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Thermodynamics as a Foundation for Kinetics

Gibbs Results for Equilibrium of Heterogenous Substances

Began Mathematical Background

3.21 Spring 2002: Lecture 02

Mathematical Background, cont'd

Fields

A field associates a physical quantity with a position, $\ensuremath{\vec{x}}= (x,y,z)$ at a time, $t$.¹A field may also be a function of time: $f(\ensuremath{\vec{x}}, t)$ where $f$ is the physical quantity that depends on location and time.

Scalar Fields

Vector Fields

Tensor Fields

Sometimes an external (laboratory) coordinate system must be specified as well as an internal (material) coordinate system, especially in those cases for which the material coordinate system refers to a symmetry direction or the orientation of a plane in the material. Tensors represent ways of connecting quantities to coordinate systems. A vector is a simple tensor--called a rank 1 tensor--that connects a single value (i.e. a magnitude) to a coordinate frame. A rank two tensor connects two coordinate systems--for instance the stress $\sigma_{ij}$ represents the magnitude of the total force in a particular direction (the $i$-direction) distributed over a plane with a particular area (magnitude) oriented with a particular normal vector in the $j$-direction.²

Fluxes

Flux is an important vector field quantity in kinetics and it is important to understand it. It represents the rate at which ``stuff'' flows through a specified unit of area--an area is oriented in space.

Figure: The vector $\vec{\Delta A}$.

Let $\vec{\Delta A}$ be an oriented patch of area, $\vec{\Delta A} = \hat{n} \Delta A = (A_x , A_y , A_z)$. If $\dot{M_i}$ is the rate at which $i$ flows through a unit area, it follows that

$$\dot{M_i}(\vec{\Delta A}) \propto \vert \vec{\Delta A} \vert$$ (02-1)

The proportionality factor must be a vector field:

$$\dot{M_i}(\vec{\Delta A}) = \ensuremath{\vec{J}}_i \cdot \vec{\Delta A}$$ (02-2)

This defines the local flux as the continuum limit of:

$$\frac{\dot{M_i}(\vec{\Delta A})}{\Delta A} = \vec{J_i}(\vec{x}) \cdot \hat{n}$$ (02-3)

Accumulation

Figure 2-2: Illustration of limiting method of finding accumulation related to the divergence of the flux.

The rate at which $i$ accumulates in a volume $\Delta V = dx dy dz$ (with outward oriented normals) during time interval $\Delta t$ is:

$$\Delta M_i = (i ) - (i ) + ( i)$$ (02-4)

$$\begin{array}{ll} \delta M_{i} = & - \vec{J}(x + dx/2, 0, 0) ... ...a t \\ & + \dot{\rho}_i(\vec{x}) \Delta t \Delta V \end{array}$$ (02-5)

where $\dot{\rho}_i(\vec{x})$ is the density of the rate of production of $i$ in $\Delta V$.

Expanding to first order in $dx, dy , dz$, subtracting, and using the continuum limit,

$$\frac{\partial c}{\partial t} = -\nabla \cdot \vec{J} + \dot{\rho}_m$$ (02-6)

The rate of accumulation of the density of an extensive quantity is minus the divergence of the flux of that quantity plus the rate of production

Note that Eq. 2-6 could have been derived directly from:

$$\begin{array}{ll} \dot{M_i} &= -\int_{\mathcal{B}(\Delta V)} ... ... \cdot \ensuremath{\vec{J}}_i + \dot{\rho_i}) dV\\ \end{array}$$ (02-7)

where $\mathcal{B}(\Delta V)$ is the oriented surface around $\Delta V$ and the

Conserved and Unconserved Quantities

Conserved quantities are those that do not vanish from or spring into existence at any time or place. Therefore, the rate of production term in the accumulation must be identically equal to zero--for conserved quantities:

$$\ensuremath{\frac{\partial{C_i}}{\partial{t}}}= - \ensuremath{\nabla}\cdot \ensuremath{\vec{J}}_{N_i}$$ (02-8)

where $C_i$ is the density (or concentration) of $i$, or the continuum limit of $N_i/\Delta V$ where $N_i$ is the amount of $i$. The number of atoms of a particular type is conserved.³

Molecules are not conserved if chemical reactions take place.

