MP350 Classical Mechanics Jon-Ivar Skullerud October 16, 2014

MP350 Classical Mechanics
Jon-Ivar Skullerud
October 16, 2014
1
Contents
1 Lagrangian mechanics
1.1
5
The principle of least action . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.1.1
Hamilton’s principle . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.2
The Euler–Lagrange equations . . . . . . . . . . . . . . . . . . . . . . . .
7
1.3
Generalised coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
1.3.1
1.4
1.5
1.6
1.7
Polar and spherical coordinates . . . . . . . . . . . . . . . . . . . 16
Lagrange multipliers [Optional] . . . . . . . . . . . . . . . . . . . . . . . .
17
1.4.1
Velocity constraints and differential constraints . . . . . . . . . .
17
1.4.2
Lagrange undetermined multipliers . . . . . . . . . . . . . . . . .
18
Canonical momenta and conservation laws . . . . . . . . . . . . . . . . .
19
1.5.1
. . . . . . . . . . . . . . . . . . . . . . . . .
20
Energy conservation: the hamiltonian . . . . . . . . . . . . . . . . . . . .
22
1.6.1
When is H (not) conserved? . . . . . . . . . . . . . . . . . . . . .
23
1.6.2
When is H (not) equal to the total energy?
24
Angular momentum
. . . . . . . . . . . .
Lagrangian mechanics — summary sheet . . . . . . . . . . . . . . . . . . 29
2 Hamiltonian dynamics
31
2.1
Hamilton’s equations of motion . . . . . . . . . . . . . . . . . . . . . . .
32
2.2
Cyclic coordinates and effective potential . . . . . . . . . . . . . . . . . .
34
2.3
Hamilton’s equations from a variational principle . . . . . . . . . . . . .
36
2.4
Phase space [Optional] . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
2.5
Liouville’s theorem [Optional] . . . . . . . . . . . . . . . . . . . . . . . . .
40
2.6
Poisson brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42
2.6.1
Properties of Poisson brackets . . . . . . . . . . . . . . . . . . . .
42
2.6.2
Poisson brackets and conservation laws . . . . . . . . . . . . . . . 44
2
2.6.3
2.7
The Jacobi identity and Poisson’s theorem . . . . . . . . . . . . .
45
Hamiltonian dynamics — summary sheet . . . . . . . . . . . . . . . . . .
47
3 Central forces
49
3.1
One-body reduction, reduced mass . . . . . . . . . . . . . . . . . . . . .
49
3.2
Angular momentum and Kepler’s second law . . . . . . . . . . . . . . . .
52
3.3
Effective potential and classification of orbits . . . . . . . . . . . . . . . .
54
3.4
Integrating the energy equation . . . . . . . . . . . . . . . . . . . . . . .
55
3.5
The inverse square force, Kepler’s first law . . . . . . . . . . . . . . . . .
56
3.6
More on conic sections . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59
3.6.1
Ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
60
3.6.2
Parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
3.6.3
Hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
3.7
Kepler’s third law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63
3.8
Kepler’s equations
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
3.9
Central forces — summary sheet . . . . . . . . . . . . . . . . . . . . . . .
67
4 Rotational motion
4.1
How many degrees of freedom do we have? . . . . . . . . . . . . . . . . . 70
4.1.1
4.2
4.3
4.4
69
Relative motion as rotation . . . . . . . . . . . . . . . . . . . . .
71
Rotated coordinate systems and rotation matrices . . . . . . . . . . . . .
72
4.2.1
Active and passive transformations . . . . . . . . . . . . . . . . .
72
4.2.2
Elementary rotation matrices . . . . . . . . . . . . . . . . . . . . 73
4.2.3
General properties of rotation matrices . . . . . . . . . . . . . . .
73
4.2.4
The rotation group [optional] . . . . . . . . . . . . . . . . . . . . .
76
Euler angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
77
4.3.1
Rotation matrix for Euler angles . . . . . . . . . . . . . . . . . . 78
4.3.2
Euler angles and angular velocity . . . . . . . . . . . . . . . . . .
79
The inertia tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
4.4.1
Rotational kinetic energy . . . . . . . . . . . . . . . . . . . . . . . 80
4.4.2
What is a tensor? Scalars, vectors and tensors. . . . . . . . . . . .
4.4.3
Rotations and the inertia tensor . . . . . . . . . . . . . . . . . . . 85
4.4.4
Angular momentum and the inertia tensor . . . . . . . . . . . . .
3
83
85
4.5
4.6
4.7
Principal axes of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . .
86
4.5.1
Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88
Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89
4.6.1
The symmetric heavy top . . . . . . . . . . . . . . . . . . . . . .
90
4.6.2
Euler’s equations for rigid bodies . . . . . . . . . . . . . . . . . .
91
4.6.3
Stability of rigid-body rotations . . . . . . . . . . . . . . . . . . .
92
Rigid body motion — summary sheet . . . . . . . . . . . . . . . . . . . .
94
4
Chapter 1
Lagrangian mechanics
1.1
The principle of least action
The starting point for the reformulation of classical mechanics is the principle of least
action, which may be somewhat flippantly paraphrased as “The world is lazy”, or in
the more flowery words of Pierre Louis Maupertuis (1744), Nature is thrifty in all its
actions:
The laws of movement and of rest deduced from this principle being precisely
the same as those observed in nature, we can admire the application of it to
all phenomena. The movement of animals, the vegetative growth of plants
. . . are only its consequences; and the spectacle of the universe becomes so
much the grander, so much more beautiful, the worthier of its Author, when
one knows that a small number of laws, most wisely established, suffice for
all movements.
This very general formulation does not in itself have any predictive power, but the
idea that nature’s “thrift” could be used to derive laws of motion had already been
successfully applied in optics for a long time:
Fermat’s principle
The path take between two points by a ray of light is the path that can be traversed
in the least time.
This principle was first formulated by Ibn al-Haytham (aka Alhacen) in his Book of
Optics from 1021, which formed one of the main foundations of geometric optics and
the scientific method in general. He proved that it led to the law of reflection. It was
restated by Pierre de Fermat in 1662, who also derived Snell’s law of refraction from
this principle.
5
1.1.1
Hamilton’s principle
In mechanics the proper mathematical formulation of Maupertuis’ principle is due to
William Rowan Hamilton1 , building on earlier work by Joseph Louis Lagrange.
We will denote the kinetic and potential energy of a particle, or of a mechanical system
in general, as
T = kinetic energy
V = potential energy
T usually depends on the velocities
V usually depends on the positions
but may also depend explicitly on time
(for example with time-varying extermal
i
≡ x˙ i
vi = dx
dt
xi
t
forces).
