Exercises — Hamilton's equations of motion

WHAT: apply $p=\partial L/\partial \dot{q}$. Only the first term contains $\dot{q}$. $p=\frac{\partial }{\partial \dot{q}}(\frac{1}{2}m{\dot{q}}^2-mgq)=m\dot{q}.$ Invert: $\dot{q}=p/m$. WHY it matters: this inversion is the step that lets us express everything in $(q,p)$ instead of $(q,\dot{q})$ — the doorway to $H$.

(a) A mass $m$ on a spring of stiffness $k$ — a simple harmonic oscillator, since $\frac{1}{2}k{q}^2$ is spring potential energy. (b) There is no explicit $t$ in $H$, so by $dH/dt=\partial H/\partial t=0$, $H$ is conserved. (c) $\frac{p^2}{2m}+\frac{1}{2}k{q}^2=\text{const}$ is the equation of an ellipse in the $(q,p)$ plane — a closed loop the state travels around forever. (The word "ellipse" is clozed as a recall cue, not a link.) See the figure below.

Hamilton gives two equations per coordinate (${\dot{q}}_i$ and ${\dot{p}}_i$), so $2n=6$ first-order equations. Lagrange gives one second-order equation per coordinate, so $3$ equations — but each is second-order. Trading order for number, the total "amount of information" ($6$ initial conditions) is the same. This $2n$-dimensional picture is exactly the phase space.

Step 1 (momentum): $p=m\dot{q}\Rightarrow \dot{q}=p/m$. Why: to eliminate $\dot{q}$. Step 2 (Hamiltonian): $H=p\dot{q}-L=p\cdot \frac{p}{m}-\left(\frac{1}{2}m\right(\frac{p}{m}{)}^2-mgq)=\frac{p^2}{m}-\frac{p^2}{2m}+mgq=\frac{p^2}{2m}+mgq.$ Why: substitute $\dot{q}=p/m$ everywhere so $H$ depends only on $(q,p)$. This equals $T+V$. Step 3 (equations): $\dot{q}=\frac{\partial H}{\partial p}=\frac{p}{m},\dot{p}=-\frac{\partial H}{\partial q}=-mg.$ Check: $\dot{p}=-mg$ means $m\ddot{q}=-mg$, i.e. $\ddot{q}=-g$ — free fall. Newton re-emerges. ✓

Step 1 (momentum). Why: we need the variable that will replace $\dot{\theta }$, so differentiate $L$ with respect to $\dot{\theta }$ — only the kinetic term $\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$ contains it. $p_{\theta }=\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }\Rightarrow \dot{\theta }=\frac{p_{\theta }}{m{\ell }^2}.$ Why invert: $H$ must depend on $(\theta ,p_{\theta })$, never on $\dot{\theta }$, so we need $\dot{\theta }$ written in terms of $p_{\theta }$. Step 2 (Hamiltonian). Why: plug $\dot{\theta }=p_{\theta }/(m{\ell }^2)$ into $H=p_{\theta }\dot{\theta }-L$ so every $\dot{\theta }$ disappears. $H=p_{\theta }\dot{\theta }-L=\underset{p_{\theta }\dot{\theta }}{\underbrace{\frac{p_{\theta }^2}{m{\ell }^2}}}-\underset{L}{\underbrace{(\frac{1}{2}\frac{p_{\theta }^2}{m{\ell }^2}+mg\ell \cos \theta )}}=\frac{p_{\theta }^2}{2m{\ell }^2}-mg\ell \cos \theta .$ Reading it: the first term is rotational kinetic energy, the second is gravitational potential energy $-mg\ell \cos \theta$ (lowest at the bottom, $\theta =0$). Step 3 (equations). Why: apply the two canonical rules — $\dot{\theta }$ reads off the $p_{\theta }$-slope of $H$, and ${\dot{p}}_{\theta }$ reads off minus the $\theta$-slope. $\dot{\theta }=\frac{\partial H}{\partial p_{\theta }}=\frac{p_{\theta }}{m{\ell }^2},{\dot{p}}_{\theta }=-\frac{\partial H}{\partial \theta }=-mg\ell \sin \theta .$ Why the last minus: the potential slope $\partial H/\partial \theta =+mg\ell \sin \theta$ pushes the momentum down the hill, exactly the "p fights q minus" rule. Check. Why: to confirm we recover known physics, eliminate $p_{\theta }$ by differentiating $\dot{\theta }$: $m{\ell }^2\ddot{\theta }={\dot{p}}_{\theta }=-mg\ell \sin \theta \Rightarrow \ddot{\theta }=-\frac{g}{\ell }\sin \theta$ — the exact pendulum equation. ✓

