Download Lagrangian Dynamics 2008/09

Lagrangian Dynamics 2008/09
Lecture 0: Introduction
Synopsis I – From the Programme Guide
The principles of classical dynamics, in the Newtonian formulation, are expressed in terms
of (vectorial) equations of motion. These principles are recapitulated and extended to cover
systems of many particles. The laws of dynamics are then reformulated in the Lagrangian
framework, in which a scalar quantity (the Lagrangian) takes centre stage. The equations
of motion then follow by differentiation, and can be obtained directly in terms of whatever
“generalised coordinates” suit the problem at hand. These ideas are encapsulated in Hamilton’s principle, a statement that the motion of any classical system is such as to extremise
the value of a certain integral. The laws of mechanics are then obtained by a method known
as the “calculus of variations”. As a problem-solving tool, the Lagrangian approach is especially useful in dealing with constrained systems, including (for example) rotating rigid
bodies, and one aim of the course is to gain proficiency in such methods. At the same time,
we examine the conceptual content of the theory, which reveals the deep connection between
symmetries and conservation laws in physics. Hamilton’s formulation of classical dynamics
(Hamiltonian Dynamics) is introduced, and some of its consequences and applications are
explored.
1
Synopsis II – Informal Commentary
• The course begins with a recap of Newton’s laws (vectorial mechanics), especially as
they apply to N -particle systems, with emphasis on concepts of energy and momentum
(both linear and angular).
• We then move on, via a discussion of constrained systems, to explore the Lagrangian
formulation of classical mechanics. The centre of attention shifts to scalar, rather
than vector, quantities from which the equations of motion follow by differentiation
(analytical mechanics).
• The Lagrangian method is equivalent to Newtonian mechanics, but is better for many
types of problem, especially those involving constrained systems and/or arbitrary
(“generalised”) coordinates, rather than the cartesian coordinates within which Newtonian mechanics is framed. It is also an alternative starting point for studying
problems in mechanics.
• Lagrangian methods are well-suited to problems in rigid body motion (covered in some
depth) and in small oscillation theory.
• In addition, the Lagrangian approach exposes some deep physical concepts, such as
the role of variational principles in physics (which we take time off from mechanics
proper to explore) and the relation between symmetries and conservation laws (which
extends far beyond the domain of classical physics).
• Lagrangian mechanics therefore provides a “way of thinking” as well as a “problemsolving toolkit”. Both aspects are equally important to this course. The lectures
will deal mainly with the former, whilst the equally-important tutorial sheets and
workshops will deal mostly with the latter.
• Although we shall work chiefly with the Lagrangian formulation of classical mechanics,
we shall also introduce and utilise the closely-related Hamiltonian approach.
• Preparation for future courses: the Hamiltonian and Lagrangian formulations of classical mechanics are closely related to the traditional (canonical) operator and pathintegral formulations of quantum theory respectively.
2
Syllabus
1. Introduction: scope of the course. Newton’s formulation: its strengths and weaknesses.
Its relation to 20th Century physics. Inertial frames. Dynamics of a particle: torque,
angular momentum, conservation laws. Examples.
2. Dynamics of a system of particles. Central and noncentral forces – the third law in
strong and weak form. Centre of mass motion, angular momentum, its independence
of origin in the centre of momentum frame. Rigid body as a collection of particles.
3. Constraints and their classification. Generalised displacements, generalised forces.
From Newton’s laws to Lagrange’s equations (the hard way - superseded later). Conservative forces, the Lagrangian function. Absence of friction. Examples. Revision of
rotating frames and fictitious forces: derivation of centrifugal and Coriolis forces as an
illustration of the Lagrangian approach.
4. Functions and functionals. Calculus of variations. The Euler-Lagrange equation and its
multi-variable extension. Examples of variational calculus from other areas of physics.
Constrained variations. Hamilton’s principle as an embodiment of Lagrangian dynamics. Its coordinate invariance. From Newton to Lagrange (the easy way).
5. Lagrangian Mechanics. Changes of variable, changes of Lagrangian. Ignorable coordinates, first integrals. Generalised momenta. The energy (or Hamiltonian) function
and its relation to the total energy of the system. Examples. Conservation of angular
and linear momentum deduced from the homogeneity and isotropy of space. General
relation between symmetries of the Lagrangian and conservation laws.
6. Extensions of the Lagrangian method. Velocity-dependent forces; the Lagrangian of
a charged particle in an external electromagnetic field, distinction between canonical
and mechanical momentum. The Lagrangian for a relativistic particle in an external
electromagnetic field (briefly).
7. Hamiltonian Dynamics. The Legendre transformation. Derivation of Hamilton’s equations of motion (from Lagrange’s equations). Poisson brackets. Hamiltonian of a
charged particle in an external electromagnetic field. The connection with Quantum
Mechanics.
8. Rigid body motion. Angular velocity, angular momentum; their noncollinearity. The
moment of inertia tensor. Principal axes. Constrained motion about a general axis.
Spherical, symmetrical and asymmetrical tops defined. Rigid body kinematics: Eulerian approach – axes in the moving frame of the body. Torque-free motion of a
symmetric top. Stability of steady motion in the asymmetric case. Limitations. Rigid
body dynamics: Lagrangian method – axes fixed in space. The Euler angles as generalised coordinates. The Lagrangian of a symmetrical top under gravity. The gyroscope:
precession, nutation. Examples.
9. Small oscillation theory. Normal modes studied by Lagrangian methods. Simultaneous
diagonalisation of two matrices. Examples.
3
Textbooks
R D Gregory, Classical Mechanics (Cambridge). A fairly new book which covers the whole
of classical mechanics up to and including the level of this course. Recommended.
H Goldstein, C P Poole, J L Safko, Classical Mechanics (Addison-Wesley). The “Bible”
of Classical Mechanics. Well balanced between formalism and examples, but a little longwinded and a bit esoteric in parts. We only cover a small part of what’s there, but it will
be useful for anyone planning to take Hamiltonian Dynamics next year.
J Leech, Classical Mechanics (Chapman and Hall). Very concise, good for revision if you
already have seen the material. Emphasises concepts rather than problem solving.
M Spiegel, Theoretical Mechanics (Schaum Series, McGraw Hill). Teaches by worked
example, emphasises problem solving rather than concepts.
V Barger and M Olsson, Classical Mechanics, A Modern Perspective (McGraw Hill,
second edition. 1995). A very unusual book. There is a concise introduction to Lagrangian
methods in Chapter 3; thereafter the book is strong on rigid body motion but weak on other
parts of the course. Worth looking at for its discussions of esoterica such as the superball,
the tippy-top, boomerangs, etc. . .
W Greiner, Classical Mechanics: Systems of Particles and Hamiltonian Dynamics (SpringerVerlag, 2003). A relatively recent book. It lies somewhere between Barger & Olsson and
Goldstein et al in both level & content.
T W B Kibble, F H Berkshire, Classical Mechanics (Imperial College Press, fourth or
fifth Edition). A well-established text which covers most aspects of classical dynamics. It’s
coverage of Lagrangian Dynamics is a bit thin, but you might already have a copy, so don’t
throw it away!
K F Riley, M P Hobson and S J Bence, Mathematical Methods for Physics and Engineering (third edition, 2006). See Chapter 20 for mathematical background on variational
calculus.
4
Lagrangian Dynamics 2008/09
Lecture 1: Mostly Revision
Section I: Newtonian Mechanics
Newton’s laws
First Law (N1): Define the velocity of a (point) particle as
v=
dr
≡ ṙ
dt
where r is the position vector relative to some origin O. [Note: In this course, f˙ is always
the total derivative df /dt not the partial derivative ∂f /∂t.]
If no forces act on a body, it remains at rest or moves with constant velocity: v̇ = 0. (N1)
Second Law (N2): Define the momentum of a particle as
p = mv
In the presence of a force F ,
ṗ = F
(N2)
and if m is constant, F = mr̈ ≡ ma.
N2 is three equations, one for each cartesian axis, eg, Fx = mẍ. In the Newtonian scheme,
the force F on a given particle is taken as known. Of course, when other particles are present
the force can depend on the positions of the other particles.
Third Law (N3): For a system of N point particles, with a = 1, 2, 3, · · · , N ,
ṗ a = F a
where F a (r1 , r2 , · · · , rN ) is the (total) force on particle a, and is presumed known.
In general Newton did not specify F a , except to say that the force F ab on a due to b obeys
F ab = −F ba
(weak form of ) (N3)
The strong form of N3 states in addition that F ab is directed along the line r ab = r a − r b
between the two particles.
There are exceptions to N3 in both forms, for example the EM force between current elements
is not directed along the line between them. (Indeed, to treat electromagnetic forces within
the Newtonian framework is rather clumsy: momentum and angular momentum are only
conserved if the contributions of the electric and magnetic fields are included.) We shall
return to this later.
In Newtonian mechanics, we need N3 to derive conservation laws. Later we shall rederive
these without appeal to N3 using Lagrangian methods.
1
Pairwise additivity of forces
Even to write down N3 we have tacitly assumed
Fa=
X
F ab
b
which says that the force on a particle can be decomposed into the sum of contributions
from other particles – two-body forces. This is not always true.
Hypothetical Example: 3 particles in one dimension with interaction potential V (x1 , x2 , x3 ) =
k x1 x2 x3 . The force on particle 1 is:
∂V
F1 = −
∂x1
!
x2 ,x3
= −k x2 x3
which cannot be written as F12 + F13 .
Realistic Example: Two permanent dipoles (1 & 3), and one induced dipole (2). The force
“on 3 from 2” depends on what 1 is doing.
Inertial frames: N1 and N2 (and N3, if applicable) hold only with respect to an inertial
frame of reference. If a given frame is found to be inertial, so that Newton’s laws are obeyed,
then they are also obeyed in any frame moving with constant velocity relative to the first
(the second frame is also inertial). The transformation between two such frames is called a
Galilean transformation: Newton’s laws are Galilean invariant (proved later).
Note that Newton’s laws are not obeyed in a frame that is accelerating (for example, rotating
at uniform angular velocity) relative to some inertial frame. However, the laws of mechanics
in such a non-inertial frame can be put in Newtonian form by introducing so-called “fictitious
forces” (see later). The earth is not an inertial frame (think of Foucault’s pendulum), but it
is a sufficiently good approximation to one for the problems we shall meet in this course.
For classical mechanics purposes, we require that an inertial frame is stationary, or moving
uniformly, with respect to “the fixed stars”, ie with respect to distant gravitational sources.1
Dynamics of a Particle: Conservation Laws
Conservation of linear momentum
Start with N2:
ṗ = F
If F = 0, then ṗ = 0 and p is conserved. This clearly holds also component by component,
eg, if Fx = 0 then ṗx = 0, (irrespective of whether Fy 6= 0 or Fz 6= 0.)
1
What distinguishes the inertial frames from the others? To get to the bottom of this really requires
general relativity (GR). Consider Mach’s famous thought experiment: a bucket of water rotating about its
axis has a curved top surface (meniscus). If we choose to rotate with the bucket, we see that the meniscus
of the (stationary) water is not flat, and thus the frame is not inertial. However, we also notice the fact
that distant gravitational bodies are rotating around us. Mach suggested that the gravity of these distant
rotating bodies “causes” the curvature. In fact, it turns out that Einstein’s equations for GR possess a
solution representing a rotating body in an otherwise empty universe and a test-particle on the surface of
the body would feel a centrifugal force, so Mach’s principle is not obeyed and the concept of a non-inertial
frame is a fundamental.
2
Moment of force, moment of momentum
Again, start with N2:
F = ṗ
Take the cross product of r with this equation, ie take r×(both sides) to obtain
r × F = r × ṗ =
d(r × p)
− ṙ × p
dt
The last term is v × mv = 0, hence
dL
(1)
dt
where G ≡ r × F is the moment of force about O (also known as the torque or couple) and
G=
L≡r×p
is the moment of momentum (usually known as the angular momentum) about O.
G is the turning effect of the force, which tries to rotate the particle about an axis through
O perpendicular to both r and F (the axis defined by the unit vector Ĝ). The magnitude
of the moment of the force is |G| = rF sin θ where θ is the angle between the force and the
position vector. Likewise L is a vector perpendicular to r and p of magnitude mvr sin θ0
where θ0 is the angle between these vectors.
G and L are both axial vectors (aka pseudovectors): they change sign in transforming from
a right to a left-handed coordinate system. Both depend on the choice of O.
Conservation of angular momentum
From equation (1), if G = 0, then L is conserved. This is clearly true componentwise, eg if
Gx = 0 then Lx is constant, (irrespective of whether Gy 6= 0 or Gz 6= 0.)
Conservation of energy
Start with N2: F = mv̇ , and define the work done by the force F in moving a particle from
point A to point B
WBA ≡
Z B
A
F · dr =
⇒
Z tB
tA
F·
Z tB
Z tB
m d(v · v)
dr
dt =
mv̇ · v dt =
dt
dt
dt
tA
tA 2
WBA =
1 2
mv
2
tB
tA
≡ TB − TA
which we identify as the change in the kinetic energy T ≡ 12 mv 2 . This may be regarded as
the definition of the kinetic energy in Newtonian Mechanics.
If F is conservative, we can (almost2 ) always write F = −∇V , where V is the potential
energy. Then
Z B
Z B
TB − TA = −
(∇V ) · dr = −
dV = VA − VB
A
A
independent of the path taken. We have thus VB + TB = VA + TA which expresses the law
of conservation of energy E ≡ T + V .
2
The important case of conservative velocity-dependent forces and “potentials” is treated in Lecture 12.
3
Example of the use of Conservation Laws: Orbits
A particle moves in a central force field described by a potential V (r) = V (r). Find an
equation for the radial velocity.
Revision: Recall that a central force can be written as the gradient of a potential which
depends only on the magnitude r of the position vector r: F (r) = F (r) r̂ = −∇V (r), where
r̂ is the unit vector r/r.
Since F k r we have G = r × F = 0, hence L is conserved – it has constant magnitude and
direction. Since r · L = r · (r × p) = 0, the motion is confined to a plane perpendicular to L.
Coordinates: Choose circular cylindrical coordinates (r, φ, z) (aka cylindrical polars) with
the z axis parallel to L, with the motion confined to the plane z = 0.
√
Clearly x = r cos φ and y = r sin φ with r = |r| = x2 + y 2 , and z = 0.
Define two orthonormal basis vectors e r and e φ
parallel and perpendicular (respectively) to the position vector r. At the point P :
er =
eφ
er
φ
cos φ i + sin φ j
y
e φ = − sin φ i + cos φ j
where i and j are the usual basis vectors in the
x and y directions. The position vector r(t) of a
particle which is at the point P at time t is then
0
r = x i + yj = r e r
r
φ
P
φ
x
Velocity: The velocity v(t) of the particle at the point P at time t is
v = ṙ =
Now
ė r =
d
(r e r ) = ṙ e r + r ė r
dt
d cos φ i + sin φ j = (−φ̇ sin φ i + φ̇ cos φ j) = φ̇ e φ
dt
so that ṙ = ṙ e r + r φ̇ e φ
Evidently, the radial component of velocity vr = ṙ, and the tangential component of velocity
vφ = rφ̇. We shall use these results a great deal, so you should know them!
Kinetic energy:
1
1
1
T = mv 2 = mṙ · ṙ = m(ṙ2 + r2 φ̇2 )
2
2
2
Alternatively, we may evaluate T by differentiating x = r cos φ and y = r sin φ:
ẋ = ṙ cos φ − rφ̇ sin φ
and
This gives
ẏ = ṙ sin φ + rφ̇ cos φ
1
1
1
T = mv 2 = m(ẋ2 + ẏ 2 ) = m(ṙ2 + r2 φ̇2 )
2
2
2
as before. We will usually use vr and vt to simply write down the kinetic energy when using
polar coordinates, but this second method is much more general and we’ll use it a lot.
4
Conservation of energy and angular momentum:
n
L = r × mṙ = m r e r × (ṙ e r + r φ̇ e φ )
o
= 0 + mr2 φ̇ e r × e φ = Le z
Conservation of angular momentum gives L = mr2 φ̇ = constant. This relation between r
and φ̇ is valid at all times, and may be used to eliminate the orbital angular velocity φ̇ from
the expression for T . Conservation of energy then gives
E = T +V =
1
L2
1
m(ṙ2 + r2 φ̇2 ) + V (r) = mṙ2 +
+ V (r) = constant
2
2
2mr2
(2)
Equation for the radial velocity: Solving equation (2) for ṙ gives:
⇒
"
L2
2
E − V (r) −
ṙ = ±
m
2mr2
!#1/2
2
L /2mr
(3)
The conserved quantities E and L are fixed by the
initial conditions. Since ṙ must be real, the quantity in square brackets in equation (3) must always
be greater than or equal to zero.
The radial motion is that of a 1-D system in the
effective potential
L2
Veff (r) ≡ V (r) +
2mr2
r2 r1
2
r2’
r
Veff(r)
E2
E1
V(r)
The sketch shows the effective potential Veff (r) for
the special case of an attractive inverse-square-law
force F (r) = −kr̂/r2 and hence V (r) = −k/r.
Effective EoM: Although it’s not strictly necessary in order to proceed, we may obtain an
effective “Newtonian” equation of motion for the radial motion using energy conservation:
d 1 2
dE
=
mṙ + Veff (r)
dt
dt 2
= mṙr̈ +
dVeff (r)
ṙ = 0
dr
(4)
This holds for all ṙ, hence
mr̈ = −
dVeff (r)
dr
(5)
Circular orbit: By definition, a circular orbit has constant radius r = r1 (say), so ṙ = 0.
From equation (3) this requires E = Veff (r1 ) ≡ E1 at all times. Thus a circular orbit is
possible where the effective potential has zero gradient, ie at a local minimum (or maximum)
of the effective potential – see figure.
Alternatively, since ṙ = 0, then r̈ = 0 too, and equation (5) gives the same result:
dVeff (r) = 0.
dr r=r1
(6)
If E = E2 > E1 , the particle is confined to the zone between the two radii r2 and r20 : at
these points E = Veff (r2 ) = Veff (r20 ) = E2 and the radial component of velocity ṙ = 0 there.
5
Perturbing circular orbits: What happens if we make a small perturbation to a circular
orbit? Let r(t) = r1 + (t). Then, since r1 is a constant in time, we have
ṙ(t) = 0 + (t)
˙
For small , we may Taylor expand the effective potential about r = r1 :
2 d2 Veff (r) dVeff (r) +
Veff (r) = Veff (r1 + ) = Veff (r1 ) + + O(3 )
2
dr r=r1
2
dr
r=r1
k
= constant + 0 + 2 + O(3 )
2
d2 Veff (r) 00
(r1 ).
where we have defined the constant k ≡
≡ Veff
dr2 r=r1
For small (and k 6= 0), we can ignore the terms of order O(3 ), and the total energy
becomes3
1
k
E = m˙2 + 2 ,
(7)
2
2
If r1 is a local minimum of the effective potential, then k > 0, and equation (7) is the
total energy of a harmonic oscillator of mass m and spring-constant
q k. Thus our small
perturbation leads to radial oscillations with angular frequency ω = k/m about r1 .
If r1 is a local maximum of the effective potential, then k < 0, and the circular orbit is
unstable. If k = 0, we have to go beyond O(2 ) in the Taylor expansion of V (r).
The Energy Method – Zones
The above method, which was based on using the energy directly to obtain an equation of
motion, is especially powerful for systems with only one or two degrees of freedom. Rather
than three second order ODEs, the application of conservation laws gives one first order
ODE – a major simplification. In the energy method, as with the Lagrangian methods
covered later, it is extremely important to write down the kinetic and potential energies
correctly(!) as a function of the chosen coordinates, whatever they may be. Practice makes
perfect – see the problem sheets.
Zones:
In our example, if 0 > E > Veff (r1 ) then the particle is confined to a zone:
rmin ≤ r ≤ rmax , where rmin,max are the roots of Veff (r) = E. Put differently, equation (3) for
ṙ is of the form
q
ṙ = ± g(r)
with, in this case, g(r) = 2(E − Veff (r))/m. Then the fact that ṙ must be real requires that
g(r) must be positive. So, whatever happens in the subsequent motion, r cannot escape the
zone rmin ≤ r ≤ rmax where these are the roots of g(r) = 0. (Eg: if E = E2 , r2 ≤ r ≤ r20 .)
The use of zones is an important method which can be applied to more complicated problems
(eg spinning tops) to find partial information without solving for the full trajectory – which
in many cases is extremely difficult. If asked to show that some coordinate q
is confined to a
zone, your first reaction should be to look for a velocity equation like ṙ = ± g(r).
3
We have dropped the constant term, Veff (r1 ) = E1 , from the RHS of the expression for E. E1 is just
the total energy of the original circular orbit, and it doesn’t affect the oscillation frequency.
6
Lagrangian Dynamics 2008/09
Lecture 2: Dynamics of a system of particles
Consider a system of N particles, labelled by a = 1, 2, · · · , N ; with constant masses ma . As
usual we start our analysis with Newton’s second law for particle a:
ma r̈ a = F a
Centre of Mass Motion
Sum N2 over all particles
X
a
ma r̈ a =
X
a
Fa =
X
a
+
F ext
a
X
F ab
a,b
is the external force on a (ie not
where F ab is the force on a due to b (with a 6= b), and F ext
a
due to the other particles). If we assume N3 (weak), ie F ab + F ba = 0, then
X
F ab =
a,b
X
(F ab + F ba ) = 0
pairs
ie all internal forces cancel pairwise. Hence
X
a
ma r̈ a =
X
a
F ext
= F ext
a
where F ext is the total external force. The total mass is M =
mass position vector R by
X
MR =
ma r a
(1)
P
a
ma . Define the centre-of-
a
Differentiating this twice with respect to time and substituting into the LHS of equation (1),
we deduce that the centre of mass moves as a particle of mass M under the resultant F ext :
M R̈ = F ext
The total linear momentum is
P =
X
a
pa =
X
a
ma ṙa = M Ṙ
so we may also write
Ṗ = F ext
1
Moments of Force and Momentum
Start with N2 as usual, ṗ a = F a , take the cross product with r a , and sum over all particles:
X
a
X
r a × ṗ a =
a
The LHS of equation (2) includes
ra × F a
(2)
dL a
d
(r a × p a ) − ṙ a × p a =
− ṙ a × p a
dt
dt
P
P
The last term is zero (since p a = mṙ a ) and so a r a × ṗ a = L̇, where L = a L a .
r a × ṗ a =
The RHS of equation (2) is
X
a
ra × F a =
X
a
+
r a × F ext
a
P
X
a,b
r a × F ab
Using N3 (weak) the last term becomes pairs (r a −r b )×F ab . This vanishes if we also assume
