Tensor derivative (continuum mechanics)

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) is the vector defined through its dot product with any vector u being

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

for all vectors u. The above dot product yields a scalar, and if u is a unit vector gives the directional derivative of f at v, in the u direction.

Properties:

1. If ${\displaystyle f(\mathbf {v} )=f_{1}(\mathbf {v} )+f_{2}(\mathbf {v} )}$ then ${\displaystyle {\frac {\partial f}{\partial \mathbf {v} }}\cdot \mathbf {u} =\left({\frac {\partial f_{1}}{\partial \mathbf {v} }}+{\frac {\partial f_{2}}{\partial \mathbf {v} }}\right)\cdot \mathbf {u} }$
2. If ${\displaystyle f(\mathbf {v} )=f_{1}(\mathbf {v} )~f_{2}(\mathbf {v} )}$ then ${\displaystyle {\frac {\partial f}{\partial \mathbf {v} }}\cdot \mathbf {u} =\left({\frac {\partial f_{1}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right)~f_{2}(\mathbf {v} )+f_{1}(\mathbf {v} )~\left({\frac {\partial f_{2}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right)}$
3. If ${\displaystyle f(\mathbf {v} )=f_{1}(f_{2}(\mathbf {v} ))}$ then ${\displaystyle {\frac {\partial f}{\partial \mathbf {v} }}\cdot \mathbf {u} ={\frac {\partial f_{1}}{\partial f_{2}}}~{\frac {\partial f_{2}}{\partial \mathbf {v} }}\cdot \mathbf {u} }$

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) is the second order tensor defined through its dot product with any vector u being

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

for all vectors u. The above dot product yields a vector, and if u is a unit vector gives the direction derivative of f at v, in the directional u.

Properties:

1. If ${\displaystyle \mathbf {f} (\mathbf {v} )=\mathbf {f} _{1}(\mathbf {v} )+\mathbf {f} _{2}(\mathbf {v} )}$ then ${\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} =\left({\frac {\partial \mathbf {f} _{1}}{\partial \mathbf {v} }}+{\frac {\partial \mathbf {f} _{2}}{\partial \mathbf {v} }}\right)\cdot \mathbf {u} }$
2. If ${\displaystyle \mathbf {f} (\mathbf {v} )=\mathbf {f} _{1}(\mathbf {v} )\times \mathbf {f} _{2}(\mathbf {v} )}$ then ${\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} =\left({\frac {\partial \mathbf {f} _{1}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right)\times \mathbf {f} _{2}(\mathbf {v} )+\mathbf {f} _{1}(\mathbf {v} )\times \left({\frac {\partial \mathbf {f} _{2}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right)}$
3. If ${\displaystyle \mathbf {f} (\mathbf {v} )=\mathbf {f} _{1}(\mathbf {f} _{2}(\mathbf {v} ))}$ then ${\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} ={\frac {\partial \mathbf {f} _{1}}{\partial \mathbf {f} _{2}}}\cdot \left({\frac {\partial \mathbf {f} _{2}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right)}$

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

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

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

Properties:

1. If ${\displaystyle f({\boldsymbol {S}})=f_{1}({\boldsymbol {S}})+f_{2}({\boldsymbol {S}})}$ then ${\displaystyle {\frac {\partial f}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}=\left({\frac {\partial f_{1}}{\partial {\boldsymbol {S}}}}+{\frac {\partial f_{2}}{\partial {\boldsymbol {S}}}}\right):{\boldsymbol {T}}}$
2. If ${\displaystyle f({\boldsymbol {S}})=f_{1}({\boldsymbol {S}})~f_{2}({\boldsymbol {S}})}$ then ${\displaystyle {\frac {\partial f}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}=\left({\frac {\partial f_{1}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)~f_{2}({\boldsymbol {S}})+f_{1}({\boldsymbol {S}})~\left({\frac {\partial f_{2}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)}$
3. If ${\displaystyle f({\boldsymbol {S}})=f_{1}(f_{2}({\boldsymbol {S}}))}$ then ${\displaystyle {\frac {\partial f}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}={\frac {\partial f_{1}}{\partial f_{2}}}~\left({\frac {\partial f_{2}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)}$

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

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

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

Properties:

1. If ${\displaystyle {\boldsymbol {F}}({\boldsymbol {S}})={\boldsymbol {F}}_{1}({\boldsymbol {S}})+{\boldsymbol {F}}_{2}({\boldsymbol {S}})}$ then ${\displaystyle {\frac {\partial {\boldsymbol {F}}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}=\left({\frac {\partial {\boldsymbol {F}}_{1}}{\partial {\boldsymbol {S}}}}+{\frac {\partial {\boldsymbol {F}}_{2}}{\partial {\boldsymbol {S}}}}\right):{\boldsymbol {T}}}$
2. If ${\displaystyle {\boldsymbol {F}}({\boldsymbol {S}})={\boldsymbol {F}}_{1}({\boldsymbol {S}})\cdot {\boldsymbol {F}}_{2}({\boldsymbol {S}})}$ then ${\displaystyle {\frac {\partial {\boldsymbol {F}}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}=\left({\frac {\partial {\boldsymbol {F}}_{1}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)\cdot {\boldsymbol {F}}_{2}({\boldsymbol {S}})+{\boldsymbol {F}}_{1}({\boldsymbol {S}})\cdot \left({\frac {\partial {\boldsymbol {F}}_{2}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)}$
3. If ${\displaystyle {\boldsymbol {F}}({\boldsymbol {S}})={\boldsymbol {F}}_{1}({\boldsymbol {F}}_{2}({\boldsymbol {S}}))}$ then ${\displaystyle {\frac {\partial {\boldsymbol {F}}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}={\frac {\partial {\boldsymbol {F}}_{1}}{\partial {\boldsymbol {F}}_{2}}}:\left({\frac {\partial {\boldsymbol {F}}_{2}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)}$
4. If ${\displaystyle f({\boldsymbol {S}})=f_{1}({\boldsymbol {F}}_{2}({\boldsymbol {S}}))}$ then ${\displaystyle {\frac {\partial f}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}={\frac {\partial f_{1}}{\partial {\boldsymbol {F}}_{2}}}:\left({\frac {\partial {\boldsymbol {F}}_{2}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)}$

If ${\displaystyle \mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}}$ are the basis vectors in a Cartesian coordinate system, with coordinates of points denoted by (${\displaystyle x_{1},x_{2},x_{3}}$), then the gradient of the tensor field ${\displaystyle {\boldsymbol {T}}}$ is given by

$${\displaystyle {\boldsymbol {\nabla }}{\boldsymbol {T}}={\cfrac {\partial {\boldsymbol {T}}}{\partial x_{i}}}\otimes \mathbf {e} _{i}}$$

Proof

The vectors x and c can be written as ${\displaystyle \mathbf {x} =x_{i}~\mathbf {e} _{i}}$ and ${\displaystyle \mathbf {c} =c_{i}~\mathbf {e} _{i}}$. Let y := x + αc. In that case the gradient is given by

$${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}{\boldsymbol {T}}\cdot \mathbf {c} &=\left.{\cfrac {d}{d\alpha }}~{\boldsymbol {T}}(x_{1}+\alpha c_{1},x_{2}+\alpha c_{2},x_{3}+\alpha c_{3})\right|_{\alpha =0}\equiv \left.{\cfrac {d}{d\alpha }}~{\boldsymbol {T}}(y_{1},y_{2},y_{3})\right|_{\alpha =0}\\&=\left[{\cfrac {\partial {\boldsymbol {T}}}{\partial y_{1}}}~{\cfrac {\partial y_{1}}{\partial \alpha }}+{\cfrac {\partial {\boldsymbol {T}}}{\partial y_{2}}}~{\cfrac {\partial y_{2}}{\partial \alpha }}+{\cfrac {\partial {\boldsymbol {T}}}{\partial y_{3}}}~{\cfrac {\partial y_{3}}{\partial \alpha }}\right]_{\alpha =0}=\left[{\cfrac {\partial {\boldsymbol {T}}}{\partial y_{1}}}~c_{1}+{\cfrac {\partial {\boldsymbol {T}}}{\partial y_{2}}}~c_{2}+{\cfrac {\partial {\boldsymbol {T}}}{\partial y_{3}}}~c_{3}\right]_{\alpha =0}\\&={\cfrac {\partial {\boldsymbol {T}}}{\partial x_{1}}}~c_{1}+{\cfrac {\partial {\boldsymbol {T}}}{\partial x_{2}}}~c_{2}+{\cfrac {\partial {\boldsymbol {T}}}{\partial x_{3}}}~c_{3}\equiv {\cfrac {\partial {\boldsymbol {T}}}{\partial x_{i}}}~c_{i}={\cfrac {\partial {\boldsymbol {T}}}{\partial x_{i}}}~(\mathbf {e} _{i}\cdot \mathbf {c} )=\left[{\cfrac {\partial {\boldsymbol {T}}}{\partial x_{i}}}\otimes \mathbf {e} _{i}\right]\cdot \mathbf {c} \qquad \square \end{aligned}}}$$