T = T (x˙ i
V = V (xi )
V = V (xi , t)
xi and x˙ i here denote all the coordinates and their time derivatives. So for example we
have
xi → x
xi → {x, y, z}
xi → {x1 , y1 , z1 , x2 , y2 , z2 , . . . , xN , yN , zN }
for a single particle in one dimension
for a particle in three dimensions
for N particles in three dimensions
We now define the lagrangian L as the difference between kinetic and potential energy,
L(xi , x˙ i , t) = T − V .
(1.1)
Note that L will be a function of the coordinates xi , the velocities x˙ i , and the time t,
although in many cases there is no explicit time dependence; ie, if we know the positions
and velocities of all the particles we know the lagrangian.
A particular path is given by specifying the coordinates xi as a function of time, xi =
xi (t). (Note that if xi (t) is known, its derivative x˙ i (t) is also known.) For a given path,
the action S is defined as
S[x] ≡
Zt2
L(x(t), x(t),
˙
t)dt .
t1
We are now in a position to formulate Hamilton’s principle of least action.
1
On a General Method in Dynamics, Phil. Trans. Roy. Soc. (1834) 247; (1835) 95.
6
(1.2)
The principle of least action:
The physical path a system will take between two points in a certain time interval is
the one that gives the smallest action S.
Comments:
1. The potential energy V is defined only for conservative forces, so the action as it
is written here is defined only for conservative forces. It is possible to generalise
this to certain non-conservative forces and obtain the correct equations of motion
(we will see examples of this later). However, all microscopic (fundamental) forces
are conservative.
2. The action S is a “function of a function” since it depends on the function(s) xi (t).
We call this a functional, and denote it by putting the function argument in square
brackets, S = S[x].
1.2
The Euler–Lagrange equations
What does ‘the path that gives the smallest action’ actually
mean, and how can we find it? To work this out, let us
consider a path x(t) and another path x′ (t) = x(t) + αh(t),
where h(t) is some arbitrary smooth function of t, and α is
a parameter that we will vary.
x’(t)
x2(t2)
x(t)
Since we are looking for the path the system will take be- x (t )
tween two specific points in a specific time interval, the
endpoints of the two paths must be the same. We therefore
have
x(t1 ) = x′ (t1 ) = x1 ; x(t2 ) = x′ (t2 ) = x2 ⇐⇒ h(t1 ) = h(t2 ) = 0 .
1 1
(1.3)
We can now write S[x + αh] = S(α), and treat it as a function of the parameter α. For
a given h(t), the minimum of S will occur when dS
= 0.
dα
This allows us to restate the principle of least action:
For any smooth hi (t) with hi (t1 ) = hi (t2 ) = 0, the physical path xi (t) is such that
d
d
S[x + αh] =
dα
dα
Zt2
L(xi + αhi , x˙ i + αh˙ i , t)dt = 0 .
(1.4)
t1
We often use the shorthands αh = δx and S[x + δx] − S[x] = δS = the variation of S,
the functional derivative of S. The principle of least action is then often
and call δS
δx
written as
δS = 0 or
δS
=0
δx
⇐⇒
d
S[x + αh] = 0 for any h(t) .
dα
7
(1.5)
Let us now calculate the variation δS. For a single particle in one dimension, we have
Z t2
d
d
˙ t)dt
S[x + αh] =
L(x + αh, x˙ + αh,
(1.6)
dα
dα t1
Z t2 ∂L ˙
∂L
h+
h dt
(1.7)
=
∂x
∂ x˙
t1
t=t2 Z t2 Z t2
∂L
d ∂L
∂L
=
h dt
(1.8)
h dt +
h
−
∂ x˙ t=t1
dt ∂ x˙
t1 ∂x
t1
Z t2 ∂L
d ∂L
=
−
h(t)dt .
(1.9)
∂x dt ∂ x˙
t1
In the first step we used that L is a function of the three variables x, x,
˙ t, but t does not
depend on α. We can then use the chain rule for a function of two variables,
∂f dx ∂f dy
d
f (x, y) =
+
.
dα
∂x dα ∂y dα
In the second step we used integration by parts,
Z
Z
∂L
,
uvdt
˙ = uv − uvdt
˙
with u =
∂x
v = h.
In the final step the boundary term vanishes since h(t1 ) = h(t2 ) = 0.
But h(t) is a completely arbitrary smooth function, and we must have δS = 0 for any
h(t). This is only possible if the term within the brackets in (1.9) is 0 for all t, ie
d ∂L ∂L
−
=0
dt ∂ x˙
∂x
The Euler–Lagrange equation
(1.10)
If we have N coordinates xi , the derivation proceeds following the same steps. Using
the chain rule for a function of 2N variables, we find
d
L(x1 + αh1 , x2 + αh2 , . . . , xN + αhN , x˙ 1 + αh˙ 1 , x˙ 2 + αh˙ 2 , . . . , x˙ N + αh˙ N )
dα
∂L
∂L
∂L
∂L ˙
∂L ˙
∂L ˙
=
h1 +
h2 + · · · +
hN +
h1 +
h2 + · · · +
hN
∂x1
∂x2
∂xN
∂ x˙ 1
∂ x˙ 2
∂ x˙ N
N X
∂L ˙
∂L
hi +
=
hi .
(1.11)
∂xi
∂ x˙ i
i=1
Using integration by parts on the second term (for each i) gives us
N
X
d
S[x + αh] =
dα
i=1
Z
t2
t1
∂L
∂xi
−
d ∂L hi (t)dt = 0 .
dt ∂ x˙ i
(1.12)
Since all the hi are independent, arbitrary functions, the expression within the brackets
must vanish for each i:
8
d ∂L
∂L
−
= 0 for all i = 1, . . . , N .
dt ∂ x˙ i ∂xi
(1.13)
Example 1.1 Particle in a potential
The kinetic energy of a single particle is
1
1
1
T = mv 2 = m(vx2 + vy2 + vz2 ) = m(x˙ 2 + y˙ 2 + z˙ 2 ) .
2
2
2
(1.14)
We take an arbitrary potential energy V = V (x, y, z, t).
The Euler–Lagrange equations are
d ∂L ∂L
−
= 0;
dt ∂ x˙
∂x
d ∂L ∂L
−
= 0;
dt ∂ y˙
∂y
d ∂L ∂L
−
= 0.
dt ∂ z˙
∂z
(1.15)
We find
∂L
∂V
∂L
= mx˙ ;
=−
∂ x˙
∂x
∂x
d ∂L ∂L
∂V
−
= m¨
x+
= 0 ⇐⇒
dt ∂ x˙
∂x
∂x
=⇒
(1.16)
m¨
x=−
∂V
.
∂x
(1.17)
Likewise, we get
m¨
y=−
∂V
,
∂y
m¨
z=−
−
→
→
→
¨r = m−
m−
a = −∇V = F
∂V
∂z
or
= Newton’s 2nd law!
So the Euler–Lagrange equations are exactly equivalent to Newton’s laws.