Step 1. Why: the time law $dH/dt=\partial H/\partial t$ says the only way $H$ can change along the motion is through an explicit $t$ in the formula — so we hunt for that explicit $t$. Here it sits inside $\cos (\Omega t)$. Step 2. Why: differentiate $H$ with respect to $t$ treating $q,p$ as fixed labels (that is what "explicit" means): $\frac{\partial H}{\partial t}=-Fq\cdot (-\Omega \sin (\Omega t))=+Fq\Omega \sin (\Omega t),$ which is not zero. Step 3. Why: feed this into the time law to get the actual rate of change along a trajectory: $\frac{dH}{dt}=\frac{\partial H}{\partial t}=Fq\Omega \sin (\Omega t)\ne 0.$ So $H$ is not conserved — the external drive pumps energy in and out. The explicit $t$ in $H$ is the tell-tale.

Consistency first. Why: the four points must lie on one energy contour, otherwise comparing their flow is meaningless. From $a=\sqrt{2E/(m{\omega }^2)}$ and $b=\sqrt{2mE}$, $\frac{b}{a}=\frac{\sqrt{2mE}}{\sqrt{2E/(m{\omega }^2)}}=\sqrt{m^2{\omega }^2}=m\omega \Rightarrow b=m\omega a.$ So we take $a>0$ and $b=m\omega a$ — not independent. Evaluate the velocity vector $(\dot{q},\dot{p})$ at each point:

• $(+a,0)$: $\dot{q}=0,\dot{p}=-m{\omega }^{2}a<0$ → moving down.
• $(0,+b)$: $\dot{q}=b/m=\omega a>0,\dot{p}=0$ → moving right.
• $(-a,0)$: $\dot{q}=0,\dot{p}=+m{\omega }^{2}a>0$ → moving up.
• $(0,-b)$: $\dot{q}=-b/m=-\omega a<0,\dot{p}=0$ → moving left.

Right at the top, down at the right, left at the bottom, up at the left — that circulation is clockwise. See the arrows.

Momentum. Why: differentiate $L$ by $\dot{r}$ to get the replacement variable; only $\frac{1}{2}m{\dot{r}}^2$ contains $\dot{r}$. $p_r=\frac{\partial L}{\partial \dot{r}}=m\dot{r}\Rightarrow \dot{r}=p_r/m.$ Hamiltonian. Why: substitute $\dot{r}=p_r/m$ into $H=p_r\dot{r}-L$ to remove $\dot{r}$. $H=p_r\dot{r}-L=\frac{p_r^2}{m}-(\frac{1}{2}\frac{p_r^2}{m}+\frac{1}{2}m{\omega }^2{r}^2)=\frac{p_r^2}{2m}-\frac{1}{2}m{\omega }^2{r}^2.$ Compare. $E$, written in $(r,p_r)$, is $E=\frac{p_r^2}{2m}-\frac{1}{2}m{\omega }^2{r}^2$ as well — here they happen to coincide, because the $\dot{\theta }$ term was absorbed into the effective potential and the Lagrangian, once reduced to the single coordinate $r$, is time-independent so $H$ is conserved. The subtle point: $H$ here is not the total kinetic energy $\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{\omega }^2{r}^2$ of the moving bead in the lab frame. The reduced $H$ is a conserved quantity (Jacobi integral), but it is not the lab-frame kinetic energy plus zero potential. This is the canonical example where "$H$ = energy" needs care: the wire does work, and $H$ measures the conserved Jacobi function, not naive $T$.

