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$$\frac{\partial (\rho e)}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j e)}{\partial x_j} = - \sum\limits_{j \in \mathcal{D}} \frac{\partial q_j}{\partial x_j} + \sum\limits_{i, j \in \mathcal{D}} \frac{\partial (\sigma _{ij} u_i)}{\partial x_j} + \sum\limits_{j \in \mathcal{D}} \rho u_j g_j \tag{3}\label{3}$$$$\frac{\partial (\rho e)}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j e)}{\partial x_j} = - \sum\limits_{j \in \mathcal{D}} \frac{\partial q_j}{\partial x_j} + \sum\limits_{(i, j) \in \mathcal{D}} \frac{\partial (\sigma _{ij} u_i)}{\partial x_j} + \sum\limits_{j \in \mathcal{D}} \rho u_j g_j \tag{3}\label{3}$$

Note the similar structure of the conservation equations: We are dealing with a property $\phi$ which changes according to $\frac{D \Phi}{D t} = \frac{\partial \Phi}{\partial t} + \frac{\partial (\phi u_i)}{\partial x_i} = s$$\frac{D \Phi}{D t} = \frac{\partial \Phi}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\phi u_j)}{\partial x_j} = s$ where $s$ is a source term. For example what we perceive as a change of momentum in a particular direction $i$ due to temporal change or transport of the momentum in every potential direction $j$ with a velocity $u_j$ is equal to our sources of momentum - the change of stress pointing in direction $i$ (not only normal but also due to shearing - thus the sum over $j$) and the specific body force in the direction $i$.

$$\sigma_{ij} = \underbrace{-p \delta_{ij}}_{\sigma_{ij}^{(0)}} + \underbrace{\sum\limits_{k,l} C_{ijkl} \epsilon_{kl} + \sum\limits_{k,l} D_{ijkl} S_{kl}}_{\tau_{ij}}$$$$\sigma_{ij} = \underbrace{-p \delta_{ij}}_{\sigma_{ij}^{(0)}} + \underbrace{\sum\limits_{(k,l) \in \mathcal{D}} C_{ijkl} \epsilon_{kl} + \sum\limits_{(k,l) \in \mathcal{D}} D_{ijkl} S_{kl}}_{\tau_{ij}}$$

$$\sigma_{ij} = -p \delta_{ij} + \underbrace{\sum\limits_{k,l} C_{ijkl} \epsilon_{kl}}_{\tau_{ij}}.$$$$\sigma_{ij} = -p \delta_{ij} + \underbrace{\sum\limits_{(k,l) \in \mathcal{D}} C_{ijkl} \epsilon_{kl}}_{\tau_{ij}}.$$

For an isotropic material, where there is no preferred direction, $C_{ijkl}$ and $D_{ijkl}$ clearly have to be a isotropic tensors of rank four. In order to derive the form of an isotropic tensor of rank four $\underline{T}$ we introduce a scalar $s$ that is obtained from vectors $\vec a$, $\vec b$, $\vec c$, $\vec d$ and a tensor $\underline{T}$ as \begin{equation} s = \sum\limits_{i,j,k,l} T_{ijkl} a_i b_j c_k d_l. \end{equation} Without

