2.6. Integral Balance Equations in reference space

2.6.1. Mass

Integal balance for a material volume \(V_t\) reads

\[\begin{split}\begin{aligned} 0 & = \dfrac{d}{dt}\int_{V_t} \rho(\vec{r}, t) \, dV = \\ & = \dfrac{d}{dt}\int_{V_0} \rho(\vec{r}(\vec{r}_0,t), t) \, J(\vec{r}_0,t) \, dV_0 = \\ & = \dfrac{d}{dt}\int_{V_0} \rho_0(\vec{r}_0,t) \, J(\vec{r}_0,t) \, dV_0 = \\ & = \int_{V_0} \dfrac{D}{D t} \left( \rho_0(\vec{r}_0,t) \, J(\vec{r}_0,t) \right) \, dV_0 \ . \end{aligned}\end{split}\]

Since the domain \(V_0\) is arbitrary, with some abuse of notation to indicate the density field as \(\rho\), hidihg the dependence on the independet fvariables \(\rho_0(\vec{r}_0,t)\), the differential balance in reference space follows

\[\dfrac{D}{Dt} \left( \rho \, J \right) = 0\]

or

\[\rho \, J = \rho^0 \ ,\]

i.e. the product \(\rho \, J\) equals the initial density field \(\rho^0\), assuming that the determinant of the transformation is \(J^0 = 1\), in the reference configuration.

2.6.2. Momentum

Integral balance for a material volume \(V_t\) reads

\[\begin{split}\begin{aligned} & \dfrac{d}{dt}\int_{V_t} \rho \vec{v} \, dV = \int_{V_t} \rho \vec{g} + \oint_{\partial V_t} \hat{n} \cdot \mathbb{T} \, dS \\ & \dfrac{d}{dt}\int_{V_0} \rho \, J \, \vec{v} \, dV_0 = \int_{V_0} \rho \, J \, \vec{g} + \oint_{\partial V_t} \hat{n}_0 \cdot \left( J \mathbb{F}^{-T} \cdot \mathbb{T} \right) \, d S_0 \\ & \dfrac{d}{dt}\int_{V_0} \rho^0 \, \vec{v} \, dV_0 = \int_{V_0} \rho^0 \, \vec{g} + \oint_{\partial V_0} \hat{n}_0 \cdot \mathbb{\Sigma}_n \, d S_0 \\ & \int_{V_0} \rho^0 \, \dfrac{D}{Dt} \vec{v} \, dV_0 = \int_{V_0} \rho^0 \, \vec{g} + \oint_{V_0} \nabla_0 \cdot \mathbb{\Sigma}_n \, d S_0 \\ \end{aligned}\end{split}\]

Since the domain \(V_0\) is arbitrary, the differential balance in reference space follows

\[\rho^0 \dfrac{D \vec{v}}{D t} = \rho^0 \, \vec{g} + \nabla_0 \cdot \mathbb{\Sigma}_n\]

Nanson’s formula

\[d \vec{r} = d \vec{r}_0 \cdot \dfrac{\partial \vec{r}}{\partial \vec{r}_0} = d \vec{r}_0 \cdot \nabla_0 \vec{r} = d \vec{r}_0 \cdot \mathbb{F}\]

\[\begin{split}\begin{aligned} dV & = J \, dV_0 \\ d \vec{r} \cdot \hat{n} dS & = J \, d \vec{r}_0 \cdot \hat{n}_0 \, dS_0 \\ d \vec{r}_0 \cdot \mathbb{F} \cdot \hat{n} \, dS & = J \, d \vec{r}_0 \cdot \hat{n}_0 \, dS_0 \\ \end{aligned}\end{split}\]

must be true for all \(\vec{r}_0\) arbitrary, so that

\[\mathbb{F} \cdot \hat{n} \, d S = J \, \hat{n}_0 \, d S_0\]

and

\[\begin{split}\begin{aligned} \hat{n} \, d S & = J \, \mathbb{F}^{-1} \cdot \hat{n}_0 \, d S_0 = \\ & = J \, \hat{n}_0 \cdot \mathbb{F}^{-T} \, d S_0 \end{aligned}\end{split}\]

Stress tensors

Cauchy stress tensor.

