Essentials of Non-Equilibrium Themodynamics of macroscopic systems

Transcription

1 Essentials of Non-Equilibrium Themodynamics of macroscopic systems Diego Frezzato Part of the course Theoretical Chemistry A.A. 2014/15 I like to thank Prof. Giorgio J. Moro for having kindly provided past notes as a guide to prepare these lectures.

2 Objective: Outlines of methods and tools used to describe macroscopic systems in non-equilibrium conditions and their dynamics (relaxation towards equilibrium) Differently from other disciplines (like electro-magnetism theory based on Maxwell equations), Themodynamics does not intend to give a general description of systems, rather to provide general methodologies to achieve the description of a particular system under given conditions [G. J. Moro] References for insights - S. R. de Groot, P. Mazur, Non-equilibrium Thermodynamics (Dover, 1984) - K. Kondepudi, I. Prigogine, Modern Thermodynamics (Wiley, 1998) - R. B. Callen, Thermodynamics and an introduction to Thermostatistics 2 Ed., Chapter 14 (Wiley, 1985) - I. Prigogine, From being to becoming. Time and complexity in the physical sciences (Trad. it. Dall essere al divenire, Einaudi 1986)

3 Equilibrium Thermodynamics: basis for the build-up of the non-equilibrium theory of macroscopic systems

4 The subject: macroscopic systems with a number of molecules of the order of N. Avogadro BASIC TERMINOLOGY Closed system: no exchanges of matter with the exterior (environment in a broad sense) [Open system: it can exchange matter with the exterior ] Isolated system: no exchanges of matter nor energy with the exterior Adiabatically isolated system: no exchanges of heat (nor matter) with the exterior Work: energy exchanged in a controlled way by involving specific degrees of freedom of the system, and by following an established protocol. Heat: energy exchanded in an uncontrolled (un-specific) way

5 Equilibrium Thermodynamics deals with characterization of equilibrium states of the system Stationary state = physical state of the system which is invariant in time Equilibrium state = physical state of the system which is invariant in time in the absence of energetic exchanges (heat/work) with the exterior The specification in the absence of energetic exchanges makes the distinction between equilibrium and stationary states. Stationary states are possible under non-equilibrium conditions (e.g., consider a system in contact with two thermostats at different temperatures, with a heat flux through the system ).

6 Bénard cells

7 A thermodynamic state is specified by a set of state variables: some measurable macroscopic properties at equilibrium. Extensive properties (they depend on the extension of the system): Mass, M Volume, V Internal energy, U Entropy, S etc. Intensive properties (independent of the extension of the system): Temperature, T Pressure, p Several densities (per unit of volume o per unit of mass) Composition variables (if we deal with mixtures) etc.

8 Densities that we shall use in what follows: Volumetric mass-density, ρ = M V Densities of U, S, V (per unit of mass) : u = U / M s = S / M v = V / M = 1/ ρ State functions: functions of the state variables. They can be state variables expressed as function of the independent state variables; e.g., ρ = ρ( T, p). Fundamental differential for a closed system. It concerns the internal energy U ( S, V ) (entropy and volume are natural variables for U): du = TdS pdv In terms of the densities above introduced: du = Tds pdv = Tds p d(1/ ρ) 2 = Tds + ( p / ρ ) d ρ

9 TRANSFORMATIONS Reversible transformation: the ideal limit case of a system which passes through intermediate states of (quasi) equilibrium; it is also called quasi-static since infinitely slow processes allow the system to remain (almost) at equilibrium. In such a limit one can specify the state variables (e.g., T e p). Irreversible transformation: any transformation of the system that takes place in a finite time. All spontaneous processes are irreversible transformations. Spontaneous process: under externally imposed constraints, the system evolves towards an equilibrium state which is compatible with these constraints. The evolution takes place over change of the uncontrolled (unconstrained) degrees of freedom. Example: a volatile liquid is initially contained in a closed bottle (initial equilibrium state) in a closed room; the bottle is opened, the liquid evaporates and diffuses up to occupy homogeneously the whole volume at disposal (spatial constraint imposed by the walls of the room): this is the final equilibrium state.

10 PRINCIPLES OF (EQUILIBRIUM) THERMODYNAMICS 1 PRINCIPLE: conservation of the total energy E of the system Let E = EKin + Epot + U be the total energy of the system (with E Kin kinetic energy, E pot potential energy due to external force-fields, U the internal energy). Variations of E are due to interactions with the exterior which imply exchanges of energy in terms of heat (q) and work (w), and it holds E = q + w If kinetic and potential energies are null or constant, we can focus only on variations of internal energy: U = q + w. This is the usal form of the 1 Principle in chemical contexts.

11 2 PRINCIPLE: entropy S and its production a) Definition of the state function called entropy For a system at equilibrium one defines that extensive function, called entropy, S, such that its variations between any two equilibrium states can be evaluated by integrating δ q ds = T rev along any reversibile transformation ( rev ) that connects the two states. b) Entropy production Let us consider a system which passes from an initial equilibrium state to a final equilibrium state. If such a transformation is an irreversible adiabatic process ( ad ), then S ad > 0 with Sad = Sfinal Sinitial the variation of the system s entropy. If the transformation is reversible, Sad = 0.

12 Adiabatically isolated system (driven/induced) transformation: - driven by an external apparatus - due to change of external constraints (free) relaxation initial equilibrium state non-equilibrium state at the end of the irreversible transformation final equilibrium state Sad = Sfinal Sinitial 0 if the transformation is reversible (the non-equilibrium states are quasi-equilibrium states): - quasi-equilibrium with the external apparatus - external constraints are slowly changed (examples?)

