Hamilton's equations of motion

1. WHAT is the Hamiltonian? (derivation from scratch)

We start from the Lagrangian $L(q_i,\dot q_i,t)$ . The Euler–Lagrange equations are $\frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}=0.$

HOW do we switch variables cleanly? We use a Legendre transform. The trick: if a function depends on a variable $\dot q$ , but we'd rather it depend on its slope $p=\partial L/\partial\dot q$ , we form a new function whose differential no longer contains $d\dot q$ .

2. DERIVING Hamilton's equations (the heart)

Take the total differential of $H=\sum_i p_i\dot q_i - L$ : $dH=\sum_i\big(\dot q_i\,dp_i+p_i\,d\dot q_i\big)-\sum_i\Big(\frac{\partial L}{\partial q_i}dq_i+\frac{\partial L}{\partial \dot q_i}d\dot q_i\Big)-\frac{\partial L}{\partial t}dt.$

After cancellation, and using $\partial L/\partial q_i=\dot p_i$ (from Euler–Lagrange, since $\dot p_i=\frac{d}{dt}\frac{\partial L}{\partial \dot q_i}=\frac{\partial L}{\partial q_i}$ ): $dH=\sum_i\dot q_i\,dp_i-\sum_i\dot p_i\,dq_i-\frac{\partial L}{\partial t}dt.$

But by definition of a differential of $H(q,p,t)$ : $dH=\sum_i\frac{\partial H}{\partial q_i}dq_i+\sum_i\frac{\partial H}{\partial p_i}dp_i+\frac{\partial H}{\partial t}dt.$

Matching coefficients of the independent differentials $dq_i,\,dp_i,\,dt$ :

WHY the minus sign? It comes purely from matching $-\dot p_i\,dq_i$ with $+\frac{\partial H}{\partial q_i}dq_i$ . The asymmetric sign is what makes phase-space flow area-preserving (Liouville) — it is a feature, not a typo.

3. Conservation of $H$ (Forecast-then-Verify)

Worked example Forecast: when is

$H$ conserved? Predict before deriving: $H$ should be conserved when it has no explicit time dependence.

Verify by computing the total time derivative along a trajectory: $\frac{dH}{dt}=\sum_i\Big(\frac{\partial H}{\partial q_i}\dot q_i+\frac{\partial H}{\partial p_i}\dot p_i\Big)+\frac{\partial H}{\partial t}.$ Substitute Hamilton's equations $\dot q_i=\partial H/\partial p_i,\ \dot p_i=-\partial H/\partial q_i$ : $\frac{dH}{dt}=\sum_i\Big(\frac{\partial H}{\partial q_i}\frac{\partial H}{\partial p_i}-\frac{\partial H}{\partial p_i}\frac{\partial H}{\partial q_i}\Big)+\frac{\partial H}{\partial t}=\frac{\partial H}{\partial t}.$ The bracketed sum cancels term-by-term. Result: $\dfrac{dH}{dt}=\dfrac{\partial H}{\partial t}$ . So if $H$ has no explicit time dependence, it is conserved. ✓ Forecast confirmed.

4. Worked Example A — 1D particle in a potential

System: $L=\tfrac12 m\dot x^2 - V(x)$ .

Momentum. $p=\dfrac{\partial L}{\partial\dot x}=m\dot x$ . Why this step? It defines the variable replacing $\dot x$ , and lets us invert $\dot x=p/m$ .
Hamiltonian. $H=p\dot x-L = p\cdot\frac{p}{m}-\big(\tfrac12 m(\frac{p}{m})^2-V\big)=\dfrac{p^2}{2m}+V(x)$ . Why this step? Substituting $\dot x=p/m$ everywhere expresses $H$ in $(x,p)$ only — here it equals total energy $T+V$ .
Equations. $\dot x=\frac{\partial H}{\partial p}=\frac{p}{m},\qquad \dot p=-\frac{\partial H}{\partial x}=-\frac{dV}{dx}.$ Why this step? The second is just $m\ddot x=-V'(x)$ — Newton's second law re-emerges, confirming consistency.

5. Worked Example B — Simple Harmonic Oscillator

$V=\tfrac12 m\omega^2 x^2$ , so $H=\dfrac{p^2}{2m}+\dfrac12 m\omega^2 x^2$ . $\dot x=\frac{p}{m},\qquad \dot p=-m\omega^2 x.$ Why interesting? Eliminating $p$ : $\ddot x=-\omega^2 x$ . In phase space $(x,p)$ the trajectory is an ellipse of constant energy — the orbit encloses constant area $\oint p\,dx = 2\pi E/\omega$ , a hint of action quantization.

6. Worked Example C — Mass on a frictionless wire bead / generic check

Show the abstract recipe always reduces to Newton when $H=T+V$ with $p=m\dot q$ ; the structure guarantees it.

Common mistake Steel-manning common errors

Error 1: " $H$ is always the total energy $E$ ." Why it feels right: In example A it literally was $T+V$ . The fix: $H=E$ only when (i) the kinetic energy is quadratic in velocities AND (ii) the coordinate transformation is time-independent (scleronomic). For a rotating frame or time-dependent constraints, $H\neq E$ . Also $H$ may be conserved yet not equal $E$ , or equal $E$ but not conserved.

