14 Classical mechanics
1.4.1 Example: Motion in a central potential
Consider a single particle in three dimensions subject to a potentialU(r) that depends
only on the particle’s distance from the origin. This meansU(r) =U(|r|) =U(r),
wherer =
√
x^2 +y^2 +z^2 and is known as acentral potential. The most natural
coordinates are not the Cartesian coordinates (x,y,z) but rather spherical polar co-
ordinates (r,θ,φ) given by
r=√
x^2 +y^2 +z^2 , θ= tan−^1√
x^2 +y^2
z, φ= tan−^1
y
x, (1.4.23)
which can be inverted to give
x=rsinθcosφ, y=rsinθsinφ, z=rcosθ. (1.4.24)The mass metric tensor is a 3×3 diagonal matrix given by
G 11 (r,θ,φ) =m
G 22 (r,θ,φ) =mr^2
G 33 (r,θ,φ) =mr^2 sin^2 θ
Gαβ(r,θ,φ) = 0 α 6 =β. (1.4.25)Returning to our example of a single particle moving in a central potential,U(r), we
find that the Lagrangian obtained by substituting eqn. (1.4.25) intoeqn. (1.4.18) is
L=
1
2
m(
r ̇^2 +r^2 θ ̇^2 +r^2 sin^2 θφ ̇^2)
−U(r). (1.4.26)In order to obtain the equations of motion from the Euler–Lagrange equations, eqn.
(1.4.6), derivatives ofLwith respect to each of the variables and their time derivatives
are required. These are given by:
∂L
∂r ̇=mr, ̇∂L
∂r=mrθ ̇^2 +mrsin^2 θφ ̇^2 −dU
dr∂L
∂θ ̇=mr^2 θ, ̇∂L
∂θ=mr^2 sinθcosθφ ̇^2∂L
∂φ ̇=mr^2 sin^2 θφ, ̇∂L
∂φ= 0. (1.4.27)
Note that in eqn. (1.4.27), the derivative∂L/∂φ= 0. The coordinateφis an example
of acycliccoordinate. In general, if a coordinateqsatisfies∂L/∂q= 0, it is called
cyclic. It is also possible to makeθa cyclic coordinate by recognizing that the quantity
l=r×p, called theorbital angular momentum, is a constant (l(0) =l(t)) when the
potential only depends onr. (Angular momentum will be discussed in more detail in
Section 1.11.) Thus, the quantitylis conserved by the motion and, therefore, satisfies
dl/dt= 0. Becauselis constant, it is always possible to simplify the problem by