13 Notice that: 1 e 2 Principles refer to transformations bewteen equilibrium states! In fact they regard state functions (E and S). The 2 Principle, even if nothing is said about the (non-equilibrium) intermediate states, nor about the system s dynamics, contains information (namely a constraint) about the evolution of the system: emergence of a time arrow. Time arrow: it gives the direction of spontaneous processes; a system which is adiabatically isolated evolves spontaneously towards states (of final equilibrium) of higher entropy, hence time elapse and entropy increase goes together. q q lower entropy pre spontaneous transf. time elapse higher entropy post [ The Universe is an isolated system under spontaneous evolution: the entropy of Universe increases (think critically to this popular statement ) ]

14 The fact that Sad is positive-valued allows one to determine the final equilibrium state that will be reached under adiabatic conditions: it will be the state corresponding to the system configuration ξ that maximizes the entropy S( ξ constrains) as function of the uncontrolled degrees of freedom ξ but under the constraints externally imposed. eq In the example of the bottle of volatile liquid: Constraints = room volume, T ξ = spatial distribution of the species ξ eq corresponds to the homogeneous distribution Example: thermal equilibrium between two bodies in contact (but adiabatically isolated from the environment) (0) T 1 (0) T 2 T m

15 Simplifying assumptions: - identical bodies: same material, same volumes ( V 1 = V 2 = const. ) taken to be constant under temperature variations, and (same) specific heat c v taken to be constant under temperature variations. From the constant-volume heat capacity C v - no work is perfomed q = w = U : = T 0 0 V it follows U1 Cv T1, U2 Cv T2 = = with imply U = U1 + U2 = 0 T = T T (0) (0) T = T T T + = 1 T2 0 T m (0) (0) T1 + T2 T1 + T = = 2 = const. 2 2

16 System s degrees of freedom: T 1, T 2 Constraint: T m = const. Single constrained degree: ξ T = T2 T1 Seach for the equilibrium state: T that maximizes S( T ) at the given m T. Entropy variation of versus temperature at constant volume: rev, T ( δ q ) S( T, V ) S( T, V ) =, δ q du = C dt ' ref T ref T dt ' T S( T, V ) = S( Tref, V ) + Cv = S( Tref, V ) + Cv ln T ' T T ref T ' rev (assumed heat capacity independent of temperature) T is any reference temperature. ref ref v For each body (i =1, 2) : T Si ( Ti ) = Si ( Tm ) + Cv ln i T m

17 The total entropy is S( T ) = S1( T1 ) + S2( T2 ). In terms T : S( T ) = S1( Tm T / 2) + S2( Tm + T / 2) T T / 2 T / 2 ( ) ln ( ) ln m + T = S T + C + S T + C Tm m 1 m v 2 m v Tm 2 2 Tm ( T / 2) 1( m) 2( m) v ln 2 Tm 2 1( m) 2( m) v ln 1 ( / 2 m ) = S T + S T + C = S T + S T + C T T S( T ) = S(0) + Cv ln 1 ( T / 2 Tm ) 2 with S(0) = S1( Tm ) + S2( Tm ) Equilibrium state spontaneously reached: T = 0, that is, the two bodies at equal temperature [see figure]. If we want to end up with an equilibrium state with T 0, the only way is to intervene by stopping the heat flux: in this case we must apply a constraint on T (it is no more an uncontrolled degree of freedom!).

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19 Remark. In the chemical contexts, the thermodynamic equilibrium is mainly characterized at constant (p, T) or at constant (V, T) (not under adiabatic conditions). The leading equilibrium condition above seen is exploited by considering the isolated super-system made of system + thermostat (+ barostat). One ends up with the following reformulations of the 2 Principle: (p, T) constant: G w wvol T constant: A w G = H T S (Gibbs free-energy of the system, with H = U + pv the entalpy) A = U T S (Helmholtz free-energy) w vol : work-of-volume performed by the external pressure, i.e., δ vol ext w = p δv. Equilibrium conditions: (p, T) constant, allowed only work-of-volume: ξ eq is point of minimum of Gξ ( ) (V, T) constant, no work: ξ eq is point of minimum of Aξ ( )

20 First steps in Non-Equilibrium Thermodynamics: 1) near-equilibrium postulate 2) volumetric densities of state functions 3) conservative and non-conservative properties 4) entropy production rate

21 Let us consider a (for simplicity) closed system at equilibrium. Let us choose, for example, the intensive variables T and ρ to specify the thermodynamics state. Densities of extensive state functions are expressed as functions of T and ρ. For example, u ( T, ρ), s ( T, ρ ), etc. eq eq (subscript eq recalls that they refer to an equilibrium state). Out-of-equilibrium, the intensive variables may vary in space and in time. We assume they are still measurable. Time-dependent scalar fields are requited to specify the local state of the system: T ( r, t), ρ ( r, t), r = ( x, y, z) (Near-equilibrium) postulate about state functions: under the assumption of local quasi-equilibrium, one introduces the following scalar fields for the densities of the state functions: u( r, t) : = u ( T ( r, t), ρ( r, t)), s( r, t) : = s ( T ( r, t), ρ( r, t)), etc. eq Near-equilibrium postulate eq