$$ s = \sum\limits_{(i,j,k,l) \in \mathcal{D}} T_{ijkl} a_i b_j c_k d_l. $$

Without making any assumptions on the fourth-order tensor $\underline{T}$ the scalar $s$ would depend linearly on the magnitude of every single vector $\in \{ \vec a, \vec b, \vec c, \vec d \}$ and their relative orientation in space. If we now assume an isotropic tensor the precise direction of the four vectors should not affect the scalar $s$ but instead only the orientation of the vectors to another given by the dot product \begin{equation} \begin{aligned} s = \sum\limits_{i,j,k,l} T_{ijkl} a_i b_j c_k d_l \stackrel{!}{=} \alpha \left(\vec a \cdot \vec b \right)\left(\vec c \cdot \vec d \right) + \beta \left(\vec a \cdot \vec c \right)\left(\vec b \cdot \vec d \right) + \gamma \left(\vec a \cdot \vec d \right)\left(\vec c \cdot \vec b \right) = \\ = \sum\limits_{i,j} \left( \alpha \, a_i b_i c_j d_j + \beta \, a_i c_i b_j d_j + \gamma \, a_i d_i c_j b_j \right) = \sum\limits_{i,j,k,l} \underbrace{\left( \alpha \, \delta_{ij} \delta_{kl} + \beta \, \delta_{ik} \delta_{jl} + \gamma \, \delta_{il} \delta_{jk} \right)}_{T_{ijkl}} a_i b_j c_k d_l. \end{aligned} \end{equation}\begin{equation} \begin{aligned} s = \sum\limits_{(i,j,k,l) \in \mathcal{D}} T_{ijkl} a_i b_j c_k d_l \stackrel{!}{=} \alpha \left(\vec a \cdot \vec b \right)\left(\vec c \cdot \vec d \right) + \beta \left(\vec a \cdot \vec c \right)\left(\vec b \cdot \vec d \right) + \gamma \left(\vec a \cdot \vec d \right)\left(\vec c \cdot \vec b \right) = \\ = \sum\limits_{(i,j) \in \mathcal{D}} \left( \alpha \, a_i b_i c_j d_j + \beta \, a_i c_i b_j d_j + \gamma \, a_i d_i c_j b_j \right) = \sum\limits_{(i,j,k,l) \in \mathcal{D}} \underbrace{\left( \alpha \, \delta_{ij} \delta_{kl} + \beta \, \delta_{ik} \delta_{jl} + \gamma \, \delta_{il} \delta_{jk} \right)}_{T_{ijkl}} a_i b_j c_k d_l. \end{aligned} \end{equation}

$$\tau_{ij} = \sum\limits_{k,l} C_{ijkl} \epsilon_{kl} = \sum\limits_{k,l} \left( \alpha \delta_{ij} \delta_{kl} + \beta \delta_{ik} \delta_{jl} + \gamma \delta_{il} \delta_{jk} \right) \epsilon_{kl}$$$$\tau_{ij} = \sum\limits_{(k,l) \in \mathcal{D}} C_{ijkl} \epsilon_{kl} = \sum\limits_{(k,l) \in \mathcal{D}} \left( \alpha \delta_{ij} \delta_{kl} + \beta \delta_{ik} \delta_{jl} + \gamma \delta_{il} \delta_{jk} \right) \epsilon_{kl}$$

$$\tau_{ji} = \sum\limits_{k,l} C_{jikl} \epsilon_{kl} = \sum\limits_{k,l} \left( \alpha \delta_{ji} \delta_{kl} + \beta \delta_{jk} \delta_{il} + \gamma \delta_{jl} \delta_{ik} \right) \epsilon_{kl}$$$$\tau_{ji} = \sum\limits_{(k,l) \in \mathcal{D}} C_{jikl} \epsilon_{kl} = \sum\limits_{(k,l) \in \mathcal{D}} \left( \alpha \delta_{ji} \delta_{kl} + \beta \delta_{jk} \delta_{il} + \gamma \delta_{jl} \delta_{ik} \right) \epsilon_{kl}$$

$$\sigma_{ij} = \sigma_{ij}^{(0)} + \sum\limits_{k,l} D_{ijkl} S_{kl} = - p \delta_{ij} + \tau_{ij}.$$$$\sigma_{ij} = \sigma_{ij}^{(0)} + \sum\limits_{(k,l) \in \mathcal{D}} D_{ijkl} S_{kl} = - p \delta_{ij} + \tau_{ij}.$$

$$\frac{\partial (\rho e)}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j e)}{\partial x_j} = - \sum\limits_{j \in \mathcal{D}} \frac{\partial q_j}{\partial x_j} + \sum\limits_{i, j \in \mathcal{D}} \frac{\partial (\sigma _{ij} u_i)}{\partial x_j} + \sum\limits_{j \in \mathcal{D}} \rho u_j g_j \tag{3}\label{3}$$

Note the similar structure of the conservation equations: We are dealing with a property $\phi$ which changes according to $\frac{D \Phi}{D t} = \frac{\partial \Phi}{\partial t} + \frac{\partial (\phi u_i)}{\partial x_i} = s$ where $s$ is a source term. For example what we perceive as a change of momentum in a particular direction $i$ due to temporal change or transport of the momentum in every potential direction $j$ with a velocity $u_j$ is equal to our sources of momentum - the change of stress pointing in direction $i$ (not only normal but also due to shearing - thus the sum over $j$) and the specific body force in the direction $i$.

$$\sigma_{ij} = \underbrace{-p \delta_{ij}}_{\sigma_{ij}^{(0)}} + \underbrace{\sum\limits_{k,l} C_{ijkl} \epsilon_{kl} + \sum\limits_{k,l} D_{ijkl} S_{kl}}_{\tau_{ij}}$$

$$\sigma_{ij} = -p \delta_{ij} + \underbrace{\sum\limits_{k,l} C_{ijkl} \epsilon_{kl}}_{\tau_{ij}}.$$

