Unproved statement from my textbook

dEdt

I've included a pic of a section of Arnold's Mathematical Methods of Classical Mechanics. The relevant passage is: "Since [itex]h_*^s[/itex] preserves [itex]L[/itex], the translation of the solution, [itex]h^s\circ \mathbf{\varphi}:\mathbb{R}\rightarrow M[/itex] also satisfies Lagrange's equation for any s."

Maybe Arnold thought it was too obvious to prove, but for whatever reason I'm having trouble seeing why it's true. I tried proving it with chain rule (I was sure the proof will be a simple cal exercise), but to no avail. It's probably really trivial to prove, but my brain doesn't seem to be working. Could someone help?

dEdt

Forgot my pic.

Attached Thumbnails

 

dEdt

Anyone?

Ben Niehoff

Attached image is unreadable. I have the book, though, what page is this?

dEdt

Page 88. What's wrong with the image? Too small?

dEdt

Maybe this'll make the image more readable: http://imgur.com/Emv3s

Also, in case it wasn't know, Lagrange's equation is [itex]\frac{d}{dt}\frac{\partial L}{\partial \mathbf{\dot{q}}} = \frac{\partial L}{\partial \mathbf{q}}[/itex], where L is a function of [itex]\mathbf{q}[/itex] and [itex]\mathbf{\dot{q}}[/itex], both of which are part of R^n. In the above equation equation, it's understood that [itex]\mathbf{q}[/itex] and [itex]\mathbf{\dot{q}}[/itex] are replaced by [itex]\mathbf{q}(t)[/itex] and [itex]\mathbf{\dot{q}}(t)[/itex] such that [itex]\mathbf{\dot{q}}(t) = \frac{d}{dt} \mathbf{q}(t)[/itex].

dEdt

Okay, so I'll show you my attempt at a solution. Something must have went terribly wrong somewhere; maybe one of you can find my mistake.

We want to show that if [itex]\mathbf{Q}(\mathbf{q},t)[/itex] is the result of applying a transformation to the system's generalized coordinates [itex]\mathbf{q}[/itex]at time [itex]t[/itex], and [itex]L(\mathbf{Q},\dot{\mathbf{Q}},t) = L(\mathbf{q},\dot{\mathbf{q}},t)[/itex], and [itex]\mathbf{q}(t)[/itex] satisfies Lagrange's equations, then so does [itex]\mathbf{Q}(\mathbf{q}(t),t)[/itex].

[tex]\frac{\partial}{\partial \mathbf{q}}L(\mathbf{q},\dot{\mathbf{q}},t) =\frac{\partial}{\partial \mathbf{q}}L(\mathbf{Q}(\mathbf{q},t),\dot{\mathbf{Q}}(\mathbf{q},\dot{ \mathbf{q}},t),t)= \frac{\partial}{\partial \mathbf{Q}}L(\mathbf{Q},\dot{\mathbf{Q}},t)\cdot \frac{\partial \mathbf{Q}}{\partial \mathbf{q}}[/tex]

[tex]\frac{d}{dt}\left(\frac{\partial}{\partial \dot{\mathbf{q}}}L(\mathbf{q},\dot{\mathbf{q}},t) \right)= \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\mathbf{q}}}L(\mathbf{Q}(\mathbf{q},t),\dot{\mathbf{Q}}(\mathbf{q} ,\dot{\mathbf{q}},t),t)\right)= \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\mathbf{Q}}} L(\mathbf{Q},\dot{\mathbf{Q}},t)\cdot \frac{\partial \dot{\mathbf{Q}}}{\partial \dot{\mathbf{q}}}\right)[/tex]

Now, the two expressions above equal each other because [itex]\mathbf{q}(t)[/itex] satisfies Lagrange's equation. From this, I'm supposed to conclude that
[tex]\frac{d}{dt}\left(\frac{\partial}{\partial \dot{\mathbf{Q}}} L(\mathbf{Q},\dot{\mathbf{Q}},t)\right)=\frac{\partial }{\partial \mathbf{Q}}L(\mathbf{Q},\dot{\mathbf{Q}},t)[/tex]
but I don't see how that's possible.

dEdt

If there's something unclear about how I formulated the question -- or if you think I should make a new question from scratch because of how scrambled this question has become -- let me know please.

voko

This is most trivially seen from the principle of least action. h does not change the Lagrangian, so it does not change the functional, so hv is a stationary point if v is stationary.

Oxvillian

I think Q = Q(q), not Q = Q(q,t). That map isn't time dependent.

dEdt

Yes, thank you! Now it makes sense.

But that said, I'm now left wondering why my (attempted) proof using the Lagrangian directly didn't work. Any ideas?

I asked whether the map needs to be time independent or not in another PF thread, and the answer I got was that it could be time dependent without affecting Noether's theorem.

Can someone else confirm who's right?

Oxvillian

Well if it's time dependent, I think we could map a solution q(t) onto pretty much any path Q(t) that we wanted. There would be no reason to expect a Q(t) so constructed to satisfy the equations of motion.

voko

One problem is that you seem to confuse the formal argument q with a particular trajectory, also denoted by q.

The derivatives Lagrange's equation are taken by the formal argument, but are evaluated at a particular trajectory. The equation should really be written this way: [tex]\left[\frac {\partial L} {\partial q}\right]_{q = Q(t)} - \frac {d}{dt} \left[\frac {\partial L}{d \dot{q}}\right]_{q = Q(t)} = 0[/tex]

It is true.

dEdt

I suspected my mistake had something to do with that. But where do I go from there?

I can't say that
[tex]\left[\frac {\partial L} {\partial q}\right]_{q = Q(t)}=\left[\frac {\partial L} {\partial q}\right]_{q = q(t)}[/tex]
because it isn't true. There's probably a manipulation I can do with chain rule, but I don't see it.

voko

The point here is not mapping an arbitrary trajectory to another trajectory. The point is being able to construct a non-trivial map that preserves the Lagrangian.

Oxvillian

The map h is a map from M to M, and so it can be used to map trajectories into other trajectories - hence the "translation of the solution" mentioned.

Note that the configuration space M has no time coordinate defined on it - so h is time-independent by definition. At least by Mr Arnold's definition, if I read him correctly.

Example - the trajectory x = 2t is a solution for a free-particle Lagrangian. If we translate it, x = 2t + 5, it's still a solution (because the free particle Lagrangian has the appropriate symmetry).

But if we use a time-dependent translation, to get say x = 2t + sin(t), it's not a solution any more.

voko

I think you could look at variations of q, then the derivatives of L can be represented as limits of variations of L, where you could apply L(hv) = L(v).