22 Volumetric densities of state functions Let us consider a spatial domain Ω, delimited by the (oriented) border Ω. Be P some extensive property, and PΩ ( t) its value in such a domain at time t. Let us assume to be able to define a volumetric density ρ ( r, t), associated to the property, such that P P ( t) = dv ρ ( r, t) Ω For example, the internal energy of the system s portion Ω is given by Ω U U ( t) = dv ρ ( r, t) Ω Ω r Ω n(r) ^ spatial domain outer normal versor Ω oriented border P

23 Conservative and non-conservative properties P Density-flux vector j ( r, t) associated to the property P : given an oriented surface element which contains the point r, of area da, and whose orientation is specified by the normal vector nˆ( r ), one has that P da nˆ( r) j ( r, t) = flux across the surface element: rate of trasfer (amount per unit of time) of the property P through the surface element, in the specified direction. Ω r n(r) ^ Ω The (net) flux across the border (a closed surface) Ω of the domain is obtained by integrating over all surface elements: da n(r) ^ j P (r,t) ˆ rate of transfer of property J P ( ) ( ) P (, ) = P Ω t = da n r j r t across the border of the domain r Ω Ω

24 Definition: An extensive property P is conservative if its variation, in a given domain, is only due to flux of the property through the border of the domain: P dpω ( t) ρ ( r, t) = dv dt t Ω Gauss Theorem P P = da n( r) j ( r, t) = dv j ( r, t) r Ω t 0 Ω conservative property dp ( ) ( ) ( ) Conservative property: Ω t P P lim Ω t + t P t = Ω J Ω ( t) dt t In the absence of fluxes at boundaries, the value of the property inside the domain does not change in time (i.e., it is conserved). divergence Ω dv P ρ ( r, t) t P + j ( r, t) = 0 must hold for anyω

25 P ρ ( r, t) P + j ( r, t) = 0 t continuity equation for a conservative property On the contrary, non-conservative extensive properties are those for which the variation can be also due to internal processes (creation/consumption) that take place inside the domain: dp ( ) Non conservative property: Ω t P = J Ω( t) + rate of internal creation/consumption dt Ω Ω Ω Ω conservative property: net change +/- only due to flux at the border non-conservative property: net change due to also to internal processes of production For example, mass, energy, components of linear and angular momenta are conservative properties. Entropy is a non-conservative property!

26 Entropy production rate Entropy is NOT a conservative property, hence the time variation of SΩ ( t) is due to both entropy-flux at the (oriented) border Ω of the domain Ω, and internal processes of entropy production. ds ( t) (, t) Sɺ ( t) Ω ρ r Ω = dv dt t Let us adopt the following decomposition for the rate of entropy variation: Ω Sɺ ( t) Sɺ ( t) Sɺ ( t) rev irrev Ω = Ω + Ω where the two addends have to be assigned. S rev The quantity dt S ( t) ( ds ) ɺ has the meaning of entropy variation if the system Ω = were subjected to a reversible transformation. rev How do we assign such a term?

27 Temperature differences across the border Ω are infinitesimal, hence the heat is transferred in a reversible way through it. By setting J( r, t) = heat density-flux vector The following assignment is made: Gauss Theorem rev Sɺ Ω ( t) = da nˆ ( r) J( r, t) / T ( r, t) = dv J( r, t) / T ( r, t) r Ω Ω divergence [ ] The irreversible contribution is obtained by subtraction: Sɺ ( t) Sɺ ( t) Sɺ ( t) irrev rev Ω = Ω Ω S ρ ( r, t) = dv + t T t t Ω [ J( r, ) / ( r, )] (*)

28 Let us introduce the rate of entropy-density production, σ ( r, t), such that S irrev ( ) S Sɺ Ω t dv σ ( r, t ) Ω σ ( r, t) dv dt is the amount of entropy irreversibly produced in the volume element dv and in the time interval dt. S (**) By comparing (*) with (**) it follows σ S S ρ ( r, t) ( r, t) + J( r, t) / T ( r, t) t [ ] The 2 Principle imposes a constraint on the rate of entropy-density production due to internal irreversible processes.

29 irrev Consider that dt S ( t) ( ds ) ɺ is the entropy variation that would occur even if Ω = irrev the reversibile heat flux at boundaries were impeded. In this case, the portion Ω would be adiabatically isolated. By considering that Sad 0 (2 Principle), i.e. ( ds ) irrev > 0, it follows that irrev ( ) S Sɺ Ω t = dv σ ( r, t ) > 0 This inequality must hold for any choice of the domain Ω, hence it must be σ S Ω ( r, t) 0 any point, any time (equality holds for the system at equilibirum) [ In practice, the term production (in entropy production) means effectively creation ]

30 Thermal conductivity as an explanatory case

31 The system: a rigid and homogeneous body at constant density ρ (hence at constant volume), at rest. [The system may be anisotropic (like crystalline materials)] The temperature T is the sole independent state variable (at equilibrium, eq ); the densities of the relevant state functions (internal energy and entropy) are dueq ( T ) dseq ( T ) ueq ( T ), seq ( T ) with = T dt dt Out-of-equilibrium, only the time-dependent temperature (scalar) field is required to specify the actual state: T ( r, t), r = ( x, y, z) Assumption of local quasi-equilibrium: u( r, t) : = u ( T ( r, t)), s( r, t) : = s ( T ( r, t)) eq eq

32 Let us consider a spatial portion (a domain) Ω of the body; then U U ( t) = dv ρ ( r, t), S ( t) = dv ρ ( r, t) Ω Ω The internal energy is a conservative property, hence U U ρ ( r, t) t ρ ( r, t) : volumetric density of internal energy U Ω Ω U + j ( r, t) = 0 j ( r, t) : the vector density-flux of internal energy. S In the absence of work, like in this case, the internal energy flux-density coincides U with the heat flux-density J( r, t) : j ( r, t) J( r, t) Entropy production rate: S S ρ ( r, t) σ ( r, t) = + J( r, t) / T( r, t) t [ ]