Example 1.2 The shortest path between two points
In deriving the Euler–Lagrange equations we did not make any use of the definition
of L: it could be any function of x, x˙ and t. Therefore, the EL equations give the
stationary points for any functional of the path between two points — for example,
the length of the path!
Consider a curve y = y(x) between two points (x1 , y1 ) and (x2 , y2 ). The length ds
of an infinitesimal segment (dx, dy) of this curve is given by Pythagoras:
ds2 = dx2 + dy 2 = dx2 + (y ′ (x)dx)2 = (1 + y ′ (x)2 )dx2
p
=⇒ ds = 1 + y ′ (x)2 dx .
(1.18)
(1.19)
If, to make life simpler for ourselves, we assume that x is monotonically increasing
9
along the curve, we find that the total length of the curve is
Z x2
Z x2 p
q
′
′
2
L(y(x), y (x), x)dx with L = 1 + y ′ 2 . (1.20)
1 + y (x) dx =
S=
x1
x1
This looks like what we had before, but with t → x; x(t) → y(x); x(t)
˙
→ y ′ (x).
The Euler–Lagrange equation becomes
d ∂L ∂L
−
= 0.
dt ∂y ′
∂y
(1.21)
We see immediately that ∂L/∂y = 0. To find ∂L/∂y ′ we use the chain rule,
To find
d ∂L
dx ∂y ′
√
dL ∂u
∂L
′2
=
with
u
=
1
+
y
;
L
=
u
∂y ′
du ∂y ′
∂L
y′
1
′
p
p
=⇒
· 2y =
.
=
∂y ′
2 1 + y′2
1 + y′2
we use the product rule and the chain rule:
1
= u−1/2
with v = y ′ , w = p
2
′
1+y
′
′
dv
dw du
dy
d ∂L
′ dw du dy
=
w
+
v
=
w
+
y
=⇒
dx ∂y ′
dx
du dx
dx
du dy ′ dx
1
1 −3/2 ′
= y ′′ p
+
y
·
−
· 2y ′ · y ′′
u
2
′
2
1+y
2
y′2 y′
1
y ′′
′′
1−
−
=y p
=p
2
1 + y′2
1 + y ′ 2 (1 + y ′ )3/2
1 + y′2
y ′′
1
=p
2 .
1 + y′2 1 + y′
d ∂L
So
= 0 =⇒ y ′′ (x) = 0 =⇒ y(x) = Ax + B .
dx ∂y ′
∂L
= vw
∂y ′
(1.22)
(1.23)
(1.24)
This describes a straight line, so we have shown that the shortest path between two
points is a straight line!
So what is the point?
1. The equations are often easier : We get rid of complicated vectors and forces, and
derive everything from scalars (energy).
2. It is easier to generalise to systems with constraints.
10
3. We can choose whichever coordinates we want.
4. The lagrangian formalism can be generalised to quantum mechanics (in the Feynman formulation: all paths are possible, but weighted by the action) and field
theory (with infinitely many degrees of freedom).
We will look at points 2 and 3 next.
1.3
Generalised coordinates
It is often advantageous to change variables from the cartesian coordinates {xi , yi , zi }
for each particle i = 1, . . . , N to some other variables {qj }, j = 1, . . . , n. These are called
generalised coordinates.
Consider for example a system of N particles. We need 3N independent coordinates to
describe the system completely: we say that there are 3N degrees of freedom.
Now, imagine that there is a constraint relating the 3N coordinates, for example:
1. Two particles are tied together with a rod of length L, so that
(x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 = L2 .
(1.25)
2. The N particles are all moving on the surface of a sphere, ie
x2i + yi2 + zi2 = R2
∀i = 1, . . . , N .
(1.26)
3. A ball in a squash court, x ≥ 0, z ≥ 0.
The first two of these can be described by M equations of the form
→
→
fj (−
x 1, . . . , −
x N , t) = 0 ,
j = 1, . . . , M .
(1.27)
Such constraints are called holonomic (or integrable) constraints, and we will focus only
on such constraints in the following. With such constraint equations, the coordinates
xi , yi , zi are no longer independent. Instead we have
M relations
=⇒
n = 3N − M real degrees of freedom.
By choosing n suitable generalised coordinates to describe these degrees of freedom, we
achieve two things:
• We eliminate the forces of constraints which are required in the newtonian formulation. No net work is done by these forces, so they can safely be eliminated.
11
• The Euler–Lagrange equations look exactly the same in the new coordinates, so
the problem is no more difficult (and probably easier) than the original one.
In the first example above, the constraint (1.25) reduces the number of degrees of freedom
from 6 to 5. The 5 coordinates can for example be chosen to be the centre of mass
coordinates X, Y, Z for the two particles, and two angles θ, φ that describe the orientation
of the rod.2
In the second example, each particle is described by 2 instead of 3 coordinates. These
can be chosen to be the ‘latitude’ θ and ‘longitude’ φ of each particle (corresponding to
spherical coordinates, see Sec. 1.3.1).
Example 1.3 Simple pendulum
Consider a simple pendulum with length ℓ, mass m in a constant gravitational field
g (see Fig. 1.1).
r
❆
❆
❆
❆
.
..........
......................
Here it is convenient to choose the angle θ as our coordinate. The x (horizontal) and z (vertical) coordinates
and their time derivatives can be written in terms of θ as
θ ❆
ℓ
❆
❆
❆
x = ℓ sin θ
z = −ℓ cos θ
❆
❆❆⑤m
x˙ = ℓθ˙ cos θ ,
z˙ = ℓθ˙ sin θ .
(1.28)
(1.29)
The kinetic energy is
1 →2 1
T = m−
v = m(x˙ 2 + z˙ 2 )
2
2
1
1 2 ˙6
= mℓ θ (cos2 θ + sin2 θ) = mℓ2 θ˙2 . (1.30)
2
2
Figure 1.1: A simple pendulum
The potential energy is
V = mgz = −mgℓ cos θ .
(1.31)
The lagrangian therefore becomes
1
L = T − V = mℓ2 θ˙2 + mgℓ cos θ .
2
(1.32)
The Euler–Lagrange equation is
∂L
d ∂L
=
∂θ
dt ∂ θ˙
=⇒
=⇒
−mgℓ sin θ =
g
θ¨ = − sin θ .
ℓ
d
mℓ2 θ˙
dt
This is the equation of motion for the pendulum.
2
In Chapter 4 we will look more at how these angles can be chosen.
12
(1.33)
(1.34)
Once we have found the equation of motion for θ, and the solution to this equation, we
can go back and calculate x and z as functions of time. However, in the example of the
simple pendulum, we are not usually interested in this.
We note that the mass m does not appear in the equation of motion. We could have
seen this already by inspecting the lagrangian: the EL equations are unchanged if the
lagrangian is multiplied by an overall constant α, L → αL. In this case, since the mass
just enters as an overall factor in the lagrangian, the EL equation will not depend on
the mass.
