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Last modified: December 9, 2021 ✓

Euler-Lagrange Equation
-----------------------

Newton's laws are relationships among vectors, which 
is why they get so messy when coordinate systems are 
changed.  The Lagrangian formulation, on the other 
hand, just uses scalars, consequently coordinate 
transformations tend to be much easier.  Also, using 
the Lagrangian approach is much simpler for many 
body (complex) systems.

Given a Lagrangian, L, which is a function of the 
location q and the velocity dq/dt where qi etc. are 
generalized coordinates, the action, A, is defined 
as:

A = ∫L dt    
          .
where L(q,q) = T - U    T = KE, U = PE 

Given particular starting and ending positions, 
the system follows a path between the start and 
end points which minimizes the "action".  Let's 
call the coordinates in space "q1" through "qn". 
To get the path of least action we use the 
EULER-LAGRANGE equation.
         .
d/dt(∂L/∂qi) = ∂L/∂qi 

The solutions to the E-L equation are differential 
equations that describe the equations of motion of 
the system.

Define: 
                              .       
Canonical momentum, Πi =  ∂L/∂qi  
                 .
Force = ∂L/∂qi = Πi

Example 1:  F = ma

L = T - U

L = (1/2)mv2 - U
          .
  = (1/2)mq2 = U

LHS of Euler-Lagrange:
    .
∂L/∂q = mv
         .
      = mq
   .        .
d(mq)/dt = mq = ma 

RHS of E-L:

∂L/∂q = -∂U/∂q = F (since F = -∂U/∂x by definition)

Equating the left and right side gives,

F = ma


Example 2.  Harmonic Oscillator

x
|  \
|  /
|  \  <-- spring
|  /
|  \
|  m
|_________

                  .
L = T - U = (1/2)mx2 - (k/2)x2 
    .     .
∂L/∂q =  mx

∂L/∂qi = -kx
         .
d/dt(∂L/∂qi) = ∂L/∂qi
      .
d/dt(mx) = -kx
 ..
mx = -kx
..
x = (-k/m)x

Let x = cosωt
.
x = -ωsinωt
..
x = -ω2cosωt = -ω2x

Which leads to,

-k/mx = - ω2x

=> ω = √(k/m)
 

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