3.4 The balance equations

Newton’s second law for the box: $$-{\breve{\mathbf{t}}}_{1}d{\breve{A}}_1+{\breve{\mathbf{t}}}_1^{+}d{\breve{A}}_1-{\breve{\mathbf{t}}}_{2}d{\breve{A}}_2+{\breve{\mathbf{t}}}_2^{+}d{\breve{A}}_2-{\breve{\mathbf{t}}}_{3}d{\breve{A}}_3+{\breve{\mathbf{t}}}_3^{+}d{\breve{A}}_3+\breve{\mathbf{b}}d\breve{V}=\frac{D}{Dt}\left(\breve{\rho }d\breve{V}\breve{\mathbf{v}}\right)=\breve{\rho }d\breve{V}\breve{\mathbf{a}}.$$ (The mass $\breve{\rho }d\breve{V}$ of the box is constant, which is why the derivative only acts on $\breve{\mathbf{v}}$.)

Expand each ${\breve{\mathbf{t}}}_j^{+}$ to first order (Taylor): $${\breve{\mathbf{t}}}_j^{+}={\breve{\mathbf{t}}}_j(\mathbf{y})+\frac{\partial {\breve{\mathbf{t}}}_j}{\partial y_j}d{y}_j+O(d{y}_j^2)(\text{no sum}).$$ Substitute. Each $-{\breve{\mathbf{t}}}_{j}d{\breve{A}}_j$ cancels against the $+{\breve{\mathbf{t}}}_{j}d{\breve{A}}_j$ of its expansion, leaving $$\frac{\partial {\breve{\mathbf{t}}}_1}{\partial y_1}d{y}_{1}d{\breve{A}}_1+\frac{\partial {\breve{\mathbf{t}}}_2}{\partial y_2}d{y}_{2}d{\breve{A}}_2+\frac{\partial {\breve{\mathbf{t}}}_3}{\partial y_3}d{y}_{3}d{\breve{A}}_3+\breve{\mathbf{b}}d\breve{V}=\breve{\rho }d\breve{V}\breve{\mathbf{a}}.$$ Every $d{y}_{j}d{\breve{A}}_j$ is $d\breve{V}$; divide by it. For the components, ${\breve{\mathbf{t}}}_j={\breve{T}}_{ij}{\mathbf{e}}_i$, so $${\mathbf{e}}_i\frac{\partial {\breve{T}}_{ij}}{\partial y_j}+\breve{\mathbf{b}}=\breve{\rho }\breve{\mathbf{a}},$$ and the dot product with ${\mathbf{e}}_i$ gives the three scalar equations.

Take moments about the corner $\breve{P}$ at $\mathbf{y}$. The moments that survive at the lowest order are those of the forces on the far faces, with arm $d{y}_j{\mathbf{e}}_j$: $$d{\breve{\mathbf{M}}}_j=d{y}_j{\mathbf{e}}_j\times {\breve{\mathbf{t}}}_j^{+}d{\breve{A}}_j(j=1,2,3).$$ The body force and the momentum act at the centre, arm $\frac{1}{2}(d{y}_1{\mathbf{e}}_1+d{y}_2{\mathbf{e}}_2+d{y}_3{\mathbf{e}}_3)$: their moments are an arm times $d\breve{V}$, fourth order, and drop out against the third-order face terms. (Beyond the lectures: the faces through $\mathbf{y}$ also have moments, since their forces act at the face centres, off $\breve{P}$. Each is third order, but it pairs with the in-plane part of the arm of the opposite face: equal arms, forces $-{\breve{\mathbf{t}}}_j$ and ${\breve{\mathbf{t}}}_j^{+}$ that differ by $O(dy)$. The pair is fourth order.)