P
N3 (strong), namely that F ab is along r ab . Hence, given N3 (strong), L = a L a obeys
X
L̇ =
a
ext
X
=
r a × F ext
a
a
= Gext
Gext
a
where we have defined G as the total external couple acting on the system. Note that
L and G are additive over particles (ie, they obey vector addition rules, as axial vectors
should). Both still depend on the origin O chosen.
Energy
For one particle, we showed previously that
1 2
mv
= dT
2
where T is the kinetic energy. For N particles, it is straightforward to show that
F · dr = d
dT =
X
a
dTa =
X
a
F a · dr a =
X
a
F ext
· dr a +
a
X
pairs
F ab · (dr a − dr b )
where we used N3 (weak) in the last step.
Now, if the internal forces are conservative there is an internal energy function U (r 1 , r 2 , · · · , r N )
such that
X
F ab · (dr a − dr b ) = − dU
pairs
(see Goldstein for a detailed derivation) in which case
dT + dU =
X
a
F ext
· dr a
a
In words: the change in KE + the change in internal energy = work done by external forces.
If the external forces are also conservative, there is a potential function V (r 1 , r 2 , · · · , r N )
such that
= − ∇a V
F ext
a
where ∇ a denotes the gradient wrt to r a , hence
P
a
F ext
· dr a = −dV . If so
a
dT + dU + dV = 0
and the total energy E = T + U + V is conserved.
2
Centre of momentum frame
This is defined as an inertial frame chosen instantaneously so that
P =
X
a
pa =
X
a
ma ṙ a = 0
(usually with its origin chosen to be at the instantaneous centre of mass).
Transformations to and from this frame are discussed below.
(Note: one may also define the centre of mass frame as the frame whose origin is at the
centre of mass for all times. This is a non-inertial frame if the centre of mass is accelerating,
ie when the total external force is non-zero.)
Intrinsic angular momentum
The quantities L (total angular momentum) and Gext (total external torque) are origindependent. To check this, consider a shift in origin O corresponding to a change from frame
S to S 0 say. Then if O0 = O + b, we have r = r0 + b, ṙ = ṙ0 , p = p0 and no change in N2
0
(hence Ṗ = F ext as usual). On the other hand,
Now
X
a
L a = r a × p a = L0a + b × p a
⇒
L = L0 + b × P
where we summed both sides over a. Hence L depends on the choice of origin unless P = 0.
Let us now define the intrinsic angular momentum J as the value of L in the centre of
momentum frame; the quantity J is automatically independent of the choice of origin.
Technical Note: In contrast, Gext = (Gext )0 + b × F ext where the extra term from shifting O
does not vanish in the centre of momentum frame. The origin for G must therefore always
be specified.
Galilean transformation
This is the name for a transformation from one inertial frame to another. Suppose frame S 0
moves at constant velocity V wrt frame S. For definiteness we choose O0 = O + b at time
t = 0.
In frame S, a position vector is r . In S 0 , the corresponding r0 obeys r = r0 + V t + b. Hence
the linear momentum of a particle obeys the transformation law:
p a = p0a + ma V
0
So that ṗ a = ṗ0a = F a . Likewise Ṗ = Ṗ = F ext . Newton’s equations are therefore unaffected
by any Galilean transformation.
3
Transformation law for angular momentum
In terms of primed coordinates, we can write the angular momentum L in frame S as follows:
L =
X
a
(r0a + b + V t) × (p0a + ma V )
Now expand and rearrange (this is a useful exercise – do it!) to get
L = L0 + (b + V t) × P 0 + M R0 × V + M (b + V t) × V
This is the general transformation law for the total angular momentum.
An extremely important special case is when S 0 is the centre-of-momentum frame, so that
P 0 = 0 (by definition), L0 = J (intrinsic angular momentum), R0 = 0 (centre of mass is at
origin in S 0 ) and b + V t = R (position of centre of mass in frame S at time t). Then
L = J +R×P
where P = M V .
This tells us that the angular momentum in S is the intrinsic value (in the c-of-mom frame)
plus the angular momentum due to the centre of mass motion in S, with the centre of mass
treated as a point particle. This is a very important and useful result, as we shall see.
Transformation law for kinetic energy
In frame S,
T =
X
a
expanding:
X1
1
ma v 2a =
ma (v 0a + V ) · (v 0a + V )
2
a 2
T = T0 +
X
a
0
1
m a v 0a · V + M V 2
2
If S is again chosen as the centre of momentum frame, the middle term vanishes (since
P
0
0
a ma v a = P = 0) to give
1
T = T 0 + MV 2
2
that is, the kinetic energy in frame S equals the value in the centre-of-momentum frame, plus
the contribution of the motion in S of the centre of mass (again treated as a point particle).
This result is a great help in writing down the kinetic energy T of a system in arbitrary
motion. We shall use it a lot in problem solving.
4
Lagrangian Dynamics 2008/09
Lecture 3: Use of centre of momentum; Summary & drawbacks of Newton’s
scheme
Example of use of Centre of Momentum Frame
A thin hoop of radius a and mass m is confined to lie in a vertical plane; it rolls without slipping on the inner surface of a hollow tube of circular cross-section (with its axis horizontal)
of radius R. Defining an angular coordinate θ as the angle to the vertical of the line from
the centre of the tube to that of the hoop, and z as a horizontal coordinate, find the angular
momentum Lz about the axis of the tube, and the kinetic energy T , each as a function of θ̇.
Method: First draw a diagram:
V, e
R
θ
ω
O
θ
a
C of Mom Frame
Lab Frame
The centre of mass (CM) of the hoop moves on a circle of radius R−a, with velocity V = vt e θ
where vt = (R − a)θ̇ (the subscript “t” stands for tangential). The angular momentum and
ez
kinetic energy associated with this motion, in the lab frame, are respectively LCM = LCM
z
1
CM
2
CM
2 2
with Lz = m(R − a) θ̇ and T
= 2 m(R − a) θ̇ where e θ and e z are the usual unit
vectors, with e z pointing out of the page.
To these we must add the contributions from the hoop in its centre-of-momentum frame.
This frame is an inertial frame instantaneously co-moving with the centre of mass of the
hoop; hence it is not rotating relative to lab coordinates (just translating with speed vt in
a direction tangential to the radius vector at which the frame is defined instantaneously).
The motion of the hoop in its c-of-mom (centre of momentum) frame consists of a rotation
about the centre of mass at some angular velocity ω. To find ω, we note that the (lab-frame)
velocity of the point on the hoop in instantaneous contact with the tube is v = vt +ωa. Since
the hoop is rolling (not sliding) with respect to the stationary tube, we must have v = 0, or
ω = −vt /a = −(R − a)θ̇/a. Note that ω is negative: the hoop is rotating clockwise.
The intrinsic angular momentum is Jz = ma2 ω = −ma(R − a)θ̇ (again, note the sign), and
the (intrinsic) contribution to T is T int = 12 ma2 ω 2 = 12 m(R − a)2 θ̇2 .
Adding these to the previous results for the angular momentum and kinetic energy for
centre of mass motion, we find the total angular momentum, Lz = LCM
+ Jz , and total
z
CM
int
kinetic energy, T = T
+ T , respectively, in the lab frame are
Lz = m(R − 2a)(R − a)θ̇
and
1
T = m(R − a)2 θ̇2 .
This calculation is not easy – it requires clear thinking, but not much more than that. Also,
it’s rather easier than a direct evaluation in the lab frame (try it if you are sceptical).
To get the equation of motion, one uses the conservation laws to get an expression for θ̇ as
a function of θ.
Noninertial/Rotating frames
This topic was covered in Junior Honours Dynamics & Relativity; it is included here for
completeness as we shall use it later in the course. See D&R notes for a more detailed
discussion.
Noninertial frames are frames which, with respect to an inertial frame, are accelerating. The
most important case is a frame rotating at constant angular velocity. We will sometimes
encounter noninertial frames in this course (although there are pitfalls in applying Lagrangian
methods in such frames). Something we will need later is the equation of motion (which is
not N2) for a particle in a rotating frame.
Consider an inertial frame S0 and a noninertial frame S rotating with constant angular
velocity ω with respect to such a frame, but instantaneously coincident with it (so the origin
and cartesian axes are lined up). We use notation [f ]X to denote a quantity measured in
frame X. Now, a particle with position vector [R]S , stationary in S, has velocity in S0
[v]S0 = ω × [R]S
where [R]S = [R]S0 since S and S0 instantaneously coincide. Adding any velocity the particle
may have in S itself, we have
(1)
[Ṙ]S0 = [Ṙ]S + ω × [R]S
We derived this for a position vector R but if it holds for one of these it must also apply to
any other vector A (since all vectors transform between frames according to the same rules):
[Ȧ]S0 = [Ȧ]S + ω × [A]S
(2)
Now for the clever bit. We choose A = m[Ṙ]S0 . We then substitute A = m[Ṙ]S0 into the
LHS and (using equation (1)), A = m([Ṙ]S + ω × [R]S ) into the RHS of the transformation
equation (2) for A itself. The result is
m[R̈]S0
"
d =m
[Ṙ]S + ω × [R]S
dt
#
S
h
+ mω × [Ṙ]S + ω × [R]S
i
S
The left hand side is the acceleration in the inertial frame S0 , and since Newton’s second
law holds in this frame, we can write this term as F , the force on the particle. Expanding
the RHS gives:
F = m[R̈]S + mω × [Ṙ]S + mω × [Ṙ]s + mω × (ω × [R]S )
Rearranging, we find the equation of motion in the rotating frame S to be:
m[R̈]S = F − mω × (ω × [R]S ) − 2mω × [Ṙ]S
2
Or in a less pedantic notation,
ma = F − mω × (ω × R) − 2mω × v
Though this equation is not N2, it can be put in the same form if we identify the last two
terms, which are really contributions to the acceleration, as fictitious forces. The first is
the centrifugal force which, in a rotating frame, appears to cause objects to fly outwards.
The second is the Coriolis force which appears to cause an object, moving in the plane
perpendicular to the axis of rotation, to be deflected at rightangles. This term is responsible
for the (anti-)cyclonic weather pattern typical in the Northern Hemisphere.
Summary: Newton’s scheme
• Newton’s second law states that in any inertial frame, ma r̈ a = F a , a = 1, 2, · · · , N .
This is a set of 3N second-order differential equations, solvable in principle if the initial
coordinates and velocities of all particles are known. Also required is an expression
for the forces, in terms of particle positions and (possibly) particle velocities; in some
cases, the expression for the force could depend explicitly on time as well. (Example:
two particles connected by a spring, one of which is subject to a periodic driving force.)
• If in addition N3 (weak) holds (F ab = −F ba ), then Ṗ = F ext . If there are no external
forces on the system, linear momentum is conserved.
If N3 (strong) holds (F ab = −F ba k r ab ), then L̇ = Gext . If there are no external
torques, angular momentum is conserved.
• If the inter-particle and external forces are conservative (derived from potential energy
functions U and V ), then the total energy of the system, E = T + V + U , is conserved.
• Non-inertial frames can be included in Newton’s scheme by invoking suitable fictitious
forces.
Drawbacks of Newton’s approach
Vectorial formulation, coordinates: The starting point is vector analysis using the
basic objects of force and acceleration. This is fine in Cartesians but becomes very messy
when other coordinate systems are used (consider, for example, the nasty expression for the
acceleration of a particle in spherical polar coordinates). The Lagrangian formulation is
more easily adapted to generalised coordinate systems since it formulates mechanics directly
in terms of scalar (energy-like) quantities.
Constraints: In solving mechanics problems we often encounter, for example, rigid rods,
inextensible strings, etc. These are constrained systems. The prime example is a rigid body
which consists of an enormous number of particles (N ' 1023 ) but an almost equally enormous number of constraints (the relative separations of all the particles are fixed). Applying
Newtonian methods, at least if done from first principles, means explicitly calculating all
the forces that enforce the constraints. For an inextensible string, this is the tension in the
string; for the rigid body there are ' 1023 constraint forces and obviously it is impractical
3
to calculate these. Even when few in number, constraint forces are often uninteresting and
end up being finally eliminated. Lagrangian methods allow one to do this elimination at the
outset.
Conservation laws: In Newton’s scheme, conservation of energy, and of linear and angular
momentum are derived consequences, not fundamental principles. Since the derivation
requires N3 (strong), the fundamental status of angular momentum conservation, in particular, is rather unclear. Even linear momentum conservation, derived via N3 (weak), assumes
pairwise additive forces – which is not very general. The Lagrangian approach completely
avoids N3 and lays bare the deep relationship between conservation laws and symmetries.
It turns out that N3 is a red herring. (This relationship extends beyond strictly classical
mechanics into relativity and quantum theory, too.)
Relation to modern physics: The real world is described (as far as we know) by relativistic quantum field theory (or perhaps (super)string theory or even M -theory). Classical
mechanics should emerge from this as the limiting case of macroscopic (ie, non-quantum),
slow-moving (ie, nonrelativistic) objects. Already you have learned something about (nonquantum) relativity, and about (non-relativistic) quantum mechanics. In the first case,
the emergence of Newtonian physics for small v/c is fairly natural. (In fact, Lagrangian
methods can be extended to include relativity and we explore this briefly). For quantum
mechanics, the limiting process is harder to tease out, but a careful pursuit of a “correspondence principle” between classical and quantum regimes leads to the “scalar” (Lagrangian
& Hamiltonian) formulations of dynamics, rather than the Newtonian one.
Section II: Dealing with constraints
Example: Rigid bodies
Consider Newton’s second law for N particles:
+
ma r̈ a = F a = F ext
a
In a rigid body, the interparticle distances are all fixed:
P
b
F ab
|r a − r b | = ρ ab = constant for each pair.
This provides a set of constraint equations (not all independent). N2 still applies but the
forces F ab take whatever values are necessary to ensure the constraints are obeyed.
In a strict Newtonian scheme, one would have to solve the above set of 3N simultaneous
equations with arbitrary F ab (t) and then use the constraint equations to eliminate the forces.
In practice, if there are many constraints (as here) it is better to formulate the problem
directly in terms of a reduced set of coordinates, one for each of the degrees of freedom that
are left in the system after the constraints are imposed. Lagrangian methods are well suited
to this procedure.
4
(N2)
Lagrangian Dynamics 2008/09
Lecture 4: Dealing with Constraints; Generalised Coordinates & Velocities
Degrees of freedom
A general rigid body is specified fully by R (the centre of mass position) and three angles
of rotation, giving the inclination of a set of axes in the body with respect to those in the
lab frame (we return to this in depth later on). Hence there are 6 degrees of freedom (below
abbreviated as dof), rather than 3N. We deduce that of the constraint equations, 3N-6 are
independent (not obvious).
z0
z
Axes fixed
in LAB
y0
y
R
O
x0
x
Axes fixed on body
Technical Note: The limit of a continuous material can be taken by sending N → ∞; for a
rigid body, the number of degrees of freedom is still 6.
Even given that we have a rigid body, there may be further constraints. For example, a disc
confined to a plane has 2 centre of mass coordinates and one rotation angle giving a total of
3 dof. For our previous example of a hoop rolling inside a cylinder, the hoop rolls without
slipping so that the tangential velocity of the hoop and the cylinder at the point of contact
must be equal. Moreover the centre of the hoop is confined to a circle. This reduces the
problem to one dof. In problem solving, the first step is very often to identify the number
and type of degrees of freedom.
What is a constraint? For all the examples we will meet in this course, a constraint is a
limiting approximation valid for the problem at hand.
Example: Two particles a, b are connected by a light inextensible rod of length ρba . In
reality the rod has a large stiffness k – there is a harmonic restoring force (Hooke’s Law)
when the rod is stretched to length rba = |r ba | :
F ba (r ba ) = −k(r ba − ρ ba ) r̂ ba
1
U (r ba ) = k (rba − ρba )2 + constant
2
⇔
U
^r
ma
ba
ρba
mb
ρba
rba
In the limit k → ∞, rba = ρ ba always, and the corresponding degree of freedom is eliminated.
1
Constraint forces and work
Consider the above system subject to an external force F ext acting along r ba .
ρ
δr
F ext
ma
mb
(held fixed)
In equilibrium
F constraint (r) = − k δr = − F ext
with δr = (r − ρ ba ) r̂ ba . The work done by the constraint force in reaching this state is
Z
1
1
F constraint (r) · dr = − k(δr)2 = − (F ext )2 /k
2
2
which vanishes in the limit of large k. In general:
δU =
Constraint forces do no work in any small instantaneous displacement of the system consistent with the constraints themselves.
A small instantaneous displacement of this kind is called a virtual displacement. The point
is that constraint forces are there to prevent certain types of virtual displacement; only if
they fail to do this can work be done. And if they fail, they are no longer constraints.
Technical Note: The above result does not mean that the constraint forces can do no work
during the actual motion of a system. Consider, for example, a particle constrained to lie
on a surface which is itself moving: there may then be a component of the actual particle
velocity in the direction of the constraint force, so that work is done. We return to this later.
Holonomic Constraints
|r ab | = ρ ab
Rigid body:
This is an example of a holonomic constrant, ie there is a relation
f (ra , rb , . . . , rN ; t) = 0
which is an algebraic equation between coordinates, ie not a differential relation; not an
inequality. Explicit time dependence is permitted, however.
Holonomic constraints reduce the number of degrees of freedom.
They are important because each holonomic constraint reduces by one the number of independent differential equations we have to solve!
y’
a
Example: Consider a cylindrical
log rolling down an inclined plane as
shown.
x’
2
θ
This system has 3 dofs: (x0 , y 0 , θ). If we impose rolling conditions (no-slide and no-bounce)
on the centre of mass of the log,
x0 = aθ + constant
and
y0 = a ,
ie two holonomic constraints, then only one dof remains. We can choose either θ or x0 as
the independent variable. The hoop rolling in the cylinder is similar.
Algebraic constraints mean that we can eliminate variables before solving the equations of
motion. This wouldn’t be true if the log started to slide, or to bounce.
Nonholonomic constraints
Common examples of these are: (i) inequalities (ii) differential equations of constraint:
Example of an inequality constraint: a particle constrained to move inside a rectangular box:
The constraints:
(
0 ≤ x ≤ a
0 ≤ y ≤ b
b
are enforced by collisions with the walls of the box,
but they obviously don’t reduce the number of dofs.
a
Example of differential constraint: A vertical disc is free to roll without slipping on a
horizontal plane. This is basically a unicycle! There are 4 degrees of freedom, (x, y, θ, φ).
The angle θ gives the orientation of the
vertical disc wrt to the coordinate axes.
Rolling constraints: as the disc rolls such
that φ → φ + dφ, then x → x + dx, and
y → y + dy with
a
φ
θ
y
dx = a cos θ dφ
dy = a sin θ dφ
x
These constraint equations cannot be integrated to eliminate a dof. The physical interpretation is that in the 4-D coordinate space, any point (x, y, θ, φ), can be reached from any other
(try it on your unicycle). Hence there can be no real reduction in the number of degrees of
freedom and variables cannot be eliminated. Differential constraint equations like these can
only be solved after a full general solution of the unconstrained problem has been found.
From now on, we consider only holonomic systems but keep an eye out for places where our
constraints might break down or cease to be holonomic.
3
Technical Note: The rolling constraint for a log (see above) can also be viewed as a differential
constraint (ẋ0 = aθ̇), however this can be integrated immediately (x0 = aθ + constant) to
give an algebraic relation, which is holonomic. Almost all the rolling constraints met later
in the course are holonomic in the same way.
Section III: From Newton to Lagrange
In this section we shall derive Lagrange’s equations in an important but relatively difficult way. Things will get easier when we actually start using the Lagrangian formalism.
Furthermore, we shall present an easier derivation later using the calculus of variations.
Generalised coordinates and velocities
As usual we begin our discussion with Newton’s Second Law:
a = 1, 2, · · · , N
ma r̈ a = F a
There are 3N Cartesian coordinates, r a = (xa , ya , za ). As a shorthand notation, we write
the entire list (x1 , y1 , z1 , x2 , y2 , z2 , · · · , xN , yN , zN ) as
xi
where i = 1, 2, · · · , 3N
with each corresponding component of the force Fi . The list of masses is
mi = (m1 , m1 , m1 , m2 , m2 , m2 , · · · , mN , mN , mN )
(think about it).
If holonomic constraints are present, not all the xi are independent. For M constraints,
there must be a set of independent variables
qi = qi (x1 , x2 , · · · , x3N ; t)
where the index i now only runs from 1 to 3N − M . Our aim will be to generate 3N − M
second order DE’s between the qi ’s; these are the Equations of Motion (EoMs) of the system.
(Actually solving them is a different problem!)
There is no reason for the qi ’s to be cartesian coordinates; they are very often angles, etc. In
general the qi ’s represent a set of generalised coordinates; for shorthand we write the full set
as {q}. The transformation to our generalised coordinates is invertible: given {q} we can use
the M constraint equations to recover the 3N original variables if the 3N − M generalised
coordinates are known, ie we may write
xi = xi ({q}, t)
The important point is that the original variables {x} cannot be varied independently without violating the constraints, where as {q} (by construction) can be so varied.
The time derivatives of the generalised coordinates (ie, the quantities {q̇}) are called the
generalised velocities.
4
Lagrangian Dynamics 2008/09
Lecture 5: From Newton to Lagrange – the hard way first
Generalised forces
Consider a system described by 3N cartesian coordinates xi , M constraints, and hence