Solutions to the equations of motion?
Now we have found the equation of motion for the simple pendulum, and we may want
to know the solutions to this equation, ie what the actual motion of the pendulum is for
different initial conditions. It is actually possible to integrate the equation (1.34) and
write down a solution, but this involves elliptic integrals and lots of other complicated
maths, and will not help us to understand the physical system. It will be more useful to
find numerical solutions, and in Computational Physics MP354 we will learn how this
can be done.
What we can do to understand the system better, is
• look at the general types of solutions we may have. We will do this when we
discuss conservation of energy;
• consider limiting cases such as small oscillations. This is what we will do now.
If θ is small, we may approximate sin θ with the first term in its power expansion (Taylor
expansion),
1
1
sin θ = θ − θ2 + θ5 + · · · ≈ θ .
(1.35)
3!
5!
In that case (1.34) simplifies to
g
θ¨ = − θ .
(1.36)
ℓ
We recognise this as the equation for a simple harmonic oscillator, x¨ + ω 2 x = 0, with
x → θ, ω 2 → g/ℓ. We therefore see that for small oscillations,
p the simple pendulum
behaves as a simple harmonic oscillator with frequency ωs = g/ℓ, ie the frequency is
inversely proportional to the square root of the length of the pendulum (and independent
of the mass).
13
Example 1.4 Double Atwood machine
Consider the double Atwood machine in Fig. 1.2. We
assume that
..............................
.....
...
...
.....
....
① .....
...
.
..
....
....
.......
.........................
1. the pulleys are light, so we can ignore their kinetic energy; and
2. the ropes do not slip (or they slide without friction).
x
ℓ1 − x
❄
Here we have two independent degrees of freedom,
which we can choose to be x and y. In terms of these,
the positions of the three blocks are
x1 = −x ,
x2 = −(ℓ1 − x + y) ,
x3 = −(ℓ1 − x + ℓ2 − y) .
T2
T3
V1
V2
V3
..........................
.......
....
...
...
...
s ❄ .....
....
...
...
....
...
.
.......
.
.
.
......................
y
ℓ2 − y
❄
m3
The kinetic and potential energy of the three blocks
are
T1
m1
1
= m1 x˙ 21
2
2 1
1 d
= m2
(ℓ1 − x − y) = m2 (y˙ − x)
˙ 2
2
dt
2
1
˙ 2
= m3 (x˙ + y)
2
= −mgx
= −m2 g(ℓ1 − x + y)
= −m2 g(ℓ1 + ℓ2 − x − y)
❄
m2
Figure 1.2: Double Atwood
machine
The lagrangian becomes
1
1
L = (m1 + m2 + m3 )x˙ 2 + (m1 + m2 )y˙ 2 + (m3 − m2 )x˙ y˙
2
2
+ (m1 − m2 − m3 )gx + (m2 − m3 )gy + m2 gℓ2 + m3 g(ℓ1 + ℓ2 ) .
Note that the last two terms are constants which do not play any role in the equations
of motion. We get two equations of motion:
∂L
d ∂L
= (m1 + m2 + m3 )¨
x + (m3 − m2 )¨
y=
= (m1 − m2 − m3 )g
dt ∂ x˙
∂x
∂L
d ∂L
= (m2 + m3 )¨
y + (m3 − m2 )¨
x=
= (m2 − m3 )g
dt ∂ y˙
∂y
14
Example 1.5 Pendulum with rotating support
Consider a pendulum mounted on the edge of a disc
with radius a, rotating with constant angular velocity
ω (see Fig. 1.3). If the support point is in the horizontal position at t = 0, the angular position of the
support point at time t is φ = ωt, and the cartesian
coordinates of the bob at time t are
........
.................... .............................
........
..........
.......
......
......
.....
.....
....
........
.
.
.
.
.
.
.
.
.
.
....
..
.....
.
.
.
.
.
.
.
...
... ......
...
.
...
..
..
.
.
...
..
....
...
...
...
..
...
..
...
..
..
..
..
..
...
..
.
.
...
..
...
.
...
...
...
...
....
...
....
....
.
.....
.
....
......
.......
......
..........
.......
.........
................
...................................
✌
ω
s
a✚✚✚❈
❈
s✚ ωt
❈
❈
❈
❈
❈
❈
❈ ℓ
.................................
x = a cos ωt + ℓ sin θ
z = a sin ωt − ℓ cos θ
θ ❈
giving the velocities
x˙ = −aω sin ωt + ℓθ˙ cos θ
z˙ = aω cos ωt + ℓθ˙ sin θ
❈
❈
❈
❈
❈
❈ m
❈⑤
Figure 1.3: Pendulum with
rotating support.
This gives us the lagrangian
1
L = T − V = m(x˙ 2 + z˙ 2 ) − mgz
2
m 2 2
=
a ω + ℓ2 θ˙2 + 2aωℓθ˙ sin(θ − ωt) − mg a sin ωt − ℓ cos θ
2
This system has only one degree of freedom θ, but the lagrangian depends explicitly
on time because of the rotation of the support point. The Euler–Lagrange equation
is
d
d ∂L
=
mℓ2 θ˙ + maωℓ sin(θ − ωt) = mℓ2 θ¨ + maωℓ(θ˙ − ω) cos(θ − ωt)
dt ∂ θ˙
dt
∂L
= maωℓθ˙ cos(θ − ωt) + mgℓ sin θ
=
∂θ
aω 2
g
=⇒ ℓθ¨ − aω 2 cos(θ − ωt) = g sin θ =⇒ θ¨ =
cos(θ − ωt) − sin θ
ℓ
ℓ
Finding the equation of motion for this system becomes a bit complicated, but it is
still far simpler than it would have been to compute the forces at each point and use
Newton’s second law.
We can check that our result is sensible by seeing what happens if there is no rotation,
ie ω = 0. In this case the system reduces to the simple pendulum, and the equation
of motion should be the same. We can immediately see that this is the case.
It is worth noting that the potential energy contains a time-dependent term
mga sin ωt, which one naively would think should contribute to the dynamics of
the system — however, it plays no role since it does not contain the coordinate.
There is also a constant term ma2 ω 2 /2 in the kinetic energy which plays no role.
15
1.3.1
Polar and spherical coordinates
When we have rotational motion, or a system with rotational (or spherical) symmetry,
it is very often most convenient to use polar coordinates (in 2 dimensions) or spherical
coordinates (in 3 dimensions). The definition of these coordinates are given in Fig. 1.4.
Since we will be using them often, we need to know what the kinetic energy of a particle
is in terms of these coordinates.
z
y
✻
✻
θ. ✁
........................ ✁
✁✕
✁
✁
r
✁
✁
.....
✟✟
✟
✟✯
✟.....✟
✟ ..... θ
✁
✁
r
✲
y
.
