Micro-macro coupling - Strength and Structure

The previous chapter was devoted to the quantification of microstructure. In this chapter, we examine how the behavior of the microstructure can be coupled with the behavior on the macroscopic level. As shown in Figure 10.1, several order in length (and time) scales need to be spanned to achieve that goal.

Figure 10.1:Illustration of the several levels of detail, as well as a possible coupling between them.

The coupling between the different levels occurs as follows. On the macroscopic scale, one imposes deformation. This deformation leads to changes in the microstructure, which in turn lead to micro-forces on that same small scale (since the structure is not in its equilibrium state anymore). These micro-forces are felt on the macroscopic scale as a stress. In summary, one thus achieves a relation between the imposed deformation and the resulting stress. However, this relation is not a one-level (macroscopic) explicit relation, but rather it involves the “detour” via the microstructure. The benefit in “detouring” via the microstructure is that, on that level, one has direct access to implement the physical insight into the modeling approach (much preferable over any phenomenological approach).

A useful tool to dynamically couple multiple levels of description is nonequilibrium thermodynamics. While there is a large “zoo” of approaches, we are going to present one that is particularly useful for our purposes.

10.1Weak formulation of GENERIC¶

The acronym GENERIC stands for: General Equation for the Non-Equilibrium Reversible-Irreversible Coupling. It is a nonequilibrium thermodynamics technique that has been developed at the end of the nineties, and been applied in many branches of modeling complex fluids and complex solids. It applies to mechanically and thermally closed (in other words: isolated) systems. Here, we do not present the full framework, but rather restrict ourselves to a so-called weak formulation. It works as follows:

Choose a set of variables, denoted by the generalized vector $\vec{X}$ , that describes your system to sufficient detail. For example, you may choose to include the deformation gradient $\T{F}$ as well as the temperature $T$ in that set.
Find an expression for the total energy $E$ of the system, expressed in terms of the variables $\vec{X}$ : $E = E[\vec{X}]$ .
Find an expression for the entropy $S$ of the system, expressed in terms of the variables $\vec{X}$ : $S = S[\vec{X}]$ .
Use the chain-rule to calculate the change in energy $E$ and entropy $S$ :
$\begin{aligned} \dot{E} &= \sum_k \frac{\delta E}{\delta X_k} \dot{X}_k \ , \\ \dot{S} &= \sum_k \frac{\delta S}{\delta X_k} \dot{X}_k \ , \end{aligned}$
(10.1)
where the summation runs over all components of the vector $\vec{X}$ . So, if you knew the evolution equations $\dot{X}_k$ , you could calculate the change in energy and entropy. However, GENERIC works differently: GENERIC postulates constraints on $\dot{E}$ and $\dot{S}$ , which in turn restricts the evolution equations for $\dot{X}_k$ , as described in the following step.
The GENERIC gives four conditions for the dynamics:
$\dot{E} = 0 \qquad \qquad \textrm{for reversible dynamics,}$
(10.2)
$\dot{E} = 0 \qquad \qquad \textrm{for irreversible dynamics,}$
(10.3)
$\dot{S} = 0 \qquad \qquad \textrm{for reversible dynamics,}$
(10.4)
$\dot{S} \ge 0 \qquad \qquad \textrm{for irreversible dynamics.}$
(10.5)
Conditions (10.2) and (10.3) tell that the total energy should not change at all during the dynamics. However, in many cases, there are different contributions to the dynamics: reversible (for example: kinematics, affine deformation), and irreversible (for example: viscous friction, or viscoplastic deformation) - this distinction will be discussed further below. Conditions (10.2) and (10.3) say that energy must not change in either of these two contributions separately, which is a stronger statement that the usual conservation of energy ( $1^{\rm st}$ law of thermodynamics). Relation (10.4) says that the entropy is unaffected by the reversible dynamics, and in the irreversible dynamics the entropy can only stay constant or increase, but not decrease. This last statement means that the rate of dissipation is never negative ( $2^{\rm nd}$ law of thermodynamics).