Piola-Kirchhoff I - transpose of normal stress tensors.

Piola-Kirchhoff II

Example 2.1 (Relation between description in physical and reference space)

\[\begin{split}\begin{aligned} \rho^0 \dfrac{D \vec{v}}{D t} & = \rho^0 \, \vec{g} + \nabla_0 \cdot \mathbb{\Sigma}_n \\ J \, \rho \, \dfrac{D \vec{v}}{D t} & = J \, \rho \, \vec{g} + \nabla_0 \cdot \mathbb{\Sigma}_n \\ \rho \, \dfrac{D \vec{v}}{D t} & = \rho \, \vec{g} + \frac{1}{J} \nabla_0 \cdot \mathbb{\Sigma}_n \\ \end{aligned}\end{split}\]

thus,

\[\begin{split}\begin{aligned} \nabla \cdot \mathbb{T} & = \frac{1}{J} \nabla_0 \cdot \mathbb{\Sigma}_n = \\ & = \frac{1}{J} \nabla_0 \cdot \left( J \, \mathbb{F}^{-T} \cdot \mathbb{T} \right) \end{aligned}\end{split}\]

todo Prove it with derivation!

2.6.3. Kinetic energy

\[\begin{split}\begin{aligned} 0 & = \vec{v} \cdot \left\{ \rho^0 \dfrac{D \vec{v}}{D t} - \rho^0 \vec{g} - \nabla_0 \cdot \mathbb{\Sigma}_n \right\} = \\ & = \rho^0 \dfrac{D}{D t}\dfrac{|\vec{v}|^2}{2} - \rho^0 \vec{v} \cdot \vec{g} - \vec{v} \cdot \nabla_0 \cdot \mathbb{\Sigma}_n = \\ & = \rho^0 \dfrac{D}{D t}\dfrac{|\vec{v}|^2}{2} - \rho^0 \vec{v} \cdot \vec{g} - \nabla_0 \cdot \left( \vec{v} \cdot \mathbb{\Sigma}_n \right) + \nabla_0 \vec{v} : \mathbb{\Sigma}_n \end{aligned}\end{split}\]

\[v_i \partial^0_k \Sigma_{ki} = \partial^0_k \left( v_i \Sigma_{ki} \right) - \partial^0_k v_i \Sigma_{ki}\]

\[\begin{split}\begin{aligned} & d V = J d V_0 \\ & dr_i n_i \, dS = J dr^0_{k} n^0_k \, dS_0 \\ & dr^0_k \dfrac{\partial r_i}{\partial r^0_k} n_i \, dS = dr^0_{k} J n^0_k \, dS_0 \\ \end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} \dfrac{\partial r_i}{\partial r^0_k} n_i \, dS & = J n^0_k \, dS_0 \\ \underbrace{\dfrac{\partial r^0_k}{\partial r_j} \dfrac{\partial r_i}{\partial r^0_k}}_{=\delta_{ij}} n_i \, dS & = J \dfrac{\partial r^0_k}{\partial r_j} n^0_k \, dS_0 \\ n_j \, dS & = J \frac{\partial r^0_k}{\partial r_j} n_k^0 \, dS_0 \end{aligned}\end{split}\]

\[\mathbb{F} = \hat{e}^0_k \hat{e}^0_i \dfrac{\partial r_i}{\partial r^0_k}\]

\[\mathbb{F}^{-1} = \hat{e}^0_j \hat{e}^0_k \dfrac{\partial r^0_k}{\partial r_j}\]

