Continuum mechanics/Balance of linear momentum

Statement of the balance of linear momentum

The balance of linear momentum can be expressed as:

$\rho \dot{𝐯}-𝜵\bullet 𝝈-\rho 𝐛=0$ 

where $\rho (𝐱,t)$ is the mass density, $𝐯(𝐱,t)$ is the velocity, $𝝈(𝐱,t)$ is the Cauchy stress, and $\rho 𝐛$ is the body force density.

Proof

Recall the general equation for the balance of a physical quantity

$\frac{d}{dt}[{\int }_{\mathrm{\Omega }}f(𝐱,t)\text{dV}]={\int }_{\partial \mathrm{\Omega }}f(𝐱,t)[u_n(𝐱,t)-𝐯(𝐱,t)\cdot 𝐧(𝐱,t)]\text{dA}+{\int }_{\partial \mathrm{\Omega }}g(𝐱,t)\text{dA}+{\int }_{\mathrm{\Omega }}h(𝐱,t)\text{dV}.$

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

$\frac{d}{dt}\left[{\int }_{\mathrm{\Omega }}\rho 𝐯\text{dV}\right]={\int }_{\partial \mathrm{\Omega }}\rho 𝐯[u_n-𝐯\cdot 𝐧]\text{dA}+{\int }_{\partial \mathrm{\Omega }}𝐭\text{dA}+{\int }_{\mathrm{\Omega }}\rho 𝐛\text{dV}.$

The surface tractions are related to the Cauchy stress by

$𝐭=𝝈\cdot 𝐧.$

Therefore,

$\frac{d}{dt}\left[{\int }_{\mathrm{\Omega }}\rho 𝐯\text{dV}\right]={\int }_{\partial \mathrm{\Omega }}\rho 𝐯[u_n-𝐯\cdot 𝐧]\text{dA}+{\int }_{\partial \mathrm{\Omega }}𝝈\cdot 𝐧\text{dA}+{\int }_{\mathrm{\Omega }}\rho 𝐛\text{dV}.$

Let us assume that $\mathrm{\Omega }$ is an arbitrary fixed control volume. Then,

${\int }_{\mathrm{\Omega }}\frac{\partial }{\partial t}(\rho 𝐯)\text{dV}=-{\int }_{\partial \mathrm{\Omega }}\rho 𝐯(𝐯\cdot 𝐧)\text{dA}+{\int }_{\partial \mathrm{\Omega }}𝝈\cdot 𝐧\text{dA}+{\int }_{\mathrm{\Omega }}\rho 𝐛\text{dV}.$

Now, from the definition of the tensor product we have (for all vectors $𝐚$)

$(𝐮\otimes 𝐯)\cdot 𝐚=(𝐚\cdot 𝐯)𝐮.$

Therefore,

${\int }_{\mathrm{\Omega }}\frac{\partial }{\partial t}(\rho 𝐯)\text{dV}=-{\int }_{\partial \mathrm{\Omega }}\rho (𝐯\otimes 𝐯)\cdot 𝐧\text{dA}+{\int }_{\partial \mathrm{\Omega }}𝝈\cdot 𝐧\text{dA}+{\int }_{\mathrm{\Omega }}\rho 𝐛\text{dV}.$

Using the divergence theorem

${\int }_{\mathrm{\Omega }}𝜵\bullet 𝐯\text{dV}={\int }_{\partial \mathrm{\Omega }}𝐯\cdot 𝐧\text{dA}$

we have

${\int }_{\mathrm{\Omega }}\frac{\partial }{\partial t}(\rho 𝐯)\text{dV}=-{\int }_{\mathrm{\Omega }}𝜵\bullet [\rho (𝐯\otimes 𝐯)]\text{dV}+{\int }_{\mathrm{\Omega }}𝜵\bullet 𝝈\text{dV}+{\int }_{\mathrm{\Omega }}\rho 𝐛\text{dV}$

or,

${\int }_{\mathrm{\Omega }}[\frac{\partial }{\partial t}(\rho 𝐯)+𝜵\bullet [(\rho 𝐯)\otimes 𝐯)]-𝜵\bullet 𝝈-\rho 𝐛]\text{dV}=0.$