Internal energy is conserved:

$$\ensuremath{\frac{\partial{u}}{\partial{t}}}= - \ensuremath{\nabla}\cdot \ensuremath{\vec{J}}_{U}$$ (02-9)

Examples of things that are not conserved include magnetization or spin density, atomic order, and polarization. Entropy is not conserved, thus there must be an expression:

$$\ensuremath{\frac{\partial{s}}{\partial{t}}}= - \ensuremath{\nabla}\cdot \ensuremath{\vec{J}}_{S} + \dot{\sigma}(\ensuremath{\vec{x}}, t)$$ (02-10)

where $\dot{\sigma}$ is the rate of entropy production per unit volume.

Introduction to Irreversible Thermodynamics

Equation 2-10 may seem sensible, but is it possible to form a physical picture of what is meant by the flux of entropy $\ensuremath{\vec{J}}_S$?

Figure 2-3: Schematic picture of entropy flux and density of entropy production.

What is meant by the continuum limit of the entropy density, $s$?

One way to find a meaningful picture of the entropy density is to assume that equilibrium thermodynamics applies locally. Then, the expression for equilibrium changes,

$$dU = T dS - P dV + \sum_i \mu_i dN_i$$ (02-11)

can be rearranged to find an expression for entropy in terms of more intuitive quantities. It is useful to write the above in terms of densities (dividing every extensive quantity by a reference unit volume $\ensuremath{{V}_\circ}$

$$T ds = du + Pdv - \sum_i \mu_i dc_i$$ (02-12)

It is useful to generalize this to other thermodynamic systems of interest and write the above equation as

$$T ds = du - \sum_i F_i d x_i$$ (02-13)

where the $F_i$ are generalized potentials and the $x_i$ are generalized displacements.

To illustrate how the assumption of local equilibrium will be used, consider a closed system that does or receives no work from its surroundings--i.e. a system where entropy can only increase according to the second law.

$$\frac{\partial }{\partial t} \int_{V_{closed}} s dV = \int_{V_{cl... ...} dV - \int_{\mathcal{B}({V_{closed}})} \ensuremath{\vec{J}}_S \cdot \hat{n} dA$$ (02-14)

where $\mathcal{B}({V_{closed}})$ is the surface of the closed volume $V_{closed}$ and $\hat{n}$ is the outward normal of that surface.

Supposing that the system does no work, only heat contributes to $\ensuremath{\vec{J}}_S = \ensuremath{\vec{J}}_U/T = \ensuremath{\vec{J}}_{Q}/T$ at the surface:

$$\frac{\partial }{\partial t} \int_{V_{closed}} s dV = \int_{V_{cl... ...{\mathcal{B}({V_{closed}})} \frac{\ensuremath{\vec{J}}_{Q} \cdot \hat{n}}{T} dA$$ (02-15)

The last term is (minus) the total heat that enters the system:

$$\frac{\partial Q_{in} }{\partial t} = -\int_{\mathcal{B}({V_{clos... ...n}}{T} dA = -\int_{V_{closed}} \frac{\nabla \cdot \ensuremath{\vec{J}}_Q}{T} dV$$ (02-16)

(It is minus because if the $\ensuremath{\vec{J}}_Q$ is in the same general direction a $\hat{n}$ then heat is leaving the system.)

If the surface is uniformly at constant temperature, then

$$\frac{\partial S_{total} }{\partial t} - \frac{1}{T} \frac{\partial Q_{in} }{\partial t} = \int_V \dot{\sigma} dV$$ (02-17)

The term on the right of Eq. 2-17 is a measure of the irreversibility.

