Explanation as to why we treat position and velocity as independent variables in the lagrangian?

Question

Although having studied calculus of variations and lagrangian mechanics, something I've never felt that I've fully justified in my mind is why the lagrangian is a function of position and velocity?

My understanding is that the lagrangian characterises the dynamics of a system in its evolution from one configuration at time $t_{1}$ to another configuration at some later time $t_{2}$. Before considering the actual physical path taken we would like the lagrangian to be able to describe all possible trajectories between the two (fixed) configurations. Therefore, for any given path $q(t) $ chosen we are free to choose any velocity $v=\dot{q} (t)$ (as we are considering all possible trajectories between the two configurations, that satisfy the boundary conditions). As such, before invoking any variational principles, we are able to treat position, $q(t) $ and velocity, $\dot{q} (t) $ as independent variables. However, once applying the principle of least action and thus choosing a specific path, i.e. The one which gives an extremum to the action, $q$ and $\dot{q}(t) $ are no longer independent (as $\dot{q}(t) $ is dependent on the chosen extremal path). Indeed, we find upon varying the path around this extremal, that they are related by $$ \delta \dot{q}(t) =\frac{d} {dt} \left(\delta q(t)\right) $$ i.e. The variation in the velocity is equal to the time derivative in the variation of the position (as one might intuitively expect).

Is this a correct understanding and description? If not could please could someone enlighten me on the matter?! Thanks.

Answer 1

The simplest reason for why we can do that is because

Given a function $f(x)$, if we can write it as $f(x,y)$ where $y = y(x)$, we can apply the identity $$ df = \frac {\partial{f}} {\partial{x}} dx + \frac {\partial{f}} {\partial{y}} dy$$

The derivation of this identity never makes the assumption that $x$ and $y$ have to be independent. The $only$ problem that can arise is that we might give $y$ values which are not allowed.

For example, if $y = x^2$, we find that $\frac {df} {dx}$ is the same as that calculated using the identity, but f(2,3) is not valid, as $y = 2^2 = 4$.

But when we talk about the Lagrangian, since at the end we use the identity that $\frac {dx} {dt} = v$, hence we are assured that we will never stumble upon any such problem.

Answer 2

As this would be too long as a comment let me try to answer.

"As such, before invoking any variational principles, we are able to treat position, $q(t)$ and velocity, $\dot q(t)$ as independent variables."

No, I don't think so because by $\dot q(t)=$ is the time derivative of $q$, which is the only degree of freedom here. Mathematically you can't change a function and its derivative independently. The notation $L(q,\dot q)$ is understood as a function of two independent arguments evaluated at not independent points. $$\dot{(\delta q)} = \delta \dot q$$ follows.

Of course you are allowed to consider the mathematical problem with $q\rightarrow q_1$, $\dot q\rightarrow q_2$, but the physics behind it changes (2 degrees of freedom instead of 1).

Edit:

The action functional is $S=S[q]$, no need to specify what it does to the function $q$ (i.e. derivatives, ...), but it is a functional, not a function. The Lagrangian is a function, i.e. maps a "number to a number", not a function to a number. It is convenient to specify derivatives because you can then use standard analysis to make the expansion. You could write $S=\int L[q(t)] dt$ but then $L$ is a differential operator which makes the notation un-explicit, namely $\delta L$ is strongly dependent on $L$, whereas the equation of motion for $L=L(q,\dot q)$ are universal, regardless of $L$.

Of course one could think of many other functional, for example not local ones of the form

$$ S=\int L[q(t),q(t')]dt dt' $$

but the physics described by such model is very different, notably the notion of causality: what happens at $t<t'$ is sensible by what will be at $t'$ !