33 The constant volume approximation allows us to turn from volumetric densities to densities per unit of mass: U ρ ( r, t) = ρ u( r, t), ρ ( r, t) = ρ s( r, t) S u(, t) ρ r t + J( r, t) = 0 ( ) S s(, t) σ (, t) ρ r r = + J( r, t) / T ( r, t) t Let us elaborate the partial derivative (, ) / T = T ( r, t) s r t t : s( r, t) seq ( T ( r, t)) dseq ( T ) T ( r, t) = t t dt t T = T ( r, t) [ ] 1 dueq ( T ) T ( r, t) 1 ueq ( T ( r, t)) 1 u( r, t) = = = T ( r, t) dt t T ( r, t) t T ( r, t) t dseq = dueq / T (no work)

34 By using ( ) : s( r, t) 1 u( r, t) 1 ρ = ρ = J( r, t) t T ( r, t) t T ( r, t) S s( r, t) σ ( r, t) = ρ + [ J( r, t) / T ( r, t) ] t 1 = J( r, t) + [ J( r, t) / T ( r, t) ] T ( r, t) 1 1 { T ( r, t) = J( r, t) + J( r, t) + J( r, t) T ( r, t) T ( r, t) In summary, the equations of the thermal diffusion: 1 u( r, t) ρ + J( r, t) = 0 t 1 S T ( r, t) σ ( r, t) = J( r, t) 0 u( r, t) = u ( T ( r, t)) eq 4 unkonwn fields: J ( r, t), J ( r, t), J ( r, t), T ( r, t) x y z Only 1 equation which links them, + 1 inequality (a constraint) We need further conditions (equalities)!

35 Simplification for the one-dimensional case : rod of conducting material xˆ Area A If A L, then it is licit to assume temperature homogeneity in the transverse planes (hence no components of heat flux on these planes): T ( r, t) T ( x, t), J ( r, t) J ( r, t) 0, J ( r, t) J ( x, t) J ( x, t) y z x x The general equations reduce to u( x, t) J ( x, t) ρ + = 0 t x 1 S T ( x, t) σ ( x, t) = J ( x, t) 0 x

36 The expression of the entropy production rate can be put in the following appealing form: σ S ( x, t ) = J ( x, t ) F ( x, t ) where F( x, t ) is named thermodynamic force : F( x, t) : = T ( x, t) x 1 Note that the thermodynamic force vanishes at equilibrium. Still we have 2 fields to be determined, that is J ( x, t ) and T ( x, t ) (or F( x, t )) but only one equation. We need a further equation! How do we proceed? There are two possibilities

37 1) Phenomenological procedure. The thermodynamic force F( x, t ) is seen as the cause, and the related flux J ( x, t ) as the effect. In this view, for slight displacements from equilibrium we assume a proportionality relation (linear regime) effect-cause: J ( x, t) = L F( x, t) where L > 0 is called transport coefficient (or dissipation coefficient). The positive sign is established since σ S 2 ( x, t ) = J ( x, t ) F ( x, t ) = L F ( x, t ) > 0 for F 0 only if L > 0 2) Formal procedure. Let us assume that, at any point and time, the value of the entropy production rate can be expressed as function of the local temperature and thermodynamic force; that is, it exists a function ˆ σ S ( T, F ) such that S σ ( x, t) = ˆ σ ( T, F) = S T T ( x, t) F = F ( x, t)

38 From the 2 Principle: a) ˆ σ S ( T, 0) = 0 (equilibrium) ˆ (, ) σ > b) ˆ S ( T, F 0) 0 (out-of-equilibrium) σ S T F has a minimum at F = 0 For small-amplitude deviations from equilibrium (small F in module) we can assume a local parabolic form; by truncating at the quadratic expansion of ˆ σ S ( T, F ) vs. F : ˆ σ S ˆ (, ) 1 ˆ (, ) (, ) ˆ S σ T F σ T F T F σ ( T,0) + F + F F F F = 0 F = 0 L( T ) F with the temperature-dependent transport coefficient defined by From S S 2 ˆ S σ ( ) : = 2 L T σ = J F it follows J = L F. 1 ( T, F) F 2 F = 0 2 > 0 S 2 to have a minumum

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40 With both procedures, one ends up with the further relation which links the two unknown fields T ( x, t ) and J ( x, t ) : J ( x, t) = L( T ( x, t)) F ( x, t) 1 T ( x, t) = L( T ( x, t)) x 1 T ( x, t) = L( T ( x, t)) 2 T ( x, t) x Coefficient of thermal (or heat) conductivity: λ ( T ) : = L( T ) T 2 Thermal (or heat) conductivity in 1D: u( x, t) J ( x, t) (#) ρ + = 0 t x (##) T ( x, t) J ( x, t) = λ( T ( x, t)) x Now we have 2 unknown fields and 2 equations (taking into account that the internal energy depends on temperature in some known way ).