\[\begin{split}\begin{aligned} \mathbb{F}^{-1} \cdot \mathbb{F} & = \left( \hat{e}^0_j \hat{e}^0_k \dfrac{\partial r^0_k}{\partial r_j} \right) \cdot \left( \hat{e}^0_l \hat{e}^0_i \dfrac{\partial r_i}{\partial r^0_l} \right) = \hat{e}^0_j \hat{e}^0_i \dfrac{\partial r^i}{\partial r_j} = \hat{e}^0_j \hat{e}^0_i \delta_{ij} \\ \mathbb{F} \cdot \mathbb{F}^{-1} & = \left( \hat{e}^0_l \hat{e}^0_i \dfrac{\partial r_i}{\partial r^0_l} \right) \cdot \left( \hat{e}^0_j \hat{e}^0_k \dfrac{\partial r^0_k}{\partial r_j} \right) = \hat{e}^0_l \hat{e}^0_k \dfrac{\partial r^k}{\partial r_l} = \hat{e}^0_l \hat{e}^0_k \delta_{lk} \\ \end{aligned}\end{split}\]

\[\frac{\partial r^0_k}{\partial r_i} \frac{\partial r_i}{\partial r^0_l} = \delta_{kl}\]

\[\mathbb{\Sigma} := \mathbb{\Sigma}_n \cdot \mathbb{F}^{-1} = J \mathbb{F}^{-T} \cdot \mathbb{T} \cdot \mathbb{F}^{-1}\]

\[\mathbb{\Sigma}_n = \mathbb{\Sigma} \cdot \mathbb{F}\]

\[\Sigma_{ik} = \Sigma_{n, ij} \left( \mathbb{F}^{-1} \right)_{jk} = \Sigma_{n, ij} \frac{\partial r^0_k}{\partial r_j}\]

\[\Sigma_{n, ij} = \Sigma_{ik} \dfrac{\partial x_j}{\partial x^0_k}\]

\[\begin{split}\begin{aligned} \nabla_0 \vec{v} : \mathbb{\Sigma}_n & = \frac{D}{Dt} \mathbb{F} : \mathbb{\Sigma}_n = \\ & = \dfrac{\partial v_j}{\partial x^0_i} \, \Sigma_{n, ij} = \\ & = \dfrac{\partial v_j}{\partial x^0_i} \, \Sigma_{ik} \, \dfrac{\partial x_j}{\partial x^0_k} = \\ & = \Sigma_{ik} \, \frac{1}{2} \left( \, \dfrac{\partial v_j}{\partial x^0_i} \, \dfrac{\partial x_j}{\partial x^0_k} + \dfrac{\partial v_j}{\partial x^0_k} \, \dfrac{\partial x_j}{\partial x^0_i} \, \right) \end{aligned}\end{split}\]

if \(\mathbb{\Sigma}\) is symmetric, \(\Sigma_{ik} = \Sigma_{ki}\), or with tensor notation

\[\begin{split}\begin{aligned} \nabla_0 \vec{v} : \mathbb{\Sigma}_n & = \dfrac{D}{D t} \mathbb{F} : \mathbb{\Sigma}_n = \\ & = \nabla_0 \vec{v} : \left( \mathbb{\Sigma} \cdot \mathbb{F} \right) = \\ & = \mathbb{\Sigma} : \frac{1}{2} \left( \dfrac{D \mathbb{F}}{D t} \cdot \mathbb{F}^T + \mathbb{F} \cdot \frac{D \mathbb{F}^T}{D t} \right) = \\ & = \mathbb{\Sigma} : \dfrac{D}{Dt} \mathbb{E} \ , \end{aligned}\end{split}\]

having recognized the time derivative (1.6) of the Green-Lagrange tensor (1.5).