In the above scheme, we see that three ingredients are needed to make use of the GENERIC procedure:

What is the energy $E$ and what can we achieve with the conditions on $\dot{E}$ ?
What is the entropy $S$ and what can we achieve with the conditions on $\dot{S}$ ?
How can we differentiate between reversible and irreversible contributions to dynamics?

These three issues are addressed in the following.

10.2Ingredient (1/3): Energy¶

While the meaning of total energy $E$ is rather clear, we give some examples on what can be achieved with the GENERIC conditions on $\dot{E}$ .

Example 10.1 (Single particle in gravitation field)

Imagine a single particle of mass $m$ , that can only move along the $z$ -direction, which is the direction parallel to the gravitation field. More explicitly, the gravitational acceleration $g$ shall point in the negative $z$ -direction. Considering the particle moving in an external, fixed field, the degrees of freedom can be summarized as $\vec{X} = (z,v_z)$ , where $v_z$ denotes the velocity of the particle. The energy of that system is given by

E[\vec{X}] = \tfrac{m}{2} v_z^2 + m \, g\, z \ ,

(10.6)

where $m \, g\, z$ is the potential energy of the particle in the gravitation field (the higher its position, the more potential energy it has). The change in energy is then

\dot{E} = \frac{\partial E}{\partial z} \dot{z} + \frac{\partial E}{\partial v_z} \dot{v}_z = \left(g + \dot{v}_z\right) \, m v_z \ ,

(10.7)

which should be zero, according to GENERIC. Obviously, this condition is fulfilled for Newton’s equation, which in this case amounts to $\dot{v}_z = -g$ .

As a word of caution, we briefly mention the case where also friction is involved. For example, if the particle moves through air or even is immersed in a fluid, a non-zero velocity leads to a friction force of the form $F^{\rm fr} = -\zeta \, v_z$ , with friction coefficient $\zeta$ . However, how does this fit into the above procedure? In this case we do not use the GENERIC condition $\dot{E} = 0$ to derive the evolution equation of the particle (which is what we have done above). Rather, we assume that we know the evolution equation already and see what additional we can learn from nonequilibrium thermodynamics. So, the evolution equation of the particle with friction is given by

m \dot{v}_z = -m g - \zeta v_z .

(10.8)

Inserting this in $\dot{E}$ above leads

\dot{E} = -\zeta \, v^2_z \ ,

(10.9)

which vanishes (for arbitrary situations $v_z$ ) only if the friction coefficient vanishes, $\zeta = 0$ . So, it seems that we run into a conflict with the GENERIC condition $\dot{E}=0$ as soon as we include friction. The reason will be discussed further below. However, we can say already here that this has to do with the fact that our set of variables is not complete: friction means that the particle looses kinetic energy, which is dissipated, meaning that the temperature of the surrounding medium should increase. But: none of our variables in $\vec{X}$ can describe this increased thermal energy in the surrounding medium, and therefore an extension of $\vec{X}$ will be needed.

Example 10.2 (Many interacting particles)

Imagine a system with $N$ particles, that have positions $\vec{r}_i$ and velocities $\vec{v}_i$ . So, the system is described by $\vec{X} = (\vec{r}_1, \ldots, \vec{r}_N, \vec{v}_1, \ldots, \vec{v}_N)$ . The energy can then be written as

E[\vec{X}] = \sum_{i=1}^N \tfrac{m}{2} |\vec{v}_i|^2 + \Phi(\vec{r}_1, \ldots, \vec{r}_N) \ ,

(10.10)

with $\Phi$ the potential energy, in other words the sum of all interaction energies.

The change in total energy can then be written as

\dot{E} = \sum_{i=1}^N \left( \frac{\delta E}{\delta \vec{r}_i} \cdot \dot{\vec{r}}_i + \frac{\delta E}{\delta \vec{v}_i} \cdot \dot{\vec{v}}_i \right) = \sum_{i=1}^N \left( \frac{\partial \Phi}{\partial \vec{r}_i} + m \dot{\vec{v}}_i \right) \cdot \vec{v}_i \ ,

(10.11)

which should be equal to zero, according to the GENERIC. This condition is fulfilled by the classical Newton’s equation of motion,

m \dot{\vec{v}}_i = -\frac{\partial \Phi}{\partial \vec{r}_i} \ .

