# Solving functions for S in a q-q* Hamilton-Jacobi diffeq

1. Jun 3, 2016

### dynamicskillingme

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
So far I have a solution for a) as

For b) I formulate the equation as

and so far for c) I have

My main idea at the moment is that as the Lagrangian was not time dependent, the Hamiltonian will not be. Following on maybe I can say that when separating variables T is constant so I can separate it so
but this is as far as I could get, as I couldn't think of how to separate them.

Any ideas, or are my problems due to a mistake in working?

2. Jun 4, 2016

### TSny

The last three terms do not have the correct dimesions. I believe you have dropped some factors of $2$ and $m$. So, check your derivation.

Alternately, you might see if you can first take the Lagrangian as given in terms of a general vector potential $\bf{A}$ and show that the Hamiltonian can be written nicely as $H = \frac{1}{2m} (\mathbf{p}-e \mathbf{A}) \cdot (\mathbf{p}-e \mathbf{A})$. Then substitute the given expression for $\mathbf{A}$.
You probably didn't mean to have an equal sign after the $T(t, \overline{x}, \overline{y})$

You cannot conclude that $T$ is a constant. But you should be able to argue that $\frac{\partial T}{\partial t}$ is a constant.
Also, you should hope that $X$ and $Y$ are not identically equal to zero. If they were, there would be no hope of finding the correct equations for $x(t)$ and $y(t)$.

Last edited: Jun 4, 2016
3. Jun 6, 2016

### dynamicskillingme

My Hamilton was a mess from a mistake early on. After redoing it I get .

I can then formulate the hamiltonian jacobi again and get

Then (and I'm mostly going from an example in my lecture notes and the question hints) I can separate the variables out as

That can be substituted as

In my notes the example only contains two separate variables e.g. X and T. He then makes every non time dependent term equal the time dependent terms and says they are both equal to a constant. He then solves X and T wrt this constant.

I'm not sure how to take this question further and solve for these functions.

4. Jun 6, 2016

### TSny

The numerical factor of -3 is not correct for the $(eBx)^2$ term. Also, I believe the last term has the wrong sign.
Yes
Here you want to use the usual type of reasoning when separating variables.
Suppose you have an equation of the form $f(x) + g(y)$ = 0 that must hold for all values of $x$ and $y$ within some domain. Here, $f(x)$ is a function only of $x$ and $g(y)$ is a function only of $y$. Then you can argue that $f(x)$ must be a constant: $f(x) = C$ for all $x$. Moreover, $g(y)$ must also be a constant which is the negative of $C$; i.e., $g(y) = -C$. Try to construct such an argument.

For your H-J equation, you have something of the form $f(x,y) + \frac{dT(t)}{dt} = 0$ that must hold for a range of values of $x, y$, and $t$. Use the type of argument mentioned above to deduce something about $\frac{dT}{dt}$.

5. Jun 6, 2016

### dynamicskillingme

Okay, so I must have assumed something wrong early on or have made a pretty bad mistake in my work (not unheard of)

6. Jun 6, 2016

### TSny

So far so good. Just a matter of simplifying.

7. Jun 6, 2016

### Fred Wright

The Hamiltonian is the Legendre transformation of the Lagrangian:
$H = \sum_{i} \dot q_i \frac {\partial L} {\partial \dot q_i} - L = m(\dot x^2 + \dot y^2) + eBx\dot y - \frac {m} {2}(\dot x^2 + \dot y^2)- eBx\dot y$
$= \frac {1} {2m} (p_x^2 + p_y^2)$
In this problem you need to find constants of motion. If you solve for the Lagrangian eqns. of motion, you will find:
$m\ddot x -eB\dot y = 0 →\frac {d} {dt}(m\dot x -eBy) = 0$ and
$m\ddot y +eB\dot x = 0→\frac {d} {dt}(m\dot y +eBx) = 0$
Thus we have the constants of motion:
$\tilde p_x =(p_x -eBy)$,
$\tilde p_y =(p_y +eBx)$.
We write the HJ equation with $T(t,\tilde x,\tilde y) = \tilde xt$ and substitute $p_y = \tilde p_y - eBx$ to get:
$(\frac {dX} {dx})^2 + ( \tilde p_y - eBx)^2 = 2m \tilde x$,
$X= \int dx\sqrt { 2m \tilde x- ( \tilde p_y - eBx)^2}$
$S= \int dx\sqrt { 2m \tilde x- ( \tilde p_y - eBx)^2} - \tilde xt$
$p_x = \frac {\partial S} {\partial x} = \sqrt { 2m \tilde x- ( \tilde p_y - eBx)^2}$
$\tilde p_x = -\frac {\partial S} {\partial \tilde x} = -m \int dx\frac {1} { \sqrt { 2m \tilde x- ( \tilde p_y - eBx)^2}} +t$
We now make the substitution in the integral:
$x' =\tilde p_y - eBx$,
$dx' = -eBdx$
$\tilde p_x = -\frac {\partial S} {\partial \tilde x} = \frac {m} {eB} \int \frac {dx'} { \sqrt { 2m \tilde x- x'^2}} +t = \frac {m} {eB}sin^{-1}(\frac{\tilde p_y - eBx} { 2m \tilde x}) + t$
You can do the algebra to solve for x in terms of $\tilde p_x ,\tilde p_y$ and $\tilde x$.
To get y just use the constant of motion equation $y = \frac {m\dot x - \tilde p_x} {eB}$