6.2.1 Restrictions imposed by the second law of thermodynamics

Substitution of the constitutive assumptions (6.6) into the Clausius-Duhem inequality (5.35) gives

$-\rho \frac{\partial \breve{\psi }}{\partial \mathbf{F}}·\stackrel{.}{\mathbf{F}}+\breve{\mathbf{T}}·\mathbf{D}-\rho \left(\frac{\partial \breve{\psi }}{\partial \text{Θ}}+\breve{\eta }\right)\stackrel{.}{\text{Θ}}-\rho \frac{\partial \breve{\psi }}{\partial \mathbf{g}}·\stackrel{.}{\mathbf{g}}-\frac{1}{\text{Θ}}\breve{\mathbf{q}}·\mathbf{g}\ge 0,$

where we have used the chain rule (refer to Section 2.5.2):

$\stackrel{.}{\psi }=\frac{\partial \breve{\psi }}{\partial \mathbf{F}}·\stackrel{.}{\mathbf{F}}+\frac{\partial \breve{\psi }}{\partial \text{Θ}}\stackrel{.}{\text{Θ}}+\frac{\partial \breve{\psi }}{\partial \mathbf{g}}·\stackrel{.}{\mathbf{g}}.$

It can be shown (refer to Problems 6.1 and 6.2) that

$\breve{\mathbf{T}}·\mathbf{D}=\breve{\mathbf{T}}·\mathbf{L},\frac{\partial \breve{\psi }}{\partial \mathbf{F}}·\stackrel{.}{\mathbf{F}}=\frac{\partial \breve{\psi }}{\partial \mathbf{F}}{\mathbf{F}}^{\text{T}}·\mathbf{L},$

so inequality (6.7) becomes

$\left(\breve{\mathbf{T}}-\rho \frac{\partial \breve{\psi }}{\partial \mathbf{F}}{\mathbf{F}}^{\text{T}}\right)·\mathbf{L}-\rho \left(\frac{\partial \breve{\psi }}{\partial \text{Θ}}+\breve{\eta }\right)\stackrel{.}{\text{Θ}}-\rho \frac{\partial \breve{\psi }}{\partial \mathbf{g}}·\stackrel{.}{\mathbf{g}}-\frac{1}{\text{Θ}}\breve{\mathbf{q}}·\mathbf{g}\ge 0.$

Inequality (6.8) must hold for all processes. In particular, it must hold for the family of processes with g = 0, $\stackrel{.}{\mathbf{g}}=\mathbf{0}$, and L = 0, but $\stackrel{.}{\text{Θ}}$ arbitrary, at particular position x and time t. (An example is the family of processes (D.1) in Appendix D with A = 0, a = 0, and g₀ = 0, but a any real number.) For members of this family of processes, (6.8) simplifies to

$-\rho \left(\frac{\partial \breve{\psi }}{\partial \text{Θ}}+\breve{\eta }\right)\stackrel{.}{\text{Θ}}\ge 0.$

The density ρ is positive, which implies

$\left(\frac{\partial \breve{\psi }}{\partial \text{Θ}}+\breve{\eta }\right)\stackrel{.}{\text{Θ}}\ge 0.$

The coefficient of $\stackrel{.}{\text{Θ}}$in (6.9) is independent of $\stackrel{.}{\text{Θ}}$. For (6.9) to hold for all members of this family, the coefficient of $\stackrel{.}{\text{Θ}}$ must vanish, i.e.,

$\eta =-\frac{\partial \breve{\psi }}{\partial \text{Θ}}.$

Thus, the partial derivative of the response function for the Helmholtz free energy ψ with respect to temperature Θ gives the value of the entropy η, so $\breve{\eta }$ is not an independent response function. Result (6.10) holds for all processes, not just the special family considered above. Hence, inequality (6.8) reduces to

$\left(\breve{\mathbf{T}}-\rho \frac{\partial \breve{\psi }}{\partial \mathbf{F}}{\mathbf{F}}^{\text{T}}\right)·\mathbf{L}-\rho \frac{\partial \breve{\psi }}{\partial \mathbf{g}}·\stackrel{.}{\mathbf{g}}-\frac{1}{\text{Θ}}\breve{\mathbf{q}}·\mathbf{g}\ge 0.$

