Hamiltonian Mechanics

Classical Mechanics Level 3

I've been looking into Hamiltonian mechanics lately. Here is an opportunity to use it to derive a classic result in kinematics: the parabolic trajectory of an object under the influence of gravity. Let the standard coordinates be $(x,y)$ . The Lagrangian is:

$L = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) - mg y$

Then we form the Hamiltonian as follows:

$H = p_x \dot{x} + p_y \dot{y} - L \\ p_x = \frac{\partial{L}}{\partial{\dot{x}}} \,\,\,\,\,\, p_y = \frac{\partial{L}}{\partial{\dot{y}}}$

Then the equations of motion are:

$\dot{x} = \frac{\partial{H}}{\partial{p_x}} \,\,\,\,\,\, \dot{p}_x = -\frac{\partial{H}}{\partial{x}} \,\,\,\,\,\, \dot{y} = \frac{\partial{H}}{\partial{p_y}} \,\,\,\,\,\, \dot{p}_y = -\frac{\partial{H}}{\partial{y}}$

What are the resulting $(\ddot{x}, \ddot{y})$ values?

$(\ddot{x}, \ddot{y}) = ?$

Hint: Before evaluating the equations of motion, replace the $\dot{x}$ and $\dot{y}$ terms in the Hamiltonian with expressions written in terms of $p_x$ and $p_y$

Note: I have read that Hamiltonian mechanics is less useful than Lagrangian mechanics for simple problems, but that it is far more useful for advanced applications such as statistical mechanics.

(-g, 0)

(0, g )

(g, 0)

(0, -g )

2 solutions

Karan Chatrath
Mar 19, 2020

My two cents on this problem:

$L = \frac{1}{2m}\left(p_x^2 + p_y^2\right) -mgy$

$H = p_x \dot{x} +p_x \dot{x} -L$ $p_x = m \dot{x}$ $p_y = m \dot{y}$

$\implies H = \frac{1}{2m}\left(p_x^2 + p_y^2\right) +mgy$

Crunching out the partial derivatives gives the answer. However, I have a comment. If we stick to the given definition of the Hamiltonian, we get:

$H =\frac{1}{2}\left(p_x \dot{x} +p_x \dot{x}\right) +mgy$

Using this expression to crunch out the partial derivatives would result in:

$\frac{\partial H}{\partial p_x} = \frac{\dot{x}}{2} \ne \dot{x}$ $\frac{\partial H}{\partial p_y} = \frac{\dot{y}}{2} \ne \dot{y}$ $\frac{\partial H}{\partial x} = 0 = -\dot{p}_x$ $\frac{\partial H}{\partial y} = mg = -\dot{p}_y$

The above result is strange. So what I understand from this is that the Hamiltonian must be only a function of generalised momenta and generalised coordinates, and must be made independent of generalised velocities.

Please correct me if I have drawn a wrong conclusion here.

Karan Chatrath - 1 year, 2 months ago

I've been looking at this book chapter:

https://scholar.harvard.edu/files/david-morin/files/cmchap15.pdf

Equation 15.21 makes this clear. In the author's notation, $(q_1, q_2)$ correspond to $( x, y)$ .

Steven Chase - 1 year, 2 months ago

Thanks for sharing this. I just read that section that you pointed me to.

In the past days, I was contemplating looking into Hamiltonian mechanics but laziness prevailed. This exercise has given me a good start to it.

Karan Chatrath - 1 year, 2 months ago

As far as I can tell, the Hamiltonian may contain $(x, y, p_x, p_y)$ , but not any of their time derivatives. So I agree with you

Steven Chase - 1 year, 2 months ago

How do you get $\dfrac{\partial H}{\partial x}=-p_x$ and similar equation for $\dfrac{\partial H}{\partial y}$ ? You are doing a fundamental mistake. By writing $\dfrac {\partial {(p_x\dot x)}}{\partial {p_x}}=\dot x$ , you are assuming that $p_x$ and $\dot x$ are independent variables, which is obviously not true.

A Former Brilliant Member - 1 year, 2 months ago

They were typos in my solution. I have corrected them. The equation is indeed:

$\frac{\partial H}{\partial x} = -\dot{p}_x$

Karan Chatrath - 1 year, 2 months ago

Regarding your quary, are you satisfied with my argument, or you have any questions?

A Former Brilliant Member - 1 year, 2 months ago

@A Former Brilliant Member – If you are referring to your previous comment, no, I do not have questions. Thank you for pointing out the typo

Karan Chatrath - 1 year, 2 months ago

A Former Brilliant Member
Mar 19, 2020

From the given definitions we get $p_x=\dfrac {\partial {L}}{\partial {\dot {x}}}=m\dot {x}$ , $p_y=\dfrac{\partial {L}}{\partial {\dot {y}}}=m\dot {y}$ . This implies $H=\dfrac {1}{2}m({\dot {x}}^2+{\dot {y}}^2)+mgy$ . So $\dot {p_x}=m\ddot {x}=-\dfrac{\partial {H}}{\partial {x}}=0, \dot {p_y}=m\ddot {y}=-\dfrac{\partial {H}}{\partial {y}}=-mg$ , and so $\boxed {\ddot {x}=0, \ddot {y}=-g}$

Hamiltonian Mechanics

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