Lagrangian formulation 15choosing a coordinate frame in which thezaxis lies along the direction ofl. In such a
frame, the motion occurs solely in thexyplane so thatθ=π/2 andθ ̇= 0. With this
simplification, the equations of motion become
m ̈r−mrφ ̇^2 =−
dU
drmr^2 φ ̈+ 2mrr ̇φ ̇= 0. (1.4.28)The second equation can be expressed in the form
d
dt(
1
2
r^2 φ ̇)
= 0, (1.4.29)
which expresses another conservation law known as the conservation ofareal velocity,
defined as the area swept out by the radius vector per unit time. Setting the quantity,
mr^2 φ ̇=λ, whereλis constant, the first equation of motion can be written as
m ̈r−λ^2
mr^3=−
dU
dr. (1.4.30)
Since the total energy
E=
1
2
m(
r ̇^2 +r^2 φ ̇^2)
+U(r) =1
2
mr ̇^2 +λ^2
2 mr^2+U(r) (1.4.31)is conserved, eqn. (1.4.31) can be inverted to give an integral expression
dt=dr
√
2
m(
E−U(r)− λ
2
2 mr^2)
t=∫rr(0)dr′
√
2
m(
E−U(r′)− λ
2
2 mr′^2), (1.4.32)
which, for certain choices of the potential, can be integrated analytically and inverted
to yield the trajectoryr(t).
1.4.2 Example: Two-particle system
Consider a two-particle system with massesm 1 andm 2 , positionsr 1 andr 2 , and
velocitiesr ̇ 1 andr ̇ 2 subject to a potentialUthat is a function of only the distance
|r 1 −r 2 |between them. Such would be the case, for example, in a diatomic molecule.
The Lagrangian for the system can be written as
L=1
2
m 1 r ̇^21 +1
2
m 2 r ̇^22 −U(|r 1 −r 2 |). (1.4.33)Although such a system can easily be treated directly in terms of theCartesian posi-
tionsr 1 andr 2 , for which the equations of motion are
m 1 ̈r 1 =−U′(|r 1 −r 2 |)
r 1 −r 2
|r 1 −r 2 |