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Mass in one Dimension, Noether's Theoren, Hamiltonian Stable and unstable Equilibrium
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A symmetry under a coordinate means that the Lagrangian is the same after
perturbation of that coordinate,
L(q, q, t˙ ) = L(˜q,
q, t˜ )
where
q˜ = q + K
and is small.
L(x, x˙) =
m x˙
2
− V (x)
Let ˜x = x +
˜x = ˙x
⇒ L(˜x,
x˜) =
m
x
2 − V (x + )
L(x, x˙) = L(˜x,
x˜)
⇒ V (x) = V (x + )
Using a Taylor expansion,
V (x + ) ≈ V (x) +
∂x
By the requirement of symmetry, V (˜x) = V (x),
∂x
d~p
dt
dp x
dt
⇒ p x
= const
Forcing the Lagrangian to by symmetric results in a conservation law. In
this case, it is conservation of linear momentum.
2 Noether’s Theorem
Force the Lagrangian to be symmetric,
L(q, q˙) = L(q + k, q˙ +
k)
L(q + k, q˙ +
k) ≈ L(q, q˙) +
q
k q
∂ q˙
q
k q
∂q
Requiring symmetry means that the first term cancels.
q
k
∂ q˙
q
k
∂q
This is Noether’s theorem.
Consider
q
k
∂L
∂ q˙
d
dt
q
k
∂ q˙
q
k
∂ q˙
q
k
d
dt
∂ q˙
From the Euler-Lagrange equations,
d
dt
∂ q˙
∂q
d
dt
q
k
∂ q˙
q
k
∂ q˙
q
k
∂q
3 Hamiltonian
Suppose we are minimizing
f (x, x
′
, t)dt
We can define another quantity,
H = x
′
∂f
∂x
′
− f
Consider,
H = ˙q
∂ q˙
dH
dt
d
dt
q ˙
∂ q˙
dL
dt
= ¨q
∂ q˙
d
dt
∂ q˙
dL
dt
dL
dt
∂ q˙
∂dt
∂ q˙
∂q
∂t
∂q
∂t
Combining these terms and canceling using the Euler Lagrange equations,
dH
dt
∂t
If we have a lagrangian that is independent of time,
L(q, q˙) ⇒
dH
dt
= 0 ⇒ H = const
m x˙
2
− V (x)
H = ˙x
∂ x˙
∂ x˙
= m x˙
H = m x˙
2
−
m x˙
2
− V (x)
m x˙
2
T + V = total energy
3.2.1 Notes
We see that if the Lagrangian is invariant to time translation, then energy
is conserved. The Hamiltonian can be used to prove other symmetries, like
charge, parity, and baryon number.
The actual definition the the Hamiltonian is
i
q ˙i
∂ q˙ i
Usually, the Hamiltonian is usually written in terms of the position and
momentum.
4 Stable and unstable equilibrium
Consider the same spherical pendulum from a previous lecture. The length
is fixed at R, and so there are only two degrees of freedom, θ and φ. We saw
that one condition for stability was
sin θ 0
= 0 ⇒ θ 0
= 0, π
Now we plug in our possible values of θ 0
θ 0
= 0 ⇒ p = ±
g
This leads to an exponentially increasing solution, thus an unstable equilib-
rium.
For the other point,
θ 0
= π ⇒ p = ±
g
= ±ıω 0
δθ = Ae
ıω 0 t
−ıω 0 t
= C cos ω 0
t + D sin ω 0
t
This is an oscillating solution, which is stable.
This pendulum example is an example of the general method for stability
analysis.
~r 1
= b sin θ 1
xˆ − b cos θ 1
yˆ
~r 2
= ~r 1
xˆ − a cos θ 2
yˆ
We take derivatives in order to get the kinetic energy.
~r 1
= b
θ 1
cos θ 1
xˆ + b
θ 1
sin θ 1
yˆ
~r 2
~r 1
θ 2
cos θ 2
xˆ + a
θ 2
sin θ 2
yˆ
~r 2
= (b
θ 1
cos θ 1
θ 2
)ˆx + (b
θ 1
sin θ 1
θ 2
sin θ 2
)ˆy
r ˙
2
1
~r 1
~r 1
= b
2 ˙ θ
2
1
r ˙
2
2
= b
2 ˙ θ
2
1
cos
2
θ 1
2 ˙ θ
2
2
cos
2
θ 2
θ 1
θ 2
cos θ 1
cosθ 2
= b
θ
2
1
θ
2
2
θ 1
θ 2 (sin θ 1 sin θ 2 + cos θ 1 cos θ 2 )
r˙
2
2
= b
2 ˙ θ
2
1
2 ˙ θ
2
2
θ 1
θ 2
cos(θ 1
− θ 2
m 1
r˙
2
1
m 2
r˙
2
2
(m 1
)b
2 ˙ θ
2
1
m 2
a
2 ˙ θ
2
2
ab
θ 1
θ 2
cos(θ 1
− θ 2
Now we find the potential energy, given by gravity,
U = m 1
gb(1 − cos θ 1
) + m 2
g(1 − b cos θ 1
− a cos θ 2
The total lagrangian is then
We have two degrees of freedom, θ 1
and θ 2
. We imagine that they will be
coupled. We start cranking the derivatives.
θ 1
= (m 1
)b
θ 1
ab
θ 2
cos(θ 1
− θ 2
∂θ 1
= −m 2
ab
θ 1
θ 2
sin(θ 1
− θ 2
) − m 1
gb sin θ 1
− m 2
gb sin θ 1
We will take the time derivative after some approximations.
4.2.1 Summary
There are two modes. The first is the low frequency mode, in which the
masses stay underneath each other, with a small angular difference. The
second is the high frequency mode, in which the two masses will be be out
of phase. We will analyze these solutions next time with eigenvalues and
eigenvectors.