✟
✟
...
..
..
.
✲
......
........
.............. ...................
.............
φ
x
❙
❙
x✠
Figure 1.4: Plane polar coordinates (r, θ) (left) and spherical coordinates (r, θ, φ) (right).
Polar coordinates
The relation between cartesian and polar coordinates is given by
x = r cos θ
=⇒
y = r sin θ
=⇒
x˙ = r˙ cos θ − rθ˙ sin θ
y˙ = r˙ sin θ + rθ˙ cos θ
(1.37)
(1.38)
This gives for the kinetic energy,
1
T = m(x˙ 2 + y˙ 2 )
2
m 2
= (r˙ cos2 θ + r2 θ˙2 sin2 θ − 2rr˙ θ˙ cos θ sin θ + r˙ 2 sin2 θ + r2 θ˙2 cos2 θ + 2rr˙ θ˙ cos θ sin θ)
2
m
= (r˙ 2 + r2 θ˙2 ) . (1.39)
2
16
Spherical coordinates
The relation between cartesian and spherical coordinates is given by
x = r sin θ cos φ
=⇒
y = r sin θ sin φ
=⇒
z = r cos θ
x˙ = r˙ sin θ cos φ + rθ˙ cos θ cos φ − rφ˙ sin θ sin φ
y˙ = r˙ sin θ sin φ + rθ˙ cos θ sin φ + rφ˙ sin θ cos φ
z˙ = r˙ cos θ − rθ˙ sin θ
=⇒
(1.40)
(1.41)
(1.42)
Using this we find that
1
1
T = m(x˙ 2 + y˙ 2 + z˙ 2 ) = m(r˙ 2 + r2 θ˙2 + r2 φ˙ 2 sin2 θ) .
2
2
(1.43)
The complete derivation is left as an exercise.
1.4
Lagrange multipliers
[Optional]
Using the constraint equations to reduce the number of coordinates is usually the most
straightforward way of handling constraints. But it is not always practical:
• It may not be straightforward to solve the constraint equations.
• The constraint equations may involve velocities.
• The constraint equations may be expressed as differential rather than algebraic
equations.
• We may want to know the forces of constraint (for example, to find out when they
become too large or too small to physically constrain the system).
1.4.1
Velocity constraints and differential constraints
A constraint relation of the form
gα (x, x,
˙ t) = 0 ,
x = {x1 , . . . , xN }
(1.44)
involving velocities x˙ i , can in general not be integrated to yield relations between coordinates. However, consider the equation
X
Ai (x, t)x˙ i + B(x, t) = 0 , .
(1.45)
i
If we can write
Ai =
∂f
∂xi
B=
∂f
∂t
17
with f = f (x, t) ,
(1.46)
then (1.45) is equivalent to an algebraic relation among the coordinates xi :
X ∂f dxi ∂f
df
+
=
=0
∂x
dt
∂t
dt
i
i
⇐⇒
f (x, t) − C = 0 .
(1.48)
This is an example of a differential constraint
X ∂fj
∂fj
dqi +
dt = 0 .
∂qi
∂t
i
1.4.2
(1.47)
(1.49)
Lagrange undetermined multipliers
Assume now that we are keeping our original coordinates qi (whatever they are) and
write the constraint equations on the differential form (1.49). We can now try to rederive
the Euler–Lagrange equations.
Consider a variation qiα (t) = qi (t) + αhi (t). We find, as before,
d
dS
=
dα
dα
Z
t2
t1
L(qiα , q˙iα , t)dt
=
Z
N t2 X
t1
i=1
d ∂L ∂L
−
hi dt = 0 .
∂qi dt ∂ q˙i
(1.50)
However, the functions hi are no longer independent, since the qiα must satisfy the
equations (1.49). In particular, we must have
X ∂fj dq α X ∂fj
i
=
hi = 0 .
(1.51)
∂q
dα
∂qi
i
i
i
Note that the second term in (1.49) does not contribute, since dt/dα = 0 as t is not
varied.
If we have m constraint equations, we can use them to eliminate m of the functions hi :
N −1
X ∂f1
∂f1
hN (t) = −
hi (t)
∂qN
∂qi
i=1
⇐⇒
hN (t) = −
X
i = 1N −1
∂f1 /∂qi
hi (t) (1.52)
∂f1 /∂qN
N −2
X ∂f2
∂f2
∂f2
hN −1 (t) = −
hi (t) −
hN (t)
∂qN −1
∂qi
∂qN
i=1
=−
N
−2
X
i=1
N −2
∂f2 /∂qN X ∂f1
∂f2 /∂qN ∂f1
∂f2
hi (t) −
hi (t) −
hN −1 (t)
∂qi
∂f1 /∂qN i=1 ∂qi
∂f1 /∂qN ∂qN −1
(1.53)
. . . etc.
This becomes messy and not very illuminating in general, so let us consider just the
case with m = 1, N = 3. Eq. (1.52) then becomes
h3 = −
∂f /∂q1
∂f /∂q2
h1 −
h2
∂f /∂q3
∂f /∂q3
18
(1.54)
and (1.50) becomes
Z t2 ∂L
∂L
d ∂L d ∂L d ∂L ∂L
h1 +
h2 +
h3 dt = 0
−
−
−
∂q1 dt ∂ q˙1
∂q2 dt ∂ q˙2
∂q3 dt ∂ q˙3
t1
Z t2 h
d ∂L ∂f /∂q1 ∂L
d ∂L i
∂L
−
h1
−
(
−
(
⇐⇒
∂q1 dt ∂ q˙1
∂f /∂q3 ∂q3 dt ∂ q˙3
t1
h ∂L
d ∂L ∂f /∂q2 ∂L
d ∂L i
+ (
−
(
−
−
h2 dt = 0
∂q2 dt ∂ q˙2
∂f /∂q3 ∂q3 dt ∂ q˙3
(1.55)
(1.56)
Since h1 and h2 are now both arbitrary functions, the terms inside the square brackets
must both be zero. So we have
∂L
∂L
d ∂L ∂f −1
d ∂L ∂f −1
(
=(
−
−
∂q1 dt ∂ q˙1 ∂q1
∂q3 dt ∂ q˙3 ∂q3
at all t .
(1.57)
−1
d ∂L
d ∂L ∂f −1
∂f
∂L
∂L
−
−
=(
(
∂q2 dt ∂ q˙2 ∂q2
∂q3 dt ∂ q˙3 ∂q3
The two sides of the equations can be any function of t; we call this function −λ(t).
This gives the equations
∂L
d ∂L
∂f
−
+ λ(t)
=0
∂qi dt ∂ q˙i
∂qi
(1.58)
The function λ(t) is called a Lagrange undetermined multiplier.
