Projectile Motion the Hard Way

The trajectory of a body moving in a constant gravitational field is one of the simplest and yet most practical of problems in classical physics. The Lagrangian for the system is

$L = \frac{m}{2} ({\overset{\cdot}{x}}^{2} + {\overset{\cdot}{y}}^{2} + {\overset{\cdot}{z}}^{2}) - m g z$

with the corresponding Hamiltonian

$H = \frac{1}{2 m} (p_{x}^{2} + p_{y}^{2} + p_{z}^{2}) + m g z$

The equations of motion for the canonical coordinates are

$\begin{array}{l} \overset{\cdot}{x} = \frac{\partial H}{\partial p_{x}} = \frac{p_{x}}{m} & {\overset{\cdot}{p}}_{x} = - \frac{\partial H}{\partial x} = 0 \\ \overset{\cdot}{y} = \frac{\partial H}{\partial p_{y}} = \frac{p_{y}}{m} & {\overset{\cdot}{p}}_{y} = - \frac{\partial H}{\partial y} = 0 \\ \overset{\cdot}{z} = \frac{\partial H}{\partial p_{z}} = \frac{p_{z}}{m} & {\overset{\cdot}{p}}_{z} = - \frac{\partial H}{\partial z} = - m g \end{array}$

Since both p_x and p_y are constant, their conjugate coordinates are linear in the temporal coordinate

$x = \frac{p_{x}}{m} t + x_{0} y = \frac{p_{y}}{m} t + y_{0}$

while the vertical coordinate affected by gravity is quadratic

$\overset{\cdot \cdot}{z} = \frac{{\overset{\cdot}{p}}_{z}}{m} = - g \to z = z_{0} + v_{0} t - \frac{1}{2} g t^{2}$

and its conjugate momentum is linear:

$p_{z} = m \overset{\cdot}{z} = m v_{0} - m g t$

The simplicity of this system makes it ideal for seeing the mechanism of Hamilton-Jacobi theory in action, without being as trivial as the equally tractable simple harmonic oscillator. While the Hamilton-Jacobi formalism is massive overkill for projectile motion, it is a convenient opportunity to understand the essence of the formalism.

Hamilton-Jacobi theory determines the classical action from the first-order nonlinear differential equation

$H [x, y, z; \frac{\partial S}{\partial x}, \frac{\partial S}{\partial y}, \frac{\partial S}{\partial z}] + \frac{\partial S}{\partial t} = 0$

where linear momenta are replaced by partial derivatives of the action with respect to their conjugate coordinates. The equation for the case of projectiles is

$\frac{1}{2 m} [(\frac{\partial S}{\partial x})^{2} + (\frac{\partial S}{\partial y})^{2} + (\frac{\partial S}{\partial z})^{2}] + m g z + \frac{\partial S}{\partial t} = 0$

Since the potential is a function of the vertical coordinate only, the equation is separable if the action is a linear sum of the form

$S = p_{x} x + p_{y} y - E t + f (z)$

where the coefficient of the explicit temporal variable is chosen in accordance with the defining equation. The Hamilton-Jacobi equation for projectiles becomes

$(\frac{\partial f}{\partial z})^{2} = 2 m E - p_{x}^{2} - p_{y}^{2} - 2 m^{2} g z$

which is easily integrated. The action in terms of canonical coordinates is thus

$S = p_{x} x + p_{y} y - E t - \frac{1}{3 m^{2} g} [2 m E - p_{x}^{2} - p_{y}^{2} - 2 m^{2} g z]^{3 / 2}$

The action is equivalently given by an indefinite integral of the Lagrangian:

$S = \int L d t + C$

To verify this, the Lagrangian must be expressed in terms of constant quantities and the temporal variable. The most efficient way to do this is to replace the potential energy part of the Lagrangian using the expression for total energy,

$\begin{array}{l} L = m ({\overset{\cdot}{x}}^{2} + {\overset{\cdot}{y}}^{2} + {\overset{\cdot}{z}}^{2}) - E \\ = \frac{1}{m} (p_{x}^{2} + p_{y}^{2}) - E + m (v_{0} - g t)^{2} \end{array}$

so that the action integral is

$\begin{array}{l} S = \frac{1}{m} (p_{x}^{2} + p_{y}^{2}) t - E t - \frac{m}{3 g} (v_{0} - g t)^{3} \\ = p_{x} x + p_{y} y - E t - \frac{1}{3 m^{2} g} p_{z}^{3} \\ - p_{x} x_{0} - p_{y} y_{0} \\ S = p_{x} x + p_{y} y - E t - \frac{1}{3 m^{2} g} [2 m E - p_{x}^{2} - p_{y}^{2} - 2 m^{2} g z]^{3 / 2} \\ - p_{x} x_{0} - p_{y} y_{0} \end{array}$

which matches the previous result up to a constant of integration.