Apply ${\dot{p}}_y=-\frac{\partial H}{\partial y}$. Since $H$ has no $y$ in it, $\frac{\partial H}{\partial y}=0$, so ${\dot{p}}_y=0$ — $p_y$ is conserved. Meanwhile ${\dot{p}}_x=-\frac{\partial H}{\partial x}=-V^{\prime }(x)\ne 0$ in general — $p_x$ is not conserved. Insight: a coordinate absent from $H$ (a cyclic coordinate) hands you a conserved momentum for free — the seed of Noether's Theorem.

Step 1 (momentum — watch the cross-term). Why: differentiate the actual $L$ by $\dot{x}$; here the linear term $eA\dot{x}$ also contributes, so $p$ is not just $m\dot{x}$. $p=\frac{\partial L}{\partial \dot{x}}=m\dot{x}+eA(x).$ So $p\ne m\dot{x}$! Why invert: we still need $\dot{x}$ in terms of $(x,p)$ to eliminate it — solve the linear relation: $\dot{x}=\frac{p-eA}{m}$. Step 2 (Hamiltonian). Why: substitute that $\dot{x}$ into $H=p\dot{x}-L$ and simplify. $H=p\dot{x}-L=p\dot{x}-\frac{1}{2}m{\dot{x}}^2-eA\dot{x}=(p-eA)\dot{x}-\frac{1}{2}m{\dot{x}}^2.$ Why group $(p-eA)$: it is exactly $m\dot{x}$, so the algebra collapses cleanly. Put $\dot{x}=(p-eA)/m$: $H=(p-eA)\frac{p-eA}{m}-\frac{1}{2}m(\frac{p-eA}{m}{)}^2=\frac{(p-eA{)}^2}{m}-\frac{(p-eA{)}^2}{2m}=\frac{(p-eA{)}^2}{2m}.$ The lesson: the canonical momentum $p=m\dot{x}+eA$ differs from the kinetic momentum $m\dot{x}$; $H$ is quadratic in the gauge-invariant combination $(p-eA)$. This is the origin of the "minimal coupling" $p\to p-eA$ used in quantum mechanics.

Derivation. Why start with the chain rule: $f$ changes only because $q$ and $p$ move, so $\frac{df}{dt}=\frac{\partial f}{\partial q}\dot{q}+\frac{\partial f}{\partial p}\dot{p}.$ Why substitute Hamilton's equations: to express the rate purely in terms of $H$. Insert $\dot{q}=\partial H/\partial p$ and $\dot{p}=-\partial H/\partial q$: $\frac{df}{dt}=\frac{\partial f}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial H}{\partial q}=\{f,H\}.$ Recompute $dH/dt$. Why set $f=H$: to test the bracket on $H$ itself. Then $\{H,H\}=\frac{\partial H}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial H}{\partial p}\frac{\partial H}{\partial q}=0.$ So (with no explicit $t$) $dH/dt=0$ — consistent with the parent note, and now packaged in one bracket. Anything whose bracket with $H$ vanishes is conserved.

Momenta. Why: differentiate $L$ by each velocity separately; the two coordinates do not mix. $p_x=m\dot{x},p_y=m\dot{y}\Rightarrow \dot{x}=p_x/m,\dot{y}=p_y/m.$ Hamiltonian. Why: substitute both inversions into $H={\sum }_i{p}_i{\dot{q}}_i-L$. $H=\frac{p_x^2+p_y^2}{2m}+\frac{1}{2}m{\omega }^2(x^2+y^2).$ Four equations. Why: one $\dot{q}=\partial H/\partial p$ and one $\dot{p}=-\partial H/\partial q$ per coordinate. $\dot{x}=\frac{p_x}{m},\dot{y}=\frac{p_y}{m},{\dot{p}}_x=-m{\omega }^{2}x,{\dot{p}}_y=-m{\omega }^{2}y.$ Conservation. Why: $H$ has no explicit $t$, so $dH/dt=\partial H/\partial t=0$. Energy conserved. ✓ (Bonus: $p_{x}y-p_{y}x$ is also conserved — angular momentum — because the potential is rotationally symmetric.)

