Local form of balance principles

The Green-Nagdhi-Rivlin theorem, in conjunction with the various localization lemmas presented earlier, allowed us to derive the balance laws associated with mass, linear momentum and angular momentum. We will now use these balance laws to develop the local version of the first and second laws of thermodynamics. We will subsequently present the Lagrangian version of all the balance principles of continuum mechanics.

Eulerian form of balance principles

Notice that all the discussion so far is concerned with a spatial description of the various continuum fields. We will now summarize them and cast them in a form that highlights their physical significance.

Local form of the first law of thermodynamics

Recall that the first law of thermodynamics for continua takes the following form: $$\begin{split} \frac{d}{dt} \int_{B_t} \rho(\mathsf{x},t)\left(e(\mathsf{x},t) + \frac{1}{2}\mathsf{v}(\mathsf{x},t)\cdot\mathsf{v}(\mathsf{x},t)\right)\,dv &= \int_{B_t} \rho(\mathsf{x},t)\mathsf{b}(\mathsf{x},t)\cdot\mathsf{v}(\mathsf{x},t)\,dv\\ &+ \int_{\partial B_t} \mathsf{t}(\mathsf{x},t,\mathsf{n}(\mathsf{x},t))\cdot\mathsf{v}(\mathsf{x},t)\,da\\ &+ \int_{B_t} \rho(\mathsf{x},t)r(\mathsf{x},t)\,dv\\ &- \int_{\partial B_t} h(\mathsf{x},t,\mathsf{n}(\mathsf{x},t))\,da. \end{split}$$ Using the relation between the traction vector and the Cauchy stress tensor, and the Reynolds' transport theorem, we see that the first law can be written as $$\int_{B_t}\left(\left(\rho\frac{De}{Dt} + \rho (\mathsf{a} - \mathsf{b})\cdot\mathsf{v} - \text{div }(\mathsf{v}\cdot\sigma)\right) - \rho r\right)dv = -\int_{\partial B_t} h\,da.$$ Note that this equation has the familiar "volume integral = surface integral" structure, which suggests a use of the Cauchy tetrahedron argument to localize it. A straightforward application of the Cauch localization lemma informs us that thre exists a vector field $\mathsf{q}(\mathsf{x},t)$ in $B_t$ such that for any $\mathsf{x} \in B_t$ and $\mathsf{n} \in T_{\mathsf{x}}B_t$, $$h(\mathsf{x},t,\mathsf{n}) = \mathsf{q}(\mathsf{x},)\cdot\mathsf{n}.$$ The vector field $\mathsf{q}$ is called the heat flux. Using the definition of the heat flux and the divergence theorem, we see that $$\begin{split} \int_{\partial B_t} h\,da &= \int_{\partial B_t} \mathsf{q}\cdot\mathsf{n} \,da\\ &= \int_{B_t} \text{div }\mathsf{q}\,dv. \end{split}$$ Substituting this in the integral form of the first law, we get $$\int_{B_t}\left(\left(\rho\frac{De}{Dt} + \rho (\mathsf{a} - \mathsf{b})\cdot\mathsf{v} - \text{div }(\mathsf{v}\cdot\sigma)\right) - (\rho r - \text{div }\mathsf{q})\right)dv = 0.$$ The localization lemma then allows us to express this as the following partial differential equation: $$\rho\frac{De}{Dt} + \rho (\mathsf{a} - \mathsf{b})\cdot\mathsf{v} - \text{div }(\mathsf{v}\cdot\sigma) = \rho r - \text{div }\mathsf{q}.$$ This equation can be simplified by noting that $$\begin{split} \text{div }(\mathsf{v}\cdot\sigma) &= \sum \frac{\partial}{\partial x_j}(v_i \sigma_{ij})\\ &= \sum v_i \frac{\partial \sigma_{ij}}{\partial x_j} + \frac{\partial v_i}{\partial x_j}\sigma_{ij}\\ &= \mathsf{v} \cdot \text{div }\sigma + \sigma \cdot \text{grad }\mathsf{v}. \end{split}$$ Using this in the local form of the first law in conjunction with the linear momentum balance relation, we get the following local form of the first law of thermodynamics: $$\rho\frac{De}{Dt} = \sigma \cdot \text{grad }\mathsf{v} + \rho r - \text{div }\mathsf{q}.$$ Using the fact that the Cauchy stress tensor is symmetric, we can write the term $\sigma \cdot \text{grad }\mathsf{v}$ as $$\begin{split} \sigma \cdot \text{grad }\mathsf{v} &= \frac{1}{2}(\sigma + \sigma^T)\cdot \text{grad }\mathsf{v}\\ &= \sigma \cdot \frac{1}{2}(\text{grad }\mathsf{v} + (\text{grad }\mathsf{v})^T)\\ &= \sigma \cdot \mathsf{d}, \end{split}$$ where $\mathsf{d}(\mathsf{x},t):T_{\mathsf{x}}B_t \to T_{\mathsf{x}}B_t$, defined as $$\mathsf{d}(\mathsf{x},t) = \frac{1}{2}\left(\text{grad }\mathsf{v}(\mathsf{x},t) + (\text{grad }\mathsf{v}(\mathsf{x},t))^T\right),$$ is the rate of deformation tensor. Using this, we obtain the final expression for the local form of the first law of thermodynamics as $$\rho(\mathsf{x},t)\frac{De(\mathsf{x},t)}{Dt} = \sigma(\mathsf{x},t)\cdot\mathsf{d}(\mathsf{x},t) + \rho(\mathsf{x},t)r(\mathsf{x},t) - \text{div }\mathsf{q}(\mathsf{x},t).$$ This is also called the energy balance equation for obvious reasons.

