Lagrangian: q and q-dot independence

[Erland]

Well, there might be such functions

On the contrary: there are no such functions. There is only one possible pair of functions which works for all paths: the one given by the chain rule. I would guess that this uniqueness is a direct consequence of your tangent bundle approach (but I need to study this field more, my knowledge about these issues are quite shallow at the present), and anyway, I have given a more elementary proof.

The one that has been derived using the chain rule is right.

It is in fact the only one. But my point is that this uniqueness is not obvious, but has to be proved.
For suppose that we don't know about this uniqueness. Suppose also that we derive the formulas $\dot x=-\cos(\varphi)\dot\varphi$ and $\dot y=\sin(\varphi)\dot\varphi$ without using the chain rule, for example by a geometric argument (like: "The norm of the velocity vector is $|\dot\varphi|$, since the particle moves along the unit circle in the $xy$-plane with angular velocity $\dot \varphi$, and its direction is obtained from the direction of the position vector $(x,y)=(\cos \varphi,\sin\varphi)$ by rotating it 90 degrees counterclockwise if $\varphi>0$, clockwise if $\varphi <0$. It follows that $\dot x=f(\varphi,\dot\varphi)=-\cos(\varphi)\dot\varphi$ and $\dot y=g(\varphi,\dot\varphi)=\sin(\varphi)\dot\varphi$.) We then plug this in the expression for L and procced by differentiating wrt $\varphi$ and $\dot\varphi$, thereby using the partial derivatives of $f$ and $g$.
But then, we suddenly hesitate and say: "Are these functions $f(u,v)=-\cos (u)\,v$ and $g(u,v)=\sin (u)\,v$ really the same functions as those we obtain from the chain rule? If not, we will get wrong result."
Of course, in this simple case it is easy to see that we will get the same functions with the chain rule, but one can imagine more complicated cases with complicated coordinate transformation formulas, where geometric or other alternative methods exist to obtain these functions, which would be considerably simpler than using the chain rule. If we then don't know that these functions are unique, we would have to check with the chain rule to see if we obatined the right ones. If we know that they are unique, this is not necessary.

In applications, we often see such alternative derivations. Wouldn't it be nice to not have to check with the chain rule every time? If we prove the uniqueness, we know that this can be avoided.

 

[rubi]

On the contrary: there are no such functions. There is only one possible pair of functions which work for all paths: the one given by the chain rule.

Maybe this is a true statement. But even if it weren't true, this wouldn't be a problem at all. It's just not relevant for Lagrangian mechanics.

I would guess that this uniqueness is a direct consequence of your tangent bundle approach (but I need to study this field more, my knowledge about these issues are quite shallow at the present), and anyway, I have given a more elementary proof.

I didn't prove any uniqueness properties using the tangent bundles. I just explained the general formalism.

It is in fact the only one. But my point is that this uniqueness is not obvious, but has to be proved.

I agree that the uniqueness of these functions isn't obvious, but I don't agree that it has to be proved in order to make the formalism work. I'm not arguing that it is wrong (in fact I don't know). I'm just arguing that no such theorem is required. Maybe it is useful in some other field of mathematics.

For suppose that we don't know about this uniquess. Suppose also that we derive the formulas $\dot x=-\cos(\varphi)\dot\varphi$ and $\dot y=\sin(\varphi)\dot\varphi$ without using the chain rule, for example by a geometric argument (like: "The norm of the velocity vector is $|\dot\varphi|$, since the particle moves along the unit circle in the $xy$-plane with angular velocity $\dot \varphi$, and its direction is obtained from the direction of the position vector $(x,y)=(\cos \varphi,\sin\varphi)$ by rotating it 90 degrees counterclockwise if $\varphi>0$, clockwise if $\varphi <0$. It follows that $\dot x=f(\varphi,\dot\varphi)=-\cos(\varphi)\dot\varphi$ and $\dot y=g(\varphi,\dot\varphi)=\sin(\varphi)\dot\varphi$.) We then plug this in the expression for L and procced by differentiating wrt $\varphi$ and $\dot\varphi$, thereby using the partial derivatives of $f$ and $g$.
But then, we suddenly hesitate and say: "Are these functions $f(u,v)=-\cos (u)\,v$ and $g(u,v)=\sin (u)\,v$ really the same functions as those we obtain from the chain rule? If not, we will get wrong result."

If your reasoning was mathematically sound, you would automatically get the same result as if you had applied the chain rule. Otherwise, mathematics would be fundamentally flawed, since there can't be contradictory answers to the same problem. You don't need any uniqueness proof for this to work. On the other hand, if you were using handwaving arguments, a uniqueness theorem wouldn't help either.

In applications, we often see such alternative derivations. Wouldn't it be nice to not have to check with the chain rule every time? If we prove the uniqueness, we know that this can be avoided.

As I said, it's not the uniqueness of these functions that makes other derivations possible. It's the consistency of mathematics that makes them work.

 

[Erland]

If your reasoning was mathematically sound, you would automatically get the same result as if you had applied the chain rule.

I don't think you can prove this without using uniqueness.

 

[rubi]

I don't think you can prove this without using uniqueness.

Having established that the chain rule gives the correct answer, everything that doesn't give the same answer as the chain rule gives a wrong answer (or it gives an equally valid answer, which would be fine as well). Assuming mathematics is non-contradictory, you can't derive a wrong answer from true statements, so you can't derive an expression for $\dot x$ that yields wrong Euler-Lagrange equations.


--
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.

 

[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 $t=0$).
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.

 

[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.