To find the trajectory of a ball thrown with an initial speed $v_0$ at an angle $\theta$ (projectile motion), we will set up our coordinate system where $x$ is the horizontal distance and $y$ is the vertical height.
we consider a particle of mass $m$ in a uniform gravitational field $\mathbf{g} = -g\hat{\mathbf{j}}$. The motion is restricted to the $xy$-plane.
I. System Specification
Initial Conditions at $t=0$:
- Position: $\mathbf{r}(0) = (0, 0)$
- Velocity: $\mathbf{v}(0) = (v_0 \cos\theta, v_0 \sin\theta)$
- Momentum: $\mathbf{p}(0) = (mv_0 \cos\theta, mv_0 \sin\theta)$
II. The Newtonian Formulation
In the Newtonian framework, the dynamics are governed by Newton’s Second Law, relating the net force to the rate of change of momentum.
1. Equations of Motion:
The only force acting on the particle is gravity, $\mathbf{F} = -mg\hat{\mathbf{j}}$. Thus:
$$m\frac{d^2x}{dt^2} = 0 \quad (1)$$
$$m\frac{d^2y}{dt^2} = -mg \quad (2)$$
2. Integration of the Horizontal Component:
Integrating (1) with respect to time:
$$\frac{dx}{dt} = v_{x,0} = v_0 \cos\theta$$
Integrating a second time with the boundary condition $x(0)=0$:
$$x(t) = (v_0 \cos\theta)t \quad (3)$$
3. Integration of the Vertical Component:
Integrating (2) with respect to time:
$$\frac{dy}{dt} = -gt + v_{y,0} = -gt + v_0 \sin\theta$$
Integrating again with the boundary condition $y(0)=0$:
$$y(t) = -\frac{1}{2}gt^2 + (v_0 \sin\theta)t \quad (4)$$
III. The Hamiltonian Formulation
The Hamiltonian approach describes the system in Phase Space using generalized coordinates $q_i$ and their conjugate momenta $p_i$.
1. Definition of the Hamiltonian ($H$):
For a conservative system, the Hamiltonian is the sum of the kinetic energy $T$ and potential energy $V$. In Cartesian coordinates:
$$T = \frac{1}{2m}(p_x^2 + p_y^2), \quad V = mgy$$
$$H(x, y, p_x, p_y) = \frac{p_x^2 + p_y^2}{2m} + mgy$$
2. Hamilton’s Canonical Equations:
The evolution of the system is given by the first-order differential equations:
$$\dot{q}_i = \frac{\partial H}{\partial p_i}, \quad \dot{p}_i = -\frac{\partial H}{\partial q_i}$$
Applying these to our coordinates $(x, y)$:
- For $x$:
$$\dot{x} = \frac{\partial H}{\partial p_x} = \frac{p_x}{m} \quad (5)$$
$$\dot{p}_x = -\frac{\partial H}{\partial x} = 0 \quad (6)$$ - For $y$:
$$\dot{y} = \frac{\partial H}{\partial p_y} = \frac{p_y}{m} \quad (7)$$
$$\dot{p}_y = -\frac{\partial H}{\partial y} = -mg \quad (8)$$
3. Solving the Canonical Equations:
From (6), we see $p_x$ is a constant of motion:
$$p_x(t) = p_{x,0} = mv_0 \cos\theta$$
Substituting this into (5):
$$\dot{x} = \frac{mv_0 \cos\theta}{m} = v_0 \cos\theta \implies x(t) = (v_0 \cos\theta)t$$
From (8), we integrate to find $p_y(t)$:
$$p_y(t) = \int -mg \, dt = -mgt + C$$
Using $p_y(0) = mv_0 \sin\theta$, we get $p_y(t) = m(v_0 \sin\theta – gt)$. Substituting this into (7):
$$\dot{y} = \frac{m(v_0 \sin\theta – gt)}{m} = v_0 \sin\theta – gt \implies y(t) = (v_0 \sin\theta)t – \frac{1}{2}gt^2$$
IV. The Trajectory Equation
To find the spatial trajectory $y(x)$, we eliminate the parameter $t$. From the expression for $x(t)$:
$$t = \frac{x}{v_0 \cos\theta}$$
Substituting $t$ into the expression for $y(t)$:
$$y = (v_0 \sin\theta)\left( \frac{x}{v_0 \cos\theta} \right) – \frac{1}{2}g\left( \frac{x}{v_0 \cos\theta} \right)^2$$
Final Result:
The trajectory is a parabola defined by the equation:
$$y(x) = x \tan\theta – \frac{gx^2}{2v_0^2 \cos^2\theta}$$
V. Conclusion
While the Newtonian method arrives at the result by considering the vector sum of forces and solving second-order equations, the Hamiltonian method reduces the problem to a set of coupled first-order equations derived from the scalar energy function of the system. The Hamiltonian approach is the standard formalism for moving toward more complex classical systems and is the direct precursor to the Schrödinger formulation of Quantum Mechanics.