Remark

Notice how the energy balance principle does not explicitly involve the body force and spatial acceleration terms.

Local form of the second law of thermodynamics

Let us now turn to the second law of thermodynamics. Recall that the Clausius-Duhem inequality reads $$\frac{d}{dt}\int_{B_t} \rho(\mathsf{x},t)\eta(\mathsf{x},t)\,dv \ge \int_{B_t} \frac{\rho(\mathsf{x},t)r(\mathsf{x},t)}{\theta(\mathsf{x},t)}\,dv - \int_{\partial B_t} \frac{h(\mathsf{x},t,\mathsf{n}(\mathsf{x},t))}{\theta(\mathsf{x},t)}\,da.$$ It is a straightforward exercise to localize the second law. Using the relation $\mathsf{h} = \mathsf{q}\cdot\mathsf{n}$ and the divergence theorem, we see that $$\frac{d}{dt}\int_{B_t} \rho(\mathsf{x},t)\eta(\mathsf{x},t)\,dv \ge \int_{B_t} \left(\frac{\rho(\mathsf{x},t)r(\mathsf{x},t)}{\theta(\mathsf{x},t)} - \text{div }\left(\frac{\mathsf{q}}{\theta}\right)\right)\,dv.$$ The localization lemma then yields the following local form of the second law of thermodynamics: $$\rho\theta\frac{D\eta}{Dt} \ge \rho r - \text{div }\left(\mathsf{q}{\theta}\right).$$ This expression can be simplified by noting the following relations: $$\begin{split} \text{div }\frac{\mathsf{q}}{\theta} &= \sum \frac{\partial}{\partial x_i}\left(\frac{q_i}{\theta}\right)\\ &= \frac{1}{\theta}\sum \frac{\partial q_i}{\partial x_i} - \sum \frac{q_i}{\theta^2}\frac{\partial \theta}{\partial x_i}\\ &= \frac{1}{\theta}\text{div }\mathsf{q} - \frac{\mathsf{q}}{\theta^2}\cdot\text{grad }\theta. \end{split}$$ We can thus express the local form of the second law as $$\rho(\mathsf{x},t)\theta(\mathsf{x},t)\frac{D\eta(\mathsf{x},t)}{Dt} \ge \rho(\mathsf{x},t)r(\mathsf{x},t) - \text{div }\mathsf{q}(\mathsf{x},t) + \frac{\mathsf{q}(\mathsf{x},t)}{\theta(\mathsf{x},t)}\cdot\text{grad }\theta(\mathsf{x},t).$$ This is also known as the entropy inequality, or the dissipation inequality.

Summary of balance principles

We have now derived all the fundamental balance principles of continuum mechanics starting from the first and second laws of thermodynamics. Recall that we obtained these balance laws by postulating the invariance of the first and second laws of thermodynamics with respect to Euclidean observer transformations. Let us now summarize these balance principles:

Balance law	Local form
Mass	$\dot{\rho}(\mathsf{x},t) + \rho(\mathsf{x},t)\,\text{div }\mathsf{v}(\mathsf{x},t) = 0$
Linear momentum	$\rho(\mathsf{x},t)\dot{\mathsf{v}}(\mathsf{x},t) = \rho(\mathsf{x},t)\mathsf{b}(\mathsf{x},t) - \text{div }\sigma(\mathsf{x},t)$
Angular momentum	$\sigma(\mathsf{x},t) = \sigma^T(\mathsf{x},t)$
Energy	$\rho(\mathsf{x},t)\dot{e}(\mathsf{x},t) = \sigma(\mathsf{x},t)\cdot\mathsf{d}(\mathsf{x},t) + \rho(\mathsf{x},t)r(\mathsf{x},t) - \text{div }\mathsf{q}(\mathsf{x},t)$
Entropy	$\rho(\mathsf{x},t)\theta(\mathsf{x},t)\dot{\eta}(\mathsf{x},t) \ge \rho(\mathsf{x},t)r(\mathsf{x},t) - \text{div }\mathsf{q}(\mathsf{x},t) + \left(\mathsf{q}(\mathsf{x},t)\cdot\text{grad }\theta(\mathsf{x},t)\right)/\theta(\mathsf{x},t)$

Remark

In all the equations displayed above, an overdot represents the material time derivative. This is a convention often used in continuum mechanics. Just to reiterate, the quantity $\dot{\rho}(\mathsf{x},t)$ denotes the material time derivative of the mass density, and not the partiat time derivative of the mass density.