3N −M generalised coordinates qi , and write xi = xi ({q}, t).
In a general small (ie infinitesimal) change in the coordinates: xi → xi + dxi , a generalised
displacement, we have
X ∂xi
∂xi
dxi =
dqj +
dt
∂t
j ∂qj
Here and below, the partial derivatives are performed with all other variables held fixed, ie,
∂xi /∂qj means that all the {q} (except qj of course), and also t, are held constant during
the differentiation.
In a virtual displacement (an instantaneous small change consistent with the constraints),
which we denote {δx}, there is no contribution proportional to dt :
δxi =
X
j
∂xi
δqj + 0
∂qj
In such a change the work done is
δW =
X
Fi δxi
i
Writing Fi = Ficonstraint + Fio , where “o” stands for “other”, we obtain
δW = 0 +
X
Fio δxi
(1)
i
since the constraint forces do no work in a virtual displacement (see Lecture 4). In terms of
{δq}, we have
XX
∂xi
δqj
(2)
Fio
δW =
∂qj
j
i
which can be written
δW =
X
Qj δqj
j
where we have defined the generalised forces {Q} by
Qj =
X
i
Fi
∂xi X o ∂xi
=
Fi
∂qj
∂qj
i
Note that the constraint forces do not contribute to this expression because they do no work
in a virtual displacement consistent with the constraints (equation (1) above).
A given Qj is called the generalised force conjugate to the corresponding generalised coordinate qj .
1
Example: A particle of mass m slides under gravity on a fixed vertical hoop of radius a.
θ a
z
x
In cartesians (x, z), there is one holonomic constraint and therefore just one degree of freedom
remains. We choose θ, with the origin at the centre of the hoop:
x = −a sin θ
and
z = a cos θ
From equation (2), the work done in a virtual displacement is
δW =
!
∂x
∂z
Fxo
+ Fzo
δθ = {0 + (−mg) (−a sin θ)} δθ ≡ Qθ δθ
∂θ
∂θ
Hence the generalised force conjugate to θ is
Qθ = mga sin θ
which has the dimensions of work. Note that qj can have any dimensions but that Qj qj (not
summed over j) will always have the dimensions of work (ie of energy).
Mathematical Digression: Functions of {q}, {q̇} and t
We shall soon meet several functions of the form
f = f ({q}, {q̇}, t)
An example is the kinetic energy
T =
X
i
1
mi ẋ2i
2
Since xi = xi ({q}, t), in generalised coordinates the expression for T can depend explicitly
on both the coordinates {q} and the velocities {q̇}, as well as on time t. Thus we have
T = T ({q}, {q̇}, t). (For example, in plane polars T = 12 m(ṙ2 + r2 φ̇2 ) which depends on r in
addition to ṙ and φ̇.)
For such functions, the increment df can be written
df =
X
j
X ∂f
∂f
∂f
dqj +
dq̇j +
dt
∂qj
∂t
j ∂ q̇j
Here the partial derivatives are performed at fixed values of all the other variables. For
example
∂f
∂qj
means the derivative of f wrt qj at fixed values of all the other {q}, and fixed values of all
the velocities {q̇}, including q̇j , and fixed t. Because all remaining variables are always held
fixed, there is no need to put subscripts on the partial derivatives.
2
Technical Note: Formally, this procedure consists of treating the coordinates {q} and the
velocities {q̇} as independent variables. This is legitimate since we are so far considering
arbitrary motions of the system – we have not yet applied (or even found!) the equations of
motion which relate the velocities to the coordinates.
Special Case: Consider a function f = f ({q}, {q̇}, t) for which there happens to be no
dependence on the velocities {q̇}. (Important examples include the cartesian coordinates
xi = xi ({q}, t).) To emphasise this independence we will sometimes write in such cases
f = f ({q}, {6 q̇}, t)
For such a function, clearly ∂f /∂ q̇j = 0, and so
X ∂f dqj
∂f
df
=
+ 0 +
f˙ ≡
dt
∂t
j ∂qj dt
⇒
X
f˙ =
j
∂f
∂f
q̇ j +
∂qj
∂t
(3)
ie f˙ = f˙({q}, {q̇}, t)
Moreover the partial derivatives of f appearing in this expression will themselves not depend
on velocities (think about it):
∂f
∂qj
=
∂f
∂t
=
∂f
({q}, {6 q̇}, t)
∂qj
∂f
({q}, {6 q̇}, t)
∂t
Thus the only explicit velocity dependence in f˙ comes from the q̇ j shown in bold in equation (3).
By partially differentiating the expression for f˙ wrt q̇j in equation (3) we therefore obtain
∂f
∂ f˙
=
+ 0
∂ q̇j
∂qj
(4)
This is called the cancellation of dots rule and holds for functions f ({q}, {6 q̇}, t) ONLY.
Another result for such functions, needed below, is that
d
dt
∂f
∂qj
!
∂
=
∂qj
df
dt
!
The proof of this commuting derivatives rule is left as an exercise (see Tutorial Sheet).
NB the above results are needed now for the derivation of Lagrange’s equations, but only
occasionally thereafter.
Lagrange’s equations
We shall now find the equations of motion in generalised coordinates. The derivation (from
Newton’s laws) is a bit technical and not very illuminating. A more attractive one (using
the calculus of variations) is given later in the course.
3
Our starting point involves the cartesian coordinates, xi = xi ({q}, {6 q̇}, t). Because there is
no dependence on {q̇}, we can apply the cancellation of dots rule from equation (4)
∂ ẋi
∂xi
=
∂ q̇j
∂qj
which holds for all i and j. Now multiply by mi ẋi and take the time derivative of each side:
!
∂ ẋi
d
mi ẋi
dt
∂ q̇j
d
∂xi
=
mi ẋi
dt
∂qj
!
Expanding the RHS and applying the commuting derivatives rule gives
d
∂ ẋi
mi ẋi
dt
∂ q̇j
!
=
d
∂xi
∂ ẋi
(mi ẋi )
+ mi ẋi
dt
∂qj
∂qj
which can be rewritten
"
#
d ∂ 1
d
∂xi
∂ 1
2
mi ẋi
=
(mi ẋi )
+
mi ẋ2i
dt ∂ q̇j 2
dt
∂qj
∂qj 2
d
(mi ẋi ) = Fi :
dt
"
#
∂ 1
d ∂ 1
∂xi
2
mi ẋi
+
mi ẋ2i
= Fi
dt ∂ q̇j 2
∂qj
∂qj 2
Now use Newton’s Second Law,
Finally, we sum on i:
"
d ∂
dt ∂ q̇j
X
i
1
mi ẋ2i
2
!#
=
X
i
∂xi
∂
Fi
+
∂qj
∂qj
X
i
1
mi ẋ2i
2
!
In this equation we identify the kinetic energy
T ({q}, {q̇}, t) =
and the generalised force
X
Qj ({q}, {q̇}, t) =
to obtain
i
1
mi ẋ2i
2
X
i
Fi
∂xi
∂qj
(5)
!
d ∂T
∂T
= Qj +
dt ∂ q̇j
∂qj
There is one such equation for each generalised coordinate: j = 1, · · · , 3N −M . These
3N −M equations, usually written
d
dt
∂T
∂ q̇j
!
−
∂T
= Qj
∂qj
are Lagrange’s equations of motion in their most general form.
They are derived from Newton’s laws and equivalent to them. However, the constraint
equations have been eliminated by reducing the number of independent variables so as to be
equal to the number of degrees of freedom; constraint forces no longer appear.
Note: You might think the constraint forces enter implicitly through the generalised forces
{Q}. But it was shown in Lecture 4 that, because constraint forces do no work in a virtual
displacement, they make no contribution to the generalised forces {Q}. Hence we may
replace Fi by Fio in equation (5) and thus forget about the constraint forces from now on.
4
Lagrangian Dynamics 2008/09
Lecture 6: The Lagrangian; Lagrangian Methods in Action
Example: We return to the particle on a hoop under gravity
θ a
As shown previously, there is one generalised coordinate, θ, and the generalised force conjugate to this is Qθ = mga sin θ. The particle’s velocity along the hoop is aθ̇, so the kinetic
energy is
1
T (θ, θ̇, t) = ma2 θ̇2
2
for which ∂T /∂θ = 0 and ∂T /∂ θ̇ = ma2 θ̇. Hence Lagrange’s equation of motion is
d
(ma2 θ̇) = mga sin θ
dt
which can easily be checked by other means – do it!
Conservative forces
In an instantaneous virtual displacement
δW =
X
Qj δqj
j
If (as in the example above) the forces are conservative, then there exists a potential energy
function V such that
δW = −δV
We suppose that V = V ({q}, {6 q̇}, t) (where we have allowed V to be explicitly timedependent, but so-called “velocity dependent potentials” will be treated separately later),
in which case this may be written
X
j
Qj δqj = −δV = −
X
j
∂V
δqj
∂qj
Because of our choice of generalised coordinates (a set of 3N − M independent variables)
the increments {δq} are independent. Hence we can vary any one coordinate with the others
constant to obtain
∂V
Qj = −
∂qj
which holds for all j. (Note that this argument would not work for the xi coordinates in a
constrained system.)
1
Accordingly, Lagrange’s equations become
d
dt
∂T
∂ q̇j
!
−
∂T
∂V
=−
∂qj
∂qj
(1)
Also, since V = V ({q}, {6 q̇}, t) we have ∂V /∂ q̇j = 0. This allows us to write
d
dt
where
∂L
∂ q̇j
!
−
∂L
= 0
∂qj
L({q}, {q̇}, t) = T ({q}, {q̇}, t) − V ({q}, t)
L is called the Lagrangian. These equations are the key to the entire course.
L=T −V.
Note the minus sign in front of the V :
Note also the total time derivative in Lagrange’s equations, and be sure you know what it
means, viz:
X ∂f
X ∂f
∂f
df
=
q̇j +
q̈j +
dt
∂t
j ∂qj
j ∂ q̇j
(This should become second nature after enough examples.) Finally, note that since L is
always differentiated to get the equations of motion, any additive constant in the potential
energy V can be neglected.
Technical Note: In the above we have allowed for explicit time-dependence in the potentials
V = V ({q}, {6 q̇}, t). These are not conservative in the usual everyday sense because total
energy is not conserved (see later).
Generalisations: a partial list.
1. Our constraints are holonomic. Any additional nonholonomic constraints would have
to be imposed explicitly after solving Lagrange’s equations. We do not pursue this,
but bear it in mind.
2. Not all forces are conservative, for example friction. We can separate {Q} into a
conservative part (derived from V ) and a non-conservative part. This gives
d
dt
∂L
∂ q̇j
!
−
∂L
= Qnon−con
j
∂qj
In any case, friction is a macroscopic phenomenon, so there is no term like this in
fundamental theories of matter (for which the Lagrangian formulation is well suited).
We do not study friction, but again bear it in mind.
3. If there are velocity-dependent forces, we must return to the most general form of
Lagrange’s equations, viz:
d
dt
∂T
∂ q̇j
!
−
∂T
= Qj ({q}, {q̇}, t)
∂qj
Example: Lorentz force on a charged particle in a magnetic field. We return to this
important topic later on.
2
Lagrangian Methods in Action
The title of this section is a dreadful pun. (See Section IV.)
The above developments are quite formal. The best way of understanding what it all means
is to get plenty of practice with examples. (See Tutorial Sheets.)
The biggest difficulties are:
1. Writing down T and V correctly in terms of a suitable number of generalised coordinates;
2. Avoiding mistakes with partial and total derivatives.
Single Particle in Cartesians: For an unconstrained particle in a potential V (r), we
have in cartesians
3
X
1
L=T −V =
mẋ2i − V ({x})
2
i=1
ie three dofs. Now make the obvious choice, qi = xi . Then Lagrange’s equations are
∂L
∂ ẋj
d
dt
!
−
∂L
∂xj
= 0
d
∂V
(mẋj ) = −
dt
∂xj
⇒
which is the usual Newtonian result. (Note that if m is time-dependent, this is handled
correctly.)
Single Particle in plane polars: Find the equations of motion of a particle moving in a
conservative potential in two dimensions, in polar coordinates.
There are two dofs and we choose generalised coordinates as suggested: q1 = r, q2 = θ.
The radial and tangential components of the particle’s velocity are vr = ṙ, vθ = rθ̇, hence
T = 12 m(ṙ2 + r2 θ̇2 ). The Lagrangian is therefore
1
L = m(ṙ2 + r2 θ̇2 ) − V (r, θ)
2
Lagrange’s equation for q1 = r is
d
dt
⇒
⇒
∂L
∂ ṙ
!
=
∂L
∂r
d
∂V
(mṙ) = mrθ̇2 −
dt
∂r
∂V
m(r̈ − rθ̇2 ) = −
∂r
The term in parentheses on the LHS is the usual expression for the radial acceleration ar ;
the RHS is the radial force component – see, eg, MP2B or FoMP lectures on central forces.
3
For q2 = θ, the Lagrange equation reads
d
dt
!
∂L
∂L
−
= 0
∂θ
∂ θ̇
∂V
d 2 ⇒
mr θ̇ = −
dt
∂θ
And taking the total time derivative (care needed here!) gives
∂V
∂θ
The left hand side is “r times maθ ” where aθ is the tangential acceleration.
m(2rṙθ̇ + r2 θ̈) = −
We have thus obtained correctly the two equations of motion for a 2-D particle in plane polar
coordinates (or a 3-D particle confined to the x−y plane in circular cylindrical coordinates).
Comments:
1. Is it easier than Newton? Even for this very simple problem, perhaps yes. In generalised
coordinates it is usually easier to write down the velocity components and hence the
scalars T and V , and then to differentiate than it is to write down by inspection (or
memory) the correct expressions for the acceleration.
2. Care is needed with differentiation (practice makes perfect).
3. Lagrangian methods help us formulate the equations of motion. We still have to solve
them!
Atwood’s machine
• This is a fancy name for a smooth pulley, a light
inextensible string (a rope!) and two masses M
& m.
• The “frictionless” pulley is attached to a fixed
support. The rope slides without friction on the
pulley.
h
M
z1
m
z2
There is one holonomic constraint: z1 + z2 + l = 2h (l is the fixed length of the string,
and we ignore the size of the pulley). One dof remains. Choose q = z1 as the independent
coordinate. Clearly ż1 + ż2 = 0, hence
1
1
1
1
1
T =
M ż12 + mż22 = M q̇ 2 + m(−q̇)2 = (M + m)q̇ 2
2
2
2
2
2
V
= M gz1 + mgz2 = (M − m)gq + const
Lagrange’s equation is:
d
dt
∂L
∂ q̇
!
=
∂L
∂q
⇒
(M + m)q̈ = −(M − m)g
Note that the constraint force (the string tension) is not mentioned at all.
Exercise: Generalise to compound pulleys (see Tutorial Sheet).
4
Lagrangian Dynamics 2008/09
Lecture 7: Lagrangian Methods in Action (continued)
Atwood’s Monkey
In an Atwood machine, one of the masses is replaced
by a monkey which proceeds to climb upward, relative
to the rope, at a known velocity φ̇(t). Find the motion.
h
This is an example of a time-dependent holonomic constraint
z1 + z2 + l − φ(t) = 2h
z1
m
z2
where h, l are constant and φ(t) is a known function.
The kinetic and potential energies, respectively, are
1
1
M ż12 + mż22
2
2
= M gz1 + mgz2
T =
V
After we impose the constraint
z1 + z2 + l − φ(t) = 2h
there will be one degree of freedom left: choose q = z1 . Differentiating the constraint wrt t
gives ż2 = −ż1 + φ̇. Hence
1
1
L(z1 , ż1 , t) = M ż12 + m(ż1 − φ̇)2 − M gz1 − mg(2h − l + φ(t) − z1 )
2
2
which is explicitly time-dependent. Lagrange’s equation is
d
dt
⇒
∂L
∂ ż1
!
−
∂L
= 0
∂z1
d M ż1 + m(ż1 − φ̇) − mg + M g = 0
dt
⇒
(M + m)z̈1 = −(M − m)g + mφ̈
This is our equation of motion. Integrating twice gives the general solution
M − m gt2
m
z1 (t) = At + B −
+
φ(t)
M +m 2
M +m
with A and B fixed by the initial conditions.
Comments:
1. The tension in the rope is never mentioned.
2. L is explicitly time-dependent.
3. The monkey does work on the system, so energy is not conserved.
1
Particle and wedge: A point particle moves without friction down a wedge that is free to
slide on a frictionless table. Find the acceleration of the wedge.
y
x
v1
m
v2
M
φ
There are two dofs; we are free to choose our generalised coordinates. Let’s take x, y as
shown in the figure (Note: these are not a conventional orthonormal cartesian pair for the
particle; we choose our set to include directly the position of the block, x, which is what
we’re asked to find. This is often a good strategy.) We have two contributions to T :
1
M ẋ2
2
i
1 2
1 h
=
m v1 + v22 = m (ẋ + ẏ cos φ)2 + (ẏ sin φ)2
2
2
Tblock =
Tparticle
(The Lagrange EoMs are not trivial in these coordinates.) For the potential energy we have
= −mgy sin φ + constant
i
1 h
1
M ẋ2 + m ẋ2 + ẏ 2 + 2ẋẏ cos φ + mgy sin φ
L =
2
2
V
Hence
(Obviously, in this equation, φ is a constant, not a generalised coordinate.)
Our first Lagrange equation is
d
dt
so that
∂L
∂ ẋ
!
=
∂L
∂x
( = 0)
d
(M ẋ + m(ẋ + ẏ cos φ)) = (M + m)ẍ + m cos φ ÿ = 0
dt
(1)
The second Lagrange equation is
d
dt
The left hand side is
∂L
∂ ẏ
!
=
∂L
∂y
d
(m [ẏ + ẋ cos φ]) = m(ẍ cos φ + ÿ)
dt
Substituting the expression for ÿ from the first Lagrange equation, (1), into (2) gives
"
#
m+M
∂L
m cos φ −
ẍ =
= mg sin φ
m cos φ
∂y
and therefore the acceleration of the block is
ẍ =
g sin φ
cos φ − (m + M )/(m cos φ)
2
(2)
Simple Pendulum:
θ
This is constrained motion of a particle in cylindrical
polars: we set r = a, the length of the pendulum. One
dof remains; this is the angle θ.
a
The Lagrangian is
g
m
L = T −V
d
dt
∂L
∂ θ̇
!
=
1 2 2
ma θ̇ + mga cos θ + constant
2
=
∂L
∂θ
ma2 θ̈ = − mga sin θ
⇒
which is the (hopefully familiar) equation of motion.
φ
Spherical Pendulum:
Such a pendulum is free to rotate about the vertical
as well as swing in a plane. There is one more dof,
described by the rotation angle φ.
The kinetic energy is
T =
θ
a
1 2
m vθ + vφ2
2
where vθ = aθ̇
vθ
and vφ = a sin θφ̇
Therefore
1 2 2 2 1 2 2
ma sin θ φ̇ + ma θ̇
2
2
= −mga cos θ + constant
T =
V
⇒
1 2 2 2 1 2 2
ma sin θ φ̇ + ma θ̇ + mga cos θ
2
2
!
∂L
∂L
=
∂θ
∂ θ̇
ma2 θ̈ = ma2 sin θ cos θ φ̇2 − mga sin θ
⇒
d
dt
The EoM for φ is:
X
g
L =
d
dt
The EoM for θ is:
a sin θ
m
vφ
∂L
∂ φ̇
!
=
(3)
∂L
∂φ
(4)
The right hand side of the latter is zero, so the EoM for φ is
d 2 2 ma sin θ φ̇ = ma2 sin2 θ φ̈ + 2ma2 sin θ cos θ θ̇φ̇ = 0
dt
(5)
We have two coupled nonlinear differential equations of second order. A direct attack is
pretty difficult. However, conservation laws come to the rescue. We deal with these
generally later on, but in the present example we notice from the φ equation
d
dt
∂L
∂ φ̇
!
=
∂L
=0
∂φ
⇒
3
∂L
= constant
∂ φ̇
Or
ma2 sin2 θ φ̇ = Lz = constant
(6)
This is telling us about conservation of angular momentum (Lz ) about the vertical axis.
Substituting for φ̇ from equation (6) into the EoM for θ (equation (3)) gives
ma2 θ̈ = ma2 sin θ cos θ
Lz
ma2 sin2 θ
Fortunately both sides can be integrated wrt θ (since
Z
2
− mga sin θ
θ̈ dθ =
Z
θ̈
dθ
1
dt = θ̇2 ) to give
dt
2
1 2 2
L2z
ma θ̇ = mga cos θ −
+E
2
2ma2 sin2 θ
where E is an integration constant. Rearranging we have
L2z
1 2 2
E = ma θ̇ +
− mga cos θ
2
2ma2 sin2 θ
where the first two terms are the θ and φ parts of the KE, and the last term is the potential
energy V . So E is the total energy and this equation tells us about conservation of energy.
Conceptual Lessons:
1. Conservation of Lz arises because L is invariant under rotations about the z axis:
∂L
∂L
=0
⇒
= Lz = const.
∂φ
∂ φ̇
We return later to the generic relation between symmetry and conservation laws.
2. Conservation of E: we show later that this arises similarly from
∂L
=0
∂t
which is also a symmetry (symmetry under time-translation).
Problem-Solving Lessons:
1. The explicit integration (done above) to get the conservation of energy was tricky and
unpleasant. In practice, unless asked for proof, one should write down this and any
other applicable conservation laws by inspection where possible. For the above example
we could (and should!) have written directly the conservation laws
ma2 sin2 θ φ̇ = Lz = constant
L2z
1 2 2
ma θ̇ +
− mga cos θ = E = constant
2
2ma2 sin2 θ
With Lz and E fixed by the initial conditions.
2. These are first order DE’s and, as we showed explicitly, are the first integrals of our
two Lagrange’s equations. Generically, each symmetry gives a first integral, which is a
big step on the way to solving the equations. In some cases, like this one, there are as
many symmetries as dofs in which case we may be able to write down a complete set
of first order equations of motion without explicitly mentioning Lagrange’s equation
at all. Such a system is said to be integrable.
4
Lagrangian Dynamics 2008/09
Lecture 8: Rotating Frames; the Calculus of Variations
Rotating Frame: As a final example, we re-derive the equation of motion for a particle in
a rotating frame via the Lagrangian route. The results should agree with those of Lecture 3.
Suppose we have cartesian coordinates r = (x, y, z) in an inertial frame S0 . What we want
is the equation of motion in terms of a new set q = (q1 , q2 , q3 ) with
q1 =
x cos(ωt) + y sin(ωt)
q2 = −x sin(ωt) + y cos(ωt)
q3 = z
which are cartesian coordinates in a rotating frame S. We have chosen the rotation axis to
be z (without loss of generality) and the angular velocity of S to be ω = (0, 0, ω).
z
Inertial frame S 0
y
q2
(Looking down)
(x, y, z)
(x, y, z)
q1
y
ωt
x
x
Clearly T = 12 m|ṙ|2 . To write this correctly in terms of the q’s and q̇’s requires care, but the
required result is given in equation (1) of Lecture 3.
ṙ = [q̇]S + ω × q = ((q̇1 − q2 ω), (q̇2 + q1 ω), q̇3 )
Hence the kinetic energy in the inertial frame S0 may be written as
i
1 h
T = m (q̇1 − q2 ω)2 + (q̇2 + q1 ω)2 + q̇32
2
and the potential is V = V (q, t) – an explicit t dependence is more than likely here if V is
time-independent in the lab frame.
Constructing L = T − V and applying Lagrange’s equations gives, for q1
∂V
d
m(q̇1 − q2 ω) = mω(q̇2 + q1 ω) −
dt
∂q1
Rewriting this, and doing the same for q2 and q3 gives
∂V
∂q1
∂V
−
∂q2
∂V
−
∂q3
mq̈1 = mω 2 q1 + 2mω q̇2 −
mq̈2 = mω 2 q2 − 2mω q̇1
mq̈3 = 0
+
0
These are, componentwise, the correct equations of motion for a particle in the frame rotating
with angular velocity ω = ω e z , as derived earlier (see Lecture 3, pages 2-3). Check this claim!
1
Some Common Mistakes:
1. Failing to specify an appropriate set of generalised coordinates {q} (usually by having
the wrong number of them).
2. Failing to write down L({q}, {q̇}, t) correctly. The main difficulty is usually in getting
T right.
3. Failing to exploit conservation laws (more later).
4. Failing to differentiate correctly to give Lagrange’s equations.
5. Writing the Lagrangian as a function of the wrong variables. Consider, for example, a
particle in one dimension (coordinate x) with T = 21 mẋ2 and a potential V (x). Since
the energy E = T + V = L + 2V is conserved, one might be tempted to write instead
of
L(x, ẋ, t) = 21 mẋ2 − V (x) (right)
L(x, ẋ, t) = E − 2V (x)
(wrong)
where E is taken as a constant. Applying Lagrange’s equation to the latter gives
d0/dt = −2∂V /∂x, which is nonsense.
The reason this fails is that the information we need lies in the functional form of the
Lagrangian, not in its numerical value. To replace it with something else of equal value
in the actual motion, but of a different functional form, is not permitted.
Section IV: The Calculus of Variations
Motivation: So far, we have Lagrange’s equations
d
dt
∂L
∂ q̇j
!
−
∂L
=0
∂qj
j = 1, · · · , 3N − M
where {q} are arbitrary generalised coordinates.
Is there any way of stating the laws of dynamics which is manifestly independent of the choice
of the q’s? The answer is yes, by means of a “variational principle”. The result (Hamilton’s
principle) has many advantages, including a more direct proof of the equivalence of the
Lagrangian and Newtonian formulations.
To study this we need to develop the general theory of “variational calculus”. Since this is
not taught in any detail elsewhere in our (Mathematical) Physics degree programme, we do
it from first principles. Though not in itself part of mechanics, the variational calculus is a
powerful and important part of the physicist’s toolkit.
y (s)
Functions and Functionals
Consider a function y(s). For every s in some range we
associate a value of y. We can represent y(s) as a curve
or path as shown.
Consider now
J=
Z s2
s1
|y(s)|2 ds
2
s1
s2
(1)
This is an example of a functional of the path y(s); we write it J[y(s)] or simply J[y]. The
square brackets remind us that the argument is already a function. J depends on the shape
of the function y throughout its range, not just on its value in a particular place.
Another example:
Z s2
L[y(s)] =
s1