Integral of the volume stress in the reference space can be recast as the volume in the physical space

\[\int_{V_0} \nabla_0 \vec{v} : \mathbb{\Sigma}_n \, dV_0 = \int_{V_0} \mathbb{\Sigma} : \dfrac{D \mathbb{E}}{D t} \, d V_0 \]

\[\Sigma_{n, ik} = J \, \dfrac{\partial x^0_i}{\partial x_k} \, T_{jk}\]

\[\begin{split}\begin{aligned} \int_{V_0} \nabla_0 \vec{v} : \mathbb{\Sigma}_n \, dV_0 & = \int_{V_0} \dfrac{\partial v_k}{\partial x^0_i} \Sigma_{n, ik} \, dV_0 = \\ & = \int_{V_0} \underbrace{\dfrac{\partial v_k}{\partial x^0_i} \left( \dfrac{\partial x^0_i}{\partial x_j} \right.}_{} \, T_{kj} \underbrace{J\, \bigg) dV_0}_{d V} = \\ & = \int_{V} \dfrac{\partial v_k}{\partial x^j} \, T_{kj} \, d V = \\ & = \int_{V} \frac{1}{2} \left( \dfrac{\partial v_k}{\partial x^j} + \dfrac{\partial v_j}{\partial x^k} \right) \, T_{kj} \, d V = \\ & = \int_{V} D_{jk} \, T_{kj} \, d V = \\ & = \int_{V} \mathbb{D} : \mathbb{T} \, d V \ . \end{aligned}\end{split}\]

2.6.3.1. Variational principles

Using an arbitrary test function \(\vec{w}(\vec{r}_0)\),

\[0 = \vec{w} \cdot \left\{ \rho^0 \dfrac{D \vec{v}}{D t} - \rho^0 \vec{g} - \nabla_0 \cdot \mathbb{\Sigma}_n \right\}\]

and using rule of product

\[\begin{split}\begin{aligned} w_i \dfrac{\partial \Sigma_{n,ji}}{\partial x^0_j} & = \dfrac{\partial }{\partial x^0_j} \left( w_i \Sigma_{n, ji} \right) - \dfrac{\partial w_i}{\partial x^0_j} \Sigma_{n, ji} = \\ \end{aligned}\end{split}\]

and the second term can be transformed using the relation bewteen normal stress and second Piola-Kirchhoff tensor

\[\begin{aligned} \dfrac{\partial w_i}{\partial x^0_j} \Sigma_{n, ji} = \dfrac{\partial w_i}{\partial x^0_j} \Sigma_{jk} \dfrac{\partial x_i}{\partial x^0_k} = \Sigma_{jk} \, \dfrac{1}{2} \, \left[ \, \dfrac{\partial x_i}{\partial x^0_k} \dfrac{\partial w_i}{\partial x^0_j} + \dfrac{\partial x_i}{\partial x^0_j} \dfrac{\partial w_i}{\partial x^0_k} \, \right] = \Sigma_{jk} \, W_{ij}(\vec{w}) \ , \end{aligned}\]

having defined the tensor

\[\mathbb{W}(\vec{w}) := \dfrac{1}{2} \left[ \, \nabla_0 \vec{w} \cdot \mathbb{F}^T + \mathbb{F} \cdot \nabla^T_0 \vec{w} \, \right] \ ,\]

with the evident analogy with the time derivative of Green-Lagrange strain tensor, namely

\[\mathbb{\epsilon} = \mathbb{W}(\vec{v}) \ ,\]

being \(\vec{v}\) the velocity field. Integrating on the domain \(V_0\) and using divergence theorem, the problem is written in its weak form

\[\int_{V_0} \left\{ \rho^0 \vec{w} \cdot \dfrac{D \vec{v}}{D t} + \mathbb{W}(\vec{w}) : \mathbb{\Sigma} \right\} = \int_{V_0} \rho^0 \vec{w} \cdot \vec{g} + \oint_{\partial V_0} \hat{n}_0 \cdot \mathbb{\Sigma}_n \cdot \vec{w} \ , \]

