# Tensor derivative (continuum mechanics)

The derivatives of scalars, vectors, and second-order tensors with respect to second-order tensors are of considerable use in continuum mechanics. These derivatives are used in the theories of nonlinear elasticity and plasticity, particularly in the design of algorithms for numerical simulations.[1]

The directional derivative provides a systematic way of finding these derivatives.[2]

## Derivatives with respect to vectors and second-order tensors

The definitions of directional derivatives for various situations are given below. It is assumed that the functions are sufficiently smooth that derivatives can be taken.

### Derivatives of scalar valued functions of vectors

Let f(v) be a real valued function of the vector v. Then the derivative of f(v) with respect to v (or at v) in the direction u is the vector defined as

$\frac{\partial f}{\partial \mathbf{v}}\cdot\mathbf{u} = Df(\mathbf{v})[\mathbf{u}] = \left[\frac{d }{d \alpha}~f(\mathbf{v} + \alpha~\mathbf{u})\right]_{\alpha = 0}$

for all vectors u.

Properties:

1) If $f(\mathbf{v}) = f_1(\mathbf{v}) + f_2(\mathbf{v})$ then $\frac{\partial f}{\partial \mathbf{v}}\cdot\mathbf{u} = \left(\frac{\partial f_1}{\partial \mathbf{v}} + \frac{\partial f_2}{\partial \mathbf{v}}\right)\cdot\mathbf{u}$

2) If $f(\mathbf{v}) = f_1(\mathbf{v})~ f_2(\mathbf{v})$ then $\frac{\partial f}{\partial \mathbf{v}}\cdot\mathbf{u} = \left(\frac{\partial f_1}{\partial \mathbf{v}} \cdot \mathbf{u} \right)~f_2(\mathbf{v}) + f_1(\mathbf{v})~\left(\frac{\partial f_2}{\partial \mathbf{v}}\cdot\mathbf{u} \right)$

3) If $f(\mathbf{v}) = f_1(f_2(\mathbf{v}))$ then $\frac{\partial f}{\partial \mathbf{v}}\cdot\mathbf{u} = \frac{\partial f_1}{\partial f_2}~\frac{\partial f_2}{\partial \mathbf{v}}\cdot\mathbf{u}$

### Derivatives of vector valued functions of vectors

Let f(v) be a vector valued function of the vector v. Then the derivative of f(v) with respect to v (or at v) in the direction u is the second order tensor defined as

$\frac{\partial \mathbf{f}}{\partial \mathbf{v}}\cdot\mathbf{u} = D\mathbf{f}(\mathbf{v})[\mathbf{u}] = \left[\frac{d }{d \alpha}~\mathbf{f}(\mathbf{v} + \alpha~\mathbf{u} ) \right]_{\alpha = 0}$

for all vectors u.

Properties:
1) If $\mathbf{f}(\mathbf{v}) = \mathbf{f}_1(\mathbf{v}) + \mathbf{f}_2(\mathbf{v})$ then $\frac{\partial \mathbf{f}}{\partial \mathbf{v}}\cdot\mathbf{u} = \left(\frac{\partial \mathbf{f}_1}{\partial \mathbf{v}} + \frac{\partial \mathbf{f}_2}{\partial \mathbf{v}}\right)\cdot\mathbf{u}$
2) If $\mathbf{f}(\mathbf{v}) = \mathbf{f}_1(\mathbf{v})\times\mathbf{f}_2(\mathbf{v})$ then $\frac{\partial \mathbf{f}}{\partial \mathbf{v}}\cdot\mathbf{u} = \left(\frac{\partial \mathbf{f}_1}{\partial \mathbf{v}}\cdot\mathbf{u}\right)\times\mathbf{f}_2(\mathbf{v}) + \mathbf{f}_1(\mathbf{v})\times\left(\frac{\partial \mathbf{f}_2}{\partial \mathbf{v}}\cdot\mathbf{u} \right)$
3) If $\mathbf{f}(\mathbf{v}) = \mathbf{f}_1(\mathbf{f}_2(\mathbf{v}))$ then $\frac{\partial \mathbf{f}}{\partial \mathbf{v}}\cdot\mathbf{u} = \frac{\partial \mathbf{f}_1}{\partial \mathbf{f}_2}\cdot\left(\frac{\partial \mathbf{f}_2}{\partial \mathbf{v}}\cdot\mathbf{u} \right)$

### Derivatives of scalar valued functions of second-order tensors

Let $f(\boldsymbol{S})$ be a real valued function of the second order tensor $\boldsymbol{S}$. Then the derivative of $f(\boldsymbol{S})$ with respect to $\boldsymbol{S}$ (or at $\boldsymbol{S}$) in the direction $\boldsymbol{T}$ is the second order tensor defined as

$\frac{\partial f}{\partial \boldsymbol{S}}:\boldsymbol{T} = Df(\boldsymbol{S})[\boldsymbol{T}] = \left[\frac{d }{d \alpha}~f(\boldsymbol{S} + \alpha~\boldsymbol{T})\right]_{\alpha = 0}$

for all second order tensors $\boldsymbol{T}$.