(i) Why check this: a term that is a total time derivative adds a constant-endpoint piece to the action and so leaves the Euler–Lagrange equations untouched. Indeed $\beta q\dot{q}=\frac{d}{dt}\left(\frac{1}{2}\beta q^2\right)$, so the motion is the same simple harmonic oscillator. (ii) Momentum. Why: differentiate the given $L$ by $\dot{q}$; the $\beta q\dot{q}$ term is linear in $\dot{q}$ and contributes $\beta q$. $p=\frac{\partial L}{\partial \dot{q}}=m\dot{q}+\beta q\Rightarrow \dot{q}=\frac{p-\beta q}{m}.$ Hamiltonian. Why: substitute $\dot{q}=(p-\beta q)/m$ into $H=p\dot{q}-L$, term by term. $H=p\dot{q}-L=p\dot{q}-\frac{1}{2}m{\dot{q}}^2+\frac{1}{2}m{\omega }^2{q}^2-\beta q\dot{q}=(p-\beta q)\dot{q}-\frac{1}{2}m{\dot{q}}^2+\frac{1}{2}m{\omega }^2{q}^2.$ Why group $(p-\beta q)$: it equals $m\dot{q}$, so the first two terms combine just like a plain kinetic term. Put $\dot{q}=(p-\beta q)/m$: $H=\frac{(p-\beta q{)}^2}{m}-\frac{(p-\beta q{)}^2}{2m}+\frac{1}{2}m{\omega }^2{q}^2=\frac{(p-\beta q{)}^2}{2m}+\frac{1}{2}m{\omega }^2{q}^2.$ Punchline: the Hamiltonian is not the naive $\frac{p^2}{2m}+\frac{1}{2}m{\omega }^2{q}^2$ — it carries the shift $p\to p-\beta q$. It is still conserved (no explicit $t$), and it still generates the same SHO motion, but its form in $(q,p)$ was changed by a mere total-derivative gauge in $L$. This is a hands-on view of what Canonical Transformations formalize: different $H$'s can describe the same physics.

Why an ellipse: the orbit is $\frac{p^2}{2mE}+\frac{q^2}{2E/(m{\omega }^2)}=1$, an ellipse with semi-axes $q_{\max }=\sqrt{\frac{2E}{m{\omega }^2}},p_{\max }=\sqrt{2mE}.$ Why the area formula: an ellipse of semi-axes $A,B$ encloses area $\pi AB$, so $\oint pdq=\pi q_{\max }{p}_{\max }=\pi \sqrt{\frac{2E}{m{\omega }^2}}\sqrt{2mE}=\pi \cdot \frac{2E}{\omega }=\frac{2\pi E}{\omega }.$ Meaning: this enclosed area is the action variable $J=\oint pdq$. Its constancy under slow parameter changes (adiabatic invariance) and its quantization $J=nh$ are gateways into Hamilton-Jacobi Theory and old quantum theory.

Fundamental bracket. Why: plug $f=q,g=p$ into $\{f,g\}=\frac{\partial f}{\partial q}\frac{\partial g}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial g}{\partial q}$; note $\partial q/\partial q=1$, $\partial q/\partial p=0$, etc. $\{q,p\}=1\cdot 1-0\cdot 0=1.$ Reproduce the equations. Why: using $\{f,H\}$ with $f=q$ then $f=p$ should hand back Hamilton's equations if the bracket formulation is equivalent. $\{q,H\}=\frac{\partial q}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial q}{\partial p}\frac{\partial H}{\partial q}=\frac{\partial H}{\partial p}=\dot{q}.✓$ $\{p,H\}=\frac{\partial p}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial p}{\partial p}\frac{\partial H}{\partial q}=-\frac{\partial H}{\partial q}=\dot{p}.✓$ The whole of Hamiltonian mechanics can thus be phrased as "time evolution = bracket with $H$," and the invariance of $\{q,p\}=1$ is exactly the condition that a change of variables is a canonical transformation.

Why does a cyclic coordinate (absent from $H$) give a conserved momentum? ::: Because ${\dot{p}}_i=-\partial H/\partial q_i=0$ when $H$ does not contain $q_i$. In L4.1, why is $p\ne m\dot{x}$? ::: The Lagrangian has a term linear in $\dot{x}$, so $p=\partial L/\partial \dot{x}=m\dot{x}+eA$. What is $\oint pdq$ for the SHO of energy $E$? ::: $2\pi E/\omega$, the action variable.