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

[Storm Butler]

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?

 

[vanhees71]

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.

 

[lugita15]

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?

 

[lugita15]

I would also like to know, what is the functional derivative of A before you do anything to η?

 

[PJC01]

The reason we treat velocity and position as independent in a Lagrangian is because they represent two fundamental aspects of motion that are not directly dependent on each other. Velocity is the rate of change of position, but it is not determined solely by the position itself. It also depends on other factors such as the forces acting on the object. Therefore, in order to fully describe the motion of an object, we need to consider both position and velocity separately.

In the derivation of the Euler-Lagrange equations, we treat position and velocity as independent variables because they are the two most basic quantities that are needed to describe the state of a system. By treating them as independent, we can take into account all the relevant factors that influence the motion of the system, such as forces and constraints.

Furthermore, treating position and velocity as independent allows us to consider a wide range of systems, including those with complex and nonlinear dynamics. If we were to treat them as dependent variables, it would limit our ability to accurately describe and analyze the behavior of these systems.

In summary, treating position and velocity as independent in a Lagrangian is necessary in order to fully describe and analyze the motion of a system, and it allows for a more comprehensive understanding of complex systems.