time and they are canonically conjugate {θ, I} = 1. Moreover, the hamiltonian when expressed in
26
action-angle variables H(θ, I) = ωI is a function of I alone, the angle variable is a cyclic coordinate.
The action variable I has dimensions of action, energy × time. As mentioned before, θ is the angle that
the phase point makes with respect to the p axis clockwise. On the other hand, we notice that the action
variable I(E) is 1/2π times the area enclosed by the
q ellipse traced√ out by the phase point during one
complete oscillation: the ellipse has semi-axes A = 2E
2mE and area πAB:
k and B =
I(E) =
1
Area(E).
2π
(131)
We could have anticipated this relation between I(E) and the area for the harmonic oscillator. It follows
from the fact that a CT on the phase plane preserves areas. The area enclosed by a closed phase trajectory
of energy E may be found in two sets of canonical coordinates and equated. On the one hand,
I
p dq
(132)
Area(E) =
where we could choose a clockwise orientation since the phase trajectory went clockwise with A > 0.
On the other hand, by preservation of areas, we must have
I
p dq =
I
I dθ =
Z
2π
I(E)dθ = 2πI(E)
(133)
0
by constancy of I along the trajectory. Again we choose θ to increase clockwise. Thus I(E) =
Area(E)/2π as we obtained by explicit calculation.
• We have found a canonical transformation to action-angle variables for the harmonic oscillator problem. What is the generator of the finite canonical transformation from q, p to θ, I ? The Hamilton-Jacobi
equation gives us a method to find it.
2.7.2
Generator of canonical transformation to action-angle variables: Hamilton-Jacobi equation
• Consider a dynamical system6 with hamiltonian H(q, p) and canonical variables {q, p} = 1. An
p2
example to keep in mind is H(q, p) = 2m
+ V(q) and more specifically the harmonic oscillator. We
are interested in trajectories with energy E . So q, p must satisfy the condition H(q, p) = E . Now
we seek the generator of a canonical transformation to new co-ordinates and momenta Q = θ, P = I
where all coordinates are cyclic. In other words, the hamiltonian depends only on the new momenta P,
H(q, p) = K(P). The new canonical variables would be action-angle variables: as Q̇ = ∂K
∂P are constant,
∂K
the Q’s evolve linearly in time. Moreover, Ṗ = − ∂Q
= 0, so the new momenta are constants of motion7 .
So we use I, θ in place of P, Q. Let us look for a generating function of the second type, depending on
old coordinates and new momenta F2 (q, I, t). In terms of F2 we have formulae for the new coordinates
and old momenta
∂F2
∂F2
∂F2
and p =
and K = H +
(134)
θ=
∂I
∂q
∂t
6
Our notation will be for one degree of freedom, though some of this generalises to n > 1 degrees of freedom. For a
hamiltonian system with one degree of freedom with conserved energy, one may usually pass to action-angle variables. But
this is not always possible for systems with n > 1 degrees of freedom.
7
Of course, there may not be n independent conserved quantities to serve as action variables. Then it is not possible to find
action-angle variables and we say the system is not integrable. Note also that the action variables must Poisson commute with
each other, so that they form a canonical family with their conjugate angle variables {θi , I j } = δi j , {Ii , I j } = {θi , θ j } = 0 .
27
We indicate below how to proceed with a generator of type F1 (q, θ) if necessary. Furthermore, let us
consider only the case where F2 is not explicitly dependent on time. This should be adequate for the
harmonic oscillator since in that case we found K = H simply by substitution and I, θ depended on time
only via their dependence on q, p, and not explicitly. So we restrict to canonical transformations that are
not explicitly dependent on time. In this case, it is more conventional to denote F2 (q, I) = W(q, I). The
equations of transformation are
θ=
∂W(q, I)
,
∂I
p=
∂W(q, I)
∂q
and
K(I(q, p)) = H(q, p).
(135)
We hope that the trajectories of energy E are simpler to describe in terms of I, θ . Whatever this generating function may be,it must
satisfy the energy constraint H(q, p) = E which is in general a non-linear
first order PDE for W ~q, ~I . For the above example with one degree of freedom, it is
!
∂W
H q,
=E
∂q
1 ∂W
2m ∂q
or
!2
+ V(q) = E.
(136)
This equation is called the time-independent Hamilton-Jacobi equation (HJ) and W is called Hamilton’s
characteristic function, it is the generator of the canonical transformation to action-angle variables.
• In general, W will generate a CT to new canonical variables. The HJ equation does not by itself pick
out the CT to action-angle variables. To impose this additional requirement, we seek solutions W(q, I)
of the HJ equation subject to the condition that the action variable I is a constant of motion. This will
˙
ensure that θ is a cyclic coordinate since ∂K
∂θ = − I = 0 so that θ evolves linearly in time.
• The time derivative of W is
dW(q, I) ∂W dq
=
= pq̇
dt
∂q dt
since
I˙ = 0.
p dx
where
(137)
Thus we must have
W(q(t), I) = W(q(t0 ), I) +
Z
t
p q̇ dt = W(q(t0 ), I) +
t0
Z
q
p(x, E) =
p
2m(E − V(x)).
q0
Any such W
p is a solution of the HJ equation for some E as we check: differentiating the integral,
∂W
=
p
=
2m(E − V(q)) so the time-independent HJ equation is satisfied. These solutions of HJ are
∂q
parametrized by a ‘constant of integration’ W(q(t0 ), I) which can be an arbitrary function of the action
variable, since it drops out of ∂W
∂q . However, W(q(t0 ), I) does affect the dependence of W on I , and
through this, it affects the angle variable θ = ∂W
∂I .
• In particular, we see that there are many sets of action-angle variables for the same system, obtained
by different choices of the constant of integration.
Rq
• The second term q p dq0 is sometimes called an abbreviated action integral8 . It depends in a specific
0
way on both q and I . The dependence on q is via the upper limit of integration, while the dependence
on I is through p. The integrand p is determined by solving H(q, p) = E for p. For the above example,
q
p(q, I) = 2m E(I) − V(q) .
(138)
8
Recall that S =
R tf
ti
(pq̇ − Hdt) was also called the action.
28
For a single degree of freedom, E must be a function of I as there is only one independent conserved
quantity. More generally, E is a function of the action variables E = K(I) = H(q, p) since the hamiltonian in the new variables depends only on the action variables.
• We could also have proceeded using a generating function of type 1. If we use F1 (q, θ) instead of
W(q, I), then the HJ equation still reads H(q, ∂F1∂q(q,θ) ) = E . Now both q and θ depend on time and
Ḟ1 = (F1 )q q̇ + (F1 )θ θ̇ = pq̇ − I θ̇ . The condition that I and θ be action-angle variables is imposed by
requiring that θ̇ be a constant. We could then integrate and find F1 as before, but with the extra term.
2.7.3
Generating function for transformation to action-angle variables for harmonic oscillator
• For the SHO, we already found action-angle variables and we now seek a generating function:
 √ 
 q km 
E
 .
I=
and θ = arctan 
(139)
ω
p
Based on the previous section, a candidate generating function for the canonical transformation to actionangle variables is provided by hamilton’s characteristic function, which is a solution of the HJ equation.
So let us consider the abbreviated action integral
Z q
Z qp
E
k
1
2m(Iω − V(x)) dx where I = , ω2 =
and V(x) = kx2 . (140)
W(q, I) =
p dx =
ω
m
2
0
0
We evaluate the integral after rescaling variables9
√
q k/2Iω
r
 r
 r

2



kq
k
k

W(q, I) = 2I
I q
1−
+ arcsin q
2Iω
2Iω
2Iω 
0
s

 r
!


q
1 2
k
 .
=
2m Iω − kq + I arcsin q
2
2
2Iω 
Z
q
1 − y2 dy =
(141)
Exercise: Verify that this function generates the desired transformation,
p
∂W ?
= p = 2m(E − V)
∂q
and
 r


∂W ?
k 

= θ = arcsin q
.
∂I
2Iω 
(142)
where V = 21 kq2 and E = Iω. So we have found a generating function of the second kind W(q, I) =
F2 (q, I) that allows us to pass to action-angle variables for the harmonic oscillator.
• The generating function takes a simpler form when expressed in terms of the old coordinates and the
new coordinates, i.e., as F1 (q, Q) of the first kind, which is obtained via a Legendre transform
F1 (q, θ) = extI W(q, I) − Iθ
Exercise: Find F1 (q, θ). For this CT we are free to use a generator of either of the first two kinds.
9
Use
R p
1 − y2 dy =
1
2
q
p
k
y 1 − y2 + arcsin y and y = x 2Iω
.
29
(143)
2.7.4
Action-angle variables for systems with one degree of freedom
• Based on our experience with the SHO, we briefly comment on a passage to action-angle variables for
a system with one degree of freedom and conserved hamiltonian, say
H(q, p) =
p2
+ V(q)
2m
(144)
We will suppose that V(q) is such that all the phase space trajectories are bounded and periodic in time
as for V(q) = 12 q2 + q4 . This is guaranteed if V(q) → ∞ sufficiently fast, as |q| → ∞. Further assume
that for a given energy there is a unique periodic trajectory in phase space10 . Then we define the action
variable as the area enclosed by that phase trajectory
I
I p
I(E) =
p(E, q) dq =
2m (E − V(q)) dq.
(145)
This also defines E as a function of I , there is only one independent conserved quantity. As before, we
look for a generating function of the second kind that allows us to transform to action-angle variables.
We do this by solving the HJ equation via an abbreviated action integral with a suitable constant of
integration which we do not indicate explicitly below
Z q
Z qp
2m (E − V(q)) dx
(146)
W(q, I) =
p(E(I), x) dx =
∂W
The angle variable is defined by θ = ∂W
∂I and moreover, p = ∂q . By virtue of having a generating
function, we are guaranteed that {θ, I} = 1, i.e., they are canonically conjugate. Moreover, since the
hamiltonian K(I) is a function of I alone, θ is cyclic and therefore evolves linear in time. Thus we have
a way of identifying action-angle variables for such a system with one degree of freedom.
2.7.5
Time dependent Hamilton-Jacobi equation
• Time evolution is particularly simple in action-angle variables where the hamiltonian is independent
of all coordinates. An even more extreme case of canonical variables is those in which the hamiltonian is
independent of all coordinates as well as momenta, i.e., if the hamiltonian is a constant. By a choice of
zero of energy, this constant may be taken as zero. Now if the hamiltonian in the new variables K = 0,
then the new coordinates and momenta are both constant in time, and are therefore determined by their
initial values: Qi (t) = Qi (0) = βi and Pi (t) = Pi (0) = αi . Time evolution is very simple in such
variables! However, it is not always possible to find canonical variables in which K = 0. But if it is
possible, then the generator of the canonical transformation to such variables must satisfy an interesting
first order PDE called the (time-dependent) Hamilton-Jacobi equation. Let us look for a generating
function of the second type F2 (q, P, t) for the transformation from (q, p, H) → (Q, P, K). For reasons
to be clarified below, it is conventional in this context to denote F2 by S (q, P, t) and call it Hamilton’s
principal function. Pi are the new constant momenta and p = ∂S
∂q . Then S must satisfy
∂S (q, P, t)
K = H(q, p, t) +
=0
∂t
10
or
!
∂S
∂S
H qi ,
,t +
= 0.
∂q j
∂t
(147)
There are potentials such as V = −q2 + q4 for which there may be more than one phase trajectory for a given energy.
30
This is the Hamilton-Jacobi equation (HJ), a first order PDE for the unknown generating function S in
n + 1 variables q1 , · · · , qn , t . We will be interested in the so-called ‘complete integrals/solutions’ of HJ,
which depend on n + 1 constants of integration. These solutions are of the form
S = S (q1 , · · · qn , α1 · · · αn , t)
(148)
We have not indicated the dependence on the n + 1st constant of integration αn+1 . αn+1 may be taken
as an additive constant in S , since only derivatives of S appear in the HJ eqn. We will choose αn+1 = 0
since it does not enter the equations of motion.
• The value of a ‘complete’ solution of the HJ equation is that it provides a way of solving the original
mechanical system, i.e., of expressing qi (t) and p j (t) in terms of their initial values. First, we are free to
take (i.e., define) the new constant momenta to equal the constants of integration, i.e., P j = α j .
• The equations of transformation in terms of the generator S read
pj =
∂S (q, α, t)
∂q j
and
βi = Q i =
∂S (q, α, t)
.
∂α
(149)
The second equation may be used to solve for qi = qi (α j , βk , t). This may be put in the first equation to
find pi = pi (α j , βk , t). Now the solution of the mechanical problem in the sense mentioned above would
be obtained if we express α j , βk in terms of the initial values of the old variables qi (0) and pi (0). To do
this, let us consider these equations at t = 0. We get
p j (0) =
∂S (q, α, 0)
∂q j
evaluated at t = 0
and
βi =
∂S (q(0), α, 0)
∂αi
(150)
We may use the first equation to express αi in terms of qi (0) and pi (0). Then the second equation gives
us βi in terms of q j (0) and pk (0).
• Using the above results, we get
qi (t) = qi (q j (0), pk (0), t) = qi (α, β, t)
and
pi (t) = pi (q j (0), p j (0), t) = pi (α, β, t)
(151)
These give the solution to the mechanical problem since they express the old coordinates and momenta
in terms of their initial values.
• It is not always possible to find a complete solution of the HJ equation. One may only find a solution
S depending on less than n constants of integration. Even this can be used to provide a partial solution
of the original mechanical problem.
• Connection to the time-independent HJ equation by separation of the time variable. If H =
H(q, p) is not explicitly dependent on time, so that the hamiltonian is a constant of motion, then we may
∂S
seek a solution of the HJ equation H(q, ∂S
∂q ) + ∂t = 0 in the form
S (q, α, t) = W(q, α) − Et
(152)
Inserting this in HJ, we find that it is satisfied provided W solves the time-independent HJ equation for
Hamilton’s characteristic function
!
∂W
E = H q,
.
(153)
∂q
• Why do we use the letter S to denote the generating function F2 in the above discussion? An
interpretation of S is got by computing its time derivative.
Z t
∂S
∂S
∂S
Ṡ =
q̇ +
= pi q̇i +
= pq̇ − H ⇒ S (t) − S (t0 ) =
[pq̇ − H]dt0
(154)
∂q
∂t
∂t
t0
31
So S is the action that appears in Hamilton’s variational principle, now regarded as a function of the
final time.
2.7.6
Example: Solving HJ by separation of variables
• As an illustration of the use of the Hamilton-Jacobi equation to solve the initial value problem for a
mechanical system, consider a particle in the vertical constant gravitational field on the Earth. We made
some observations on the nature of trajectories using conservation laws at the beginning of this course.
Now we will solve for trajectories using the Hamilton-Jacobi method. The hamiltonian is
H=
1 2
p x + p2y + p2z + mgz
2m
(155)
We seek a generating function of second kind S (x, y, z, Pi , t) of a CT to new variables Qi = βi and
P j = α j which are both constant in time so that the new hamiltonian K = 0 by a choice of zero of
energy. The HJ equation is