Along the trajectory of the ball, the Hamiltonian $H$ has a very specific and fundamental property: it is a constant of motion.
In classical mechanics, if the Hamiltonian does not explicitly depend on time ($\frac{\partial H}{\partial t} = 0$), the total energy of the system is conserved. For our projectile, because there is no air resistance (non-conservative forces), the value of the Hamiltonian remains fixed at its initial value throughout the entire flight.
Here is the formal proof and analysis of the Hamiltonian along the trajectory.
1. Calculation of $H$ as a function of time
Let us take the expressions for momentum and position derived previously and substitute them back into the Hamiltonian:
$$H = \frac{p_x^2 + p_y^2}{2m} + mgy$$
Substitute the components:
- $p_x(t) = mv_0 \cos\theta$
- $p_y(t) = m(v_0 \sin\theta – gt)$
- $y(t) = (v_0 \sin\theta)t – \frac{1}{2}gt^2$
Step-by-step substitution:
- Kinetic Term ($T$):
$$\frac{p_x^2 + p_y^2}{2m} = \frac{(mv_0\cos\theta)^2 + (m(v_0\sin\theta – gt))^2}{2m}$$
$$\dots = \frac{m^2v_0^2\cos^2\theta + m^2(v_0^2\sin^2\theta – 2v_0gt\sin\theta + g^2t^2)}{2m}$$
$$\dots = \frac{1}{2}mv_0^2(\cos^2\theta + \sin^2\theta) – mv_0gt\sin\theta + \frac{1}{2}mg^2t^2$$
$$T(t) = \frac{1}{2}mv_0^2 – mv_0gt\sin\theta + \frac{1}{2}mg^2t^2$$ - Potential Term ($V$):
$$V(t) = mgy(t) = mg(v_0t\sin\theta – \frac{1}{2}gt^2)$$
$$V(t) = mv_0gt\sin\theta – \frac{1}{2}mg^2t^2$$ - Total Hamiltonian ($H = T + V$):
$$H = \left[ \frac{1}{2}mv_0^2 – mv_0gt\sin\theta + \frac{1}{2}mg^2t^2 \right] + \left[ mv_0gt\sin\theta – \frac{1}{2}mg^2t^2 \right]$$
Result:
$$H = \frac{1}{2}mv_0^2$$
All time-dependent terms cancel out perfectly. The Hamiltonian is equal to the initial kinetic energy of the ball.
2. Formal Proof via Time Evolution
In a more advanced textbook treatment, we prove that $H$ is constant by taking the total derivative of $H$ with respect to time:
$$\frac{dH}{dt} = \frac{\partial H}{\partial t} + \sum_{i} \left( \frac{\partial H}{\partial q_i}\dot{q}_i + \frac{\partial H}{\partial p_i}\dot{p}_i \right)$$
Using Hamilton’s equations ($\dot{q}i = \frac{\partial H}{\partial p_i}$ and $\dot{p}_i = -\frac{\partial H}{\partial q_i}$): $$\frac{dH}{dt} = \frac{\partial H}{\partial t} + \sum{i} \left( \frac{\partial H}{\partial q_i}\frac{\partial H}{\partial p_i} – \frac{\partial H}{\partial p_i}\frac{\partial H}{\partial q_i} \right)$$
The terms in the parentheses cancel to zero. Therefore:
$$\frac{dH}{dt} = \frac{\partial H}{\partial t}$$
Since our Hamiltonian $H = \frac{p^2}{2m} + mgy$ has no explicit “t” variable (gravity is constant and doesn’t change with time), $\frac{\partial H}{\partial t} = 0$. Thus, $dH/dt = 0$, proving $H$ is a constant of motion.
3. Physical Interpretation along the Path
As the ball moves along its parabolic trajectory:
- At the launch point ($y=0$): $H$ is entirely kinetic energy.
- During the ascent: Kinetic energy ($T$) is converted into potential energy ($V$). The particle slows down, but $H$ remains the same.
- At the apex (peak): The vertical momentum $p_y = 0$. Here, $V$ is at its maximum and $T$ is at its minimum (only horizontal motion remains), but the sum is still $\frac{1}{2}mv_0^2$.
- During descent: $V$ is converted back into $T$.
Summary
While the coordinates $(x, y)$ and momenta $(p_x, p_y)$ are changing at every microsecond of the flight, the Hamiltonian is the “invariant” of the system—the value that defines the surface in phase space upon which the trajectory must stay.