with the proper boundary conditions and the corresponding conditions on the test function \(\vec{w}\). As an example, if the boundary is composed of two different regions, \(S_{D,0} \cup S_{N,0} = \partial V_0\), \(S_D \cap S_N = \emptyset\) where either position (called \(S_D\) from Dirichlet boundary) and stress (called \(S_N\) from Neumann boundary) are prescribed

\[\begin{split}\begin{aligned} \vec{r} = \vec{r}_b \qquad , \qquad & \vec{w} = \vec{0} & \qquad \text{(on $ S_{D,0} $ Dirichlet - essential - boundary)} \\ \hat{n}_0 \cdot \mathbb{\Sigma}_n = \vec{t}_{0, \hat{n}_0} \qquad , \qquad & & \text{(on $ S_{N,0} $ Neumann - natural - boundary)} \\ \end{aligned}\end{split}\]

the weak form of the equation reads

\[\int_{V_0} \left\{ \rho^0 \vec{w} \cdot \dfrac{D \vec{v}}{D t} + \mathbb{W}(\vec{w}) : \mathbb{\Sigma} \right\} = \int_{V_0} \rho^0 \vec{w} \cdot \vec{g} + \int_{S_{n,0}} \hat{n}_0 \cdot \vec{t}_{\hat{n}_0} \]

2.6.4. Total energy

Using Nanson’s relation \(\hat{n} \, dS = \hat{n}_0 \cdot \left(J \mathbb{F}^{-T}\right) \, d S_0\),

\[\begin{split}\begin{aligned} \dfrac{d}{d t} \int_{V_t} \rho e^t \, dV & = \int_{V_t} \rho \vec{g} \cdot \vec{v} \, dV + \oint_{\partial V_t} \vec{t}_{\hat{n}} \cdot \vec{v} \, dS - \oint_{\partial V_t} \hat{n} \cdot \vec{q} + \int_{V_t} \rho \, r \, dV = \\ & = \int_{V_t} \rho \vec{g} \cdot \vec{v} \, dV + \oint_{\partial V_t} \hat{n} \cdot \mathbb{T} \cdot \vec{v} \, dS - \oint_{\partial V_t} \hat{n} \cdot \vec{q} \, dS + \int_{V_t} \rho \, r \, dV \\ & = \int_{V_0} \rho^0 \, \vec{g} \cdot \vec{v} \, dV_0 + \oint_{\partial V_0} \hat{n}_0 \cdot \left( J \mathbb{F}^{-T} \cdot \mathbb{T} \right) \cdot \vec{v} \, dS_0 - \oint_{\partial V_0} \hat{n}_0 \cdot \left( J \mathbb{F}^{-T} \cdot \vec{q} \right) \, dS_0 + \int_{V_0} \rho^0 \, r \, dV_0 \ . \end{aligned}\end{split}\]

and the differential form reads

\[ \rho^0 \dfrac{D e^t}{Dt} = \rho^0 \vec{g} \cdot \vec{v} + \nabla_0 \cdot \left( J \mathbb{F}^{-T} \cdot \mathbb{T} \cdot \vec{v} \right) - \nabla_0 \cdot \left( J \mathbb{F}^{-T} \cdot \vec{q} \right) + \rho^0 \, r \ . \]

or

\[ \rho^0 \dfrac{D e^t}{Dt} = \rho^0 \vec{g} \cdot \vec{v} + \nabla_0 \cdot \left( \mathbb{\Sigma}_n \cdot \vec{v} \right) - \nabla_0 \cdot \vec{q}_0 + \rho^0 \, r \ . \]

and dividing by \(J\) and using the relation (see below) \(\nabla_0 \cdot ( J \mathbb{F}^{-T} \cdot \vec{a} ) = J \, \nabla \cdot \vec{a}\),

\[\rho \dfrac{D e^t}{D t} = \rho \vec{g} \cdot \vec{v} + \nabla \cdot \left( \mathbb{T} \cdot \vec{v} \right) - \nabla \cdot \vec{q} + \rho r \ .\]