(10.12)

We have thus re-derived Newton’s equation of motion with the GENERIC procedure.

10.3Ingredient (2/3): Entropy¶

The GENERIC procedure also requires knowledge of the entropy $S$ as a function of the system variables, $S = S[\vec{X}]$ . In this section, we derive the entropy expression from a known expression for the Helmholtz free energy $A$ . In the context of thermodynamics, the entropy can be written (or even defined) as the negative of the partial derivative of $A$ with respect to the temperature $T$ , while keeping all other variables constant,

S = -\frac{\partial A}{\partial T} \ .

(10.13)

Therefore, the entropy quantifies the rate at which $A$ decreases when increasing $T$ . A graphical representation of this relation is given in Figure 10.2. The energy is defined thereafter, as

E = A + T \, S = A - T \frac{\partial A}{\partial T} \ .

(10.14)

Illustration of the relation between the Helmholtz free energy A, the entropy S and the energy E. — Figure 10.2:Illustration of the relation between the Helmholtz free energy $A$ , the entropy $S$ and the energy $E$ .

In the extreme case that $A$ is independent of $T$ , one finds $S = 0$ , which holds to a first approximation for metals. In this case, when studying elastic behavior, one speaks of “energy elasticity” (since $A=E$ ). In the other extreme that $A$ is linear in $T$ , one finds $E = 0$ , which holds to a first approximation for rubbers. In this case, when studying elastic behavior, one speaks of “entropy elasticity” (since $A=-TS$ ).

10.4Ingredient (3/3): Distinction “reversible” / “irreversible”¶

The distinction between reversible and irreversible can only be done on the level of evolution equations. The question is how the solution to the equation behaves if the time-axis is reversed.

We will illustrate this with one explicit example. More examples can be found on the lecture sheets.

The evolution equation of a single particle in the gravitation field can be written in the form

\begin{aligned} \tfrac{d}{dt} z &= v_z , \\ m \tfrac{d}{dt} v_z &= -m g . \end{aligned}

(10.15)

While $z$ does not change sign under time-reversal, both $dz/dt$ and $v_z$ do, and so the first of the above equations is invariant under time-reversal. In the second of the above equations, both sides to not change sign, and so also this second equation is invariant under time-reversal ( $v_z$ changes sign, but $dv_z/dt$ does not; $g$ also does not change sign). In summary, this implies that the entire dynamics consists of purely reversible (time-reversal invariant) contributions.

If we include friction, then the above set of equations is modified. The first equation remains, however, the second equation becomes

m \tfrac{d}{dt} v_z = -m g - \zeta v_z .

(10.16)

Upon time-reversal, one gets $dv_z/dt \to -d(-v_z)/dt=dv_z/dt$ and $-m g \to -m g$ as before. However, the last term changes sign upon time-reversal: $- \zeta v_z \to + \zeta v_z$ . Therefore, the dynamics associated with $-m g$ is reversible, while the dynamics due to the friction force $- \zeta v_z$ is irreversible.

In summary:

if reversible effects (time-reversal invariant) should be modeled, then conditions (10.2) and (10.4) are relevant;
if irreversible effects (sign change under time-reversal) should be modeled, then conditions (10.3) and (10.5) are relevant.

10.5Examples¶

The two examples below combine all three GENERIC ingredients: energy, entropy, and the reversible/irreversible distinction into complete models.

Example 10.3 (Particle in gravitation field, with friction and heat bath)

As in Example 10.1, we consider again a single particle in the gravitation field. We have seen there that problems arise when incorporating friction if the thermal energy of the heat bath is not properly accounted for. Particularly, the friction leads to a loss of energy of the particle, while at the same time the thermal energy of the surrounding medium increases. Only the sum of particle plus surrounding medium is a closed system, and is therefore suitable for the GENERIC framework.

