54.3 Electron Equilibrium in Neutron Stars 417 where without lack of generality it is assumed that only ν k is negative. (Of course we choose the smallest possible integral values for the |ν i |). The concentration ratios c i then result from (54.3) and (54.4) according to the principle that the entropy should not change in course of the equilibrium reaction. In this way from the related free enthalpy condition one obtains the Law of Mass Action : c ν 1 1 · c ν 2 2 · · c ν k−1 k−1 c −ν k k ≡ f(T,p) , (54.5) where from the ideal gas equation (see the preceding subsection) for the yield function f(T,p) with a pressure unit p 0 the following general expression results: f(T,p)= k i=1 c ν i i ≡ k i=1 p p 0 −ν i · e −ν i ˜g (0) i (T ) k B T . (54.6) For further simplification we have written g (0) i (T,p)=˜g (0) (T )+k B T · ln p p 0 , which agrees with s i = k B ln 5k B T 2p + . By varying pressure and/or temperature one can thus systematically shift the reaction yield in accordance with (54.6) (e.g., increasing the pressure leads to an increased fraction of components with negative ν i ). As an application and generalization we shall presently treat the osmotic pressure and the decrease in boiling and solidification temperatures of a liquid by addition of sugar and de-icing salt. Firstly, however, an unusual topic. 54.3 Electron Equilibrium in Neutron Stars Consider the quantitative answer to the following question: How many elec- trons (or protons!) are there in a neutron star? Firstly we need to know the fraction: N electrons = N protons ∼ = 10 −5 ·N neutrons ; i.e., the ratio is extremely small, although different from zero. Of course, all absolute numbers would be extremely large 3 . 3 How large is N e ? One can estimate this number by inserting typical values for the radius of a neutron star (≈ 10 km) and of a neutron (≈ 10 −13 cm). 418 54 Applications II: Phase Equilibria in Chemical Physics The above result is then obtained. Consider the relevant astrophysical equilibrium reaction, which is the equilibrium inverse β-decay process: pp + ee ↔ nn + νν e Here pp stands for the proton 4 , nn for the neutron, ee for the electron and νν e for the electron-neutrino. (In a β-decay process a neutron emitting an electron plus an electron-antineutrino νν e decays as follows: nn → pp + ee + νν e ;theinverseβ-decay is the fusion of a proton and an electron under very high pressure, as high as is typical for a neutron star, or even higher, into a neutron plus an outgoing electron-neutrino, i.e., essentially according to the ← part of the previous reaction equation. The electron-neutrino and electron-antineutrino are particles and antiparticles with (almost) vanishing rest mass, i.e., negligible for the present thermodynamics (see below).) In any case, the first reaction can actually be “equilibrated” in both directions; in neutron stars there is thermodynamic equilibrium as above, and additionally the temperature can be neglected, since one is dealing with a degenerate Fermi gas. The equilibrium condition is i ν i · μ i (T,p,N 1 , ,N k ) ! =0, where the μ i (T,p,N 1 , ) are the respective chemical potentials, i.e., the free enthalpies per particle for the particle considered, and the ν i the reaction numbers (not to be confused with the neutrinos). The chemical potentials of neutrino and anti-neutrino, as already men- tioned, can be neglected in the present context. However, for the other parti- cles under consideration, the chemical potential at zero temperature (and in this approximation, our neutron stars – and also “ white dwarfs ”, see above, can always be treated at the temperatures considered, viz as a degenerate Fermi gas ) is identical with the nonrelativistic kinetic energy per