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The [[directional derivative|derivatives]] of [[Scalar (mathematics)|scalars]], [[Euclidean vector|vectors]], and second-order [[tensor]]s 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 (physics)|plasticity]], particularly in the design of [[algorithms]] for [[numerical simulation]]s.<ref name=Simo98>J. C. Simo and T. J. R. Hughes, 1998, ''Computational Inelasticity'', Springer</ref>
 
The [[directional derivative]] provides a systematic way of finding these derivatives.<ref name=Marsden00>J. E. Marsden and T. J. R. Hughes, 2000, ''Mathematical Foundations of Elasticity'', Dover.</ref> 
 
== 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
 
:<math>\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}</math>
 
for all vectors '''u'''.
 
''Properties:''
 
1) If <math>f(\mathbf{v}) = f_1(\mathbf{v}) + f_2(\mathbf{v})</math> then <math>\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}</math>
 
2) If <math>f(\mathbf{v}) = f_1(\mathbf{v})~ f_2(\mathbf{v})</math> then <math>\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)</math>
 
3) If <math>f(\mathbf{v}) = f_1(f_2(\mathbf{v}))</math> then <math>\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}</math>
 
===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
:<math> \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}</math>
for all vectors '''u'''.
 
:'''Properties:'''
 
:1) If <math>\mathbf{f}(\mathbf{v}) = \mathbf{f}_1(\mathbf{v}) + \mathbf{f}_2(\mathbf{v})</math> then <math>\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} </math>
 
:2) If <math>\mathbf{f}(\mathbf{v}) = \mathbf{f}_1(\mathbf{v})\times\mathbf{f}_2(\mathbf{v})</math> then <math>\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)</math>
 
:3) If <math>\mathbf{f}(\mathbf{v}) = \mathbf{f}_1(\mathbf{f}_2(\mathbf{v}))</math> then <math>\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)</math>
 
===Derivatives of scalar valued functions of second-order tensors===
Let <math>f(\boldsymbol{S})</math> be a real valued function of the second order tensor <math>\boldsymbol{S}</math>.  Then the derivative of <math>f(\boldsymbol{S})</math> with respect to <math>\boldsymbol{S}</math> (or at <math>\boldsymbol{S}</math>) in the direction <math>\boldsymbol{T}</math> is the ''' second order tensor''' defined as
:<math>\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}</math>
for all second order tensors <math>\boldsymbol{T}</math>.
 
:'''Properties:'''
 
:1) If <math>f(\boldsymbol{S}) = f_1(\boldsymbol{S}) + f_2(\boldsymbol{S})</math> then <math> \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} </math>
 
:2) If <math>f(\boldsymbol{S}) = f_1(\boldsymbol{S})~ f_2(\boldsymbol{S})</math> then <math> \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) </math>
 
:3) If <math>f(\boldsymbol{S}) = f_1(f_2(\boldsymbol{S}))</math> then <math> \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) </math>
 
===Derivatives of tensor valued functions of second-order tensors===
Let <math>\boldsymbol{F}(\boldsymbol{S})</math> be a second order tensor valued function of the second order tensor <math>\boldsymbol{S}</math>. Then the derivative of <math>\boldsymbol{F}(\boldsymbol{S})</math> with respect to <math>\boldsymbol{S}</math>  (or at <math>\boldsymbol{S}</math>) in the direction <math>\boldsymbol{T}</math> is the ''' fourth order tensor''' defined as
:<math>\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}</math>
for all second order tensors <math>\boldsymbol{T}</math>.
 
:'''Properties:'''
 
:1) If <math>\boldsymbol{F}(\boldsymbol{S}) = \boldsymbol{F}_1(\boldsymbol{S}) + \boldsymbol{F}_2(\boldsymbol{S})</math> then <math> \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} </math>
 
:2) If <math>\boldsymbol{F}(\boldsymbol{S}) = \boldsymbol{F}_1(\boldsymbol{S})\cdot\boldsymbol{F}_2(\boldsymbol{S})</math> then <math> \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) </math>
 
:3) If <math>\boldsymbol{F}(\boldsymbol{S}) = \boldsymbol{F}_1(\boldsymbol{F}_2(\boldsymbol{S}))</math> then <math> \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) </math>
 
:4) If <math>f(\boldsymbol{S}) = f_1(\boldsymbol{F}_2(\boldsymbol{S}))</math> then <math> \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) </math>
 
== Gradient of a tensor field ==
The [[gradient]], <math>\boldsymbol{\nabla}\boldsymbol{T}</math>, of a tensor field <math>\boldsymbol{T}(\mathbf{x})</math> in the direction of an arbitrary constant vector '''c''' is defined as:
:<math>  \boldsymbol{\nabla}\boldsymbol{T}\cdot\mathbf{c} = \left.\cfrac{d}{d\alpha}~\boldsymbol{T}(\mathbf{x}+\alpha\mathbf{c})\right|_{\alpha=0}</math>
The gradient of a tensor field of order ''n'' is a tensor field of order ''n''+1.
 
