3Hamilton's principle
IB Variational Principles
3.1 The Lagrangian
The concept of generalized coordinates was first introduced by Lagrange in 1788.
He then showed that
ξ
(
t
) obeys certain complicated ODEs which are determined
by the kinetic energy and the potential energy.
In the 1830s, Hamilton made Lagrange’s mechanics much more pleasant. He
showed that the solutions of these ODEs are extremal points of a new “action”,
S[ξ] =
Z
L dt
where
L = T − V
is the Lagrangian, with T the kinetic energy and V the potential energy.
Law
(Hamilton’s principle)
.
The actual path
ξ
(
t
) taken by a particle is the path
that makes the action S stationary.
Note that
S
has dimensions
ML
2
T
−1
, which is the same as the 18th century
action (and Plank’s constant).
Example.
Suppose we have 1 particle in Euclidean 3-space. The configuration
space is simply the coordinates of the particle in space. We can choose Cartesian
coordinates x. Then
T =
1
2
m|
˙
x|
2
, V = V (x, t)
and
S[x] =
Z
t
B
t
A
1
2
m|
˙
x|
2
− V (x, t)
dt.
Then the Lagrangian is
L(x,
˙
x, t) =
1
2
m|
˙
x|
2
− V (x, t)
We apply the Euler-Lagrange equations to obtain
0 =
d
dt
∂L
∂
˙
x
−
∂L
∂x
= m
¨
x + ∇V.
So
m
¨
x = −∇V
This is Newton’s law
F
=
ma
with
F
=
−∇V
. This shows that Lagrangian
mechanics is “the same” as Newton’s law. However, Lagrangian mechanics has
the advantage that it does not care what coordinates you use, while Newton’s
law requires an inertial frame of reference.
Lagrangian mechanics applies even when
V
is time-dependent. However, if
V is independent of time, then so is L. Then we can obtain a first integral.
As before, the chain rule gives
dL
dt
=
∂L
∂t
+
˙
x ·
∂L
∂x
+
¨
x ·
∂
∂
˙
x
=
∂L
∂t
+
˙
x ·
∂L
∂x
−
d
dt
∂L
∂
˙
x
| {z }
δS
δx
= 0
+
˙
x ·
d
dt
∂L
∂
˙
x
+
¨
x ·
∂L
∂
˙
x
| {z }
d
dt
˙
x ·
∂L
∂
˙
x
So we have
d
dt
L −
˙
x ·
∂L
∂
˙
x
=
∂L
∂t
.
If
∂L
∂t
= 0, then
˙
x ·
∂L
∂
˙
x
− L = E
for some constant E. For example, for one particle,
E = m|
˙
x|
2
−
1
2
m|
˙
x|
2
+ V = T + V = total energy.
Example.
Consider a central force field
F
=
−∇V
, where
V
=
V
(
r
) is inde-
pendent of time. We use spherical polar coordinates (r, θ, φ), where
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ.
So
T =
1
2
m|
˙
x|
2
=
1
2
m
˙r
2
+ r
2
(
˙
θ
2
+ sin
2
θ
˙
φ
2
)
So
L =
1
2
m ˙r
2
+
1
2
mr
2
˙
θ
2
+ sin
2
θ
˙
φ
2
− V (r).
We’ll use the fact that motion is planar (a consequence of angular momentum
conservation). So wlog θ =
π
2
. Then
L =
1
2
m ˙r
2
+
1
2
mr
2
˙
φ
2
− V (r).
Then the Euler Lagrange equations give
m¨r − mr
˙
φ
2
+ V
0
(r) = 0
d
dt
mr
2
˙
φ
= 0.
From the second equation, we see that
r
2
˙
φ
=
h
is a constant (angular momentum
per unit mass). Then
˙
φ = h/r
2
. So
m¨r −
mh
2
r
3
+ V
0
(r) = 0.
If we let
V
eff
= V (r) +
mh
2
2r
2
be the effective potential, then we have
m¨r = −V
0
eff
(r).
For example, in a gravitational field, V (r) = −
GM
r
. Then
V
eff
= m
−
GM
r
+
h
2
2r
2
.