particle: μ i ≈ ε kin. i = 2 2M n 2 3 V , as was shown for the electron gas in metals (n V is the number density and M the mass of the considered particles). Thus − (n V ) 2/3 neutron M neutron + (n V ) 2/3 proton M proton + (n V ) 2/3 electron m electron ! =0. (54.7) 4 We write pp (instead of p) for the proton, nn (instead of n) for the neutron, ee for the electron, and νν for the neutrino, to avoid confusion with the pressure p, the particle density n or the particle number N as well as with the elementary charge e and the reaction numbers ν i mentioned above. 54.4 Gibbs’ Phase Rule 419 However, the electron mass, m e , is 2000 times smaller than the proton mass M P (≈ neutron mass M N ), whereas the number densities of electrons and protons are equal. Thus the second term of the previous equation can be neglected. As a consequence (n V ) electron (n V ) neutron ≈ m e M N 3 2 ≈ 10 −5 , as stated. 54.4 Gibbs’ Phase Rule In an earlier section we considered the case of a single component (K =1, e.g., H 2 O) which could exist in three different phases (P = 3, solid, liquid or vapor.) In contrast, in the second-from-last subsection we treated the case of two or more components (K ≥ 2) reacting with each other, but in a single phase (P = 1), according to the reaction equation k i=1 ν i A i =0. We shall now consider the general case, and ask how many degrees of freedom f, i.e., arbitrary real variables, can be chosen, if K different compo- nents are in thermodynamic equilibrium for P different phases. The answer is found in Gibbs’ phase rule: f ≡ K − P +2. (54.8) As a first application we again consider the (p, T ) phase diagram of H 2 O with the three phases: solid, liquid and vapor. Within a single phase one has f ≡ 2(=1− 1 + 2) degrees of freedom, e.g., G = G(T,p); on the boundary lines between two phases there remains only 1 degree of freedom (= 1−2+2) (e.g., the saturation pressure is a unique function of the temperature only, and cannot be varied by choosing additional variables), and finally at the triple point we have f = 0, i.e., at this point all variables, temperature and pressure are completely fixed (f =1−3+2≡ 0). As a second application consider a system with K = 2, e.g., two salts. In one solvent (i.e., for P = 1) in thermal equilibrium one would have f = 2 − 1 + 2 = 3 degrees of freedom. For example, one could vary T , p and c 1 ,whereasc 2 =1− c 1 would then be fixed. If there is thermodynamic equilibrium with P = 2 phases, e.g., a solid phase plus a fluid phase, the number of degrees of freedom is reduced to f = 2; e.g., only T and p can be varied independently, in contrast to c i . If there is thermodynamic equilibrium of the two components in three phases, then one has only one free variable (f =1). 420 54 Applications II: Phase Equilibria in Chemical Physics We shall now proceed to a proof of Gibbs’ phase rule: a) Firstly, consider a system with P =1.ThenK + 1 variables can be freely chosen, e.g., T , p, c 1 , c 2 , ,c k−1 ,whereasc k =1−c 1 −c 2 − −c k−1 is dependent; thus the Gibbs’ phase rule is explicitly satisfied: f = K −1+2. b) Now let P ≥ 2 ! Then at first one should consider that for every compo- nent, k =1, ,K,thereareP −1 additional degrees of freedom, i.e. the ratios c (2) k /c (1) k ,c (3) k /c (1) k , , c (P ) k /c (1) k . Thus one has (P −1)·K additional variables. But there are also (P −1)·K additional constraints, i.e. μ (2) k ! = μ (1) k , , μ (P ) k ! = μ (1) k . c) A last sequence of constraints must be considered: p (2) k ! = p (1) k , , p (P ) k ! = p (1) k . This gives P − 1 constraints on the pressure. Hence f = K − 1+2+(P −1) · K − (P − 1) ·K −(P − 1) ≡ K −P +2, q.e.d. 54.5 Osmotic Pressure Again assume, as in section 54.2, that we