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Next: Exercises Up: Hamiltonian Dynamics Previous: Constrained Lagrangian Dynamics

Hamilton's Equations

Consider a dynamical system with ${\cal F}$ degrees of freedom which is described by the generalized coordinates $q_i$, for $i=1,{\cal F}$. Suppose that neither the kinetic energy, $K$, nor the potential energy, $U$, depend explicitly on the time, $t$. Now, in conventional dynamical systems, the potential energy is generally independent of the $\dot{q}_i$, whereas the kinetic energy takes the form of a homogeneous quadratic function of the $\dot{q}_i$. In other words,
K = \sum_{i,j = 1,{\cal F}} m_{ij}\,\dot{q}_i\,\dot{q}_j,
\end{displaymath} (744)

where the $m_{ij}$ depend on the $q_i$, but not on the $\dot{q}_i$. It is easily demonstrated from the above equation that
\sum_{i=1,{\cal F}} \dot{q}_i\,\frac{\partial K}{\partial \dot{q}_i} = 2\,K.
\end{displaymath} (745)

Recall, from Section 9.8, that generalized momentum conjugate to the $i$th generalized coordinate is defined

p_i = \frac{\partial L}{\partial \dot{q}_i} = \frac{\partial K}{\partial \dot{q}_i},
\end{displaymath} (746)

where $L=K-U$ is the Lagrangian of the system, and we have made use of the fact that $U$ is independent of the $\dot{q}_i$. Consider the function
H = \sum_{i=1,{\cal F}} \dot{q}_i\,p_i - L = \sum_{i=1,{\cal F}} \dot{q}_i\,p_i -K + U.
\end{displaymath} (747)

If all of the conditions discussed above are satisfied then Equations (745) and (746) yield
H = K+ U.
\end{displaymath} (748)

In other words, the function $H$ is equal to the total energy of the system.

Consider the variation of the function $H$. We have

\delta H = \sum_{i=1,{\cal F}} \left(\delta\dot{q}_i\,p_i + ...{q}_i - \frac{\partial L}{\partial q_i}\,\delta q_i\right).
\end{displaymath} (749)

The first and third terms in the bracket cancel, because $p_i=
\partial L/\partial \dot{q}_i$. Furthermore, since Lagrange's equation can be written $\dot{p}_i = \partial L/\partial q_i$ (see Section 9.8), we obtain
\delta H = \sum_{i=1,{\cal F}} \left(\dot{q}_i\,\delta p_i - \dot{p}_i\,\delta q_i\right).
\end{displaymath} (750)

Suppose, now, that we can express the total energy of the system, $H$, solely as a function of the $q_i$ and the $p_i$, with no explicit dependence on the $\dot{q}_i$. In other words, suppose that we can write $H=H(q_i,p_i)$. When the energy is written in this fashion it is generally termed the Hamiltonian of the system. The variation of the Hamiltonian function takes the form
\delta H =\sum_{i=1,{\cal F}} \left(\frac{\partial H}{\parti...
...ta p_i +
\frac{\partial H}{\partial q_i}\,\delta{q}_i\right).
\end{displaymath} (751)

A comparison of the previous two equations yields
$\displaystyle \dot{q}_i$ $\textstyle =$ $\displaystyle \frac{\partial H}{\partial p_i},$ (752)
$\displaystyle \dot{p}_i$ $\textstyle =$ $\displaystyle -\frac{\partial H}{\partial q_i},$ (753)

for $i=1,{\cal F}$. These $2{\cal F}$ first-order differential equations are known as Hamilton's equations. Hamilton's equations are often a useful alternative to Lagrange's equations, which take the form of ${\cal F}$ second-order differential equations.

Consider a one-dimensional harmonic oscillator. The kinetic and potential energies of the system are written $K = (1/2)\,m\,\dot{x}^{\,2}$ and $U=(1/2)\,k\,x^2$, where $x$ is the displacement, $m$ the mass, and $k>0$. The generalized momentum conjugate to $x$ is

p = \frac{\partial K}{\partial\dot{x}} = m\,\dot{x}.
\end{displaymath} (754)

Hence, we can write
K = \frac{1}{2}\,\frac{p^{\,2}}{m}.
\end{displaymath} (755)

So, the Hamiltonian of the system takes the form
H = K + U = \frac{1}{2}\,\frac{p^{\,2}}{m} + \frac{1}{2}\,k\,x^2.
\end{displaymath} (756)

Thus, Hamilton's equations, (752) and (753), yield
$\displaystyle \dot{x}$ $\textstyle =$ $\displaystyle \frac{\partial H}{\partial p} = \frac{p}{m},$ (757)
$\displaystyle \dot{p}$ $\textstyle =$ $\displaystyle - \frac{\partial H}{\partial x} = - k\,x.$ (758)

Of course, the first equation is just a restatement of Equation (754), whereas the second is Newton's second law of motion for the system.

Consider a particle of mass $m$ moving in the central potential $U(r)$. In this case,

K = \frac{1}{2}\,m\,(\dot{r}^{\,2} + r^2\,\dot{\theta}^{\,2}),
\end{displaymath} (759)

where $r,\theta$ are polar coordinates. The generalized momenta conjugate to $r$ and $\theta $ are
$\displaystyle p_r$ $\textstyle =$ $\displaystyle \frac{\partial K}{\partial \dot{r}} = m\,\dot{r},$ (760)
$\displaystyle p_\theta$ $\textstyle =$ $\displaystyle \frac{\partial K}{\partial \dot{\theta}} = m\,r^2\,\dot{\theta},$ (761)

respectively. Hence, we can write
K = \frac{1}{2\,m}\left(p_r^{\,2} + \frac{p_\theta^{\,2}}{r^2}\right).
\end{displaymath} (762)

Thus, the Hamiltonian of the system takes the form
H = \frac{1}{2\,m}\left(p_r^{\,2} + \frac{p_\theta^{\,2}}{r^2}\right)
+ U(r).
\end{displaymath} (763)

In this case, Hamilton's equations yield
$\displaystyle \dot{r}$ $\textstyle =$ $\displaystyle \frac{\partial H}{\partial p_r} = \frac{p_r}{m},$ (764)
$\displaystyle \dot{\theta}$ $\textstyle =$ $\displaystyle \frac{\partial H}{\partial p_\theta} = \frac{p_\theta}{m\,r^2},$ (765)

which are just restatements of Equations (760) and (761), respectively, as well as
$\displaystyle \dot{p}_r$ $\textstyle =$ $\displaystyle -\frac{\partial H}{\partial r} = \frac{p_\theta^{\,2}}{m\,r^3}-\frac{\partial U}{\partial r},$ (766)
$\displaystyle \dot{p}_\theta$ $\textstyle =$ $\displaystyle -\frac{\partial H}{\partial \theta} = 0.$ (767)

The last equation implies that
\frac{p_\theta}{m} =r^2\,\dot{\theta} =h,
\end{displaymath} (768)

where $h$ is a constant. This can be combined with Equation (766) to give
\frac{\dot{p}_r}{m} = \ddot{r} = \frac{h^2}{r^3} - \frac{\partial V}{\partial r},
\end{displaymath} (769)

where $V = U/m$. Of course, Equations (768) and (769) are the conventional equations of motion for a particle moving in a central potential--see Chapter 5.

next up previous
Next: Exercises Up: Hamiltonian Dynamics Previous: Constrained Lagrangian Dynamics
Richard Fitzpatrick 2011-03-31