dy
1 +
ds
!2 1/2

ds
(2)
This is the length of the curve y(s). Why? (Convince yourself – we shall use this later.)
The following are not functionals of y(s), but merely functions:
y 0 (s) ≡
sin y(s)
dy(s)
ds
These are merely functions because each depends on the value of y only in some infinitesimal
neighbourhood (near the point s).
Euler’s Equation: We are often interested in functionals of the form
I[y(s)] =
Z b
a
F (y(s), y 0 (s), s) ds
where F is (merely) a function of the variables y, y 0 and s. As far as F is concerned, these
variables are treated as independent.
The examples of functionals in equations (1) and (2) above are of this form, with respectively
F = |y|2 ,
F = [1 + y 02 ]1/2
The calculus of variations involves asking:
What is the path that maximises or minimises the integral I[y(s)] subject to specified values
y(a), y(b) at the end points?
y
To address this we consider the effect of a small change
in path
y(s) → y(s) + η(s)
y (s)
where η(s) is everywhere small and vanishes at the endpoints:
η(a) = η(b) = 0
Consider
I[y(s) + η(s)] =
Taylor expand for small η :
I[y + η] =
Z b
a
Z b
∂F
F (y, y 0 , s) + η(s)
∂y
Defining δI = I[y + η] − I[y] we obtain
Z b"
a
s1 = a
s2 = b
F (y(s) + η(s), y 0 (s) + η 0 (s), s) ds
a

δI =
η (s)
!
∂F
+ η 0 (s)
∂y 0
y0 , s
#
∂F
∂F
η(s)
+ η 0 (s) 0 ds
∂y
∂y
3
!
y, s


ds
where we no longer need subscripts on the partial derivatives (since all other variables are
held constant). The second term is now integrated by parts to give:
δI =
Z b
a
"
∂F
∂F
η(s)
ds + η(s) 0
∂y
∂y
#b
a
−
Z b
a
d
η(s)
ds
∂F
∂y 0
!
ds
The boundary terms vanish because η is zero at the endpoints. Thus
δI =
Z b"
∂F
a
d
−
∂y
ds
∂F
∂y 0
!#
η(s) ds
For I[y] to be stationary as required, δI must vanish for any arbitrary small change η(s).
This is only possible if the square bracket vanishes identically. Therefore
∂F
d
−
∂y
ds
∂F
∂y 0
!
=0
This is Euler’s equation. Note the striking resemblance to Lagrange’s equations!
First Integrals of Euler’s Equation
d
ds
∂F
∂y 0
!
∂F
∂y
=
The obvious one: If F (y, y 0 , s) does not depend explicitly on y, so that ∂F/∂y = 0, then
∂F
= constant (independent of s)
∂y 0
The less-obvious one: If F (y, y 0 , s) does not depend explicitly on s, so that ∂F/∂s = 0,
then
∂F
y 0 0 − F = constant (independent of s)
∂y
Proof: The total derivative of F (y, y 0 , s) with respect to s is:
dF
ds
=
∂F 0 ∂F 00 ∂F
y + 0y +
∂y
∂y
∂s
=
"
d
ds
d
=
ds
∂F
∂y 0
!#
y0 +
∂F 00 ∂F
y +
∂y 0
∂s
(using Euler’s equation)
!
∂F
∂F 0
y +
0
∂y
∂s
which we may rewrite as
d
∂F
y0 0 − F
ds
∂y
!
= −
∂F
∂s
So if ∂F/∂s = 0 as assumed, the left hand side is also zero. This proves the result.
This first integral is very important but much less obvious.
4
Lagrangian Dynamics 2008/09
Lecture 9: Simple Applications of Variational Calculus; Hamilton’s Principle
Examples
To fix these ideas we do some examples. These do not directly involve Lagrangian mechanics,
but they are an important part of the course.
y
A Theorem of Euclid:
(x 1 , y1)
Find the shortest path between (x1 , y1 ) and
(x2 , y2 ) in the plane.
y (x)
(x 2 , y2)
x
Let the path be described by a function y(x). Its length is
L=
Z x2
x1