!2
!2
!2 
∂S
∂S
∂S 
1  ∂S
+
+
= 0.
(156)

 + mgz +
2m  ∂x
∂y
∂z 
∂t
Since H does not depend explicitly on time, we separate out the time dependence by writing S (xi , αi ; t) =
W(xi , αi ) − Et . Then W must satisfy the time-independent HJ equation

!2
!2
!2 
1  ∂W
∂W
∂W 
+
+
(157)

 + mgz = E.
2m  ∂x
∂y
∂z 
Now we try to separate variables further by writing W = W1 (x) + W2 (y) + W3 (z)
1 0 2
W1 (x) + W20 (y)2 + W30 (z)2 + mgz = E.
2m
(158)
The RHS E is a constant while the LHS is a sum of three terms that depends on different independent
variables x, y, z. The only way this equation can be satisfied is if each of these terms is a constant, i.e.,
1 0 2
W (x) = α1 ,
2m 1
1 0 2
W (y) = α2 ,
2m 2
and
1 0 2
W (z) + mgz = α3
2m 3
where
E = α1 + α2 + α3 . (159)
So we have reduced the search for a complete solution of the HJ equation to three separate ODEs whose
solutions are (up to additive constants in S which may be ignored)
s
p
p
2
2
W1 (x) = ± 2mα1 x, W2 (y) = ± 2mα2 y, and W3 (z) = ∓
(α3 − mgz)3/2 .
(160)
3 mg2
Thus we have found a complete integral of HJ S = −Et + W1 (x) + W2 (y) + W3 (z) by separation of
variables. The solution indeed depends on three independent constants of integration αi which are taken
∂S
as the new constant momenta Pi . The new constant coordinates are given by Qi = βi = ∂α
or
i
r
β1 = ±
m
x − t,
2α1
r
β2 = ±
m
y − t,
2α2
32
s
and
β3 = ±
2(α3 − mgz)
y − t.
mg2
(161)
Solving these equations for x, y, z we find the trajectory of the particle in terms of the constants αi , βi
r
r
α3 g
2α1
2α2
x = ±(t + β1 )
, y = ±(t + β2 )
, and z =
− (β3 + t)2 .
(162)
m
m
mg 2
We may express the αi , βi in terms of the initial position and momentum by putting t = 0. We see that
x, y are linear in time as expected since there is no force in the x and y directions. The projection of the
trajectory on the x-y plane is a straight line describing free particle rectilinear motion. z is a quadratic
function of time. From this we get the expected parabolic trajectory of a projectile moving under Earth’s
gravity.
33
3
3.1
Rigid body mechanics
Lab and co-rotating frames
• A rigid body in mechanics is a system of particles such that the distances between the particles is
fixed. E.g. four mass points arranged at the vertices of a regular tetrahedron and pairwise connected
by light rigid rods. A continuous distribution of mass is also possible, as in a stone. In this case, the
masses of individual particles is replaced by the mass ρdV in any elemental volume dV where ρ is the
mass density. Examples of rigid bodies include a top, spaceship, gyroscope, tennis ball and plate. Our
approach is based on the treatment of Landau and Lifshitz.
Anisotropic rigid body with 8 particles
Anisotropic rigid body with continuous mass distribution
m1
dV
rho(x)
m2
m4
m3
• For example, a top fixed at a point on the ground in the uniform gravitational field of the earth executes
a motion which involves rotation about its own axis, precession about a vertical axis parallel to the
gravitational field and a wabble about its axis called nutation. The three periodic motions have their own
periods, which if incommensurate prevent the top from returning to its position under time evolution.
Figure 1: Three periodic motions of a top.
• The motion of points in the rigid body may be described using the lab (‘inertial’ or ‘space’ or ‘fixed’)
frame. We will use a system of cartesian co-ordinates X, Y, Z for the lab frame. The centre of mass of
the body is the point R = (X̄, Ȳ, Z̄) whose coordinates are
X̄ =
1 X
1 X
1 X
ma Xa , Ȳ =
ma Ya , Z̄ =
ma Za ,
M
M
M
where
M=
X
ma .
(163)
a
The sum on a is over all the mass points that make up the body. The lab frame coordinates (or coordinates
in ‘space’) of a point in the body change as the body moves.
• We may also use a moving system of cartesian coordinates r1 = x, r2 = y, r3 = z referred to the
so-called co-rotating frame whose axes are rigidly fixed in the body and participate in the motion. The
34
origin O, of the co-rotating frame is most conveniently chosen to lie at the centre of mass of the body:
X
ma ra = 0.
(164)
a
The moving frame coordinates r of a point in the body are independent of time.
• A general (i.e. non-collinear) rigid body has six degrees of freedom. We need three coordinates to
locate the position of the centre of mass, which we denote R. We are then free to rotate the rigid body
about its center of mass in any manner. The orientation of the moving frame relative to the lab frame is
determined by such a rotation of 3D space. A rotation is determined by an axis and an angle of rotation.
We need two angles to define an axis and one angle to specify the amount of rotation. So rotations are a
three parameter family. Thus a general rigid body has three translational and three rotational degrees of
freedom. Moreover, rotations are parametrized by special orthogonal matrices. Thus, the configuration
space of the rigid body is R3 × SO(3).
• In the special case where the rigid body is concentrated along a line (e.g. a pair of mass points or a
thin wire/pen), the number of degrees of freedom is five. Such a rigid body is called a rigid rotator. As
before, a rigid rotator has three translational degrees of freedom which may be regarded as specifying
the location of the center of mass. On the other hand, the orientation of the pen is determined by the
location of the nib, which can lie at any point on the surface of a two dimensional sphere11 (S 2 ), once
the center of mass has been fixed. Thus, the configuration space of a rigid rotator is R3 × S 2 .
3.2
Infinitesimal displacement and angular velocity of rigid body
• We begin with some kinematical aspects and in particular the concept of angular velocity. An infinitesimal displacement of a rigid body in a time dt may be expressed as a sum of an infinitesimal translation
of the center of mass to its final location (keeping the orientation of the body fixed) and an infinitesimal rotation about the center of mass O that orients the moving frame appropriately. Moreover, any
infinitesimal rotation about O is a rotation about some axis passing through O.
z
Z
Omega
v
P
y
r
rho
O
V
R
x
Y
X
Figure 2: Lab and co-rotating frames. ‘rho’ corresponds to
r
r in the text.
• Let us denote by the radius vector of a point P in the rigid body, referred to the lab frame. Suppose
the same point has the (fixed) radius vector r in the moving frame. Then = R + r where R is the radius
vector of the centre of mass in the lab frame. Now, a small displacement δ of P may be written as
r
r
r
δ = δR + δφ × r
11
Note that the 2 -sphere S 2 = {~x ∈ R3 such that ||~x|| = 1} is not a group while SO (3) is a group.
35
(165)
delta r
delta phi
theta
r
r sin theta
Figure 3: Infinitesimal displacement due to a rotation
Here δR is the displacement of the center of mass. δφ × r is the infinitesimal change in the radius vector
r (relative to the lab frame) due to a counter-clockwise rotation about the axis δφ by an angle δφ = |δφ|.
To see why, consult fig. 3. The change in r under an infinitesimal rotation by δφ about the axis δφ is in
magnitude
|δr| = |r| sin θ |δφ|
(166)
δr is perpendicular to both r and δφ. So δr = δφ × r.
• Thus dividing by the time δt in which the infinitesimal motion took place,
r
δ
δR δφ
=
+
×r
δt
δt
δt
(167)
v
r
Now we let δt → 0. If we denote the velocity of P in the lab frame by = ddt , the translational velocity
dφ
of the center of mass by V = dR
dt and the angular velocity by Ω = dt then we have
v=V+Ω×r
or
v=V+v
where v = Ω × r.
(168)
Ω points along the axis of rotation passing through the center of mass (direction of Ω is along axis of
right-handed rotation). Ω may of course change with time both in direction and magnitude.
• To summarize, for each mass point ma in the rigid body we use the following notation (not quite that
of L & L). R is the radius vector of the center of mass and ra the radius vector of the point a in the
co-moving frame (i.e., relative to the center of mass).
r
a
= R + ra ,
and differentiating
v
a
= V + va
where V = Ṙ and va = ṙa = Ω × ra .
(169)
va = ṙa is the velocity relative to the center of mass. Multiplying by the mass ma we get
p
a
= ma V + pa
where pa = ma va = ma Ω × ra .
(170)
The vector sum of all the momenta in the lab frame coincides with the center of mass momentum
X
X
P=
ma ra = MV.
(171)
a = MV + Ω ×
p
a
a
The second sum vanishes as the center of mass lies at the origin of the co-moving frame.
36
3.3
Kinetic energy and inertia tensor or matrix
• The Lagrangian of the rigid body is L = T − U where U is the external potential that the body moves
in. Let us express the kinetic energy of the rigid body in terms of the center of mass velocity and angular
velocity. The kinetic energy is just a sum of free particle KE of each constituent mass point labelled a,
whose velocity we have denoted a in the lab frame. So
v
T=
X1
2
ma
v
2
a
≡
X1
2
v
m
2
(172)
where the sum is over mass points. We will often suppress the index a that labels the points. Note that
this index appears on r, but not on Ω or V, which are properties of the body as a whole. Thus,
v
T
=
=
X1
X
X1
m (V + Ω × r)2 =
mV2 +
mV · Ω × r +
m(Ω × r)2
2
2
2
X
i
1
1X h 2 2
MV2 + (V × Ω) ·
mr +
m Ω r − (Ω · r)2
2
2
X1
(173)
We simplified the second term using the cyclic symmetry of the formula for the volume of a parallelepiped V · Ω × r = r · V × Ω. But the second term vanishes since the center of mass lies at the origin of
P
the moving system: mr = 0. Ω, r are the magnitudes of Ω, r. Thus the kinetic energy may be written
as a sum of the translational kinetic energy of a body of mass M located at the center of mass and the
kinetic energy of rotation about the center of mass
T=
X 1
1
1
1
MV2 + Ωi Ω j
m r2 δi j − ri r j = MV2 + Ii j Ωi Ω j .
2
2
2
2
The rotational kinetic energy involves a 3 × 3 matrix called the inertia matrix/tensor
X $
2
Ii j =
m r δi j − ri r j =
ρ(r) r2 δi j − ri r j d3 r.
(174)
(175)
We have written the formula for a rigid body with continuous mass distribution as well.
• The inertia matrix is real and symmetric matrix Ii j = I ji . It is a positive matrix since the associated
quadratic form (‘rotational kinetic energy’) is manifestly non-negative
T rot =
X1
1
1
Ii j Ωi Ω j = Ωt IΩ =
m(Ω × r)2 ≥ 0.
2
2
2
We may write out the components of the inertia matrix,
 2

