Page 197 - Intro to Tensor Calculus

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This equation is called the Lagrange’s form of the equations of motion.

EXAMPLE 2.2-1. (Equations of motion in spherical coordinates) Find the Lagrange’s form of
the equations of motion in spherical coordinates.
1
2
3
Solution: Let x = ρ, x = θ, x = φ then the element of arc length squared in spherical coordinates has
the form
2 2 2 2 2 2 2
ds =(dρ) + ρ (dθ) + ρ sin θ(dφ) .
The element of arc length squared can be used to construct the kinetic energy. For example,

2
1 ds 1 h 2 2 ˙ 2 2 2 i
˙ 2
T = M = M (˙ρ) + ρ (θ) + ρ sin θ(φ) .
2 dt 2
The Lagrange form of the equations of motion of a particle are found from the relations (2.2.23) and are
calculated to be:

d ∂T ∂T h 2 i
˙ 2
˙ 2
Mf 1 = Q 1 = − = M ¨ − ρ(θ) − ρ sin θ(φ)
ρ
dt ∂ ˙ρ ∂ρ

d ∂T ∂T d 2 ˙ 2
˙ 2
Mf 2 = Q 2 = − = M ρ θ − ρ sin θ cos θ(φ)
dt ∂θ ˙ ∂θ dt

d ∂T ∂T d 2 2 ˙
Mf 3 = Q 3 = − = M ρ sin θφ .
dt ∂φ ˙ ∂φ dt
In terms of physical components we have
h 2 i
˙ 2
˙ 2
Q ρ = M ¨ρ − ρ(θ) − ρ sin θ(φ)

M d 2 ˙ 2
˙ 2
Q θ = ρ θ − ρ sin θ cos θ(φ)
ρ dt

M d 2 2 ˙
Q φ = ρ sin θφ .
ρ sin θ dt
Euler-Lagrange Equations of Motion
Starting with the Lagrange’s form of the equations of motion from equation (2.2.23), we assume that
the external force Q r is derivable from a potential function V as specified by the equation (2.2.20). That is,
we assume the system is conservative and express the equations of motion in the form

d ∂T ∂T ∂V
− = − = Q r , r =1,... ,N (2.2.24)
dt ∂ ˙x r ∂x r ∂x r
The Lagrangian is defined by the equation

1 N 1 N 1 N i i
L = T − V = T (x ,... ,x , ˙x ,..., ˙x ) − V (x ,... ,x )= L(x , ˙x ). (2.2.25)
Employing the defining equation (2.2.25), it is readily verified that the equations of motion are expressible
in the form
d ∂L ∂L
− =0, r =1,... ,N, (2.2.26)
dt ∂ ˙x r ∂x r
which are called the Euler-Lagrange form for the equations of motion.