Legendre transform swaps conjugate variables

Statement

Let $X(x,y)$ be a thermodynamic potential (a state function) with $x$ one of its natural variables and $y$ collecting the remaining variables. Assume $X$ is differentiable and define the conjugate variable $p := \left(\frac{\partial X}{\partial x}\right)_y$ and the (partial) Legendre transform in $x$ by

where $x=x(p,y)$ is chosen so that $p = \left(\frac{\partial X}{\partial x}\right)_y$.

Then the differential of $Y$ is

so the variable $x$ has been replaced by its conjugate $p$ as an independent variable, and the conjugacy is recovered by

In thermodynamics, this “swap” is the mechanism behind passing from the internal energy $U(S,V,N)$ to standard thermodynamic potentials such as the Helmholtz free energy $F(T,V,N)=U-TS$, the Gibbs free energy $G(T,P,N)=U-TS+PV$, and the grand potential $\Omega(T,V,\mu)=U-TS-\mu N$.

Key hypotheses

• $X(x,y)$ is a well-defined state function (depends only on the thermodynamic state ).
• $X$ is differentiable in $x$ and the map $x \mapsto p=(\partial X/\partial x)_y$ can be locally inverted (often ensured by appropriate convexity/concavity assumptions; see Legendre transform ).

Key conclusions

• The Legendre-transformed potential $Y$ has $p$ (not $x$) as a natural variable.
• Conjugate variables are recovered by differentiation:
  • $p = (\partial X/\partial x)_y$ in the original potential,
  • $x = -(\partial Y/\partial p)_y$ in the transformed potential.
• In thermodynamic examples, the transform replaces an extensive variable (e.g. $S$) by its conjugate intensive variable (e.g. temperature $T$), or replaces $V$ by pressure $P$, or $N$ by chemical potential $\mu$, with the corresponding sign conventions.

Cross-links to definitions

• Legendre transform
• internal energy , thermodynamic entropy
• temperature , pressure , chemical potential
• Helmholtz free energy , Gibbs free energy , grand potential

Proof idea / significance

Differentiate $Y=X-px$:

Using $p=(\partial X/\partial x)_y$ gives $dX = p\,dx + \sum_i (\partial X/\partial y_i)_x\,dy_i$, so the $p\,dx$ terms cancel and the claimed form of $dY$ follows.

Significance: the Legendre transform produces potentials whose natural variables match common experimental controls (e.g. fixing $T$ rather than $S$), while preserving access to the conjugate extensive quantities through derivatives (e.g. $S=-\partial F/\partial T$ at fixed $V,N$).