2
−mxy
−mxz 
X m(y + z )

 −myx
m(z2 + x2 )
−myz 
I=


−mzx
−mzy
m(x2 + y2 )
(176)
(177)
it is evidently the sum of the inertia matrices of each mass point in the body. The diagonal entries
P
I11 = m(y2 + z2 ) etc are called the moments of inertia about the corresponding axes of the rotating
P
frame. In general, given any axis n̂, the moment of inertia about n̂ is defined as In̂ = a ma ρ2a where ρa
is the distance of point a from the axis.
• Being a real symmetric matrix, the inertia matrix may be diagonalized by an orthogonal transformation
S that rotates the axes of the co-moving frame: S −1 IS = D where D is the diagonal matrix of eigenvalues. As the matrix is positive, the eigenvalues are non-negative, they are called the principal moments of
37
inertia, which we may order as 0 ≤ I1 ≤ I2 ≤ I3 . The eigenvectors of the inertia matrix may be chosen
orthonormal and are called the principal axes of inertia. If the axes of the moving frame are chosen along
the principal axes of inertia, then the inertia matrix is diagonal12
 
P

0
0
 I1 0 0 
 a ma (y2a + z2a )
P
 


2
2
0
0
I = 
(178)
 =  0 I2 0  .
a ma (za + xa ) P

2
2
0
0
0 0 I3
a ma (xa + ya )
P
Note that the off diagonal entries vanish due to cancellations a ma xa ya = 0 even though ma xa ya is,
in general, non-zero for various particles in the body. In the principal axis basis, the rotational kinetic
energy is particularly simple
1
I1 Ω21 + I2 Ω22 + I3 Ω23 .
(179)
T rot =
2
If all three principal moments of inertia are unequal, we call it an anisotropic rigid body. If one pair
coincide, it is called a symmetrical top. For e.g. if I1 = I2 then the corresponding two principal axes
may be chosen to be any pair of mutually perpendicular unit vectors in the corresponding x-y eigenplane.
If all three eigenvalues coincide it is called a spherical top and the principal axes of inertia can be chosen
as any orthonormal frame.
• Note that the inertia matrix is a constant matrix, it does not change with time since the coordinates of
mass points are time-independent in the co-moving frame. The matrix I is an intrinsic property of the
mass distribution of the rigid body, the chosen origin and axes of the co-moving frame.
• If the body is concentrated along a straight line, say the z-axis, then it has no rotational inertia when
spinning about the z-axis. Such a body is called a rigid rotator. Examples include a very thin pencil/dumbell or wire. Argue that the configuration space of such a rigid body is R3 × S 2 . It has only
two rotational degrees of freedom which may be parametrized by points on a two-sphere which specify
the direction in which the body is pointing. Note that the mass distribution need not be uniform along
the z-axis. Since x = y = 0 for all particles, the centre of mass lies on the z-axis and we must have
P
I3 = 0 and I1 = I2 = a ma z2a . Note that two of the triangle inequalities are saturated I1 + I3 = I2 and
I2 + I3 = I1 . The principal axes of inertia point along the z-axis and any pair of mutually orthogonal
directions in the x-y plane.
• Consider a rigid body that is concentrated on a plane (say the x-y plane), like a flat plate or sheet of
cardboard. z = 0 for all points on the body so the centre of mass lies on the x-y plane. It is clear from
the explicit matrix representation of I that it is block diagonal and that the z-axis is one of the principal
axes. The other two lie in the x-y. Let us choose the x and y axes to point along these principal axes of
P
P
P
inertia. The principal moments of inertia are I1 = my2 , I2 = mx2 and I3 = m(x2 + y2 ). Notice
that the triangle inequality is saturated I1 + I2 = I3 , this relation is called the perpendicular axis theorem.
3.4
Angular momentum of a rigid body
• The angular momentum of a system of particles is defined with respect to an origin. If we use the
origin of the lab frame, then the radius vector of point labelled a in the rigid body is a = R + ra where
R is the location of the centre of mass. If a is the lab-frame momentum of the same particle, we must
have a = ma a = ma V + ma Ω × ra . Then the ‘total’ angular momentum about the origin of the lab
frame is
X
X
X
X
Ltot =
R× a+
ma~ra × V +
ma ra × (Ω × ra )
(180)
a× a =
p
p
v
r p
a
12
r
p
a
a
a
The principal moments of inertia satisfy a triangle inequality Ii + I j ≥ Ik where i, j, k are distinct indices from 1, 2, 3 .
38
Furthermore let P =
P
p
be the total momentum (‘centre of mass momentum’) of the body, then
X
X
X
X
P=
ma a =
ma V + Ω ×
ma ra = MV
(181)
a =
a
a
p
v
a
a
a
a
by the definition of center of mass. Thus the total angular momentum is