Properties:
1) If $f(\boldsymbol{S}) = f_1(\boldsymbol{S}) + f_2(\boldsymbol{S})$ then $\frac{\partial f}{\partial \boldsymbol{S}}:\boldsymbol{T} = \left(\frac{\partial f_1}{\partial \boldsymbol{S}} + \frac{\partial f_2}{\partial \boldsymbol{S}}\right):\boldsymbol{T}$
2) If $f(\boldsymbol{S}) = f_1(\boldsymbol{S})~ f_2(\boldsymbol{S})$ then $\frac{\partial f}{\partial \boldsymbol{S}}:\boldsymbol{T} = \left(\frac{\partial f_1}{\partial \boldsymbol{S}}:\boldsymbol{T}\right)~f_2(\boldsymbol{S}) + f_1(\boldsymbol{S})~\left(\frac{\partial f_2}{\partial \boldsymbol{S}}:\boldsymbol{T} \right)$
3) If $f(\boldsymbol{S}) = f_1(f_2(\boldsymbol{S}))$ then $\frac{\partial f}{\partial \boldsymbol{S}}:\boldsymbol{T} = \frac{\partial f_1}{\partial f_2}~\left(\frac{\partial f_2}{\partial \boldsymbol{S}}:\boldsymbol{T} \right)$

### Derivatives of tensor valued functions of second-order tensors

Let $\boldsymbol{F}(\boldsymbol{S})$ be a second order tensor valued function of the second order tensor $\boldsymbol{S}$. Then the derivative of $\boldsymbol{F}(\boldsymbol{S})$ with respect to $\boldsymbol{S}$ (or at $\boldsymbol{S}$) in the direction $\boldsymbol{T}$ is the fourth order tensor defined as

$\frac{\partial \boldsymbol{F}}{\partial \boldsymbol{S}}:\boldsymbol{T} = D\boldsymbol{F}(\boldsymbol{S})[\boldsymbol{T}] = \left[\frac{d }{d \alpha}~\boldsymbol{F}(\boldsymbol{S} + \alpha~\boldsymbol{T})\right]_{\alpha = 0}$

for all second order tensors $\boldsymbol{T}$.

Properties:
1) If $\boldsymbol{F}(\boldsymbol{S}) = \boldsymbol{F}_1(\boldsymbol{S}) + \boldsymbol{F}_2(\boldsymbol{S})$ then $\frac{\partial \boldsymbol{F}}{\partial \boldsymbol{S}}:\boldsymbol{T} = \left(\frac{\partial \boldsymbol{F}_1}{\partial \boldsymbol{S}} + \frac{\partial \boldsymbol{F}_2}{\partial \boldsymbol{S}}\right):\boldsymbol{T}$
2) If $\boldsymbol{F}(\boldsymbol{S}) = \boldsymbol{F}_1(\boldsymbol{S})\cdot\boldsymbol{F}_2(\boldsymbol{S})$ then $\frac{\partial \boldsymbol{F}}{\partial \boldsymbol{S}}:\boldsymbol{T} = \left(\frac{\partial \boldsymbol{F}_1}{\partial \boldsymbol{S}}:\boldsymbol{T}\right)\cdot\boldsymbol{F}_2(\boldsymbol{S}) + \boldsymbol{F}_1 (\boldsymbol{S}) \cdot\left(\frac{\partial \boldsymbol{F}_2}{\partial \boldsymbol{S}}:\boldsymbol{T} \right)$
3) If $\boldsymbol{F}(\boldsymbol{S}) = \boldsymbol{F}_1(\boldsymbol{F}_2(\boldsymbol{S}))$ then $\frac{\partial \boldsymbol{F}}{\partial \boldsymbol{S}}:\boldsymbol{T} = \frac{\partial \boldsymbol{F}_1}{\partial \boldsymbol{F}_2}:\left(\frac{\partial \boldsymbol{F}_2}{\partial \boldsymbol{S}}:\boldsymbol{T} \right)$
4) If $f(\boldsymbol{S}) = f_1(\boldsymbol{F}_2(\boldsymbol{S}))$ then $\frac{\partial f}{\partial \boldsymbol{S}}:\boldsymbol{T} = \frac{\partial f_1}{\partial \boldsymbol{F}_2}:\left(\frac{\partial \boldsymbol{F}_2}{\partial \boldsymbol{S}}:\boldsymbol{T} \right)$