Comparison with equation in physical space (dividing by \(J\)) suggests the identity

\[\dfrac{1}{J} \nabla_0 \cdot \left( J \mathbb{F}^{-T} \cdot \vec{a} \right) = \nabla \cdot \vec{a} \ ,\]

and thus

\[\nabla_0 \cdot ( J \mathbb{F}^{-T} ) = \vec{0} \ ,\]

since

\[ \nabla_0 \cdot \left( J \mathbb{F}^{-T} \cdot \vec{a} \right) = \nabla_0 \cdot \left( J \mathbb{F}^{-T} \right) \cdot \vec{a} + J \mathbb{F}^{-T} : \nabla_0 \vec{a} = \nabla_0 \cdot \left( J \mathbb{F}^{-T} \right) \cdot \vec{a} + J \nabla \cdot \vec{a} \]

Proof.

\[J = \left| \dfrac{\partial r_k}{\partial r^0_i} \right| = \varepsilon_{i_1, i_2, i_3} \dfrac{\partial r_{i_1}}{\partial r^0_1} \dfrac{\partial r_{i_2}}{\partial r^0_2} \dfrac{\partial r_{i_3}}{\partial r^0_3}\]

\[\left( \mathbb{F}^{-T} \right)_{ij} = \dfrac{\partial r^0_i}{\partial r_j}\]

\[\begin{split}\begin{aligned} \left\{ \nabla_0 \cdot \left( J \, \mathbb{F}^{-T} \right) \right\}_j & = \dfrac{\partial }{\partial r_i^0} \left( \varepsilon_{i_1, i_2, i_3} \dfrac{\partial r_{i_1}}{\partial r^0_1} \dfrac{\partial r_{i_2}}{\partial r^0_2} \dfrac{\partial r_{i_3}}{\partial r^0_3} \, \dfrac{\partial r^0_i}{\partial r_j}\right) = \\ & = \varepsilon_{i_1, i_2, i_3} \dfrac{\partial }{\partial r_i^0} \left(\dfrac{\partial r_{i_1}}{\partial r^0_1} \right)\dfrac{\partial r_{i_2}}{\partial r^0_2} \dfrac{\partial r_{i_3}}{\partial r^0_3} \, \dfrac{\partial r^0_i}{\partial r_j} + \varepsilon_{i_1, i_2, i_3} \dfrac{\partial r_{i_1}}{\partial r^0_1} \dfrac{\partial }{\partial r_i^0} \left( \dfrac{\partial r_{i_2}}{\partial r^0_2} \right) \dfrac{\partial r_{i_3}}{\partial r^0_3} \, \dfrac{\partial r^0_i}{\partial r_j} + \\ & + \varepsilon_{i_1, i_2, i_3} \dfrac{\partial r_{i_1}}{\partial r^0_1} \, \dfrac{\partial r_{i_2}}{\partial r^0_2} \dfrac{\partial }{\partial r_i^0} \left(\dfrac{\partial r_{i_3}}{\partial r^0_3}\right) \, \dfrac{\partial r^0_i}{\partial r_j} + \varepsilon_{i_1, i_2, i_3}\dfrac{\partial r_{i_1}}{\partial r^0_1} \dfrac{\partial r_{i_2}}{\partial r^0_2} \dfrac{\partial r_{i_3}}{\partial r^0_3} \, \dfrac{\partial }{\partial r_i^0} \left( \dfrac{\partial r^0_i}{\partial r_j} \right) = \\ & = \varepsilon_{i_1, i_2, i_3} \, \underbrace{\dfrac{\partial r^0_i}{\partial r_j}\dfrac{\partial }{\partial r_i^0}}_{= \frac{\partial }{\partial r_j}} \left(\dfrac{\partial r_{i_1}}{\partial r^0_1} \right)\dfrac{\partial r_{i_2}}{\partial r^0_2} \dfrac{\partial r_{i_3}}{\partial r^0_3} + \varepsilon_{i_1, i_2, i_3} \dfrac{\partial r_{i_1}}{\partial r^0_1} \underbrace{ \dfrac{\partial r^0_i}{\partial r_j} \dfrac{\partial }{\partial r_i^0}}_{= \frac{\partial }{\partial r_j}} \, \left(\dfrac{\partial r_{i_2}}{\partial r^0_2} \right) \dfrac{\partial r_{i_3}}{\partial r^0_3} + \\ & + \varepsilon_{i_1, i_2, i_3} \dfrac{\partial r_{i_1}}{\partial r^0_1} \, \dfrac{\partial r_{i_2}}{\partial r^0_2} \underbrace{\dfrac{\partial r^0_i}{\partial r_j}\dfrac{\partial }{\partial r_i^0}}_{=\frac{\partial}{\partial r_j}} \, \left(\dfrac{\partial r_{i_3}}{\partial r^0_3}\right) + \varepsilon_{i_1, i_2, i_3}\dfrac{\partial r_{i_1}}{\partial r^0_1} \dfrac{\partial r_{i_2}}{\partial r^0_2} \dfrac{\partial r_{i_3}}{\partial r^0_3} \, \dfrac{\partial }{\partial r_j} \underbrace{\left( \dfrac{\partial r^0_i}{\partial r^0_i} \right)}_{=3} = \\ & = 0 \ , \end{aligned}\end{split}\]