So, with $d{y}_{j}d{\breve{A}}_j=d\breve{V}$, $$({\mathbf{e}}_1\times {\breve{\mathbf{t}}}_1^{+}+{\mathbf{e}}_2\times {\breve{\mathbf{t}}}_2^{+}+{\mathbf{e}}_3\times {\breve{\mathbf{t}}}_3^{+})d\breve{V}=0.$$ Expand ${\breve{\mathbf{t}}}_j^{+}={\breve{\mathbf{t}}}_j+\frac{\partial {\breve{\mathbf{t}}}_j}{\partial y_j}d{y}_j$: $$({\mathbf{e}}_1\times {\breve{\mathbf{t}}}_1+{\mathbf{e}}_2\times {\breve{\mathbf{t}}}_2+{\mathbf{e}}_3\times {\breve{\mathbf{t}}}_3)d\breve{V}+({\mathbf{e}}_1\times \frac{\partial {\breve{\mathbf{t}}}_1}{\partial y_1}d{y}_1+{\mathbf{e}}_2\times \frac{\partial {\breve{\mathbf{t}}}_2}{\partial y_2}d{y}_2+{\mathbf{e}}_3\times \frac{\partial {\breve{\mathbf{t}}}_3}{\partial y_3}d{y}_3)d\breve{V}=0.$$ The second bracket is fourth order. What remains is ${\mathbf{e}}_j\times {\breve{\mathbf{t}}}_j=0$. With ${\breve{\mathbf{t}}}_j={\breve{T}}_{ij}{\mathbf{e}}_i$: $${\mathbf{e}}_j\times {\breve{T}}_{ij}{\mathbf{e}}_i={\breve{T}}_{ij}{\mathbf{e}}_j\times {\mathbf{e}}_i={\mathbf{e}}_1({\breve{T}}_{32}-{\breve{T}}_{23})+{\mathbf{e}}_2({\breve{T}}_{13}-{\breve{T}}_{31})+{\mathbf{e}}_3({\breve{T}}_{21}-{\breve{T}}_{12})=0,$$ using ${\mathbf{e}}_1\times {\mathbf{e}}_2={\mathbf{e}}_3$, ${\mathbf{e}}_2\times {\mathbf{e}}_3={\mathbf{e}}_1$, ${\mathbf{e}}_3\times {\mathbf{e}}_1={\mathbf{e}}_2$ and ${\mathbf{e}}_j\times {\mathbf{e}}_i=-{\mathbf{e}}_i\times {\mathbf{e}}_j$. Each bracket vanishes.

Linear momentum. By Cauchy–Poisson, ${\breve{\mathbf{t}}}_n={\breve{\mathbf{t}}}_j{\breve{n}}_j$. By the divergence theorem (a surface integral of a field dotted with the outward normal is the volume integral of its divergence), $${\int }_{\partial {\breve{B}}_0}{\breve{n}}_j{\breve{\mathbf{t}}}_{j}d\breve{A}={\int }_{{\breve{B}}_0}\frac{\partial {\breve{\mathbf{t}}}_j}{\partial y_j}d\breve{V}.$$ Substitute: $${\int }_{{\breve{B}}_0}(\frac{\partial {\breve{\mathbf{t}}}_j}{\partial y_j}+\breve{\mathbf{b}}-\breve{\rho }\breve{\mathbf{a}})d\breve{V}=0.$$ This holds for every part ${\breve{B}}_0$, so (if the integrand is continuous) the integrand is zero at every point: if it were not zero somewhere, a small ball around that point would give a non-zero integral. That is Cauchy’s equation of motion.

Angular momentum. With ${\breve{\mathbf{t}}}_n={\breve{\mathbf{t}}}_j{\breve{n}}_j$ and the divergence theorem, $${\int }_{\partial {\breve{B}}_0}\mathbf{y}\times {\breve{\mathbf{t}}}_j{\breve{n}}_{j}d\breve{A}={\int }_{{\breve{B}}_0}\frac{\partial }{\partial y_j}(\mathbf{y}\times {\breve{\mathbf{t}}}_j)d\breve{V}={\int }_{{\breve{B}}_0}{\mathbf{e}}_j\times {\breve{\mathbf{t}}}_{j}d\breve{V}+{\int }_{{\breve{B}}_0}\mathbf{y}\times \frac{\partial {\breve{\mathbf{t}}}_j}{\partial y_j}d\breve{V},$$ using the product rule and $\partial \mathbf{y}/\partial y_j={\mathbf{e}}_j$. Substitute: $${\int }_{{\breve{B}}_0}{\mathbf{e}}_j\times {\breve{\mathbf{t}}}_{j}d\breve{V}+{\int }_{{\breve{B}}_0}\mathbf{y}\times (\frac{\partial {\breve{\mathbf{t}}}_j}{\partial y_j}+\breve{\mathbf{b}}-\breve{\rho }\breve{\mathbf{a}})d\breve{V}=0.$$ The second integral is zero by the linear momentum equation. The first must vanish for every ${\breve{B}}_0$, so ${\mathbf{e}}_j\times {\breve{\mathbf{t}}}_j=0$ and $\breve{\mathbf{T}}={\breve{\mathbf{T}}}^T$, as in Theorem 3.4.3.