Principle of Least Action Derivation

[tharchin]

There is one step I am having trouble understanding in the derivation of the principle of least action which leads to the Euler-Lagrange equations.

When you have the variation of the action:

$$\delta S = \int_{t_1}^{t_2} [ L(q+\delta q, \dot q + \delta \dot q, t ) - L(q, \dot q, t ) ] \, dt$$

The next step is to expand the first term in the integral using a Taylor expansion and throw out any terms higher than the first order. I follow the math steps all the way through but nowhere have a found a justification for not including the higher order terms? Thanks.

 

[JohnSimpson]

The higher order terms immediately go as $$(\delta q)^2$$ or $$(\delta \dot{q})^2$$ which are much, much smaller than the terms of order $$\delta q$$. This is why we ignore them, and not because the other derivatives of the Lagrangian have any particular properties.

 

[tharchin]

The higher order terms immediately go as $$(\delta q)^2$$ or $$(\delta \dot{q})^2$$ which are much, much smaller than the terms of order $$\delta q$$. This is why we ignore them, and not because the other derivatives of the Lagrangian have any particular properties.

It is always possible that in some cases these terms may have an effect that needs to be considered. Since the Lagrangian is one of the basic elements of QFT maybe on the quantum level these second order effects are noticeable?

 

[lalbatros]

A variational principle is about finding an extremum, not about Taylor expansion.

To find an extremum of a function, you need two conditions:

• the first order term of the Taylor expansion must cancel (first derivative = 0)
• the second order term must be positive or negative definite in a finite neighborhood (second derivative not changing sign)

When using a variational principle, these conditions remain essentially the same.
The only difference is that there is more than one single unknown.
There is even more than a countable set of unknowns.
Despite these technicalities, the principle remains:

• the first order term locates the possible extrema
• the second order term must be used to check for an actual extremum,
  to exclude saddle points

This is all simply related to the basic definition of what an extremum is.
A variational principle is about finding an extremum, not about Taylor expansion.
The Taylor expansion is simply the easiest way to get the end result: the Lagrange equations.
Without recourse to the Taylor expansion, one would need exotic developments about the functional derivatives and maybe the functional Jacobian!