All the balance equations displayed above can be written in the following form: $$\frac{\partial \chi(\mathsf{x},t)}{\partial t} + \text{div }\phi(\mathsf{x},t) = \varsigma(\mathsf{x},t).$$ Here, $\chi$ is a physical field variable, $\phi$ is the flux associated with the field $\chi$, and $\varsigma$ is a source term for the field $\chi$. The entropy inequality can also be written in the following form: $$\frac{\partial \chi(\mathsf{x},t)}{\partial t} + \text{div }\phi(\mathsf{x},t) \ge \varsigma(\mathsf{x},t).$$ Notice the formal similarity of this inequality with that of the other balance equalities. The various field, flux and source terms are listed below:

Physical variable	Field	Flux	Source
Mass	$\rho$	$\rho\mathsf{v}$	0
Linear momentum	$\rho\mathsf{v}$	$\rho\,\mathsf{v}\otimes\mathsf{v} - \sigma$	$\rho\mathsf{b}$
Angular momentum	$\mathsf{x} \times \rho\mathsf{v}$	$\mathsf{x} \times (\rho\,\mathsf{v}\otimes\mathsf{v} - \sigma)$	$\mathsf{x}\times\rho\mathsf{b}$
Energy	$\rho\left(e + \frac{1}{2}\mathsf{v}\cdot\mathsf{v}\right)$	$\rho\left(e + \frac{1}{2}\mathsf{v}\cdot\mathsf{v}\right)\,\mathsf{v} + \mathsf{q} - \mathsf{v}\cdot\sigma$	$\rho r + \rho \mathsf{v}\cdot\mathsf{b}$
Entropy	$\rho\eta$	$\rho\eta\,\mathsf{v} + q/\theta$	$\rho r/\theta$

This form of expressing the balance principles is known as the master balance equation/inequality. This form is especially convenient to provide a nice physical interpretation to the balance principles. The master form of the balance principles informs us that the rate at which a continuum field variable evolves in a small volume inside the continuum is a consequence of the influx of a corresponding flux variable across the surface of the volume element, and as a consequence of an internal source term for this field variable.

Remark

The master form of the balance principles is especially useful for the control volume methods in fluid dynamics.

Lagrangian form of balance principles

Our study of the balance principles of continuum mechanics has thus far been in the Eulerian setting. When we are intereted in the mechanics of solids, it helps to reformulate the balance principles in the Lagrangian setting. The reason why a Lagrangian formulation is more useful in the study of solids is that the magnitude of the deformation is often small enough in deformation processes involving solids in comparison to that of a fluid that we can effectively tag particles based on their reference positions.

Throughout this discussion, we will suppose that we are equipped with an initial configuration $\chi_0:\mathcal{B}\to\mathbb{R}^3$ that places the body $\mathcal{B}$ in the region $B_0 \subseteq\mathbb{R}^3$. The current configuration $\chi_t:\mathcal{B}\to\mathbb{R}^3$ places the body at $B_t \subseteq \mathbb{R}^3$ at the current instant of time, and $\varphi_t:B_0 \to B_t$ defined as $\varphi_t = \chi_t \circ \chi_0^{-1}$ denotes the deformation map.

Mass balance

Let us start with the mass continuity equation. We studied the mass continuity equation in the context of our discussion of kinematics; let us revisit this now. Let us define the Lagrangian form of the material density, $\rho_0(\cdot,t):B_0 \to \mathbb{R}$, using the the physical requirement that the mass in $B_0$ is identical to the mass in $B_t$: $$\int_{B_0} \rho_0(\mathsf{X},t)\,dV = \int_{B_t} \rho(\mathsf{x},t)\,dv.$$ Taking time derivatives on both sides, we see that $$\begin{split} \frac{d}{dt}\int_{B_0} \rho_0(\mathsf{X},t)\,dV &= \frac{d}{dt}\int_{B_t} \rho(\mathsf{x},t) \, dv\\ &= \int_{B_t} \left(\frac{D\rho(\mathsf{x},t)}{Dt} + \rho(\mathsf{x},t)\text{div }\mathsf{v}(\mathsf{x},t)\right)\,dv\\ &= 0. \end{split}$$ This informs us, using the standard localization argument, that $$\frac{\partial \rho_0(\mathsf{X},t)}{\partial t} = 0 \quad\Rightarrow\quad \rho_0(\mathsf{X},t) = \rho_0(\mathsf{X},0) = \rho_0(\mathsf{X}).$$ In the last step, the quantity $\rho_0(\mathsf{X})$ is introduced as a compact notation for $\rho_0(\mathsf{X},0$. The Lagrangian form of the mass balance thus recovers the obvious result that the initial mass density does not evolve with time.