## Gradient of a tensor field

The gradient, $\boldsymbol{\nabla}\boldsymbol{T}$, of a tensor field $\boldsymbol{T}(\mathbf{x})$ in the direction of an arbitrary constant vector c is defined as:

$\boldsymbol{\nabla}\boldsymbol{T}\cdot\mathbf{c} = \left.\cfrac{d}{d\alpha}~\boldsymbol{T}(\mathbf{x}+\alpha\mathbf{c})\right|_{\alpha=0}$

The gradient of a tensor field of order n is a tensor field of order n+1.

### Cartesian coordinates

Note: the Einstein summation convention of summing on repeated indices is used below.

If $\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3$ are the basis vectors in a Cartesian coordinate system, with coordinates of points denoted by ($x_1, x_2, x_3$), then the gradient of the tensor field $\boldsymbol{T}$ is given by

$\boldsymbol{\nabla}\boldsymbol{T} = \cfrac{\partial{\boldsymbol{T}}}{\partial x_i}\otimes\mathbf{e}_i$

Since the basis vectors do not vary in a Cartesian coordinate system we have the following relations for the gradients of a scalar field $\phi$, a vector field v, and a second-order tensor field $\boldsymbol{S}$.

\begin{align} \boldsymbol{\nabla}\phi & = \cfrac{\partial\phi}{\partial x_i}~\mathbf{e}_i \\ \boldsymbol{\nabla}\mathbf{v} & = \cfrac{\partial (v_j \mathbf{e}_j)}{\partial x_i}\otimes\mathbf{e}_i = \cfrac{\partial v_j}{\partial x_i}~\mathbf{e}_j\otimes\mathbf{e}_i\\ \boldsymbol{\nabla}\boldsymbol{S} & = \cfrac{\partial (S_{jk} \mathbf{e}_j\otimes\mathbf{e}_k)}{\partial x_i}\otimes\mathbf{e}_i = \cfrac{\partial S_{jk}}{\partial x_i}~\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_i \end{align}

### Curvilinear coordinates

Note: the Einstein summation convention of summing on repeated indices is used below.

If $\mathbf{g}^1,\mathbf{g}^2,\mathbf{g}^3$ are the contravariant basis vectors in a curvilinear coordinate system, with coordinates of points denoted by ($\xi^1, \xi^2, \xi^3$), then the gradient of the tensor field $\boldsymbol{T}$ is given by (see [3] for a proof.)

$\boldsymbol{\nabla}\boldsymbol{T} = \cfrac{\partial{\boldsymbol{T}}}{\partial \xi^i}\otimes\mathbf{g}^i$

From this definition we have the following relations for the gradients of a scalar field $\phi$, a vector field v, and a second-order tensor field $\boldsymbol{S}$.

\begin{align} \boldsymbol{\nabla}\phi & = \cfrac{\partial\phi}{\partial \xi^i}~\mathbf{g}^i \\ \boldsymbol{\nabla}\mathbf{v} & = \cfrac{\partial (v^j \mathbf{g}_j)}{\partial \xi^i}\otimes\mathbf{g}^i = \left(\cfrac{\partial v^j}{\partial \xi^i} + v^k~\Gamma_{ik}^j\right)~\mathbf{g}_j\otimes\mathbf{g}^i = \left(\cfrac{\partial v_j}{\partial \xi^i} - v_k~\Gamma_{ij}^k\right)~\mathbf{g}^j\otimes\mathbf{g}^i\\ \boldsymbol{\nabla}\boldsymbol{S} & = \cfrac{\partial (S_{jk}~\mathbf{g}^j\otimes\mathbf{g}^k)}{\partial \xi^i}\otimes\mathbf{g}^i = \left(\cfrac{\partial S_{jk}}{\partial \xi_i}- S_{lk}~\Gamma_{ij}^l - S_{jl}~\Gamma_{ik}^l\right)~\mathbf{g}^j\otimes\mathbf{g}^k\otimes\mathbf{g}^i \end{align}

where the Christoffel symbol $\Gamma_{ij}^k$ is defined using

$\Gamma_{ij}^k~\mathbf{g}_k = \cfrac{\partial \mathbf{g}_i}{\partial \xi^j} \quad \implies \quad \Gamma_{ij}^k = \cfrac{\partial \mathbf{g}_i}{\partial \xi^j}\cdot\mathbf{g}_k = -\mathbf{g}_i\cdot\cfrac{\partial \mathbf{g}^k}{\partial \xi^j}$