Inequality (6.11) must hold for all processes, in particular the family of processes with g = 0 and L = 0, but $\stackrel{.}{\mathbf{g}}$ arbitrary, at x and t (e.g., the family (D.1) in Appendix D with A = 0 and g₀ = 0, but a any vector). For members of this family, (6.11) simplifies to

$-\rho \frac{\partial \breve{\psi }}{\partial \mathbf{g}}·\stackrel{.}{\mathbf{g}}\ge 0.$

The density is positive; hence,

$\frac{\partial \breve{\psi }}{\partial \mathbf{g}}·\stackrel{.}{\mathbf{g}}\le 0.$

The coefficient of $\stackrel{.}{\mathbf{g}}$ in (6.12) is independent of $\stackrel{.}{\mathbf{g}}$ . Hence, the rate $\stackrel{.}{\mathbf{g}}$ may be chosen to violate the inequality (6.12) unless its coefficient is the zero vector. Therefore,

$\frac{\partial \breve{\psi }}{\partial \mathbf{g}}=\mathbf{0}.$

Thus, if the response function for ψ depended on the temperature gradient g, the second law would be violated for some processes. Since the second law must be obeyed for all processes, we conclude that a thermoelastic response function for ψ must be independent of g. Inequality (6.11) thus reduces to

$\left(\breve{\mathbf{T}}-\rho \frac{\partial \breve{\psi }}{\partial \mathbf{F}}{\mathbf{F}}^{\text{T}}\right)·\mathbf{L}-\frac{1}{\text{Θ}}\breve{\mathbf{q}}·\mathbf{g}\ge 0.$

which must hold for all processes.

Consider those processes with g = 0 but L arbitrary at x and t (e.g., the family (D.1) in Appendix D with g₀ = 0, F₀ fixed and invertible, and A any tensor). For these processes, (6.13) simplifies to

$\left(\breve{\mathbf{T}}-\rho \frac{\partial \breve{\psi }}{\partial \mathbf{F}}{\mathbf{F}}^{\text{T}}\right)·\mathbf{L}\ge 0.$

The coefficient of the rate L in (6.14) is independent of rates. Hence, we obtain

$\mathbf{T}=\rho \frac{\partial \breve{\psi }}{\partial \mathbf{F}}{\mathbf{F}}^{\text{T}}.$

Thus, the partial derivative of the response function for the Helmholtz free energy ψ with respect to strain F gives the value of the stress T, i.e., the Helmholtz free energy is a strain energy. Recall that the class of materials for which the stress-strain response is derived from a strain energy potential is called hyperelastic. The inequality (6.13) thus reduces to

$-\frac{1}{\text{Θ}}\breve{\mathbf{q}}·\mathbf{g}\ge 0$

or

$\breve{\mathbf{q}}·\mathbf{g}\le 0,$

since the absolute temperature Θ is strictly positive. Inequality (6.15) cannot be simplified any further since $\breve{\mathbf{q}}$ depends on g. Thus, the Clausius-Duhem inequality has been reduced for thermoelastic materials to an intuitive statement of the second law: heat flows against the temperature gradient, from hot to cold.

We now pause and take stock of our accomplishments. By demanding that the Clausius-Duhem inequality holds for all thermoelastic processes, we have reduced the constitutive assumption (6.6) to

$\psi =\breve{\psi }\left(\mathbf{F},\text{Θ}\right),\mathbf{q}=\breve{\mathbf{q}}\left(\mathbf{F},\text{Θ},\mathbf{g}\right)\text{with}\breve{\mathbf{q}}·\mathbf{g}\le 0,$

and

$\mathbf{T}=\rho \frac{\partial \breve{\psi }}{\partial \mathbf{F}}{\mathbf{F}}^{\text{T}},\eta =-\frac{\partial \breve{\psi }}{\partial \text{Θ}}.$

Thus, as a consequence of the second law, we have found that:

(1) Only two response functions ( $\breve{\psi }$ and $\breve{\mathbf{q}}$ ) are needed to characterize a thermoelastic material, rather than the four initially supposed. (Nice! Fewer response functions means fewer unknowns in the characterization problem.)

(2) The response function for ψ is independent of g. (Again, fewer experiments to perform.)

(3) The response function for q is restricted by the second law through the inequality $\breve{\mathbf{q}}·\mathbf{g}\le 0$, i.e., heat flows against the temperature gradient.