dy
1 +
dx
!2 1/2

dx
which is of the required form (setting y 0 ≡ dy/dx) with
F (y, y 0 , x) = [1 + y 02 ]1/2
Hence from Euler’s equation
d
dx
∂F
∂y 0
!
d
=
dx
y0
[1 + y 02 ]1/2
!
=
∂F
=0
∂y
F is independent of y, so we can use the “obvious” first integral to obtain
√
y0
= constant ≡ p
1 + y 02
⇒
y0 = √
p
= constant, ie independent of x
1 − p2
Hence the shortest path is a straight line. Obvious! But note the method !
Brachistochrone
(x 1 , y1 )
y
A smooth wire connects (x1 , y1 ) to (x2 , y2 ). Find
ψ (x)
the path y(x) such that a bead sliding down the
wire (starting from rest) does so in the least possiy (x)
ble time.
(x 2 , y2)
This is the problem for which Euler invented the
variational calculus. Brachistochrone is Greek for
x
“shortest time”.
We need to minimise the transit time, τ [y(x)]. At height y(x) the speed v along the curve
is determined by conservation of energy:
1
0 + mgy1 = mv 2 + mgy(x)
2
1
⇒
v=
q
2g(y1 − y(x))
The time dτ required to fall an increment of arc-length ds = (1 + y 02 )1/2 dx along the curve
is dτ = ds/v, hence the transit time is
τ [y] =
where
Z x2
x1
Z x2
ds
F (y, y 0 , x) dx
=
v
x1
1 + y 02
2g(y1 − y)
0
F (y, y , x) =
!1/2
[Here y1 is just a constant.]
Now, this F has no explicit x dependence, hence, from the non-obvious first integral:
y0
∂F
− F = constant
∂y 0
For this form of F , we have (do the algebra for yourself):
y0
⇒
F y 02
∂F
=
∂y 0
1 + y 02
q
⇒
1
2g(y1 − y) (1 +
y 02 )
y0
∂F
− F = − F/(1 + y 02 ) = constant
∂y 0
= constant
(y1 − y)(1 + y 02 ) = constant
⇒
As far as variational calculus is concerned, this is the “answer”: an explicit differential
equation for the required curve. Euler recognised it as the equation of a cycloid :
x = a(2ψ − sin(2ψ)) + x1
y = a(cos(2ψ) − 1) + y1
where ψ is the angle of the tangent of the curve to the upward vertical, and a depends on
the coordinates x2 , y2 of the other end of the wire. The cycloid starts off vertical at (x1 , y1 )
and arrives at (x2 , y2 ) at a finite angle; this gives a large acceleration at the beginning, but
not too long a curve.
Soap film: Two wire rings of radius a, are held at separation 2d. What axially symmetric
shape ρ(z) minimises the surface area A of the soap film?
ds
ρ (z)
z
d
a
dz
d
We have
A[ρ(z)] =
Z d
z=−d
2πρ(z) ds
where
2
h
ds = 1 + (dρ/dz)2
i1/2
dz
Hence A = 2π
Z d
−d
F (ρ, ρ0 , z) dz where
F = ρ(1 + ρ02 )1/2
This has no explicit z dependence, hence we have the first integral
∂F
− F = constant ≡ C
∂ρ0
ρρ0
− ρ(1 + ρ02 )1/2 = C
ρ0
(1 + ρ02 )1/2
ρ0
⇒
which simplifies to (check this!)
ρ = −C(1 + ρ02 )1/2
⇒
dρ
=
dz
s
ρ
C
2
−1
The solution of this differential equation is (exercise, use separation of variables)
ρ(z) = C cosh((z − z0 )/C)
where z0 is a constant of integration. Applying the boundary condition that ρ(±d) = a gives
z0 = 0 and
a/C = cosh(d/C)
Technical Note: This has solutions only for d/a < 0.663 . . . (You can check this graphically.) If the rings are moved further apart than this, the soap film makes in theory an
infinitely thin tube along the axis connecting two flat discs. (In practice, it breaks).
Constrained Variation: This topic is mentioned for completeness; it’s not particularly
difficult, but it currently lies outside our syllabus.
Often one would like to minimise some functional I[y] subject to a constraint that some
other functional J[y] of the same curve is held constant. For example, an inextensible chain,
mass m per unit length, is held under gravity with its ends a certain distance
apart – how
Z
does it hang? Here one wants to minimise the gravitational PE, I[y(x)] = −mgy ds with
the constraint of fixed length, J[y(x)] =
Z
ds = L.
In general this involves a version of the method of Lagrange multipliers: the procedure is to
minimise I − λJ with λ an undetermined multiplier. This gives an Euler equation involving
λ, whose value is finally fixed by imposing the constraint itself. For examples, see text books.
Calculus of Variations for Many Variables: We now consider
I[{y(s)}] =
Z b
a
F ({y}, {y 0 }, s) ds
where {y} = y1 (s), y2 (s), · · · , yp (s) is a set of p linearly independent functions of the (single)
variable s.
3
For a small variation in the paths {y(s)} → {y(s)} + {η(s)} we can expand
F ({y + η}, {y 0 + η 0 }, s) = F ({y}, {y 0 }, s) +
p
X
p
X
∂F
∂F 0
ηj (s) +
η (s)
0 j
j=1 ∂yj
j=1 ∂yj
Demanding δI = 0, and following exactly the derivation in the one dimensional case (and
using the fact that all the variations {η} must vanish at the endpoints) one obtains a set of
p equations, the Euler-Lagrange equations
d
ds
∂F
∂yj0
!
−
∂F
=0
∂yj
j = 1, · · · , p
The details are left as an exercise (see Tutorial Sheet).
First Integrals of the Euler-Lagrange Equations
The obvious ones: If F does not depend explicitly on a particular variable yj
d
ds
∂F
∂yj0
!
∂F
= const.
∂yj0
⇒
=0
This is identical to the single-variable case. There is one such equation for each variable that
F does not depend on.
The less obvious one: Now consider the case when F does not depend explicitly on s.
Not surprisingly this leads to only one first integral (not one for each of our p independent
functions {y}). The result, whose proof is left as an exercise (see Tutorial Sheet), is
X
∂F
yj0 0 − F = const.
∂yj
j
Hamilton’s Principle
The above results show that Lagrange’s equations for dynamics, namely
d
dt
∂L
∂ q̇j
!
−
∂L
=0
∂qj
j = 1 · · · , 3N − M
are neither more nor less than the Euler-Lagrange equations found by demanding that the
following integral is stationary:
S[{q(t)}] =
Z t2
t1
L({q}, {q̇}, t) dt
To make the correspondence, we just switch notation from {y(s)} to {q(t)} (bearing in mind
that dq/dt is conventionally written q̇ rather than q 0 ).
Hence Lagrange’s equations are equivalent to Hamilton’s Principle:
The quantity S =
Z t2
t1
δS = 0
L dt is usually called the Action.
Hamilton’s principle can be used as an alternative starting point for mechanics. As a
postulate, it is equivalent to those of Newton or Lagrange, as we shall show in the next
lecture.
4
Lagrangian Dynamics 2008/09
Lecture 10: Hamilton’s Principle; Conservation Laws, the Energy Function
In the last lecture we demonstrated that Lagrange’s equations are equivalent to Hamilton’s
Principle:
δS = 0
ie the Action S[{q(t)}] =
Z t2
t1
L({q}, {q̇}, t) dt is stationary under small variations.
Technical Note: Hamilton’s principle involves choosing two fixed times t1 and t2 , and
(since the end points of the paths are not allowed to vary) specifying the initial and the final
coordinates, {q(t1 )} and {q(t2 )}. It is formally under these conditions that S is stationary.
This is a legitimate, but unusual choice of boundary conditions in a system with one second
order DE for each degree of freedom (which is the case in classical mechanics). More normal
would be to specify the initial coordinates and velocities, {q(t1 )} and {q̇(t1 )}. Fortunately,
starting from Hamilton’s principle, we can derive Lagrange’s equations and then solve them
for any set of boundary conditions we choose. Hence Hamilton’s principle is true of any
possible motion of the system (ie any that obeys Lagrange’s equations of motion), between
any pair of times (t1 , t2 ), regardless of what boundary conditions are, in practice, applied.
Coordinate Invariance: Hamilton’s principle states that
δ
Z
L dt = 0
where L = T − V . The choice of generalised coordinates {q} need not be specified. Hence
Hamilton’s principle is a coordinate-invariant statement of the laws of mechanics. As
such it is very powerful. For example it allows, for conservative systems, a much nicer proof
of the equivalence of Newtonian and Lagrangian schemes, as follows.
From Newton to Lagrange via Hamilton’s Principle
1. Consider first a system of N particles with no constraints (3N dofs). The kinetic
energy is (in cartesians)
X1
mi ẋ2i
T =
2
i
Observe that ∂T /∂ ẋi = mi ẋi and that ∂T /∂xi = 0. Using the first of these, Newton’s
equations may be written
!
∂V
d ∂T
=−
dt ∂ ẋi
∂xi
where V = V ({x}, {6 ẋ}, t) is a potential (which obeys ∂V /∂ ẋi = 0).
We are now free to write Newton’s equations (a bit perversely) as follows:
d
dt
∂T
∂ ẋi
!
∂V
∂T
d
=−
+
+
∂xi ∂xi dt
∂V
∂ ẋi
!
where we have added two terms to the RHS, both of which are zero.
1
Defining L = T − V , we thereby obtain Lagrange’s equations in cartesians for our 3N
variables:
!
d ∂L
∂L
−
=0
dt ∂ ẋi
∂xi
2. We construct the action, S =
Z
L({x}, {ẋ}, t) dt, and deduce that δS = 0 in cartesians.
However, δS = 0 is a coordinate-invariant statement and therefore holds for any choice
of generalised coordinates, say {q̃}. Since we have not yet applied any constraints,
there must be 3N of these coordinates at this stage.
3. Now we consider the constraints, of which there are M . As usual we take these to be
holonomic, in which case they can be expressed (in cartesians) as a set of M algebraic
equalities between the coordinates:
where α = 1, · · · , M .
fα ({x}, t) = 0
4. (The clever bit.) We are free now to specify our choice of 3N coordinates {q̃} =
(q̃1 , q̃2 , · · · , q̃3N ) as we wish. We make the following choice
{q̃} = (q1 , q2 , · · · , q3N −M , f1 , f2 , · · · , fM ) = {q, f }
In other words, we choose M of our coordinates to be precisely those functions that
are held constant by the constraints. What remains is a set of 3N − M generalised
coordinates {q} which are linearly independent of each other and of the constraints.
5. Hamilton’s principle states that, in an unconstrained motion, δS = 0 or
δ
Z
L({q, f }, {q̇, f˙}, t) dt = 0
To describe a constrained motion, we now demand that all the f ’s (and therefore the
f˙’s) are zero identically. Accordingly these “variables” can just be dropped.
Indeed, defining L({q}, {q̇}, t) = L({q, 0}, {q̇, 0}, t), we are left with
δ
Z
L({q}, {q̇}, t) dt = 0
where {q} are our set of 3N − M of independent generalised coordinates (one for each
dof that remains). Writing the Euler-Lagrange equations for this variational principle
gives Lagrange’s equations of motion for the {q} variables in their usual form.
Lesson: This shows the conceptual power of the variational formulation of mechanics. To
most people, the above proof is more appealing than the one starting from the “cancellation
of dots rule” of Lecture 5 though it is slightly less general (it doesn’t include forces, such as
friction, which can’t be written in terms of derivatives of a potential).
Technical Note: In Step 5 we could, if desired, impose the constraints explicitly, by adding
P
to the Lagrangian a term V constraint = 21 α kα fα2 . Constraint forces will then appear in the
Lagrange equations for the {f } variables in such a way that, if we let each kα → ∞ (see
Lecture 4) the constraints will take hold and fα → 0. The trick here is that, by choosing
our generalised coordinates appropriately, we have arranged that V constraint has no effect on
the Lagrange equations for any of the remaining {q} variables.
2
Section V: Lagrangian Dynamics: Symmetries & Conversation Laws
In this section we shall explore the general implications of Lagrangian Mechanics, and focus
on symmetry, invariance properties, conservations laws, etc. We shall also introduce the
energy function h and explain its significance. Our starting point is Lagrange’s equations
d
dt
∂L
∂ q̇j
!
−
∂L
=0
∂qj
or, equivalently, Hamilton’s principle:
j = 1, · · · , 3N − M
δS = δ
Z
L dt = 0
Invariance Properties of the Lagrangian
Changes of Variable: Consider two sets of generalised coordinates, {q} and {q 0 }. The
Lagrangians are numerically equal, but are different functions of their arguments
L({q}, {q̇}, t) = T − V = L0 ({q 0 }, {q̇ 0 }, t)
It is a worthwhile mathematical exercise to prove (starting from the cancellation of dots
rule) that in general
d
dt
∂L
∂ q̇j
!
"
∂L X d
−
=
∂qj
dt
i
∂L0
∂ q̇i0
!
#
∂L0 ∂qi0
− 0
∂qi ∂qj
If the LHS = 0, then [· · ·] = 0 ∀i. Hence, if Lagrange’s equations hold for one set of
coordinates, they do for any other. Happily, the algebraic proof can be entirely bypassed by
invoking Hamilton’s principle, whose coordinate invariance guarantees the result.
Changes of Lagrangian: If two Lagrangians L1 and L2 are related as follows
d
F ({q}, {6 q̇}, t)
dt
then L1 and L2 are equivalent Lagrangians, ie they give the same equations of motion. There
are two proofs (direct and via Hamilton’s principle) – see Tutorial Sheet .
L1 ({q}, {q̇}, t) = L2 ({q}, {q̇}, t) +
Generalised Momenta: The generalised momentum pj conjugate to a generalised coordinate qj is defined by
∂L
pj ≡
∂ q̇j
so that Lagrange’s equations may be written
dpj
∂L
=
dt
∂qj
The quantity pj is also called the canonical momentum conjugate to qj . This usually has
an obvious interpretation, eg for particle(s) in cartesians {x}
L = T −V
=
1X
mi ẋ2i − V ({x}, {6 ẋ}, t)
2 i
⇒ pi = mi ẋi = the ith component of linear momentum
3
Similarly, if qj is an angle, T contains a term 12 Ij q̇j2 , and if there is no other q̇j dependence
in L, then
pj = Ij q̇j
which is the ith component of the usual angular momentum.
Technical Note: Things are more complicated if there are velocity-dependent forces. In
this case there is an important distinction between the full canonical momentum defined
above and the naive mechanical momentum. We return to this issue later on.
Momentum Conservation: If L({q}, {q̇}, t) does not depend explicitly on a certain qj ,
then
∂L
dpj
=
= 0
⇒
pj is conserved
dt
∂qj
The corresponding qj is called a cyclic or ignorable coordinate. A conserved momentum
(or other conserved quantity) is called a constant of the motion. In the language of Lecture 8,
a conserved momentum corresponds to an “obvious” first integral.
The statement that ∂L/∂qj = 0 can be viewed as a symmetry property of L:
L(q1 , · · · , qj , · · · , q3N −M , {q̇}, t) = L(q1 , · · · , qj + λ, · · · , q3N −M , {q̇}, t)
In other words, L is invariant under changes in qj . For the examples above, these changes
correspond to a translation (xi → xi + λ) or, for an angle, a rotation.
The energy function: We now define the energy function
X ∂L
h({q}, {q̇}, t) =
q̇j − L({q}, {q̇}, t)
j ∂ q̇j
Or equivalently:
h =
X
pj q̇j − L
j
Conservation of h: If the Lagrangian has no explicit time dependence,
dh
= 0
so h is conserved
L = L({q}, {q̇}, 6 t)
⇒
dt
ie h is a constant of the motion. This is an example of the “less obvious” first integral of
the Euler-Lagrange equations (see Lecture 8), whose proof is on the Tutorial Sheet. Hence
we obtain one of the most important results of the whole course:
Time Translational Symmetry of L
⇒
Conservation of h
The energy function h is a generalisation of the notion of energy. For many (but not all)
conservative systems we will show that h is identical to the total energy E = T + V .
Technical Note: h is numerically equal to the Hamiltonian (see later). There one expresses
the velocities in terms of the canonical momenta and coordinates, so that the Hamiltonian
becomes a function of the coordinates and their conjugate momenta:
H({q}, {p}, t) =
X
j
pj q̇j − L({q}, {q̇}, t)
Beware: Prior to 2006/07, the energy function h was called the Hamiltonian function,
denoted by H. Both names can be confusing – we now use the more conventional one.
4
Lagrangian Dynamics 2008/09
Lecture 11: More on the Energy Function, Conservation Laws and Symmetry.
The energy function and the total energy
For many (but not all) conservative systems, the kinetic energy is quadratic in the velocities,
so that L = T ({q}, {q̇}, t) − V ({q}, {6 q̇}, t) with
1X
T ({q}, {q̇}, t) =
Tij ({q}, {6 q̇}, t) q˙i q˙j
2 ij
where Tij is a symmetric matrix whose entries can depend on the coordinates {q}, and on
time, but not the velocities. An obvious example is a set of particles in cartesians for which
P
T = 12 i mi ẋ2i which is of the required form (with a diagonal, constant matrix Tij = mi δij ).
For quadratic T it follows that the momentum pi conjugate to the coordinate qi is
X
∂T
=
pi =
Tij q̇j
∂ q̇i
j
⇒
h =
X
i
so that h = E
pi q̇i − L =
X
ij
Tij q̇i q̇j − T + V = 2T − T + V = T + V
where we define E as the total energy E ≡ T + V as usual.
For these systems, time-translational symmetry of L results in conservation of the total
energy E. However, not all systems have T of the required form: in particular this will
not happen if there are time-dependent constraints (example below). Hence in the general
case, it is the energy function h, rather than the total energy E which is conserved as a
consequence of time-translational symmetry in L. The distinction is important!
Distinguishing between E and h – via two examples
(1) Particle on a rod
A particle slides on a massless rod which is pivoted
at one end and constrained to rotate in a plane. The
potential V (r) has rotational symmetry about the pivot.
Find two constants of the motion.
This is really just a particle in (cylindrical) polars. We
have two degrees of freedom, r & θ. The Lagrangian is
1
L = T − V = m(ṙ2 + r2 θ̇2 ) − V (r)
2
.
vθ = r θ
m
r
θ
The generalised momenta which are canonically conjugate to r and θ are
∂L
pr =
= mṙ
= radial component of linear momentum
∂ ṙ
∂L
pθ =
= mr2 θ̇ = angular momentum about the rotation axis
∂ θ̇
What are the conservation laws for this case?
1
.
vr = r
1. L does not depend explictly on θ (rotational symmetry) hence pθ is conserved. This is
conservation of angular momentum about an axis of symmetry.
2. L does not depend explicitly on t. Hence
h = pr ṙ + pθ θ̇ − L
is conserved. Since T is quadratic in the velocities, this coincides with conservation of
the total energy E = 12 m(ṙ2 + r2 θ̇2 ) + V (r).
Particle on a Rod in Forced Motion:
The same system is now forced to rotate at
fixed angular velocity θ̇ = ω. Find a constant of the motion.
We now have a time-dependent constraint. There is only one dof left, namely r. We can
still write
1
L = T − V = m(ṙ2 + r2 θ̇2 ) − V (r)
2
except that the angle θ is no longer a dof, ie no longer one of the q’s. It obeys the holonomic
constraint θ = ωt + constant. To emphasise this it is much better to write
1
L(r, ṙ, t) = m(ṙ2 + r2 ω 2 ) − V (r)
2
where ω is more obviously constant. Note that T is no longer quadratic in the velocities –
the second term in T is not dependent on any generalised velocity.
The canonical momentum is
pr =
∂L
= mṙ
∂ ṙ
as usual. However, we note that
∂L
∂V
= mrω 2 −
∂r
∂r
so that pr is not conserved (even when the potential V = 0.) The first term is easily identified
as the centrifugal force due to the forced rotation.
On the other hand, L still has no explicit t dependence, ∂L/∂t = 0. (This is because the
forced rotation rate ω is constant in time.) Thus the energy function
1
h = pr ṙ − L = m(ṙ2 − r2 ω 2 ) + V (r)
2
is a constant of the motion. Evidently h is not the total energy
1
E = T + V = m(ṙ2 + r2 ω 2 ) + V (r)
2
Indeed, since r is not a constant, E is not a conserved quantity.
Discussion: To understand why E is not conserved,
consider the constraint force F constr (say) which keeps the
bead on the rod.
The tangential acceleration is (in cylindrical polars)
aθ = rθ̈+2ṙθ̇ ⇒ F constr = maθ = 2mṙω (since θ̈ = ω̇ = 0)
2
vθ = r ω
F constr
ωt
This constraint force acts perpendicular to the wire, in the direction of the velocity vθ = rω.
Hence the constraint force does work on the bead at a rate
d
F constr vθ = 2mω 2 rṙ = (mr2 ω 2 )
dt
This is just
d
dE
(E − h) =
dt
dt
(since h is conserved) which precisely accounts for the change in E.
Technical Query: But didn’t we say earlier in Lecture 4 that constraint forces do no
work? Not quite, we said that they do no work in any virtual displacement consistent with
the constraints. The only virtual (that is, instantaneous) displacement possible in this system
is for the bead to move along r, which is perpendicular to F constr and hence no work is done.
This does not rule out constraint forces doing work in a non-instantaneous change, including
the actual motion, as the above example shows.
More on Symmetries and Conservation Laws
We now go back to first principles. Consider an isolated system of N particles in cartesians
r a = (xa , ya , za ), where a = 1, · · · , N . (This explicit notation is more suited to our current purposes than the {x} notation.) We now prove the conservation of the total linear
momentum and total angular momentum for this system.
Homogeneity of Space: We demand that space is homogeneous, ie it’s the same ‘everywhere’. If all particles are subjected to the same small displacement (keeping the velocities
ṙ a fixed):