=== Cartesian coordinates ===
{{Einstein_summation_convention}}
 
If <math>\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3</math> are the basis vectors in a [[Cartesian coordinate]] system, with coordinates of points denoted by (<math>x_1, x_2, x_3</math>), then the gradient of the tensor field <math>\boldsymbol{T}</math> is given by
:<math>   \boldsymbol{\nabla}\boldsymbol{T} = \cfrac{\partial{\boldsymbol{T}}}{\partial x_i}\otimes\mathbf{e}_i </math>
 
:{| class="toccolours collapsible collapsed" width="60%" style="text-align:left"
!Proof
|-
|
The vectors '''x''' and '''c''' can be written as <math> \mathbf{x} = x_i~\mathbf{e}_i </math> and <math>\mathbf{c} = c_i~\mathbf{e}_i </math> . Let '''y''' := '''x''' + α'''c'''. In that case the gradient is given by
:<math>
  \begin{align}
  \boldsymbol{\nabla}\boldsymbol{T}\cdot\mathbf{c} & = \left.\cfrac{d}{d\alpha}~\boldsymbol{T}(x_1+\alpha c_1, x_2 + \alpha c_2, x_3 + \alpha c_3)\right|_{\alpha=0} \equiv \left.\cfrac{d}{d\alpha}~\boldsymbol{T}(y_1, y_2, y_3)\right|_{\alpha=0} \\
    & = \left [\cfrac{\partial{\boldsymbol{T}}}{\partial y_1}~\cfrac{\partial y_1}{\partial \alpha} + \cfrac{\partial{\boldsymbol{T}}}{\partial y_2}~\cfrac{\partial y_2}{\partial \alpha} +
        \cfrac{\partial{\boldsymbol{T}}}{\partial y_3}~\cfrac{\partial y_3}{\partial \alpha} \right]_{\alpha=0} =
        \left [\cfrac{\partial{\boldsymbol{T}}}{\partial y_1}~c_1 +          \cfrac{\partial{\boldsymbol{T}}}{\partial y_2}~c_2 +
        \cfrac{\partial{\boldsymbol{T}}}{\partial y_3}~c_3          \right]_{\alpha=0} \\
    & = \cfrac{\partial{\boldsymbol{T}}}{\partial x_1}~c_1 + \cfrac{\partial{\boldsymbol{T}}}{\partial x_2}~c_2 +
        \cfrac{\partial{\boldsymbol{T}}}{\partial x_3}~c_3 \equiv \cfrac{\partial{\boldsymbol{T}}}{\partial x_i}~c_i = \cfrac{\partial{\boldsymbol{T}}}{\partial x_i}~(\mathbf{e}_i\cdot\mathbf{c})
        = \left[\cfrac{\partial{\boldsymbol{T}}}{\partial x_i}\otimes\mathbf{e}_i\right]\cdot\mathbf{c} \qquad \square
  \end{align}
</math>
|}
Since the basis vectors do not vary in a Cartesian coordinate system we have the following relations for the gradients of a scalar field <math>\phi</math>, a vector field '''v''', and a second-order tensor field <math>\boldsymbol{S}</math>.
:<math>  \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}
</math>
 
=== Curvilinear coordinates ===
{{main|Curvilinear coordinates}}
{{Einstein_summation_convention}}
 
If <math>\mathbf{g}^1,\mathbf{g}^2,\mathbf{g}^3</math> are the [[Covariance and contravariance of vectors|contravariant]] [[basis vector]]s in a [[curvilinear coordinates|curvilinear coordinate]] system, with coordinates of points denoted by (<math>\xi^1, \xi^2, \xi^3</math>), then the gradient of the tensor field <math>\boldsymbol{T}</math> is given by (see <ref>Ogden, R. W., 2000, '''Nonlinear Elastic Deformations''', Dover.</ref> for a proof.)
:<math>
  \boldsymbol{\nabla}\boldsymbol{T} = \cfrac{\partial{\boldsymbol{T}}}{\partial \xi^i}\otimes\mathbf{g}^i
</math>
 
From this definition we have the following relations for the gradients of a scalar field <math>\phi</math>, a vector field '''v''', and a second-order tensor field <math>\boldsymbol{S}</math>.
:<math>\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} </math>
 
where the [[Christoffel symbol]] <math>\Gamma_{ij}^k</math> is defined using
:<math>  \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}
</math>
 
==== Cylindrical polar coordinates ====
In [[cylindrical coordinate]]s, the gradient is given by
:<math>\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} </math>
 
== Divergence of a tensor field ==
The [[divergence]] of a tensor field <math>\boldsymbol{T}(\mathbf{x})</math> is defined using the recursive relation
:<math>
  (\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})
</math>
where '''c''' is an arbitrary constant vector and '''v''' is a vector field. If <math>\boldsymbol{T}</math> is a tensor field of order ''n'' > 1 then the divergence of the field is a tensor of order ''n''−1.
 