For an isotropic material, where there is no preferred direction, $C_{ijkl}$ and $D_{ijkl}$ clearly have to be a isotropic tensors of rank four. In order to derive the form of an isotropic tensor of rank four $\underline{T}$ we introduce a scalar $s$ that is obtained from vectors $\vec a$, $\vec b$, $\vec c$, $\vec d$ and a tensor $\underline{T}$ as \begin{equation} s = \sum\limits_{i,j,k,l} T_{ijkl} a_i b_j c_k d_l. \end{equation} Without making any assumptions on the fourth-order tensor $\underline{T}$ the scalar $s$ would depend linearly on the magnitude of every single vector $\in \{ \vec a, \vec b, \vec c, \vec d \}$ and their relative orientation in space. If we now assume an isotropic tensor the precise direction of the four vectors should not affect the scalar $s$ but instead only the orientation of the vectors to another given by the dot product \begin{equation} \begin{aligned} s = \sum\limits_{i,j,k,l} T_{ijkl} a_i b_j c_k d_l \stackrel{!}{=} \alpha \left(\vec a \cdot \vec b \right)\left(\vec c \cdot \vec d \right) + \beta \left(\vec a \cdot \vec c \right)\left(\vec b \cdot \vec d \right) + \gamma \left(\vec a \cdot \vec d \right)\left(\vec c \cdot \vec b \right) = \\ = \sum\limits_{i,j} \left( \alpha \, a_i b_i c_j d_j + \beta \, a_i c_i b_j d_j + \gamma \, a_i d_i c_j b_j \right) = \sum\limits_{i,j,k,l} \underbrace{\left( \alpha \, \delta_{ij} \delta_{kl} + \beta \, \delta_{ik} \delta_{jl} + \gamma \, \delta_{il} \delta_{jk} \right)}_{T_{ijkl}} a_i b_j c_k d_l. \end{aligned} \end{equation}

$$\tau_{ij} = \sum\limits_{k,l} C_{ijkl} \epsilon_{kl} = \sum\limits_{k,l} \left( \alpha \delta_{ij} \delta_{kl} + \beta \delta_{ik} \delta_{jl} + \gamma \delta_{il} \delta_{jk} \right) \epsilon_{kl}$$

$$\tau_{ji} = \sum\limits_{k,l} C_{jikl} \epsilon_{kl} = \sum\limits_{k,l} \left( \alpha \delta_{ji} \delta_{kl} + \beta \delta_{jk} \delta_{il} + \gamma \delta_{jl} \delta_{ik} \right) \epsilon_{kl}$$

$$\sigma_{ij} = \sigma_{ij}^{(0)} + \sum\limits_{k,l} D_{ijkl} S_{kl} = - p \delta_{ij} + \tau_{ij}.$$

$$\frac{\partial (\rho e)}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j e)}{\partial x_j} = - \sum\limits_{j \in \mathcal{D}} \frac{\partial q_j}{\partial x_j} + \sum\limits_{(i, j) \in \mathcal{D}} \frac{\partial (\sigma _{ij} u_i)}{\partial x_j} + \sum\limits_{j \in \mathcal{D}} \rho u_j g_j \tag{3}\label{3}$$

Note the similar structure of the conservation equations: We are dealing with a property $\phi$ which changes according to $\frac{D \Phi}{D t} = \frac{\partial \Phi}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\phi u_j)}{\partial x_j} = s$ where $s$ is a source term. For example what we perceive as a change of momentum in a particular direction $i$ due to temporal change or transport of the momentum in every potential direction $j$ with a velocity $u_j$ is equal to our sources of momentum - the change of stress pointing in direction $i$ (not only normal but also due to shearing - thus the sum over $j$) and the specific body force in the direction $i$.

$$\sigma_{ij} = \underbrace{-p \delta_{ij}}_{\sigma_{ij}^{(0)}} + \underbrace{\sum\limits_{(k,l) \in \mathcal{D}} C_{ijkl} \epsilon_{kl} + \sum\limits_{(k,l) \in \mathcal{D}} D_{ijkl} S_{kl}}_{\tau_{ij}}$$

$$\sigma_{ij} = -p \delta_{ij} + \underbrace{\sum\limits_{(k,l) \in \mathcal{D}} C_{ijkl} \epsilon_{kl}}_{\tau_{ij}}.$$