With m constraint equations, we get m functions λk (t), ie
m
∂L
d ∂L X ∂fk
−
+
λk (t) = 0
∂qi dt ∂ q˙i k=1 ∂qi
(1.59)
The Euler–Lagrange equations with Lagrange multipliers
We now have N + m unknown functions qi (t), λk (t), but we also have N + m equations:
the N EL equations and the m constraint equations. This will therefore completely
determine the system once the initial conditions are given.
If we know the Lagrange multipliers, we can find the (generalised) constraint forces Fi .
These are given by
X ∂fk
Fi =
λk .
(1.60)
∂qi
k
1.5
Canonical momenta and conservation laws
Assume the lagrangian L does not depend explicitly on the coordinate qi . Such coordinates are called cyclic. The Euler–Lagrange equations for the cyclic coordinate qi
19
becomes
∂L
d ∂L
=
=0
dt ∂ q˙i
∂qi
∂L
≡ pi = constant .
∂ q˙i
=⇒
(1.61)
We call the qunatity pi the canonical momentum conjugate to (or corresponding to) qi .
Why momentum?
Consider the ’usual’ case where
1. we use cartesian coordinates qi = xi ;
2. there are no constraints; and
3. the potential depends only on the coordinates, V = V (x).
In this case we have
1 X 2
x˙ j − V (x)
L=T −V = m
2
j
=⇒
∂L
= mx˙ i = pi = ordinary momentum.
∂ x˙ i
So we have found the law of conservation of momentum pi if the potential V does not
depend on the coordinate xi — ie, if the system is translationally invariant in the idirection. Note that if V does not depend on xi this implies that there are no net forces
in the i-direction.
We may in a similar way demonstrate conservation of total momentum for a system of
n particles if the potential energy does not depend on the centre of mass coordinate.
But the concept of canonical momenta is much more general and powerful than this,
and can be used to derive a whole host of other conservation laws. One of the most
important is angular momentum, which we will look at next.
1.5.1
Angular momentum
Consider a one-particle rotationally symmetric 2-dimensional system, and let us use
polar coordinates (r, θ) to describe the parrticle. Rotational symmetry then means that
the potential energy V (r, θ) = V (r), independent of the angle θ. The lagrangian is then
1
L = T − V = m(r˙ 2 + r2 θ˙2 ) − V (r) .
2
(1.62)
We see that θ is a cyclic coordinate, and the canonical momentum pθ is therefore conserved. What is this canonical momentum?
We straightforwardly find
pθ =
∂ 1 2 ˙2 ∂L
=
mr θ = mr2 θ˙ .
˙
˙
2
∂θ
∂θ
20
(1.63)
But θ˙ is the same as the angular velocity ω, and we know that the velocity vθ in the
˙ so pθ = r(mvθ ). But
angular direction (perpendicular to the radius r) is vθ = rω = rθ,
this is exactly equal to the angular momentum of the particle,
→
→
Lz = (−
r ×−
p )z = rvθ .
(1.64)
So the canonical momentum conjugate to the angle θ is the angular momentum, which
is conserved if the system is rotationally symmetric, ie the lagrangian does not depend
on θ.
Angular momentum in spherical coordinates
In section 1.3.1 we found that the kinetic energy in spherical coordinates (see Fig. 1.4 is
1
T = m(r˙ 2 + r2 θ˙2 + r2 sin2 θφ˙ 2 ) .
2
(1.43)
The angle φ corresponds to rotations about the z-axis: if a particle rotates about the
z-axis, φ changes while r and θ are unchanged. If the potential energy does not depend
on φ, we have rotational symmetry about the z-axis, and the canonical momentum pφ
is conserved. From (1.43) we find
˙
pφ = mr2 sin2 θφ˙ = r(mr sin θφ)(sin
θ) .
(1.65)
We now want to show that this is equal to the z-component of the angular momentum,
→
→
Lz = (−
r ×−
p )z .
→
ˆ φ)
ˆ at the point −
We can put a coordinate system (ˆ
r, θ,
r
z
and decompose the velocity in its (r, θ, φ) components,
✻
rˆ
−
→
ˆ ✄✗✄✁✁✕
φ
v = vr rˆ + vθ θˆ + vφ φˆ
(1.66)
...
..
.... φ
....
✄✁
.......
....................................
✁
✕
❅
The unit vector rˆ denotes the radial direction, ie the
✁ ❅
❘
❅
θ.
direction where r changes, while θ, φ are unchanged.
.................... ✁
θˆ
....
✁
ˆ
Similarly, θ denotes the direction where θ changes while
✁
r, φ are unchanged, and φˆ denotes the direction where φ
✁
changes while r, θ are unchanged. The three vectors are
orthogonal, and φˆ is also orthogonal to zˆ, since motion Figure 1.5: Unit vectors in
spherical coordinates.
in the φ-direction does not change z.
The velocity component vφ is the rotational velocity about the z-axis, which again is
equal to the distance from the axis times the angular velocity about the axis. Since φ is
˙ The distance
the rotational angle about the z-axis, the angular velocity is dφ/dt = φ.
from the axis is r sin θ, so
vφ = (r sin θ)φ˙ .
(1.67)
→
→
We can now work out the vector product −
r × m−
v in the spherical coordinate system.
−
→
Since r = rˆ
r, we need the cross product of rˆ with each unit vector. Using the right-hand
rule we find
rˆ × θˆ = φˆ ,
rˆ × φˆ = −θˆ ,
rˆ × rˆ = 0 .
(1.68)
21
The z-component of the angular momentum is therefore
→
→
ˆ z = mr[vθ φˆ − vφ θ]
ˆ z = −mrvφ θˆ · zˆ . (1.69)
Lz = (−
r × m−
v )z = mr[ˆ
r × (vr rˆ + vθ θˆ + vφ φ)]
We now need to work out the scalar product θˆ · zˆ. Looking at the
→
figure on the right, we see that since θ is the angle of −
r with the
−
→
z-axis, and θˆ is orthogonal to r (but still in the z − r plane), the
projection of θˆ onto the z-axis is θˆ · zˆ = − sin θ.
z
✻θ ✁✁✕rˆ
..................✁
..
✁
✁ ...
❍❍
.
... θ
Therefore we find that the z-component of the angular momentum
❍❍ θˆ
❥
❍
-sin θ
is
→
→
˙ · sin θ = pφ .
Lz = (−
r × m−
v )z = −mrvφ (− sin θ) = r · (mr sin θφ)
(1.70)
So the canonical momentum pφ is indeed the angular momentum about the z-axis, and
it is conserved if we have rotational symmetry about the z-axis.
If we have full spherical symmetry, this means we have rotational symmetry about all 3
axes, so by the same argument as above Lx and Ly must also be conserved. Therefore,
−
→
→
→
for a spherically symmetric system, the angular momentum vector L = −
r ×−
p is
conserved.