X
X

Ltot = R × P +  ma ra  × V +
ma ra × (Ω × ra ) = Lcm + Lrot
a
(182)
a
The middle term is zero by the definition of center of mass. We see that the total angular momentum
about the origin of lab frame splits into a centre of mass part Lcm = R × P and a rotational part. We are
primarily interested in the latter. Lrot is in fact the angular momentum about the center of mass (origin
of co-moving frame). It is the angular momentum resulting from motion relative to the center of mass.
P
Lrot = a ra ×pa where, pa is the momentum relative to the center of mass, i.e., pa = ma va = ma (Ω×ra ).
So we define
X
X
X
L ≡ Lrot =
ra × pa =
ma ra × (Ω × ra ) =
ma (ra2 Ω − (Ω · ra ) ra )
a
a
Xa 2
⇒
Li =
m r δi j − ri r j Ω j = Ii j Ω j .
(183)
Thus the ‘rotational’ angular momentum with respect to the centre of mass is related to the angular
velocity via the inertia matrix. This is loosely analogous to how the translational momentum of a point
particle is related to its velocity via the mass p = mv. Momentum always points in the same direction
as velocity. But in general, angular momentum points in the same direction as angular velocity only for
an isotropic rigid body, for which the inertia matrix is a multiple of the identity. For a non-isotropic rigid
body, the angular momentum points in the same direction as angular velocity only if a principal axis of
inertia can be taken to point along the angular velocity vector.
3.5
Equations of motion of a rigid body
• The equations of motion for the rigid body may be formulated as equations for the centre of mass
momentum P and for the angular momentum about the centre of mass L. We will obtain these equations
in the lab frame or any frame inertially related to it, the equations have the same form in all such frames.
Note that the lab frame and the co-rotating frame are not related by galilean transformations, the corotating frame is in general a non-inertial frame. So the eom will take different forms in these two
frames. We will transform the eom to the co-rotating frame in the next section.
ṗ
p
• Let us work in the fixed lab frame. If fa is the force on the ath particle, then a = fa is Newton’s
P
equation. Adding these up for all the particles we get Ṗ = F where P = a a = MV is the total
P
momentum and F = fa is the total force acting on the body. Here we need only include the external
forces acting on the particles since the inter-particle forces balance out and cancel. Ṗ = F is the equation
of motion in the lab frame. By Galilean relativity, it also takes the same form in any frame that is
inertially related (via a translation/rotation/boost or some combination of these) to the lab frame.
• This equation of motion may also be obtained from the Lagrangian13
L=
1
1
MV2 + Ii j Ωi Ω j − U(R, φ)
2
2
13
where V = Ṙ.
(184)
The external potential energy U could depend both on the location R of the center of mass as well as the orientation of
the body, specified, say, by angular variables φ which could specify an axis and an angle that determine a rotation that would
bring the body to a reference orientation.
39
• The corresponding equations of motion for the coordinate R are seen to reproduce what we got above
d ∂L
∂L
∂U
=
⇒ M V̇ ≡ Ṗ = −
= F = force.
dt ∂Ṙ ∂R
∂R
(185)
Here we used the fact that the change in potential energy under a translation of the center of mass by δR
P
P
is δU = a ∂∂Ua · δR since all particles are translated by δR So δU = − a fa · δR = −F · δR.
P
• Next we compute the time derivative of the angular momentum about the centre of mass L = a ra ×pa
where r is the radius vector of the ath particle and p its momentum relative to the center of mass. We
want the equation of motion for L in the lab frame, not the co-rotating frame. But the equation of motion
will take the same form in any frame that is related to the lab frame by a Galilean transformation. So
for convenience, let us work in an inertial frame that at the instant considered is moving with velocity
V with respect to the lab frame and has origin at the center of mass. In this frame, the center of mass
is instantaneously at rest and the mass point a has radius vector ra , velocity va = ṙa = Ω × ra and
momentum pa = mṙa . Moreover, ṗa = fa is the force on this particle
X
(ṙa × pa + ra × ṗa ) .
So
L̇ =
(186)
r
a
The first term vanishes as ṙa × mṙa = 0. In the second term, ra × ṗa = ra × fa = ka is the torque on
P
the ath particle about the centre of mass and so the second term is K = a ka which is the total torque
on the body about the centre of mass. Thus, the equation for evolution of the angular momentum of the
rigid body is L̇ = K. Here both the angular momentum and torque are defined with respect to the centre
of mass of the rigid body and this equation is written in an inertial frame moving at velocity V relative
to the lab frame. Moreover, L = I Ω, so we could also say I Ω̇ = K. In the absence of any external
torque about the centre of mass (e.g. if there are no external forces), the angular momentum of the rigid
body about its CM is independent of time L̇ = 0. This equation of motion takes the same form in the lab
frame, by Galilean relativity.
• The equation for L̇ also follows from the above Lagrangian. We think of the components of angular
~ specifying the orientation of the rigid body,
velocity Ω as the rate of change φ̇ of angular variables φ
for instance, via an axis and an angle to specify a rotation from a reference orientation. We will be more
explicit about these angular variables later on. Then the LHS of Lagrange’s equations for the angular
variables is
d ∂L
d ∂L
=
= I Ω̇.
(187)
˙~ dt ∂Ω
dt ∂φ
∂L
∂U
As for the RHS, let us show that ∂L
∂φ = K. Now ∂φ = − ∂φ is (minus) the change in potential energy due
to an infinitesimal rotation δφ. The change in potential energy δU due to an infinitesimal rotation δφ is
minus the work done by the external forces
δU = −
X
a
fa · δ
r
a
=−
X
fa · (δφ × ra ) = −δφ ·
X
a
a
δU
= −K.
δφ→0 δφ
ra × fa = −δφ · K ⇒ lim
(188)
So the Euler-Lagrange equations imply the law of evolution of angular momentum of the rigid body
d ∂L ∂L
∂U
=
⇒ I Ω̇ ≡ L̇ = −
= K = total torque about CM.
dt ∂φ̇ ∂φ
∂φ
40
(189)
3.6
Force-free motion of rigid bodies
• Now consider a rigid body in the absence of any external forces. If the centre of mass was initially at
rest, it will remain so. More generally, the center of mass will move along a straight line since R̈ = 0.
Let us choose our inertial frame to be such that its origin always lied at the CM, i.e., the inertial frame
moves at constant velocity V relative to the fixed lab frame. So we have ensured that the CM is at rest in
the inertial frame. Of course, the body could rotate while the centre of mass remains at rest. Since there
are no external forces, there are no external torques either, about any point. So the angular momentum
about any point must be conserved. In particular, the angular momentum about the CM must be constant
in time, provided it is measured with respect to an inertial system. It is important to realize that the same
angular momentum, with respect to the co-moving frame is in general not constant in time, since the
frame is rotating relative to the inertial frame. For now let us consider the angular momentum vector and
angular velocity vector relative to the lab frame.
• Let us illustrate some consequences of conservation of angular momentum and the formula L = IΩ,
for force-free motion of simple rigid bodies. For a spherical top the principal moments of inertia are all
equal. I is a multiple of the identity in any basis, We may write L = I1 Ω, so the angular velocity is
just a multiple of the angular momentum, both point in the same direction and are both constant in time.
In particular, force free motion of a spherical top consists of uniform rotation about some axis that is
fixed in the lab frame. The direction of this axis and the rate of rotation |Ω| are determined by initial
2
conditions. The conserved energy E = 21 I1 (Ω21 + Ω22 + Ω23 ) = 12 I1 |Ω|2 = |L|
2I1 is also determined by initial
conditions.
3.6.1
Free motion of rigid rotator
• Next consider a rigid rotator (collinear rigid body). Suppose the third principal axis points along the
axis of the rotator. In this case the principal moments of inertia are I1 = I2 and I3 = 0. Since all the
P
masses lie along the axis of the rotator, in the formula for angular momentum ~L = a ~ra × ~pa , ~ra is along
the axis of the rotator. So ~L must be orthogonal to the axis of the rotator at all times. We could have
reached this conclusion by a different argument as well. From L = IΩ written in the principal axis basis,
we see that the component of angular momentum in the direction of the axis of the rotator must vanish
( L3 = I3 Ω3 = 0) irrespective of what Ω3 is. So the angular momentum must always point in a direction
orthogonal to the axis of the rotator. Since L is a constant vector, the axis of the rotator must always lie
in the plane orthogonal to ~L. In other words, the rotator must rotate in a fixed plane with respect to the
lab frame. Let us for simplicity call this the X -Y plane (in the lab system) so that L points along Z .
~ ∝ Ẑ . It follows that the angular velocity Ω =
• So for an infinitesimal rotation, δφ
Ẑ . So both L and Ω point along Ẑ .
dφ
dt
also points along
• If the z is chosen along the axis of the rod, x = y = 0 for all mass points and so the inertia tensor
becomes diagonal I = diag(I1 , I2 , 0) irrespective of how we choose the x, y in the plane orthogonal to
z. It follow that L1 = I1 Ω1 , L2 = I1 Ω2 and L3 = 0 where 1, 2, 3 refer to components in the co-moving
principal axis frame. It follows that L = I2 Ω so Ω is a constant vector in space, just like L is. Moreover,
the rate at which the rod rotates (axis ‘precesses’ about L) is independent of time and equal to |Ω| = L/I1
where L is the magnitude of the angular momentum vector. So the most general free motion of a rigid
rotator is uniform rotation in a plane fixed with respect to the lab frame and orthogonal to the direction
of angular momentum.
• We may also specify a convenient body fixed frame. z was taken along the axis of the rotator ( I3 = 0).
41
We may take x along the fixed Z axis (direction of L), which is a direction that is fixed both in the body
and in space. Then y must lie in the XY plane in such a way that xyz is a right-handed system.
3.6.2
Free motion of symmetrical top
Figure 4: Fig 46 of symmetric top from Landau & Lifshitz. M = L is vertical and axis is along x̂3 = ẑ.
• Conservation of angular momentum and the relation between angular momentum and angular velocity
allow us to understand some aspects of the free motion of a symmetrical top as well. Following Landau
and Lifshitz (see fig. 4), consider a symmetrical top with 0 < I1 = I2 , I3 , 0. The angular momentum
in space is of course a constant vector L. Let us choose the Z axis of the lab frame along L. The axis
of the top is along the third principal axis x̂3 , i.e. the ẑ-axis of the co-rotating frame. The first two
principal axes of inertia x̂1 , x̂2 may be freely chosen (orthogonal to the z-axis) since the corresponding
eigenvalues are equal I1 = I2 . However, not every such choice will lead to principal axes that are fixed
in the body. However, in this section we do not need to choose a co-moving frame. So we will choose a
convenient set of principal axes with respect to which the inertia tensor is diagonal. This principal axis
frame will not qualify as a co-moving frame.
• Let us choose the second principal axis of inertia x̂2 to be always perpendicular to the plane spanned
by L and the axis of the top. Then the x̂1 principal axis must lie in the same plane as L and the x̂3 in
such a way that the x1 , x2 , x3 axes form a right handed orthonormal system. Since the x2 axis is always
orthogonal to L, it follows that the component of angular momentum along the x2 axis, L2 = 0. This
implies Ω2 = L2 /I2 = 0 as I2 , 0. So Ω always lies in the plane spanned by L and the axis of the top.
It follows that the velocity of any point on the axis of the top, v = Ω × r always points perpendicular to
that plane. So the axis of the top precesses about the direction of angular momentum in space, sweeping
out a cone. Let us denote the angle between the axis of the top and L by the symbol θ . Let us argue
that θ is time independent so that the cone has a fixed opening angle. v for a point on the axis of the
top would have to have a component in the plane containing L and the axis of the top for the angle θ
to change. However, as noted above, v (for points on the axis of the top) always points in a direction
perpendicular to this plane. So θ must be fixed by the initial conditions. We will see below that the rate
of precession is also constant in time. We will also show that the length of Ω is constant in time. In
42
addition to precessing about the Z axis, the top also spins about its own axis. We wish to find the angular
speeds of precession Ωpr and spin Ωspin . Let us find expressions for these two angular velocities in terms
of the angle θ and the constant magnitude of angular momentum |L| = L and the principal moments of
inertia. It helps to expand L and Ω in the principal axis basis. Ω = Ω1 x̂1 + Ω3 x̂3 and L = L1 x̂1 + L3 x̂3
where L1 = L sin θ and L3 = L cos θ where θ is the angle between L and ẑ. Since the components of
L in the principal axis frame are independent of time, L is a constant vector both in the principal axis
frame and in space. Moreover, Ω1 = L sin θ/I1 and Ω3 = L cos θ/I3 are also constant in time. So Ω is a
constant vector in the principal axis frame. But Ω is not a constant vector in space, the plane in which it