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

$${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\phi &={\cfrac {\partial \phi }{\partial x_{i}}}~\mathbf {e} _{i}=\phi _{,i}~\mathbf {e} _{i}\\{\boldsymbol {\nabla }}\mathbf {v} &={\cfrac {\partial (v_{j}\mathbf {e} _{j})}{\partial x_{i}}}\otimes \mathbf {e} _{i}={\cfrac {\partial v_{j}}{\partial x_{i}}}~\mathbf {e} _{j}\otimes \mathbf {e} _{i}=v_{j,i}~\mathbf {e} _{j}\otimes \mathbf {e} _{i}\\{\boldsymbol {\nabla }}{\boldsymbol {S}}&={\cfrac {\partial (S_{jk}\mathbf {e} _{j}\otimes \mathbf {e} _{k})}{\partial x_{i}}}\otimes \mathbf {e} _{i}={\cfrac {\partial S_{jk}}{\partial x_{i}}}~\mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{i}=S_{jk,i}~\mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{i}\end{aligned}}}$$

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

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

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

$${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\phi &={\frac {\partial \phi }{\partial \xi ^{i}}}~\mathbf {g} ^{i}\\{\boldsymbol {\nabla }}\mathbf {v} &={\frac {\partial \left(v^{j}\mathbf {g} _{j}\right)}{\partial \xi ^{i}}}\otimes \mathbf {g} ^{i}=\left({\frac {\partial v^{j}}{\partial \xi ^{i}}}+v^{k}~\Gamma _{ik}^{j}\right)~\mathbf {g} _{j}\otimes \mathbf {g} ^{i}=\left({\frac {\partial v_{j}}{\partial \xi ^{i}}}-v_{k}~\Gamma _{ij}^{k}\right)~\mathbf {g} ^{j}\otimes \mathbf {g} ^{i}\\{\boldsymbol {\nabla }}{\boldsymbol {S}}&={\frac {\partial \left(S_{jk}~\mathbf {g} ^{j}\otimes \mathbf {g} ^{k}\right)}{\partial \xi ^{i}}}\otimes \mathbf {g} ^{i}=\left({\frac {\partial S_{jk}}{\partial \xi _{i}}}-S_{lk}~\Gamma _{ij}^{l}-S_{jl}~\Gamma _{ik}^{l}\right)~\mathbf {g} ^{j}\otimes \mathbf {g} ^{k}\otimes \mathbf {g} ^{i}\end{aligned}}}$$

where the Christoffel symbol ${\displaystyle \Gamma _{ij}^{k}}$ is defined using

$${\displaystyle \Gamma _{ij}^{k}~\mathbf {g} _{k}={\frac {\partial \mathbf {g} _{i}}{\partial \xi ^{j}}}\quad \implies \quad \Gamma _{ij}^{k}={\frac {\partial \mathbf {g} _{i}}{\partial \xi ^{j}}}\cdot \mathbf {g} ^{k}=-\mathbf {g} _{i}\cdot {\frac {\partial \mathbf {g} ^{k}}{\partial \xi ^{j}}}}$$

In cylindrical coordinates, the gradient is given by

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

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

$${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\cdot \mathbf {v} &={\frac {\partial v_{i}}{\partial x_{i}}}=v_{i,i}\\{\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}&={\frac {\partial S_{ik}}{\partial x_{i}}}~\mathbf {e} _{k}=S_{ik,i}~\mathbf {e} _{k}\end{aligned}}}$$

where tensor index notation for partial derivatives is used in the rightmost expressions. Note that

$${\displaystyle {\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}\neq {\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}^{\textsf {T}}.}$$

For a symmetric second-order tensor, the divergence is also often written as[4]

$${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}&={\cfrac {\partial S_{ki}}{\partial x_{i}}}~\mathbf {e} _{k}=S_{ki,i}~\mathbf {e} _{k}\end{aligned}}}$$

The above expression is sometimes used as the definition of ${\displaystyle {\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}}$ in Cartesian component form (often also written as ${\displaystyle \operatorname {div} {\boldsymbol {S}}}$). Note that such a definition is not consistent with the rest of this article (see the section on curvilinear co-ordinates).

