## Unproved statement from my textbook

I've included a pic of a section of Arnold's Mathematical Methods of Classical Mechanics. The relevant passage is: "Since $h_*^s$ preserves $L$, the translation of the solution, $h^s\circ \mathbf{\varphi}:\mathbb{R}\rightarrow M$ 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?
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 Forgot my pic. Attached Thumbnails
 Anyone?

Recognitions:

## Unproved statement from my textbook

Attached image is unreadable. I have the book, though, what page is this?
 Page 88. What's wrong with the image? Too small?
 Maybe this'll make the image more readable: http://imgur.com/Emv3s Also, in case it wasn't know, Lagrange's equation is $\frac{d}{dt}\frac{\partial L}{\partial \mathbf{\dot{q}}} = \frac{\partial L}{\partial \mathbf{q}}$, where L is a function of $\mathbf{q}$ and $\mathbf{\dot{q}}$, both of which are part of R^n. In the above equation equation, it's understood that $\mathbf{q}$ and $\mathbf{\dot{q}}$ are replaced by $\mathbf{q}(t)$ and $\mathbf{\dot{q}}(t)$ such that $\mathbf{\dot{q}}(t) = \frac{d}{dt} \mathbf{q}(t)$.
 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 $\mathbf{Q}(\mathbf{q},t)$ is the result of applying a transformation to the system's generalized coordinates $\mathbf{q}$at time $t$, and $L(\mathbf{Q},\dot{\mathbf{Q}},t) = L(\mathbf{q},\dot{\mathbf{q}},t)$, and $\mathbf{q}(t)$ satisfies Lagrange's equations, then so does $\mathbf{Q}(\mathbf{q}(t),t)$. $$\frac{\partial}{\partial \mathbf{q}}L(\mathbf{q},\dot{\mathbf{q}},t) =\frac{\partial}{\partial \mathbf{q}}L(\mathbf{Q}(\mathbf{q},t),\dot{\mathbf{Q}}(\mathbf{q},\dot{ \mathbf{q}},t),t)= \frac{\partial}{\partial \mathbf{Q}}L(\mathbf{Q},\dot{\mathbf{Q}},t)\cdot \frac{\partial \mathbf{Q}}{\partial \mathbf{q}}$$ $$\frac{d}{dt}\left(\frac{\partial}{\partial \dot{\mathbf{q}}}L(\mathbf{q},\dot{\mathbf{q}},t) \right)= \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\mathbf{q}}}L(\mathbf{Q}(\mathbf{q},t),\dot{\mathbf{Q}}(\mathbf{q} ,\dot{\mathbf{q}},t),t)\right)= \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\mathbf{Q}}} L(\mathbf{Q},\dot{\mathbf{Q}},t)\cdot \frac{\partial \dot{\mathbf{Q}}}{\partial \dot{\mathbf{q}}}\right)$$ Now, the two expressions above equal each other because $\mathbf{q}(t)$ satisfies Lagrange's equation. From this, I'm supposed to conclude that $$\frac{d}{dt}\left(\frac{\partial}{\partial \dot{\mathbf{Q}}} L(\mathbf{Q},\dot{\mathbf{Q}},t)\right)=\frac{\partial }{\partial \mathbf{Q}}L(\mathbf{Q},\dot{\mathbf{Q}},t)$$ but I don't see how that's possible.
 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.
 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.
 I think Q = Q(q), not Q = Q(q,t). That map isn't time dependent.

 Quote by 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.
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?

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

 Quote by dEdt But that said, I'm now left wondering why my (attempted) proof using the Lagrangian directly didn't work. Any ideas?
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: $$\left[\frac {\partial L} {\partial q}\right]_{q = Q(t)} - \frac {d}{dt} \left[\frac {\partial L}{d \dot{q}}\right]_{q = Q(t)} = 0$$

 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.
It is true.

 Quote by 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: $$\left[\frac {\partial L} {\partial q}\right]_{q = Q(t)} - \frac {d}{dt} \left[\frac {\partial L}{d \dot{q}}\right]_{q = Q(t)} = 0$$.
I suspected my mistake had something to do with that. But where do I go from there?

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

 Quote by 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.
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.

 Quote by 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.
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.

 Quote by dEdt I suspected my mistake had something to do with that. But where do I go from there? I can't say that $$\left[\frac {\partial L} {\partial q}\right]_{q = Q(t)}=\left[\frac {\partial L} {\partial q}\right]_{q = q(t)}$$ because it isn't true. There's probably a manipulation I can do with chain rule, but I don't see it.
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).