#### Cylindrical polar coordinates

In cylindrical coordinates, the gradient is given by

\begin{align} \boldsymbol{\nabla}\phi &=\cfrac{\partial \phi}{\partial r}~\mathbf{e}_r + \cfrac{1}{r}~\cfrac{\partial \phi}{\partial \theta}~\mathbf{e}_\theta +\cfrac{\partial \phi}{\partial z}~\mathbf{e}_z \\ \boldsymbol{\nabla}\mathbf{v} &= \cfrac{\partial v_r}{\partial r}~\mathbf{e}_r\otimes\mathbf{e}_r + \cfrac{1}{r}\left(\cfrac{\partial v_r}{\partial \theta} -v_\theta \right)~\mathbf{e}_r \otimes \mathbf{e}_\theta + \cfrac{\partial v_r}{\partial z}~\mathbf{e}_r\otimes\mathbf{e}_z +\cfrac{\partial v_\theta}{\partial r}~\mathbf{e}_\theta\otimes\mathbf{e}_r +\cfrac{1}{r}\left(\cfrac{\partial v_\theta}{\partial \theta} + v_r \right)~\mathbf{e}_\theta\otimes\mathbf{e}_\theta \\ &\quad + \cfrac{\partial v_\theta}{\partial z}~\mathbf{e}_\theta \otimes\mathbf{e}_z + \cfrac{\partial v_z}{\partial r}~\mathbf{e}_z\otimes\mathbf{e}_r + \cfrac{1}{r}\cfrac{\partial v_z}{\partial \theta}~\mathbf{e}_z \otimes\mathbf{e}_\theta + \cfrac{\partial v_z}{\partial z}~\mathbf{e}_z\otimes\mathbf{e}_z \\ \boldsymbol{\nabla}\boldsymbol{S} & = \frac{\partial S_{rr}}{\partial r}~\mathbf{e}_r\otimes\mathbf{e}_r\otimes\mathbf{e}_r + \cfrac{1}{r}\left[\frac{\partial S_{rr}}{\partial \theta} - (S_{\theta r}+S_{r\theta})\right]~\mathbf{e}_r\otimes\mathbf{e}_r\otimes\mathbf{e}_\theta + \frac{\partial S_{rr}}{\partial z}~\mathbf{e}_r \otimes \mathbf{e}_r\otimes\mathbf{e}_z + \frac{\partial S_{r\theta}}{\partial r}~\mathbf{e}_r\otimes\mathbf{e}_\theta\otimes\mathbf{e}_r \\ &\quad+ \cfrac{1}{r}\left[\frac{\partial S_{r\theta}}{\partial \theta} + (S_{rr}-S_{\theta\theta}) \right]~\mathbf{e}_r\otimes\mathbf{e}_\theta\otimes\mathbf{e}_\theta + \frac{\partial S_{r\theta}}{\partial z}~\mathbf{e}_r\otimes\mathbf{e}_\theta\otimes\mathbf{e}_z + \frac{\partial S_{rz}}{\partial r}~\mathbf{e}_r \otimes \mathbf{e}_z \otimes \mathbf{e}_r + \cfrac{1}{r}\left[\frac{\partial S_{rz}}{\partial \theta} -S_{\theta z} \right]~\mathbf{e}_r\otimes\mathbf{e}_z\otimes\mathbf{e}_\theta \\ &\quad+ \frac{\partial S_{rz}}{\partial z}~\mathbf{e}_r \otimes \mathbf{e}_z\otimes\mathbf{e}_z + \frac{\partial S_{\theta r}}{\partial r}~\mathbf{e}_\theta \otimes \mathbf{e}_r \otimes \mathbf{e}_r + \cfrac{1}{r}\left[\frac{\partial S_{\theta r}}{\partial \theta} + (S_{rr}-S_{\theta\theta}) \right]~\mathbf{e}_\theta\otimes\mathbf{e}_r\otimes\mathbf{e}_\theta + \frac{\partial S_{\theta r}}{\partial z}~\mathbf{e}_\theta\otimes\mathbf{e}_r\otimes\mathbf{e}_z \\ &\quad+ \frac{\partial S_{\theta\theta}}{\partial r}~\mathbf{e}_\theta\otimes\mathbf{e}_\theta\otimes\mathbf{e}_r +\cfrac{1}{r}\left[\frac{\partial S_{\theta\theta}}{\partial \theta} + (S_{r\theta}+S_{\theta r})\right]~\mathbf{e}_\theta\otimes\mathbf{e}_\theta\otimes\mathbf{e}_\theta + \frac{\partial S_{\theta\theta}}{\partial z}~\mathbf{e}_\theta \otimes \mathbf{e}_\theta \otimes \mathbf{e}_z + \frac{\partial S_{\theta z}}{\partial r}~\mathbf{e}_\theta \otimes \mathbf{e}_z\otimes\mathbf{e}_r \\ &\quad+ \cfrac{1}{r}\left[\frac{\partial S_{\theta z}}{\partial \theta} + S_{rz} \right]~\mathbf{e}_\theta \otimes\mathbf{e}_z\otimes\mathbf{e}_\theta + \frac{\partial S_{\theta z}}{\partial z}~\mathbf{e}_\theta\otimes\mathbf{e}_z\otimes\mathbf{e}_z + \frac{\partial S_{zr}}{\partial r}~\mathbf{e}_z\otimes\mathbf{e}_r\otimes\mathbf{e}_r + \cfrac{1}{r}\left[\frac{\partial S_{zr}}{\partial \theta} - S_{z\theta} \right]~\mathbf{e}_z \otimes \mathbf{e}_r \otimes\mathbf{e}_\theta \\ &\quad+ \frac{\partial S_{zr}}{\partial z}~\mathbf{e}_z\otimes\mathbf{e}_r\otimes\mathbf{e}_z + \frac{\partial S_{z\theta}}{\partial r}~\mathbf{e}_z \otimes \mathbf{e}_\theta \otimes \mathbf{e}_r + \cfrac{1}{r}\left[\frac{\partial S_{z\theta}}{\partial \theta} + S_{zr} \right]~\mathbf{e}_z \otimes\mathbf{e}_\theta \otimes \mathbf{e}_\theta + \frac{\partial S_{z\theta}}{\partial z}~\mathbf{e}_z\otimes\mathbf{e}_\theta\otimes\mathbf{e}_z \\ &\quad+ \frac{\partial S_{zz}}{\partial r}~\mathbf{e}_z\otimes\mathbf{e}_z\otimes\mathbf{e}_r + \cfrac{1}{r}~\frac{\partial S_{zz}}{\partial \theta }~\mathbf{e}_z \otimes \mathbf{e}_z \otimes\mathbf{e}_\theta + \frac{\partial S_{zz}}{\partial z}~\mathbf{e}_z\otimes\mathbf{e}_z\otimes\mathbf{e}_z \end{align}