=== Cartesian coordinates ===
{{Einstein_summation_convention}}
In a Cartesian coordinate system we have the following relations for a vector field '''v''' and a second-order tensor field <math>\boldsymbol{S}</math>.
:<math>\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}</math>
 
Note that in the case of the second-order tensor field, we have<ref>http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_III/Chapter_1_Vectors_Tensors/Vectors_Tensors_14_Tensor_Calculus.pdf</ref>
 
:<math>\boldsymbol{\nabla}\cdot\boldsymbol{S} \neq \operatorname{div}\boldsymbol{S} = \boldsymbol{\nabla}\cdot\boldsymbol{S}^\mathrm{T}.</math>
 
=== Curvilinear coordinates ===
{{main|Curvilinear coordinates}}
{{Einstein_summation_convention}}
In curvilinear coordinates, the divergences of a vector field '''v''' and a second-order tensor field <math>\boldsymbol{S}</math> are
:<math>
  \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}
</math>
 
==== Cylindrical polar coordinates ====
In [[cylindrical coordinates|cylindrical polar coordinates]]
:<math>
  \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}
</math>
 
== Curl of a tensor field ==
The [[curl (mathematics)|curl]] of an order-''n'' > 1 tensor field <math>\boldsymbol{T}(\mathbf{x})</math> is also defined using the recursive relation
:<math>(\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})</math>
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
:<math>
  \mathbf{v} \times \mathbf{c} = e_{ijk}~v_j~c_k~\mathbf{e}_i
</math>
where <math>e_{ijk}</math> is the [[permutation symbol]].  Then,
:<math>
  \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}
</math>
Therefore
:<math>\boldsymbol{\nabla}\times\mathbf{v} = e_{ijk}~v_{j,i}~\mathbf{e}_k</math>
 
=== Curl of a second-order tensor field ===
For a second-order tensor <math>\boldsymbol{S}</math>
:<math>  \mathbf{c}\cdot\boldsymbol{S} = c_m~S_{mj}~\mathbf{e}_j
</math>
Hence, using the definition of the curl of a first-order tensor field,
:<math>  \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}
</math>
Therefore, we have
:<math>  \boldsymbol{\nabla}\times\boldsymbol{S} = e_{ijk}~S_{mj,i}~\mathbf{e}_k\otimes\mathbf{e}_m
</math>
 
=== Identities involving the curl of a tensor field ===
The most commonly used identity involving the curl of a tensor field, <math>\boldsymbol{T}</math>, is
:<math>
  \boldsymbol{\nabla}\times(\boldsymbol{\nabla}\boldsymbol{T}) = \boldsymbol{0}
</math>
This identity hold for tensor fields of all orders.  For the important case of a second-order tensor, <math>\boldsymbol{S}</math>, this identity implies that
:<math>
  \boldsymbol{\nabla}\times\boldsymbol{S} = \boldsymbol{0} \quad \implies \quad S_{mi,j} - S_{mj,i} = 0
</math>
 
==Derivative of the determinant of a second-order tensor==
The derivative of the determinant of a second order tensor <math>\boldsymbol{A}</math> is given by
:<math>
  \frac{\partial }{\partial \boldsymbol{A}}\det(\boldsymbol{A}) = \det(\boldsymbol{A})~[\boldsymbol{A}^{-1}]^T ~.
</math>
In an orthonormal basis, the components of <math>\boldsymbol{A}</math> can be written as a matrix '''A'''. In that case, the right hand side corresponds the cofactors of the matrix.
 
:{| class="toccolours collapsible collapsed" width="60%" style="text-align:left"
!Proof
|-
|
Let <math>\boldsymbol{A}</math> be a second order tensor and let <math>f(\boldsymbol{A}) = \det(\boldsymbol{A})</math>.  Then, from the definition of the derivative of a scalar valued function of a tensor, we have
:<math>\begin{align}
    \frac{\partial f}{\partial \boldsymbol{A}}:\boldsymbol{T} & =  \left.\cfrac{d}{d\alpha} \det(\boldsymbol{A} + \alpha~\boldsymbol{T}) \right|_{\alpha=0} \\
    & = \left.\cfrac{d}{d\alpha} \det\left[\alpha~\boldsymbol{A}\left(\cfrac{1}{\alpha}~\boldsymbol{\mathit{I}} + \boldsymbol{A}^{-1}\cdot\boldsymbol{T}\right) \right] \right|_{\alpha=0} \\
    & = \left.\cfrac{d}{d\alpha} \left[\alpha^3~\det(\boldsymbol{A})~\det\left(\cfrac{1}{\alpha}~\boldsymbol{\mathit{I}} + \boldsymbol{A}^{-1} \cdot \boldsymbol{T}\right)\right]\right|_{\alpha=0}.
  \end{align}</math>
Recall that we can expand the determinant of a tensor in the form of  a characteristic equation in terms of the invariants <math>I_1,I_2,I_3</math> using (note the sign of λ)
:<math>  \det(\lambda~\boldsymbol{\mathit{I}} + \boldsymbol{A}) =      \lambda^3 + I_1(\boldsymbol{A})~\lambda^2 + I_2(\boldsymbol{A})~\lambda + I_3(\boldsymbol{A}).</math>
 
