Course:PHYS350/Hamilton's Equations

Hamilton's equations provide a new and equivalent way of looking at classical mechanics. Generally, these equations do not provide a new, more convenient way of solving a particular problem but rather they provide deeper insights into the structure of classical mechanics in general and its connection to quantum mechanics.

Deriving Hamilton's Equations

We can derive Hamilton's equations by looking at how the Lagrangian changes as you change the time and the positions and velocities of particles.

$dL=\sum _{i}\left({\frac {\partial L}{\partial q_{i}}}dq_{i}+{\frac {\partial L}{\partial {{\dot {q}}_{i}}}}d{{\dot {q}}_{i}}\right)+{\frac {\partial L}{\partial t}}dt$

Now the generalized momenta were defined as $p_{i}={\frac {\partial L}{\partial {{\dot {q}}_{i}}}}$ and Lagrange's equations tell us that ${\frac {d}{dt}}{\frac {\partial L}{\partial {{\dot {q}}_{i}}}}-{\frac {\partial L}{\partial q_{i}}}=F_{i}$ where $F_{i}$ is the generalized force. We can rearrange this to get ${\frac {\partial L}{\partial q_{i}}}={\dot {p}}_{i}-F_{i}$ and substitute the result into the variation of the Lagrangian

$dL=\sum _{i}\left[\left({\dot {p}}_{i}-F_{i}\right)dq_{i}+p_{i}d{{\dot {q}}_{i}}\right)+{\frac {\partial L}{\partial t}}dt$

We can rewrite this as

$dL=\sum _{i}\left[\left({\dot {p}}_{i}-F_{i}\right)dq_{i}+d\left(p_{i}{{\dot {q}}_{i}}\right)-{{\dot {q}}_{i}}dp_{i}\right)+{\frac {\partial L}{\partial t}}dt$

and rearrange again to get

$d\left(\sum _{i}p_{i}{{\dot {q}}_{i}}-L\right)=\sum _{i}\left[\left(F_{i}-{\dot {p}}_{i}\right)dq_{i}+{{\dot {q}}_{i}}dp_{i}\right)-{\frac {\partial L}{\partial t}}dt$

The term on the left-hand side is just the Hamiltonian that we have defined before, so we find that

$dH=\sum _{i}\left[\left(F_{i}-{\dot {p}}_{i}\right)dq_{i}+{{\dot {q}}_{i}}dp_{i}\right)-{\frac {\partial L}{\partial t}}dt=\sum _{i}\left[{\frac {\partial H}{\partial q_{i}}}dq_{i}+{\frac {\partial H}{\partial p_{i}}}dp_{i}\right]+{\frac {\partial H}{\partial t}}dt$

where the second equality holds because of the defintion of the partial derivatives.

Hamilton's Equations

The result encourages us to think of the Hamiltonian not just a function of coordinates, velocities and time, but alternatively as a function of coordinates, momenta and time. We have

${\frac {\partial H}{\partial p_{i}}}={\dot {q}}_{i},{\frac {\partial H}{\partial q_{i}}}=F_{i}-{\dot {p}}_{i},{\frac {\partial H}{\partial t}}=-{\frac {\partial L}{\partial t}}$

The first two relativions give 2n first-order differential equations called Hamilton's canonical equations of motion.

Using Hamilton's Equations

1) First write out L = T - V. Express T and V as though you we re going to use Lagrange's equation.

2) Calculate the momenta by differentiating the Lagrangian.

3) Express the velocities in terms of the momenta by inverting the expressions in step (2).

4) Calculate the Hamiltonian using the usual definition $H=\sum _{i}p_{i}{{\dot {q}}_{i}}-L$ Substitute for the velocities using the results in step (3).

5) Apply Hamilton's equations.

Hamilton's Equations of a Pendulum

A pendulum of mass m is suspended by a string of length l. Let's use Hamilton's equations to find the equations of motion.

We have $L={\frac {1}{2}}ml^{2}{\dot {\theta }}^{2}+mgl\cos \theta$ so $p_{\theta }=ml^{2}{\dot {\theta }}$ and ${\dot {\theta }}=p_{\theta }/(ml^{2}).$

Now let's calculate the Hamiltonian

$H=p_{\theta }{\dot {\theta }}-L{\frac {1}{2}}ml^{2}{\dot {\theta }}^{2}-mgl\cos \theta ={\frac {p_{\theta }^{2}}{2ml^{2}}}+mgl\cos \theta$

and apply Hamilton's equations. There are no external forces so we have

${\dot {p}}_{\theta }=-{\frac {\partial H}{\partial \theta }}=-mgl\sin \theta ,{\dot {\theta }}={\frac {\partial H}{\partial p_{\theta }}}={\frac {p_{\theta }}{ml^{2}}}$

Notice how the second relation just gave us back what we already knew from the inversion in the previous step. I could combine these expressions by take the time derivative of the second expression and substituting in the first one. I get a single second order differential equation

${\ddot {\theta }}=-{\frac {g}{l}}\sin \theta$

that is the same as the one I would have gotten had I used the Lagrange technique. Hey, I never promised that Hamilton's equations would make solving specific problems easier.