It can be shown (refer to Problem 6.3) that if the heat flux vector $\breve{\mathbf{q}}$ in (6.6) is a continuous function of temperature gradient g at g = 0, restriction (3) above has an additional implication: $\breve{\mathbf{q}}$ evaluated at g = 0 is zero, i.e., heat does not flow in the absence of a temperature gradient.

6.2.2 Restrictions imposed by invariance under superposed rigid body motions and conservation of angular momentum

Our appeal to the second law has proven fruitful. We now appeal to invariance requirements under superposed rigid body motions (refer to Sections 5.2.1 and 5.4.1) and obtain further restrictions. (Note that any restrictions on the response functions are very helpful, as they reduce the number of experiments needed to characterize a particular material.) In particular, we must have

${\psi }^{+}=\psi ,{\mathbf{q}}^{+}=\mathbf{Qq}$

when

${\mathbf{F}}^{+}=\mathbf{QF},{\text{Θ}}^{+}=\text{Θ},{\mathbf{g}}^{+}=\mathbf{Qg}$

for all proper orthogonal tensors Q(t). It can be shown (refer to Problems 6.4 and 6.5) that these invariance requirements demand

$\psi =\breve{\psi }\left(\mathbf{F},\text{Θ}\right)=\stackrel{\prime }{\psi }\left(\mathbf{C},\text{Θ}\right)$

and

$\mathbf{q}=\breve{\mathbf{q}}\left(\mathbf{F},\text{Θ},\mathbf{g}\right)=\mathbf{F}{\mathbf{q}}^{‵}\left(\mathbf{C},\text{Θ},{\mathbf{g}}_{\text{R}}\right).$

Note that $\breve{\psi }$ and $\stackrel{‵}{\psi }$ denote two different response functions for ψ. Equation (6.17) implies that the response function for ψ can depend on F only through the combination C = F^TF, where C is Green's deformation tensor; the dependence on Θ is unrestricted. Equation (6.18) implies that the dependence of the response function for q on g must be through g_R = F^Tg (where g_R is the referential temperature gradient), there must be a linear dependence of q on F, and any further dependence on F must be through the combination C = F^TF.

Now that we have established $\psi =\stackrel{‵}{\psi }\left(\mathbf{C},\text{Θ}\right)$, it can be shown, through a change of independent variable from F to C, that (6.16b)₁ becomes

$\mathbf{T}=\rho \frac{\partial \breve{\psi }\left(\mathbf{F},\text{Θ}\right)}{\partial \mathbf{F}}{\mathbf{F}}^{\text{T}}=\rho \mathbf{F}\left(\frac{\partial \stackrel{‵}{\psi }\left(\mathbf{C},\text{Θ}\right)}{\partial \mathbf{C}}+\frac{\partial \stackrel{\prime }{\psi }\left(\mathbf{C},\text{Θ}\right)}{\partial {\mathbf{C}}^{\text{T}}}\right){\mathbf{F}}^{\text{T}}.$

If $\stackrel{‵}{\psi }\left(\mathbf{C},\text{Θ}\right)$ is a symmetric function of C, then (6.19) becomes

$\mathbf{T}=2\rho \mathbf{F}\frac{\partial \stackrel{‵}{\psi }\left(\mathbf{C},\text{Θ}\right)}{\partial \mathbf{C}}{\mathbf{F}}^{\text{T}}.$

It can be shown that (6.19) and (6.20) ensure that the Cauchy stress T is symmetric, so conservation of angular momentum (5.25) is satisfied.

To summarize the results of this section thus far: A thermoelastic material has a perfect memory of its reference state. On physical grounds, it is assumed that such a material is characterized by at most four response functions, each of which could conceivably depend on the deformation gradient F (with respect to some reference configuration), temperature Θ, and temperature gradient g, all evaluated at present time t (i.e., the response at present time t depends on the present values of F, Θ, and g). The four response functions and their lists of arguments are

$\mathbf{T}=\breve{\mathbf{T}}\left(\mathbf{F},\text{Θ},\mathbf{g}\right),\mathbf{q}=\breve{\mathbf{q}}\left(\mathbf{F},\text{Θ},\mathbf{g}\right),\psi =\breve{\psi }\left(\mathbf{F},\text{Θ},\mathbf{g}\right),\eta =\breve{\eta }\left(\mathbf{F},\text{Θ},\mathbf{g}\right).$

Note that the dependence on deformation does not involve its history, but rather just the present value of the deformation gradient from some reference configuration.