The complete set of variables for the particle with friction is therefore given by

\V{X} = (z, v_z, E_{\rm th}) \ ,

(10.17)

with $E_{\rm th}$ the thermal energy of the surrounding medium. The total energy and entropy of the system can then be written as

\begin{aligned} E[\V{X}] &= \tfrac{m}{2} v_z^2 + m \, g \, z + E_{\rm th} \ , \\ S[\V{X}] &= S_{\rm th}(E_{\rm th}) \ , \end{aligned}

(10.18)

where we have assumed that the particle itself possesses no entropy and the entropy of the surrounding medium $S_{\rm th}$ is independent of particle position $z$ and velocity $v_z$ .

Using a calculation analogous to Example 10.1, we obtain for the changes in energy and entropy

\dot{E} = -\zeta v_z^2 + \dot{E}_{\rm th} \, ,

(10.19)

\dot{S} = T^{-1} \dot{E}_{\rm th} \ ,

(10.20)

where we have made use of the fundamental thermodynamic relation $dS_{\rm th}/dE_{\rm th} = T^{-1}$ , with $T > 0$ the temperature.

In reversible dynamics, friction is absent, and hence the GENERIC condition (10.2) imposed on (10.19) requires that in this case the thermal energy stays constant. According to (10.20), this implies that also the entropy does not change, in agreement with the condition (10.4).

The situation gets more interesting for irreversible dynamics, with non-zero friction coefficient. From (10.19) we see that $\dot{E}_{\rm th} = \zeta v_z^2$ , and with the help of (10.20) one obtains for the entropy change

\dot{S}_{\rm th} = \frac{\zeta}{T} v_z^2 \ .

(10.21)

If we now use the GENERIC condition (10.5), saying that the entropy change can not be negative, and since $T>0$ , we conclude that

\zeta > 0 \ .

(10.22)

In other words, nonequilibrium thermodynamics requires that the friction coefficient is positive, meaning that the friction force must be opposite to the velocity.

Example 10.4 (Pseudo-1D, nonisothermal elasto-viscoplasticity)

In this example, we study elasto-viscoplastic behavior in simple shear (deformation direction $\V{e}_x$ , and gradient direction $\V{e}_y$ ). The system is described by the variables

\V{X} = (u_x(y), v_x(y), \gstrn_p(y), T(y)) \ ,

(10.23)

with the displacement in $x$ -direction $u_x$ , the velocity in $x$ -direction $v_x$ , the plastic part of the shear strain $\gstrn_p$ , and the temperature $T$ .

For the total energy and entropy one can write

E[\V{X}] = \int \left[ \tfrac{\rho_0}{2} v_z^2 + e\left(\gstrn_e,T\right) \right] dy \ ,

(10.24)

S[\V{X}] = \int s\left(\gstrn_e,T\right) dy \ ,

(10.25)

where $\gstrn_e$ denotes the elastic part of the shear strain, that is related as follows to the system variables $u_x$ and $\gstrn_p$ :

\gstrn_e = \gstrn - \gstrn_p = (\partial u_x/\partial y) - \gstrn_p \ .

(10.26)

In (10.24), the symbol $\rho_0$ denotes the constant mass density. The functions $e$ and $s$ are the energy density and entropy density, respectively.

Studying the effect of the GENERIC conditions (10.2) and (10.4) for reversible dynamics, one can show (not done here) that the shear stress is given by the relation

\sigma_{xy} = \frac{\partial a}{\partial (\partial u_x/\partial y)} \ ,

(10.27)

with the Helmholtz free energy density $a = e - T s$ .