First law of thermodynamics

Let us now look at the balance of energy in integral form. This can be written in integral form, using the expressions derived earlier, as follows: $$\int_{B_t} \rho\frac{De}{Dt}\,dv = \int_{B_t} \sigma \cdot \mathsf{d} \, dv + \int_{B_t} \rho r \, dv - \int_{\partial B_t} \mathsf{q}\cdot\mathsf{n}\,da.$$ To cast this in Lagrangian form, we need to define the Lagrangian version of the various quantities involved. To begin with, let us define the Lagrangian version of the energy density $E(\cdot,t):B_0 \to \mathbb{R}$ as $$E(\mathsf{X},t) = e(\mathsf{x},t),$$ where $\mathsf{X} \in B_0$ and $\mathsf{x} = \varphi(\mathsf{X},t) \in B_t$. Using this, we see that $$\int_{B_t} \rho\frac{De}{Dt}\,dv = \int_{B_0} \rho \frac{DE}{Dt} JdV = \int_{B_0} \rho_0\frac{\partial E}{\partial t}\,dV.$$ Notice how we have used the volume transformation rule and the mass continuity equation $\rho_0 = \rho J$ in deriving this.

The term in the energy equation involving the heat source can be similarly expressed in Lagrangian form as $$\int_{B_t} \rho r \, dv = \int_{B_0} \rho_0 R \, dV,$$ where we have introduced the Lagrangian heat source term using the definition $R(\mathsf{X},t) = r(\mathsf{x},t)$.

Let us now turn to the term $$\int_{B_t} \sigma \cdot \mathsf{d}\,dv = \int_{B_t} \sigma \cdot \text{grad }\mathsf{v}\,dv.$$ To compute the Lagrangian version of this, begin by recalling from our discussion of kinematics that $$\dot{\mathsf{F}} = (\text{grad }\mathsf{v})\mathsf{F} \quad\Rightarrow\quad \text{grad }\mathsf{v} = \dot{\mathsf{F}}\mathsf{F}^{-1}.$$ We thus see that $$\begin{split} \int_{B_t} \sigma(\mathsf{x},t)\cdot\mathsf{d}(\mathsf{x},t)\,dv &= \int_{B_0} \sigma(\varphi_t(\mathsf{X}),t)\cdot\dot{\mathsf{F}}(\mathsf{X},t)\mathsf{F}^{-1}(\varphi_t(\mathsf{X}),t)\,J(\mathsf{X},t)dV\\ &= \int_{B_0} J(\mathsf{X},t)\sigma(\varphi_t(\mathsf{X}),t)\mathsf{F}^{-T}(\mathsf{X},t)\cdot\dot{\mathsf{F}}(\mathsf{X},t)\,dV. \end{split}$$ To understand the last step, note that $$\begin{split} \sigma \cdot \dot{\mathsf{F}}\mathsf{F}^{-1} &= \mathsf{F}^{-T}\dot{\mathsf{F}}^T\cdot\sigma^T\\ &= \dot{\mathsf{F}}^T\cdot\mathsf{F}^{-1}\sigma\\ &= \sigma\mathsf{F}^{-T}\cdot\dot{\mathsf{F}}. \end{split}$$ Notice how the symmetry of the Cauchy stress tensor is used here. Let us now introduce the first Piola-Kirchhoff stress tensor, $\mathsf{P}(\mathsf{X},t):T_{\mathsf{X}B_0} \to T_{\mathsf{x}}B_t$ as follows: $$\mathsf{P}(\mathsf{X},t) = J(\mathsf{X},t)\sigma(\varphi_t(\mathsf{X}),t)\mathsf{F}^{-T}(\mathsf{X},t).$$ Before proceeding further, let us quickly consider the Cartesian coordinate representation of the first Piola-Kirchhoff stress tensor. If $(\mathsf{e}_i)$ and $(\mathsf{E}_I)$ are orthonormal bases of $T_{\mathsf{x}}B_t$ and $T_{\mathsf{X}}B_0$, respectively, then $$\mathsf{P}(\mathsf{X},t) = \sum P_{iJ}(\mathsf{X},t)\,\mathsf{e}_i \otimes \mathsf{E}_J, \qquad \sigma(\mathsf{x},t) = \sum \sigma_{ij}(\mathsf{x},t)\,\mathsf{e}_i \otimes \mathsf{e}_j.$$ The components of the first Piola-Kirchhoff and Cauchy stress tensors are related as follows: $$P_{iJ} = \sum J\sigma_{ij}F^{-T}_{jJ}.$$ The arguments of the various quantities in the equation above are suppressed for notational convenience. Note that since $\mathsf{P}(\mathsf{X},t)$ is a linear map between two different vector spaces , it is not equivalent to a tensor in the conventional sense of the term. It is sometimes called a two-point tensor. Note also that for this same reason, it doesn't make sense to

