Why does thermodynamics use the negative of the Legendre transform? So I see how the negative Legendre transform is very helpful in interchanging dependencies and giving us the four different major thermodynamic potentials, from internal energy to Helmholtz, Gibbs, and enthalpy.
But what I'm unclear on is why we use the negative Legendre transform. Is it more that people derived the four above thermodynamic potentials by physical arguments, and then noticed that you have to use the negative Legendre transform to move between them, or is there some principled physical reason for why the negative Legendre transform makes more physical sense than the usual Legendre transform?
Any thoughts appreciated.
 A: The choice of the sign of the Legendre transform is purely conventional. There is no obliged way of defining it. The mathematical definition, used in physics in the case of the passage from Lagrangian to Hamiltonian function, has some advantages with respect to the opposite sign convention used in thermodynamics. Probably one of the most important is the conservation of the kind of convexity: if $f(x)$ is a convex function, its Legendre transform $\phi(p)=\sup_x (xp-f(x))$ remains a convex function, while with the opposite choice for the sign it would become concave. 
However, permanence of convexity is just one of the possible reasons for a choice. Another, which probably played a major role in Thermodynamics, although I am not aware of explicit mentioning of such a point in books or papers, is to preserve an energy-like character of the other thermodynamic potentials. By energy-like character I mean that a positive heat transfer to the system, or a positive work done on the system would result in an increase of the corresponding thermodynamic potential.. For example, if enthalpy would had been defined as 
$$
\tilde H(P,S) = -PV -U
$$
its differential would be
$$
d  \tilde H = -VdP -TdS
$$
with the consequence that adding positive heat  $TdS$ to the system, at constant pressure,  would decrease such a redefined enthalpy.
Of course, nothing would be wrong with such an alternative choice, and the physics would remain exactly the same.
A: The point is to preserve the intuition we have about energy. The Helmholtz free energy, Gibbs free energy, enthalpy, and internal energy all can be described as some kind of energy (subject to various other things being held constant) that can be used up to do work, and is minimized at equilibrium. If you added random sign flips, this intuition wouldn't work anymore. It's like defining your $z$-axis to point downward; totally mathematically allowed, but physically confusing. 
A: I'm certain this is not the way we normally reason about the potentials, but one could also argue the following. Thermodynamics is actually using the normal mathematical convention for the Legendre transformation, it's just that not energy, but entropy is the function being transformed. For example, starting from the entropy as $S(E,V,N)$, the Legendre transformation w.r.t. E and V is
$$
S^*(e,v,N) = eE + vV - S.
$$
Substituting the usual definitions for the derivatives of S we get:
$$
\begin{aligned}
S^* &= \left(\frac{∂S}{∂E}\right)_{V,N}E + \left(\frac{∂S}{∂V}\right)_{E,N}V - S \\
    &= \frac{E}{T} + \frac{p}{T}V - S \\
    &= \frac{1}{T} (E + PV - TS) = \frac{G}{T}
\end{aligned}
$$
where $G(T,P,N)$ is the Gibbs free energy.
So, we could say all the thermodynamic potentials are $T$-times Legendre transformation of the entropy, with $T$ because historical definitions involved energy-like quantities.
