# Thread: Hamiltonian mechanics- show that the following are extremals...

1. ## Hamiltonian mechanics- show that the following are extremals...

The problem statement :

Consider the variational problem with lagrangian function $L(t,x,\dot x)= \sqrt{(t^2+x^2)(1+\dot x^2)}$

Show that the extremals are given by the two-parameter family of curves,
$t^2cos\beta + 2tsin\beta -x^2cos\beta=\alpha$, where alpha and beta are the parameters, by means of the following 3 methods:

a) transform L to polar coordinates,

b) using canonical variables,

c)using the Hamilton-Jacobi equation.

Hint for c): Let $S(t,x)=\frac{1}{2}(At^2+2Btx+Cx^2)$ be a solution of the hamilton-jacobi equation. This will show that $S(x,t) =\frac{1}{2}(t^2sin\beta-2txcos\beta-x^2sin\beta)$ is a solution of the hamilton jacobi equation. Now prove that $\frac{\partial S}{\partial \beta} =\alpha$ is a solution of the corresponding hamilton's equations.

My (limited)thoughts:

For a) using $x=rcos\theta$ and finding $\dot x$ i substitute and get L in polar form. That's easy, but now how to check that the given extremal satisfies this polar form? I think i check that the extremal satisfies the Euler - lagrange equation for L polar ? But i'm not sure of some things, like do i first solve for x in the extremal and use this as my general coordinate in the euler-lagrange equation? I also don't know how to get $\frac{\partial L}{\partial x}$ where x would then be the expression for the extremal? Please help me on how to think about this !

for b) really not sure here, but do i just show that for the given extremal, Hamilton's canonical equations $\frac{dp}{dt} = -\frac{\partial H}{\partial x}\ \ and\ \ \frac{dx}{dt} = \frac{\partial H}{\partial p}$ are satisfied ? Again should i be solving for x in the given extremal to use in hamilton's equations? If so how do i get the partial with respect to that expression ?

for c) I sub The given form of S into the Hamilton-Jacobi equation to get the 2nd form of S ( in the hints) . Now will me proving $\frac{\partial S}{\partial \beta} =\alpha$ somehow yield the given extremal? I would think so, but again the math eludes me.

Any help,especially on the thinking behind all this, is much appreciated.

2. For part a), I have a question: if you're required to change to polar coordinates, then precisely what is $x$ in your original Lagrangian? Is it merely the x-coordinate, like you say? Or is it the vector $x=\langle r\cos(\theta),r\sin(\theta)\rangle$? Either would make sense in your original Lagrangian. If $x$ is merely the x-coordinate, then how come there isn't a $y$ in the original Lagrangian? It just seems a bit strange.

But i'm not sure of some things, like do i first solve for x in the extremal and use this as my general coordinate in the euler-lagrange equation?
Normally, you wouldn't do that. The beauty of the Euler-Lagrange equations is that they apply to any set of generalized coordinates. Let's say I have a Lagrangian

$L(x,\dot{x},y,\dot{y},t),$ and I transform it into polar coordinates:

$L(r,\dot{r},\theta,\dot{\theta},t).$

The Euler-Lagrange equations before the transformation would have been

$\displaystyle{\frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}=\frac{\partial L}{\partial y}-\frac{d}{dt}\frac{\partial L}{\partial\dot{y}}=0.}$

The Euler-Lagrange equations after the transformation are

$\displaystyle{\frac{\partial L}{\partial r}-\frac{d}{dt}\frac{\partial L}{\partial\dot{r}}=\frac{\partial L}{\partial\theta}-\frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}}=0.}$

So you see that the form does not change, even when you do a change of variables. What you want to do is apply this second system of equations to your transformed Lagrangian, without reference to back-solving for x or anything like that. Just stay in the new variable system.

For part b), the same thing applies. Hamilton's equations look the same for a coordinate and its conjugate momentum, regardless of what coordinates you choose. So do the transformation to whatever variables you want to use, find the conjugate momenta for each new coordinate, plug those momenta back into your Lagrangian to eliminate the coordinate velocities, form the sum that defines the Hamiltonian, and then write down Hamilton's equations. At least, that's the usual procedure. There might be some fancy shortcut somewhere to help you out. But this is what is normally done.

For part c), I'm afraid I can't help you much. I haven't studied the Hamilton-Jacobi equation in enough detail to really know what I'm talking about.

3. Thanx Adrian, as for part a), i am just given the information about the problem as i put it up here, so i just assumed it to be cartesian? I unfortunately have to abandon Hamiltonian mechanics and study Analysis for the next month or so... so will have to pick up this question again later! As for the heat equation with the exponent w that was so baffling, the memo was sent out with solutions as though the w never existed, but still lost my 5 marks though !! Thanks for the help, undoubtedly of all 5 of my subjects i am taking this year, 3 of them 3rd year applied math courses, Hamilton-Jacobi theory is both the most interesting, and the most infuriating subject, just like a woman ! Thanks for all your help!

4. You're welcome for whatever help I was able to provide. Let me know when you'd like some more!

Regards.