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Up to date as of January 14, 2010

From Wikiversity

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 ~. }

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