Micromechanics of composites/Conservation of mass

Statement of the balance of mass
The balance of mass can be expressed as: $$  \dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v} = 0 $$ where $$\rho(\mathbf{x},t)$$ is the current mass density, $$\dot{\rho}$$ is the material time derivative of $$\rho$$, and $$\mathbf{v}(\mathbf{x},t)$$ is the velocity of physical particles in the body $$\Omega$$ bounded by the surface $$\partial{\Omega}$$.

Proof
We can show how this relation is derived by recalling that the general equation for the balance of a physical quantity $$f(\mathbf{x},t)$$ is given by

\cfrac{d}{dt}\left[\int_{\Omega} f(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} f(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA} + \int_{\partial{\Omega}} g(\mathbf{x},t)~\text{dA} + \int_{\Omega} h(\mathbf{x},t)~\text{dV} ~. $$ To derive the equation for the balance of mass, we assume that the physical quantity of interest is the mass density $$\rho(\mathbf{x},t)$$. Since mass is neither created or destroyed, the surface and interior sources are zero, i.e., $$g(\mathbf{x}, t) = h(\mathbf{x},t) = 0$$. Therefore, we have

\cfrac{d}{dt}\left[\int_{\Omega} \rho(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} \rho(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA}~. $$ Let us assume that the volume $$\Omega$$ is a control volume (i.e., it does not change with time). Then the surface $$\partial{\Omega}$$ has a zero velocity ($$u_n = 0$$) and we get

\int_{\Omega} \frac{\partial \rho}{\partial t}~\text{dV} = - \int_{\partial{\Omega}} \rho~(\mathbf{v}\cdot\mathbf{n})~\text{dA}~. $$ Using the divergence theorem

\int_{\Omega} \boldsymbol{\nabla} \bullet \mathbf{v}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\cdot\mathbf{n}~\text{dA} $$ we get

\int_{\Omega} \frac{\partial \rho}{\partial t}~\text{dV} = - \int_{\Omega} \boldsymbol{\nabla} \bullet (\rho~\mathbf{v})~\text{dV}. $$ or,

\int_{\Omega} \left[\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \bullet (\rho~\mathbf{v})\right]~\text{dV} = 0 ~. $$ Since $$\Omega$$ is arbitrary, we must have

\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \bullet (\rho~\mathbf{v}) = 0 ~. $$ Using the identity

\boldsymbol{\nabla} \bullet (\varphi~\mathbf{v}) = \varphi~\boldsymbol{\nabla} \bullet \mathbf{v} + \boldsymbol{\nabla} \varphi\cdot\mathbf{v} $$ we have

\frac{\partial \rho}{\partial t} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} = 0 ~. $$ Now, the material time derivative of $$\rho$$ is defined as

\dot{\rho} = \frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} ~. $$ Therefore,

{ \dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}= 0 ~. } $$