## Divergence of a tensor field

The divergence of a tensor field $\boldsymbol{T}(\mathbf{x})$ is defined using the recursive relation

$(\boldsymbol{\nabla}\cdot\boldsymbol{T})\cdot\mathbf{c} = \boldsymbol{\nabla}\cdot(\mathbf{c}\cdot\boldsymbol{T}) ~;\qquad\boldsymbol{\nabla}\cdot\mathbf{v} = \text{tr}(\boldsymbol{\nabla}\mathbf{v})$

where c is an arbitrary constant vector and v is a vector field. If $\boldsymbol{T}$ is a tensor field of order n > 1 then the divergence of the field is a tensor of order n−1.

### Cartesian coordinates

Note: the Einstein summation convention of summing on repeated indices is used below.

In a Cartesian coordinate system we have the following relations for a vector field v and a second-order tensor field $\boldsymbol{S}$.

\begin{align} \boldsymbol{\nabla}\cdot\mathbf{v} &= \cfrac{\partial v_i}{\partial x_i} \\ \boldsymbol{\nabla}\cdot\boldsymbol{S} &= \cfrac{\partial S_{ki}}{\partial x_i}~\mathbf{e}_k \end{align}

Note that in the case of the second-order tensor field, we have[4]

$\boldsymbol{\nabla}\cdot\boldsymbol{S} \neq \operatorname{div}\boldsymbol{S} = \boldsymbol{\nabla}\cdot\boldsymbol{S}^\mathrm{T}.$

### Curvilinear coordinates

Note: the Einstein summation convention of summing on repeated indices is used below.