Since $\mathrm{\Omega }$ is arbitrary, we have

$\frac{\partial }{\partial t}(\rho 𝐯)+𝜵\bullet [(\rho 𝐯)\otimes 𝐯)]-𝜵\bullet 𝝈-\rho 𝐛=0.$

Using the identity

$𝜵\bullet (𝐮\otimes 𝐯)=(𝜵\bullet 𝐯)𝐮+(𝜵𝐮)\cdot 𝐯$

we get

$\frac{\partial \rho }{\partial t}𝐯+\rho \frac{\partial 𝐯}{\partial t}+(𝜵\bullet 𝐯)(\rho 𝐯)+𝜵(\rho 𝐯)\cdot 𝐯-𝜵\bullet 𝝈-\rho 𝐛=0$

or,

$[\frac{\partial \rho }{\partial t}+\rho 𝜵\bullet 𝐯]𝐯+\rho \frac{\partial 𝐯}{\partial t}+𝜵(\rho 𝐯)\cdot 𝐯-𝜵\bullet 𝝈-\rho 𝐛=0$

Using the identity

$𝜵(\phi 𝐯)=\phi 𝜵𝐯+𝐯\otimes (𝜵\phi )$

we get

$[\frac{\partial \rho }{\partial t}+\rho \mathrm{\nabla }\bullet 𝐯]𝐯+\rho \frac{\partial 𝐯}{\partial t}+[\rho \mathrm{\nabla }𝐯+𝐯\otimes (\mathrm{\nabla }\rho )]\cdot 𝐯-\mathrm{\nabla }\bullet 𝝈-\rho 𝐛=0$

From the definition

$(𝐮\otimes 𝐯)\cdot 𝐚=(𝐚\cdot 𝐯)𝐮$

we have

$[𝐯\otimes (𝜵\rho )]\cdot 𝐯=[𝐯\cdot (𝜵\rho )]𝐯.$

Hence,

$[\frac{\partial \rho }{\partial t}+\rho 𝜵\bullet 𝐯]𝐯+\rho \frac{\partial 𝐯}{\partial t}+\rho 𝜵𝐯\cdot 𝐯+[𝐯\cdot (𝜵\rho )]𝐯-𝜵\bullet 𝝈-\rho 𝐛=0$

or,

$[\frac{\partial \rho }{\partial t}+𝜵\rho \cdot 𝐯+\rho 𝜵\bullet 𝐯]𝐯+\rho \frac{\partial 𝐯}{\partial t}+\rho 𝜵𝐯\cdot 𝐯-𝜵\bullet 𝝈-\rho 𝐛=0.$

The material time derivative of $\rho$ is defined as

$\dot{\rho }=\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\rho \cdot 𝐯.$

Therefore,

$[\dot{\rho }+\rho 𝜵\bullet 𝐯]𝐯+\rho \frac{\partial 𝐯}{\partial t}+\rho 𝜵𝐯\cdot 𝐯-𝜵\bullet 𝝈-\rho 𝐛=0.$

From the balance of mass, we have

$\dot{\rho }+\rho 𝜵\bullet 𝐯=0.$

Therefore,

$\rho \frac{\partial 𝐯}{\partial t}+\rho 𝜵𝐯\cdot 𝐯-𝜵\bullet 𝝈-\rho 𝐛=0.$

The material time derivative of $𝐯$ is defined as

$\dot{𝐯}=\frac{\partial 𝐯}{\partial t}+𝜵𝐯\cdot 𝐯.$

Hence,

$\rho \dot{𝐯}-\mathrm{\nabla }\bullet 𝝈-\rho 𝐛=0.$

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