Free entropy

Dependence of the potentials on the natural variables

Entropy

${\displaystyle S=S(U,V,\{N_i\})}$

By the definition of a total differential,

${\displaystyle dS=\frac{\mathrm{\partial }S}{\mathrm{\partial }U}dU+\frac{\mathrm{\partial }S}{\mathrm{\partial }V}dV+\sum _{i=1}^s\frac{\mathrm{\partial }S}{\mathrm{\partial }{N}_i}d{N}_i.}$

From the equations of state,

${\displaystyle dS=\frac{1}{T}dU+\frac{P}{T}dV+\sum _{i=1}^s\left(-\frac{{\mu }_i}{T}\right)d{N}_i.}$

The differentials in the above equation are all of extensive variables, so they may be integrated to yield

${\displaystyle S=\frac{U}{T}+\frac{PV}{T}+\sum _{i=1}^s\left(-\frac{{\mu }_{i}N}{T}\right)+\text{constant}.}$

Massieu potential / Helmholtz free entropy

${\displaystyle \mathrm{\Phi }=S-\frac{U}{T}}$

${\displaystyle \mathrm{\Phi }=\frac{U}{T}+\frac{PV}{T}+\sum _{i=1}^s\left(-\frac{{\mu }_{i}N}{T}\right)-\frac{U}{T}}$

${\displaystyle \mathrm{\Phi }=\frac{PV}{T}+\sum _{i=1}^s\left(-\frac{{\mu }_{i}N}{T}\right)}$

Starting over at the definition of ${\displaystyle \mathrm{\Phi }}$ and taking the total differential, we have via a Legendre transform (and the chain rule)

${\displaystyle d\mathrm{\Phi }=dS-\frac{1}{T}dU-Ud\frac{1}{T},}$

${\displaystyle d\mathrm{\Phi }=\frac{1}{T}dU+\frac{P}{T}dV+\sum _{i=1}^s\left(-\frac{{\mu }_i}{T}\right)d{N}_i-\frac{1}{T}dU-Ud\frac{1}{T},}$

${\displaystyle d\mathrm{\Phi }=-Ud\frac{1}{T}+\frac{P}{T}dV+\sum _{i=1}^s\left(-\frac{{\mu }_i}{T}\right)d{N}_i.}$

The above differentials are not all of extensive variables, so the equation may not be directly integrated. From ${\displaystyle d\mathrm{\Phi }}$ we see that

${\displaystyle \mathrm{\Phi }=\mathrm{\Phi }(\frac{1}{T},V,\{N_i\}).}$

If reciprocal variables are not desired,

${\displaystyle d\mathrm{\Phi }=dS-\frac{TdU-UdT}{T^2},}$

${\displaystyle d\mathrm{\Phi }=dS-\frac{1}{T}dU+\frac{U}{T^2}dT,}$

${\displaystyle d\mathrm{\Phi }=\frac{1}{T}dU+\frac{P}{T}dV+\sum _{i=1}^s\left(-\frac{{\mu }_i}{T}\right)d{N}_i-\frac{1}{T}dU+\frac{U}{T^2}dT,}$

${\displaystyle d\mathrm{\Phi }=\frac{U}{T^2}dT+\frac{P}{T}dV+\sum _{i=1}^s\left(-\frac{{\mu }_i}{T}\right)d{N}_i,}$

${\displaystyle \mathrm{\Phi }=\mathrm{\Phi }(T,V,\{N_i\}).}$

Planck potential / Gibbs free entropy

${\displaystyle \mathrm{\Xi }=\mathrm{\Phi }-\frac{PV}{T}}$

${\displaystyle \mathrm{\Xi }=\frac{PV}{T}+\sum _{i=1}^s\left(-\frac{{\mu }_{i}N}{T}\right)-\frac{PV}{T}}$

${\displaystyle \mathrm{\Xi }=\sum _{i=1}^s\left(-\frac{{\mu }_{i}N}{T}\right)}$

Starting over at the definition of ${\displaystyle \mathrm{\Xi }}$ and taking the total differential, we have via a Legendre transform (and the chain rule)

${\displaystyle d\mathrm{\Xi }=d\mathrm{\Phi }-\frac{P}{T}dV-Vd\frac{P}{T}}$

${\displaystyle d\mathrm{\Xi }=-Ud\frac{2}{T}+\frac{P}{T}dV+\sum _{i=1}^s\left(-\frac{{\mu }_i}{T}\right)d{N}_i-\frac{P}{T}dV-Vd\frac{P}{T}}$

${\displaystyle d\mathrm{\Xi }=-Ud\frac{1}{T}-Vd\frac{P}{T}+\sum _{i=1}^s\left(-\frac{{\mu }_i}{T}\right)d{N}_i.}$

The above differentials are not all of extensive variables, so the equation may not be directly integrated. From ${\displaystyle d\mathrm{\Xi }}$ we see that

${\displaystyle \mathrm{\Xi }=\mathrm{\Xi }\left(\frac{1}{T},\frac{P}{T},\{N_i\}\right).}$

If reciprocal variables are not desired,

${\displaystyle d\mathrm{\Xi }=d\mathrm{\Phi }-\frac{T(PdV+VdP)-PVdT}{T^2},}$

${\displaystyle d\mathrm{\Xi }=d\mathrm{\Phi }-\frac{P}{T}dV-\frac{V}{T}dP+\frac{PV}{T^2}dT,}$

${\displaystyle d\mathrm{\Xi }=\frac{U}{T^2}dT+\frac{P}{T}dV+\sum _{i=1}^s\left(-\frac{{\mu }_i}{T}\right)d{N}_i-\frac{P}{T}dV-\frac{V}{T}dP+\frac{PV}{T^2}dT,}$

${\displaystyle d\mathrm{\Xi }=\frac{U+PV}{T^2}dT-\frac{V}{T}dP+\sum _{i=1}^s\left(-\frac{{\mu }_i}{T}\right)d{N}_i,}$

${\displaystyle \mathrm{\Xi }=\mathrm{\Xi }(T,P,\{N_i\}).}$