Returning back to the term involving the Cauchy stress tensor in the energy equation, we see that $$\int_{B_t} \sigma(\mathsf{x},t)\cdot\mathsf{d}(\mathsf{x},t)\,dv = \int_{B_0} \mathsf{P}(\mathsf{X},t)\cdot\dot{\mathsf{F}}(\mathsf{X},t)\,dV.$$ There is yet another way to express this term in Lagrangian form. To derive this, note that $$\begin{split} \mathsf{d} &= \frac{1}{2}\left(\text{grad }\mathsf{v} + (\text{grad }\mathsf{v})^T\right)\\ &= \frac{1}{2}\left(\dot{\mathsf{F}}\mathsf{F}^{-1} + \mathsf{F}^{-T}\dot{\mathsf{F}}^T\right). \end{split}$$ It follows that $$\mathsf{F}^T\mathsf{d}\mathsf{F} = \frac{1}{2}\left(\mathsf{F}^T\dot{\mathsf{F}} + \dot{\mathsf{F}}^T\mathsf{F}\right) = \mathsf{D},$$ where $\mathsf{D} = \dot{\mathsf{C}}/2$ is the Lagrangian rate of deformation tensor. Inverting this relation, we get $$\mathsf{d} = \mathsf{F}^{-T}\mathsf{D}\mathsf{F}^{-1}.$$ Using this in the term involving the stress tensor in the energy equation, we see that $$\begin{split} \int_{B_t} \sigma \cdot \mathsf{d}\,dv &= \int_{B_0} \sigma \cdot \mathsf{F}^{-T}\mathsf{D}\mathsf{F}^{-1}J\,dV\\ &= \int_{B_0} J\mathsf{F}^{-1}\sigma\mathsf{F}^{-T}\cdot\mathsf{D}\,dV. \end{split}$$ Introducing the second Piola-Kirchhorff stress tensor $\mathsf{S}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{X}}B_0$ as $$\mathsf{S}(\mathsf{X},t) = J(\mathsf{X},t)\mathsf{F}^{-1}(\varphi_t(\mathsf{X}),t)\sigma(\varphi_t(\mathsf{X}),t)\mathsf{F}^{-T}(\mathsf{X},t),$$ it follows that $$\int_{B_t} \sigma\cdot\mathsf{d}\,dv = \int_{B_0} \mathsf{S}\cdot\mathsf{D}\,dV.$$ Note that the second Piola-Kirchhoff stress tensor can be expressed in terms of the Cartesian coordinate system introduced earlier as $$\mathsf{S}(\mathsf{X},t) = \sum S_{IJ}(\mathsf{X},t)\,\mathsf{E}_I \otimes \mathsf{E}_J,$$ where the component fields can be computed in terms of the Cauchy stress tensor and the deformation gradient using a procedure similar to the one employed in the context of the first Piola-Kirchhoff stress tensor. Note that the second Piola-Kirchhoff stress tensor is symmetric: $\mathsf{S}^T = \mathsf{S}$.

Let us collect together the results involving the first and second Piola-Kirchhoff stress tensors that we have derived so far: $$\int_{B_t} \sigma \cdot \mathsf{d}\,dv = \int_{B_0} \mathsf{P} \cdot \dot{\mathsf{F}}\,dV = \int_{B_0} \mathsf{S}\cdot\mathsf{D}\,dV.$$ In terms of the Cartesian coordinate system introduced earlier, this can be written as follows: $$\sum \int_{B_t} \sigma_{ij}d_{ij}\,dv = \sum \int_{B_0} P_{iJ}\dot{F}_{iJ}\,dV = \sum \int_{B_0} S_{IJ}D_{IJ}\,dV.$$ We see that there is a natural pairing between specific stress and strain rate measures: $(\sigma,\mathsf{d})$, $\mathsf{P},\dot{\mathsf{F}})$ and $(\mathsf{S},\mathsf{D})$. The stress and strain rate measures in each pair are said to be conjugate to each other. This observation will turn out to be crucial when we discuss constitutive relations later on.

