Micromechanics of composites/Balance of linear momentum

Statement of the balance of linear momentum
The balance of linear momentum can be expressed as: $$ \rho~\dot{\mathbf{v}} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 $$ where $$\rho(\mathbf{x},t)$$ is the mass density, $$\mathbf{v}(\mathbf{x},t)$$ is the velocity, $$\boldsymbol{\sigma}(\mathbf{x},t)$$ is the Cauchy stress, and $$\rho~\mathbf{b}$$ is the body force density.

Proof
Recall the general equation for the balance of a physical quantity

\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} ~. $$ In this case the physical quantity of interest is the momentum density, i.e., $$f(\mathbf{x},t) = \rho(\mathbf{x},t)~\mathbf{v}(\mathbf{x},t)$$. The source of momentum flux at the surface is the surface traction, i.e., $$g(\mathbf{x},t) = \mathbf{t}$$. The source of momentum inside the body is the body force, i.e., $$h(\mathbf{x},t) = \rho(\mathbf{x},t)~\mathbf{b}(\mathbf{x},t)$$. Therefore, we have

\cfrac{d}{dt}\left[\int_{\Omega} \rho~\mathbf{v}~\text{dV}\right] = \int_{\partial{\Omega}} \rho~\mathbf{v}[u_n - \mathbf{v}\cdot\mathbf{n}]~\text{dA} + \int_{\partial{\Omega}} \mathbf{t}~\text{dA} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} ~. $$ The surface tractions are related to the Cauchy stress by

\mathbf{t} = \boldsymbol{\sigma}\cdot\mathbf{n} ~. $$ Therefore,

\cfrac{d}{dt}\left[\int_{\Omega} \rho~\mathbf{v}~\text{dV}\right] = \int_{\partial{\Omega}} \rho~\mathbf{v}[u_n - \mathbf{v}\cdot\mathbf{n}]~\text{dA} + \int_{\partial{\Omega}} \boldsymbol{\sigma}\cdot\mathbf{n}~\text{dA} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} ~. $$ Let us assume that $$\Omega$$ is an arbitrary fixed control volume. Then,

\int_{\Omega} \frac{\partial }{\partial t}(\rho~\mathbf{v})~\text{dV} = - \int_{\partial{\Omega}} \rho~\mathbf{v}~(\mathbf{v}\cdot\mathbf{n})~\text{dA} + \int_{\partial{\Omega}} \boldsymbol{\sigma}\cdot\mathbf{n}~\text{dA} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} ~. $$ Now, from the definition of the tensor product we have (for all vectors $$\mathbf{a}$$)

(\mathbf{u}\otimes\mathbf{v})\cdot\mathbf{a} = (\mathbf{a}\cdot\mathbf{v})~\mathbf{u} ~. $$ Therefore,

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

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

\int_{\Omega} \frac{\partial }{\partial t}(\rho~\mathbf{v})~\text{dV} = - \int_{\Omega} \boldsymbol{\nabla} \bullet [\rho~(\mathbf{v}\otimes\mathbf{v})]~\text{dV} + \int_{\Omega} \boldsymbol{\nabla} \bullet \boldsymbol{\sigma}~\text{dV} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} $$ or,

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

\frac{\partial }{\partial t}(\rho~\mathbf{v}) + \boldsymbol{\nabla} \bullet [(\rho~\mathbf{v})\otimes\mathbf{v})] - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0~. $$ Using the identity

\boldsymbol{\nabla} \bullet (\mathbf{u}\otimes\mathbf{v}) = (\boldsymbol{\nabla} \bullet \mathbf{v})\mathbf{u} + (\boldsymbol{\nabla}\mathbf{u})\cdot\mathbf{v} $$ we get

\frac{\partial \rho}{\partial t}~\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + (\boldsymbol{\nabla} \bullet \mathbf{v})(\rho\mathbf{v}) + \boldsymbol{\nabla} (\rho~\mathbf{v})\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 $$ or,

\left[\frac{\partial \rho}{\partial t} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \boldsymbol{\nabla} (\rho~\mathbf{v})\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 $$ Using the identity

\boldsymbol{\nabla} (\varphi~\mathbf{v}) = \varphi~\boldsymbol{\nabla}\mathbf{v} + \mathbf{v}\otimes(\boldsymbol{\nabla} \varphi) $$ we get

\left[\frac{\partial \rho}{\partial t} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \left[\rho~\boldsymbol{\nabla}\mathbf{v} + \mathbf{v}\otimes(\boldsymbol{\nabla} \rho)\right]\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 $$ From the definition

(\mathbf{u}\otimes\mathbf{v})\cdot\mathbf{a} = (\mathbf{a}\cdot\mathbf{v})~\mathbf{u} $$ we have

[\mathbf{v}\otimes(\boldsymbol{\nabla} \rho)]\cdot\mathbf{v} = [\mathbf{v}\cdot(\boldsymbol{\nabla} \rho)]~\mathbf{v} ~. $$ Hence,

\left[\frac{\partial \rho}{\partial t} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} +[\mathbf{v}\cdot(\boldsymbol{\nabla} \rho)]~\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 $$ or,

\left[\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~. $$ The material time derivative of $$\rho$$ is defined as

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

\left[\dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~. $$ From the balance of mass, we have

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

\rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~. $$ The material time derivative of $$\mathbf{v}$$ is defined as

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

{ \rho~\dot{\mathbf{v}} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~. } $$