= Solution
{c}
For an arbitrary <material volume> $V(t)$, <Newton's second law> equates the rate of change of <linear momentum> to the resultant contact <force>, since there is no <body force>:
$$
\frac d{dt}\int_{V(t)}\rho u_i\,dV
=\int_{\partial V(t)}t_i(\mathbf n)\,dS
=\int_{\partial V(t)}\sigma_{ij}n_j\,dS.
$$
Use the <Reynolds transport theorem> on the left. Expanding the product and using the <continuity equation> leaves
$$
\partial_t(\rho u_i)+\partial_j(\rho u_i u_j)
=\rho(\partial_tu_i+u_j\partial_ju_i)
+u_i[\partial_t\rho+\partial_j(\rho u_j)]
=\rho\frac{Du_i}{Dt}.
$$
On the right, the <divergence theorem> applied to each row of the <Cauchy stress tensor> gives
$$
\int_{\partial V(t)}\sigma_{ij}n_j\,dS
=\int_{V(t)}\partial_j\sigma_{ij}\,dV.
$$
Consequently the <volume integral> of $\rho Du_i/Dt-\partial_j\sigma_{ij}$ vanishes on every <material volume>. Localizing as in part (i) gives the <Cauchy momentum equation>:
$$
\boxed{\rho\left(\frac{\partial\mathbf u}{\partial t}
+(\mathbf u\cdot\nabla)\mathbf u\right)
=\mathbf e_i\frac{\partial\sigma_{ij}}{\partial x_j}.}
$$
The left side is <mass density> times <material acceleration>; the right side is the <divergence> of the <Cauchy stress tensor>.
Back to article page
