Visual walkthrough — Hamiltonian — definition H = Σpᵢq̇ᵢ − L

We build one idea: how to swap the variable "velocity" for the variable "momentum". That swap is the whole game, and it is called a Legendre Transform.

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Step 1 — A curve, and two honest ways to describe it

WHAT. Forget physics for a moment. Take any curve that bends upward — a bowl shape. Call the horizontal axis $x$ and the height of the curve $f(x)$. "$f(x)$" just means "the height of the curve above the point $x$".

WHY. Before we swap variables we must ask: what does it even mean to describe a curve? The usual way is the list of points $(x,f(x))$. But there is a second, equally complete way — describe the curve by its slopes. This second description is exactly what turns $L$ into $H$, so we start here.

PICTURE.

The blue bowl is $f(x)$. The amber dot sits at one chosen $x$. The cyan line touching the bowl there is the tangent — the straight line that grazes the curve at that single point. Its steepness is the slope, written $p$:

$p=\underset{\text{slope of the tangent line at }x}{\underbrace{f^{\prime }(x)}}.$

Here $f^{\prime }(x)$ (read "$f$-prime of $x$") is the derivative — a number that answers "if I nudge $x$ a tiny bit right, how fast does the height climb?" A steep bowl gives a big $p$; a flat bottom gives $p=0$.

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Step 2 — Reading the curve by its slope instead of its height

WHAT. As you slide the amber dot left to right along the bowl, the tangent's slope $p$ steadily changes. For a bowl (a convex curve — one that only ever bends one way) each position $x$ gives a different slope $p$. No two points share a slope.

WHY. That "no two points share a slope" property is the key that lets us run the description backwards: if someone hands you a slope $p$, there is exactly one point on the bowl with that slope. So the slope $p$ can serve as a label for points, just as good as $x$ itself. This is why we insist the curve is convex — otherwise two points would share a slope and the label would be ambiguous.

PICTURE.

Two amber dots, two cyan tangents, two clearly different slopes. The arrow reminds you: slope is a legitimate name-tag for position on a convex curve. We are about to trade the name-tag $x$ for the name-tag $p$.

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Step 3 — The gap between the tangent line and the curve

WHAT. Pick a point at $x$ with slope $p$. Extend its tangent line all the way back to the vertical axis (to $x=0$). Read off where it crosses. Call the crossing height $-g$ (the minus sign is a convention that will pay off). Then

$g=px-f(x).$

Let us name every symbol right here:

• $px$ — slope times horizontal distance = how much height the straight tangent line gains from $0$ up to $x$.
• $f(x)$ — the actual height of the curve at $x$.
• $g=px-f$ — the vertical gap you'd fall from the tangent's back-projection to the axis; equivalently the negative of the tangent's intercept.

WHY. This one algebraic combination $px-f$ is the entire Legendre transform. Everything after this is bookkeeping. We build it geometrically first so the formula is never a rabbit out of a hat.

PICTURE.

Follow the cyan tangent line back to the axis. The amber bracket is $g=px-f$. Notice: as we slide $x$, both $p$ and $f$ change, so $g$ changes too — and we will now think of $g$ as a function of the slope $p$, not of $x$.

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Step 4 — Prove the swap really removed $x$ (the differential trick)

WHAT. We must check that $g$ genuinely depends on $p$ and not secretly on $x$. Take a tiny change of everything (a differential, written $d(\cdot )$ = "an infinitesimal change of"):

$dg=d(px)-df=\underset{\text{product rule on }px}{\underbrace{pdx+xdp}}-\underset{df=f^{\prime }dx=pdx}{\underbrace{pdx}}.$

The two $pdx$ terms cancel, leaving

$\boxed{dg=xdp}.$

WHY. Look what happened: $dg$ contains only $dp$ — no leftover $dx$. That is the mathematical proof that $g$ is a clean function of $p$. And the coefficient of $dp$ is $x$, so $x=dg/dp$: the transform is its own inverse, slopes and positions have simply traded roles.

PICTURE.

The cartoon shows the cancellation: a green $+pdx$ arrow and a red $-pdx$ arrow annihilate, and only the amber $xdp$ survives. This exact cancellation is what happens in the physics derivation below — watch for it.

