Visual walkthrough — Hamilton's equations of motion

We only assume you know from Lagrangian Mechanics one thing: a system has a function $L$ called the Lagrangian, and its motion obeys the Euler–Lagrange rule. We rebuild everything else.

Step 1 — A curve, and the slope of that curve

WHAT. Fix a position $q$ and time $t$, and let the velocity $\dot{q}$ (the rate the position changes, "how fast") be the only thing we vary. Then $L$ becomes an ordinary curve: horizontal axis $\dot{q}$, vertical axis $L$.

WHY. Everything Hamilton does is a trick about slopes of this curve. Before we can trade $\dot{q}$ for something else, we must look at what the curve of $L$ versus $\dot{q}$ even is. For a free particle $L=\frac{1}{2}m{\dot{q}}^2$, so it is a simple upward bowl (a parabola).

PICTURE. Look at the pale-yellow curve. At one chosen point on it, the chalk-blue line is tangent — it just grazes the curve. Its steepness (rise over run) is the number we are about to name.

Step 2 — Reading a point on the curve two ways

WHAT. The bowl-shaped curve can be described by listing every point $(\dot{q},L)$ — OR by listing every tangent line: its slope $p$ and where that tangent hits the vertical axis.

WHY. This is the deep idea behind the Legendre Transform (see Legendre Transform). A convex curve is completely pinned down by its family of tangent lines. So instead of "the height $L$ at velocity $\dot{q}$," we may hand someone "the slope $p$ and the intercept." No information is lost — we just changed which variable we hold in our hand from $\dot{q}$ to $p$.

PICTURE. The blue tangent line, extended, crosses the vertical axis at a negative height. Call the positive distance from the origin down to that intercept our new quantity $H$.

Step 3 — Why the velocity vanishes: watch the tangent slide

WHAT. Nudge $\dot{q}$ a tiny bit. The curve height changes by $dL=pd\dot{q}$ (slope times step). At the same time $p\dot{q}$ changes by $\dot{q}dp+pd\dot{q}$. Subtract, as $H$ demands.

WHY. We want $H$ to forget $\dot{q}$ entirely. Let's see the cancellation happen, not just claim it.

PICTURE. Two nearby tangent lines: as $\dot{q}$ steps right by $d\dot{q}$, the extra bit $pd\dot{q}$ the curve climbs is exactly the extra bit the term $p\dot{q}$ climbs. They erase each other, leaving only the change coming from $p$ moving.

$dH=\underset{\text{from }p\text{ moving}}{\underbrace{\dot{q}dp}}+\underset{\text{cancels}\to }{\underbrace{pd\dot{q}}}-\underset{\leftarrow dL\text{ piece}}{\underbrace{pd\dot{q}}}-\frac{\partial L}{\partial q}dq-\frac{\partial L}{\partial t}dt.$

The two middle terms are equal and opposite because $\partial L/\partial \dot{q}=p$ by definition (Step 1). The velocity's differential is gone.

Step 4 — Feeding in Euler–Lagrange to name the leftover slopes

WHAT. Two pieces remain from Step 3: the term with $dq$ and the term with $dp$. Euler–Lagrange lets us relabel the $dq$ term.

WHY. Right now the $dq$ coefficient is $-\partial L/\partial q$ — written in Lagrangian language. We want it in momentum language so both equations look symmetric.

PICTURE. Euler–Lagrange says: "the rate the momentum-slope changes in time equals the position-slope of $L$." Follow the chalk-pink arrow from the time-derivative of $p$ across to $\partial L/\partial q$.

$\dot{p}\equiv \frac{dp}{dt}=\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\stackrel{\text{E–L}}{=}\frac{\partial L}{\partial q}.$

So the leftover from Step 3 becomes $dH=\dot{q}dp-\dot{p}dq-\frac{\partial L}{\partial t}dt.$

Each term now speaks only of $q$, $p$, and $t$. Good — that matches what $H$ is allowed to depend on.

Step 5 — Matching two expressions for the same $dH$

WHAT. We have two formulas for the same small change $dH$. One we just derived (Step 4). The other is what calculus always says for a function of $(q,p,t)$.

WHY. If two expressions equal the same $dH$ for every independent wiggle $dq$, $dp$, $dt$, then the coefficient sitting on each wiggle must match separately. That is how we read off the equations.

PICTURE. Line up the two rows. The coefficient of $dp$ on top must equal the coefficient of $dp$ on the bottom; likewise for $dq$ and $dt$. Matching colours show the pairings.

Step 6 — WHY the minus sign is not optional: the flow must swirl

WHAT. Draw the velocity field in phase space (axes $q$ horizontal, $p$ vertical). At each point the state moves with velocity $(\dot{q},\dot{p})=({\partial }_{p}H,-{\partial }_{q}H)$.

WHY. The minus flips one component of the gradient by $90^{\circ }$. Instead of flowing downhill toward lowest $H$ (which a plus-plus rule would do, and everything would collapse to a point), the state flows around the hills of constant $H$. Motion is perpetual, energy is preserved — this is the seed of Phase Space and Liouville's Theorem.

PICTURE. Left panel: the wrong plus-plus field — every arrow points to the pit, motion dies. Right panel: the correct Hamilton field — arrows circle the contours of $H$, tracing a closed loop.

Step 7 — Degenerate & edge cases (never leave a gap)

WHAT & WHY & PICTURE together, because each is a special reading of the same figure below.

• Free particle, $V=0$. $H=p^2/2m$ has no $q$ inside, so $\dot{p}=-{\partial }_{q}H=0$: momentum is constant. Phase-space lines are horizontal (top panel) — the state slides right at fixed height.
• At an equilibrium (${\partial }_{q}H=0$ and ${\partial }_{p}H=0$). Both velocities vanish: the state is a fixed point and never moves (centre dot).
• Curve not convex (inflection, ${\partial }^{2}L/\partial {\dot{q}}^2=0$). The slope $p$ no longer uniquely picks a $\dot{q}$ — the Legendre step of Step 2 cannot be inverted. This is the singular case where the whole Hamiltonian recipe stalls (bottom panel: two velocities share one slope).
• $L$ explicitly time-dependent. Then $\partial L/\partial t\ne 0$, so $\partial H/\partial t\ne 0$ and $H$ drifts: $dH/dt=\partial H/\partial t$. Energy is no longer conserved even though the two canonical equations still hold unchanged.

The one-picture summary

Everything on this page is one journey: curve → its slope → intercept → strip out $\dot{q}$ → match differentials → circulate. The final board compresses it.

Active-Recall