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44 The final step: work out an independent equation for the temperature field. Consider a location x at two vicinal times t and t+ t. Having assumed constant density (contraction/dilatation neglected), the internal energy density changes only because the temperature varies in time: By recalling that the body s contraction/dilatation is neglected (constant volume approximation), from du ( T ) = c dt (c v is the specific heat, here assumed constant) we have u( x, t + t) u( x, t) = u ( T ( x, t + t)) u ( T ( x, t)) eq v eq [ ] u ( T ( x, t + t)) u ( T ( x, t)) = c T ( x, t + t) T ( x, t) eq eq v eq Taking the limit t 0 we get [ (, + ) (, )] u( x, t + t) u( x, t) cv T x t t T x t = t t T ( x, t) 1 u( x, t) (#) 1 J ( x, t) (##) 1 T ( x, t) λ( T ( x, t)) = = = t c t ρ c x ρ c x x v v v

45 If also λ is constant (not dependent on temperature), then T ( x, t) λ T ( x, t) = t ρ c 2 x v 2 Thermal diffusion coefficient (m 2 /s): D : = λ ρ c v Equation of thermal diffusion (Fourier): T ( x, t) T ( x, t) = D t 2 x 2 Notice the similariry with Schrödinger equation in Quantum Mechanins (even the tools for its solution are similar ). The solution depends on the specific initial and boundary condition.

46 Example. Analytic solution for an infinitely long cylinder, initially at temperature T L, and then at contact with a thermostat at temperature T 0 on one side (x = 0): T L T 0 at time t > 0 x = 0 T ( x, t) T ( T T ) erf = 0 + L 0 x Dt u 2 2 y erf ( u) : = dy e (Error function) π [ The solution holds in the limit L ] 0

47 Representation in dimensionless (scaled) units T 0 t τ * = scale t / = τ L scale 2 / D stationary state: linear variation of T along the rod

48 Generalization to the three-dimensional (3D) case u( r, t) ρ + J( r, t) = 0 t 1 S T ( r, t) σ ( r, t) = J( r, t) 0 u( r, t) = u ( T ( r, t)) eq The thermodynamic force: F( r, t) : = T ( r, t) 1 The entropy production rate expressed in terms of temperature and thermodynamic force: S σ ( r, t) ˆ σ ( T, F ) = = S T T ( r, t) F F( r, t ) S σ ( r, t) = J( r, t) F( r, t)

49 Taylor expansion (up to 2 nd order terms): ˆ σ S S ˆ (, ) 1 ˆ (, ) (, ) ˆ S σ T F σ T F T F = σ ( T, 0 ) + F + F F j j k F 2,, j F j x y z j, k x, y, z j F = = k F= 0 F= 0 2 S S T ˆ σ (, F ) = F L ( T ) F Dissipation matrix: L jk ( T ) : = 2 ˆ S 1 σ ( T, F) 2 F F j k F= 0 From this definition it follows: 1) L is a symmetric matrix ( L jk = Lkj from Schwartz identity on second derivatives ).

50 2) L must be a positive-definite matrix (i.e., its eigenvalues are strictly positive), so that ˆ σ S ( T, ) = ( T ) > 0 L v ( α) ( α) ( α) F 0. F F L F for any The eigenvectors form an orthogonal base (recall that L is symmetric), then we can use it to get the following decomposition of the thermodynamic force α = l v, α = 1, 2,3 α ( ) ( ) ( α ) F = F v v ( )( ) ( )( ) ( ) ( α ) ( β ) ( α ) ( β ) ( α ) ( β ) ( β ) ( α ) ( β ) F L F = F v F v v Lv = F v F v l v v α, β α, β 2 ( α ) ( α ) ( α ) = F v l > 0 for any F 0 l > 0 α α Eigenvalues of L must be positive. δ α, β 3) The symmetry of matrix L must reflect the physical symmetry of the system, since the derivative are computed at equilibrium (F = 0). Examples: L 0 0 L = 0 L L isotropic system (reference to any system of axes) L 0 0 L = 0 L L // uni-axial system (on the principal system of axes)

51 Consequent linear relation flux-force: J( r, t) = L( T( r, t)) F( r, t) 1 T ( r, t) L( T ( r, t)) T ( r, t) J( r, t) = L( T ( r, t)) = 2 T ( r, t) Thermal conductivity matrix: λ ( T ) = jk L jk 2 T ( T ) J( r, t) = λ( T ( r, t)) T ( r, t) Let us assume that, under temperature variations, - the density change is negligible (the volume is constant) - the specific heat c v is constant u( r, t) T ( r, t) = cv t t u(, t) ρ r t + J( r, t) = 0 T ( r, t) 1 T ( r, t) λ( T ( r, t)) = 0 t ρ c v

52 Thermal diffusion matrix: ( T (, t)) D( r, t) : = λ r ρ c v Let us consider the case of constant D( r, t ) = D (negligible spatial/time dependence due to weak temperature dependence of matrix λ ( T ) ): D = λ ρ c v T ( r, t) T ( r, t) = D t Thermal diffusion equation (Fourier) in 3D For isotropic systems D jk = D δ jk, hence T ( r, t) 2 2 = D T ( r, t), + + t x y z 2 2 2

53 As already seen for thermal conductivity in 1D, the perspective can be reversed by adopting a phenomenological approach : 1) assume first-order relation flux/force: J( r, t) = L F( r, t) 2) get the entropy production rate as σ ( r, t) = J( r, t) F( r, t) = F( r, t) L F( r, t) S By following such a phenomenological approach, the structure of L (symmetric?) is not immediate! Decomposition of L into symmetric (L s ) and anti-symmetric (L a ) parts: s a s s L + L a a L L L = L + L with L jk = Lkj =, L jk = Lkj = 2 2 jk kj jk kj One can only state that the anti-symmetric part does not contribute to the entropy production rate: S ˆ σ ( T, F) = F L( T ) F s a = F L ( T ) F + F L ( T ) F x A x s, F L ( T ) F Given any vector x and a anti-symmetric matrix A: = x x A = j k j k = x x A + j k jk j k jk j, k > j j, k < j jk x x A = x x A x x A x x A x x A = 0 j k jk j k kj j k jk k j jk j, k > j j, k < j j, k > j k, j< k