Na¨ıvely one would think that if we have full rotational symmetry, the angle θ should
also be irrelevant, and the canonical momentum pθ should also be conserved. However,
this is not the case: although the potential energy does not depend on θ, the kinetic
energy does, through the term 12 mr2 sin2 θφ˙ 2 . This θ-dependence is an artefact of how
we have chosen the coordinate system, but it is an unavoidable artefact: no matter
how we choose our spherical coordinate angles, these coordinates must break the full
spherical symmetry somehow.
We realise the full symmetry by noting that we could have chosen the coordinates
differently, eg we could have chosen θ to be the angle with the x-axis and φ to correspond
to rotations about the x-axis — which would have led us to find that Lx is conserved.
Similarly, if we choose θ to be the angle with the y-axis we will find that Ly is conserved.
1.6
Energy conservation: the hamiltonian
We know that when we have conservative forces, the potential energy depends only
on positions, and not on time, and the total energy is conserved. We have derived
conservation of linear and angular momentum in lagrangian mechanics, so we may ask
ourselves if we can also derive energy conservation within the same framework?
The answer to this is that not only can we do this, but the energy conservation theorem
we arrive at is more general than the one we already know!
To see how this works, let us take the (total) time derivative of the lagrangian L =
22
L(qi (t), q˙i (t), t). Using the chain rule and the Euler–Lagrange equations we get
X ∂L
∂L
dL X ∂L
=
q˙i +
q¨i +
dt
∂qi
∂ q˙i
∂t
i
i
X d ∂L
X ∂L dq˙i ∂L
=
q˙i +
+
dt
∂
q
˙
∂
q
˙
dt
∂t
i
i
i
i
d X ∂L
∂L
=
q˙i +
dt
∂ q˙i
∂t
Xi
∂L
dH ∂L
∂L
d
q˙i − L +
≡
+
= 0,
⇐⇒
dt
∂
q
˙
∂t
dt
∂t
i
i
(1.71)
(1.72)
where we have defined
H=
X ∂L
i
∂ q˙i
q˙i − L =
X
i
pi q˙i − L = the hamiltonian
(1.73)
So we find that if the lagrangian does not depend explicitly on time, then the hamiltonian
or energy function H is conserved.
To see how this relates to energy conservation as we know it from before, consider a
system of particles in cartesian coordinates, described by the lagrangian
L=T −V =
1X
mi q˙i2 − V (q) .
2 i
The hamiltonian for this system is
H=
X ∂L
q˙i − L =
h1 X
i
X
(mi q˙i )q˙i −
mi q˙i2 − V (q)
2 i
i
∂ q˙i
1X
=
mi q˙i2 + V (q) = T + V .
2 i
i
(1.74)
So we find that the hamiltonian is equal to the total energy, so conservation of the
hamiltonian is the same as energy conservation in this particular (most common) case.
1.6.1
When is H (not) conserved?
We found that H is conserved if L does not depend explicitly on time, ie L(q, q,
˙ t) =
L(q, q).
˙ We would like to understand in what circumstances an explicit time dependence
could appear in the lagrangian. One possibility would be that the potential energy
depends explicitly on time in the first place. But there are also other possibilites. The
kinetic energy, written in terms of the original cartesian (or, for that sake, ordinary
23
polar or spherical coordinates) does not have any explicit time dependence. But time
dependence could still appear in either the kinetic or the potential energy when we write
it in terms of generalised coordinates.
To see how this can happen, let us recall why we introduced generalised coordinates in
the first place:
1. Constraints: There are fewer actual degrees of freedom in the system because of
constraints. We use generalised coordinates to denote the real (relevant) degrees
of freedom. An example of this would be the pendulum, where the original x and
z coordinates are reduced to the single coordinate θ.
2. Symmetries: There are symmetries in the system which mean that using noncartesian coordinates may give a simpler description. An example of this would
be using polar coordinates for a system with rotational symmetry.
Explicit time-dependence can appear in both those types of cases, leaving us with three
possibilites for how explicit time-dependence could appear in the lagrangian:
1. The potential energy is explicity time-dependent, V = V (x, t). Physically, this
means that there are external or non-conservative forces, so the energy of the
system is not conserved.
2. The constraints are time-dependent. An example of that would be Example 5,
the pendulum with rotating support. In such cases, external forces are usually
required to maintain the constraint, so the energy of the system is not conserved.
3. We have chosen to use time-dependent transformations xi = fi (q, t) between the
old coordinates x and the new coordinates q because this may simplify the description of the system. In this case, the hamiltonian may be not conserved even
if the total energy is conserved.
1.6.2
When is H (not) equal to the total energy?
We showed the hamiltonian H is equal to the total energy E when
L=T −V =
1X
mi q˙i2 − V (q) .
2 i
More generally, it is the case when
1. V is independent of the velocitiec q˙i , V = V (q, t), and
2. T is a homegeneous quadratic function of q,
˙
X
T =
aij q˙i q˙j , .
ij
24
Proof
Take
L=T =V
X
ij
aij q˙i q˙j − V (q, t)
(1.75)
We note that we can always arrange it so that aij = aji , since q˙i q˙j = q˙j q˙i . The canonical
momenta are
X
X
X
∂L
=
aik q˙i +
akj q˙j = 2
akj q˙j .
(1.76)
pk =
∂ q˙k
i
j
j
The two terms appear because we get a contribution both from the k = j term and
from the k = i term in the sum. The hamiltonian is then
X
X
X
H =
pi q˙i − L = 2
aij q˙i q˙j −
aij q˙i q˙j + V (q, t) = T + V = E , (1.77)
i
ij
ij
which completes the proof.
So when does T (not) have this form?
Let us start from cartesian coordinates, where
1X
mi x˙ 2i i = 1, . . . , 3N .
T =
2 i
(1.78)
We now introduce generalised coordinates qj , which are related to the xi via general,
time-dependent transformations,
xi = xi (q1 , . . . , qm , t)
(1.79)
Using the chain rule, we find
m
dxi X ∂xi
∂xi
x˙ i =
=
,
q˙j +
dt
∂qj
∂t
j=1
X
X
m
m
∂xi
∂xi
∂xi
∂xi
2
x˙ i =
q˙j +
q˙k +
∂qj
∂t
∂qk
∂t
j=1
k=1
2
m
m
X
X
∂xi
∂xi ∂xi
∂xi ∂xi
q˙j +
q˙j
q˙k + 2
.
=
∂qj ∂qk
∂qj ∂t
∂t
j=1
j,k=1
(1.80)
(1.81)
The kinetic enery is therefore
3N
1X
T =
mi x˙ 2i
2 i=1
2 (1.82)
m 3N
3N
m X
3N
X
1 X X ∂xi ∂xi
∂xi ∂xi
∂xi
1X
=
mi
q˙j q˙k +
mi
mi
q˙j +
.