Using this expansion we can write
:<math> \begin{align}
    \frac{\partial f}{\partial \boldsymbol{A}}:\boldsymbol{T}
    & = \left.\cfrac{d}{d\alpha} \left[\alpha^3~\det(\boldsymbol{A})~
      \left(\cfrac{1}{\alpha^3} + I_1(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\cfrac{1}{\alpha^2} +
      I_2(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\cfrac{1}{\alpha} + I_3(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})\right)
          \right] \right|_{\alpha=0} \\
    & = \left.\det(\boldsymbol{A})~\cfrac{d}{d\alpha} \left[
          1 + I_1(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha +
            I_2(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha^2 + I_3(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha^3
          \right] \right|_{\alpha=0} \\
    & = \left.\det(\boldsymbol{A})~\left[I_1(\boldsymbol{A}^{-1}\cdot\boldsymbol{T}) +
            2~I_2(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha + 3~I_3(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})~\alpha^2
          \right] \right|_{\alpha=0} \\
    & = \det(\boldsymbol{A})~I_1(\boldsymbol{A}^{-1}\cdot\boldsymbol{T}) ~.
  \end{align}
</math>
Recall that the invariant <math>I_1</math> is given by
:<math>  I_1(\boldsymbol{A}) = \text{tr}{\boldsymbol{A}}.</math>
Hence,
:<math>\frac{\partial f}{\partial \boldsymbol{A}}:\boldsymbol{T} = \det(\boldsymbol{A})~\text{tr}(\boldsymbol{A}^{-1}\cdot\boldsymbol{T})= \det(\boldsymbol{A})~[\boldsymbol{A}^{-1}]^T : \boldsymbol{T}.</math>
Invoking the arbitrariness of <math>\boldsymbol{T}</math> we then have
:<math>
  \frac{\partial f}{\partial \boldsymbol{A}} = \det(\boldsymbol{A})~[\boldsymbol{A}^{-1}]^T ~.
</math>
|}
 
==Derivatives of the invariants of a second-order tensor==
The principal invariants of a second order tensor are
:<math>
  \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}
</math>
The derivatives of these three invariants with respect to <math>\boldsymbol{A}</math> are
:<math>
  \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}
</math>
 
:{| class="toccolours collapsible collapsed" width="60%" style="text-align:left"
!Proof
|-
|From the derivative of the determinant we know that
:<math>
  \frac{\partial I_3}{\partial \boldsymbol{A}} = \det(\boldsymbol{A})~[\boldsymbol{A}^{-1}]^T ~.
</math>
 