ra → ra + then L is invariant. This is symmetry under translations of the entire system – a bit different
from varying one coordinate on its own. If so, we must have (using Taylor’s theorem to
expand ∆L):
!
∆L =
X
a
∇ (r ) L · ∆r a =
a
X
a
∇ (r ) L · = 0
a
where ∇ (r ) L is the gradient of L wrt r a , ie its components are (∂L/∂xa , ∂L/∂ya , ∂L/∂za )
a
and the dot product is the usual three-dimensional one. Since is arbitrary, we have
X
a
∇ (r ) L
a
= 0
Writing Lagrange’s equations in the form
dp a
= ∇ (r ) L ,
a
dt
summing over all particles a, and using equation (1) we obtain the result
X
a
dp a
dP
=
=0
dt
dt
This is the conservation of P , the total momentum of our system of particles.
Homogeneity of space ⇒ Conservation of total momentum
3
(1)
Isotropy of space: We now demand that space is isotropic: ie, that L is invariant
under rotations of the entire system of N particles about an arbitrary axis. For a rotation
b through a small angle Φ about an axis Φ,
b we have
Φ ≡ ΦΦ
∆r a = Φ × r a
∆ṙ a = Φ × ṙ a
which are the usual transformation laws for vectors under infinitesimal rotation. Now
∆L =
X
a
∇ (r ) L · ∆r a +
a
X
a
∇ (ṙ ) L · ∆ṙ a = 0
a
where ∇ (ṙ ) L has components (∂L/∂ ẋa , ∂L/∂ ẏa , ∂L/∂ ża ).
a
Substituting for ∇ (r ) L via Lagrange’s equations, and for ∆r a & ∆ṙ a , we obtain
a
X
a
X
d
∇ (ṙ ) L · (Φ × r a ) +
∇ (ṙ ) L · (Φ × ṙ a ) = 0
a
a
dt
a
Using the result that A · (B × C) = (C × A) · B we get
"
X
a
d
ra ×
∇ (ṙ ) L
a
dt
or
"
X
+
a
ṙ a × ∇ (ṙ ) L
a
!#
d X
r × ∇ (ṙ ) L
a
dt a a
#
·Φ = 0
·Φ = 0
This holds for any small rotation Φ, hence the term in square brackets vanishes identically.
Hence
X
X
=
r a × p a ≡ L = constant
r a × ∇ (ṙ ) L
a
a
a
Isotropy of space ⇒ Conservation of total angular momentum
The above proofs make no use of Newton’s third law in either form! The third law is thereby
exposed as a red herring.
Symmetries and Conservation Laws: Summary
L invariant under:
qj → qj + λ
Conserved Quantity
pj
P
t→t+λ
h=
Global Translations
P
Global Rotations
L
Poincaré Transformations
Energy-momentum Tensor
Gauge Transformations
Electric charge
j
pj q̇j − L
In the last two examples we have strayed rather beyond the course boundaries. . .
4
Lagrangian Dynamics 2008/09
Lecture 12: Velocity-dependent Forces, Lorentz Force, Special Relativity
Section VI: From Lagrange to Hamilton, via EM & Relativity
So far we have concentrated on Lagrange’s equations in their usual form,
d
dt
!
∂L
∂L
−
=0
∂ q̇i
∂qi
These were derived as a special case of the most general form
d
dt
∂T
∂ q̇i
!
−
∂T
= Qi
∂qi
(where Qi is the generalised force conjugate to qi ) by assuming there exists a potential
V = V ({q}, {6 q̇}, t) for which
∂V
Qi = −
∂qi
Velocity-dependent forces
In general, if velocity-dependent forces are present, we must return to the most general form
and put in generalised forces
Qi ({q}, {q̇}, t)
in which there will now be an explicit dependence on the velocities. Suppose, however, we
can find a function Ṽ ({q}, {q̇}, t) such that
d
Qi =
dt
∂ Ṽ
∂ q̇i
!
−
∂ Ṽ
∂qi
If so, by construction we have the usual Lagrange equations
d
dt
!
∂L
∂L
−
=0
∂ q̇i
∂qi
but now with L = T − Ṽ . Clearly, Ṽ is not a “conventional” potential energy. However, the
formal structure of the theory, including Hamilton’s principle and any relevant conservation
laws, carries over.
Lorentz Force: The construction above works for a very important special case: a particle
of charge e in an EM field:
mr̈ = F = e(E + ṙ × B)
where E & B are the electric and magnetic fields respectively.
The E & B fields can both be written as derivatives of potentials as follows
E = −∇φ −
∂A
∂t
and
1
B =∇×A
Here φ is the electrostatic potential, and A is the magnetic vector potential. Since this is not
a course on electromagnetism, we take these results on trust (see EM lectures or handout).
We shall show explicitly that the following function
Ṽ (r, ṙ, t) = e [φ(r, t) − ṙ · A(r, t)]
does exactly the job we require.
Proof: The derivation is a little complicated, but the result is sufficiently important to
include it here. We will work in Cartesians using suffix notation, r = xi e i (i = 1, 2, 3), and
(for this section only) use the summation convention. We also need the standard result
ijk klm = δil δjm − δim δjl
where ijk is the usual Levi-Civita or epsilon symbol (so (a × b)i = ijk aj bk ).
Writing E & B in terms of the potentials, the ith component of the Lorentz force becomes
!
∂A
Fi = e −∇φ −
+ ṙ × (∇ × A)
∂t
i
The last term in parentheses may be simplified:
(ṙ × (∇ × A))i = ijk ẋj (∇ × A)k = ijk ẋj klm
∂Am
∂xl
∂Am
= (δil δjm − δim δjl ) ẋj
= ẋj
∂xl
so the Lorentz force becomes
"
∂Ai
∂φ
−
+ ẋj
Fi = e −
∂xi
∂t
∂Ai
∂Aj
−
∂xi
∂xj
∂Aj
∂Ai
−
∂xi
∂xj
!
!#
(1)
Our task is to identify a function Ṽ (r, ṙ, t) such that
d
Fi =
dt
∂ Ṽ
∂ ẋi
!
−
∂ Ṽ
∂xi
The first and third terms in equation (1) may be written
"
#
∂φ
∂
+
(ẋj Aj )
e −
∂xi ∂xi
= −
∂ Ṽ
∂xi
with
Ṽ ≡ e (φ(r, t) − ẋj Aj (r, t))
(2)
Now for the less-obvious bit. Recalling that r(t) is the position of a moving particle, so that
Ai (r, t) really means Ai (r(t), t), we may verify that
d
dt
∂ Ṽ
∂ ẋi
!
"
d ∂
= e
φ(r, t) − ẋj Aj (r, t)
dt ∂ ẋi
d
= e
0 − Ai (r, t)
dt
"
#
∂Ai ∂Ai
= −e
+
ẋj
∂t
∂xj
#
which we identify with the second and fourth terms in equation (1). We have thus shown
that Ṽ as defined in equation (2) has the desired property that:
!
∂A
e −∇φ −
+ ṙ × (∇ × A)
∂t
i
d
=
dt
∂ Ṽ
∂ ẋi
This completes the proof.
2
!
−
∂ Ṽ
∂xi
with
Ṽ = e (φ − ẋj Aj )
Mechanical Momentum versus Canonical Momentum
We have shown that the Lagrangian of a charged particle in an EM field may be written
L(r, ṙ, t) = T − Ṽ =
1
m|ṙ|2 − e[φ(r, t) − ṙ · A(r, t)]
2
Writing this out explicitly in components (i = 1, 2, 3) gives
"
X
1 X 2
L= m
ẋi − e φ −
ẋi Ai
2
i
i
#
The j th component of the canonical momentum is, by definition:
pj ≡
∂L
= mẋj + eAj
∂ ẋj
Because of the presence of velocity-dependent forces, the canonical momentum
p = mv + eA
is not the same as the mechanical momentum of the particle alone
p mech = mv
It is essential to remember that the canonical momentum is the one that matters in conservation laws, etc.
Energy Function: For the particle in an EM field
h = p · ṙ − L = (mv + eA) · v −
⇒
h =
1 2
mv + eφ(r, t)
2
1 2
mv − e(φ − v · A)
2
This is conserved only if L has no explicit time dependence, which requires that both φ and
A are not explicitly time dependent (which in turn implies E & B fields which don’t have
explicit time dependence.)
Technical Note: Under such conditions, h can be viewed as the sum of the kinetic and
electrostatic potential energies. This is because the magnetic force v × B is perpendicular
to the velocity and therefore does no work. If B is time dependent, the latter is still true,
but now E = −∇φ − ∂A/∂t, which includes a part that does not derive from φ.
Example: A particle moves in a steady “toroidal” B field for which φ = 0 and
A = (0, 0, Az ) = (0, 0, B(x2 + y 2 )1/2 ) = (0, 0, Bρ)
with ρ = (x2 + y 2 )1/2 .
Find two constants of the motion.
Though we don’t need it here, the magnetic field takes the form
!
!
y
−x
y
x
B = ∇ × A = (Bx , By , Bz ) = B
, 2
,0 =B
,− ,0
2
2
1/2
2
1/2
(x + y )
(x + y )
ρ
ρ
3
so the magnetic field has constant magnitude B. The field lines are circles in the x−y plane:
z
A
y
x
B
The Lagrangian is
Since Az = B(x2 + y 2 )1/2
1
L = m(ẋ2 + ẏ 2 + ż 2 ) + eżAz
2
has no explicit z dependence, then nor does L, and so
∂L
= mż + eB(x2 + y 2 )1/2
∂ ż
is conserved. This is our first constant of the motion.
pz =
On the other hand, there is explicit x and y dependence (via Az ) so px and py are not
conserved.
But since L has no explicit time dependence h = 21 m(ẋ2 + ẏ 2 + ż 2 ) + 0 is conserved (as
discussed above); this is the second constant of the motion.
Special Relativity
Lagrangian methods can be used in relativity. We have time for a brief treatment of this
important topic here.
Lagrangian for a Relativistic Particle
Suppose we postulate the following Lagrangian for a particle of mass m moving with velocity
u in a potential V (r)
L = − mc2 [1 − u2 /c2 ]1/2 − V (r) = −
1
mc2 − V (r)
γ(u)
with u2 = ẋ2 + ẏ 2 + ż 2 . Then
px =
or in vector notation
∂L
mẋ
=
= γ(u)mẋ
∂ ẋ
[1 − u2 /c2 ]1/2
p = γ(u) mu
This is the correct relativistic expression for momentum. Lagrange’s equations are
ṗ = ∇L = −∇V
which is the expected result. Since this Lagrangian gives the correct equations of motion, it
is an acceptable choice for a relativistic particle.
4
Lagrangian Dynamics 2008/09
Lecture 13: Special Relativity, Hamiltonian Dynamics, Quantum Mechanics
Charged Relativistic Particle in an EM Field: We want a Lagrangian that gives the
equation of motion
dp
= F Lorentz = e(E + u × B)
dt
with p = γ(u) mu. We have already shown that the right hand side can be derived from Ṽ .
Hence L is simply
1
L=−
mc2 − e[φ − u · A]
(1)
γ(u)
This gives the correct equations of motion and is therefore a suitable Lagrangian.
Covariant form: For a free particle, we may rewrite
L = −mc2 [1 − u2 /c2 ]1/2 = −mc2
dτ
dt
(2)
where dτ is the proper time interval defined by dt = γ(u)dτ .
µ
Recall from Dynamics & Relativity the four-velocity uµ = dx
= γ(u)(c, u), and define the
dτ
µ
electromagnetic four-vector potential A = (φ/c, A). Relativity then tells us the form of
the Lagrangian because the only Lorentz-invariant quantity we can construct from these two
four-vectors is
dt
(3)
uµ Aµ = γ(u)(φ − u · A) =
(φ − u · A)
dτ
Hence, from equations (1), (2) and (3), the action S may be written in the simple form
S =
Z
L dt =
Z
2 dτ
dτ
−mc
− eu Aµ
dt
dt
µ
!
dt = −
Z mc2 + eAµ uµ dτ
where the integral is along the particle’s
world line. S is manifestly Lorentz invariant. For
R
2
a free particle we have S = −mc dτ , which is just −mc2 × the “length” of the world line.
Because relativity can be cast as a Lagrangian theory, the fundamental connections between
symmetries of the Lagrangian and conservation laws still apply.
Hamiltonian Dynamics – from Lagrange to Hamilton
We only have time for an elementary introduction to Hamiltonian Dynamics. However, this
should be sufficient to give a flavour of its formulation and to illustrate its connection to
quantum mechanics. For a full treatment, see the year 4/5 course Hamilton Dynamics.
For a general system with 3N −M dofs, we defined the energy function
h({q}, {q̇}, t) =
where
pi =
X
i
pi q̇i − L({q}, {q̇}, t)
∂L({q}, {q̇}, t)
∂ q̇i
1
(4)
The Hamiltonian, H, is numerically equal to the energy function but is chosen to be a
function of the variables {q} and {p}, rather than of {q} and {q̇}.
We first solve equation (4) for the generalised velocities in terms of the generalised coordinates
and their conjugate momenta, ie q̇i = q̇i ({q}, {p}). We then define the Hamiltonian as a
function of ({q}, {p}) by the so-called Legendre transformation:
H({q}, {p}, t) =
X
i
pi q̇i − L({q}, {q̇}, t) .
(5)
The change of variables looks harmless but has deep consequences. The (3N −M ) generalised
coordinates {q} and their (3N − M ) conjugate momenta {p} now enter on an equal footing,
and can more obviously be varied independently.
Example: For a particle moving in one dimension with a conservative potential V (x),
H = E = T + V and the Hamiltonian is (exercise)
H(x, p) =
p2
+ V (x) .
2m
Hamilton’s Equations: Consider the variation of H due to arbitrary small variations in
its arguments.
From the LHS of equation (5), we obtain, by Taylor’s theorem
X
δH =
i
"
#
∂H
∂H
∂H
δqi +
δpi +
δt
∂qi
∂pi
∂t
(6)
and from the RHS of equation (5)
δH =
X
i
=
X
i
"
#
∂L
∂L
∂L
δqi −
δ q̇i −
q̇i δpi + pi δ q̇i −
δt
∂qi
∂ q̇i
∂t
[q̇i δpi − ṗi δqi ] −
∂L
δt
∂t
(7)
where we used the definition of the canonical momenta
∂L
pi =
∂ q̇i
(8)
to cancel the 2nd and 4th terms in the square bracket, and the Euler-Lagrange equations
in the form ṗi = ∂L/∂qi to obtain the final result. Comparing equations (6) and (7), which
must hold for all δqi and δpi , we obtain Hamilton’s Equations of motion
q̇i =
∂H
,
∂pi
ṗi = −
∂H
∂qi
and
−
∂L
∂H
=
∂t
∂t
Essentially, we have converted 3N −M Lagrange’s equations (2nd order DE’s for {q}) into
exactly twice as many Hamilton’s equations (1st order DE’s for {q} and {p}), plus the
(almost trivial) last equation for ∂H/∂t.
Hamilton’s equations may also be derived directly by applying Hamilton’s Principle, δS = 0,
to the action in the form (tutorial)
S =
Z (X
i
pi q̇i − H({q}, {p}, t)
2
)
dt
Poisson Brackets: Consider an arbitrary function A = A({q}, {p}, t) of the generalised
coordinates and momenta, which may also have explicit time dependence. Its total derivative
wrt time is
"
#
X ∂A
dA
∂A
∂A
=
q̇i +
ṗi +
dt
∂qi
∂pi
∂t
i
=
X
i
"
#
∂A
∂A ∂H ∂A ∂H
−
+
.
∂qi ∂pi
∂pi ∂qi
∂t
where we used Hamilton’s equations in the last step.
Now consider two such arbitrary functions A({q}, {p}, t) and B({q}, {p}, t), and define the
Poisson Bracket of A and B as
"
#
X ∂A ∂B
∂A ∂B
{A, B}PB ≡
−
.
∂qi ∂pi ∂pi ∂qi
i
The derivative wrt time of any quantity A({q}, {p}, t) may then be written in terms of the
Poisson bracket of A with the Hamiltonian
dA
∂A
= {A, H}PB +
.
(9)
dt
∂t
In particular, Hamilton’s equations may be expressed as
q̇i = {qi , H}PB
and ṗi = {pi , H}PB .
Notice the similarity in form to the Heisenberg equations of motion in quantum mechanics.
Lorentz Force: For a non-relativistic charged particle in an EM field, we showed that the
energy function expressed in terms of the particle’s position and velocity is
1
h(r, ṙ, t) = p · ṙ − L = (mṙ + eA) · v − [ mv 2 − e(φ − v · A)]
2
1
2
m|v| + eφ
=
2
To obtain the Hamiltonian, we recall that the canonical momentum is p = mv + eA We then
1 solve for v in terms of p to obtain (trivially) v =
p − eA , so the Hamiltonian is simply
m
1
H(r, p, t) =
|p − eA|2 + eφ
2m
which is probably not what you would first think of for H. It is left as a tutorial problem to
check that Hamilton’s equations lead to the correct Lorentz force.
From Classical to Quantum: H({q}, {p}, t) corresponds very closely to the Hamiltonian
operator in quantum mechanics. Specifically, to get quantum mechanics for a system with
one generalised coordinate q we replace
!
∂
H(q, p, t) → Ĥ q, −ih̄ , t
∂q
keeping the functional form of H the same. Thereby a classical function is converted into a
quantum operator which acts on the wavefunction ψ(q, t) obeying the Schrödinger equation.1
1
For a system with more than one degree of freedom, the naive quantisation prescription pi → −ih̄∂/∂qi
works for Cartesian coordinates, but it must be modified slightly for general curvilinear coordinates.
3
A familiar example is the 1-D particle
!
p2
h̄2 ∂ 2
H=
+ V (x) → Ĥ = −
+ V (x)
2m
2m ∂x2
For the charged particle in the EM field, the Hamiltonian is H(r, p, t) =
A suitable Ĥ is therefore (tutorial):
Ĥ =
1
|p
2m
− eA|2 + eφ
−h̄2 ∇2
eh̄
e2
−
[2A · ∇ + (∇ · A)] +
|A|2 + eφ
2m
2im
2m
Canonical Quantisation: A more general form of quantisation begins with the Poisson
brackets for the classical quantities A and B, say, from which the canonical commutation
relations between the quantum operators  and B̂ are obtained by the following prescription
→
ih̄{A, B}PB
[Â, B̂]
Using this method, the quantum equation of motion for the operator (in the Heisenberg
picture – see SH Quantum Theory) may be postulated almost trivially from equation (9)
ih̄
h
i
dÂ
∂ Â
= Â, Ĥ + ih̄
dt
∂t
The same equation gives directly the expectation values of operators in the Schrödinger
picture – see Junior Honours Quantum Mechanics, Section 6, Page 1.
Feynman Path Integrals: Finally, there is a formulation of quantum mechanics due to
Feynman which is based on the Lagrangian rather than the Hamiltonian. It says. . .
The quantum amplitude for a particle in one dimension to propagate from position xi at
time ti to position xf at time tf is the weighted sum over all paths between the two points:
• Draw all possible continuous paths the particle can take between xi at ti and xf at tf .
• Calculate the classical action S =
Z tf
ti
L(x, ẋ) dt for each path.
• The amplitude hxf , tf |xi , ti i for the particle to propagate from position xi at time ti to
position xf at time tf is the weighted sum or integral over all possible paths:
hxf , tf |xi , ti i = N
X
exp (iS[x(t)]/h̄) = N
all paths
Z
Dx(t) exp
i Z tf
L(x, ẋ) dt
h̄ ti
where N is a normalisation constant, and the path integral is over all continuous functions
x(t). All paths are treated on an equal footing – one simply sums the phases for all paths. The
particle does whatever it likes and the αth path contributes a factor Zα = exp(iS[xα (t)]/h̄)
to the amplitude. Hamilton’s principle emerges in the limit h̄ → 0, but to actually show this
is beyond the scope of this course. See Senior Honours Quantum Theory for details.
4
Lagrangian Dynamics 2008/09
Lecture 14: Rigid Body Motion – Introduction & Euler’s Equations
Section VII: Rigid Body Motion
Nomenclature: the motion of rigid bodies was studied in depth by Euler and Lagrange.
Their names are attached to equations (eg Euler’s equations of motion for a top) which are
unrelated to those met previously (Euler’s equation for stationarity of an integral). This is
confusing, but not as confusing as trying to give these equations new names.
Recap: A rigid body is viewed as a system of N particles, a = 1, · · · , N , with constraints
|ra − rb | = ρab = constant
for each pair (a, b). These are not all linearly independent, however. In three dimensions,
we end up with 6 degrees of freedom (see below).
For such a system, we showed
M R̈ = F ext
(centre of mass motion), and
L̇ = G ext
We also showed, for angular momentum L about some stated axis and for kinetic energy T
L = J +R×P
1
T = Tc-of-mom + M V 2
2
(1)
(2)
ie, the value in a general (inertial) frame is the value in the centre of momentum (c-of-mom)
frame plus a contribution from the motion of the centre of mass, treated as though it were
a point particle.
NB Remember that in the c-of-mom frame, the intrinsic angular momentum J does not
depend on the origin chosen (though, because of the second term in equation (1) this will
not be true for L in a general frame). Recall that the c-of-mom frame is an inertial frame (not
a rotating frame) in which the centre of mass is stationary, but not necessarily at the origin.
In such a frame, at least instantaneously, the motion of a rigid body is purely rotational.
Euler’s Theorem: Any displacement of a rigid body with one point fixed in space can be
described as a rotation about some single axis
This is surely obvious(!), and the formal proof is not illuminating. To specify a finite rotation
requires an axis (unit vector, two degrees of freedom) and a magnitude of the rotation (one
dof). Hence we will need 3 generalised coordinates to specify the centre of mass motion
and three for the angular motion. In practice, there are many ways to choose three angular
coordinates: we do not make an explicit choice at this stage.
The velocity of the system is usually specified by Ṙ(t) (the centre of mass motion) and ω(t)
(the angular velocity about the centre of mass), ie, also six independent components.
1
Technical Note: In choosing angular coordinates, thereR is some subtlety since finite rotations about different axes do not commute. Hence Ω(t) = 0t ω(t0 ) dt0 is not a valid generalised
coordinate; the same value of this quantity can be obtained for different sequences of rotations and hence different actual orientations of the system. Hence the need for “Euler
angles” introduced later.
The Inertia Tensor: Working in the c-of-mom frame and choosing our origin at the centre
of mass, the instantaneous velocity of the ath particle is
ṙa = ω(t) × ra
The intrinsic angular momentum is therefore
J =
=
X
X
a
r a × pa =
a
ma (ra × (ω × ra )) =
X
a
ma (ra × ṙa )
X
a
ma ra2 ω −
X
a
ma ra (ra · ω)
(3)
(using a × (b × c) = (a · c)b − (a · b)c). The first term is a vector parallel to ω , the second
term is a vector that is, in general, not parallel to ω.
The intrinsic angular momentum J is thus a vector which depends linearly on ω (for example,
J(2ω) = 2J(ω)) but need not be parallel to it. So there must be a relationship
J = Iω
(4)
where I is a rank-2 tensor (which we represent in cartesian coordinates by a 3 × 3 matrix),
called the inertia tensor. In Cartesian coordinates, equation (4) becomes Ji = Iij ωj . From
equations (3) and (4)), it is a straightforward exercise to show that
Iij =
X
a
n
ma ra2 δij − xa,i xa,j
o
or, in terms of explicit cartesians r a = (xa , ya , za ), we have
I=
X
a


