Balance University of Osnabrück, Germany Lecture presented at APS, Nankai University, China http://www.home.uni-osnabrueck.de/phertel Spring 2012
Linear and angular momentum and First and Second Law
point A material point is a piece of matter which is tiny from an engineer s point of view but huge from a physicist s point of view one mm 3 of an ideal gas under normal conditions still contains 2.7 10 16 particles ideal gas: N 2 N 2 = N i.e. relative fluctuation is δn N 1 10 8 the material point is always in thermodynamic equilibrium we usually suppress the t, x arguments
Content and flow think of an additive and transportable quantity Y mass, charge, momentum, energy etc. the content of Y in a region V is Q(Y ; t, V) = dv ϱ(y ; t, x) V the density ϱ(y ) = ϱ(y ; t, x) is a field the flow of Y through an area A is I(Y ; t, A) = da j(y ; t, x) A the current density j(y ) = j(y ; t, x) has a strength and a direction
Production the rate of production for Y in a region V is Π(Y ; t, V) = dv π(y ; t, x) V the volumetric production rate π(y ) = π(y ; t, x) describes: how much quantity Y is produced (or vanishes) per unit time and unit volume e.g. particles are produced in chemical reactions momentum is produced by external forces internal energy may be produced by friction entropy S is produced by irreversible effects π(s) 0 is the of thermodynamics
content of Y in V changes because Y is redistributed i.e. flows through the surface V or because Y is produced inside V as an d Q(Y ; t, V) = I(Y ; t, V) + Π(Y ; t, V) dt Gauss theorem dv f = da f V V generic ϱ(y ) + j(y ) = π(y ) must hold true for all times t and at all locations x
Number of particles Y = N a denotes the number of particles of species a = 1, 2,... particle density n a = ϱ(n a ) particle current density j a = j(n a ) velocities defined by j a = n a v a vanish or are generated in chemical reactions there are Γ r chemical reactions of type r per unit time and unit volume volumetric production rates are π a = Γ r ν ra r the ν ra are stoichiometric coefficients ν ra particles of species a are produced in a reaction of type r particle t n a + i n a vi a = Γ r ν ra r
Mass M particles of species a carry a mass m a mass density ϱ = ϱ(m) = a ma n a in each reaction, mass is conserved meaning Γ ra m a = 0 a consequently π(m) = Γ ra m a = 0 a r mass current density can be written as j(m) = ϱ(m)v mass continuity reads t ϱ + i ϱv i = 0
Charge Q e particles of species a carry a charge q a charge is conserved in each chemical reaction consequently π(q e ) = 0 charge density ϱ e = q a n a a current is split into convection and conduction part j(q e ) = j e = ϱ e v + J e charge continuity reads t ϱ e + i {ϱ e v i + J e i } = 0 without proof: only conduction contributions J are proper vector fields e.g. Ohm s refers to J e however, j e appears in Maxwell s s
P k is an additive transportable quantity its density has three components ϱ(p k ) = ϱv k mass times velocity per unit volume current densities for each component j i (P k ) = ϱv k v i T ki conduction part T ki is stress tensor minus sign is a convention volumetric momentum production rate is force f k per unit volume f k = ϱg k + ϱ e E k momentum t ϱv k + i {ϱv k v i T ki } = f k
Substantial time derivative time change as noted by a co-moving observer (D t f)(t, x) = i.e. D t f = t f + v f f(t + dt, x + dt v) f(t, x) dt specific quantities σ(y ) defined by ϱ(y ) = ϱ σ(y ) eqations may also be written as ϱ D t σ(y ) = i J i (Y ) + π(y ) in particular momentum ϱ D t v k = i T ki + f k
ctnd. angular momentum requires T ki = T ik at each location x there is a coordinate system such that the stress tensor is orthogonal: T 1 0 0 T ki = T ik = 0 T 2 0 0 0 T 3 a particular material can support only stress (positive T ) and pressure (negative T ) within certain limits the field of structural mechanics : work out the stress tensor, diagonalize it locally, and check whether its eigenvalues are within the allowed limits before building the bridge
Kinetic energy the kinetic energy density is ϱ(e k ) = ϱ 2 v2 it follow that ϱ 2 D t σ(e k ) = i J i (E k ) + π(e k ) where J i (E k ) = v k T ki and π(e k ) = T ik G ik + v k f k the volicity gradient is G ik = iv k + k v i 2
Potential energy quasistatic gravitational and electric potentials f k = ϱ k Φ g ϱ e k Φ e density of potential energy ϱ(e p ) = ϱφ g + ϱ e Φ e it follow that j i (E p ) = ϱj i (M) + ϱ e j i (Q) with j i (M) = ϱv i and j i (Q) = ϱ e + J e i π(e p ) = v i f i J e i E i
E i = U J e i E i T ij G ij E p v i f i E k Volumetric production rates for kinetic, potential and internal energy. Arrows pointing towards an energy form indicate a plus sign.
Internal energy total energy is the sum of kinetic, potential, and internal energy E i = U total energy is conserved pi(e) = π(e k + E p + U) = π(e k ) + π(e p ) + π(u) = 0 internal energy production rate per unit volume π(u) = T ik G ik + J e i E i specific internal energy is denotet by u conduction current density for internal energy... is the heat current density J u i stress tensor consists of the reversible part T ik and an irreversible part T ik likewise J e i and J e i
the for internal energy is... the of thermodynamics ϱ D t u = heat conduction i J u i deformation work +T ik G ik internal friction +T ik G ik polarization work +J e i E i and Joule s heat +J e i E i
each matierial point is always in equilibrium it undergoes a reversable change entropy S and temperature T introduced by du = T ds + dw were dw is the work done on the system entropy ϱ D t s = i J s i + π(s) entropy current j s = ϱsv + 1 T J u a µ a T J a chemical potentials µ a diffusion current densities J a
the second main of thermodynamics says 0 π(s) = heat conduction, Ji u 1 i T diffusion J a µ a i i T a friction +T ik i v k + k v i 2 Joule s heat 1 T J e i i Φ e chemical reactions + 1 Γ r A r T r
chemical affinity of reaction r is defined by A r = ν ra µ a a all contributions are of type flux times non-equilibrium indicator which would vanish in global equilibrium. there are five and only five contributions to entropy production First and are partial differential s there are much more fields than s. therefore: additional s required