For the derivatives of the other two invariants, let us go back to the characteristic equation
:<math>
    \det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A}) =
      \lambda^3 + I_1(\boldsymbol{A})~\lambda^2 + I_2(\boldsymbol{A})~\lambda + I_3(\boldsymbol{A}) ~.
</math>
Using the same approach as for the determinant of a tensor, we can show that
:<math>
    \frac{\partial }{\partial \boldsymbol{A}}\det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A}) = 
      \det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A})~[(\lambda~\boldsymbol{\mathit{1}}+\boldsymbol{A})^{-1}]^T ~.
</math>
Now the left hand side can be expanded as
:<math>
    \begin{align}
    \frac{\partial }{\partial \boldsymbol{A}}\det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A}) & = 
    \frac{\partial }{\partial \boldsymbol{A}}\left[
    \lambda^3 + I_1(\boldsymbol{A})~\lambda^2 + I_2(\boldsymbol{A})~\lambda + I_3(\boldsymbol{A}) \right] \\
    & =
    \frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda +
    \frac{\partial I_3}{\partial \boldsymbol{A}}~.
    \end{align}
</math>
Hence
:<math>
    \frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda +
    \frac{\partial I_3}{\partial \boldsymbol{A}} =
      \det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A})~[(\lambda~\boldsymbol{\mathit{1}}+\boldsymbol{A})^{-1}]^T
</math>
or,  
:<math>
    (\lambda~\boldsymbol{\mathit{1}}+\boldsymbol{A})^T\cdot\left[
    \frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda +
    \frac{\partial I_3}{\partial \boldsymbol{A}}\right] =
      \det(\lambda~\boldsymbol{\mathit{1}} + \boldsymbol{A})~\boldsymbol{\mathit{1}} ~.
</math>
Expanding the right hand side and separating terms on the left hand side gives
:<math>
    (\lambda~\boldsymbol{\mathit{1}} +\boldsymbol{A}^T)\cdot\left[
    \frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 + \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda +
    \frac{\partial I_3}{\partial \boldsymbol{A}}\right] =
      \left[\lambda^3 + I_1~\lambda^2 + I_2~\lambda + I_3\right]
      \boldsymbol{\mathit{1}}
</math>
or,
:<math>
  \begin{align}
    \left[\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^3 \right.&
    \left.+ \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda^2 +
    \frac{\partial I_3}{\partial \boldsymbol{A}}~\lambda\right]\boldsymbol{\mathit{1}} +
        \boldsymbol{A}^T\cdot\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 +
        \boldsymbol{A}^T\cdot\frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda +
        \boldsymbol{A}^T\cdot\frac{\partial I_3}{\partial \boldsymbol{A}} \\
    & =
      \left[\lambda^3 + I_1~\lambda^2 + I_2~\lambda + I_3\right]
      \boldsymbol{\mathit{1}} ~.
  \end{align}
</math>
If we define <math>I_0 := 1</math> and <math>I_4 := 0</math>, we can write the above as
:<math>
  \begin{align}
    \left[\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^3 \right.&
    \left.+ \frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda^2 +
    \frac{\partial I_3}{\partial \boldsymbol{A}}~\lambda + \frac{\partial I_4}{\partial \boldsymbol{A}}\right]\boldsymbol{\mathit{1}} +
    \boldsymbol{A}^T\cdot\frac{\partial I_0}{\partial \boldsymbol{A}}~\lambda^3 +
    \boldsymbol{A}^T\cdot\frac{\partial I_1}{\partial \boldsymbol{A}}~\lambda^2 +
    \boldsymbol{A}^T\cdot\frac{\partial I_2}{\partial \boldsymbol{A}}~\lambda +
    \boldsymbol{A}^T\cdot\frac{\partial I_3}{\partial \boldsymbol{A}} \\
    &=
      \left[I_0~\lambda^3 + I_1~\lambda^2 + I_2~\lambda + I_3\right]
      \boldsymbol{\mathit{1}} ~.
  \end{align}
</math>
Collecting terms containing various powers of λ, we get
:<math>
    \begin{align}
    \lambda^3&\left(I_0~\boldsymbol{\mathit{1}} - \frac{\partial I_1}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} -
                  \boldsymbol{A}^T\cdot\frac{\partial I_0}{\partial \boldsymbol{A}}\right) +
    \lambda^2\left(I_1~\boldsymbol{\mathit{1}} - \frac{\partial I_2}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} -
                  \boldsymbol{A}^T\cdot\frac{\partial I_1}{\partial \boldsymbol{A}}\right) + \\
    &\qquad \qquad\lambda\left(I_2~\boldsymbol{\mathit{1}} - \frac{\partial I_3}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} -
                  \boldsymbol{A}^T\cdot\frac{\partial I_2}{\partial \boldsymbol{A}}\right) +
    \left(I_3~\boldsymbol{\mathit{1}} - \frac{\partial I_4}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} -
                  \boldsymbol{A}^T\cdot\frac{\partial I_3}{\partial \boldsymbol{A}}\right)  = 0 ~.
    \end{align}
</math>
Then, invoking the arbitrariness of λ, we have
:<math>\begin{align}
I_0~\boldsymbol{\mathit{1}} - \frac{\partial I_1}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_0}{\partial \boldsymbol{A}} & = 0 \\
I_1~\boldsymbol{\mathit{1}} - \frac{\partial I_2}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - I_2~\boldsymbol{\mathit{1}} - \frac{\partial I_3}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_2}{\partial \boldsymbol{A}} & = 0 \\
I_3~\boldsymbol{\mathit{1}} - \frac{\partial I_4}{\partial \boldsymbol{A}}~\boldsymbol{\mathit{1}} - \boldsymbol{A}^T\cdot\frac{\partial I_3}{\partial \boldsymbol{A}} & = 0 ~.
    \end{align}</math>
This implies that
:<math>\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}} & = 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}</math>
|}
 
==Derivative of the second-order identity tensor==
Let <math>\boldsymbol{\mathit{1}}</math> be the second order identity tensor. Then the derivative of this tensor with respect to a second order tensor <math>\boldsymbol{A}</math> is given by
:<math>
  \frac{\partial \boldsymbol{\mathit{1}}}{\partial \boldsymbol{A}}:\boldsymbol{T} = \boldsymbol{\mathsf{0}}:\boldsymbol{T} = \boldsymbol{\mathit{0}}
</math>
This is because <math>\boldsymbol{\mathit{1}}</math> is independent of <math>\boldsymbol{A}</math>.
 
==Derivative of a second-order tensor with respect to itself==
Let <math>\boldsymbol{A}</math> be a second order tensor.  Then
:<math>
  \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}
</math>
Therefore,
:<math>
  \frac{\partial \boldsymbol{A}}{\partial \boldsymbol{A}} = \boldsymbol{\mathsf{I}}
</math>
Here <math>\boldsymbol{\mathsf{I}}</math> is the fourth order identity tensor. In index
notation with respect to an orthonormal basis
:<math>
  \boldsymbol{\mathsf{I}} = \delta_{ik}~\delta_{jl}~\mathbf{e}_i\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_l
</math>
 
