= Solution
{c}
The first equation is the local <balance of linear momentum>. Use $\mathsf M,\mathsf L,\mathsf T$ for the dimensions of <mass>, <length>, and <time>, to distinguish dimensional time from the <Cauchy stress tensor> $\mathbf T$.
* $\mathbf x(\mathbf X,t)$ is the current position of the particle labelled by reference position $\mathbf X$; both have dimension $\mathsf L$. The <deformation map> specifies the particle's trajectory.
* $\ddot{\mathbf x}=\partial_t^2\mathbf x|_{\mathbf X}$ is its <material acceleration>, with dimension $\mathsf L\mathsf T^{-2}$. The dots follow the particle and are not spatially fixed time derivatives.
* $\rho$ is current <mass density>, with dimension $\mathsf M\mathsf L^{-3}$. For a homogeneous material with <incompressibility>, <mass conservation> makes it constant in space and time.
* $\mathbf b$ is <body force> per unit <mass>, such as gravitational <acceleration>, with dimension $\mathsf L\mathsf T^{-2}$. Hence $\rho\mathbf b$ is <force> per unit <volume>.
* $\mathbf T$ is the <Cauchy stress tensor>, mapping a unit <normal vector> $\mathbf n$ to surface <traction> $\mathbf T\mathbf n$. Each component has dimension $\mathsf M\mathsf L^{-1}\mathsf T^{-2}$, or <force> per unit <area>.
* $\operatorname{div}$ differentiates in the <current configuration>, with dimension $\mathsf L^{-1}$. In <Cartesian coordinates>, $(\operatorname{div}\mathbf T)_i=\partial T_{ij}/\partial x_j$. Thus $\operatorname{div}\mathbf T$, $\rho\mathbf b$, and $\rho\ddot{\mathbf x}$ all have dimension $\mathsf M\mathsf L^{-2}\mathsf T^{-2}$.
* The superscript $T$ denotes the dimension-preserving <matrix transpose>. The relation $\mathbf T=\mathbf T^T$ expresses <angular momentum conservation> in a continuum without body couples or couple <stresses>.
Here $t$ has dimension $\mathsf T$. The hypothesis $\partial_t\mathbf b=\mathbf0$ means that the prescribed spatial <body force> field is the same for every member of the static family and the resulting motion.
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