54 Thus one realized that in the 3D case (contrary to the 1D case), the dissipation matrices which enter force-flux: (, t) = ( T ) (, t) J r L F r (*) S T entropy production rate expansion: ˆ σ (, ) = '( T ) are different! Namely one has s ( T ) + ( T ) L '( T ) = L ( T ) = L L 2 tr F F L F (**) (in the 1D case, the unique coefficient from both approaches coincides ) The phenomenological approach brings a matrix with lower symmetry: are there some constraints on L? Constraints come from Onsager s Theory of reciprocity relations. a In the present case of thermal conductivity: L = 0 and L '( T ) L ( T )

55 Onsagers s reciprocity relations (statements*) * For demonstration and insights see: R. B. Callen, Thermodynamics and an introduction to Thermostatistics 2 Ed. (Wiley, 1985), Pages S. R. de Groot, P. Mazur, Non-equilibrium Thermodynamics (Dover, 1984), Pages

56 1) Physical conditions that specify fluctuations at equilibrium: - Invariance under time-shift (dynamics without memory : fluctuations is a Markov process) - Invariance under time-reverse (by reversing all velocities, the system retraces backward the same trajectory: there is no time-arrow for equilibrium fluctuations!) 2) Onsager s assumption For a given set of state variables that describe the system, the equilibrium fluctuations and the macroscopic relaxation (towards equilibrium) are governed by the same dynamic law. [Relaxation to equilibrium is nothing but the tail of a large-amplitude fluctuation!]

57 3) Consider a set of N relevant extensive properties, and take the corresponding densities (per unit of mass) as state variables x. Divide the system into constitutive elements ( particles ). These state variables (densities) are of two kinds: a) Even functions of particles velocities ( α 1, α 2,... densities of properties 1 2 example: internal energy density u( r, t). A, A,...); b) Odd functions of particles velocities ( β 1, β 2,... densities of properties 1 2 example: components of the linear momentum density. B, B,...); Any of the state variables x i belongs to the set α or to the set β.

58 4) Assume linearity between fluxes and forces: X j i : density-flux vector (dependent on space/time) associated to the extensive property X i x A1 j α1 A 2 α2 j A α 3 3 j =, J β1 B j 1 β2 B2 β3 j B3... j... J = L F dissipative matrix made of N N elements (each element is a 3 3 block if the system extends in the 3D space) The matrix L may depend, in all generality, on - local value of intensive state variables (e.g., T in the thermal conductivity case ) - external parameters (e.g, external fields) In the case of thermal conductivity we had a single extensive property ( N = 1 with variable x 1 = u of kind a) ), a single density-flux j U J (heat flux being at constant volume), and the matrix L( T ) was a 3 3 matrix dependent on local temperature.

59 Special form of Onsager s reciprocity relations: If only state variables of kind a) are accounted for, then L tr = L General form of Onsager s reciprocity relations for variables of both kinds a), b) and in presence of an external magnetic field* B : L Lαα ( B) Lαβ( B) = ( ) ( ) Lβα B Lββ B tr L ( B) = L ( B) (symmetric matrix when B = 0) αα αα tr L ( B) = L ( B) (symmetric matrix when B = 0) ββ ββ tr L ( B) = L ( B) (and vanish when B = 0) αβ βα * The derivarion considers the Lorentz formula for forces of charged particles moving in an electromagnetic field: f = q( E + ν B) The force is unchanged when both B and velocity v are reversed.

60 The general recipe to build a non-equilibrium description of a macroscopic system: find a set of relevant extensive * properties (then take the densities per unit of mass as state variables) and of intensive state variables ** * e.g., the internal energy density field in the case of thermal conductivity ** e.g., the temperature field in the of thermal conductivity consider the fluxes of the extensive properties, and the related thermodynamic forces (named also affinities in the old literature) assume the fluxes linearly dependent on forces impose the correct symmetry to the phenomenological matrix L (Onsager relations). Exploit the system s symmetry to further simplify L (Curie s Principle) impose that L s be positive-definite in order to have entropy production for any value of the forces out-of-equilibrium try to express the dissipative parameters (entering L) on molecular basis, by exploiting the matching between out-of-equilibrium relaxation and equilibrium fluctuations (Onsager s assumption) Warning: the linear relation flux-forces may not be appropriate in normal conditions! As example, consider the case of chemical reactions kinetics ( )

61 Application to thermo / fluid-dynamics of single-component isotropic fluids

62 The system: an out-of-equilibrium isotropic fluid with its constitutive elements in motion. Pure substance (no mixtures). State variables (their fields) of the system: T ( r, t) ρ( r, t) vx( r, t) v( r, t) = vy ( r, t) vz ( r, t) v( r, t) is the velocity-vector field to be defined on microscopic grounds. Think to divide the space (into which the fluid moves) in cells of small volume V which contain the location r; the single molecules inside each cell have their own instantaneous velocity v i ( t ). The fluid velocity at location r and time t is defined as the average v( r, t) N ( t) 1 v( r, t) : = vi ( t) N( t) i= 1 (average over the N molecules which, at time t, are inside the cell containing r) at time t spatial cell at location r

63 At equilibrium, T ( r, t) = T, ρ( r, t) = ρ (or ρ( r ) ), v( r, t ) = v. The fluid, even at equilibrium, may be subjected to global advection motion (as a whole). Non-uniform velocity field v( r, t) v means that internal motion (convective dynamics) takes place in out-of-equilibrium conditions. Again, we apply the near-equilibrium postulate for the densities of extensive properties. Consider a generic extensive property A. Then its density (per unit of mass) a is a( r, t) a ( T,, v ) T T t eq ρ (, ) ρ = r == ρ ( r, t ) v v( r, t ) The usual partial derivative a( r, t) a(, t + t) a(, t) = lim r r t t t is not suitable in the case of fluid motion, that is through the elemental cell containing the point r: the cell contents is an open system!