In curvilinear coordinates, the divergences of a vector field v and a second-order tensor field $\boldsymbol{S}$ are

\begin{align} \boldsymbol{\nabla}\cdot\mathbf{v} & = \left(\cfrac{\partial v^i}{\partial \xi^i} + v^k~\Gamma_{ik}^i\right)\\ \boldsymbol{\nabla}\cdot\boldsymbol{S} & = \left(\cfrac{\partial S_{ik}}{\partial \xi_i}- S_{lk}~\Gamma_{ii}^l - S_{il}~\Gamma_{ik}^l\right)~\mathbf{g}^k \end{align}

#### Cylindrical polar coordinates

\begin{align} \boldsymbol{\nabla}\cdot\mathbf{v} & = \cfrac{\partial v_r}{\partial r} + \cfrac{1}{r}\left(\cfrac{\partial v_\theta}{\partial \theta} + v_r \right) + \cfrac{\partial v_z}{\partial z}\\ \boldsymbol{\nabla}\cdot\boldsymbol{S} & = \frac{\partial S_{rr}}{\partial r}~\mathbf{e}_r + \frac{\partial S_{r\theta}}{\partial r}~\mathbf{e}_\theta + \frac{\partial S_{rz}}{\partial r}~\mathbf{e}_z \\ & + \cfrac{1}{r}\left[\frac{\partial S_{\theta r}}{\partial \theta} + (S_{rr}-S_{\theta\theta})\right]~\mathbf{e}_r + \cfrac{1}{r}\left[\frac{\partial S_{\theta\theta}}{\partial \theta} + (S_{r\theta}+S_{\theta r})\right]~\mathbf{e}_\theta +\cfrac{1}{r}\left[\frac{\partial S_{\theta z}}{\partial \theta} + S_{rz}\right]~\mathbf{e}_z \\ & + \frac{\partial S_{zr}}{\partial z}~\mathbf{e}_r + \frac{\partial S_{z\theta}}{\partial z}~\mathbf{e}_\theta + \frac{\partial S_{zz}}{\partial z}~\mathbf{e}_z \end{align}

## Curl of a tensor field

The curl of an order-n > 1 tensor field $\boldsymbol{T}(\mathbf{x})$ is also defined using the recursive relation

$(\boldsymbol{\nabla}\times\boldsymbol{T})\cdot\mathbf{c} = \boldsymbol{\nabla}\times(\mathbf{c}\cdot\boldsymbol{T}) ~;\qquad (\boldsymbol{\nabla}\times\mathbf{v})\cdot\mathbf{c} = \boldsymbol{\nabla}\cdot(\mathbf{v}\times\mathbf{c})$

where c is an arbitrary constant vector and v is a vector field.

### Curl of a first-order tensor (vector) field

Consider a vector field v and an arbitrary constant vector c. In index notation, the cross product is given by

$\mathbf{v} \times \mathbf{c} = e_{ijk}~v_j~c_k~\mathbf{e}_i$

where $e_{ijk}$ is the permutation symbol. Then,

$\boldsymbol{\nabla}\cdot(\mathbf{v} \times \mathbf{c}) = e_{ijk}~v_{j,i}~c_k = (e_{ijk}~v_{j,i}~\mathbf{e}_k)\cdot\mathbf{c} = (\boldsymbol{\nabla}\times\mathbf{v})\cdot\mathbf{c}$

Therefore

$\boldsymbol{\nabla}\times\mathbf{v} = e_{ijk}~v_{j,i}~\mathbf{e}_k$

### Curl of a second-order tensor field

For a second-order tensor $\boldsymbol{S}$

$\mathbf{c}\cdot\boldsymbol{S} = c_m~S_{mj}~\mathbf{e}_j$

Hence, using the definition of the curl of a first-order tensor field,

$\boldsymbol{\nabla}\times(\mathbf{c}\cdot\boldsymbol{S}) = e_{ijk}~c_m~S_{mj,i}~\mathbf{e}_k = (e_{ijk}~S_{mj,i}~\mathbf{e}_k\otimes\mathbf{e}_m)\cdot\mathbf{c} = (\boldsymbol{\nabla}\times\boldsymbol{S})\cdot\mathbf{c}$

Therefore, we have

$\boldsymbol{\nabla}\times\boldsymbol{S} = e_{ijk}~S_{mj,i}~\mathbf{e}_k\otimes\mathbf{e}_m$

### Identities involving the curl of a tensor field

The most commonly used identity involving the curl of a tensor field, $\boldsymbol{T}$, is

$\boldsymbol{\nabla}\times(\boldsymbol{\nabla}\boldsymbol{T}) = \boldsymbol{0}$

This identity hold for tensor fields of all orders. For the important case of a second-order tensor, $\boldsymbol{S}$, this identity implies that

$\boldsymbol{\nabla}\times\boldsymbol{S} = \boldsymbol{0} \quad \implies \quad S_{mi,j} - S_{mj,i} = 0$

## Derivative of the determinant of a second-order tensor

The derivative of the determinant of a second order tensor $\boldsymbol{A}$ is given by

$\frac{\partial }{\partial \boldsymbol{A}}\det(\boldsymbol{A}) = \det(\boldsymbol{A})~[\boldsymbol{A}^{-1}]^T ~.$

In an orthonormal basis, the components of $\boldsymbol{A}$ can be written as a matrix A. In that case, the right hand side corresponds the cofactors of the matrix.