lies goes round and round the Z axis as seen in the figure.
• The rate at which the top spins on its axis is the component of Ω along the direction of the axis of the
θ
top, i.e. Ωspin = Ω · x̂3 = Ω3 = LI33 = L cos
I3 . We argued above that θ is independent of time, so the top
spins at a constant rate on its axis. Moreover, a symmetric top must spin on its axis, as long as the axis is
inclined at anything other than right angles to the direction of L.
• On the other hand, the precession rate is obtained by writing Ω as a (non-orthogonal in general!)
~ = Ω(3) x̂3 + Ωpr Ẑ . The first of these (which is not equal to Ω3 x̂3 in
linear combination of ẑ and Ẑ , Ω
general) does not produce any displacement of the axis of the top. The second component gives the
precession rate. To find the precession rate Ωpr we simply take the dot product with x̂1 and use the fact
that x̂1 · x̂3 = 0:
~ · x̂1 = Ωpr Ẑ · x̂1 = Ωpr sin θ ⇒ Ωpr = Ω1 = L1 = L .
Ω1 = Ω
sin θ I1 sin θ I1
(190)
So we see that the precession rate is constant. The axis of the top rotates uniformly about L in space.
Since Ω lies in the same plane as L and the axis of the top, it follows that Ω also precesses about L at
the same rate Ωpr .
3.7
Euler angles
• Euler angles are a way of parametrizing the rotational degrees of freedom of a rigid body. They give
us a way of specifying the orientation of the co-rotating frame x = x1 , y = x2 , z = x3 with respect to the
inertial frame X, Y, Z , both of which are right-handed systems. Alternatively, Euler angles parametrize
points on the rotation group SO(3). Suppose both these orthonormal frames have the same origin O, see
fig. 5. Now, the XY and xy planes intersect along a line ON which is called the line of nodes. This line
is of course orthogonal to both the Ẑ and ẑ axes, and the direction of ON is chosen along Ẑ × ẑ. Now the
orientation of the xyz axes relative to the fixed XYZ axes is specified as follows. θ is the angle between
the Z and z axes. The line of nodes ON on the XY plane is fixed by saying that it makes an angle φ
with the X axis. This fixes the xy plane as it must be perpendicular to z and contain the line of nodes.
Finally, the x axis is fixed by saying it makes an angle ψ with the line of nodes. Note that there are
other conventions for specifying the Euler angles. The Euler angles θ, φ, ψ are generalised coordinates
that fix the angular orientation of a rigid body. The corresponding generalised velocities are their time
derivatives θ̇, φ̇, ψ̇.
• Digression: (A mistake in the order of rotations was made in the lecture!) The body fixed frame x̂, ŷ, ẑ may
be obtained from the lab frame X̂, Ŷ, Ẑ by a sequence of three rotations by the Euler angles about suitably chosen
axes. Suppose the body-fixed frame initially coincides with the lab frame (and has the same origin). Then we first
rotate the body frame by an angle φ counter clockwise about the Z axis. As a result, the rotated x̂ will now point
along the intended line of nodes while ẑ continues to coincide with Ẑ . Next we rotate the body frame by an angle
43
Figure 5: Euler angles and their time derivatives, from Landau and Lifshitz, Mechanics (fig. 47).
θ counter clockwise about the new x̂ axis (line of nodes). As a result of this, the new ẑ = x̂3 axis has reached its
desired orientation. Finally, we rotate the body frame by an angle ψ counter-clockwise about the new ẑ axis. As a
result, x̂ moves off the XY plane and makes an angle ψ with the line of nodes, and reaches its desired orientation.
~ = ~θ˙ + φ
~˙ + ψ
~˙ . Its components with respect to the co-rotating
• Now consider the angular velocity vector Ω
frame xyz are denoted Ω1 , Ω2 , Ω3 . We wish to express the components of angular velocity in terms of
the generalised velocities θ̇, φ̇, ψ̇. First, from their definitions and the figure we see that a small change in
˙
θ holding φ, ψ fixed is a rotation about the line of nodes. So we say that ~θ points along the line of nodes
˙
ˆ Similarly, φ
~˙ points along Ẑ so φ
~˙ = φ̇Ẑ and ψ
~˙ = ψ̇ẑ. Now let us denote by θ̇1 , θ̇2 , θ̇3 the
ON : ~θ = θ̇ ON.
components of the vector ~θ̇ in the three directions of the co-rotating xyz frame and similarly for φ̇i and
˙
ψ̇i ; i.e., θ1 = ~θ · x̂ etc. Then by some trigonometry we determine the components.
θ̇1 = θ̇ cos ψ, θ̇2 = −θ̇ sin ψ, and θ̇3 = 0;
ψ̇1 = ψ̇2 = 0, and ψ̇3 = ψ̇;
φ̇1 = φ̇ sin θ sin ψ, φ̇2 = φ̇ sin θ cos ψ, φ̇3 = φ̇ cos θ
(191)
~ ∝ Ẑ onto the
In the last equation, we used the fact that Ẑ is ⊥ to the line of nodes. So the projection of φ̇
xy plane must also be perpendicular to the line of nodes. It follows that this projection makes an angle
~ + ψ̇)
~ · x̂ = θ̇ + ψ̇ + φ̇ etc:
~ · x̂ = (~θ̇ + φ̇
ψ with the y axis. Combing these, we get Ω1 = Ω
1
1
1
Ω1 = θ̇ cos ψ + φ̇ sin θ sin ψ,
Ω2 = −θ̇ sin ψ + φ̇ sin θ cos ψ,
Ω3 = ψ̇ + φ̇ cos θ.
(192)
We may also express the generalized velocities in terms of the angular velocity components
Ω1 sin ψ + Ω2 cos ψ
Ω1 sin ψ + Ω2 cos ψ
, ψ̇ = Ω3 −
.
(193)
sin θ
tan θ
Now, if x, y, z are taken to point along the principal axes of inertia, the rotational kinetic energy of the
P
rigid body is T = 21 j I j Ω2j where I j are the principal moments of inertia. We may substitute for Ω1,2,3
from above to express the kinetic energy in terms of the Euler angles and their time derivatives:
I1 2
θ̇ cos2 ψ + φ̇2 sin2 θ sin2 ψ + 2θ̇φ̇ sin θ sin ψ cos ψ
T =
2
θ̇ = Ω1 cos ψ − Ω2 sin ψ,
φ̇ =
44
+
I3 I2 2 2
φ̇ sin θ cos2 ψ + θ̇2 sin2 ψ − 2θ̇φ̇ sin θ cos ψ sin ψ +
ψ̇2 + φ̇2 cos2 θ + 2ψ̇φ̇ cos θ (194)
.
2
2
• For a symmetric top, with I1 = I2 , I3 this simplifies
T=
I1 2 I1 I3 I1 2
θ̇ + φ̇2 sin2 θ +
ψ̇ + φ̇ cos θ =
θ̇2 + φ̇2 sin2 θ +
ψ̇2 + φ̇2 cos2 θ + 2ψ̇φ̇ cos θ . (195)
2
2
2
2
We could also obtain this formula by a judicious choice of co-rotating frame. Let us choose the z axis
along the third principal axis of inertia corresponding to I3 . For a symmetrical top I1 = I2 . Now we use
the freedom of choosing the first two principal axes as any pair of orthogonal vectors perpendicular to z.
Let us choose (at the instant considered), x to point along the line of nodes ON . Then ψ = 0. So
Ω1 = θ̇,
Ω2 = φ̇ sin θ,
Ω3 = ψ̇ + φ̇ cos θ.
(196)
Thus we recover the above expression T rot for a symmetric top. Let us now use these formulae in terms
of Euler angles to find the rates of precession and spin of a freely rotating symmetrical top. Suppose the
constant angular momentum vector points along the Z -axis of the lab frame. Let us choose the z axis of
the co-rotating frame to point along the axis of the top. Then the rate at which the top spins on its axis is
the component of Ω in the z direction, i.e., Ω3 . And the rate at which the axis of the top ( ẑ) precesses
about the angular momentum vector is equal to the rate at which the line of nodes goes round the Ẑ axis,
namely φ̇. We wish to express these rates in terms of L and θ . To introduce L we first note that the
components of angular momentum along the co-rotating axes are
L1 = I1 Ω1 = I1 θ̇, L2 = I2 Ω2 = I2 φ̇ sin θ and L3 = I3 Ω3 = I3 ψ̇ + φ̇ cos θ
(197)
On the other hand, since L points along Ẑ , its components along the principal axes must be
L1 = L · x̂ = 0,
L2 = L · ŷ = L sin θ
and
L3 = L · ẑ = L cos θ.
(198)
L1 vanishes since x has been chosen along the line of nodes, which is perpendicular to the Z axis.
Comparing the two formulae, we express the relations between angular velocities and angular momenta
as relations between the time derivatives of Euler angles and the magnitude of angular momentum
!
1
1
θ̇ = 0,
I2 φ̇ = L
and
I3 ψ̇ + φ̇ cos θ = L cos θ
or
ψ̇ = L cos θ
−
. (199)
I3 I1
These relations are enough to give us an expression for the rate at which the top spins on its axis
rate of spin = Ω · ẑ = Ω3 =
L3
L
= cos θ.
I3
I3
(200)
We also find the rate of precession, which turns out to be a constant:
precession rate = Ω pr = φ̇ =
L2
L sin θ
L
=
= .
I2 sin θ I2 sin θ I1
(201)
These formulae agree with what we obtained by other means in the last section. For instance, 0 = L1 =
I1 θ̇ so θ̇ = 0. So the axis of the symmetric top remains at a constant angle relative to ~L.
45
3.8
Euler equations for a rigid body in body-fixed frame
• As we have seen above, the eom of a rigid body may be formulated as equations for the centre of mass
momentum P = MV and for the angular momentum L = I Ω about the centre of mass.
dP
= F and
dt
dL
=K
dt
(202)
P
P
where F = a fa and K = a ra × fa are the external force and external torque about the centre of mass.
Here, the rates of change of both P and L are measured with respect to the fixed inertial lab frame. Now
we wish to write these equations for the time evolution of P and L in the co-rotating non-inertial frame.
The rotation of the frame imparts a time dependence even to a vector that may be fixed in the lab frame,
so we should expect the equations to look a bit different. In particular, the angular momentum which is
fixed in space is in general not a constant vector with respect to the co-moving frame.
• As we have seen in the case of the force free motion of a symmetrical top, even if the angular momentum is constant in the lab frame, the angular velocity vector (in space) need not be constant in time,
indeed the top could precess about the axis defined by the constant angular momentum vector.
• The relation between angular momentum and angular velocity is simplest in the principal axis frame
where the inertia matrix is diagonal. However, the principal axis frame is a co-rotating frame. So to
exploit its simplicity, we must write the above equations for time evolution in the body-fixed frame.
• Let A be a vector such as angular momentum or linear momentum of the rigid body or of a mass
point. We wish to relate its time dependence with respect to the lab frame to that in a frame rotating with
instantaneous angular velocity Ω. If the vector is fixed in the rotating frame ( Ȧrot = 0) then its time
dependence in the lab frame arises purely from the rotation and is given by Ȧlab = Ω × A. The reason is
the same as the one we gave in deriving the second term of the equation v = V + Ω × r at the beginning
of our study of rigid bodies. More generally, the vector A may be changing with respect to the rotating
frame as well. Combining these two,
!
!
dA
dA
=
+Ω×A
(203)
dt lab
dt rot
Thus in the body-fixed frame we have a system of six equations:
Ṗ + Ω × P = F and
L̇ + Ω × L = K where L = IΩ.
(204)
The equations for evolution of angular momentum in the body fixed frame are (the last two equations are
got by cyclic permutation of indices)
L̇1 + (Ω2 L3 − Ω3 L2 ) = K1 ,
L̇2 + (Ω3 L1 − Ω1 L3 ) = K2 ,
L̇3 + (Ω1 L2 − Ω2 L1 ) = K3 .
(205)
• If we choose the axes of the co-rotating frame to point along the principal axes of inertia, then the
equations for angular momentum take their simplest form since Li = Ii Ωi for each i = 1, 2, 3. The
resulting equations were derived by Euler and bear his name. Assuming none of the principal moments
of inertia vanish and defining ai j = Ii−1 − I −1
j ,
!
1
1
−
L2 L3 = K1 ,
L̇1 +
I2 I3
or L̇1 + a23 L2 L3 = K1 ,
!
!
1
1
1
1
L̇2 +
−
L3 L1 = K2 and L̇3 +
−
L1 L2 = K3
I3 I1
I1 I2
L̇2 + a31 L3 L1 = K2 , and L̇3 + a12 L1 L2 = K3 .
(206)
46
It is also of interest to find the time evolution of the components of angular velocity with respect to the
body fixed principal axes. Rather than try to extend the above formula to the case A = Ω, we simply
write Li = Ii Ωi in the Euler equations and obtain
!
!
!
I3 − I2
I1 − I3
I2 − I1
Ω̇1 +
Ω2 Ω3 = K1 , Ω̇2 +
Ω3 Ω1 = K2 and Ω̇3 +
Ω1 Ω2 = K3 .
(207)
I1
I2
I3
• The Euler equations could be written as second order ODEs for the Euler angles by substituting for Ωi
in terms of θ, φ, ψ using our formulae from the previous section. But we could also regard them as first
order equations specifying the evolution of the components of the angular momentum vector. Note that
P
the torque on the RHS of the Euler equations, K = a ra × fa depends on the instantaneous location and
orientation of the body (which depends for instance on R which must be determined from Ṙ = P by
solving the momentum equations Ṗ + Ω × P = F. Thus the Euler equations in general are a complicated
system that couple the rotational and translational motion. However, they simplify in the absence of
external forces, in which case the equations for L decouple from those for P. So for force free motion,
the three Euler equations for L1 , L2 , L3 are a self-contained system of quadratically non-linear ordinary
differential equations for the rotational dynamics in the principal axis frame.
3.8.1
Euler equations for force-free motion of symmetric top
• Let us consider the Euler equations in the case of the free motion of a symmetric top, for which
0 < I1 = I2 . So the axis of the top is the third principal axis, i.e., the z-axis of the co-moving frame.
Euler’s equations become
L̇1 + aL2 L3 = 0,
L̇2 − aL3 L1 = 0
and
L̇3 = 0
where
a=
1
1
1
1
− = − .
I2 I3 I1 I3
(208)
So L3 is a constant while L1 , L2 obey the coupled equations
L̇1 + ωL2 = 0
and
L̇2 − ωL1 = 0
where
ω = aL3 = L3
!
1
1
−
.
I1 I3
(209)
The general solutions depend on a multiplicative constant C and an additive phase δ
L1 = C cos (ωt + δ)
and
L2 = C sin (ωt + δ)
(210)
The motion of the angular momentum vector with respect to a frame fixed in the top is periodic in time.
The component of angular momentum along the axis of the symmetric top is fixed wile the component
orthogonal to it rotates at an angular speed ω. Since L1 = I1 Ω1 , L2 = I1 Ω2 , L3 = I3 Ω3 , the same holds
for the angular velocity vector, it rotates about the axis of the top at the constant rate ω. In fact
Ω3 =
C
