Thermodynamics

Thermodynamic potentials

We shall write below the I principle of thermodynamics, assuming reversible transformations, so that {$dQ=TdS$}.

We are assuming for the moment that the thermodynamic variables are four, e.g. {$V,P,T,S$}, as in a gas. A similar treatment may be extended to the case of more than four variables. Since there is an equation of state relating them, this fixes one of them. The first principle itself is an equation for the differentials, so that only two variable are left independent, plus the value of entropy at a given point (fixed somehow by the third principle). The choice of which two are considered independent is arbitrary, or, better, fixed by experimental conditions.

Therefore the first principle

{$$d E = TdS - PdV $$}

states that internal energy is a function of entropy and volume, {$E = E(S,V)$}. This means that if we want to measure directly internal energies we shoeld devise an experiment where we directly control volume and entropy.

The Helmholtz free energy is defined as

{$$F = E-TS$$}

to remove the contribution of the so-called lost work. Differentiating {$$F$$} and substituting the first principle one gets

{$$dF = dE -TdS -SdT = -PdV-SdT$$}

and, by the same token, the Helmholtz free energy is a function of volume and temperature, {$F=F(V,T)$}. Experiments allowing direct control of these two variables are directly detemined by Hemholtz free energy.

Gibbs free energy, that removes the effects of work on the rest of the world, is defined as

{$$G = F + PV$$}

Its differential is {$dG = dF +PdV + VdP = VdP-SdT$}, hence {$G=G(P,T)$}. {$G$} and {$F$} are most commonly used in condensed matter variational problems. Finally enthalpy is defined as

{$H = E+PV$}

to remove work, keeping lost work. The differential id {$dH = TdS + VdP$}, hence {$H=H(S,P)$} turns out to be the most relevant for chemical reactions, where volume and temperature are controlloed by the reactions itself and cannot be independently varied.

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External fields

It is useful to consider coupling of condensed matter systems in external fields. This is the typical case in which the thermodynamic system acquires additional degrees of freedom, i.e. additional thermodynamic variables, other than {$P,V,T,S$}. Typically we have the external field electric displacement {$\mathbf D$} coupled to dielectric polarization {$\boldsymbol {\cal P}$}, yielding additionally the electric field {$\mathbf E = (\mathbf D + \boldsymbol {\cal P})/\epsilon_0$}. Similarly the external magnetic field {$\mathbf H$} couples to magnetization {$\mathbf M$} to yield the magnetic induction {$\mathbf B=\mu_0(\mathbf H+\mathbf M)$}.

Fields are thermodynamic variables according to the assumptions of electromagnetism in continuous matter: we treat the mean fields (averages over a much larger scale than the atomic one), in the same way that we treat the average force applied by gas on a piston, i.e. pressure times area.

In the simple ideal geometry of an infinite solenoid we have {$H=\frac N L I$} by Ampere's law (see figure). Hence in the magnetic case, for instance, infinitesimal work is {$dW = \epsilon I dt= - NAdBI$} where the minus sign comes from Lenz's law. Recalling that {$dB = \mu_0(H +M)$} and that {$dW_{ps}= -A\mu_0dH HL= \mu_0VH^2$} is the work done in the power supply, we have the following expression for the work done by the system

{$$dW = -\mu_0V H dM$$}

Ignoring work done by pressure in a solid we have therefore {$dE = TdS - dW = TdS + \mu_0VHdM$}, whence in the magnetic case the internal energy is {$E=E(S,M$} and the Gibbs energy is

{$$G(T,H) = E(S,M) -TS-HM$$}

with {$dG = -SdT -\mu_0VMdH$}.

Maxwell relations

The figure shows the summary of the differentials of the four thermodynamic potentials and their independent thermodynamic variables. By comparing their standard differential to these expression one obtains the following definitions of the thermodynamic variables: from {$dE$}

{$$T=\left(\frac {\partial E}{\partial S}\right)_V\qquad P=-\left(\frac {\partial E}{\partial V}\right)_S$$}

from {$dF$}

{$$S=-\left(\frac {\partial F}{\partial T}\right)_V\qquad P=-\left(\frac {\partial F}{\partial V}\right)_T$$}

from {$dG$}

{$$V=\left(\frac {\partial G}{\partial P}\right)_T\qquad S=-\left(\frac {\partial G}{\partial T}\right)_P$$}

from {$dH$}

{$$V=\left(\frac {\partial H}{\partial P}\right)_S\qquad T=\left(\frac {\partial H}{\partial S}\right)_T$$}

Other useful thermodynamic relations

From the definition of specific heat, {$c=dQ/dT$}, since {$dQ=TdS$} one has, starting from the differential of {$E$} and using the Maxwell relations

{$$c_V = \left(\frac {\partial E}{\partial T}\right)_V = T\left(\frac {\partial S}{\partial T}\right)_V = -T\left(\frac {\partial^2 F}{\partial T^2}\right)_V $$}

Viceversa, startng from the differential of {$H$} one has

{$$c_P = \left(\frac {\partial H}{\partial T}\right)_P = T\left(\frac {\partial S}{\partial T}\right)_P = -T\left(\frac {\partial^2 G}{\partial T^2}\right)_P $$}

We further recall the definitions of the thermal expansion coefficient

{$$\beta_{P,S} = \frac 1 V \left(\frac {\partial V}{\partial T}\right)_{P,S}$$}

from which the linear expansion coefficient is {$\alpha = \beta/3$} and the bulk modulus (inverse compressibility) is

{$$ B_{T,S} = -\frac 1 V \left(\frac{\partial V}{\partial P}\right)_{T,S}$$}

(*) Notice that we follow Khomski and Ashcroft, calling {$B$} the bulk modulus. Wikipedia calls it {$K$} and {$\beta$} its inverse. This wiki defined the bulk modulus as {$\kappa$} and its inverse like Wikipedia in section Termodinamica and Onde