are dealing with a semipermeable membrane (as commonly occurs in biological cells). Let the semipermeable membrane be nonpermeable for the solute,2,(ˆ= salt or sugar), but permeable for the solvent, 1. Now consider a U-tube, which is separated into two parts at the center by a semipermeable membrane and filled with a liquid (water) to different heights in the respective parts. In the left-hand part of the U-tube the water level is h; in the enriched right-hand part (enriched by salt or sugar, Δc 2 > 0) the level is enhanced, h + Δh,withΔh > 0. This corresponds to a pressure difference 5 Δp, the so-called osmotic pres- sure p osmotic p. ,whichisgivenby Δp ≡ p osmotic p. = Δc 2 · Nk B T V = ΔN 2 V k B T. (54.9) 5 The solvent concentration is only slightly diminished on the right-hand side, in favor of the enhanced solute concentration on this side, see the text. 54.5 Osmotic Pressure 421 Fig. 54.2. Osmotic pressure (schematically). The volume V is divided into two parts by a semiper- meable membrane (the Y -axis between 0 and 1) . The membrane is permeable for the solvent (− sym- bols), but non-permeable for the solute molecules ( symbols). As a result, the pressure on the r.h.s. is enhanced by an amount called the osmotic pres- sure Δp. (This expression is analogous to an effective ideal gas of ΔN 2 solute molecules suspended in a solvent. For a true ideal gas the molecules would be suspended in a vacuum; but there the pressure does not depend on the mass of the molecule. This is also true for the present situation.) Figure 54.2 above schematically shows a semi-permeable membrane through which solute molecules cannot pass. In the above derivation, it does not matter that the vacuum mass of the molecule is replaced by an effective mass, for which, however, the value is unimportant if the solute concentrations c 2 (on the l.h.s.) and c 2 + Δc 2 (on the r.h.s.) are small enough, such that only the interactions with the solvent come into play. (These arguments make a lot of ‘microscopic’ calculations unnecessary.) A precise proof again uses the entropy of mixing and the partial pres- sure p i : s i (T,p i )=s i (T,p) −k B ·ln c i . Thus we have the molecular free enthalpy g i (T,p i )=g (0) i (T,p)+k B T · ln c i , and because of the equality of chemical potential and the molecular free enthalpy: μ i (T,p,c i )=μ i (T,p)+k B T · ln c i . The two equilibrium conditions (not three!) for each side of the semiper- meable membrane are: (i) μ 1 (T,p,c 1 (= 1 − c 2 )) ≡ μ 1 (T,p + Δp, c 1 − Δc 2 ) , and (ii) T 1 ≡ T 2 (= T ) . Thus g (0) 1 (T,p)+k B T ·ln(1−c 2 ) ≡ g (0) 1 (T,p)+Δp· ∂g (0) 1 ∂p +k B T ·ln(1−c 2 −Δc 2 ) . 422 54 Applications II: Phase Equilibria in Chemical Physics With ∂g (0) 1 ∂p = v 1 and the linearizations ln(1 − c 2 ) ≈−c 2 and ln(1 −c 2 − Δc 2 ) ≈−c 2 − Δc 2 we obtain for the osmotic pressure: Δp = k B T v 1 Δc 2 , and finally with v 1 := V N 1 ≈ V N 1 + N 2 , for N 2 N 1 : Δp ≈ ΔN 2 V k B T. This derivation is, in principle, astonishingly simple, and becomes even simpler by using the “effective mass” argument above; a microscopic statistical- mechanical calculation would, in contrast, be unnecessarily complicated. 54.6 Decrease of the Melting Temperature Due to “De-icing” Salt In the preceding section the addition of sugar on the nonpermeable side (nonpermeable for the sugar but not for the solvent) of a semipermeable membrane led to a pressure difference, i.e., a higher pressure, higher by the osmotic pressure, on the sugar-enriched side. However the temperatures were identical on both sides of the interface. In contrast, we now consider the changes in the melting temperature of ice (and the boiling temperature of a liquid) by addition of soluble substances, e.g., again some kind of salt or sugar, to the liquid phase 6 , i.e., the interface is here the surface of the liquid. We thus consider, e.g., the phase equilibria A) solid-liquid and B) liquid- vapor, i.e., a) without addition and b) with addition of the substance consid- ered. For simplicity we only treat case A). In case Aa) we have: μ solid 1 (T,p,c 1 =1) ! = μ liquid 1 (T,p,c 1 =1), whereas for Ab): μ solid 1 (T −ΔT,p,c 1 =1) ! = μ liquid 1 (T −ΔT,p,c 1 =1−Δc 2 ) . 