64 It is more convenient to consider the following material derivative, named also barycentric derivative : Da( r, t) a( + t, t + t) a(, t) : = lim r v r Dt t t actual v( r, t) r + v(r,t) t time t + t forward r time t The time-derivative is taken by moving together with the fluid : in this way, the elemental portion is a closed system.

65 Algebraic elaboration: Da( r, t) a( r + v t, t + t) a( r + v t, t) a( r + v t, t) a( r, t) = lim + lim Dt t t t a( r, t + t) a( r, t) a( r + v t, t) a( r, t) lim + lim t t t a( r, t) = + t lim t t [ ] t a( r, t) + tv a( r, t) / a( r, t) t a( r, t) a( r, t) = + v t D a ( r, t ) = + v a ( r, t ) Dt t The barycentric derivative is a combination of first-order derivatives, hence the usual laws of derivation apply to it: D D D [ c1a 1( r, t) + c2a2 ( r, t) ] = c1 a1 ( r, t) + c2 a2( r, t) Dt Dt Dt D D D a1 ( r, t) a2( r, t) = a2( r, t) a1 ( r, t) + a1 ( r, t) a2( r, t) Dt Dt Dt D df ( a) D f ( a( r, t)) = a( r, t) Dt da Dt

66 STEP 1 Make the list of extensive properties to be considered: - Mass - Linear momentum - Angular momentum - Total Energy - Entropy Total energy at equilibrium E = Epot + Ekin + U. Its density per unit of mass of fluid: 1 e = v v + ψ + 2 u with ψ the density of potential energy (due to force-fields interacting with the fluid particles), which generally depends on the spatial coordinates. Near equilibrium postulate e( r, t) Remark. All quantities in what follows depend on r and t; for sake of notation, from now on these arguments will be omitted (but don t forget them!)

67 Extensive property, A mass linear momentum angular momentum ρ Densities per unit volume, L j ρ P j total energy E 1 entropy M ρ ρ A ρ = ρ v, j = x, y, z j Densities per unit mass, a 1 v, j = x, y, z ( r v ), j x, y, z ( r v ), j =,, j = ρ = ρ = ρ v v + ρ ψ + ρ u 2 ρ S j = ρ s j 1 2 v v + ψ + s x y z u STEP 2 For each of those properties A, consider the related density-flux vector j ( r, t) and write the evolution equation for the density (per unit of mass) in terms of barycentric derivatives. A

68 Example: continuity equation for mass (conservative property); The density-flux vector of mass: ρ + = t ( ρ v) 0 In terms of barycentric derivatives: j M = ρ v Dρ ρ ρ = + v Dt t ρ ρ ρ = ( ρ v) + v = ρ v v + v = ρ v M ρ ρ Dρ = ρ Dt v Notice that if the density of the fluid can be taken to be ρ =const. at any point and any time (so-called incompressible fluid model), then Dρ / Dt = 0 implies that incompressible fluid v = 0 any r, any t The divergence of the velocity vector is zero at all points and all times.

69 For general extensive properties (conservative or non-conservative): The evolution equation (already seen): ρ t A A A + j = σ = rate of "internal" production Turn from volumetric densities ( ρ a) A A t = σ j ( ρ a) Da D Dρ ρ a Dt Dt Dt ( ρ a) A ρ to densities per unit of mass a : ( ρ a) = + v ( ρ a) + aρ v = + v ( ρ a) + aρ v t t = + = + t A = σ A A ( ρ a v) σ j ( ρ a v) A ( j ρ a v) Dρ = ρ v Dt ρ A = ρ a

70 Barycentric density-flux: A, bar A j : = j ρ a v It is the density-flux of the given extensive property, at a certain point and time, as it would be experienced by moving together with the fluid. Da Dt ρ A = σ j A, bar For each of the relevant extensive properties listed above, write down the expressions for Da Dt (some algebra ).

71 By combining the resulting expressions for energy and linear momentum, one ends up ( ) with the following barycentric derivative for the internal energy density: Du ρ Dt q, bar = j j, k P j, k v k j ( 1 Principle of Thermodynamics) q, bar j is the barycentric density-flux vector of heat. P is the stress tensor (3 3 matrix); its j-th column contains the components of the baricentric density-flux vector of the j-th component of the linear momentum: P ( P, ) ( ) j bar Pj j j = = ρ v v ij i j i i Given a portion of fluid with changing border Ω ( t), the stress tensor is directly related to the flux (across the border) of the mechanical work due to surface forces which move/deform the border. In terms of barycentric density-flux vector: j w, bar = P v For atomic fluids P is a symmetric matrix