ma 
(ya2 + za2 )
−xa ya
−xa za
−xa ya
(x2a + za2 )
−ya za
(5)

−xa za
−ya za 

2
2
(xa + ya )
(6)
The expression for I explicitly involves the particle coordinates ra , which are measured
relative to the centre of mass. For a rigid body, these coordinates can be taken as constants,
though this applies only in a frame of reference in which the body is completely stationary
(ie, ṙ a = 0, ∀a). For a body that is actually rotating, this is a noninertial frame, the body
frame which rotates and translates with the body. Despite this inconvenience, one defines
the inertia tensor I to be evaluated in such a frame, that is, with respect to a set of x, y, z
axes fixed in the body. The inertia tensor is therefore a time independent characteristic of
how the mass is distributed in the body.
Kinetic Energy: In the centre of momentum frame, the KE is
X1
T =
ma (ω × ra ) · (ω × ra )
a 2
and (using (a × b) · c = a · (b × c))
X
1
1
1
T = ω·
ma ra × (ω × ra ) = ω · J = ω · Iω
2
2
2
a
2
Principal Axes: In this course, we will generally not need the explicit form of I found
above; we take I as given. But it is important that (however irregular the body) the tensor
I is symmetric. Hence there exist three real eigenvalues, I1 , I2 , I3 , and three mutually
perpendicular eigenvectors e 1 , e 2 , e 3 .
Choosing these to define new axes, the Principal Axes (PA), fixed in the body, we have the
diagonal form


I1 0 0


I =  0 I2 0 
0 0 I3
Where (I1 , I2 , I3 ) are the Principal Moments of Inertia. Moreover, in this coordinate system
J = (I1 ω1 , I2 ω2 , I3 ω3 )
and
T =
1 2
I1 ω1 + I2 ω22 + I3 ω32
2
Here lies the central problem: These “nice” axes are fixed in the body. But the body is
rotating. Hence in lab coordinates (or any other inertial frame), the principal axes have timedependent orientations. This is why rigid body motion is not a straightforward application
of linear algebra.
Shift of Origin: We have placed the origin of our coordinate system for the r’s at the
centre of mass O of the body. This is the best choice for a body undergoing arbitrary motion,
or whose centre of mass is constrained to be fixed in space. In systems where some other
point O0 is constrained to be fixed instead (for a spinning top, this is usually the point of
contact with the table) one can use the usual formula L = J + M R × P to calculate the
angular momentum L with respect to O0 .
Centre
of Mass
O
Pivot at O’
However, by Euler’s Theorem, the result is the same as found by writing
L = I0 ω
where I 0 is found by replacing O by O0 as the coordinate origin, in the explicit expressions
for I given in equations (5) and (6) above. The kinetic energy, including that of the centre
of mass motion, becomes, in the PA basis
1
1
T = ω · I 0 ω = (I10 ω12 + I20 ω22 + I30 ω32 )
2
2
0
The tensor I is called the inertia tensor about the point O0 , as opposed to the “inertia tensor
about the centre of mass” which is, strictly speaking, the proper name for I. If O0 lies a
distance l from the centre of mass along (say) the 3-axis, then, again in the PA basis, one
has
I10 = I1 ;
I20 = I2 ;
I30 = I3 + M l2
This is a simple example of the parallel-axes theorem proved in FoMP/T&F. In our discussion
of spinning tops, we will just write I, meaning I 0 relative to the pivot where appropriate.
3
Euler versus Lagrange: At this point there are two divergent approaches:
Eulerian Approach: Study equations in a noninertial frame (the body frame) rotating
with the body, so that principal axes are constant in time. Difficulty:
J˙ = G ext
where G ext is (usually) specified wrt axes fixed in space (ie in the lab). We can’t translate
G back into the body frame until after we have found the motion! In practice, the Eulerian
approach is therefore useful only for kinematics (motion without external forces): G ext = 0.
Lagrangian Approach: Choose a convenient set of generalised coordinates (in practice,
three angles specifying the orientation of the principal body axes with respect to another set,
fixed in space). Then construct L and differentiate as usual to obtain equations of motion.
This is more powerful but less intuitive; we follow Euler’s approach first.
Euler’s Equations Of Motion: These describe motion in the principal axes frame S
rotating with the body – this is not an inertial frame. Consider S0 , an inertial frame instaneously coinciding with S. As we saw previously, if the instantaneous angular velocity of
the body is ω, then for any vector A
[Ȧ]S0 = [Ȧ]S + ω × [A]S
Now, in the inertial frame S0 , we have the usual equation of motion
[L̇]S0 = G ext ≡ G
where G is the external torque on the system. (The label ext is implicit from now on.)
Accordingly in the body frame S
G = [L̇]S + ω × [L]S
Now, since S is the frame of the principal axes (see above)
[L]S = J = (I1 ω1 , I2 ω2 , I3 ω3 )
In this coordinate frame, the equation for G therefore reads
(G1 , G2 , G3 ) = (I1 ω̇1 , I2 ω̇2 , I3 ω̇3 ) + (ω1 , ω2 , ω3 ) × (I1 ω1 , I2 ω2 , I3 ω3 )
or, expanding this out
G1 = I1 ω̇1 + (I3 − I2 ) ω2 ω3
G2 = I2 ω̇2 + (I1 − I3 ) ω3 ω1
G3 = I3 ω̇3 + (I2 − I1 ) ω1 ω2
These are Euler’s Equations of Motion.
For the reasons stated above, the useful case is when G = 0; then Euler’s equations become
I1 ω̇1 = (I2 − I3 ) ω2 ω3
I2 ω̇2 = (I3 − I1 ) ω3 ω1
I3 ω̇3 = (I1 − I2 ) ω1 ω2
4
Lagrangian Dynamics 2008/09
Lecture 15: The Symmetric Top – Precession; the Tennis Racquet Theorem
Technical Note: Euler’s equations apply in a frame rotating with the body. So why isn’t
ω identically zero – surely there is no motion to describe, if we’re working in this frame?
This is a good question. The physical answer is that since the body frame is non-inertial,
an observer in this frame is able to sense that she is rotating, and indeed could perform
experiments which would reveal exactly the value of ω (any suggestions?). Mathematically,
ω was defined as the rotation velocity of the body frame frame relative to an inertial one;
so this is what Euler’s equations involve.
The Symmetric Top: If the body has an axis of symmetry (call this axis 3), then I1 = I2 .
Assume (for definiteness later) that I3 is smaller than the other two; then Euler’s equations
become:
I1 ω̇1 = (I1 − I3 ) ω2 ω3
I1 ω̇2 = (I3 − I1 ) ω3 ω1
I3 ω̇3 = 0
Hence ω3 is a constant, and we can write
I1 − I3
ω3 ω2 ≡ Ωω2
I1
= −Ωω1
ω̇1 =
(1)
ω̇2
(2)
Differentiating wrt t again, we easily obtain
ω̈1 = −Ω2 ω1
ω̈2 = −Ω2 ω2
Both of these are equations for SHM at angular frequency Ω. Since ω̇2 = −Ωω1 , there is a
π/2 phase difference between ω1 and ω2 (eg, if ω2 = a cos(Ωt + α), then ω1 = a sin(Ωt + α)).
This corresponds to circular motion of ω in the
(1,2) plane, with a constant ω3 .
Hence ω precesses about the 3 axis with precession frequency Ω = (I1 − I3 )ω3 /I1 . This precession
of ω in the body frame traces out the body cone.
Since ω3 is constant, and so is ω12 + ω22 (circular
motion in the 1, 2 plane) we have |ω| = constant.
Since J = (I1 ω1 , I2 ω2 , I3 ω3 ), and I1 = I2 , |J| is
constant for the same reason, and so is J · ω.
3
ω
ζ
1
2
The fact that J · ω, J · e 3 and ω · e 3 are all constants means that J, ω and e 3 make constant
angles with each other, and are coplanar because I1 = I2 . Hence ω and J precess together
at frequency Ω about the 3 axis at inclination angles ζ = cos−1 (ω3 /|ω|), τ = cos−1 (I3 ω3 /|J|)
which are fixed by the IC’s. Here, τ > ζ because I3 < I1 = I2 .
1
e3
ω
ζ
J
τ
Note: If all three principal moments are equal (isotropic body), Euler’s equations reduce to
ω̇ = 0. This is trivially true because J and ω are parallel for this case.
Symmetric Top in the Lab Frame: We now have to translate these results into the
(inertial) lab frame. We assumed G = 0 and therefore, in the lab, J is constant. Clearly, J,
ω and e3 remain coplanar in this frame. Hence we have the picture:
J
space
cone
ω
e3
τ
ζ
body
cone
So, ω traces out the surface of a cone (the space cone) about J. Since ω simultaneously
executes a cone about e3 (see above), we find that the body cone rolls without slipping
around the outside of the space cone – a nice picture if you like geometry! If I3 > I1 = I2 ,
the body cone rolls inside the space cone.
An example is the free precession of the earth (wobble): ω3 = 2π per day; the asphericity
(I1 − I3 )/I3 ' 0.3% giving a precession period of about 300 days. The observed value is 430
days because the earth is not quite a rigid body (it has a molten core), and the amplitude
is about 10m. (Note: this phenomena is not precession of the equinoxes.)
Asymmetric Top: the Tennis Racquet theorem: When all three principal moments of
inertia are different, the general solution of Euler’s equations for torque-free motion involves
elliptic integrals and is not illuminating.
We instead ask the following: Can one have motions where ω is constant in the body frame?
Inspection of Euler’s equations shows that this is only possible if two of the components of
ω (say, ω1 and ω2 ) are zero. Hence
An asymmetric body can only rotate with constant angular velocity if the rotation axis is one
of the principal axes.
Is the resulting motion stable? To investigate, write ω1 = 1 , ω2 = 2 and ω3 = ω + 3 where
the ’s are small. Expanding Euler’s equations to first order in ’s gives
I1 ˙1 = (I2 − I3 )ω 2
I2 ˙2 = (I3 − I1 )ω 1
2
I3 ˙3 = 0
Eliminating 2 (say) gives the second order equation
I1 I2 ¨1 = (I2 − I3 )(I3 − I1 )ω 2 1
The same equation holds for ¨2 . The solution is
¨1 = AeiΩt + Be−iΩt
with
(I1 − I3 )(I2 − I3 ) 2
ω
I1 I2
We conclude that if I3 is either the smallest or the largest moment of inertia, then the RHS
is positive and Ω is real. Hence perturbations to steady motion about this axis oscillate but
do not grow. In contrast, if I3 is the middle axis, Ω is imaginary and perturbations grow
exponentially. This is the tennis racquet theorem (easily demonstrated with said object or
similar):
Ω2 =
Steady rotation about the intermediate axis is unstable; about the major or minor axis, it is
stable.
Lagrangian Dynamics of Rigid Bodies
We now turn to the Lagrangian approach, which is suitable for systems in which there is an
external torque acting.
Choice of Origin: We will be concerned almost exclusively with spinning objects (tops)
with a point O held fixed:
3
Centre
of Mass
l
Pivot at O, fixed
The fixed point is usually not the centre of mass. From now on we denote by I the inertia
tensor about O (before this was called I 0 ). The angular momentum (about O) of such a
body is
L = Iω
and its kinetic energy (including centre of mass motion – remember Euler’s theorem) is
T =
1
1
ω · Iω = (I1 ω12 + I2 ω22 + I3 ω32 )
2
2
where I1,2,3 are the eigenvalues of I. We take these as arbitrary.
3
Choice of Coordinates – Euler Angles: We need a set of three generalised coordinates
to specify the orientation of the body. (According to Euler’s theorem, since O is fixed, there
are only three degrees of freedom left.)
Three Euler Angles (θ, φ, ψ) are defined by a sequence of rotations. We shall use one of
numerous possible choices for three angular coordinates. (Even among those normally called
“Euler Angles” there are several conventions, of which one used here is called the y convention – see Goldstein if interested.) The body axes and fixed space axes share a common
origin at the pivot point. In the figures below, the body axes have been displaced for clarity.
z
z
3
3
2
θ
2
1
1
y
y
x
x
The First Rotation is through angle θ
in the (z, x) plane. The handle remains
in the (z, x) plane.
Reference State: The body axes 1, 2
& 3 are aligned with the fixed space axes
x, y & z. For reference, we introduce a
“handle” which lies in the (z, x) plane.
z
z
θ
θ
3
3
2
1
1
x
φ
2
ψ
y
x
The Second Rotation is through angle
φ about the z axis. The 3 axis is now
specified by (θ, φ) as in spherical polars.
The handle still lies in the vertical plane,
φ
y
The Third Rotation is through angle
ψ about the 3 axis which is fixed to the
body. The handle is rotated by ψ away
from the vertical.
This sequence defines (θ, φ, ψ) the three Euler angles specifying the orientation of an arbitrary
rigid body. The main advantage of the Euler angles over other choices is that the various
conservation laws emerge more tidily.
4
Lagrangian Dynamics 2008/09
Lecture 16: The Lagrangian for a Top, EoMs and Conservation Laws
The Lagrangian for a Top: What follows is a rather long derivation, but it’s worth
doing. . .
Z
θ
3
l
We want to find L = T ({q}, {q̇}, t)−V ({q}, t) with {q} = (θ, φ, ψ), ie we want
L = T (θ, φ, ψ, θ̇, φ̇, ψ̇, t) − V (θ, φ, ψ, t)
The potential term V is relatively straightforward: we will consider only tops under gravity,
for which there is no time dependence and hence
V = M gl cos θ
The kinetic energy is much harder. We know
T =
1
1
ω · Iω = (I1 ω12 + I 2 ω 22 + I 3 ω32 )
2
2
and therefore we require ω(({q}, {q̇}, t), ie an explicit expression for
ωi = ωi (θ, φ, ψ, θ̇, φ̇, ψ̇, t) with i = 1, 2, 3.
Let’s define axes e 1 , e 2 and e 3 , in the body as usual, but also an intermediate set e01 , e02
and e03 as the axes after the second but before the third Euler rotation. Clearly (since this is
a rotation about the 3 axis)
e03 = e 3
whereas looking down the 3 axis we have
e2
e’2
e
ψ
1
e’1
e 1 · e01 =
cos ψ
e 2 · e01 = − sin ψ
1
e 1 · e02 = sin ψ
e 2 · e02 = cos ψ
e’3
^z
e’
2
θ
y
φ
e’1
x
The primed set are an orthonormal basis of unit vectors along the θ, φ and r directions
respectively (just as one would define by ordinary spherical polar coordinates for a single
particle); e02 is horizontal whereas the e01 − e03 plane is vertical.
Now let’s try to get the components of ω in the primed basis set. From the definition of the
Euler angles, we identify three contributions to the angular velocity
θ̇ = rate of rotation about the 20
axis
φ̇ = rate of rotation about the z
axis
ψ̇ = rate of rotation about the 30 = 3 axis
The problem here is that the 20 , 30 and z axes are not all orthogonal and so these three terms
are not directly the vector components of ω in any orthonormal basis. Nonetheless we can
add the three contributions to get
ω = θ̇ e02 + φ̇ ẑ + ψ̇ e03
and decompose
ẑ = (ẑ · e03 ) e03 + (ẑ · e01 ) e01 = e03 cos θ − e01 sin θ
(see diagram above) to give
ω = − φ̇ sin θ e01 + θ̇ e02 + (ψ̇ + φ̇ cos θ)e03 ≡
X
i
ωi0 e0i
Finally, since e03 = e 3 we have e 1 · e03 = 0, and e 2 · e03 = 0. Hence (using also the results for
e 1 · e01 , etc, given above)
ω 1 = e 1 · ω = −φ̇ sin θ cos ψ + θ̇ sin ψ
ω2 = e2 · ω =
ω3 = e3 · ω =
φ̇ sin θ sin ψ + θ̇ cos ψ
φ̇ cos θ + ψ̇
This completes the construction of the angular velocity components referred to body axes;
these can be substituted into the usual expression
1
T = (I 1 ω 21 + I 2 ω 22 + I3 ω32 )
2
to give
T (θ, φ, ψ, θ̇, φ̇, ψ̇, t) =
1
I 1 (−φ̇ sin θ cos ψ + θ̇ sin ψ)2
2
1
1
+ I 2 (φ̇ sin θ sin ψ + θ̇ cos ψ)2 + I3 (φ̇ cos θ + ψ̇)2
2
2
2
Note that this is quadratic in the velocities
T =
X
Tij q̇i q̇j
ij
but, unlike earlier examples, the matrix Tij is explicitly dependent on the coordinates {q}.
Symmetric Top: Life is a lot simpler if the 3 axis that passes through the pivot is a
symmetry axis of the body. Then I 1 = I 2 = A (say). Denoting I3 = C we have
i
1h
A(ω 21 + ω 22 ) + Cω32
2
i
1h
2
2
=
A(ω 0 1 + ω 0 2 ) + Cω32
2
i
1h
=
A(θ̇2 + φ̇2 sin2 θ) + C(φ̇ cos θ + ψ̇)2
2
T =
where we noted that the 1 and 2 axes differ from 10 and 20 by rotation in the (1, 2) plane, so
ω12 + ω22 = ω 0 1 2 + ω 0 2 2 . Hence the degeneracy of the two moments of inertia, I1 = I2 , means
that the explicit transformation from e01, 2 to e 1, 2 is not required in this case. An explicit
calculation gives the same result of course – check it! Equivalently, we may assert that T is
independent of ψ by symmetry, so that we may evaluate it for ψ = 0.
Accordingly for the symmetric top pivoted at O
φ
Z
ψ
θ
3
l
i
1h
A(θ̇2 + φ̇2 sin2 θ) + C(φ̇ cos θ + ψ̇)2 − M gl cos θ
(1)
2
Note: In examination questions, this result would either be given, or you would be asked to
quote it. Although a full derivation along the lines given above will not normally be asked
for, an explanation or interpretation of the various contributions may be required.
L=
Conservation Laws and Equations of Motion: We consider the generalised coordinates in turn:
1. L does not depend explicitly on ψ. Hence
pψ ≡
∂L
= C(ψ̇ + φ̇ cos θ) = I3 ω3
∂ ψ̇
3
is a conserved quantity. The reason for this conservation law is the symmetry of L
under rotations of the body about its 3 axis. Since the centre of mass lies on the 3
axis, the gravitational torque cannot change ω3 . (This is a good reason for choosing ψ
as one of our Euler angles.) It is conventional to write this as conservation of
n ≡ ω3 = (ψ̇ + φ̇ cos θ)
where n is called the spin of the top.
2. L does not depend explicitly on φ. Hence
pφ = Aφ̇ sin2 θ + C(ψ̇ + φ̇ cos θ) cos θ = Aφ̇ sin2 θ + Cn cos θ
is our second conserved quantity. This corresponds to the symmetry of L under rotations of the system about the z axis; the constant value of pφ is
pφ = L z
the z component of the “orbital” angular momentum of the top.
3. On the other hand, L does depend explicitly on θ. This is true even without gravity.
We have pθ = Aθ̇ and
∂L
ṗθ =
∂θ
as the Lagrange equation for θ, which gives
Aθ̈ = Aφ̇2 sin θ cos θ − C(ψ̇ + φ̇ cos θ)φ̇ sin θ + M gl sin θ
Using the conservation of pψ we write this as
Aθ̈ = Aφ̇2 sin θ cos θ − Cnφ̇ sin θ + M gl sin θ
(2)
4. L does not depend explicitly on time t. Hence
h=
2
X
i=1
q̇i pi − L
is a third conserved quantity. For this system, T is quadratic in the velocities (see
above), in which case h coincides with E = T +V , the total energy, as proved generally
in Lecture 11. (Anyway it is clear that E is conserved!) To get information about the
θ motion, we first deduce, using (say) equation (1), that
A
C
E = (θ̇2 + φ̇2 sin2 θ) + n2 + M gl cos θ
2
2
This can be written
2E − Cn2 2M gl cos θ
θ̇2 + φ̇2 sin2 θ =
−
A
A
2
and setting Aφ̇ sin θ + Cn cos θ = Lz one obtains
2E − Cn2 2M gl cos θ (Lz − Cn cos θ)2
2
θ̇ =
−
−
A
A
A2 sin2 θ
Thus, after accounting for all three conservation laws, we are left with only one nontrivial degree of freedom, ie θ, described by this nonlinear first order DE.
Technical Note: This is, in effect, the first integral of the Lagrange equation for θ̈ above; but
it would have been extremely difficult to spot the required integrating factor without invoking
the various conservation laws. Here, we have three symmetries, thus three conservation laws
and three first integrals.
4
Lagrangian Dynamics 2008/09
Lecture 17: Tops: Zones, Steady Precession, Nutation, Gyroscopes, . . .
A Common Mistake: The energy equation in the form used above depends on using the
conservation laws for n and Lz to write E = E(n, Lz , θ, θ̇). It would however be completely
wrong to “second guess” these conservation laws by trying to put them into the Lagrangian
directly. For example, there is a temptation to write
L=
C
A 2
(θ̇ + φ̇2 sin2 θ) + n2 − M gl cos θ
2
2
and then obtain Lagrange’s equation for θ by differentiation as usual, treating n as constant.
(A similar mistake was discussed in Lecture 8, but this case is more dangerous: the result is
wrong, but not obviously so.) Of course, you can always use n as shorthand for (ψ̇ + φ̇ cos θ)
in Lagrange’s equations, but then it is vital to remember that some of the ∂n/∂qi and ∂n/∂ q˙i
are non-zero in order to get the right answers. To be safe, it is probably best to write out L
longhand in terms of the chosen generalised coordinates and velocities.
Zone Function: The fact that θ̇ is real requires θ̇2 > 0 (remember Lecture 1). This defines
for given n, Lz , E a zone of allowed values of the quantity
u ≡ cos θ
⇒
u̇ = −θ̇ sin θ
Specifically, we can write
2
u̇ = f (u) ≡
!
!
(Lz − Cnu)2
2M gl
2E − Cn2
(1 − u2 ) −
−
u(1 − u2 )
2
A
A
A
(Check this as an exercise.)
The zone function f (u) is a cubic in u:
For applications of this method, see the
Tutorial Sheet 5.
u2
−1
1
allowed zone
1
u
Steady Precession: From the conservation of energy we found
1 2
Cn2
(Lz − Cn cos θ)2
Aθ̇ = E −
− M gl cos θ −
≡ E 0 − U (θ)
2
2
2A sin2 θ
where E 0 = E − Cn2 /2 is a constant and the rest of the formula defines U (θ). This corresponds to the energy equation for a particle of “mass” A moving with a one dimensional
cartesian coordinate θ in an effective potential U (θ) with total “energy” E 0 :
U (θ)
min
E’
Umin
φ
Z
θ
Zone
o
3
θ min
π
steady precession
θ
nutation (see below)
Note that U (θ) becomes very large as θ → 0, π, and that it is bounded from below, as in the
figure. Obviously, the “potential” U is different for each value of Lz and n. This “particle
analogy” gives a physically appealing alternative to the zone-function approach.
We deduce that steady precession (a motion with θ̇ = 0) is possible, but that this requires
E 0 = Umin (Lz , n) and θ = θmin (Lz , n). For this case θ = constant, and since
Lz = Aφ̇ sin2 θ + Cn cos θ
n = ψ̇ + φ̇ cos θ
are both conserved, then φ̇ and ψ̇ are also constants for this particular motion.
To find θ0 ≡ θmin , we set dU/dθ = 0, giving (after some algebra and substitutions):
h
sin θ0 φ̇2 A cos θ0 − φ̇Cn + M gl
i
= 0
(1)
This result may also be obtained trivially from the Lagrange equation for θ (equation (2) of
the previous lecture) by noting that θ̇ = 0 implies θ̈ = 0.
Equation (1) has two solutions:
1. sin θ0 = 0: an upright top. This is called the sleeping top; we return to it later.
2. θ0 6= 0, in which case, the term in square brackets in equation (1) is zero and
√
Cn ± C 2 n2 − 4AM gl cos θ0
φ̇ =
2A cos θ0
so that if cos θ0 > 0 (corresponding as usual to a top that does not hang down below
its support), the existence of steady precession requires
C 2 n2 ≥ 4AM gl cos θ0
In other words, the top must be spinning fast enough if it is to show steady precession
at an upward tilting angle θ0 .
2
Fast Top (Gyroscope): For large enough spin we can binomial expand our equation for
the angular velocity φ̇ of steady precession
Cn ± [C 2 n2 − 4AM gl cos θ0 ]
φ̇ =
2A cos θ0
"
1/2
2AM gl cos θ0
Cn
1± 1−
'
2A cos θ0
C 2 n2
#!
There are two roots:
1. Slow Precession (minus sign) which is due to gravity, and has
φ̇ =
M gl
Cn
2. Free Precession (plus sign) where
φ̇ =
Cn
A cos θ0
These limiting forms can readily be checked by other methods: for slow precession, one can
balance the gravitational torque against a constant L̇ ; for free precession the result can be
obtained from Euler’s equations of motion (with zero torque). Note that in real gyroscopes,
the free precession is rapidly damped by air friction so the slow precession is what dominates
the observed behaviour.
Nutation: Suppose now we increase the energy (at fixed Lz ) so that E 0 is slightly above
Umin ; in this case (by analogy with our 1-D particle) we have oscillations superposed on the
steady precession.
The path traced by the 3 axis looks like
Z
Z
Zone
Path traced
by 3 axis
OR
etc.
depending on the relative frequencies/phases of the φ, θ oscillations. For small amplitudes
we have SHM in the θ variable with frequency
1/2
Ωnut = (U 00 (θ0 )/A)
If the spin is large enough (the formal requirement is Cn (4AM gl cos θ0 )1/2 ) one has
Ωnut φ̇ and there are many nutational “wobbles” per precession period (ie, per complete
circuit of the 3 axis about the vertical).
3
Sleeping Top: We finally return to the case θ0 = 0, that is, we look for steady motion of
an upright top. Lagrange’s equation for the θ motion is (see Lecture 16, page 4, item 3)
Aθ̈ = Aφ̇2 sin θ cos θ − Cnφ̇ sin θ + M gl sin θ
Now expand for small θ
Aθ̈ = (Aφ̇2 − Cnφ̇ + M gl)θ
(2)
Clearly, θ = 0 is a solution, but is it stable?
Since the top starts off upright, Lz = Cn initially. Since both Lz and Cn are conserved,
Cn = Lz ≡ Aφ̇ sin2 θ + Cn cos θ
So that, expanding to order θ2 ,
Cnθ2 /2 = Aφ̇θ2
Hence the sleeping top has
Cn
2A
This is the precession frequency of a sleeping top which is perturbed infinitesimally away
from the vertical. Substituting back into equation (2) for θ̈ gives
φ̇ =
Aθ̈ =
!
−C 2 n2
+ M gl θ
4A
For n2 > 4M glA/C 2 , this is SHM and the sleeping top is stable. If the spin drops below
this value, any perturbation in θ will grow exponentially and steady motion in an upright
position is unstable.
The sleeping top will therefore spin happily in an upright position until such time as its spin
drops below the threshold for stability. (The spin decreases slowly due to air friction.) Then,
since there are always perturbations from somewhere, it suddenly “wakes up” and performs
an elaborate motion – it goes “haywire”!
4
Lagrangian Dynamics 2008/09
Lecture 18: Section VIII: Small Oscillation Theory
As with rigid body motion, this is an area where Lagrangian methods are well suited to
setting up the equations of motion. Below we also address the general way of solving these,
although for all but the hardest examples this is rather a sledgehammer approach.
This section extends the normal mode methods which were covered in some detail in FoMP
Differential Equations and Dynamics and MP2B Dynamics but it is intended to be relatively
self-contained.
Static Equilibrium Consider a system with D degrees of freedom in a time-independent
conservative potential
V = V ({q}, {6 q̇}, 6 t) = V ({q})
{q} = (q1 , q2 , · · · , qD )
Static equilibrium (no motion) is possible only if all the generalised forces Qj vanish together
at some set of coordinates {q}eq . This defines the equilibrium position. (There may be more
than one such position.)
Let us redefine our coordinates: {q} → {q} − {q}eq so that the equilibrium position is at
the origin; then
∂V ({q}) = 0
∀j
Qj ({0}) = −
∂qj {q}=0
If we Taylor expand V about this point, all terms linear in {q} vanish, and we have
1 X ∂ 2V
V ({q}) = V ({0}) +
qi qj + O({q 3 })
2 ij ∂qi ∂qj
where V ({0}) is just a constant (which we can ignore from now on). We write
1
V ({q}) = qi Vij qj
2
where from now on we use the summation convention (repeated indices are summed over)
and we have defined
∂ 2V
({0})
Vij =
∂qi ∂qj
which is a constant D × D matrix of second derivatives (evaluated at the origin). Since
V is assumed differentiable, this matrix must be symmetric. Moreover if {q = 0} is a
local minimum, then the matrix Vij is positive semi-definite: all the eigenvalues are zero
or positive. (Otherwise we could lower the potential energy by moving slightly along any
direction in q-space corresponding to a negative eigenvalue.)
1
Kinetic Energy: We assume that the kinetic energy is a time-independent quadratic form
in the velocities (thus ruling out some systems with time-dependent constraints). Then
1
T = T ({q}, {q̇}, 6 t) = q̇i Tij q̇j
2
where in general the matrix Tij will depend in some arbitrary way on the coordinates {q}.
However, for small displacements about the equilibrium position {q}eq = {0} we can again
take Tij ≡ Tij ({q = 0}) to be a constant matrix. It is convenient to choose Tij to be
symmetric. This is always possible since any asymmetric form T̃ij can be symmetrised to
(T̃ij + T̃ji )/2 without changing the above result for T . (Check for yourself.) Since kinetic
energies are generally positive, the matrix Tij is positive definite: all the eigenvalues are
greater than zero.
The Lagrangian: For small displacements, we have the Lagrangian
1
L = T − V = (q̇i Tij q̇j − qi Vij qj )
2
Lagrange’s equations become
d
dt
⇒
∂L
∂ q̇i
!
=
∂L
∂qi
Tij q̈j = −Vij qj
i = 1, · · · , D
By assuming small displacements they have been linearised; accordingly they can be solved
by matrix methods.
Normal Modes: These are solutions of the equations of motion in which all the generalised
coordinates oscillate with the same angular frequency ω:
qj = aj cos(ωt + j )
In such a motion, the equations of motion read
or, setting λ ≡ ω 2 ,
−ω 2 Tij aj + Vij aj = 0
(Vij − λTij )aj = 0
This is a generalised eigenvalue/eigenvector problem, ie it’s not the standard one, which
would be to solve (Mij − λδij )aj = 0 where Mij is symmetric and δij is the ij th identity
of the identity matrix – see, eg MP2B or FoMP lecture notes. Fortunately, the method of
solution is the same:
1. Find the eigenvalues λ(1) , λ(2) , · · · , λ(D) by solving
det (V − λT ) = 0
where the D × D matrices V and T have components Vij and Tij respectively.
2. For a given λ(k) , write
(k)
(Vij − λ(k) Tij )ej = 0
This defines an (un-normalised) eigenvector e(k) for the k th mode, with j th component
(k)
ej . There is a special feature here: since Tij is not δij , these eigenvectors are in general
not orthogonal. However, they are complete (proved later).
2
Example: A symmetric linear triatomic molecule consists of masses m, M, m connected
by springs of stiffness K. Find the normal modes and solve for the general motion.
Set up coordinates as follows:
x1
x2
x3
K
K
m
m
M
1
1
m(ẋ21 + ẋ23 ) + M ẋ22
2
2
1
1
1
=
K(x1 − x2 )2 + K(x3 − x2 )2 = K(x21 + 2x22 + x23 − 2x1 x2 − 2x3 x2 )
2
2
2
T =
V
Choosing (q1 , q2 , q3 ) = (x1 , x2 , x3 ), T and V take the required matrix form with