6 We assume that the added substance is only soluble in the liquid phase, but this can be changed, if necessary. 54.7 The Vapor Pressure of Spherical Droplets 423 Forming a suitable difference we obtain − ∂μ solid 1 ∂T ΔT = − ∂μ liquid 1 ∂T ΔT + k B T ·ln(1 − Δc 2 ) , andwithln(1−x) ≈−x we have: ΔT = − k B T ·Δc 2 ∂μ l. 1 ∂T − ∂μ s. 1 ∂T . Hence one obtains, with μ = G N and ∂G ∂T = −S, as well as S N ≡ s : ΔT = − k B T · Δc 2 s s. 1 − s l. 1 ≡ + k B T 2 · Δc 2 l s.→l. , (54.10) where additionally the molecular heat of melting l s.→l. := T · s l. 1 − s s. 1 has been introduced. As a result we can state that the melting temperature (and analogously the boiling temperature) of the ice (and of the heated liquid) is decreased by the addition of salt (or sugar) to the liquid phases. In the boiling case the relevant equation is μ l. 1 (T −ΔT,p,c 1 − Δc 2 ) ! = μ vap or 1 (T − ΔT,p,c 1 )) . In this case too it is essentially the enhancement of the entropy by mixing which is responsible for the effect. 54.7 The Vapor Pressure of Spherical Droplets Now consider an ensemble of spherical droplets of radius R with a sufficiently large value of R such that the number N of particles within a droplet is 1. Outside such droplets the saturation pressure p R is enhanced w.r.t. the value p ∞ for a planar surface, i.e., for R →∞. For the Helmholtz free energy of a droplet we obtain dF = −p ·dV + σ ·dO −S · dT, with dV =4πR 2 ·dR and dO =8πR · dR, where dV is an infinitesimally small increment of volume and dO an incre- ment of surface area of the droplet. 424 54 Applications II: Phase Equilibria in Chemical Physics The above relation for dF defines the surface tension σ,anenergy per surface area. This is best explained in the framework of the physics of soap bubbles. 7 We thus obtain for the Helmholtz free energy of the droplet: F droplet (T,R)=N droplet ·f(T,R)+4πR 2 σ(T,R) . Here f(T,R) represents the volume part of the Helmholtz free energy per atom, and σ(T,R) is the above-mentioned surface tension, which we now want to calculate. Thermal phase equilibrium gives μ liquid droplet = ∂G ∂N = f + p R ·v l. +4πσ · ∂R 2 ∂N ! = μ vap or . With N ≡ 4πR 3 3v l. , hence dN =4πR 2 dR v l. and dR 2 =2R ·dR, it follows that: μ liquid = f + v l. · p R + 2σ R ! = μ vap or ≡ k B T ·ln p R p 0 + B(T ) , 8 where p 0 is an arbitrary unit pressure and B(T ) a temperature-dependent constant of physical dimension energy per atom. For R<∞ we thus have: f + v l. · p R + 2σ R = k B T ·ln p R p 0 + B(T ) , and for R = ∞: f + v l. ·p ∞ = k B T · ln p ∞ p 0 + B(T ) . By subtracting the second equation from the first, we obtain: k B T · ln p R p ∞ = v l. · 2σ R + p R − p ∞ , and neglecting (p R − p ∞ ) 9 : p R p ∞ ≈ e v l. k B T · 2σ R . (54.11) 7 For the physics of soap bubbles see almost any textbook on basic experimental physics. 8 The last term on the r.h.s. is reminiscent of the analogously defined free enthalpy of mixing: μ vapor = k B T · ln c i + ,withthe partial pressure p i = c i · p. 9 The result reminds us (not coincidentally) of the Clausius-Clapeyron equation. 