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Question

# Write the Hamiltonian function and find Hamilton's canonical equations for the three-dimensional motion of a projectile in a uniform gravitational field with no air resistance. Show that these equations lead to the same equations of motion as found.

Solution

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For three dimensional motion of a projectile in a uniform gravitational field with no air resistance we have:

\begin{align} T &= \frac{m}{2} (\dot x^2 + \dot y^2 + \dot z^2) \\ V &= m \, g \, z \\ L &= T - V = \frac{m}{2} (\dot x^2 + \dot y^2 + \dot z^2) - m \, g \, z \end{align}

Now the generalized momentums $p_i$ are found using:

$\begin{equation} p_i = \frac{\partial L}{\partial \dot q_i} \end{equation}$

We write:

\begin{align} p_x &= \frac{\partial L}{\partial \dot x} = m\dot x \quad => \dot x = \frac{p_x}{m} \\ p_y &= \frac{\partial L}{\partial \dot y} = m \dot y \quad => \dot y = \frac{p_y}{m} \\ p_z &= \frac{\partial L}{\partial \dot z} = m \dot z \quad => \dot z =\frac{p_z}{m} \end{align}

The Hamiltonian is most generally found using:

$\begin{equation} H(p_i, q_i) = \sum\limits_{i}p_i \dot q_i (p_i, q_i) - L(p_i, q_i) \end{equation}$

but since our kinetic energy is a homogeneous quadratic function of generalized speeds (the $\dot q$'s), and our potential energy is a function of generalized coordinates (the $q$'s) only, we know the Hamiltonian is equal to the total energy:

\begin{align} H(p_i,q_i) &= T(p_i,q_i) + V(p_i, q_i) \\ H(p_i,q_i) &= \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{p_z^2}{2m} + m g z \end{align}

where we have used (5), (6), (7)

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