Lagrangian formulation 11
It can be easily verified that substitution of eqn. (1.4.5) into eqn. (1.4.6) gives eqn.
(1.2.10):
∂L
∂r ̇i
=mir ̇i
d
dt
(
∂L
∂r ̇i
)
=mi ̈ri
∂L
∂ri
=−
∂U
∂ri
=Fi
d
dt
(
∂L
∂r ̇i
)
−
∂L
∂ri
=mi ̈ri−Fi= 0, (1.4.7)
which is just Newton’s second law of motion.
As an example of the application of the Euler–Lagrange equation, consider the
one-dimensional harmonic oscillator discussed in the previous section. The Hooke’s
law forceF(x) =−kxcan be derived from a potential
U(x) =
1
2
kx^2 , (1.4.8)
so that the Lagrangian takes the form
L(x,x ̇) =
1
2
mx ̇^2 −
1
2
kx^2. (1.4.9)
Thus, the equation of motion is derived as follows:
∂L
∂x ̇
=mx ̇
d
dt
(
∂L
∂x ̇
)
=mx ̈
∂L
∂x
=−kx
d
dt
(
∂L
∂x ̇
)
−
∂L
∂x
=mx ̈+kx= 0, (1.4.10)
which is the same as eqn. (1.3.5).
It is important to note that when the forces in a particular system are conserva-
tive, then the equations of motion satisfy an important conservation law, namely the
conservation of energy. The total energy is given by the sum of kinetic and potential
energies:
E=
∑N
i=1
1
2
mir ̇^2 i+U(r 1 ,...,rN). (1.4.11)