54.7 The Vapor Pressure of Spherical Droplets 425 The saturation vapor pressure outside a spherical droplet of radius R is thus greater than above a planar surface. With decreasing radius R the tendency of the particles to leave the liquid increases. A more detailed calculation shows that a critical droplet radius R c exists, such that smaller droplets, R<R c , shrink, whereas larger droplets, R>R c , increase in size. Only at R c is there thermal equilibrium. For R c the following rough estimate applies: σ ·4πR 2 c ≈ k B T. In this context the analogy with the Ising model (cf. (42.5)) is again help- ful. For a cubic lattice with nearest-neighbor separation a and ferromagnetic nearest-neighbour interaction J(> 0)) in an external magnetic field h 0 (> 0) one has: Ising model: H = −J |r l −r m |=a s l s m − h 0 l s l , (54.12) with s l and s m = ±1, corresponding to ↑ and ↓ spins. We shall now consider the following situation: embedded in a ferromag- netic “lake” of ↓ spins is a single “island” (approximately spherical) of ↑ spins. The volume V of the 3d-“island” is assumed to be V ≈ 4πR 3 3 , while the surface area of the island is O ≈ 4πR 2 . The energy of formation ΔE of the island in the lake is then given by ΔE ≈ O a 2 ·2J − V a 3 ·2h 0 , as one can easily see from a sketch. The first term on the r.h.s. of this equation (∝ J) corresponds to the surface tension, while the second term (∝ h) corresponds to the difference of the chemical potentials, μ R − μ ∞ .FromΔE ! = 0 it follows that R c =3aJ/h 0 . In this way one can simultaneously illustrate nucleation processes (e.g., nucleation of vapor bubbles and of condensation nuclei) in the context of over- heating or supercooling, e.g., in the context of the van der Waals equation. The analogy is enhanced by the above lattice-gas interpretation. 10 10 A reminder: In the lattice-gas interpretation s l = ±1 means that the site l is either occupied, (+), or unoccupied, (-). 55 Conclusion to Part IV For all readers (not only those who understand German 1 or wish to prac- tise their knowledge of the language) for all four parts of our compendium there are (translated) exercises to complement this book. These are sepa- rately documented on the internet, see [2], and purposely not integrated into this script. In Thermodynamics and Statistical Physics both types of learning mate- rial, i.e., textbook and corresponding exercises, have been centered around a) Phenomenological Thermodynamics, with the four quantities F (T,V,N, )(Helmholtz free energy) , U(T,V,N, )(internal energy) , S(T,V,N, )(entropy) and the absolute (or Kelvin) temperature T ,and b) Statistical Physics, for historical reasons usually called Statistical Mechan- ics, although this name is unnecessarily restrictive. a) and b) are closely interdependent, due essentially to the fundamental relation F (T,V,N, ) ≡−k B T ·ln Z(T,V,N, ) , (55.1) where Z(T,V,N, )= i e − E i (V,N, ) k B T (55.2) is the partition function. This relation between a) and b), (55.1), applies for so-called canonical ensembles, i.e., when the particle number N is fixed and the heat bath only exchanges energy with the system considered – defining the reciprocal Kelvin temperature β = 1 k B T as conjugate parameter (conjugate to the energy). 1 Sometimes even a partial understanding may indeed be helpful. . temperature (and in this approximation, our neutron stars – and also “ white dwarfs ”, see above, can always be treated at the temperatures considered, viz as a degenerate Fermi gas ) is identical. outside a spherical droplet of radius R is thus greater than above a planar surface. With decreasing radius R the tendency of the particles to leave the liquid increases. A more detailed calculation. is a unique function of the temperature only, and cannot be varied by choosing additional variables), and finally at the triple point we have f = 0, i.e., at this point all variables, temperature