Remark

Noting that $\mathsf{D} = \dot{\mathsf{C}}/2 = \dot{\mathsf{E}}$, where $\mathsf{E}$ is the Lagrangian strain tensor, we see that $$\mathsf{P}\cdot\dot{\mathsf{F}} = \mathsf{S}\cdot\dot{\mathsf{E}}.$$ But note that it is not true that the rate of deformation tensor $\mathsf{d}$ is the material time derivative of the Eulerian strain tensor $\mathsf{e}$. In fact, it can be shown that $$\dot{\mathsf{e}} = \mathsf{d} - (\text{grad }\mathsf{v})^T\mathsf{e} - \mathsf{e}\,\text{grad }\mathsf{v}.$$ It is left as an exercise to verify this.

It is possible to gain some physical intuition regarding the meaning of the first and second Piola-Kirchhoff stress tensors by considering the force acting on an elementary area $\Delta a$ with normal $\mathsf{n}$ centered at $\mathsf{x} \in B_t$. Let $\Delta A$ be the elementary area at $\mathsf{X} = \varphi_t(\mathsf{x}) \in B_0$ with normal $\mathsf{N}$. The force acting on the elementary area $\Delta a$ is given by $\sigma \mathsf{n} \Delta a$. Using the Piola transform, namely $\mathsf{n}\Delta a = J\mathsf{F}^{-T}\mathsf{N}\Delta A$, we see that $$\sigma \mathsf{n} \Delta a = J\sigma \mathsf{F}^{-T}\mathsf{N}\Delta A = \mathsf{P}\mathsf{N}\Delta A.$$ We thus see that the first Piola-Kirchhoff stress tensor is the force acting on an area element in the current configuration per unit referential area. To understand what the second Piola-Kirchhoff stress tensor means in this setting, note that $\mathsf{F}^{-1}\mathsf{P}\mathsf{N}\Delta A = \mathsf{S}\mathsf{N}\Delta A$ is the force acting on the area element $\Delta A$ per unit referential area. The idea behind this conclusion is that if $\mathsf{t}$ is the force acting on the area element $\Delta a$, then the pull-back of this force that yields the force acting on the area $\Delta A$ is given by $\mathsf{F}^{-1}\mathsf{T}$, or, equivalently, $\mathsf{t} = \mathsf{F}\mathsf{T}$.

Remark

Note that it is sufficient to think of the physical content of the stress tensor in terms of the interpretation of the Cauchy stress tensor given earlier, and to view the first and second Piola-Kirchhoff stress tensors as special mathematical transforms of the Cauchy stress tensor, along the lines of the foregoing calculations.

Let us now turn to the final term in the energy equation, namely $$\int_{\partial B_t} \mathsf{q}\cdot\mathsf{n}\,da.$$ To obtain the Lagrangian version of this equation, let us use the Piola transform introduced in the context of area transformation rules eariler to get $$\begin{split} \int_{\partial B_t} \mathsf{q}\cdot\mathsf{n}\,da &= \int_{\partial B_0} \mathsf{q} \cdot J\mathsf{F}^{-T}\mathsf{N}\,dA\\ &= \int_{\partial B_0} J\mathsf{F}^{-1}\mathsf{q}\cdot\mathsf{N}\,dA. \end{split}$$ Let us introduce the Lagrangian heat flux vector field $\mathsf{Q}(\mathsf{X},t):T_{\mathsf{X}}B_0 \to T_{\mathsf{X}}B_0$ as follows: $$\mathsf{Q}(\mathsf{X},t) = J(\mathsf{X},t)\mathsf{F}^{-1}(\varphi_t(\mathsf{X}),t)\mathsf{q}(\varphi_t(\mathsf{X}),t).$$ Recall that $\mathsf{Q}$ is just the Piola transform of $\mathsf{q}$.

Remark

Note that the symbol $\mathsf{Q}$ has been used earlier to denote an orthogonal map. In practice, the specific meaning of a symbol like $\mathsf{Q}$ is always evident from the context.

Using this definition, we can write the heat flux term as $$\int_{\partial B_t} \mathsf{q}\cdot\mathsf{n}\,da = \int_{B_0}\mathsf{Q}\cdot\mathsf{N}\,dA.$$ Using the divergence theorem, we see that $$\int_{\partial B_0} \mathsf{Q}\cdot\mathsf{N}\,dA = \int_{B_0} \text{DIV }\mathsf{Q}\,dV.$$ Putting all this together and using the localization principle, we finally obtain the Lagrangian form of the energy equation as $$\rho_0(\mathsf{X})\frac{\partial \mathsf{E}(\mathsf{X},t)}{\partial t} = \mathsf{P}(\mathsf{X},t)\cdot\dot{\mathsf{F}}(\mathsf{X},t) + \rho_0(\mathsf{X},t)R(\mathsf{X},t) - \text{DIV }\mathsf{Q}(\mathsf{X},t),$$ or, equivalently, as $$\rho_0(\mathsf{X})\frac{\partial \mathsf{E}(\mathsf{X},t)}{\partial t} = \mathsf{S}(\mathsf{X},t)\cdot\mathsf{D}(\mathsf{X},t) + \rho_0(\mathsf{X},t)R(\mathsf{X},t) - \text{DIV }\mathsf{Q}(\mathsf{X},t).$$

Remark

Notice how the nonlinear material time derivative that appeared in the spatial formulation gets replaced by a simpler and linear partial derivative in the Lagrangian form.