Because these response functions must be consistent with conservation of angular momentum, the second law of thermodynamics for all thermomechanical processes, and invariance requirements under any superposed rigid body motion, we find that only two of the response functions are independent, i.e.,

$\psi =\stackrel{‵}{\psi }\left(\mathbf{C},\text{Θ}\right),\mathbf{q}=\mathbf{F}\stackrel{‵}{\mathbf{q}}\left(\mathbf{C},\text{Θ},{\mathbf{g}}_{\text{R}}\right),$

from which

$\mathbf{T}=2\rho \mathbf{F}\frac{\partial \stackrel{‵}{\psi }\left(\mathbf{C},\text{Θ}\right)}{\partial \mathbf{C}}{\mathbf{F}}^{\text{T}},\eta =-\frac{\partial \stackrel{‵}{\psi }\left(\mathbf{C},\text{Θ}\right)}{\partial \text{Θ}}$

can be deduced, with the restrictions

$\stackrel{‵}{\mathbf{q}}\left(\mathbf{C},\text{Θ},\mathbf{0}\right)=\mathbf{0},\mathbf{q}·\mathbf{g}\le 0.$

Note that the Helmholtz free energy, Cauchy stress, and entropy are independent of temperature gradient. Also note that the Clausius-Duhem inequality reduces to the intuitive second law statements (6.21c)₁ and (6.21c)₂, i.e., heat does not flow in the absence of a temperature gradient, and in the presence of a temperature gradient, heat flows opposite the temperature gradient, from hot to cold.

6.2.3 Restrictions imposed by material symmetry: isotropy

In this section, we apply material symmetry considerations (refer to Section 5.2.2) to further reduce and simplify the form of the response functions. In particular, we consider thermoelastic materials that are isotropic. Loosely speaking, isotropic materials have properties that are identical in all directions.

Recall from Section 6.2.1 that by demanding the Clausius-Duhem inequality hold for all thermoelastic processes, we found that only two response functions are necessary to characterize a thermoelastic material, i.e.,

$\psi =\breve{\psi }\left(\mathbf{F},\text{Θ}\right),\mathbf{q}=\breve{\mathbf{q}}\left(\mathbf{F},\text{Θ},\mathbf{g}\right),$

where the dependence of the deformation gradient F on a particular reference configuration κ is understood. Here, κ is chosen to be the stress-free configuration. By analogy to (5.22), we have

$\breve{\psi }\left(\mathbf{F},\text{Θ}\right)=\breve{\psi }\left(\mathbf{F}{\mathbf{H}}^{-1},\text{Θ}\right)\text{for all}\mathbf{H}\in \mathcal{G},$

where $\mathcal{G}$ is the symmetry group of the material relative to the stress-free reference configuration κ, and H is a symmetry transformation with respect to κ. Since a thermoelastic material is a solid, we have $\mathcal{G}\subseteq \mathcal{O}$, where $\mathcal{O}$ is the full orthogonal group. Then (6.22) becomes

$\breve{\psi }\left(\mathbf{F},\text{Θ}\right)=\breve{\psi }\left(\mathbf{F}{\mathbf{H}}^{\text{T}},\text{Θ}\right)\text{for all}\mathbf{H}\in \mathcal{G}.$

Recall from Section 6.2.2 that in order to satisfy invariance under superposed rigid body motions and conservation of angular momentum, we must specify ψ to be a function of F in the particular combination C = F^TF, i.e.,

$\psi =\stackrel{\prime }{\psi }\left(\mathbf{C},\text{Θ}\right),$

where C is Green's deformation tensor. The condition on $\stackrel{‵}{\psi }$ corresponding to (6.23) is

$\stackrel{‵}{\psi }\left(\mathbf{C},\text{Θ}\right)=\stackrel{‵}{\psi }\left(\mathbf{HC}{\mathbf{H}}^{\text{T}},\text{Θ}\right)\text{for all}\mathbf{H}\in \mathcal{G}.$

For isotropic thermoelastic solids, the symmetry group $\mathcal{G}$ relative to the stress-free reference configuration κ is the full orthogonal group $\mathcal{O}$. Therefore, (6.24) becomes

$\stackrel{‵}{\psi }\left(\mathbf{C},\text{Θ}\right)=\stackrel{‵}{\psi }\left(\mathbf{HC}{\mathbf{H}}^{\text{T}},\text{Θ}\right)\text{for all}\mathbf{H}\in \mathcal{O}.$