Error 2: "Just substitute $\dot q$ for $p/m$ as a constant rule." Why it feels right: It worked for free particles. The fix: You must invert $p_i=\partial L/\partial\dot q_i$ for the specific Lagrangian. With cross terms (e.g. magnetic $L=\tfrac12 m\dot{\mathbf r}^2+q\mathbf A\cdot\dot{\mathbf r}$ ), $p=m\dot{\mathbf r}+q\mathbf A \ne m\dot{\mathbf r}$ .

Error 3: Forgetting the minus sign in $\dot p=-\partial H/\partial q$ . Why it feels right: $\dot q=+\partial H/\partial p$ has a plus, so symmetry "suggests" a plus everywhere. The fix: The minus is forced by the Legendre algebra and is essential — without it energy would not be conserved.

Recall Feynman: explain to a 12-year-old

Imagine you describe a swinging ball not by "where it is and how fast," but by "where it is and how much push it carries" (that push is momentum). Hamilton found two beautifully simple rules: how the position changes depends on how the energy changes with push, and how the push changes depends on how the energy changes with position — with a minus sign so nothing runs away. Plot position sideways and push upward, and the ball traces a closed loop forever, like a clock hand. The size of the loop never changes — that's energy staying constant.

Active-Recall Flashcards

What is the generalized momentum conjugate to

q_i

p_i=\partial L/\partial\dot q_i

Define the Hamiltonian via a Legendre transform.

H=\sum_i p_i\dot q_i - L

, with

\dot q_i

expressed in terms of

(q,p)

State Hamilton's two canonical equations.

\dot q_i=\partial H/\partial p_i

and

\dot p_i=-\partial H/\partial q_i

Why does

d\dot q_i

vanish in

dH

Because

\partial L/\partial\dot q_i=p_i

, so

p_i d\dot q_i

cancels

\frac{\partial L}{\partial\dot q_i}d\dot q_i

— the Legendre transform.

What is

dH/dt

along a trajectory?

dH/dt=\partial H/\partial t

; hence

H

conserved if no explicit

t

When does

H

equal total energy

E

When KE is quadratic in velocities AND constraints are time-independent (scleronomic).

For

H=p^2/2m+V(x)

, what are the equations of motion?

\dot x=p/m

\dot p=-V'(x)

, equivalent to

m\ddot x=-V'

Why the minus sign in

\dot p=-\partial H/\partial q

It is forced by matching differentials in

dH

; it makes phase-space flow energy-preserving.

What is the relation between

\partial H/\partial t

and

\partial L/\partial t

\partial H/\partial t=-\partial L/\partial t

How many first-order equations replace the

n

second-order Lagrange equations?

2n

first-order equations in phase space.

Connections

Lagrangian Mechanics — the starting point we transform from.
Legendre Transform — the mathematical engine swapping $\dot q\leftrightarrow p$ .
Phase Space and Liouville's Theorem — where the $2n$ first-order flow lives.
Poisson Brackets — compact rewriting $\dot f=\{f,H\}+\partial_t f$ .
Canonical Transformations — symmetries preserving Hamilton's form.
Hamilton-Jacobi Theory — next abstraction, bridge to optics & QM.
Noether's Theorem — cyclic coordinate $\Rightarrow$ conserved $p_i$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Lagrangian mechanics mein hum position $q$ aur velocity $\dot q$ use karte hain, aur humein $n$ second-order equations milti hain. Hamilton ne ek smart chaal chali: velocity ko hatao, uski jagah momentum $p=\partial L/\partial \dot q$ le aao. Phir ek naya function banao — Hamiltonian $H=\sum p\dot q - L$ — jo sirf $(q,p)$ par depend karta hai. Yeh swap ek Legendre transform se hota hai, aur uska kamaal yeh hai ki $d\dot q$ wala term apne aap cancel ho jaata hai kyunki $\partial L/\partial\dot q$ to $p$ hi hai.

Iske baad do simple equations nikalti hain: $\dot q = +\partial H/\partial p$ aur $\dot p = -\partial H/\partial q$ . Yeh canonical equations kehlati hain. Note karo minus sign sirf $p$ wale equation mein aata hai — yeh koi galti nahi, yahi minus energy ko conserve rakhta hai. Yaad rakhne ka tareeka: "q follows p plus, p fights q minus."

Kyun important hai? Kyunki ab system ka pura state ek single point ban jaata hai $(q,p)$ phase space mein, aur uska future ek unique line par chalta hai. Simple harmonic oscillator mein yeh line ek ellipse hoti hai jiska area constant rehta hai — yahi aage chal kar Liouville's theorem aur quantum mechanics tak le jaata hai.

Ek warning: $H$ hamesha total energy $E$ ke barabar nahi hota. Sirf jab kinetic energy velocity mein quadratic ho aur constraints time pe depend na karein, tab $H=E$ . Aur agar $H$ mein explicit time nahi hai, to $H$ conserve hota hai — yeh yaad rakhna exam ke liye gold hai.

1. WHAT is the Hamiltonian? (derivation from scratch)

2. DERIVING Hamilton's equations (the heart)

3. Conservation of $H$ (Forecast-then-Verify)

4. Worked Example A — 1D particle in a potential

5. Worked Example B — Simple Harmonic Oscillator

6. Worked Example C — Mass on a frictionless wire bead / generic check

Active-Recall Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Analytical Mechanics

Connections