Linear momentum balance

We have derived the Lagrangian version of the mass and energy balance equations so far. Let us now turn to the balance of linear momentum. We will need the following additional definition: the Lagrangian body force density $\mathsf{B}(\cdot,t):B_0 \to TB_t$ in the usual manner as $$\mathsf{B}(\mathsf{X},t) = \mathsf{b}(\varphi_t(\mathsf{X}), t).$$ Let us now write the inegral form of the linear momentum balance as follows: $$\begin{split} \int_{B_t} \rho \frac{D\mathsf{v}}{Dt}\,dv &= \int_{B_t} \rho\mathsf{b}\,dv + \int_{B_t} \text{div }\sigma\,dv\\ &= \int_{B_t} \rho\mathsf{b}\,dv + \int_{\partial B_t} \sigma\mathsf{n}\,da. \end{split}$$

Note

Notice that the equation above involves the inegration of vector fields. As remarked earlier, integration of vector fields is well-defined only in the Euclidean setting. Since we deal only with Euclidean spaces in this course, we are justified writing an integral equation involving vector fields as above.

Let us now cast this equation in Lagrangian form using the standard procedure: $$\begin{split} \int_{B_0} \rho_0\frac{\partial \mathsf{V}}{\partial t}\,dV &= \int_{B_0} \rho_0\mathsf{B}\,dV + \int_{\partial B_0} J\sigma\mathsf{F}^{-T}\mathsf{N}\,dA\\ &= \int_{B_0} \rho_0\mathsf{B}\,dV + \int_{\partial B_0} \mathsf{P}\mathsf{N}\,dA\\ &= \int_{B_0} \rho_0\mathsf{B}\,dV + \int_{B_0} \text{DIV }\mathsf{P}\,dV. \end{split}$$ The localization lemma then yields the local form of the linear momentum balance equation in Lagrangian form as $$\rho_0(\mathsf{X})\frac{\partial \mathsf{V}(\mathsf{X},t)}{\partial t} = \rho_0(\mathsf{X})\mathsf{B}(\mathsf{X},t) + \text{DIV }\mathsf{P}(\mathsf{X},t).$$ Notice again how the time derivative on the left hand side is a simple partial derivative in sharp contrast to the nonlinear material time derivative in the spatial form of the linear momentum balance equation.

Angular momentum balance

Let us now look at the Lagrangian form of the angular momentum balance principle. Recall that this takes a particularly simple form in the Eulerian setting, namely $\sigma = \sigma^T$. Using the definitions of the first and second Piola-Kirchhoff stress tensors, it is clear that $$J^{-1}\mathsf{P}\mathsf{F}^T = J^{-1}\mathsf{F}\mathsf{P}^T \quad\Rightarrow\quad \mathsf{P}\mathsf{F}^T = \mathsf{F}\mathsf{P}^T,$$ and $$J^{-1}\mathsf{F}\mathsf{S}\mathsf{F}^T = J^{-1}\mathsf{F}\mathsf{S}^T \quad\Rightarrow\quad \mathsf{S} = \mathsf{S}^.$$ We thus see that the second Piola-Kirchhoff stress tensor is symmetric. Note that it doesn't make sense to talk about the symmetry or antisymmetry of the first Piola-Kirchhoff stress tensor since as a linear map it acts between different vector spaces.

