Download Hamilton's Principle, Lecture Notes - Physics and more Study notes Engineering Physics in PDF only on Docsity!
Adrian Down
September 13, 2005
1 Hamilton’s principle
Definition. The action S is
S[q(t)] =
t 2
t 1
dt
p
2
2 m
− V (q)
Theorem (Hamilton’s Principle). The physical path is that along which the
action integral is minimized.
1.1 Connection to quantum mechanics
1.1.1 Path integral formulation
This principle can be seen as a consequence of the path-integral formulation
of quantum mechanics. In this construct, a system moving between two
points takes all possible paths. Each path is weighted by a probability. The
probability is given by
dtq(t)e
ı
S(t)
ℏ
2
The S(t) in the exponent depends on the path, so the integral is really a sum
over all paths.
1.1.2 Method of stationary phase
The integral can be evaluated using the method of stationary phase. The
integral of a rapidly oscillating sine curve averages out to 0. The integral of
a slowly oscillating sine curve can be finite, however. The oscillation of the
curve is smallest when the argument of the exponential is smallest, i.e. when
the action is minimized. The integral over all other paths is much smaller
than that over the physical path.
1.2 The Lagrangian
Definition. The Lagrangian for a system is defined as
L = T − V =
m q˙
2
i
− V (qi, q˙i, t)
Hamilton’s principle says is that the integral of the Lagrangian over time
is minimized along the physical path. Previously, we saw that minimizing L
is equivalent to satisfying the Euler-Langrage equations,
d
dt
∂L
∂ q˙ i
∂L
∂q i
Note that there is one equation for each coordinate, q i
, but there is only one
Lagrangian.
In newtonian mechanics, the task was to examine the forces. In analytic
mechanics, the hard part is finding the Lagrangian.
2 Examples
2.1 1D particle of mass m
2.1.1 Find the Lagrangian
The first task is to find the Lagrangian.
L = T − V
T =
m q˙
2
L =
m q˙
2
− V (q)
We want
~r
2
, so we square it. Many of the cross terms cancel out and we use
cos
2 α + sin
2
α = 1.
~r
2
= R
2 ˙ θ
2
2 ˙ φ sin
2
θ
T =
m
~r
2
=
mR
2 ˙ θ
2
mR
2 ˙ φ
2
sin
2
θ
When dealing with ugly equations like these, try to use physical sense to
see which terms should cancel out. We could expect that there would be no
φ terms, since the pendulum is symmetric about the azimuth. If |r| were
changing, we would also have a centripetal term in T.
Potential term Now we need the potential. The only relevant potential
is gravity.
V = mgz = mgR cos θ
The total Lagrangian, then, is
L = T − V =
mR
2 ˙ θ
2
mR
2 ˙ φ
2
sin
2
θ − mgR cos θ
2.2.3 Evaluate the Euler-Langrage equations
There are two degrees of freedom, φ and θ, so there will be two Euler Langrage
equations.
θ coordinate We do θ first
∂L
∂θ
= mR
θ
2
sin θ cos θ + mgR sin θ
∂L
θ
= mR
2 ˙ θ
d
dt
∂L
θ
= mR
2 ¨ θ
Set the first equation equal to the third. Cancel the mass and divide by R
2
to get the first equation of motion,
θ =
φ
2
sin θ cos θ +
g
R
sin θ
φ coordinate Note that there is no φ dependence in L. We will see later
that this is because the problem is symmetric in φ.
∂L
∂φ
∂L
φ
= mR
2 ˙ φ sin
2
θ
Normally we would take the time derivative of the second equation, and set
it equal to the first. However, since the first is 0, we know that the second
equation must be equal to a constant, call it pφ.
∂L
φ
= p φ
pφ = mR
φ sin
2
θ = const
If we did not recognize this and took the time derivative of the second
equation, we would have a
φ term, and a
θ term. This would lead to a set of
coupled second order differential equations.
Combined equation of motion We want to eliminate the φ dependence
from the first equation of motion, so square p φ
and plug into the first equation.
θ =
p
2
φ
m
2 R
4 sin
4 θ
sin θ cos θ +
g
R
sin θ
The equation of motion is
θ =
p
2
φ
m
2 R
4
cos θ
sin
3
θ
g
R
sin θ
2.2.4 Interpretation
To understand messy differential equations, we usually do instability analysis
or take limits. We consider two cases.
φ = 0
p φ
= mR
φ sin
2 θ = 0
θ =
g
R
sin θ
2.4 Impose constraints
That the particle is constrained to move on the cone gives a relation between
r and z.
r = z tan α
We can eliminate one, say z.
z = r cot α
~r = r cos θˆx + r sin θyˆ + z zˆ
2.4.1 Find the Lagrangian
Kinetic term
~r = ( ˙r cos θ − r
θ sin θ)ˆx + ( ˙r sin θ + r
θ cos θ)ˆy + ˙z ˆz
When we square this, there is much happy cancellation.
~r
2
= ˙r
2
θ
2
2
T =
m r˙
2
mr
2 ˙ θ
2
m r˙
2
cot
2
α
m r˙
2
csc
2
α +
mr
2 ˙ θ
2
Potential term The only relevant potential is gravity,
V = mgr cot α
L = T − V =
m
r ˙
2
csc
2
α + r
2 ˙ θ
2
− mgr cot α
2.4.2 Evaluate the Euler-Langrage equations
θ coordinate Note right away that L does not depend on θ. We use
the same strategy that we used in the previous problem when we have a 0
derivative.
∂L
∂θ
∂L
θ
= const = p θ
= mr
θ
r coordinate
∂L
∂ r˙
= m r˙ csc
2
α
d
dt
∂L
∂ r˙
= mr¨ csc
2
α
∂L
∂r
= mr
θ
2
− mg cot α
¨r csc
2
α = r
θ
2
− g cot α
¨r = r
θ
2
sin
2
α − g sin α cos α
Combined equation of motion We could go back and solve for
θ;
θ
2
=
p
2
θ
m
2 r
4
We would plug this into our equation for ¨r, but it would be messy.
2.4.3 Interpretation
Now we see what happens in interesting cases.
r˙ = 0 Here, the particle is at a fixed height moving in a circle.
r = r 0
r 0
θ
2
= g cos α
θ
2
tan α =
g
r 0
= ω
2
0
θ is like the angular frequency of the particle performing circular motion.
2.5 General strategy
- Draw a picture that illustrates the relevant physics.
- Impose the constraints.
- Get the Lagrangian
- Kinetic term (involves finding
~r)
is small, so we Taylor expand.
V (x + ) ≈ V (x) +
∂V
∂x
+ O(
2
)
V (x) ≈ V (x) +
∂V
∂x
∂V
∂x
∂V
∂x
is the force, so we get that F = 0.
d
dt
(m x˙) = F = 0
m x˙ = const = p x
When the Lagrangian is symmetric about some coordinate, it always intro-
duces a constant of the motion. All conserved quantities arise from symme-
tries in the Lagrangian. This is a statement of Noether’s theorem.