This result implies that
:<math>
  \frac{\partial \boldsymbol{A}^T}{\partial \boldsymbol{A}}:\boldsymbol{T} = \boldsymbol{\mathsf{I}}^T:\boldsymbol{T} = \boldsymbol{T}^T
</math>
where
:<math>
  \boldsymbol{\mathsf{I}}^T = \delta_{jk}~\delta_{il}~\mathbf{e}_i\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_l
</math>
 
Therefore, if the tensor <math>\boldsymbol{A}</math> is symmetric, then the derivative is also symmetric and
we get
:<math>
  \frac{\partial \boldsymbol{A}}{\partial \boldsymbol{A}} = \boldsymbol{\mathsf{I}}^{(s)}
    = \frac{1}{2}~(\boldsymbol{\mathsf{I}} + \boldsymbol{\mathsf{I}}^T)
</math>
where the symmetric fourth order identity tensor is
:<math>
  \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
</math>
 
==Derivative of the inverse of a second-order tensor==
Let <math>\boldsymbol{A}</math> and <math>\boldsymbol{T}</math> be two second order tensors, then
:<math>
  \frac{\partial }{\partial \boldsymbol{A}} \left(\boldsymbol{A}^{-1}\right) : \boldsymbol{T} = - \boldsymbol{A}^{-1}\cdot\boldsymbol{T}\cdot\boldsymbol{A}^{-1}
</math>
In index notation with respect to an orthonormal basis
:<math>
  \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}
</math>
We also have
:<math>
\frac{\partial }{\partial \boldsymbol{A}} \left(\boldsymbol{A}^{-T}\right) : \boldsymbol{T} = - \boldsymbol{A}^{-T}\cdot\boldsymbol{T}^{T}\cdot\boldsymbol{A}^{-T}
</math>
In index notation
:<math>
  \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}
  </math>
If the tensor <math>\boldsymbol{A}</math> is symmetric then
:<math>
\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)
</math>
 
:{| class="toccolours collapsible collapsed" width="60%" style="text-align:left"
!Proof
|-
|Recall that
:<math>
  \frac{\partial \boldsymbol{\mathit{1}}}{\partial \boldsymbol{A}}:\boldsymbol{T} = \boldsymbol{\mathit{0}}
</math>
Since <math>\boldsymbol{A}^{-1}\cdot\boldsymbol{A} = \boldsymbol{\mathit{1}}</math>, we can write
:<math>
  \frac{\partial }{\partial \boldsymbol{A}}(\boldsymbol{A}^{-1}\cdot\boldsymbol{A}):\boldsymbol{T} = \boldsymbol{\mathit{0}}
</math>
Using the product rule for second order tensors
:<math>
  \frac{\partial }{\partial \boldsymbol{S}}[\boldsymbol{F}_1(\boldsymbol{S})\cdot\boldsymbol{F}_2(\boldsymbol{S})]:\boldsymbol{T} =
  \left(\frac{\partial \boldsymbol{F}_1}{\partial \boldsymbol{S}}:\boldsymbol{T}\right)\cdot\boldsymbol{F}_2 +
  \boldsymbol{F}_1\cdot\left(\frac{\partial \boldsymbol{F}_2}{\partial \boldsymbol{S}}:\boldsymbol{T}\right)
</math>
we get
:<math>
  \frac{\partial }{\partial \boldsymbol{A}}(\boldsymbol{A}^{-1}\cdot\boldsymbol{A}):\boldsymbol{T} =
  \left(\frac{\partial \boldsymbol{A}^{-1}}{\partial \boldsymbol{A}}:\boldsymbol{T}\right)\cdot\boldsymbol{A} +
  \boldsymbol{A}^{-1}\cdot\left(\frac{\partial \boldsymbol{A}}{\partial \boldsymbol{A}}:\boldsymbol{T}\right)
  = \boldsymbol{\mathit{0}}
</math>
or,
:<math>
  \left(\frac{\partial \boldsymbol{A}^{-1}}{\partial \boldsymbol{A}}:\boldsymbol{T}\right)\cdot\boldsymbol{A} = -
  \boldsymbol{A}^{-1}\cdot\boldsymbol{T}
</math>
Therefore,
:<math>
  \frac{\partial }{\partial \boldsymbol{A}} \left(\boldsymbol{A}^{-1}\right) : \boldsymbol{T} = - \boldsymbol{A}^{-1}\cdot\boldsymbol{T}\cdot\boldsymbol{A}^{-1}
</math>
|}
 