The requirement (6.25) on the functional form of $\stackrel{‵}{\psi }$ is equivalent to the condition that $\stackrel{‵}{\psi }$ depend on C only through the principal scalar invariants of C, i.e.,

$\psi =\overline{\psi }\left(I_1,I_2,I_3,\text{Θ}\right),$

where

$I_1=\text{tr}\mathbf{C},I_2=\frac{1}{2}\left[{\left(\text{tr}\mathbf{C}\right)}^2-\text{tr}\left({\mathbf{C}}^2\right)\right],I_3=\text{det}\mathbf{C}.$

Recalling that the Cauchy stress T is obtained from the Helmholtz free energy ψ by

$\mathbf{T}=2\rho \mathbf{F}\frac{\partial \stackrel{‵}{\psi }\left(\mathbf{C},\text{Θ}\right)}{\partial \mathbf{C}}{\mathbf{F}}^{\text{T}},$

we have, through a change of independent variable,

$\mathbf{T}=2\rho \mathbf{F}\left(\frac{\partial \overline{\psi }}{\partial I_1}\frac{d{I}_1}{d\mathbf{C}}+\frac{\partial \overline{\psi }}{\partial I_2}\frac{d{I}_2}{d\mathbf{C}}+\frac{\partial \overline{\psi }}{\partial I_3}\frac{d{I}_3}{d\mathbf{C}}\right){\mathbf{F}}^{\text{T}}.$

Recall from Section 2.5.1 the results

$\frac{d{I}_1}{d\mathbf{C}}=\mathbf{I},\frac{d{I}_2}{d\mathbf{C}}=I_1\mathbf{I}-\mathbf{C},\frac{d{I}_3}{d\mathbf{C}}=I_3{\mathbf{C}}^{-1}.$

Substitution of (6.28) into (6.27) gives

$\mathbf{T}={\alpha }_0\mathbf{I}+{\alpha }_1\mathbf{B}+{\alpha }_2{\mathbf{B}}^2,$

where B = FF^T is the Finger deformation tensor and

${\alpha }_0=2\rho I_3\frac{\partial \overline{\psi }}{\partial I_3},{\alpha }_1=2\rho \left(\frac{\partial \overline{\psi }}{\partial I_1}+I_1\frac{\partial \overline{\psi }}{\partial I_2}\right),{\alpha }_2=-2\rho \frac{\partial \overline{\psi }}{\partial I_2}.$

Note that since B and C have the same eigenvalues, they also have the same principal invariants (refer to Sections 2.3 and 3.3.5). Use of the Cayley-Hamilton theorem (2.75) gives the alternative form

$\mathbf{T}={\beta }_0\mathbf{I}+{\beta }_1\mathbf{B}+{\beta }_{-1}{\mathbf{B}}^{-1},$

where

${\beta }_0=2\rho \left(I_2\frac{\partial \overline{\psi }}{\partial I_2}+I_3\frac{\partial \overline{\psi }}{\partial I_3}\right),{\beta }_1=2\rho \frac{\partial \overline{\psi }}{\partial I_1},{\beta }_{-1}=-2\rho I_3\frac{\partial \overline{\psi }}{\partial I_2}.$

In nonlinear elasticity it is common to speak of the strain energy W per unit reference volume, i.e., the strain energy density, defined by

$W={\rho }_{\text{R}}\psi ,$

rather than the Helmholtz free energy. For isotropic thermoelastic materials, we have

$W=\overline{W}\left(I_1,I_2,I_3,\text{Θ}\right).$

It follows that (6.30b) becomes

${\beta }_0=2I_3^{-\frac{1}{2}}\left(I_2\frac{\partial \overline{W}}{\partial I_2}+I_3\frac{\partial \overline{W}}{\partial I_3}\right),{\beta }_1=2I_3^{-1/2}\frac{\partial \overline{W}}{\partial I_1}{\beta }_{-1}=-2I_3^{1/2}\frac{\partial \overline{W}}{\partial I_2},$

where we have used

$\frac{\rho }{{\rho }_{\text{R}}}=\frac{1}{J}=\frac{1}{\text{det}\mathbf{F}}=I_3^{-1/2},$

which follows from conservation of mass (4.56a). Recall that ρ is the density in the present configuration, ρ_R is the density in the reference configuration, and J is the determinant of the deformation gradient F.