L3
= constant, while Ω1 = cos (ωt + δ)
I3
I1
and Ω2 =
C
sin (ωt + δ)
I1
(211)
In particular, Ω3 is time-independent as is L3 . Seen from the body fixed frame, the angular velocity
~ precesses about the axis of the top at the angular rate ω, sweeping out a cone of opening angle
vector Ω
arctan(C/I1 Ω3 ). Similarly, the angular momentum vector ~L precesses about the axis of the top at the
angular frequency ω and sweeps out a cone of opening angle arctan(C/L3 ).
• Earlier we found that L3 = L cos θ or Ω3 = IL3 cos θ , where the Euler angle θ is the angle between
the (constant) angular momentum vector in space and the instantaneous direction of the axis of the top.
47
So we deduce that θ = arctan(C/L3 ) is independent of time. The axis of the top precesses at a constant
angle around the fixed direction of angular momentum in the lab frame. If the angular velocity vector
pointed along the axis of the top (i.e., if C = 0), then we would say that the top simply spun on its axis.
For C , 0, the above solution describes the force-free motion of a top that is spinning on its own axis
and precessing about an axis fixed in space.
• Furthermore, the square of the length of the angular velocity vector is constant in time. To see this, we
compute it in the co-moving frame
Ω2 = Ω21 + Ω22 + Ω23 =
C 2 C 2 L32
+ 2 + 2 = constant.
I12
I2
I3
(212)
The magnitude of Ω is of course the same in the co-rotating and lab frames, since rotations do not alter
lengths of vectors.
• Let us interpret the angular frequency ω. From the above solution, L3 is constant while L1 =
C cos (ωt + ϕ) and L2 = C sin (ωt + ϕ). So ω is the rate at which the L vector precesses about the
z axis of the co-moving frame. Above we found
!
!
1
1
1
1
ω = L3
−
−
= L cos θ
= −ψ̇.
(213)
I1 I3
I1 I3
The last equality follows from what we found in the section on Euler angles. So −ψ̇ is the rate at which
the L vector precesses around the z axis of the co-rotating frame, which is also evident from fig. 5.
• The limiting case of a rigid rotator may be obtained by letting I3 → 0 holding I1 fixed. In other words,
we first let I2 → I1 and then let I3 → 0 so that the symmetric top becomes infinitesimally thin. The
θ
only way for ω to remain finite is for θ → π2 in such a way that cos
I3 approaches a finite limit. Thus
we find that the axis of a rigid rotator must always be perpendicular to the angular momentum in space.
Since the latter is a fixed vector L ∝ Ẑ ∝ Ω, the axis of the rotator sweeps out a disc in the XY plane.
Ω3
θ
What is more, the limiting value of cos θ/I3 must be zero. This is because cos
I3 = L , and Ω3 = 0 since
x3 is along the axis of the rotator, which is always in a plane orthogonal to Ω. It follows that in this
limit, ω → 0 so that Ω1 , Ω2 are both constants. We may choose x̂ along Ẑ and ŷ in the plane swept
out by the axis of the rotator which points along ẑ. This corresponds to the choice of phase δ = 0. Then
Ω1 = C/I1 = Ω, C = L and Ω2 = Ω3 = 0.
3.9
Ellipsoid of inertia and qualitative description of free motion of rigid body
• The rotational kinetic energy T = 21 Ii j Ωi Ω j takes a simple form in the principal axis basis
T=
L2
L12
L2
+ 2 + 3.
2I1 2I2 2I3
(214)
• For torque free motion, we use Euler’s equations to show that the rotational kinetic energy and square
of angular momentum L2 = L12 + L22 + L32 are constant in time.
!
L1 L̇1 L2 L̇2 L3 L̇3
a23 a31 a12
Ḣ =
+
+
= −L1 L2 L3
+
+
= 0.
(215)
I1
I2
I3
I1
I2
I3
Similarly, we show
1 dL2
= L1 L̇1 +L2 L̇2 +L3 L̇3 = −L1 (a23 L2 L3 )−L2 (a31 L3 L1 )−L3 (a12 L1 L2 ) = −L1 L2 L3 (a23 +a31 +a12 ) = 0.
2 dt
48
Figure 6: Trajectories of Euler equations on an energy level surface (from Landau & Lifshitz)
• Thus we have two conserved quantities. We may use them to visualise the trajectories in the phase
space of angular momenta. The phase space is R3 whose coordinates are L1 , L2 , L3 . The conservation
of total angular momentum implies that the trajectory must lie on the angular momentum sphere L12 +
L22 + L32 = L2 . The radius of the sphere is determined by the initial magnitude of the angular momentum
vector. In addition, conservation of energy implies that the trajectory must lie on an ellipsoid of constant
√
P L2
energy H = i 2Iii = E . The three semi-axes of this ellipsoid of inertia are 2EIi for i = 1, 2, 3. E is
of course determined by the initial conditions. Without loss of generality let us assume that I1 ≤ I2 ≤ I3
so that I1−1 ≥ I2−1 ≥ I3−1 . Then we see that the energy satisfies the following inequalities
L2
L2
≤E≤
.
2I3
2I1
(216)
p
p
2EI1 ≤ L ≤ 2EI3 .
(217)
As a consequence
These inequalities imply that the energy ellipsoid and angular momentum sphere always have non-empty
intersection. In other words, the radius of the angular momentum sphere lies between the smallest and
largest semi-axes of the ellipsoid of inertia. The two quadratic surfaces generically intersect along a
(union of) closed curves or points. Each such intersection set is a union of possible trajectories of the
angular momentum vector, for given total angular momentum and energy.
• Let us get a qualitative picture of the types of curves traced out by the tip of the angular momentum
vector. There are six stationary points corresponding to rotation about the principal axes of inertia:
p
2
L1,2,3 = ± 2EI1,2,3 with others vanishing and with energy E = 2IL1,2,3 . Check that these are solutions of
the Euler equations.
• More generally, suppose we keep the energy E fixed and imagine varying the magnitude
√ of total
angular momentum L. When L2 = 2EI1 , there are two points of intersection, at L1 = ± 2EI1 , L2 =
L3 = 0. So the angular momentum vector is static in the co-moving frame. It always points along (or
opposite to) the principal
q axis ( x-axis) corresponding to the smallest principal moment of inertia I1 .
2E
It follows that Ω1 =
I1 , Ω2 = Ω3 = 0 so the angular velocity is constant and always along the x
principal axis of the body-fixed frame. The locus of the angular velocity vector is called a polhode curve.
Here the polhodes are a pair of points. We will see now that this motion is stable in the sense that a small
change in energy/angular momentum results in a trajectory that always remains close to this one.
• As the angular momentum is increased slightly, the angular momentum sphere intersects the inertia
ellipsoid in a pair of small closed curves encircling the x axis. So the static solutions of the previous
paragraph (rotation of the body about the ±x-axis) are stable. The present solutions correspond to a timedependent L in the body fixed frame, whose terminus precesses around the ±x-principal axis. Thus, the
49
L vector relative to the body is periodic in time, it sweeps out a conical surface (not right circular). The
angular momentum vector relative to space is of course constant.
• As L2 → 2EI2 , the curves of intersection become larger and at L2 = 2EI2 , are a√pair of large ellipses
that intersect at the points where the inertia ellipsoid meets the y axis. L2 = ± 2EI2 , L1 = L3 = 0
are static solutions corresponding to rotation about the ±y-principal axis corresponding to the middle
principal moment of inertia. However, rotation about the middle principal √
axis is unstable, since a small
change in E or L2 results in a trajectory that isn’t always close to L2 = ± 2EI2 .
• As L2 goes from 2EI2 to 2EI3 , the curves of intersection shrink to a pair
√ of curves around the z-axis
and eventually end at the pair of static solutions L1 = L2 = 0, L3 = ± 2EI3 corresponding to stable
rotation about the ±z-axis.
• Almost all trajectories are closed curves, so the motion on phase space of angular momenta is periodic.
• A rapidly spun tennis racquet (or chalkboard duster) displays the above stability while rotating about
its principal axes corresponding to its largest and smallest principal moments of inertia. The above
instability manifests itself when we try to spin the racquet about its middle principal axis. Of course, a
spinning tennis racquet is not free but subject to the gravitational force of the earth. But if the kinetic
energy of rotation is large, we would expect to be able to ignore the effects of gravity on the qualitative
rotational behavior discussed above.
4
4.1
Dynamics of continuous media
Lagrangian and Eulerian descriptions
• Continuum dynamics deals with the classical dynamics of continuous media such as gases, water (fluids), elastic rods/beams, stretched strings etc. All of these are made of a very large number of molecules
and we will treat them as continuous mass distributions with an infinite number of degrees of freedom.
• There are two principal formalisms for treating mechanics of continuous media, the so-called Lagrangian and Eulerian descriptions. The former is closer to our treatment of systems of particles, we
follow the motion of each molecule or fluid element/bit of string. For example, if a fluid element occupied the location ~r0 at t = 0, then we seek the trajectory ~r(~r0 , t) of this fluid element, which should be
determined by Lagrange’s equations (ironically, this treatment was originally attempted by Euler). The
Lagrangian description is particularly useful if we have some way of keeping track of which material
element is where. This is usually not possible in a flowing liquid or gas, but is possible in a vibrating
string since the bits of string are ordered and may be labeled by their location along the string or by
their horizontal coordinate for small vertical vibrations of a string that does not ‘bend over’. Due to
the difficulty of following the motion of individual fluid elements, Euler then developed the so-called
Eulerian description, which attempts to understand the dynamics of quantities such as density ρ(~r, t),
pressure p(~r, t) and velocity ~v(~r, t) in a fluid at the observation point ~r at time t . We will begin with the
Lagrangian description of the vibration of a stretched string.
50
4.2
4.2.1
Vibrations of a stretched string
Wave equation for transverse vibrations of a stretched string
• Perhaps the simplest physically interesting mechanical system with a continuously infinite number of
degrees of freedom is a vibrating homogeneous stretched string. We will consider the special case where,
in equilibrium, the string is stretched between two clamps located at x = 0 and x = L. We ignore the
effects of gravity since the tensional forces in the string often dominate. We shall call the direction in
which the string is stretched the horizontal direction. The string is free to move only in one direction
(vertical) transverse to the direction in which the string is stretched. We assume the string has a mass per
unit length of ρ. The instantaneous configuration of the string is specified by giving the height u(x, t)
of the string above the horizontal position x at time t . Since the string is stretched, there are tension
forces that act on any segment of the string, tangentially at either end of the segment, tending to elongate
the segment. We assume the tension in the string is a constant τ. When the string is horizontal, the
tensions at either end of any segment are horizontal, equal and opposite in direction so that the string is
in equilibrium. At the end points, the tension is balanced by the force applied by the clamps.
• Note that the length of the string is not fixed, it can stretch to a length more than L, for instance
when it is plucked as in a Veena. When the string is displaced from equilibrium by small vertical displacements, tensional forces on the ends of a small segment are not necessarily horizontal. But to a
good approximation, the horizontal components of tension are equal and opposite, ensuring that there
is no longitudinal/horizontal movement of the string. Moreover, the vertical components of tension are
in general unequal and result in a vertical acceleration of the segment. We estimate this. Consider a
small segment of string between horizontal locations x and x + dx with corresponding heights u(x) and
u(x + dx) ≈ u + du. We suppose that the tangent to the string at any point x makes a counter-clockwise
angle θ(x) with respect to the horizontal. Draw a diagram! Then since we assume the inclination angles
are small,
θ(x)2
∂u
cos θ(x) ≈ 1 −
≈ 1 and sin θ(x) ≈ tan θ(x) ≈
≡ u0 (x).
(218)
2
∂x
Then the horizontal components of tension at the right and left ends of the segment are τ cos θ(x + dx)
and −τ cos θ(x) which by assumption are equal and opposite. The vertical components of tension at the
right and left ends of the segment are
τ sin θ(x + dx)ẑ ≈ τu0 (x + dx)ẑ and
− τ sin θ(x)ẑ ≈ −τu0 (x)ẑ.
(219)
Thus the net upward force on the segment is
Fup = τu0 (x + dx) − τu0 (x) ≈
∂τu0 (x)
∂2 u
dx ≈ τ 2 dx
∂x
∂x
(220)
So Newton’s second law for the segment, whose mass is ρ dx is
Fup = τ
∂2 u
∂2 u
dx
=
ρ
dx
.
∂x2
∂t2
Thus the equation of motion for small transverse (1D) vibrations of the stretched string is
s
r
∂2 u
1 ∂2 u
τ
tension
the ‘wave equation’:
= 2 2 where c =
=
.
2
ρ
mass per unit length
∂x
c ∂t
51
(221)
(222)
c has dimensions of speed and will be seen to be the speed at which waves propagate on the string.
• We are interested in solving the initial-boundary value problem for the string. The wave equation is
second order in time and requires two initial conditions (say at t = 0), just like Newton’s equation. These
are the initial height u(x, t = 0) and the initial velocity of the string u̇(x, t = 0). The boundary conditions
corresponding to a string clamped at the end points are u(x = 0, t) = u(x = L, t) = 0. Other boundary
conditions are also of interest. For example, we might have an end (say at x = 0) of the string free to
