05 May 2026

Legendre Transform in Mechanics

The Legendre transform as the bridge from the Lagrangian to the Hamiltonian description.

msc semester-i classical-mechanics legendre-transform

In Lagrangian mechanics the natural variables are $q$ and $\dot q$. In Hamiltonian mechanics the natural variables are $q$ and $p$. The Legendre transform is the mathematical operation that changes from a variable to its conjugate derivative variable.

The need for this transform appears whenever the slope or momentum is more useful than the original velocity-like variable.

Basic idea

For a function $f(x)$, define the conjugate variable

\[y=\frac{df}{dx}.\]

The Legendre transform is

\[\boxed{ g(y)=xy-f(x), }\]

where $x$ is finally expressed as a function of $y$.

The differential is

\[dg=x\,dy,\]

so

\[\frac{dg}{dy}=x.\]

Thus the transform replaces $x$ by its conjugate variable $y$.

Geometrically, a convex curve may be described either by its point coordinate $x$ or by the slope $y=df/dx$ of the tangent at that point. The Legendre transform rewrites the information in terms of the slope.

From Lagrangian to Hamiltonian

In mechanics,

\[p_i=\frac{\partial L}{\partial\dot q_i}\]

is conjugate to $\dot q_i$. The Hamiltonian is the Legendre transform of $L$ with respect to all velocities:

\[\boxed{ H(q,p,t)=\sum_i p_i\dot q_i-L(q,\dot q,t). }\]

After the transform, $\dot q_i$ must be eliminated in favor of $(q_i,p_i,t)$.

Example: quadratic kinetic energy

For

\[L=\frac12m\dot x^2-V(x),\]

the momentum is

\[p=m\dot x.\]

So

\[\dot x=\frac{p}{m}.\]

The Hamiltonian is

\[H=p\dot x-L.\]

Substituting,

\[H=\frac{p^2}{m}-\left(\frac{p^2}{2m}-V(x)\right),\]

hence

\[\boxed{ H=\frac{p^2}{2m}+V(x). }\]

Invertibility condition

The transform is regular when the Hessian is nonzero:

\[\det\left(\frac{\partial^2L}{\partial\dot q_i\partial\dot q_j}\right)\ne0.\]

This condition allows velocities to be solved in terms of momenta.

Relation with generating functions

Generating functions in canonical transformations use the same idea. One may change from variables $(q,Q)$ to $(q,P)$ by a Legendre transform, producing the familiar type-$2$ generating function $F_2(q,P,t)$.

Transform of a quadratic function

Find the Legendre transform of

\[f(x)=\frac12ax^2, \qquad a>0.\]

The conjugate variable is

\[y=\frac{df}{dx}=ax.\]

Hence

\[x=\frac{y}{a}.\]

The Legendre transform is

\[g(y)=xy-f(x).\]

Substituting $x=y/a$,

\[g(y)=\frac{y^2}{a} -\frac12a\left(\frac{y}{a}\right)^2.\]

Therefore

\[\boxed{ g(y)=\frac{y^2}{2a}. }\]

The same algebra is used in mechanics when $m\dot q$ is replaced by $p$.

Common caution

After writing $g=xy-f(x)$ or $H=p\dot q-L$, the old variable must be eliminated. The final expression for $g$ should contain $y$, not $x$; similarly, the final Hamiltonian should contain $(q,p,t)$, not $\dot q$.

Main points

• Legendre transform replaces a variable by its conjugate derivative variable.
• In mechanics, velocity is replaced by momentum.
• The Hamiltonian is $H=\sum_i p_i\dot q_i-L$.
• A regular transform requires an invertible velocity-momentum relation.
• Generating functions use the same transformation logic.

Practice questions

1. Find the Legendre transform of $f(x)=ax^2$.
2. Show that $H=p\dot q-L$ gives $H=p^2/(2m)+V(x)$ for $L=m\dot x^2/2-V(x)$.
3. Why must $\dot q$ be eliminated after forming $H$?
4. State the regularity condition for the Legendre transform in mechanics.
5. Explain why $F_2(q,P,t)$ is related to $F_1(q,Q,t)$ by a Legendre transform.