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Step 5 — Rename the axes: this IS the Hamiltonian

WHAT. Now change the dictionary. Replace:

geometry word	physics word
horizontal $x$	velocity $\dot{q}$ ("$\dot{q}$" = rate of change of coordinate $q$)
height $f(x)$	[[Lagrangian Mechanics
slope $p=f^{\prime }(x)$	momentum $p=\partial L/\partial \dot{q}$
transform $g=px-f$	Hamiltonian $H=p\dot{q}-L$

WHY. The Lagrangian $L$ is a bowl-shaped function of velocity $\dot{q}$ (for a normal kinetic term $\frac{1}{2}m{\dot{q}}^2$ it literally is a parabola). Its slope in the velocity direction is the momentum — that is not a coincidence, it is the definition $p=\partial L/\partial \dot{q}$ (here $\partial$ means "slope while holding the other variables fixed"). So the Legendre transform of $L$ in the $\dot{q}$ direction is exactly $H$.

PICTURE.

Same bowl as Step 1, relabelled. The horizontal axis now reads $\dot{q}$, the curve is $L$, the amber gap is $H=p\dot{q}-L$. Nothing new happened — we only changed the words on the axes.

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Step 6 — Do it for a real particle (bowl → parabola)

WHAT. Take $L=\frac{1}{2}m{\dot{x}}^2-V(x)$. The velocity-bowl is the parabola $\frac{1}{2}m{\dot{x}}^2$.

• Slope (momentum): $p=\partial L/\partial \dot{x}=m\dot{x}$.
• Invert: $\dot{x}=p/m$ — read the position off the slope.
• Assemble: $H=p\dot{x}-L=\frac{p^2}{m}-(\frac{p^2}{2m}-V)=\frac{p^2}{2m}+V(x).$

WHY. This is Step 3's $g=px-f$ with an actual parabola, so you see the number $\frac{p^2}{2m}$ appear as the intercept-gap of the tangent to a parabola. This recovers kinetic-plus-potential energy — the canonical energy of the system.

PICTURE.

The cyan parabola $\frac{1}{2}m{\dot{x}}^2$; the amber gap read at velocity $\dot{x}=p/m$ equals $\frac{p^2}{2m}$. The tangent's slope is the momentum $p=m\dot{x}$. Every quantity in the boxed answer is visible in the drawing.

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Step 7 — The degenerate case: a flat curve breaks the swap

WHAT. What if $L$ is not a bowl in $\dot{q}$? Suppose $L$ is linear in the velocity, e.g. $L=c\dot{q}+(\text{stuff without }\dot{q})$. Then the slope is the constant $p=c$ — it does not depend on $\dot{q}$ at all.

WHY. Now the swap fails: the equation $p=c$ cannot be inverted for $\dot{q}$ (many $\dot{q}$ give the same slope). Geometrically the "bowl" has degenerated into a straight ramp — every point shares one slope, so slope is no longer a valid name-tag (contrast Step 2). This is the singular Legendre case; it is exactly why the parent's condition "$T$ must be quadratic in $\dot{q}$" appears. It flags constrained/gauge systems handled by more advanced machinery.

PICTURE.

Left: healthy bowl, slopes fan out, unique labels. Right: a straight ramp — the tangent line is the curve, one slope for all points, the inversion $\dot{q}(p)$ has no answer. The amber "no unique $\dot{q}$" mark warns you.

Recall When can you trust the swap?

Legendre transform is invertible iff the curve is strictly convex in the swapped variable ::: iff $L$ curves (is not flat/linear) in $\dot{q}$, i.e. ${\partial }^{2}L/\partial {\dot{q}}^2\ne 0$.

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The one-picture summary

The blueprint compresses all seven steps: the velocity-bowl $L(\dot{q})$, its tangent whose slope is the momentum $p$, and the amber intercept-gap that is the Hamiltonian $H=p\dot{q}-L$. Read the picture, read the physics.

Recall Feynman retelling (say it out loud)

"The Lagrangian is a bowl-shaped hill drawn against velocity. At any velocity, the steepness of that hill is what we call momentum. Now here's the trick: instead of labelling each spot on the hill by its velocity, label it by its steepness — because on a bowl every spot has its own unique steepness. To do the relabelling, draw the tangent line at your spot and slide it back to the axis; the little gap it leaves, $p\dot{q}-L$, is the Hamiltonian. When I take a tiny step, the velocity-change term cancels perfectly and only the momentum-change survives — that cancellation is the proof that the Hamiltonian truly forgot about velocity and only remembers momentum. If the hill is a real parabola, this gap works out to $p^2/2m+V$, the good old energy. And if the hill were a flat ramp instead of a bowl, every spot would share the same steepness, the relabelling would be impossible, and the whole trick would break — which is the physicist's warning sign of a constrained system."

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Related: Legendre Transform · Hamilton's Canonical Equations · Phase Space · Poisson Brackets · Conservation Laws & Noether's Theorem · back to the parent topic.