# Continuum mechanics/Balance of angular momentum

## Statement of the balance of angular momentum

The balance of angular momentum in an inertial frame can be expressed as:

$\boldsymbol{\sigma}= \boldsymbol{\sigma}^T$


### Proof

We assume that there are no surface couples on $\partial{\Omega}$ or body couples in $\Omega$. Recall the general balance equation

$\cfrac{d}{dt}\left[\int_{\Omega} f(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} f(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA} + \int_{\partial{\Omega}} g(\mathbf{x},t)~\text{dA} + \int_{\Omega} h(\mathbf{x},t)~\text{dV} ~.$

In this case, the physical quantity to be conserved the angular momentum density, i.e., $f = \mathbf{x}\times(\rho~\mathbf{v})$. The angular momentum source at the surface is then $g = \mathbf{x}\times\mathbf{t}$ and the angular momentum source inside the body is $h = \mathbf{x}\times(\rho~\mathbf{b})$. The angular momentum and moments are calculated with respect to a fixed origin. Hence we have

$\cfrac{d}{dt}\left[\int_{\Omega} \mathbf{x}\times(\rho~\mathbf{v})~\text{dV}\right] = \int_{\partial{\Omega}} [\mathbf{x}\times(\rho~\mathbf{v})] [u_n - \mathbf{v}\cdot\mathbf{n}]~\text{dA} + \int_{\partial{\Omega}} \mathbf{x}\times\mathbf{t}~\text{dA} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~.$

Assuming that $\Omega$ is a control volume, we have

$\int_{\Omega} \mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right]~\text{dV} = - \int_{\partial{\Omega}} [\mathbf{x}\times(\rho~\mathbf{v})][\mathbf{v}\cdot\mathbf{n}]~\text{dA} + \int_{\partial{\Omega}} \mathbf{x}\times\mathbf{t}~\text{dA} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~.$

Using the definition of a tensor product we can write

$[\mathbf{x}\times(\rho~\mathbf{v})][\mathbf{v}\cdot\mathbf{n}] = [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}]\cdot\mathbf{n} ~.$

Also, $\mathbf{t} = \boldsymbol{\sigma}\cdot\mathbf{n}$. Therefore we have

$\int_{\Omega} \mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right]~\text{dV} = - \int_{\partial{\Omega}} [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}]\cdot\mathbf{n} ~\text{dA} + \int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~.$

Using the divergence theorem, we get

$\int_{\Omega} \mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right]~\text{dV} = - \int_{\Omega} \boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}]~\text{dV} + \int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~.$

To convert the surface integral in the above equation into a volume integral, it is convenient to use index notation. Thus,

$\left[\int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA}\right]_i = \int_{\partial{\Omega}} e_{ijk}~x_j~\sigma_{kl}~n_l~\text{dA}= \int_{\partial{\Omega}} A_{il}~n_l~\text{dA}= \int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA}$

where $[~]_i$ represents the $i$-th component of the vector. Using the divergence theorem

$\int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA} = \int_{\Omega} \boldsymbol{\nabla} \bullet \boldsymbol{A}~\text{dV} = \int_{\Omega} \frac{\partial A_{il}}{\partial x_l}~\text{dV} = \int_{\Omega} \frac{\partial }{\partial x_l}(e_{ijk}~x_j~\sigma_{kl})~\text{dV}~.$

Differentiating,

$\int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA} = \int_{\Omega} \left[ e_{ijk}~\delta_{jl}~\sigma_{kl} + e_{ijk}~x_j~\frac{\partial \sigma_{kl}}{\partial x_l}\right]~\text{dV} = \int_{\Omega} \left[ e_{ijk}~\sigma_{kj} + e_{ijk}~x_j~\frac{\partial \sigma_{kl}}{\partial x_l}\right]~\text{dV} = \int_{\Omega} \left[ e_{ijk}~\sigma_{kj} + e_{ijk}~x_j~[\boldsymbol{\nabla} \bullet \boldsymbol{\sigma}]_l\right]~\text{dV} ~.$

Expressed in direct tensor notation,

$\int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA} = \int_{\Omega} \left[ [\mathcal{E}:\boldsymbol{\sigma}^T]_i + [\mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})]_i\right]~\text{dV}$

where $\mathcal{E}$ is the third-order permutation tensor. Therefore,

$\left[\int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA}\right]_i = = \int_{\Omega} \left[ [\mathcal{E}:\boldsymbol{\sigma}^T]_i + [\mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})]_i\right]~\text{dV}$

or,

$\int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA} = = \int_{\Omega} \left[ \mathcal{E}:\boldsymbol{\sigma}^T + \mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})\right]~\text{dV} ~.$

The balance of angular momentum can then be written as

$\int_{\Omega} \mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right]~\text{dV} = - \int_{\Omega} \boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}]~\text{dV} + \int_{\Omega} \left[ \mathcal{E}:\boldsymbol{\sigma}^T + \mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})\right]~\text{dV} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~.$