== Integration by parts ==
[[File:StressMeasures.png|thumb|400px|Domain <math>\Omega</math>, its boundary <math>\Gamma</math> and the outward unit normal <math>\mathbf{n}</math>]]
Another important operation related to tensor derivatives in continuum mechanics is integration by parts. The formula for integration by parts can be written as
:<math>
  \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
</math>
where <math>\boldsymbol{F}</math> and <math>\boldsymbol{G}</math> are differentiable tensor fields of arbitrary order, <math>\mathbf{n}</math> is the unit outward normal to the domain over which the tensor fields are defined, <math>\otimes</math> represents a generalized tensor product operator, and <math>\boldsymbol{\nabla}</math> is a generalized gradient operator.  When <math>\boldsymbol{F}</math> is equal to the identity tensor, we get the [[divergence theorem]]
:<math>
  \int_{\Omega}\boldsymbol{\nabla}\boldsymbol{G}\,{\rm d}\Omega = \int_{\Gamma} \mathbf{n}\otimes\boldsymbol{G}\,{\rm d}\Gamma \,.
</math>
We can express the formula for integration by parts in Cartesian index notation as
:<math>
  \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 \,.
</math>
For the special case where the tensor product operation is a contraction of one index and the gradient operation is a divergence, and both <math>\boldsymbol{F}</math> and <math>\boldsymbol{G}</math> are second order tensors, we have
:<math>
  \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 \,.
</math>
In index notation,
:<math>
  \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 \,.
</math>
 
== References ==
<references />
 
== See also ==
* [[Tensor derivative]]
* [[Directional derivative]]
* [[Curvilinear coordinates]]
* [[Continuum mechanics]]
 
[[Category:Solid mechanics]]
[[Category:Mechanics]]

Revision as of 17:57, 1 December 2013

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

fvu=Df(v)[u]=[ddαf(v+αu)]α=0

for all vectors u.

Properties:

1) If f(v)=f1(v)+f2(v) then fvu=(f1v+f2v)u

2) If f(v)=f1(v)f2(v) then fvu=(f1vu)f2(v)+f1(v)(f2vu)

3) If f(v)=f1(f2(v)) then fvu=f1f2f2vu

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

fvu=Df(v)[u]=[ddαf(v+αu)]α=0

for all vectors u.

Properties:
1) If f(v)=f1(v)+f2(v) then fvu=(f1v+f2v)u
2) If f(v)=f1(v)×f2(v) then fvu=(f1vu)×f2(v)+f1(v)×(f2vu)
3) If f(v)=f1(f2(v)) then fvu=f1f2(f2vu)

Derivatives of scalar valued functions of second-order tensors

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

fS:T=Df(S)[T]=[ddαf(S+αT)]α=0

for all second order tensors T.

Properties:
1) If f(S)=f1(S)+f2(S) then fS:T=(f1S+f2S):T
2) If f(S)=f1(S)f2(S) then fS:T=(f1S:T)f2(S)+f1(S)(f2S:T)
3) If f(S)=f1(f2(S)) then fS:T=f1f2(f2S:T)

Derivatives of tensor valued functions of second-order tensors

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

FS:T=DF(S)[T]=[ddαF(S+αT)]α=0

for all second order tensors T.

Properties:
1) If F(S)=F1(S)+F2(S) then FS:T=(F1S+F2S):T
2) If F(S)=F1(S)F2(S) then FS:T=(F1S:T)F2(S)+F1(S)(F2S:T)
3) If F(S)=F1(F2(S)) then FS:T=F1F2:(F2S:T)
4) If f(S)=f1(F2(S)) then fS:T=f1F2:(F2S:T)

Gradient of a tensor field

The gradient, T, of a tensor field T(x) in the direction of an arbitrary constant vector c is defined as:

Tc=ddαT(x+αc)|α=0

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

Cartesian coordinates

Template:Einstein summation convention

If e1,e2,e3 are the basis vectors in a Cartesian coordinate system, with coordinates of points denoted by (x1,x2,x3), then the gradient of the tensor field T is given by

T=Txiei

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

ϕ=ϕxieiv=(vjej)xiei=vjxiejeiS=(Sjkejek)xiei=Sjkxiejekei

Curvilinear coordinates

Mining Engineer (Excluding Oil ) Truman from Alma, loves to spend time knotting, largest property developers in singapore developers in singapore and stamp collecting. Recently had a family visit to Urnes Stave Church. Template:Einstein summation convention

If g1,g2,g3 are the contravariant basis vectors in a curvilinear coordinate system, with coordinates of points denoted by (ξ1,ξ2,ξ3), then the gradient of the tensor field T is given by (see [3] for a proof.)

T=Tξigi

From this definition we have the following relations for the gradients of a scalar field ϕ, a vector field v, and a second-order tensor field S.

ϕ=ϕξigiv=(vjgj)ξigi=(vjξi+vkΓikj)gjgi=(vjξivkΓijk)gjgiS=(Sjkgjgk)ξigi=(SjkξiSlkΓijlSjlΓikl)gjgkgi

where the Christoffel symbol Γijk is defined using

Γijkgk=giξjΓijk=giξjgk=gigkξj

Cylindrical polar coordinates

In cylindrical coordinates, the gradient is given by

ϕ=ϕrer+1rϕθeθ+ϕzezv=vrrerer+1r(vrθvθ)ereθ+vrzerez+vθreθer+1r(vθθ+vr)eθeθ+vθzeθez+vzrezer+1rvzθezeθ+vzzezezS=Srrrererer+1r[Srrθ(Sθr+Srθ)]erereθ+Srrzererez+Srθrereθer+1r[Srθθ+(SrrSθθ)]ereθeθ+Srθzereθez+Srzrerezer+1r[SrzθSθz]erezeθ+Srzzerezez+Sθrreθerer+1r[Sθrθ+(SrrSθθ)]eθereθ+Sθrzeθerez+Sθθreθeθer+1r[Sθθθ+(Srθ+Sθr)]eθeθeθ+Sθθzeθeθez+Sθzreθezer+1r[Sθzθ+Srz]eθezeθ+Sθzzeθezez+Szrrezerer+1r[SzrθSzθ]ezereθ+Szrzezerez+Szθrezeθer+1r[Szθθ+Szr]ezeθeθ+Szθzezeθez+Szzrezezer+1rSzzθezezeθ+Szzzezezez