move up and down (though not horizontally), so that the slope of the string vanishes at the end point.
This leads to the free/open boundary condition ∂u
∂x = 0 at x = 0. We could also consider an infinite string
with say, u(x, t) → 0 as |x| → ∞.
• As a consequence of considering small vibrations and small angles θ the equation of motion is linear,
however, it is a partial differential equation unlike Newton’s ordinary differential equations encountered
in the mechanics of finitely many particles. Above, u(x) is the analogue of the generalised coordinate and
x labels the particles in the string. The configuration space is the set of possible instantaneous locations of
the string segments, i.e. the space of twice differentiable functions u(x) on the interval [0, L] that vanish
at the end-points. This is an infinite dimensional space reflecting the fact that a string has infinitely many
degrees of freedom. The equations of continuum mechanics (e.g. fluid mechanics, electrodynamics) are
typically systems of partial differential equations and the wave equation is a prototype. We may regard a
partial differential equation such as the wave equation as a large (infinite) system of ODEs, one ODE for
each value of x.
4.2.2
Separation of variables and normal modes of a vibrating string
• On account of the linearity of the wave equation, the superposition principle applies. Linear combinations of solutions are again solutions. The solution space forms a linear vector space. This suggests that
if we can find a basis of linearly independent solutions (called normal modes of oscillation), we will be
able to express the general solution as a linear combination of the normal modes of oscillation.
• The wave equation utt = c2 u xx is a partial differential equation for an unknown height function u
dependent on two independent variables t, x. Let us look for solutions which are a product of a function
of t alone and a function of x alone: u(x, t) = X(x)T (t). We hope that solutions of this separable type
form a basis for the space of all solutions of interest. We also hope that X and T will be determined by
simpler ODEs compared to the PDE for u. Indeed, we find, wherever the quotients make sense,
X(x)T̈ (t) = c2 T (t)X 00 (x)
⇒
T̈ (t)
X 00 (x)
= c2
= −ω2 .
T (t)
X(x)
(223)
Now LHS is a function of t alone while RHS is a function of x alone. Thus, both must equal the
same constant which we called −ω2 . We anticipate that the constant must be negative for physically
interesting vibrational motion. This is because −ω2 = üu is the ratio of the acceleration of the string
element to its displacement from the mean position. As in Hooke’s law, this quotient must be negative
for a restoring force. Alternatively, the second derivative operator ∂2x is a negative operator on functions
X(x) that vanish at the end points of the interval. Thus, our PDE has reduced to a pair of ODEs
ω
is the angular wave number.
(224)
T̈ (t) = −ω2 T (t) and X 00 (x) = −k2 X(x) where k =
c
These ODEs are in fact eigenvalue problems. The first is the same as Newton’s equation for the harmonic
oscillator and the second is essentially the same, so we can write their general solutions as
T (t) = A cos ωt + B sin ωt
and
52
X(x) = C cos kx + D sin kx.
(225)
The clamping of end points of the string (Dirichlet boundary conditions) implies X(0) = X(L) = 0 so we
must have C = 0 and sin kL = 0. So kL = nπ where n is an integer, it suffices to take n ≥ 1 since the
negative values give (linearly dependent) solutions that only differ by a sign and n = 0 gives the trivial
solution. So we may write any separable solution of the wave equation as
u(x, t) = (A cos ωn t + B sin ωn t) sin
nπx
L
where ωn =
nπc
L
for some
n = 1, 2, 3, . . .
(226)
Each of these solutions for n = 1, 2, 3, . . . is called a normal mode of oscillation. The mode n = 1 is
called the fundamental or first harmonic n = 2 the second harmonic or first overtone etc. A normal mode
of oscillation has a definite angular wave number kn and spatial wave length λn = 2π/kn = 2L/n. It
has a definite angular frequency ωn = ckn and also a definite time period of oscillation T n = 2π/omn .
νn = ω/2π is the frequency at which every point along the string vibrates about its mean position. As
opposed to a normal mode, a more general motion of a stretched string will not have such a definite
wave length and time period, indeed it need not even be periodic in time! Moreover, these normal modes
do not necessarily satisfy the prescribed initial conditions. But since the wave equation is linear, we
can take linear combinations of normal modes to produce new solutions. The most general such linear
combination is a Fourier series
u(x, t) =
∞
X
(An cos ωn t + Bn sin ωn t) sin
n=1
nπx
L
(227)
The coefficients An , Bn must decay sufficiently fast as n → ∞ to ensure that the sum converges. The
theorems of Fourier series tell us that we can represent any continuous function that vanishes at the end
points of the interval [0, L] as a Fourier sine series. So we may try to fit the initial conditions by a suitable
choice of constants An , Bn for 1 ≤ n ≤ ∞. They are fixed by the initial height and velocity of the string
u(x, 0) =
∞
X
n=1
An sin
nπx
L
and
u̇(x, 0) =
∞
X
Bn ωn sin
n=1
nπx
L
where ωn =
nπc
.
L
(228)
Using the orthogonality of sin(nπx/L) on the interval [0, L] for n = 1, 2, 3, . . . and the fact that the
average value of the square of the sine function is a half, we find
Z
Z l
nπx nπx 2 l
2
An =
u(x, 0) sin
dx and Bn =
u̇(x, 0) sin
dx.
(229)
L 0
L
nπc 0
L
Thus we have solved the initial-boundary value problem for the motion of a stretched string clamped at
the end points. It is instructive to plot a movie of the time evolution of one such solution on a computer.
• We see that a general vibration of a stretched string involves a superposition of several normal modes
and does not possess a definite wave number or time period. However, we will see that in general, higher
harmonics cost more energy to excite. We might anticipate this since the restoring force was found to
be proportional u00 (x). Higher harmonics sin(nπx/L), for n 1 are rapidly oscillating functions with
large second derivatives, so they involve significant forces on the string segments. We would expect
much energy to be stored in the oscillatory motion of a higher harmonic.
4.2.3
Right and left-moving waves and d’Alembert’s solution
• The height u(x, t) (measured relative to the equilibrium height) of a stretched string executing small
transverse vibrations must satisfy the wave equation u = ( c12 ∂2t − ∂2x )u = 0. In other words, it must
53
be annihilated by the wave operator or d’Alembertian . d’Alembert’s approach to solving the wave
equation arises from factorizing the wave operator into a pair of first order operators. Let us consider
the wave equation on an infinite interval −∞ < x < ∞ subject to the initial height and initial velocity
u(x, t = 0) = h(x)
and u̇(x, 0) = v(x).
The wave equation may be factorized as
c−2 ∂2t − ∂2x u = c−1 ∂t − ∂ x c−1 ∂t + ∂ x u = c−1 ∂t + ∂ x c−1 ∂t − ∂ x u = 0
(230)
(231)
It follows that if u is annihilated by either ∂− = c−1 ∂t − ∂ x or ∂+ = c−1 ∂t + ∂ x , then it will satisfy the
wave equation. Let us consider these first order equations. We notice that any differentiable function
u(x, t) = f (x − ct) satisfies (c−1 ∂t + ∂ x )u = 0 while any differentiable function u(x, t) = g(x + ct) is
annihilated by c−1 ∂t − ∂ x . Thus, for any differentiable functions f and g,
u(x, t) = f (x − ct) + g(x + ct)
(232)
is a solution of the wave equation. A little thought shows that for c > 0, f (x − ct) is a right-moving wave
with speed c and initial profile (at t = 0) given by the function f (x). The shape of the wave f (x − ct)
is unaltered as it travels to the right. So f (x − ct) is called a right-moving wave. Similarly, for c > 0,
g(x + ct) is a left-moving wave. Thus we have found that any superposition of a right- and left-moving
wave is a solution of the wave equation.
• One wonders whether such superpositions of right and left moving waves are adequate to solve the
initial value problem for a stretched string. We will see that this is indeed the case on an infinite domain.
To solve the IVP, we wish to fix f and g in terms of the initial data.
1
u(x, 0) = f (x) + g(x) = h(x) and u̇(x, 0) = −c f 0 (x) + cg0 (x) = v(x) or − f 0 (x) + g0 (x) = v(x).
c
Integrating the latter equation with integration constant K we get
f (x) + g(x) = h(x)
and
− f (x) + g(x) =
1
c
Z
x
v(ξ) dξ + K.
(233)
x0
Adding and subtracting we solve for f, g in terms of initial data
!
!
Z
Z
1 x
1
1 x
1
h(x) −
v(ξ) dξ − K
and g(x) =
h(x) +
v(ξ) dξ + K
f (x) =
2
c x0
2
c x0
(234)
Adding these, we express the solution of the wave equation entirely in terms of initial height and velocity
"
#
Z
1
1 x+ct
u(x, t) =
h(x − ct) + h(x + ct) +
v(ξ) dξ .
(235)
2
c x−ct
It is instructive to plot a movie of this solution, for instance in the case of zero initial velocity and a
2
simple initial height profile such as h(x) = e−x /2 . One finds two little waves moving away from x = 0.
The height at xo at time to depends on the initial (t = 0) height at points xo − cto and xo + cto . So the
initial height only at points a distance cto from the observation point xo can affect the height at the point
of observation. This indicates that these ‘signals’ travel at the speed c 14 . The initial velocity v(x) only
at points within a distance cto from the observation point can affect the height at the observation point.
14
Note that the speed at which these signals travel is quite distinct from the instantaneous velocity u̇ of a point on the string.
54
4.2.4
Conserved energy of small oscillations of a stretched string
• Since we have not incorporated any dissipative effects and are not supplying any energy to the string at
any time t > 0, we expect the energy of the vibrating string to the conserved. Let us derive an expression
for the conserved energy in the same way as we did for Newton’s equation. Recall that we multiplied
∂V
mq̈i + ∂q
= 0 by the integrating factor q̇i and summed over the degrees of freedom i. The resulting
i
expression was the statement that the time derivative of energy is zero.
• So let us begin with Newton’s equation for a string in its pristine form and multiply by ut
ρutt dx = τu xx dx
⇒
ρut utt dx − τut u xx dx = 0
⇒
1 2
ρ(u )t dx − τut u xx dx = 0.
2 t
Now we sum over the degrees of freedom by integrating over x ∈ [a, b]
Z b
Z b
1 2
ρut dx −
τut u xx = 0.
∂t
a
a 2
(236)
(237)
The first term is the time derivative of what looks like a kinetic energy by analogy with a point particle
Z b
mX 2
1
(238)
q̇i →
ρ u2t .
2 i
a 2
So we would like to express the second term as the time derivative of a potential energy. To do so we
first integrate by parts
Z b
Z b
1 2
b
∂t
ρut dx − [τut u x ]a +
τu x utx dx = 0
(239)
a 2
a
The boundary term vanishes if we use Dirichlet or free boundary conditions (u = 0 or u x = 0 at x = a, b)
or even periodic boundary conditions (u(a, t) = u(b, t), u x (a, t) = u x (b, t)) and we get
Z b
Z b 2
ut
1 2
dx + ∂t
(240)
(u x ) dx = 0.
∂t
2
2
a
a 2c
Thus the conserved energy is a sum of kinetic and potential energies (check the dimensions!)
#
Z b"
1 2 1 2
dE
E =T +V =
ρut + τu x dx ⇒
= 0.
2
2
dt
a
(241)
The KE is proportional to the sum of squares of speeds of the bits of string as expected. The PE is an
energy of ‘bending’, it is proportional to the square of the gradient (slope) of the string profile. T and
V are separately non-negative and so E ≥ 0 with equality Rwhen the string is in equilibrium u(x, t) =
constant. The integrand is called the energy density E = E(x, t)dx. In general, the energy density
‘moves around the string’ in such a way that the total energy is conserved. We also see that for fixed
A, B, higher (n 1) normal modes of oscillation un = [A sin(nπct/L) + B cos(nπct/L)] sin(nπx/L) store
more energy.
4.3
4.3.1
Fluid dynamics
Material derivative
• In the Eulerian description of a fluid we are interested in the time development of various fluid variables
like velocity, pressure, density and temperature at a given location (point of observation) ~r = (x, y, z) in
55
the container. This is reasonable if we are interested in predicting the weather changes over CMI over
r)
the course of time. The change in say, density, at a fixed location is ∂ρ(~
∂t . However, different fluid
particles will arrive at the point ~r as time elapses. It is also of interest to know how the corresponding
dynamical variables evolve, not at a fixed location but for a fixed small fluid element, as in a Lagrangian
description15 . This is especially important since the dynamical laws of mechanics apply directly to the
fluid particles, not to the point of observation. So we may ask how a variable changes along the flow,
so that the observer is always attached to a fixed fluid element. For instance, the change in density of a
fixed fluid element in a small time dt as it moves from location ~r to location ~r + d~r is
dρ = ρ(~r + d~r, t + dt) − ρ(~r, t) ≈ d~r · ∇ρ +
∂ρ
dt
∂t
(242)
r
We divide by dt and take the limit dt → 0 and observe that ~v = d~
dt is the velocity of the fluid. Thus the
instantaneous rate of change of density of a fluid element that is located at ~r at time t is
Dρ dρ ∂ρ
≡
=
+ ~v · ∇ρ = ∂t + v x ∂ x + vy ∂y + vz ∂z ρ
Dt
dt
∂t
(243)
= dtd ≡ ∂t∂ + ~v · ∇ is called the material (also total, substantial, convective) derivative. It can be used
to express the rate of change of a physical quantity (velocity, pressure, temperature etc.) associated to a
fixed fluid element, i.e., along the flow specified by the velocity field ~v . This formula for the material