2 j,k=1 i=1
∂qj ∂qk
∂qj ∂t
2 i=1
∂t
j=1 i=1
25
This can be written on the form
T =
X
ajk q˙j q˙k +
X
bj q˙k + c ,
(1.83)
j
jk
with
3N
ajk
1 X ∂xi ∂xi
q˙j
,
mi
=
2 i=1
∂qj ∂qk
bj =
3N
X
i=1
∂xi ∂xi
,
mi
∂qj ∂t
2
3N
1X
∂xi
c=
mi
.
2 i=1
∂t
If the transformations do not depend on time, xi = xi (q1 , . . . , qm ), then
bj = 0, c = 0 and therefore,
X
T =
aij q˙i q˙j .
∂xi
∂t
(1.84)
= 0, so
(1.85)
ij
So the kinetic energy will be a homogeneous quadratic function of the generalised velocities q˙i whenever the constraints or coordinate transformations do not depend on time.
Conversely, we may have H 6= E if
1. the potential V depends on velocities, or
2. the constraints of coordinate tranformations are time-dependent.
Example 1.6 Spring mounted on moving cart
Consider a body with mass m sitting at the end of a horizontal spring with spring
constant k, with the other end attached to a cart moving with a constant velocity
v. Since one end of the spring is fixed to the moving cart, the equilibrium point xo
of the body on the spring is also moving with velocity v. If we say that x0 = 0 when
t = 0, we therefore have x0 = vt.
The potential energy of the body is given by the displacement x−x0 from equilibrium,
V = 21 k(x−x0 )2 = 12 k(x−vt)2 . The kinetic energy is the usual one, so the lagrangian
is
1
1
(1.86)
L = T − V = mx˙ 2 − k(x − vt)2 .
2
2
The canonical momentum is px = mx,
˙ which gives us the hamiltonian
1
1
H = px x˙ − L = mx˙ 2 + k(x − vt)2 = T + V = E .
2
2
(1.87)
Since the lagrangian depends explicitly on time, the hamiltonian (and the total
energy) is not conserved. We can understand this by noting that the motor driving
the cart will have to do work to maintain a constant velocity; in the absence of this
the cart will undergo oscillations along with the body attached to the spring.
We can now introduce a new coordinate
=⇒
q = x − vt =⇒ q˙ = x˙ − v
1
1
1
mv 2
1
.
L(q, q,
˙ t) = m(q˙ + v)2 − kq 2 = mq˙2 + mv q˙ − kq 2 +
2
2
2
2
2
26
(1.88)
(1.89)
The canonical momentum is p = m(q˙ + v), and the hamiltonian is
1
mv 2
1
1
mv 2
1
= mq˙2 + kq 2 −
. (1.90)
Hq = pq˙ −Lmq˙2 +mv q˙ − mq˙2 −mv q˙2 + kq 2 −
2
2
2
2
2
2
When written in terms of q, the lagrangian does not depend explicitly on time, and
therefore the hamiltonian (1.90) is conserved! However, it is not equal to the total
energy.
Example 1.7 Electrodynamics
One case where the distinction between ordinary and canonical momentum is im→
portant is electrodynamics. A particle with charge Q moving with velocity −
v in an
−
→
−
→
electric field E and a magnetic field B is
−
→
−
→ → −
→
F = Q( E + −
v × B)
(1.91)
Using Maxwell’s laws, we can introduce the electrostatic and vector (‘magnetic’)
−
→
potentials φ, A :
−
→
∇· B =0
⇐⇒
−
→
−
→
∂B
∇× E =−
∂t
⇐⇒
−
→
−
→
B =∇× A ,
−
→
−
→
∂A
E = −∇φ −
.
∂t
(1.92)
(1.93)
It is now possible (although the proof is not straightforward, so we will not present
it here) to derive the force (1.91) from a potential
−
→ →
U = Qφ − Q A · −
v .
(1.94)
The lagrangian is then
3
3
X
1 X 2
L=T −U = m
x˙ i − Qφ + Q
Ai x˙ i .
2 i=1
i=1
(1.95)
The canonical momentum is
pi =
∂L
= mx˙ i + QAi .
∂ x˙ i
(1.96)
This is not the ordinary momentum, a distinction which becomes quite important
in quantum mechanics, where it is the canonical momentum that enters into the
−
→
commutation relations that are used to quantise the system. Note that A and/or
φ must depend on x, otherwise the problem is trivial (there are no forces), so the
momentum is not conserved.
27
The hamiltonian of the particle is
H=
3
X
i=1
pi x˙ i − L =
3
3
X
3
X
1 X 2
x˙ i + Qφ − Q
Ai x˙ i
(mx˙ i + QAi )x˙ i − m
2
i=1
i=1
i=1
3
(1.97)
1 X 2
x˙ + Qφ .
= m
2 i=1 i
We see that the vector (magnetic) potential does not contribute to the hamiltonian.
Physically, this is because no net work is done by the magnetic field.
28
1.7
Lagrangian mechanics — summary sheet
1. Lagrangian L = T − V = kinetic energy − potential energy.
L = L(q, q,
˙ t) is a function of the coordinates qi , their time derivatives
q˙i and time t.
2. Generalised coordinates
For a system of N particles, we may instead of cartesian coordinates
−
→
ri = (xi , yi , zi ), i = 1 . . . N , use any set of coordinates
→
qj = fj (−
r1 , . . . , −
r→
N) ,
j = 1...M .
M is the number of degrees of freedom of the system. For an unconstrained system M = 3N , but if there are constraints then M < 3N .
3. Principle of least action
Nature “chooses” the path q(t) that minimises the action
Z t1
L(q(t), q(t),
˙
t)dt
S=
t0
with q(t0 ) = q0 , q(t1 ) = q1 kept fixed, or
S[q(t)] − S[q(t) + αh(t)]
=0
α→0
α
δS = lim
for arbitrary h(t) with h(t0 ) = h(t1 ) = 0. This leads to
4. Euler–Lagrange equations
d ∂L ∂L
−
=0
dt ∂ q˙i ∂qi
5. Canonical momentum
pi =
∂L
∂ q˙i
(a) Linear momentum P
If qi = xi and L = 21 i mi x˙ 2i − V (x), then pi = mx˙ i .
(b) Angular momentum
If qi is a rotational angle φ about some axis, then pi is the angular
−
→ →
→
momentum [ L = −
r × (m−
v )] about that axis.
29
6. Conservation laws
From the Euler–Lagrange equations we see that if L does not depend
explicitly on the coordinate qi then
dpi
=0
dt
⇐⇒
pi is conserved.
7. Hamiltonian
H=
X
i
pi q˙i − L
If there are no time-dependent constraints or velocity-dependent
forces (or potentials) then H = T + V = total energy.
∂L
dH
=−
,
dt
∂t
so if the lagrangian L does not explicitly depend on time, then the
hamiltonian H is conserved.
30