Adding total time derivative to Lagrangian/Canonical Transformations

[PhDeezNutz]

TL;DR

- The Lagrangian is unique up to a total time derivative of function $F(q,t)$
- Adding a total time derivative of function $F$ which can be a function of some combination of old and new $p$ and $q$ (conjugate momenta and generalized coordinates) to the Lagrangian can be used to generate canonical transformations
(Via Generating Functions)
- The latter two statements seem to contradict each other

Based off of these MIT Notes: https://ocw.mit.edu/courses/8-09-cl...a346a0868efb7430582c_MIT8_09F14_Chapter_4.pdf

1) This set of notes starts with the premise that $L’ = L + \frac{dF(q,t)}{dt} = L + \frac{\partial F}{\partial q} \dot q + \frac{\partial F}{\partial t}$ produces the same Euler Lagrange Set of Equations. I’m going to prove this.

$\frac{d}{dt} \left( \frac{\partial L’}{\partial \dot q}\right) - \frac{\partial L’}{\partial q} = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot q} + \frac{\partial F}{\partial q}\right) - \left( \frac{\partial L}{\partial q} + \frac{\partial^2 F}{\partial q^2} \dot q + \frac{\partial F}{\partial q \partial t} \right)$

$= \frac{d}{dt} \left( \frac{\partial L}{\partial \dot q} \right) + \frac{\partial^2 F}{\partial q^2} \dot q + \frac{\partial^2 F}{\partial t \partial q} - \frac{\partial L}{\partial q} - \frac{\partial^2 F}{\partial q^2} \dot q - \frac{\partial F}{\partial q \partial t}$

All the $F$ terms cancel above

And we recover the usual Euler Lagrange Equation

$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot 1}\right) - \frac{\partial L}{\partial q} = 0$

2) In the notes above on page 2 they apply the variational principle to two different Lagrangians and say they differ by a total derivative of some function $F$ which depends on some combination of old and new $p$ and $q$

$\delta \int p \dot q - H(q,p,t) \, dt = 0$
$\delta \int P \dot Q - K(Q,P,) \, dt = 0$

3) They then use the conclusion is 1) and say the Lagrangians must differ by $\dot F(q,t)$ (or $\dot F(Q,t)$)

$p \dot q - H = P \dot Q - K + \dot F$

4) Then on page 3 they change the reasoning to add $\dot F$ where $F$ depends on some combination of old and new generalized momenta and coordinates. They use this to generate canonical transformations

5) My overall question is the following. How are the two statements below reconcilable? (Let’s assume an $F(q,P)$ for the second part)

$L’ = L + \frac{dF(q,t)}{dt}$ produces the same equations of motions
$L’ = L + \frac{dF(q,P)}{dt}$ also produces the same equations of motions

Thanks for any help in advance.

 

[PhDeezNutz]

On a side note (or maybe not a side note?). I think I see the motivation for defining the Poisson Bracket.

If we add a total time derivative $\frac{dF(q,p,t)}{dt}$ to the Lagrangian and ASSUME that Hamilton’s equations hold, that is

$\dot q = \frac{\partial H}{\partial p}$

$\dot p = - \frac{\partial H}{\partial q}$

Now add a total time derivative to the Lagrangian

$p \dot q - H - \frac{\partial F}{\partial q} \dot q - \frac{\partial F}{\partial p} \dot p - \frac{\partial F}{\partial t}$

Plugging in $\dot q$ and $\dot p$ in the middle we get (After factoring out the minus sig)

$\frac{\partial F}{\partial q} \frac{\partial H}{\partial p} - \frac{\partial F}{\partial p} \frac{\partial H}{\partial q}$

Looks a lot like a Poisson Bracket. And if memory recalls we can add a PARTIAL time derivative of $F(q,p,t)$ to the Hamiltonian and preserve Hamilton’s equations of motions.

So I guess the conclusion

- ISN’T that we can add a total time derivative of $\frac{dF}{dt}$ to the Hamiltonian

BUT

- we can add rather a partial time derivative $\frac{\partial F}{\partial t}$ to the Hamiltonian and preserve the equations of motions IF the Poisson Bracket $\left[F(q,pt),H(q,p,t)\right] = 0$

 

[wrobel]

Speaking formally, the theory of Lagrangian systems and the theory of Hamiltonian systems are independent. It is a bad idea to treat one though another at least in the beginning. In particular an expression $p\dot q-H(p,q,t)$ is not a Lagrangian because the Lagrangian by its definition does not depend on $p$.

Moreover I see several strange points in the text you quote above. For example in formula (4.10) the condition $\delta p\mid_{t_1}^{t_2}$ is bad (you can not fix both ends of the trajectory at arbitrary points of the phase space). There is no such a condition in the standard statement on the Stationary Action principle for Hamiltonian equations. Perhaps it would be better to learn the Hamiltonian equations theory by Arnold's Math. methods of Classical Mech.