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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.

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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.

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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 $t'$ !

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  • $\begingroup$ I guess I was just trying to understand why we treat the lagrangian as a function of two variables, both position and velocity. Like you say, there is only one degree of freedom, so I initially thought that you need only specify the path, $q(t) $, as you can determine $\dot{q} (t) $ from this. However, if one considers that $\mathcal{L} (q, \dot{q})$ describes all possible trajectories between two configurations, then in principle one can choose any velocity, $\dot{q}(t)$ for a given path $q(t)$, enabling us to treat them as independent variables in the lagrangian?! $\endgroup$ – Will Sep 28 '14 at 18:13
  • $\begingroup$ @Will I think not. You do consider one path only, and nothing else. The variation is simply $q=q_{\mbox{equilibrium}}+\delta q$, which is a kind of reparametrization. The notation $L(q,\dot q)$ simply enforces the absence of higher derivatives, making the equation of motions of second order. It does not suddenly make $\dot q$ independent of $q$, like $f(x,y(x))$ in a sense. $\endgroup$ – Frédéric Sep 28 '14 at 18:18
  • $\begingroup$ I just don't see otherwise why we consider the lagrangian as a function of both $q(t)$ and $\dot{q}(t)$, other than arguing that we wish to be able to specify the configuration of a system and how this configuration evolves in time, thus to do so implies we need a quantity that is a function of position and velocity, such that, if we specify a particular set of positions and velocities, then we can predict the evolution of the system we wish to describe?! I can't seem to find any notes that provide a satisfactory explanation to this when introducting the concept of analytical mechanics?! $\endgroup$ – Will Sep 28 '14 at 18:23
  • $\begingroup$ The answer to that question would be, as far as I am concerned, empirical. Before analytical mechanics we had Newton's Law and others which are all second order, hence the requirement on the Lagrangian. $\endgroup$ – Frédéric Sep 28 '14 at 18:30
  • $\begingroup$ I suppose my real question is, why (before considering any variation) do we even consider the Lagrangian as a function of $q$ and $\dot{q}$ (and possibly higher order derivatives, neglecting physical arguments for the time being), why not just a function of $q$. Even in the more abstract sense of calculus of variations the integrand is considered a function of variables including first order derivatives (and, in general, higher)? I'm trying to get an intuition as to why this is exactly?! Sorry for the (perhaps) trivial questions, but I'm struggling to rationalise it in my mind. $\endgroup$ – Will Sep 28 '14 at 21:03

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