Continuum mechanics/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,

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

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