Lagrangian mechanics - Euler Lagrange Equation

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Discussion Overview

The discussion centers around the Euler Lagrange Equation in the context of Lagrangian mechanics, specifically questioning the dependence of the function f on higher-order derivatives such as y'' or y^{(3)}. Participants explore the implications of this dependence for variational problems and mechanics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why the function f is limited to three variables: x, y(x), and y'(x), suggesting that higher-order derivatives could be relevant.
  • Another participant asserts that in mechanics, it is generally accepted that forces depend only on position and velocity, implying that higher-order derivatives are unnecessary for the variational formulation.
  • A later reply discusses the relationship between conservative systems and the Lagrangian formalism, noting that the definitions of both are interdependent but not circular.
  • Another participant proposes that if higher-order derivatives are needed, one could introduce additional variables to maintain the formalism of the Lagrangian.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of higher-order derivatives in the Lagrangian framework. While some argue that they are unnecessary in mechanics, others suggest that they could be relevant in variational problems.

Contextual Notes

The discussion highlights the limitations of the traditional approach to Lagrangian mechanics, particularly regarding the assumptions about the dependence of forces on derivatives of position.

brisingr7
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Euler Lagrange Equation : if y(x) is a curve which minimizes/maximizes the functional :

F\left[y(x)\right] = \int^{a}_{b}f(x,y(x),y'(x))dx
then, the following Euler Lagrange Differential Equation is true.

\frac{\partial}{\partial x} - \frac{d}{dx}(\frac{\partial f}{\partial y'})=0

Well...
I don't understand why the function f has only three variables x, y(x) and the derivative of that.
what about y'' or y^{(3)}? I think it could be possible.(physically) All files related to this topic states that the function f as a function of variables x, f(x), and f'(x).
i.e. can function f be like : f(x,y(x),y'(x),y''(x),...) ? or... is it unnecessary to think about the second derivative and furthermore?
 
Last edited:
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oops!
the partial symbol should be on the left!
also, the lower/upper limits should change seats with the integral...
 
For variational problems in general maths, you are perfectly correct. You can perfectly well create higher-order "Euler-Lagrange"-like equations!
What they mean, however, is quite another thing!

For mechanics problems though, it is sort of axiomatic that forces are, at most, only dependent on an object's position&velocity (0th and 1st derivatives), and not on higher-order derivatives.

Thus, it is sufficient (given this "axiom") for the variational formulation to use an integrand merely dependent on x, y and y'.Exceptions probably exist, though, but I'm not familiar with them.
 
Translate this back into "pre-Euler-Lagrange mechanics", as a statement about the work done on a system and its potential and kinetic energy.

Those three quantities are (almost always!) functions of position and velocity, and nothing else.
 
The normal "baby-steps" into this are by defining a conservative system as one that you can identify a potential for, and the conditions for defining a potential is that your forces only depend on x, y, and y'. However, the formal definition seems to involve circular reasoning: the Lagrangian formalism only applies to conservative systems, and a conservative system is (formally) defined as one that follows Lagrangian mechanics. It isn't actually circular reasoning, it is just that the two are defined simultaneously, and the mathematical relationship between force and potential naturally attaches the f(x, y, y') limitation.

The Work-Energy theorem from baby physics is nothing more than the integral of the Lagrangian.

Hope that helps!
Randy
 
brisingr7,

There is no real need to introduce a y" in the Lagrangian.
If you needed that, you could just introduce an additional variable:

z = y'

and write a new Lagrangian including this additional variable:

L(x,y,y',z,z')

Additional variables keep the formalism the same and handles the extra y" you are thinking of.
 
Wow! thanks!
I think I get it!
 

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