Divergence of a tensor field

The divergence of a tensor field T(x) is defined using the recursive relation

(T)c=(cT);v=tr(v)

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

Cartesian coordinates

Template:Einstein summation convention In a Cartesian coordinate system we have the following relations for a vector field v and a second-order tensor field S.

v=vixiS=Skixiek

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

S÷S=ST.

Curvilinear coordinates

Mining Engineer (Excluding Oil ) Truman from Alma, loves to spend time knotting, largest property developers in singapore developers in singapore and stamp collecting. Recently had a family visit to Urnes Stave Church. Template:Einstein summation convention In curvilinear coordinates, the divergences of a vector field v and a second-order tensor field S are

v=(viξi+vkΓiki)S=(SikξiSlkΓiilSilΓikl)gk

Cylindrical polar coordinates

In cylindrical polar coordinates

v=vrr+1r(vθθ+vr)+vzzS=Srrrer+Srθreθ+Srzrez+1r[Sθrθ+(SrrSθθ)]er+1r[Sθθθ+(Srθ+Sθr)]eθ+1r[Sθzθ+Srz]ez+Szrzer+Szθzeθ+Szzzez

Curl of a tensor field

The curl of an order-n > 1 tensor field T(x) is also defined using the recursive relation

(×T)c=×(cT);(×v)c=(v×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

v×c=eijkvjckei

where eijk is the permutation symbol. Then,

(v×c)=eijkvj,ick=(eijkvj,iek)c=(×v)c

Therefore

×v=eijkvj,iek

Curl of a second-order tensor field

For a second-order tensor S

cS=cmSmjej

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

×(cS)=eijkcmSmj,iek=(eijkSmj,iekem)c=(×S)c

Therefore, we have

×S=eijkSmj,iekem

Identities involving the curl of a tensor field

The most commonly used identity involving the curl of a tensor field, T, is

×(T)=0

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

×S=0Smi,jSmj,i=0

Derivative of the determinant of a second-order tensor

The derivative of the determinant of a second order tensor A is given by

Adet(A)=det(A)[A1]T.

In an orthonormal basis, the components of 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

I1(A)=trAI2(A)=12[(trA)2trA2]I3(A)=det(A)

The derivatives of these three invariants with respect to A are

I1A=1I2A=I11ATI3A=det(A)[A1]T=I21AT(I11AT)=(A2I1A+I21)T

Derivative of the second-order identity tensor

Let 1 be the second order identity tensor. Then the derivative of this tensor with respect to a second order tensor A is given by

1A:T=0:T=0

This is because 1 is independent of A.

Derivative of a second-order tensor with respect to itself

Let A be a second order tensor. Then

AA:T=[α(A+αT)]α=0=T=I:T

Therefore,

AA=I

Here I is the fourth order identity tensor. In index notation with respect to an orthonormal basis

I=δikδjleiejekel

This result implies that

ATA:T=IT:T=TT

where

IT=δjkδileiejekel

Therefore, if the tensor A is symmetric, then the derivative is also symmetric and we get

AA=I(s)=12(I+IT)

where the symmetric fourth order identity tensor is

I(s)=12(δikδjl+δilδjk)eiejekel

Derivative of the inverse of a second-order tensor

Let A and T be two second order tensors, then

A(A1):T=A1TA1

In index notation with respect to an orthonormal basis

Aij1AklTkl=Aik1TklAlj1Aij1Akl=Aik1Alj1

We also have

A(AT):T=ATTTAT

In index notation

Aji1AklTkl=Ajk1TlkAli1Aji1Akl=Ali1Ajk1

If the tensor A is symmetric then

Aij1Akl=12(Aik1Ajl1+Ail1Ajk1)

Integration by parts

File:StressMeasures.png
Domain Ω, its boundary Γ and the outward unit normal 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

ΩFGdΩ=Γn(FG)dΓΩGFdΩ

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

ΩGdΩ=ΓnGdΓ.

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

ΩFijk....Glmn...,pdΩ=ΓnpFijk...Glmn...dΓΩGlmn...Fijk...,pdΩ.

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

ΩF(G)dΩ=Γn(GFT)dΓΩ(F):GTdΩ.

In index notation,

ΩFijGpj,pdΩ=ΓnpFijGpjdΓΩGpjFij,pdΩ.

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

See also