The differential equation defining the classical action, known as the Hamilton-Jacobi equation, is derived by treating the action as a generating function that induces a transformation of the Hamiltonian to generalized coordinates in which it is identically zero. If the Hamiltonian is identically zero,

${\overset{\cdot}{Q}}_{i} = \frac{\partial H}{\partial P_{i}} \equiv 0 {\overset{\cdot}{P}}_{i} = - \frac{\partial H}{\partial Q_{i}} \equiv 0$

then the new generalized coordinates are all constant. The new momenta are traditionally designated with alphas and for projectile motion can be taken as

$\begin{array}{l} P_{1} = α_{1} = E \\ P_{2} = α_{2} = p_{x} \\ P_{3} = α_{3} = p_{y} \end{array}$

The new conjugate coordinates, traditionally designated with betas, are given by partial derivatives of the action:

$Q_{i} = β_{i} = \frac{\partial S}{\partial α_{i}}$

Explicit temporal behavior is determined in the Hamiltonian-Jacobi formalism by evaluating these derivatives and inverting them for the original variables. For projectile motion,

$\begin{array}{l} β_{1} = \frac{\partial S}{\partial E} = - t - \frac{1}{m g} [2 m E - p_{x}^{2} - p_{y}^{2} - 2 m^{2} g z]^{1 / 2} \\ β_{2} = \frac{\partial S}{\partial p_{x}} = x + \frac{p_{x}}{m^{2} g} [2 m E - p_{x}^{2} - p_{y}^{2} - 2 m^{2} g z]^{1 / 2} \\ β_{3} = \frac{\partial S}{\partial p_{y}} = y + \frac{p_{y}}{m^{2} g} [2 m E - p_{x}^{2} - p_{y}^{2} - 2 m^{2} g z]^{1 / 2} \end{array}$

which do not yet look much like the simple equations given at the outset. The first equation gives a replacement for the square root,

$[2 m E - p_{x}^{2} - p_{y}^{2} - 2 m^{2} g z]^{1 / 2} = - m g (t + β_{1})$

so that the second and third equations become

$\begin{array}{l} β_{2} = x - \frac{p_{x}}{m} (t + β_{1}) \\ β_{3} = y - \frac{p_{y}}{m} (t + β_{1}) \end{array} \to \begin{array}{l} x = \frac{p_{x}}{m} t + β_{2} + \frac{p_{x}}{m} β_{1} \\ y = \frac{p_{y}}{m} t + β_{3} + \frac{p_{y}}{m} β_{1} \end{array}$

Square the remaining equation and rearrange:

$z = \frac{1}{2 m^{2} g} (2 m E - p_{x}^{2} - p_{y}^{2}) - \frac{g}{2} (t + β_{1})^{2}$

The quantity in the first set of parentheses is the sum of three constants. From the definition of total energy it is equal to

$2 m E - p_{x}^{2} - p_{y}^{2} = p_{z}^{2} + 2 m^{2} g z = (p_{z}^{2} + 2 m^{2} g z)_{t = 0}$

and as constant can be evaluated at any convenient point. Put this initial evaluation into the remaining equation:

$\begin{array}{l} z = z_{0} + \frac{v_{0}^{2}}{2 g} - \frac{g}{2} (t + β_{1})^{2} \\ = z_{0} + \frac{v_{0}^{2} - g^{2} β_{1}^{2}}{2 g} - g β_{1} t - \frac{1}{2} g t^{2} \end{array}$

Thus with the identifications

$\begin{array}{l} β_{1} = - \frac{v_{0}}{g} \\ β_{2} = x_{0} + \frac{p_{x} v_{0}}{m g} \\ β_{3} = y_{0} + \frac{p_{y} v_{0}}{m g} \end{array}$

one has the explicit temporal behavior

$\begin{array}{l} x = \frac{p_{x}}{m} t + x_{0} \\ y = \frac{p_{y}}{m} t + y_{0} \\ z = z_{0} + v_{0} t - \frac{1}{2} g t^{2} \end{array}$

which is exactly that given at the outset, after a detour through Hamilton-Jacobi theory. The identifications of the constant betas are not arbitrary choices, and can be verified by putting these equations into the partial derivatives of the action.

Happy Pi Day!

Uploaded 2017.03.14 — Updated 2017.04.13 analyticphysics.com