## Derivatives of the invariants of a second-order tensor

The principal invariants of a second order tensor are

\begin{align} I_1(\boldsymbol{A}) & = \text{tr}{\boldsymbol{A}} \\ I_2(\boldsymbol{A}) & = \frac{1}{2} \left[ (\text{tr}{\boldsymbol{A}})^2 - \text{tr}{\boldsymbol{A}^2} \right] \\ I_3(\boldsymbol{A}) & = \det(\boldsymbol{A}) \end{align}

The derivatives of these three invariants with respect to $\boldsymbol{A}$ are

\begin{align} \frac{\partial I_1}{\partial \boldsymbol{A}} & = \boldsymbol{\mathit{1}} \\ \frac{\partial I_2}{\partial \boldsymbol{A}} & = I_1~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T \\ \frac{\partial I_3}{\partial \boldsymbol{A}} & = \det(\boldsymbol{A})~[\boldsymbol{A}^{-1}]^T = I_2~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T~(I_1~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T) = (\boldsymbol{A}^2 - I_1~\boldsymbol{A} + I_2~\boldsymbol{\mathit{1}})^T \end{align}

## Derivative of the second-order identity tensor

Let $\boldsymbol{\mathit{1}}$ be the second order identity tensor. Then the derivative of this tensor with respect to a second order tensor $\boldsymbol{A}$ is given by

$\frac{\partial \boldsymbol{\mathit{1}}}{\partial \boldsymbol{A}}:\boldsymbol{T} = \boldsymbol{\mathsf{0}}:\boldsymbol{T} = \boldsymbol{\mathit{0}}$

This is because $\boldsymbol{\mathit{1}}$ is independent of $\boldsymbol{A}$.

## Derivative of a second-order tensor with respect to itself

Let $\boldsymbol{A}$ be a second order tensor. Then

$\frac{\partial \boldsymbol{A}}{\partial \boldsymbol{A}}:\boldsymbol{T} = \left[\frac{\partial }{\partial \alpha} (\boldsymbol{A} + \alpha~\boldsymbol{T})\right]_{\alpha = 0} = \boldsymbol{T} = \boldsymbol{\mathsf{I}}:\boldsymbol{T}$

Therefore,

$\frac{\partial \boldsymbol{A}}{\partial \boldsymbol{A}} = \boldsymbol{\mathsf{I}}$

Here $\boldsymbol{\mathsf{I}}$ is the fourth order identity tensor. In index notation with respect to an orthonormal basis

$\boldsymbol{\mathsf{I}} = \delta_{ik}~\delta_{jl}~\mathbf{e}_i\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_l$

This result implies that

$\frac{\partial \boldsymbol{A}^T}{\partial \boldsymbol{A}}:\boldsymbol{T} = \boldsymbol{\mathsf{I}}^T:\boldsymbol{T} = \boldsymbol{T}^T$

where

$\boldsymbol{\mathsf{I}}^T = \delta_{jk}~\delta_{il}~\mathbf{e}_i\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_l$

Therefore, if the tensor $\boldsymbol{A}$ is symmetric, then the derivative is also symmetric and we get

$\frac{\partial \boldsymbol{A}}{\partial \boldsymbol{A}} = \boldsymbol{\mathsf{I}}^{(s)} = \frac{1}{2}~(\boldsymbol{\mathsf{I}} + \boldsymbol{\mathsf{I}}^T)$

where the symmetric fourth order identity tensor is

$\boldsymbol{\mathsf{I}}^{(s)} = \frac{1}{2}~(\delta_{ik}~\delta_{jl} + \delta_{il}~\delta_{jk}) ~\mathbf{e}_i\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_l$

## Derivative of the inverse of a second-order tensor

Let $\boldsymbol{A}$ and $\boldsymbol{T}$ be two second order tensors, then

$\frac{\partial }{\partial \boldsymbol{A}} \left(\boldsymbol{A}^{-1}\right) : \boldsymbol{T} = - \boldsymbol{A}^{-1}\cdot\boldsymbol{T}\cdot\boldsymbol{A}^{-1}$

