•   Wikis

# Continuum mechanics/Balance of linear momentum: Wikis

Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.

# Study guide

Up to date as of January 14, 2010

## 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 Ω 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 Ω 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 ρ 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,

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