Why do we treat velocity and position as independent in a lagrangian

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Discussion Overview

The discussion centers on the treatment of position and velocity as independent quantities in the context of deriving the Euler-Lagrange equations from Hamilton's principle of least action. It explores theoretical aspects of Lagrangian mechanics and the implications of functional derivatives.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions why position (x) and velocity (x dot) are treated as independent in the derivation of the Euler-Lagrange equations.
  • Another participant explains that the independence arises from the need to calculate functional derivatives of the action with respect to q(t), emphasizing the role of Hamilton's principle.
  • There is a discussion about the definition of the functional derivative and the conditions under which it is evaluated, including the constraints on the function η.
  • A participant expresses uncertainty regarding the rigor of a common derivation method, suggesting that the approach they are familiar with may differ from the one presented.
  • Another participant seeks clarification on the functional derivative of the action before any modifications to η are made.

Areas of Agreement / Disagreement

Participants express differing views on the rigor of certain derivation methods and the treatment of η, indicating that multiple perspectives exist without a clear consensus on the best approach.

Contextual Notes

The discussion includes assumptions about the independence of variables in the context of functional derivatives, and the implications of boundary conditions in Hamilton's principle, which remain unresolved.

Storm Butler
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I was wondering why when we derive the euler lagrange equations and when we use them we treat x and x dot as independent quantities?
 
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Behind the Euler-Lagrange Equations of a given Lagrangian is Hamilton's principle of least action (in the Lagrange version). Thus you calculate functional derivatives of the action with respect to q(t). The Euler-Lagrange Equations follow from the principle by setting the first functional derivative to 0.

To evaluate this derivative, you have to expand the arbitrary path around the trajectory, i.e., you calculate the functional derivative from its definition. If you have the action in standard form,

A[q]=\int_{t_0}^{t_1} \mathrm{d} t L[q(t),\dot{q}(t)],

then the functional derivative is defined by

\frac{\delta A}{\delta q(t)}=\lim_{\epsilon \rightarrow 0} \frac{A[q+\epsilon \eta]-A[q]}{\epsilon}=\left. \frac{\mathrm{d} A[q+\epsilon \eta]}{\mathrm{d} \epsilon} \right|_{\epsilon \rightarrow 0}..

Here \eta is a function with support at a very small region around t. Particularly you must have \eta(t_0)=\eta(t_1)=0 according to the constraints that the boundary points of the paths are not varied in Hamilton's principle (by definition).

After taking the derivative you take as another limit the support of \eta to the single point t, i.e., in a sense \eta(t') \propto \delta(t'-t). From these considerations, after one integration by parts you get

\frac{\delta A}{\delta q(t)}=\frac{\partial L}{\partial x} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}},

where the q and \dot{q} have to be taken as independent variables when taking the partial derivatives, but then have to be interpreted as \dot{q}=\mathrm{d} q/\mathrm{d} t again when taking the time derivative.
 
vanhees71 said:
After taking the derivative you take as another limit the support of \eta to the single point t, i.e., in a sense \eta(t') \propto \delta(t'-t).
Hmm, that's not how I usually see it done. In the derivations I've seen, they say something like "Since the integral of \eta times blah is zero and \eta was arbitrary, it follows that blah is zero." Is that not fully rigorous?
 
I would also like to know, what is the functional derivative of A before you do anything to η?
 

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