Lagrangian: q and q-dot independence

[Erland]

By the way, having thought about this for a minute, I'm not sure what the statement is that you want to prove. Is it $\forall t (f(q(t),\dot q(t))=f'(q(t),\dot q(t))) \Rightarrow f = f'$? If it holds for all $t$, it also holds for $t=0$, so if $q(0)$ and $\dot q(0)$ can be arbitrary and cover the whole domain of $f$ and $f'$, this is trivially true and if they don't cover the whole domain of $f$ or $f'$ (e.g. $q$ is an angle between $0$ and $2\pi$), it's easy to find counterexamples by using piecewise defined functions for example.

I meant $\forall q\forall t (f(q(t),\dot q(t))=f'(q(t),\dot q(t))) \Rightarrow f = f'$, and my proof of this is essentially the same as your above (for every point in the domain of $f$ and $f'$ and every allowed $t$, there is a path such that $(q(t),\dot q(t))$ equals this point, but you are right that it is sufficient to consider $t=0$, provided that we, as here, don't have explicit $t$-dependence).
I don't agree that this is trivial, but it is easy when you think of it. All I want is that textbook authors point this out.

 

[rubi]

I meant $\forall q\forall t (f(q(t),\dot q(t))=f'(q(t),\dot q(t))) \Rightarrow f = f'$, and my proof of this is essentially the same as your above (for every point in the domain of $f$ and $f'$ and every allowed $t$, there is a path such that $(q(t),\dot q(t))$ equals this point, but you are right that it is sufficient to consider $t=0$, provided that we, as here, don't have explicit $t$-dependence).

Whether this statement is true or false depends on the domains of $f$ and $f'$ as well as the space of functions $q(t)$ that you consider. If the values that $q(t)$ and $\dot q(t)$ can take, don't cover the whole domain of $f$ and $f'$, you can't conclude that the functions are equal.

All I want is that textbook authors point this out.

As I said, it's not relevant for Lagrangian mechanics, so I don't see why it should be mentioned in textbooks.
Maybe it's relevant somewhere in the theory of differential equations, but I'm sure that it is pointed out in the corresponding books then.

 

[UltrafastPED]

When we say q and q-dot are independent we mean that they are linearly independent functions.

You can show linear independence of functions via the Wronskian determinant; if you have two functions you just make two columns, with each function at the head of its own column. The next element in each column is the derivative of the element above it. Then if the determinant of that matrix is non-zero over some domain you know that the two functions that you started with are linearly independent over that domain.

For more details of the method see http://en.wikipedia.org/wiki/Wronskian

For q and q-dot the first column is [q, q-dot]; the second column is [q-dot, q-dot-dot], and the determinant is:
W = q * q-dot-dot - q-dot * q-dot.

Since the relationship between a function and its derivative is not algebraic, it is clear that for an arbitrary path you would expect W=0 only very rarely. Hence you may wish to test your solutions for these conditions.

 

[rubi]

When we say q and q-dot are independent we mean that they are linearly independent functions.

Er, no. We don't mean that and there are definitely cases in which this claim is wrong (for example $q(t)=t$ and $\dot q(t) = 1$). The relation between $q$ and $\dot q$ is completely irrelevant, since neither $q$ nor $\dot q$ gets differentiated with respect to the other variable at any point. When we say $q$ and $\dot q$ are independent, we really don't refer to any relation between the two at all. It is in fact a wrong statement. The only reason why it gets taught to students is because many professors are just too lazy to explain it properly. (This is not the only point in physics education, where they lie to you by the way. It happens all the time.) What we really mean by that phrase is that the Lagrangian $L$ is a function with two distinct slots, so you can perform partial derivatives with respect to them.

 

[UltrafastPED]

@rubi -do you have a source for that?

 

[rubi]

@rubi -do you have a source for that?

What kind of source are you looking for? The correct derivation is in every text on variational calculus, but the derivations that are given in physics books don't depend on any relation between $q$ and $\dot q$ as well (although they might lack rigor in other ways). I've also explained it in post #8, #17 and #19 in this thread.