Lagrangian formulation 11It 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 ̇id
dt(
∂L
∂r ̇i)
=mi ̈ri∂L
∂ri=−
∂U
∂ri=Fid
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=−kxd
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=11
2
mir ̇^2 i+U(r 1 ,...,rN). (1.4.11)