since

\[\dfrac{\partial }{\partial r_j} \left( \dfrac{\partial r_{i_k}}{\partial r^0_k} \right) = \dfrac{\partial }{\partial r^0_k} \left( \dfrac{\partial r_{i_k}}{\partial r^j} \right) = \dfrac{\partial}{\partial r^0_k} \delta_{i_k j} = 0 \ .\]

Thus

\[\dfrac{1}{J}\dfrac{\partial}{\partial r^0_i} \left( J \dfrac{\partial r^0_i}{\partial r_j} a_j \right) = \dfrac{1}{J}\dfrac{\partial}{\partial r^0_i} \left( J \dfrac{\partial r^0_i}{\partial r_j} \right) a_j + \dfrac{1}{J} \, J \, \dfrac{\partial r^0_i}{\partial r_j} \, \dfrac{\partial a_j}{\partial r^0_i} = \dfrac{\partial a_j}{\partial r_j} = \nabla \cdot \vec{a} \ . \]

2.6.5. Internal energy

\[\begin{split}\begin{aligned} & \dfrac{d}{d t} \int_{V_t} \rho e \, dV = \int_{V_t} \mathbb{T} : \nabla \vec{v} \, dV - \oint_{\partial V_t} \hat{n} \cdot \vec{q} \, dS + \int_{V_t} \rho \, r \, dV = \\ & \dfrac{d}{d t} \int_{V_0} \rho^0 e \, dV_0 = \int_{V_0} J \, \mathbb{T} : \nabla \vec{v} \, dV_0 - \oint_{\partial V_0} \hat{n}_0 \cdot \left( J \mathbb{F}^{-T} \cdot \vec{q} \right) \, dS_0 + \int_{V_0} \rho^0 \, r \, dV_0 = \\ \end{aligned}\end{split}\]

and the differential form reads

\[\rho^0 \dfrac{D e}{Dt} = \mathbb{\Sigma}_n : \nabla_0 \vec{v} - \nabla_0 \cdot \vec{q}^0 + \rho^0 r \ .\]

todo pay attention at the definition - choose one and keep using it! - of the product \(\mathbb{A} : \mathbb{B}\), in components

\[\mathbb{A} : \mathbb{B} = A_{ij} B_{ij} \qquad \text{or} \qquad = A_{ij} B_{ji}\]