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 $h_*^s$ 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?

 

[dEdt]

Forgot my pic.

 

[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 $\frac{d}{dt}\frac{\partial L}{\partial \mathbf{\dot{q}}} = \frac{\partial L}{\partial \mathbf{q}}$, 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 $\mathbf{\dot{q}}(t) = \frac{d}{dt} \mathbf{q}(t)$.

 

[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 $\mathbf{Q}(\mathbf{q},t)$ 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.

 

[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]

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?

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?

 

[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]

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.

 

[dEdt]

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.

 

[voko]

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.

 

[Oxvillian]

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.

 

[voko]

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).

 

[voko]

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.

He does not consider non-holonomic constraints, either.

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

And it does not preserve the Lagrangian, too. But if you find one that does, even in a time dependent fashion, then it will necessarily map a solution to a solution.