Since $\Omega$ is an arbitrary volume, we have

$\mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right] = - \boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}] + \mathcal{E}:\boldsymbol{\sigma}^T + \mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma}) + \mathbf{x}\times(\rho~\mathbf{b})$

or,

${\mathbf{x}}\times {\left[\frac{\partial }{\partial t}(\rho~\mathbf{v}) - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - \boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}] + \mathcal{E}:\boldsymbol{\sigma}^T ~.$

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

$\boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}] = (\boldsymbol{\nabla} \bullet \mathbf{v})[\mathbf{x}\times(\rho~\mathbf{v})] + (\boldsymbol{\nabla} [\mathbf{x}\times(\rho~\mathbf{v})])\cdot\mathbf{v} ~.$

The second term on the right can be further simplified using index notation as follows.

\begin{align} \left[(\boldsymbol{\nabla} [\mathbf{x}\times(\rho~\mathbf{v})])\cdot\mathbf{v}\right]_i = \left[(\boldsymbol{\nabla} [\rho~(\mathbf{x}\times\mathbf{v})])\cdot\mathbf{v}\right]_i & = \frac{\partial }{\partial x_l}(\rho~e_{ijk}~x_j~v_k)~v_l \\ & = e_{ijk}\left[ \frac{\partial \rho}{\partial x_l}~x_j~v_k~v_l+ \rho~\frac{\partial x_j}{\partial x_l}~v_k~v_l + \rho~x_j~\frac{\partial v_k}{\partial x_l}~v_l\right] \\ & = (e_{ijk}~x_j~v_k)~\left(\frac{\partial \rho}{\partial x_l}~v_l\right)+ \rho~(e_{ijk}~\delta_{jl}~v_k~v_l) + e_{ijk}~x_j~\left(\rho~\frac{\partial v_k}{\partial x_l}~v_l\right) \\ & = [(\mathbf{x}\times\mathbf{v})(\boldsymbol{\nabla} \rho\cdot\mathbf{v}) + \rho~\mathbf{v}\times\mathbf{v} + \mathbf{x}\times(\rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v})]_i \\ & = [(\mathbf{x}\times\mathbf{v})(\boldsymbol{\nabla} \rho\cdot\mathbf{v}) + \mathbf{x}\times(\rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v})]_i ~. \end{align}

Therefore we can write

$\boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}] = (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) + (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) + \mathbf{x}\times(\rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v})] ~.$

The balance of angular momentum then takes the form

${\mathbf{x}}\times {\left[\frac{\partial }{\partial t}(\rho~\mathbf{v}) - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) - (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) - \mathbf{x}\times(\rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v}) + \mathcal{E}:\boldsymbol{\sigma}^T$

or,

${\mathbf{x}}\times {\left[\frac{\partial }{\partial t}(\rho~\mathbf{v}) + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) - (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) + \mathcal{E}:\boldsymbol{\sigma}^T$

or,

${\mathbf{x}}\times {\left[\rho\frac{\partial \mathbf{v}}{\partial t} + \frac{\partial \rho}{\partial t}~\mathbf{v} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) - (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) + \mathcal{E}:\boldsymbol{\sigma}^T$

The material time derivative of $\mathbf{v}$ is defined as

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

Therefore,

${\mathbf{x}}\times {\left[\rho~\dot{\mathbf{v}} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - \mathbf{x}\times\cfrac{\partial \rho}{\partial t}~\mathbf{v} + - (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) - (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) + \mathcal{E}:\boldsymbol{\sigma}^T ~.$

Also, from the conservation of linear momentum

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

Hence,

\begin{align} 0 & = \mathbf{x}\times\cfrac{\partial\rho}{\partial t}~\mathbf{v} + (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) + (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) - \mathcal{E}:\boldsymbol{\sigma}^T \\ & = \left(\frac{\partial \rho}{\partial t} + \rho\boldsymbol{\nabla} \bullet \mathbf{v} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} \right)(\mathbf{x}\times\mathbf{v}) - \mathcal{E}:\boldsymbol{\sigma}^T ~. \end{align}

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

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

Hence,

$(\dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times\mathbf{v}) - \mathcal{E}:\boldsymbol{\sigma}^T = 0 ~.$

From the balance of mass

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

Therefore,

$\mathcal{E}:\boldsymbol{\sigma}^T = 0 ~.$

In index notation,

$e_{ijk}~\sigma_{kj} = 0 ~.$

Expanding out, we get

$\sigma_{12} - \sigma_{21} = 0 ~;~~ \sigma_{23} - \sigma_{32} = 0 ~;~~ \sigma_{31} - \sigma_{13} = 0 ~.$

Hence,

${ \boldsymbol{\sigma} = \boldsymbol{\sigma}^T }$