2.4.2 · D5Thermodynamics & Statistical Mechanics (Advanced)
Question bank — Legendre transforms connecting them
Prerequisites worth having open: Internal Energy and the First Law, Maxwell Relations, Thermodynamic Stability and Convexity.
0. The toolkit these traps assume (read first)
This bank uses five objects. Here they are defined from scratch, so no symbol below is a mystery.
True or false — justify
A Legendre transform loses no information because it is an involution (doing it twice returns the original).
True. From the slope relation , transforming again has slope and rebuilds ; since you can always get back, every bit of the original data still sits inside — it was only re-encoded as (slope, intercept).
is the correct definition of enthalpy.
False. The slope of in is (from ), so Legendre subtracts , i.e. we add : .
Writing as a function of by substituting gives a valid fundamental relation.
False. Fundamental means exact in its natural variables; that substitution keeps 's value but forgets the intercept, so you cannot recover without integrating. Only keeps everything, via .
The Gibbs free energy is the "lab favourite" because both its natural variables and are knobs an experimenter controls.
True. A beaker on a bench sits at fixed (room/bath) and (atmosphere); since , those are exactly 's natural variables, making equilibrium a clean -minimization.
Every Legendre-generated potential automatically supplies a Maxwell relation.
True. Each is an exact differential, so by the mixed-partial test its two second partials must be equal; applying that test to gives , and each of hands you one this way.
The Legendre transform requires the function to be convex, which is a serious restriction for thermodynamics.
False (the "serious restriction" part). Invertibility of needs convexity, and stability supplies it: e.g. (upward-opening in ). So for stable equilibria the transform is always well-defined — a feature, not a limitation.
and contain exactly the same physical information.
True. is a Legendre transform and hence an involution, so it is invertible; and are two encodings of one and the same fundamental surface, just parametrized by different independent variables.
Because carries a minus sign, entropy can be negative here.
False. The minus sign is a feature of the slope relation (here , ); it does not make negative. still holds; simply decreases as rises at fixed .
Spot the error
"To get from , we swap for ."
Wrong pair. swaps the conjugate pair , not . Swapping instead gives .
"The intercept of the tangent line at is just ."
No — is the tangent's value at , not at . The -intercept is , which is the whole point of the construction (see the figure above).
"Since and had slope , also has slope ."
The slope of is , not . Differentiating , the term vanishes leaving .
" tells us 's natural variables are and ."
Backwards. The differentials appearing tell us the natural variables are the extensive ; and are the conjugate slopes, not the natural variables.
"We add to make because pressure is positive."
The sign of is irrelevant to the rule. We add because the slope is negative, and Legendre subtracts slope×variable: .
" collects both minus signs from the two swaps."
The -swap adds , not subtracts it. Correct is , giving .
"Maxwell relations come from the equation of state."
They come from exactness of (mixed partials commute), independent of any specific equation of state. That's why they hold for every substance.
Why questions
Why is the minus sign in the "heart" of the transform for thermodynamics?
Because it converts a differential driven by into one driven by : the new potential's independent variable becomes the slope , exactly the lab-controllable quantity we wanted.
Why can't we build a potential natural in simultaneously?
and are a conjugate pair — one is the slope of the potential with respect to the other. A fundamental relation uses one member of each pair, never both together.
Why does stability guarantee the Legendre inversion is single-valued?
Stability forces convexity (), so is strictly monotonic; a strictly increasing slope function is one-to-one and thus invertible everywhere.
Why does the naive substitution fail to be fundamental while succeeds?
Geometrically, many curves share the same set of slopes if you forget intercepts. stores the intercept via the term, so it pins down the unique curve; the substituted does not.
Why does give the cleanest equilibrium criterion for a bench-top reaction?
At fixed , , so equilibrium reduces to balancing chemical potentials — see Chemical Potential and Phase Equilibrium. No other potential makes its natural variables.
Why is called a "free" energy?
The term subtracts off the energy tied up in disordered thermal motion, leaving the portion "free" to do work at constant .
Why does the same Legendre machinery appear in mechanics as the Hamiltonian?
There it swaps velocity for its conjugate momentum : — see Lagrangian to Hamiltonian (Legendre in Mechanics). Same (slope, intercept) re-encoding, different names.
Edge cases
What happens to the Legendre transform at a point where (an inflection, zero curvature)?
The slope stops being locally invertible (its derivative vanishes), so the transform becomes singular there — this is exactly where a phase transition / loss of stability shows up.
At a first-order phase transition a curve has a kink (slope jumps). What does this mean for ?
jumps discontinuously — the two phases have different volumes. The kink in is the geometric signature of the transition; the transform's slope is multi-valued at that single pressure.
If a region of is concave instead of convex (unstable), what does the Legendre transform do?
Concave means , so the slope is no longer one-to-one and the naive transform gives a multi-valued / non-physical branch; the physical fix is the convex-hull (common-tangent) construction, which replaces unstable branches with flat coexistence lines — see Thermodynamic Stability and Convexity.
In the limit , what does reduce to, and why is that consistent?
as , since the term vanishes; consistent because at absolute zero there is no thermal energy to "free up," so free energy and internal energy coincide.
For with , why is its transform concave (opens downward)?
is convex (), and a convex always transforms to a concave (here ); the sign flip comes straight from the slope relation and verifies the involution when you transform back.
Recall One-line self-test before you leave
Cover this and answer aloud. State the three things every Legendre step does to a variable. ::: Swaps an extensive variable for its conjugate intensive slope, adds a term to preserve the intercept, and yields a new potential natural in that slope.