derivative bears a close resemblance to the rigid body formula
relating the time derivatives of a vector
dA
dA
relative to the lab and co-rotating frames: dt
= dt
+ Ω × A. A quantity f (could be a
lab
co−mov
scalar or a vector) is said to be conserved along the flow or dragged by the flow if its material derivative
vanishes DDtf = 0.
D
Dt
D
is a first order partial differential operator, Leibnitz’s product rule of differentiation holds
• Since Dt
Dfg
Dg
Df
~ , we check that
Dt = f Dt + Dt g for scalar functions f, g. Similarly for a scalar f and vector field w
the Leibnitz rule holds
~) Df
D( f w
D~
w
~+f
=
w
.
(244)
Dt
Dt
Dt
4.3.2
Incompressibility of a fluid flow
• Let us briefly introduce the concept of compressibility/incompressibility of a fluid flow. The concept
refers to a flow of a fluid, not the fluid alone. We define a flow to be incompressible if the volume occupied by any fixed fluid element (not necessarily small) remains constant in time. This is approximately
true for water flowing in a hose pipe. To quantify this, we ask how the volume V of a region V occupied
dV
by a given set of fluid molecules changes with time, we seek an expression for dV
dt . Here dt does not
refer to the material derivative, but just the instantaneous time rate of change of volume occupied by
the fluid element. Suppose the region V is bounded by a surface S with outward area element dS and
outward unit normal n̂ such that dS = n̂ dS . In a small time dt the region containing the fluid changes
by a movement of its bounding surface16 . At a point ~r on the boundary, the surface element dS moves
15
By a small fluid element, we mean a collection of molecules that is sufficiently numerous so that concepts such as ‘volume
occupied by the element’ make sense and yet small by macroscopic standards so that the velocity, density e.t.c., are roughly constant over its extent. For instance, we may divide a container with 1023 molecules into 10000 fluid elements, each containing
1019 molecules.
16
Here we neglect the infinitesimal change in the area dS of the surface element due to the flow. The change in volume due
to such a change is of second order in infinitesimals.
56
out by a distance ~v · n̂ dt where ~v is the fluid velocity at the point ~r . Thus the change in volume dV(dS )
due to the area element dS moving out a bit is ~v · n̂ dt dS . To include the contributions of all surface
area elements, we integrate over the entire bounding surface,
Z
Z
dV
~v · n̂ dS =
rate of change of volume of fixed fluid element =
=
∇ · v d3 r.
(245)
dt
S
V
The last equality uses the divergence theorem of vector calculus, to transform the surface integral into a
volume integral. Since this is true for any volume V of fluid (above molecular sizes), let us specialize to
a small fluid element (so that ∇ · v is roughly constant over its extent) at location ~r having volume δV .
Then
dδV
D δV
=
≈ (∇ · v) (δV)
dt
Dt
1 dV
d log V
= lim
.
V→0 V dt
V→0
dt
or ∇ · v = lim
(246)
So the divergence of the velocity field is the fractional rate of change of volume of a small fluid element.
• A flow is incompressible if each fluid element maintains its volume during the flow, i.e., dV
dt = 0 for all
V (above molecular scales). It follows that a flow is incompressible iff the velocity field is divergencefree ∇ · v = 0.
• An intuitive and simple example of an incompressible flow is one where the density of the fluid is
the same everywhere and at all times. We will shortly see why this is true. A more general example
is one where the density of a given fluid element is constant in time, though different fluid elements
may have different densities. This happens for horizontal flows in the atmosphere, where the density
is stratified by height though the flow is horizontal. Note that the same fluid (eg. air) under different
conditions may exhibit incompressible and compressible flows. The study of compressible flows is
usually termed aerodynamics while the study of incompressible flows is often termed hydrodynamics.
This is because typical air flows are compressible while flow of water under ordinary conditions is usually
incompressible. To relate the above divergence-free condition to the density of the fluid, we need the
continuity equation, which we turn to next.
4.3.3
Conservation of mass: continuity equation
• The total mass of fluid particles in a given fluid element remains constant in time, since the same
molecules occupy the identified element. Consider a small fluid element of volume δV in the vicinity of
the point ~r where the fluid density is ρ(~r). Then the mass of the fluid element is δm = ρδV . The material
derivative of δm must vanish and we compute it using the Leibnitz rule and result of the previous section:
D δm D(ρ δV) Dρ
D δV Dρ
0=
=
=
δV + ρ
=
+ ρ∇ · v δV.
(247)
Dt
Dt
Dt
Dt
Dt
Thus we have the continuity equation expressing conservation of mass
Dρ
+ ρ ∇ · v = 0.
Dt
(248)
We immediately see that if the density is constant along the flow (i.e. Dρ
Dt = 0), then the flow is divergence
free (∇ · v = 0) and incompressible. Expanding the substantial derivative, we get
∂ρ
+ v · ∇ρ + ρ ∇ · v = 0
∂t
If ρ = ρ0 is constant in both time and space, then the flow must be incompressible.
57
(249)
• If the flow is incompressible, i.e., ∇ · v = 0, then the density is constant along the flow,
∂ρ
∂t + (v · ∇)ρ = 0. We say the density is advected or transported by an incompressible flow.
Dρ
Dt
=
• Another way to write the continuity equation is in local conservation law form
∂ρ
+ ∇ · (ρv) = 0.
∂t
(250)
Here ~j = ρ~v is called the mass current density of the fluid. The continuity equation says that the rate of
change of density at a point is balanced by the divergence of the current density. We may also write this
in integral form, by integrating over a volume V fixed in space and applying the divergence theorem
Z
Z
Z
Z
∂ρ 3
∂
3
3
ρd r+
ρv · dS~ = 0.
(251)
d r+
∇ · (ρv) d r = 0 or
∂t V
S =∂V
V ∂t
V
The first term is the rate of increase of mass inside a fixed volume V (not a volume that moves with the
flow). The second term gives the outward flux of mass across the bounding surface S . So mass is neither
created nor destroyed.
4.3.4
Euler equation for inviscid flow
• We consider flow of an ideal fluid. This means we ignore viscous (dissipative) forces and consider
only forces due to the pressure exerted by various parts of the fluid on each other, and any external forces,
such as that due to gravity.
• The stresses (forces per unit area) acting on the surface that surrounds a fluid element can be decomposed into two components: those that act perpendicular to the surface and tend to compress/expand the
element and those that act tangential to the surface, tending to shear the element. We assume there aren’t
any shearing stresses (zero shear viscosity) and that the compressional stresses are entirely due to the
hydrostatic pressure (zero bulk viscosity).
• Now consider a small fluid element of mass δm = ρ δV occupying a volume δV and having instantaneous velocity v. Let us write Newton’s second law for this fluid element. Its acceleration is Dv
Dt and
so by Newton’s second law, the force acting on it must equal ρ δV Dv
.
We
consider
two
sorts
of
forces
Dt
acting on the fluid element. There can be an external force such as gravity (called body force) acting on
the fluid. Its value is fδV where f is the body force per unit volume. In addition, we have the force due to
pressure exerted on the fluid element by the fluid surrounding the element. To calculate this, assume the
fluid element is a cuboid with sides dx, dy, dz. The net pressure force in the x̂ direction is the product of
the area dydz and pressure differential between the two faces: δF x = − ∂p
∂x dx × dydz. The − sign arises
because if p is greater on the right face of the element compared to the left face, then the net force would
be leftward. Thus the total pressure force on the fluid element is δF = −(∇p)δV . Thus, Newton’s 2nd
law for the rate of change of momentum of the fluid element reads
ρ δV
Dv
= −(∇p) δV + f δV
Dt
(252)
Dividing through by δV we get Euler’s celebrated equation of motion for an inviscid fluid. It must be
considered in conjunction with the continuity equation
∂v
1
f
+ v · ∇v = − ∇p +
∂t
ρ
ρ
58
and
Dρ
+ ρ∇ · v = 0.
Dt
(253)
However, these are still only four equations for five unknown functions (density, pressure and three
components of the velocity field). This system must be supplemented by some condition that determines
the pressure. This is often supplied by an equation of state that expresses the pressure as a function of
density. The equation
of state depends on the specific fluid that is considered. E.g. for an isentropic fluid
γ
flow, PP0 = ρρ0 where γ = cP /cV is the adiabatic index.
• We also need initial conditions for velocity, density and pressure fields (respecting the equation of
state). In addition, boundary conditions must be specified. For an inviscid fluid, the boundary condition
simply says that the fluid is confined to the container. If the container is at rest, the component of the
velocity field normal to the bounding surface must vanish.
• More generally, if dissipative (viscous) effects are included, the Euler equation is modified leading to
the Navier-Stokes equation
∂v
1
f
+ v · ∇v = − ∇p + + ν∇2 v.
(254)
∂t
ρ
ρ
∂
∂
∂
Here ∇2 = ∂x
2 + ∂y2 + ∂z2 is the Laplace operator. ν is called the kinematic viscosity, it has dimensions
of L2 /T . The study of fluid flow including the Navier-Stokes and Euler equations is a major branch of
engineering, physics and mathematics. Unlike the linear wave equation for a string, these equations are
non-linear PDEs due to the v·∇v advection term from the material derivative, leading to many interesting
phenomena and complications.
2
4.3.5
2
2
Flow visualization: Streamlines, streak-lines and path-lines
• If the velocity vector field at every point of observation is independent of time, we say the flow is
steady, v(r, t) = v(r). To aid in the visualization of a flow we define the concepts of stream-lines, streaklines and path-lines. All these concepts coincide for a steady flow. For steady flow, they are the field
lines of the steady velocity field. The field lines are the integral curves of the velocity vector field, i.e.,
curves that are everywhere tangent to the velocity vector field. They are the trajectories of test particles
moving in the steady flow, i.e., solutions of the ODEs and boundary conditions
dr
= v(r(s)),
ds
and r(so ) = ro .
(255)
Here s is the parameter along the integral curve, it is the time that parametrizes the trajectory of the test
particle moving in the steady flow. If we write these in component-form with r(s) = (x(s), y(s), z(s)) and
v(r) = (v x (r), vy (r), vz (r)), then the ODEs for field lines become
dx
= vx ,
ds
dy
= vy
ds
and
dz
= vz
ds
or
dx dy dz
=
=
= ds
vx
vy
vz
(256)
• More generally, consider a possibly non-steady flow. Streamlines at time to are defined as the field
lines of the velocity field at that instant of time v(r, to ). The streamline through any point of observation
P at a given time to is tangent to the velocity vector v(r(P), to ). Since the flow may not be steady,
the streamlines will in general change with time. Streamlines of the velocity field are analogous to the
field lines of a (generally time-dependent) electric or magnetic field. In particular, for an incompressible
(divergence-free ∇ · v = 0) flow, streamlines cannot begin or end abruptly just as magnetic field lines
close on themselves. At a given instant of time, stream-lines cannot intersect. At a given instant of time,
streamlines provide information on the current flow pattern. For example, draw the stream lines of the air
velocity vector field over the Indian peninsula at the onset of the south-west monsoon on June 1, 2012.
59
These streamlines changed with time and almost reversed themselves during the ‘receeding’ north-east
monsoon in November 2012.
• Path-lines are the trajectories of individual fluid particles. For example, if we introduced a small
glowing speck of dust into the fluid and took a movie of its trajectory, we would get its path line. At any
point P along a path-line, it is tangent to the velocity vector at P at the time the particle passed through
P. Pathlines can intersect themselves or even retrace themselves, for instance if a fluid particle goes
round and round in a container. Two path lines can intersect if the point of intersection corresponds to a
different time on each of the two trajectories. For example, two different dust particles may pass over the
same point at the CMI entrance on two different days.
• Suppose a small quantity of ink/dye is continuously injected into a fluid flow at a fixed point of
observation P. The dye is so chosen that the dye particles do not diffuse in the fluid. Rather a dye
particle sticks to the first fluid particle it encounters and flows along with it. So the dye released at time
t sticks to the fluid particle that passes through P at time t and is then carried by that particle. The
resulting high-lighed curve is the streak-line through P. So at a given instant of time to , a streak-line
is the locus of all current locations of particles that passed through P at some time t ≤ to in the past.
Unlike streamlines, streak-lines provide information on the history of the flow. Streak-lines are specified
by three parameters, the point of injection P, the time of observation tobs and the time when the injection
of dye began ti . Such a streak-line always begins at P and extends to a point determined by ti .
60