In order to discuss irreversible dynamics, it is useful to first think about which variables do not change irreversibly. If we are concentrating on viscoplastic deformation only, then we can state the following:

The displacement $u_x$ does not change irreversibly, since its evolution equation is given by the kinematic relation $\dot{u}_x = v_x$ . This entire equation is invariant upon time-reversal, and so it is purely reversible.
The evolution equation for the velocity $v_x$ is in principle the momentum balance, which in this example takes the form $\rho_0 \dot{v}_z = \tfrac{\partial}{\partial y} \sigma_{xy}$ , with the shear stress $\sigma_{xy}$ given above. This entire evolution equation is invariant with respect to time-reversal, and so it is purely reversible.
For the plastic part of the shear strain, $\gstrn_p$ , we expect well that it changes during viscoplastic deformation (by definition of what viscoplastic deformation represents).

With all these arguments, the change in energy and entropy due to the irreversible (in this case: viscoplastic) dynamics becomes

\dot{E} = \int \left( \frac{\partial e}{\partial \gstrn_p} \dot{\gstrn}_p + \frac{\partial e}{\partial T} \dot{T} \right) dy \ ,

(10.28)

\dot{S} = \int \left( \frac{\partial s}{\partial \gstrn_p} \dot{\gstrn}_p + \frac{\partial s}{\partial T} \dot{T} \right) dy \ .

(10.29)

If we use the GENERIC condition (10.3) on the energy, we obtain from (10.28) the change in temperature, $\dot{T} = -(\partial e/\partial T)^{-1} (\partial e/\partial \gstrn_p) \dot{\gstrn}_p$ . Inserting this expression into (10.29), one obtains for the change in entropy

\dot{S} = \int \left( \frac{\partial s}{\partial \gstrn_p} \dot{\gstrn}_p -T^{-1} \frac{\partial e}{\partial \gstrn_p} \dot{\gstrn}_p \right) dy = \int \frac{\dot{\gstrn}_p}{T}\left( -\frac{\partial a}{\partial \gstrn_p} \right) dy \ge 0 \ .

(10.30)

where we have used (i) the thermodynamic relation $(\partial s/\partial T) (\partial e/\partial T)^{-1} =T^{-1}$ (see also previous example), (ii) $a = e-Ts$ , and (iii) the GENERIC condition (10.5) about the non-negativity of the entropy production.

To understand the consequences of the dissipation inequality (10.30), we start with rewriting $\partial a/\partial \gstrn_p$ . According to (10.26), and using the chain rule as well as the fact that $a$ depends on $u_x$ and $\gstrn_p$ only through the elastic part $\gstrn_e$ , one can write

-\frac{\partial a}{\partial \gstrn_p} = -\frac{\partial a}{\partial \gstrn_e} \frac{\partial \gstrn_e}{\partial \gstrn_p} = \frac{\partial a}{\partial \gstrn_e} \frac{\partial \gstrn_e}{\partial \gstrn} = \frac{\partial a}{\partial \gstrn} = \frac{\partial a}{\partial (\partial u_x/\partial y)} = \sigma_{xy} \ ,

(10.31)

and we can therefore re-write the dissipation inequality (10.30) to become

\dot{S} = \int \frac{\dot{\gstrn}_p}{T} \sigma_{xy} dy \ge 0 \ .

(10.32)

Since $T>0$ , this condition implies that the sign of the shear stress $\sigma_{xy}$ and the plastic strain rate $\dot{\gstrn}_p$ must be equal at all times. One can fulfill this condition $\dot{\gstrn}_p \sigma_{xy} \ge 0$ by the ansatz

\dot{\gstrn}_p = \frac{1}{\eta(\sigma_{xy})} \sigma_{xy} \ , \qquad \eta(\sigma_{xy}) > 0 \ .

(10.33)

In summary, the nonequilibrium thermodynamics procedure puts a constraint on the relation between the shear stress $\sigma_{xy}$ and the plastic strain rate $\dot{\gstrn}_p$ . Particularly, the Eyring viscosity mentioned in earlier chapters fulfills this condition.

10 Micro-macro coupling

10.1Weak formulation of GENERIC¶

10.2Ingredient (1/3): Energy¶

10.3Ingredient (2/3): Entropy¶

10.4Ingredient (3/3): Distinction “reversible” / “irreversible”¶

10.5Examples¶