Generating function and Lagrangian invariance

1. Mar 8, 2015

Coffee_

To make my explanation easier open the ''Generating function approach'' section on this wiki article:

http://en.wikipedia.org/wiki/Canonical_transformation

The function $\frac{dG}{dt}$ represents the function that always can be added to the Lagrangian without changing the mechanical equations.

However, in the Lagrangian mechanics formulation the function $G$ is only allowed to be a function of the coordinates and time $G(q,t)$.

If I look further down in the wiki article I find that $G$ is allowed to have other variables like $G(p,Q,t)$. It isn't very clear why it is trivial or easy to see that this won't give any problems.

2. Mar 10, 2015

optophotophys

Let's start from the Type 1 generating function described in the provided link.
This $G_1$ is a function of the old and new general coordinates.
Then, we can get the same equations of motion. This can be proved by the argument shown in
https://en.wikibooks.org/wiki/Classical_Mechanics/Lagrange_Theory#Is_the_Lagrangian_unique.3F
because the boundary value of the old and new coordinates are fixed.

Sofar we have proved that if there are the transformations $Q(q,p,t),\, P(q,p,t)$ satisfying
\begin{align} \mathbf{p} &= \frac{\partial G_{1}}{\partial \mathbf{q}} \\ \mathbf{P} &= -\frac{\partial G_{1}}{\partial \mathbf{Q}} \\ K &= H + \frac{\partial G_{1}}{\partial t}, \end{align}
then the resultant equations of motion are identical.

Next, we will consider another transformations $Q(q,p,t),\, P(q,p,t)$ satisfying
\begin{align} \mathbf{p} &= \frac{\partial G_{2}(q,P,t)}{\partial \mathbf{q}} \\ \mathbf{Q} &= \frac{\partial G_{2}(q,P,t)}{\partial \mathbf{P}} \\ K &= H + \frac{\partial G_{2}(q,P,t)}{\partial t}. \end{align}
By Legendre transformation $G'(q,Q,t)=G_2(q,P,t)-PQ$, we achieve
\begin{align} \mathbf{p} &= \frac{\partial G'}{\partial \mathbf{q}} \\ \mathbf{P} &= -\frac{\partial G'}{\partial \mathbf{Q}} \\ K &= H + \frac{\partial G'}{\partial t}. \end{align}
You can prove these equations by carefully handling the partial derivatives.
Apparently, these equations result in the same equations of motion as the original one.

The similar arguments can be applied to$G_3,\, G_4$.

3. Mar 11, 2015

Coffee_

Thanks for the reply. While I agree with everything you say you never mentioned the exact problem I had. Let's take a look at the wiki page expression as well and say that we take only one variable to make it easy writing:

$p\frac{dq}{dt} - H(p,q,t) = P \frac{dP}{dt} - K(Q,P,t) + \frac{dG}{dt}$

My point is that this is exactly equal to :

$L(q,\frac{dq}{dt},t) = L'(Q,\frac{dQ}{dt},t) + \frac{dG}{dt}$

So clearly this function $G$ is the same frunction that one can add a total derivative of and keep the Lagrangian invariant. However it is known that such a function can only explicitly depend of position and time $G(q,t)$ and so be explicitly independent of $\frac{dq}{dt}$

However writing $G(q,P,t)$ and likewise makes it not so clear anymore that it isn't explicitly independent of \frac{dq}{dt}.

4. Mar 12, 2015

optophotophys

Although I cannot give a qualitative explanation, here is my proof (based on the principle of least action):
\begin{align} 0&=\delta \int dt (p\dot{q}-H)=\delta\int dt (P\dot{Q}-K+\frac{dG_1(q,Q)}{dt})\\ &=\int dt \delta P\dot{Q}+\delta\dot{Q}P-\delta P\frac{\partial K}{\partial P}-\delta Q\frac{\partial K}{\partial Q}+\frac{d \delta G_1}{dt}\\ &=\int dt \delta P\dot{Q}+\frac{d}{dt}(P\delta Q)-\dot{P}\delta Q-\delta P\frac{\partial K}{\partial P}-\delta Q\frac{\partial K}{\partial Q}+\frac{d \delta G_1}{dt}\\ &=\int dt \delta P \left( \dot{Q}-\frac{\partial K}{\partial P}\right)+\delta Q\left(-\frac{\partial K}{\partial Q}-\dot{P}\right)+\frac{d}{dt}(P\delta Q)+\frac{d \delta G_1}{dt}. \end{align}
Here, we note that
\begin{align} \delta G_1(q(Q,P),Q)&=\delta P \frac{\partial G_1(q,Q)}{\partial q}\frac{\partial q(Q,P)}{\partial P}+\delta Q\left[ \frac{\partial G_1(q,Q)}{\partial Q}+\frac{\partial G_1(q,Q)}{\partial q}\frac{\partial q(Q,P)}{\partial Q}\right]\\ &=\delta P p\frac{\partial q(Q,P)}{\partial P}+\delta Q\left[ -P+p\frac{\partial q(Q,P)}{\partial Q}\right]. \end{align}
Then,
\begin{align} 0&=\int dt \delta P \left( \dot{Q}-\frac{\partial K}{\partial P}\right)+\delta Q\left(-\frac{\partial K}{\partial Q}-\dot{P}\right)+\frac{d}{dt}\left[ \delta P p\frac{\partial q(Q,P)}{\partial P}+\delta Qp\frac{\partial q(Q,P)}{\partial Q}\right]\\ &=\int dt \delta P \left( \dot{Q}-\frac{\partial K}{\partial P}\right)+\delta Q\left(-\frac{\partial K}{\partial Q}-\dot{P}\right)+\frac{d}{dt}\left(p\delta q\right)\\ &=p_2\delta q_2-p_1\delta q_1+\int dt \delta P \left( \dot{Q}-\frac{\partial K}{\partial P}\right)+\delta Q\left(-\frac{\partial K}{\partial Q}-\dot{P}\right). \end{align}
We always fix the boundary values, when we vary the action. So,
\begin{align} 0=\int dt \delta P \left( \dot{Q}-\frac{\partial K}{\partial P}\right)+\delta Q\left(-\frac{\partial K}{\partial Q}-\dot{P}\right). \end{align}
This gives the Hamilton's equations.

5. Mar 12, 2015

vanhees71

The point is that Hamilton's principle in the Hamiltonian form is extended compared to the one in Lagrangian form. In the Hamilton form you vary the configuration variables and their conjugate momenta independently from each other. The configuration variables are fixed at the endpoints, the momenta are free. The consequence is that you get the invariance of the equations of motion (i.e., the variation of the action) under canonical transformations (local symplectomorphisms on phase space) rather than only configuration-coordinate transformations (local diffeomorphisms on configuration space).