This leads to a fundamental postulate of irreversible thermodynamics: or

$$\dot{\sigma} \equiv \frac{\partial s }{\partial t} + \nabla \cdot \ensuremath{\vec{J}}_S \geq 0$$ (02-18)

Rewrite Eq. 2-18 using the assumption of local equilibrium (2-13):

$$\dot{\sigma} = \frac{1}{T} \frac{\partial u }{\partial t} - \sum_... ...F}_i}{T} \frac{\partial X_i }{\partial t} + \nabla \cdot \ensuremath{\vec{J}}_s$$ (02-19)

To simplify writing, it is useful to introduce the ``summation convention'' where any repeated index becomes an implied sum. For instance, the dot-product can be written as $\vec{a} \cdot \vec{b} = \sum_i^3 a_i b_i = a_i b_i$ where in the final term the repeated index $i$ is summed over all of its possible values; for instance, $T ds = du - F_i d X_i = dU - \sigma_{ij} d \epsilon_{ij} - \phi dq - H_i dB_i - \mu_i dc_i$ where $\sigma_{ij}$ are the nine (3 $\times$ 3) components of the stress tensor, $\epsilon_{ij}$ are the nine components of the strain tensor,⁴ $\phi$ is the electrostatic potential and $q$ is charge density, $H_i$ and $B_i$ are the three (i.e., $x,y,z$) components of the applied and induced magnetic fields, and $\mu_i$ and $c_i$ are the $N$ chemical potentials and concentrations of the $i$ independent chemical species.

Therefore, using the summation convention:

$$\dot{\sigma} = \frac{1}{T} \frac{\partial u }{\partial t} - \frac{{F}_i}{T} \frac{\partial X_i }{\partial t} + \nabla \cdot \ensuremath{\vec{J}}_s$$ (02-20)

Using a version of the vector chain rule: $\nabla \cdot A \vec{B} = \ensuremath{\nabla}{A} \cdot \vec{B} + A \nabla \cdot \vec{B}$:

$$\frac{{F}_i}{T} \frac{\partial X_i }{\partial t} = \nabla \cdot \... ...}{T} \right) - \ensuremath{\vec{J}}_i \cdot \nabla \left( \frac{F_i}{T} \right)$$ (02-21)

and

$$\frac{1}{T} \nabla \cdot \ensuremath{\vec{J}}_U = \nabla \cdot \l... ..._U}{T} \right) - \ensuremath{\vec{J}}_i \cdot \nabla \left( \frac{1}{T} \right)$$ (02-22)

in Eq. 2-20:

$$\dot{\sigma} = \ensuremath{\vec{J}}_u \cdot \nabla \left(\frac{1}... ...c{J}}_s - \frac{\ensuremath{\vec{J}}_u - F_i \ensuremath{\vec{J}}_i}{T} \right)$$ (02-23)

Define:

$$\ensuremath{\vec{J}}_s \equiv \frac{\ensuremath{\vec{J}}_u - F_i \ensuremath{\vec{J}}_i}{T} = \frac{\ensuremath{\vec{J}}_{Q}}{T}$$ (02-24)

so that,

$$\begin{array}{ll} \dot{\sigma} & = \ensuremath{\vec{J}}_u \cd... ...- \ensuremath{\vec{J}}_i \cdot \nabla F_i \geq 0\\ \end{array}$$ (02-25)

Because $T$ is always positive, this implies a relation between the fluxes and the gradients of the potentials: Naively (but not necessarily), ${\ensuremath{\vec{J}}_{Q}}$, must be antiparallel to $\nabla T$; ${\ensuremath{\vec{J}}_i}$, must be antiparallel to $\nabla F_i$ for the entropy production to be everywhere positive.

Consider quantity on the right-hand-side of Eq. 2-25 term-by-term:

Conjugate Forces, Fluxes and Empirical Flux Laws for Unconstrained Components
Quantity	Flux	Conjugate Force	Empirical Flux Law
Heat	$\vec{J_Q}$	$- \frac{1}{T} \nabla T$	Fourier's	$\vec{J_Q} = -\kappa \nabla T$
Mass	$\vec{J_{i}}$	$-\nabla \mu_i$	Modified5Fick's form	$\vec{J_{c_i}} = -M_i c_i \nabla \mu_i$
Charge	$\vec{J_{q}}$	$- \nabla \phi$	Ohm's	$\vec{J_{q}} = -\rho \nabla \phi$