72 STEP 3 Express the entropy production rate (, t) S σ r. As previously seen, entropy flux only comes from heat transfer (which takes place under quasi-reversible conditions if the temperature field is smooth enough ): 1 j = j T s, bar q, bar Da Dt ρ A = σ j A, bar Ds ρ = σ j = σ Dt ( j / T ) s s, bar s q, bar σ s q bar Ds (, j / T ) = + ρ Dt to be elaborated

73 The derivative Ds p 2 du = Tds + d ρ ρ Dt is obtained by the fundamental differential : Du Ds p Dρ Ds 1 Du p Dρ = T + = Dt Dt 2 ρ Dt Dt T Dt 2 T ρ Dt [ ρ ] p(, t) = p T (, t), (, t), (, t) r r r v r is the local pressure. eq s ( q, bar / ) ρ Du p Dρ σ = j T + T Dt T ρ Dt Dρ = ρ v Dt Du ρ Dt q, bar = j j, k P j, k v k j σ ( ) 1 1 v p = j / T j + v r r r s q, bar q, bar P k j, k T T, j T j k 1 1 q, bar q, bar T 1 q, bar 1 vk Pj, k j, k j 1 q, bar T 1 vk p j Pj, k T v r j, k j T p = j + j j + v T T T T = +

74 Viscous stress tensor: Π = P p1, 1 = 3 3 identity matrix Clearly also Π (as P) is a symmetric tensor (for atomic fluids) σ 1 s q, bar T 1 v = j Π k j, k T j, k j

75 Let us introduce (0) J = Tr { Π} (1) q, bar /3 ("Tr" is the trace of the viscous stress tensor: a scalar) J = j (vector) (2) J = Π 1 Tr Π / 3 (symmetric matrix for atomic fluids) { } These quantities have physical meaning of fluxes which are grouped according to tensor ranks (0, 1, 2). The related thermodynamic forces are F F F (0) (1) (2) 1 v j 1 = = v T T j 1 T = 1 v v = k T + j j, k j j k s (0) (0) i (1) (1) (2) (2) i i ik ik i, k σ = J F + J F + J F

76 STEP 4 Assume a linear relation fluxes thermodynamic forces. Totally we have = 9 independent fluxes and forces: the matrix L is a priori 9 9. Consider Onsagers s relations to set the anti-symmetric part of matrix L, and exploit the symmetry of the system to get the symmetric part of matrix L. Strong simplification from Curie principle: for isotropic fluids, only fluxes/forces of the same rank are related, and the linear coupling is scalar (i.e., a single multiplicative coefficient per each rank): J J J (0) (0) = 9η v F (1) 2 (1) = λt F = 2ηT F (2) (2) η v = bulk viscosity η = shear viscosity λ = thermal conductivity coefficient PHENOMENOLOGICAL DISSIPATION PARAMETERS s In order to have σ > 0 for any applied set of thermodynamic forces, it must be η > 0, η > 0, λ > 0 v

77 Shear viscosity of liquid water

78 Shear viscosity of some liquids

79 Shear viscosity of some glass-forming liquids

80 STEP 5 Insert the fluxes, constructed at STEP 4, in the expressions of the barycentric derivatives of the linear momentum density and of the internal energy density. For example, for the internal energy density we get Du q, bar ρ = j Dt j, k P j, k v k j v v J = (1) 2 (1) P k k j, k λt F Pj, k r, j r r j k j, k j 1 2 T vk T j, k λ j, k r j, k j j, k v = λt P = P r r r k j If λ is taken to be constant (i.e., slightly dependent on temperature, hence also on spatial coordinates) then Du T v v ρ = λ = λ Dt ( ) 2 P k j, k T Pj, k j, k j j, k k j

81 Summary of the resulting expressions (set of coupled differential equations): Dρ = ρ v mass balance Dt Dv ψ p 2 ρ = ρ + ( ηv + η / 3 ) v + η v Navier - Stokes equations Dt Du 2 ρ = λ T p v + η v + η v v thermal effects Dt 2 ( s) ( s) v 2 ij ij i, j ( s) 1 v v i j vij = + 2 rj r 2 = + + x y z i (Laplace operator) x The system of equations (1+3+1 = 5 equations totally) are solved with respect to the independent fields T ( r, t), ρ( r, t), v ( r, t), v ( r, t), v ( r, t), once the internal-energydensity fields and the pressure fields are given as input (from equilibrium thermodynamics): Equilibrium u (,, ), (,, ) Thermodynamics eq T ρ v peq T ρ v y z [ ρ ] [ ρ ] u( r, t) = ueq T ( r, t), ( r, t), v( r, t) p( r, t) = p T ( r, t), ( r, t), v( r, t) eq

82 Simplifications 1) Limit case: fluid at rest and constant density The equations reduce to the equation of thermal conductivity already seen: ρ Du u Dt ρ = t λ 2 T 2) Incompressible fluid (constant density) All terms with factors v = 0 Dv ψ p 2 ρ = ρ + η v Dt Du 2 ρ = λ T + 2η v v Dt i, j ( s) ( s) ij ij disappear:

83 3) Incompressible fluid and pressure weakly dependent on position If the density is taken (approximatively) constant, the dependence of pressure on local position is due to the dependence of temperature on local position. If p ( T, ρ ) is smooth with respect to T, then p 0 any r, any t eq Dv ψ 2 ρ ρ + η v (simplified Navier-Stokes equations) * Dt Du 2 ( s) ( s) ρ = λ T + 2η vij vij Dt ** i, j Equation * is autonomus: is can be solved to get the velocity field v( r, t). After that, such a solution is plugged into equation ** to be solved with respect to T ( r, t).