For an isotropic material, where there is no preferred direction, $C_{ijkl}$ and $D_{ijkl}$ clearly have to be a isotropic tensors of rank four. In order to derive the form of an isotropic tensor of rank four $\underline{T}$ we introduce a scalar $s$ that is obtained from vectors $\vec a$, $\vec b$, $\vec c$, $\vec d$ and a tensor $\underline{T}$ as

$$ s = \sum\limits_{(i,j,k,l) \in \mathcal{D}} T_{ijkl} a_i b_j c_k d_l. $$

Without making any assumptions on the fourth-order tensor $\underline{T}$ the scalar $s$ would depend linearly on the magnitude of every single vector $\in \{ \vec a, \vec b, \vec c, \vec d \}$ and their relative orientation in space. If we now assume an isotropic tensor the precise direction of the four vectors should not affect the scalar $s$ but instead only the orientation of the vectors to another given by the dot product \begin{equation} \begin{aligned} s = \sum\limits_{(i,j,k,l) \in \mathcal{D}} T_{ijkl} a_i b_j c_k d_l \stackrel{!}{=} \alpha \left(\vec a \cdot \vec b \right)\left(\vec c \cdot \vec d \right) + \beta \left(\vec a \cdot \vec c \right)\left(\vec b \cdot \vec d \right) + \gamma \left(\vec a \cdot \vec d \right)\left(\vec c \cdot \vec b \right) = \\ = \sum\limits_{(i,j) \in \mathcal{D}} \left( \alpha \, a_i b_i c_j d_j + \beta \, a_i c_i b_j d_j + \gamma \, a_i d_i c_j b_j \right) = \sum\limits_{(i,j,k,l) \in \mathcal{D}} \underbrace{\left( \alpha \, \delta_{ij} \delta_{kl} + \beta \, \delta_{ik} \delta_{jl} + \gamma \, \delta_{il} \delta_{jk} \right)}_{T_{ijkl}} a_i b_j c_k d_l. \end{aligned} \end{equation}

$$\tau_{ij} = \sum\limits_{(k,l) \in \mathcal{D}} C_{ijkl} \epsilon_{kl} = \sum\limits_{(k,l) \in \mathcal{D}} \left( \alpha \delta_{ij} \delta_{kl} + \beta \delta_{ik} \delta_{jl} + \gamma \delta_{il} \delta_{jk} \right) \epsilon_{kl}$$

$$\tau_{ji} = \sum\limits_{(k,l) \in \mathcal{D}} C_{jikl} \epsilon_{kl} = \sum\limits_{(k,l) \in \mathcal{D}} \left( \alpha \delta_{ji} \delta_{kl} + \beta \delta_{jk} \delta_{il} + \gamma \delta_{jl} \delta_{ik} \right) \epsilon_{kl}$$

$$\sigma_{ij} = \sigma_{ij}^{(0)} + \sum\limits_{(k,l) \in \mathcal{D}} D_{ijkl} S_{kl} = - p \delta_{ij} + \tau_{ij}.$$

added second Kronecker delta
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$$C_{ijkl} = \lambda \delta_{ijkl} + \mu (\delta_{ik} \delta_{jl} + \delta_{il} \delta_{jk})$$$$C_{ijkl} = \lambda \delta_{ij} \delta_{kl} + \mu (\delta_{ik} \delta_{jl} + \delta_{il} \delta_{jk})$$

$$C_{ijkl} = \lambda \delta_{ijkl} + \mu (\delta_{ik} \delta_{jl} + \delta_{il} \delta_{jk})$$

$$C_{ijkl} = \lambda \delta_{ij} \delta_{kl} + \mu (\delta_{ik} \delta_{jl} + \delta_{il} \delta_{jk})$$

added missing summation symbols
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$$\sigma_{ij} = \underbrace{-p \delta_{ij}}_{\sigma_{ij}^{(0)}} + \underbrace{C_{ijkl} \epsilon_{kl} + D_{ijkl} S_{kl}}_{\tau_{ij}}$$$$\sigma_{ij} = \underbrace{-p \delta_{ij}}_{\sigma_{ij}^{(0)}} + \underbrace{\sum\limits_{k,l} C_{ijkl} \epsilon_{kl} + \sum\limits_{k,l} D_{ijkl} S_{kl}}_{\tau_{ij}}$$

$$\sigma_{ij} = -p \delta_{ij} + \underbrace{C_{ijkl} \epsilon_{kl}}_{\tau_{ij}}.$$$$\sigma_{ij} = -p \delta_{ij} + \underbrace{\sum\limits_{k,l} C_{ijkl} \epsilon_{kl}}_{\tau_{ij}}.$$