In index notation with respect to an orthonormal basis

$\frac{\partial A^{-1}_{ij}}{\partial A_{kl}}~T_{kl} = - A^{-1}_{ik}~T_{kl}~A^{-1}_{lj} \implies \frac{\partial A^{-1}_{ij}}{\partial A_{kl}} = - A^{-1}_{ik}~A^{-1}_{lj}$

We also have

$\frac{\partial }{\partial \boldsymbol{A}} \left(\boldsymbol{A}^{-T}\right) : \boldsymbol{T} = - \boldsymbol{A}^{-T}\cdot\boldsymbol{T}^{T}\cdot\boldsymbol{A}^{-T}$

In index notation

$\frac{\partial A^{-1}_{ji}}{\partial A_{kl}}~T_{kl} = - A^{-1}_{jk}~T_{lk}~A^{-1}_{li} \implies \frac{\partial A^{-1}_{ji}}{\partial A_{kl}} = - A^{-1}_{li}~A^{-1}_{jk}$

If the tensor $\boldsymbol{A}$ is symmetric then

$\frac{\partial A^{-1}_{ij}}{\partial A_{kl}} = -\cfrac{1}{2}\left(A^{-1}_{ik}~A^{-1}_{jl} + A^{-1}_{il}~A^{-1}_{jk}\right)$

## Integration by parts

Domain $\Omega$, its boundary $\Gamma$ and the outward unit normal $\mathbf{n}$

Another important operation related to tensor derivatives in continuum mechanics is integration by parts. The formula for integration by parts can be written as

$\int_{\Omega} \boldsymbol{F}\otimes\boldsymbol{\nabla}\boldsymbol{G}\,{\rm d}\Omega = \int_{\Gamma} \mathbf{n}\otimes(\boldsymbol{F}\otimes\boldsymbol{G})\,{\rm d}\Gamma - \int_{\Omega} \boldsymbol{G}\otimes\boldsymbol{\nabla}\boldsymbol{F}\,{\rm d}\Omega$

where $\boldsymbol{F}$ and $\boldsymbol{G}$ are differentiable tensor fields of arbitrary order, $\mathbf{n}$ is the unit outward normal to the domain over which the tensor fields are defined, $\otimes$ represents a generalized tensor product operator, and $\boldsymbol{\nabla}$ is a generalized gradient operator. When $\boldsymbol{F}$ is equal to the identity tensor, we get the divergence theorem

$\int_{\Omega}\boldsymbol{\nabla}\boldsymbol{G}\,{\rm d}\Omega = \int_{\Gamma} \mathbf{n}\otimes\boldsymbol{G}\,{\rm d}\Gamma \,.$

We can express the formula for integration by parts in Cartesian index notation as

$\int_{\Omega} F_{ijk....}\,G_{lmn...,p}\,{\rm d}\Omega = \int_{\Gamma} n_p\,F_{ijk...}\,G_{lmn...}\,{\rm d}\Gamma - \int_{\Omega} G_{lmn...}\,F_{ijk...,p}\,{\rm d}\Omega \,.$

For the special case where the tensor product operation is a contraction of one index and the gradient operation is a divergence, and both $\boldsymbol{F}$ and $\boldsymbol{G}$ are second order tensors, we have

$\int_{\Omega} \boldsymbol{F}\cdot(\boldsymbol{\nabla}\cdot\boldsymbol{G})\,{\rm d}\Omega = \int_{\Gamma} \mathbf{n}\cdot(\boldsymbol{G}\cdot\boldsymbol{F}^T)\,{\rm d}\Gamma - \int_{\Omega} (\boldsymbol{\nabla}\boldsymbol{F}):\boldsymbol{G}^T\,{\rm d}\Omega \,.$

In index notation,

$\int_{\Omega} F_{ij}\,G_{pj,p}\,{\rm d}\Omega = \int_{\Gamma} n_p\,F_{ij}\,G_{pj}\,{\rm d}\Gamma - \int_{\Omega} G_{pj}\,F_{ij,p}\,{\rm d}\Omega \,.$

## References

1. ^ J. C. Simo and T. J. R. Hughes, 1998, Computational Inelasticity, Springer
2. ^ J. E. Marsden and T. J. R. Hughes, 2000, Mathematical Foundations of Elasticity, Dover.
3. ^ Ogden, R. W., 2000, Nonlinear Elastic Deformations, Dover.
4. ^ http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_III/Chapter_1_Vectors_Tensors/Vectors_Tensors_14_Tensor_Calculus.pdf