0
$\begingroup$

I would like to continue discussion from my previous post on time dependence of lagrangian Time dependence of the Lagrangian of a free particle?. I have also read this old post Why does calculus of variations work?

I think I understand how Lagrangian treats position and velocity as independent variables from the point of view of calculus of variations. I think velocity only really deals with "change in position" and not the position itself, and so my intuition is like this should be one of the reasons we have to treat position and velocity as independent from the physics point of view as well, and not just so because calculus of variations treats Lagrangian as a function of some variables (x,y,z). Even the change in velocity is just related to change in position in the relation below and not the position itself. $$\delta \dot q = {d \over dt} \delta q$$

So I think that not only position and velocity initial conditions are independent, but position and velocity are independent along any trajectory too. Is this why Lagrangian is specifically defined with position and velocity independent ? Also, position forms an affine space and so it doesn't make sense to me how can anything be dependent on it unless we define a frame (or origin). And we treat Lagrangian as frame independent too. Does this make any sense ? I just want to make things precise here. I hope my mind is not messing with me here.

Thank you

$\endgroup$

marked as duplicate by Qmechanic Aug 8 '15 at 5:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ The substance of matter in this question (v1) is essentially a duplicate of the Phys.SE post that OP links to. $\endgroup$ – Qmechanic Mar 5 '15 at 11:46
2
$\begingroup$

You say

but position and velocity are independent along any trajectory too.

No, a trajectory is defined as a given vector-function $\vec r(t)$ from which follows $\vec v(t) = d\vec r(t)/dt$. So, given a trajectory, the vector-function $\vec v(t)$ depends clearly on $\vec r(t)$ as being its derivative.

Now, forget the trajectories. You have an integral

$$\int_{\vec r_1}^{\vec r_2} L(\vec r, \vec v, t) dt.$$

Did someone until this point said that the integral is take about a precise trajectory, a known trajectory? No!

1) Then pick a time $t$.

2) Do you know at this time the vectors $\vec r$? No! Then pick at random a and $\vec r$.

3) Do you know for this $\vec r$, the $\vec v$? No! At the position $\vec r$?, the vector $\vec v$ may point in whatever direction. Then pick a vector $\vec v$.

By these three selections you chose a segment of trajectory which passes at the time $t$ through the position $\vec r$ ang continues for an interval $dt$ in the direction $\vec v$.

So, you have a 7-fold infinity of choices of trajectories. So, to summarize, only if you knew a well defined trajectory, $\vec v$ were dependent on $\vec r$, and both on $t$.

From now on, I recommend to look in your reference at the answer of grizzly adam. You can see that the concept of trajectory is not even introduced in the beginning of the proof of the E-L equation, only later.

About frame, yes, is it tacitly understood, at least in classical mechanics, that we take a specific frame. Anyway, it is better to understand the things classically, and only then to pass to relativity if you need.

$\endgroup$
  • $\begingroup$ Thanks for the response. Your first line is confusing because clearly velocity is dependent on "rate of change of position" and not position itself....these are two different entities. Your other argument seems good to me. I think I have my doubt clear now. The velocity is dependent on the trajectory (once we have chosen the trajectory) and not the position at an instant. Since Lagrangian is calculated at an instant and is not the function of trajectory (it is not a functional), we can say velocity and position are independent. $\endgroup$ – singularity Mar 5 '15 at 12:17
  • $\begingroup$ @stringcosmologyapplicant I modified my 1st line, but the meaning is that for a given trajectory the velocity depends on position and time in the form $vec v(\vec r, t) = d \vec r/dt$. $\endgroup$ – Sofia Mar 5 '15 at 12:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.