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

AI Thread Summary
In the discussion on treating velocity and position as independent in the Lagrangian framework, the Euler-Lagrange equations arise from Hamilton's principle of least action, requiring functional derivatives of the action with respect to the generalized coordinates. The action is expressed in standard form, and the functional derivative is calculated by expanding the path around the trajectory, ensuring boundary conditions are met. The independence of position and velocity is crucial when taking partial derivatives, but they are later related through time derivatives. The conversation also questions the rigor of certain derivation methods and seeks clarification on the functional derivative of the action before any variations are applied. Overall, the treatment of position and velocity as independent is essential for deriving the equations of motion accurately.
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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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