K −K
0


Vij =  −K 2K −K 
0 −K
K
m 0 0


Tij =  0 M 0 
0 0 m
Our eigenproblem is then


K − λm
−K
0

2K − λM
−K 
det  −K
=0
0
−K
K − λm
Expanding the determinant (exercise) gives
λ (K − λm) (λmM − (M + 2m)K) = 0
which has roots
λ(1) = 0
λ(2) = K/m
λ(3) = K(2m + M )/M m
Now to find the eigenvectors e(k) we solve
(k)
(Vij − λ(k) Tij ) ej = 0
The resulting unnormalised eigenvectors are (another exercise)
e(1) = (1, 1, 1)
e(2) = (1, 0, −1)
e(3) = (1, −2m/M, 1)
3
These correspond to:
1. Uniform translation of the entire molecule : this is called a “zero mode” for which
there is no restoring force and hence zero oscillation frequency, ω1 = 0;
2. Oscillation at ω2 = (K/m)1/2 with the two outer masses in antiphase and the central
one stationary;
3. Oscillation at ω3 = [(K(2m + M )/(mM )]1/2 with the two outer masses moving in the
same direction and the central mass moving in the opposite sense (at fixed centre of
mass for the system).
The general solution for the motion is then formally obtained as usual by superposing normal
modes:
q(t) = A1 e(1) cos(ω1 t + 1 ) + A2 e(2) cos(ω2 t + 2 ) + A3 e(3) cos(ω3 t + 3 )
or, in terms of components
(1)
(2)
(3)
qi (t) = A1 ei cos(ω1 t + 1 ) + A2 ei cos(ω2 t + 2 ) + A3 ei cos(ω3 t + 3 )
The A’s and the ’s are fixed by the initial conditions (6 integration constants for our 3,
second order DE’s, as you would expect).
Technical Note: In this particular case, ω1 = 0, but A1 could be arbitrarily large. By
taking the limit properly one can recover (exercise)
lim [A1 cos(ω1 t + 1 )] = A + vt
ω→0
where A and v are integration constants. To avoid this pathology one normally restricts attention to positive definite Vij , for which all the frequencies are strictly positive. Zero modes,
which usually correspond to conserved momenta associated with symmetries (displacement
of the entire molecule, in this case) can usually be spotted and the relevant coordinates
treated separately.
Simultaneous Diagonalisation of the Matrices T and V : The matrix T is positive
definite. We now show that a coordinate transformation exists that diagonalises both T and
V (both symmetric) simultaneously. The argument is formal but quite powerful.
1. We apply an orthogonal transformation C to T and V , such that T becomes diagonal:
T 0 = CT T C
V 0 = CT V C
where C T is the transpose of C (recall CijT ≡ Cji , and, since C is orthogonal, we also
have C T = C −1 ). This diagonalisation is always possible because T is symmetric. The
matrix T 0 is now


T1 .
.
.
 . T
.
. 

2
T0 = 


 .
. Tk . 
.
.
. TD
4
2. Now apply a second transformation matrix (this is not an orthogonal transformation)




K=
(T1 )−1/2
.
.
.
.
(T2 )−1/2
.
.
.
.
(Tk )−1/2
.
.
.
.
(TD )−1/2





This is possible so long as T is positive definite, as assumed. After this has been done,
we have
T 00 = K T C T T CK = I
V 00 = K T C T V CK
so that T has become the unit matrix, and V 00 is still symmetric (exercise: prove it).
3. Finally, apply a second orthogonal transformation D to diagonalise V . Since T 00 = I
is already the unit matrix, this is unaffected and we have T 000 = I also, whereas



V 000 = 

λ1 .
.
.
. λ2 .
.
.
. λk .
.
.
. λD





After all this, we have transformed our matrices to
T 000 = M T T M = I
V 000 = M T V M
where M = CKD. Equivalently, we have transformed our coordinates according to
qi → q̃i
with qi = Mij q̃j
which is not an orthogonal transformation (due to the second step) but is invertible.
Hence the set {q̃} are an acceptable set of generalised coordinates: no information
has been lost during the transformations. (This is the proof promised earlier of the
completeness of such a set.)
Normal Coordinates: In terms of the new set of transformed q’s (we now drop the tilde),
we have Tij = δij and V is diagonal with eigenvalues λ(k) . Hence the Lagrangian reads
X 1
1
1
1
L = q̇i Tij q̇j − qi Vij qj =
q̇i2 − λi qi2
2
2
2
2
i
where in the last form, and from now on, the summation convention is abandoned. This is
called the normal form and the transformed coordinates are normal coordinates. Lagrange’s
equations read simply
q̈i = −λi qi
i = 1, · · · , D
with general solution
qi (t) = Ai cos(ωi t + i )
1/2
where ωi = λi , and Ai and i are integration constants (fixed by the initial conditions).
There is no summation over modes left to do here: each normal coordinate executes pure
SHM independent of the others. All of the effort in solving for the motion is thereby moved
5
formally into the task of finding the transformation laws between these normal coordinates
and some “natural” ones (ie, whatever set we would like to have the answer in terms of).
In practice, one does not usually go through this formal sequence of transformations separately for each problem encountered. The transformation between natural ({q}) and normal
({q̃}) coordinates boils down to qi = Mij q̃j , where (as would be expected from examples of
single rather than double diagonalisation) M is the matrix of eigenvectors with components
Mij :
(j)
Mij = ei
Hence the task of “finding the normal coordinates and the normal form of L” is neither
more, nor less, than finding the eigenvalues and eigenvectors of (Vij − λTij ) aj = 0 as was
done for the triatomic molecule above.
6