$$\tau_{ij} = C_{ijkl} \epsilon_{kl} = \left( \alpha \delta_{ij} \delta_{kl} + \beta \delta_{ik} \delta_{jl} + \gamma \delta_{il} \delta_{jk} \right) \epsilon_{kl}$$$$\tau_{ij} = \sum\limits_{k,l} C_{ijkl} \epsilon_{kl} = \sum\limits_{k,l} \left( \alpha \delta_{ij} \delta_{kl} + \beta \delta_{ik} \delta_{jl} + \gamma \delta_{il} \delta_{jk} \right) \epsilon_{kl}$$

$$\tau_{ji} = C_{jikl} \epsilon_{kl} = \left( \alpha \delta_{ji} \delta_{kl} + \beta \delta_{jk} \delta_{il} + \gamma \delta_{jl} \delta_{ik} \right) \epsilon_{kl}$$$$\tau_{ji} = \sum\limits_{k,l} C_{jikl} \epsilon_{kl} = \sum\limits_{k,l} \left( \alpha \delta_{ji} \delta_{kl} + \beta \delta_{jk} \delta_{il} + \gamma \delta_{jl} \delta_{ik} \right) \epsilon_{kl}$$

$$\sigma_{ij} = \sigma_{ij}^{(0)} + D_{ijkl} S_{kl} = - p \delta_{ij} + \tau_{ij}.$$$$\sigma_{ij} = \sigma_{ij}^{(0)} + \sum\limits_{k,l} D_{ijkl} S_{kl} = - p \delta_{ij} + \tau_{ij}.$$

$$\sigma_{ij} = \underbrace{-p \delta_{ij}}_{\sigma_{ij}^{(0)}} + \underbrace{C_{ijkl} \epsilon_{kl} + D_{ijkl} S_{kl}}_{\tau_{ij}}$$

$$\sigma_{ij} = -p \delta_{ij} + \underbrace{C_{ijkl} \epsilon_{kl}}_{\tau_{ij}}.$$

$$\tau_{ij} = C_{ijkl} \epsilon_{kl} = \left( \alpha \delta_{ij} \delta_{kl} + \beta \delta_{ik} \delta_{jl} + \gamma \delta_{il} \delta_{jk} \right) \epsilon_{kl}$$

$$\tau_{ji} = C_{jikl} \epsilon_{kl} = \left( \alpha \delta_{ji} \delta_{kl} + \beta \delta_{jk} \delta_{il} + \gamma \delta_{jl} \delta_{ik} \right) \epsilon_{kl}$$

$$\sigma_{ij} = \sigma_{ij}^{(0)} + D_{ijkl} S_{kl} = - p \delta_{ij} + \tau_{ij}.$$

$$\sigma_{ij} = \underbrace{-p \delta_{ij}}_{\sigma_{ij}^{(0)}} + \underbrace{\sum\limits_{k,l} C_{ijkl} \epsilon_{kl} + \sum\limits_{k,l} D_{ijkl} S_{kl}}_{\tau_{ij}}$$

$$\sigma_{ij} = -p \delta_{ij} + \underbrace{\sum\limits_{k,l} C_{ijkl} \epsilon_{kl}}_{\tau_{ij}}.$$

$$\tau_{ij} = \sum\limits_{k,l} C_{ijkl} \epsilon_{kl} = \sum\limits_{k,l} \left( \alpha \delta_{ij} \delta_{kl} + \beta \delta_{ik} \delta_{jl} + \gamma \delta_{il} \delta_{jk} \right) \epsilon_{kl}$$

$$\tau_{ji} = \sum\limits_{k,l} C_{jikl} \epsilon_{kl} = \sum\limits_{k,l} \left( \alpha \delta_{ji} \delta_{kl} + \beta \delta_{jk} \delta_{il} + \gamma \delta_{jl} \delta_{ik} \right) \epsilon_{kl}$$

$$\sigma_{ij} = \sigma_{ij}^{(0)} + \sum\limits_{k,l} D_{ijkl} S_{kl} = - p \delta_{ij} + \tau_{ij}.$$

added simple derivation of isotropic tensor of rank four
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added summation symbol in the macroscopic energy term
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changed the order of the terms in the stress tensor
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changed formatting with $$ instead of $ and \label{} and \eqref{}
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improved the conservation equations to be form consistent in form, added notes on the reference frames
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math formatting
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