next up previous contents
Next: Thermodynamics Up: Classical Mechanics Previous: Lagrange's Equations   Contents


Hamilton's Equations of Motion

Hamilton's equation of motion depend on forming the Hamilton, which can be formed from the Lagrangian by

\begin{displaymath}
H = \sum_j p_j \dot{q}_j - L.
\end{displaymath} (4.9)


$p_j$ is called the conjugate momentum to the coordinate $q_j$. The conjugate momentums can be calculated from the Lagrangian by
$\displaystyle p_j$ $\textstyle =$ $\displaystyle \frac{\partial L}{\partial \dot{q}_j}$ (4.10)
$\displaystyle \dot{p}_j$ $\textstyle =$ $\displaystyle \frac{\partial L}{\partial q_j}$ (4.11)

Hamilton's equations of motion are then

$\displaystyle \dot{q}_k$ $\textstyle =$ $\displaystyle \frac{\partial H}{\partial p_k}$ (4.12)
$\displaystyle -\dot{p}_k$ $\textstyle =$ $\displaystyle \frac{\partial H}{\partial q_k}$ (4.13)

Examining a particle with

\begin{displaymath}
T = \frac{1}{2} m \dot{x}^2_i     U = U(x_i).
\end{displaymath} (4.14)


gives
\begin{displaymath}
p_i = \frac{\partial L}{\partial \dot{x}_i} = m \dot{x}_i.
\end{displaymath} (4.15)


The Hamiltonian becomes
\begin{displaymath}
H = \sum_j p_j \dot{q}_j - L
= \sum_j m \dot{x}_j \dot{x}_...
...ot{x}^2_j + U
= \sum_j \frac{1}{2} m\dot{x}^2_j + U
= T + U.
\end{displaymath} (4.16)

It can be shown that if $H$ has no explicit time dependence (e.g. $\frac{\partial H}{\partial t} = 0$) then energy is conserved.


next up previous contents
Next: Thermodynamics Up: Classical Mechanics Previous: Lagrange's Equations   Contents
Ben Breech 2003-01-14