Second law of thermodynamics

Finally, let us work out the Lagrangian form of the second law of thermodynamics. Following the same procedure as before to define the Lagrangian version of the various scalar fields, let us define the Lagrangian entropy density $N(\cdot,t):B_0 \to \mathbb{R}$ and the Lagrangian absolute temperature $\Theta(\cdot,t):B_0 \to \mathbb{R}$ as $$N(\mathsf{X},t) = \eta(\mathsf{x},t), \qquad \Theta(\mathsf{X},t) = \theta(\mathsf{x},t),$$ where $\mathsf{X} \in B_0$ and $\mathsf{x} = \varphi(\mathsf{X},t) \in B_t$. The integral form of the second law, namely the Clausius-Duhem inequality, reads $$\int_{B_t} \rho\frac{D\eta}{Dt} \,dv \ge \int_{B_t} \frac{\rho r}{\theta} \,dv - \int_{\partial B_t} \frac{\mathsf{q}}{\theta}\cdot\mathsf{n}\,da.$$ Using the defintion of the Lagrangian version of all the quantities involved in this equation, it is straightforward to verify that this can be cast in the following Lagrangian form: $$\int_{B_0} \rho_0\frac{\partial N}{\partial T} \,dV \ge \int_{B_0} \frac{\rho_0 R}{\Theta}\,dV - \int_{\partial B_0} \frac{\mathsf{Q}}{\Theta}\cdot\mathsf{N}\,dA.$$ The standard localization argument then results in the following Lagrangian form of the second law of thermodynamics: $$\rho_0(\mathsf{X})\frac{\partial N(\mathsf{X},t)}{\partial t} \ge \frac{\rho_0(\mathsf{X})R(\mathsf{X},t)}{\Theta(\mathsf{X},t)} - \text{DIV }\left(\frac{\mathsf{Q}(\mathsf{X},t)}{\Theta(\mathsf{X},t)}\right),$$ or, equivalently as $$\rho_0\Theta\frac{\partial N}{\partial t} \ge \rho_0 R - \text{DIV }\mathsf{Q} + \frac{\mathsf{Q}}{\Theta}\cdot\text{GRAD }\Theta.$$ This latter form is especially useful in applications.

Summary of Lagrangian balance laws

Let us now summarize the Lagrangian version of the balance principles of continuum mechanics that we just derived:

Balance law	Local form
Mass	$\dot{\rho}(\mathsf{X},t) = 0$
Linear momentum	$\rho_0 \dot{V} = \rho_0 B + \text{DIV }\mathsf{P}$
Angular momentum	$\mathsf{P}\mathsf{F}^T = \mathsf{F}^T\mathsf{P}$
	$\mathsf{S} = \mathsf{S}^T$
Energy	$\rho\dot{E} = \mathsf{P}\cdot\dot{\mathsf{F}} + \rho_0 R - \text{DIV }\mathsf{Q}$
	$\rho_0\dot{E} = \mathsf{S}\cdot\dot{\mathsf{E}} + \rho_0 R - \text{DIV }\mathsf{Q}$
Entropy	$\rho_0 \Theta \dot{N} \ge \rho_0 R - \text{DIV }\mathsf{Q} + \frac{\mathsf Q}{\Theta}\cdot\text{GRAD }\Theta$

In all the equations displayed above, an overdot represents the material time derivative, which in the case of Lagrangian variables is just the partial time derivative. It is worth reiterating that the nonlinearity present in the convective part of the material time derivative in the Eulerian form of the balance laws is conspicuously absent in the Lagrangian formulation. This has significant implications on whether, and how easily, we can solve these equations when provided with the appropriate initial and boundary conditions.

As before, we can also rewrite this in the form a master balance equation, $$\frac{\partial \chi_0}{\partial t} + \text{DIV }\Phi = \Sigma,$$ or an inequality of the form $$\frac{\partial \chi_0}{\partial t} + \text{DIV }\Phi \ge \Sigma,$$ where $\chi_0$, $\Phi$ and $\Sigma$ are the Lagrangian versions of the field, flux and source variables. The master form of the balance laws are summarized in the table below:

Physical variable	Field	Flux	Source
Mass	$\rho_0$	0	0
Linear momentum	$\rho_0 \mathsf{V}$	$-\mathsf{P}$	$\rho_0 \mathsf{B}$
Angular momentum	$\varphi \times \rho_0 V$	$-\varphi \times \mathsf{P}$	$\varphi \times \rho_0 \mathsf{B}$
Energy	$\rho_0 \left(E + \frac{1}{2}\mathsf{V}\cdot\mathsf{V}\right)$	$\mathsf{Q} - \mathsf{V}\cdot\mathsf{P}$	$\rho_0 R + \rho_0 \mathsf{V} \cdot \mathsf{B}$
Entropy	$\rho_0 N$	$\mathsf{Q}/\Theta$	$\rho_0 R/\Theta$

Notice how the Lagrangian version of the master balance laws are significantly simpler than their Eulerian counterparts.

Remark

The cross product term $\varphi \times \mathsf{P}$ in the angular momentum balance law displayed above means the following: for any constant $\mathsf{C} \in T_{\mathsf{X}}B_0$, $$(\varphi(\mathsf{X},t)\times\mathsf{P}(\mathsf{X},t))\mathsf{C} = \varphi(\mathsf{X},t) \times (\mathsf{P}(\mathsf{X},t)\mathsf{C}).$$ Note that $\varphi \times \mathsf{P}$ is also a two-point tensor, like $\mathsf{P}$.