--- jupytext: text_representation: extension: .md format_name: myst format_version: 0.13 kernelspec: display_name: Python 3 language: python name: python3 --- # Lecture 2 — Degrees of Freedom, Newton's Equations, and Phase Space *Source: [NPTEL Classical Physics, Mod-01 Lec-02](https://www.youtube.com/watch?v=8X1x9RLaaxc&list=PL5E4E56893588CBA8&index=2), Prof. V. Balakrishnan.* ## Counting degrees of freedom A single particle moving in three-dimensional space needs three independent coordinates to fix its position, so it has 3 degrees of freedom. $N$ unconstrained particles therefore have $3N$ degrees of freedom in total. These coordinates need not be Cartesian — they can be angular, or a mix — so we call them **generalized coordinates** and write them $q_1, \dots, q_{3N}$, reserving the symbol $q$ precisely because it carries no commitment to a particular coordinate system. A **constraint** removes degrees of freedom only if it is an *equality* relating coordinates (a **holonomic** constraint) — for instance fixing the distance between two particles, $r_{12} = \text{const}$. An *inequality*, such as confining a particle to one quadrant of a plane, does not reduce the count at all: it is a **non-holonomic** constraint, and the particle still has as many degrees of freedom as before, just a restricted range for them. This distinction matters because it is tempting, but wrong, to guess the naive formula for a rigid body by simply subtracting one constraint per pairwise distance held fixed. For $N$ points with every pairwise distance $r_{ij}$ fixed, there are $3N$ coordinates and $\binom{N}{2} = N(N-1)/2$ such constraints, so the naive count is $$ 3N - \frac{N(N-1)}{2}, $$ which turns *negative* once $N \gtrsim 6$ — an impossibility. The resolution is that most of these constraints are redundant: once a handful of them hold the body rigid, the rest follow automatically. Counting the true number of independent degrees of freedom two different, independent ways settles it: - **Way 1.** Three coordinates fix the center of mass $\vec R$. The orientation of a body-fixed frame relative to a space-fixed frame then takes exactly three Euler angles. Total: $3 + 3 = 6$. - **Way 2.** Three coordinates fix the center of mass. The rotation axis is a unit vector on a sphere, needing two angles (latitude/longitude), and the rotation angle about that axis is a third. Total: $3 + 2 + 1 = 6$. Both routes agree: **a rigid body has exactly 6 independent degrees of freedom, no matter how many particles it is built from** — 3 translational and 3 orientational. This is also why a rigid diatomic molecule has 2 rotational degrees of freedom (there is no moment of inertia about the bond axis) rather than 3. ## Newton's equations need velocities too Drop a piece of chalk from rest and it falls straight down; give it a little sideways push and it traces a parabolic (really, elliptical) arc; push harder and it goes into orbit; harder still and it escapes on a hyperbolic trajectory. The initial *position* and the *force law* alone did not fix which of these happens — the initial *velocity* did. Newton's equation, $$ m\,\ddot q = F(q, \dot q, t), $$ is second order in time, so the state that must be specified to get a unique future is the pair $(q, \dot q)$, not $q$ alone. A system with an explicitly time-dependent force is called **non-autonomous**; without that explicit $t$-dependence, it is **autonomous** — and it is autonomous systems that occupy most of this course. Because $q$ and $\dot q$ are independent initial data, the natural arena for dynamics is not configuration space $\{q\}$ but **phase space** $\{q, \dot q\}$: the space of positions *and* velocities together. ## Phase trajectories cannot cross themselves This single geometric fact carries a lot of weight. If a phase trajectory of an autonomous system crossed itself (or crossed another trajectory), the crossing point would be a single initial condition from which the future would have two different continuations — which contradicts the determinism of Newton's equations. Hence: > For an autonomous system, no phase trajectory can intersect itself, and no two > distinct phase trajectories can intersect each other. The one loophole is a > trajectory that closes on itself — which is exactly **periodic motion**. ## Worked example: the harmonic oscillator For a conservative force $F(q) = -dV/dq$, energy conservation collapses the two first-order equations $\dot q = v,\ \dot v = -V'(q)/m$ down to a single algebraic curve in phase space: $$ E = \tfrac{1}{2}mv^2 + V(q). $$ For the harmonic oscillator, $V(q) = \tfrac12 m\omega^2 q^2$, so $$ v^2 + \omega^2 q^2 = \frac{2E}{m}, $$ an **ellipse** for every $E > 0$, traversed clockwise (start at maximum $q$ with $v=0$; releasing from rest sends it back toward the origin, i.e. $v$ goes negative first). Larger $E$ gives a larger concentric ellipse, so — in Balakrishnan's phrase — *the phase plane is laminated by these ellipses*, with the one exceptional trajectory being the single point at the origin ($E=0$), the equilibrium. The figure below is exactly this picture, animated: the background shows the lamination, and the marker traces one representative trajectory in real time. ```{code-cell} ipython3 :tags: [hide-input] import numpy as np import plotly.graph_objects as go from IPython.display import HTML def show_fig(fig): """Embed a Plotly figure as self-contained HTML+JS (interactive with no running kernel needed). include_mathjax=False is essential: Plotly's default HTML bundles a legacy MathJax v2 loader that collides with Sphinx's MathJax v3 and silently breaks all math typesetting on the page. """ return HTML(fig.to_html(full_html=False, include_mathjax=False, config={"responsive": True})) class PhaseSpaceExplorer: """ Exact phase-space geometry and time evolution for a 1D particle of mass m in the potential V(q) = sign * 1/2 * m * omega^2 * q^2. sign = +1 -> stable harmonic oscillator (bound orbits, ellipses) sign = -1 -> inverted (unstable) oscillator (hyperbolic escape / separatrices) """ def __init__(self, omega=1.0, m=1.0, sign=1): self.omega = omega self.m = m self.sign = sign def velocity_branch(self, q, E): """v(q) on the constant-energy curve E = 1/2 m v^2 + sign * 1/2 m omega^2 q^2.""" v2 = (2.0 / self.m) * E - self.sign * self.omega**2 * q**2 return np.sqrt(np.where(v2 >= 0, v2, np.nan)) def family_curve(self, E, q_range=4.0, n=400): """Both branches (+v and -v) of the constant-energy curve, for a static backdrop.""" q = np.linspace(-q_range, q_range, n) return q, self.velocity_branch(q, E) def trajectory(self, E, t, phase=0.0): """Exact q(t), v(t) for one initial condition of energy E.""" omega, m, sign = self.omega, self.m, self.sign xi = omega * t + phase if sign > 0: # bound orbit: q = A cos(xi) A_q = np.sqrt(2 * E / (m * omega**2)) A_v = np.sqrt(2 * E / m) return A_q * np.cos(xi), -A_v * np.sin(xi) if E >= 0: # unbound: crosses over the top of the barrier A_v = np.sqrt(2 * E / m) return (A_v / omega) * np.sinh(xi), A_v * np.cosh(xi) # E < 0: turns around before reaching the barrier top A_v = np.sqrt(-2 * E / m) return (A_v / omega) * np.cosh(xi), A_v * np.sinh(xi) def animated_phase_portrait(explorer, energies, highlight_E, t_values, title, q_range=4.0): fig = go.Figure() # Static background: the family of constant-energy curves ("lamination") for E in energies: q, v = explorer.family_curve(E, q_range=q_range) fig.add_trace(go.Scatter( x=q, y=v, mode="lines", line=dict(color="rgba(70,130,180,0.35)", width=1.5), name=f"E={E:g}", hoverinfo="skip", showlegend=False, )) fig.add_trace(go.Scatter( x=q, y=-v, mode="lines", line=dict(color="rgba(70,130,180,0.35)", width=1.5), hoverinfo="skip", showlegend=False, )) # Highlighted trajectory (full path) for the animated marker q_full, v_full = explorer.trajectory(highlight_E, t_values) fig.add_trace(go.Scatter( x=q_full, y=v_full, mode="lines", line=dict(color="firebrick", width=2.5), name=f"trajectory, E={highlight_E:g}", )) # Animated marker: current phase-space point fig.add_trace(go.Scatter( x=[q_full[0]], y=[v_full[0]], mode="markers", marker=dict(color="firebrick", size=12, symbol="circle"), name="current state (q, v)", )) marker_trace_index = len(fig.data) - 1 frames = [] for k, t in enumerate(t_values): q_t, v_t = explorer.trajectory(highlight_E, t) frames.append(go.Frame( data=[go.Scatter(x=[q_t], y=[v_t])], traces=[marker_trace_index], name=f"{k}", )) fig.frames = frames fig.update_layout( title=title, xaxis_title="position q", yaxis_title="velocity v = q̇", yaxis=dict(scaleanchor="x", scaleratio=1), width=680, height=520, legend=dict(yanchor="top", y=0.99, xanchor="left", x=0.01), updatemenus=[dict( type="buttons", showactive=False, y=1.12, x=1.0, xanchor="right", buttons=[ dict(label="▶ Play", method="animate", args=[None, dict(frame=dict(duration=40, redraw=True), fromcurrent=True, transition=dict(duration=0))]), dict(label="⏸ Pause", method="animate", args=[[None], dict(frame=dict(duration=0, redraw=False), mode="immediate")]), ], )], sliders=[dict( active=0, x=0.08, len=0.9, currentvalue=dict(prefix="t = ", visible=True), steps=[dict(method="animate", label=f"{t:.2f}", args=[[f"{k}"], dict(mode="immediate", frame=dict(duration=0, redraw=True))]) for k, t in enumerate(t_values)], )], ) return fig sho = PhaseSpaceExplorer(omega=1.0, sign=+1) t_period = np.linspace(0, 2 * np.pi, 60) # one full period, omega = 1 fig_sho = animated_phase_portrait( sho, energies=[0.5, 1.0, 1.5, 2.0, 2.5, 3.0], highlight_E=2.0, t_values=t_period, title="Phase portrait — simple harmonic oscillator (laminated by ellipses)", ) show_fig(fig_sho) ``` Drag the slider (or hit Play): the point moves clockwise at constant angular rate $\omega$ regardless of which ellipse it is on — the hallmark of the harmonic oscillator being **isochronous** (period independent of amplitude, i.e. of energy). Every other trajectory in the lamination is a scaled copy of this one; none of them intersect, exactly as the non-crossing theorem requires. ## Worked example: the inverted oscillator Now flip the sign of the potential, $V(q) = -\tfrac12 m\omega^2 q^2$ — a hilltop at the origin rather than a well. This is precisely the "complete the phase portrait" exercise the lecture leaves open, for the three cases $E<0$, $E=0$, $E>0$: $$ v^2 - \omega^2 q^2 = \frac{2E}{m}. $$ - **$E > 0$:** $v$ never reaches zero — the particle has enough energy to cross the hilltop, so the trajectory runs monotonically from $q=-\infty$ to $q=+\infty$ (or back). These are hyperbola branches opening *up/down*. - **$E = 0$:** $v = \pm\omega q$ — straight lines through the origin, the **separatrices**, approached only asymptotically as $t\to\pm\infty$. - **$E < 0$:** the particle cannot reach the hilltop; $|q|$ is bounded below by $\sqrt{2|E|/m}/\omega$, and the particle rolls up, turns around at $v=0$, and rolls back. These are hyperbola branches opening *left/right*, confined to one side or the other — the region between the two turning points is dynamically forbidden (it would require negative kinetic energy). ```{code-cell} ipython3 :tags: [hide-input] inverted = PhaseSpaceExplorer(omega=1.0, sign=-1) t_escape = np.linspace(-2.2, 2.2, 60) # crossing the hilltop, omega = 1 fig_inv = animated_phase_portrait( inverted, energies=[-2.0, -1.0, 1.0, 2.0], # E < 0 (turn back) and E > 0 (cross over) highlight_E=1.0, t_values=t_escape, title="Phase portrait — inverted oscillator (separatrices + escape/return branches)", q_range=4.0, ) # Draw in the E = 0 separatrices explicitly, since E = 0 is excluded from the family above q_sep = np.linspace(-4, 4, 50) fig_inv.add_trace(go.Scatter(x=q_sep, y=inverted.omega * q_sep, mode="lines", line=dict(color="black", width=1.5, dash="dash"), name="separatrix, E=0")) fig_inv.add_trace(go.Scatter(x=q_sep, y=-inverted.omega * q_sep, mode="lines", line=dict(color="black", width=1.5, dash="dash"), hoverinfo="skip", showlegend=False)) show_fig(fig_inv) ``` The highlighted, animated branch has $E=1>0$: the marker sails straight across the hilltop without pausing, gaining speed as it passes $q=0$ — the mirror image of the harmonic oscillator's closed, periodic ellipses. Compare it against the $E<0$ branches in the background, which turn around before ever reaching $q=0$, and against the dashed $E=0$ separatrices they asymptote to. Unlike the harmonic oscillator, there is no periodic motion anywhere in this portrait except the unstable equilibrium point at the origin itself.