The difference stems from whether the differentiation is performed with respect to the rows or columns of ${\displaystyle {\boldsymbol {S}}}$, and is conventional. This is demonstrated by an example. In a Cartesian coordinate system the second order tensor (matrix) ${\displaystyle \mathbf {S} }$ is the gradient of a vector function ${\displaystyle \mathbf {v} }$.

$${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\cdot \left({\boldsymbol {\nabla }}\mathbf {v} \right)&={\boldsymbol {\nabla }}\cdot \left(v_{i,j}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\right)=v_{i,ji}~\mathbf {e} _{i}\cdot \mathbf {e} _{i}\otimes \mathbf {e} _{j}=\left({\boldsymbol {\nabla }}\cdot \mathbf {v} \right)_{,j}~\mathbf {e} _{j}={\boldsymbol {\nabla }}\left({\boldsymbol {\nabla }}\cdot \mathbf {v} \right)\\{\boldsymbol {\nabla }}\cdot \left[\left({\boldsymbol {\nabla }}\mathbf {v} \right)^{\textsf {T}}\right]&={\boldsymbol {\nabla }}\cdot \left(v_{j,i}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\right)=v_{j,ii}~\mathbf {e} _{i}\cdot \mathbf {e} _{i}\otimes \mathbf {e} _{j}={\boldsymbol {\nabla }}^{2}v_{j}~\mathbf {e} _{j}={\boldsymbol {\nabla }}^{2}\mathbf {v} \end{aligned}}}$$

The last equation is equivalent to the alternative definition / interpretation[4]

$${\displaystyle {\begin{aligned}\left({\boldsymbol {\nabla }}\cdot \right)_{\text{alt}}\left({\boldsymbol {\nabla }}\mathbf {v} \right)=\left({\boldsymbol {\nabla }}\cdot \right)_{\text{alt}}\left(v_{i,j}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\right)=v_{i,jj}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\cdot \mathbf {e} _{j}={\boldsymbol {\nabla }}^{2}v_{i}~\mathbf {e} _{i}={\boldsymbol {\nabla }}^{2}\mathbf {v} \end{aligned}}}$$

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

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

More generally,

$${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}&=\left[{\cfrac {\partial S_{ij}}{\partial q^{k}}}-\Gamma _{ki}^{l}~S_{lj}-\Gamma _{kj}^{l}~S_{il}\right]~g^{ik}~\mathbf {b} ^{j}\\[8pt]&=\left[{\cfrac {\partial S^{ij}}{\partial q^{i}}}+\Gamma _{il}^{i}~S^{lj}+\Gamma _{il}^{j}~S^{il}\right]~\mathbf {b} _{j}\\[8pt]&=\left[{\cfrac {\partial S_{~j}^{i}}{\partial q^{i}}}+\Gamma _{il}^{i}~S_{~j}^{l}-\Gamma _{ij}^{l}~S_{~l}^{i}\right]~\mathbf {b} ^{j}\\[8pt]&=\left[{\cfrac {\partial S_{i}^{~j}}{\partial q^{k}}}-\Gamma _{ik}^{l}~S_{l}^{~j}+\Gamma _{kl}^{j}~S_{i}^{~l}\right]~g^{ik}~\mathbf {b} _{j}\end{aligned}}}$$

In cylindrical polar coordinates

$${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\cdot \mathbf {v} =\quad &{\frac {\partial v_{r}}{\partial r}}+{\frac {1}{r}}\left({\frac {\partial v_{\theta }}{\partial \theta }}+v_{r}\right)+{\frac {\partial v_{z}}{\partial z}}\\{\boldsymbol {\nabla }}\cdot {\boldsymbol {S}}=\quad &{\frac {\partial S_{rr}}{\partial r}}~\mathbf {e} _{r}+{\frac {\partial S_{r\theta }}{\partial r}}~\mathbf {e} _{\theta }+{\frac {\partial S_{rz}}{\partial r}}~\mathbf {e} _{z}\\{}+{}&{\frac {1}{r}}\left[{\frac {\partial S_{\theta r}}{\partial \theta }}+(S_{rr}-S_{\theta \theta })\right]~\mathbf {e} _{r}+{\frac {1}{r}}\left[{\frac {\partial S_{\theta \theta }}{\partial \theta }}+(S_{r\theta }+S_{\theta r})\right]~\mathbf {e} _{\theta }+{\frac {1}{r}}\left[{\frac {\partial S_{\theta z}}{\partial \theta }}+S_{rz}\right]~\mathbf {e} _{z}\\{}+{}&{\frac {\partial S_{zr}}{\partial z}}~\mathbf {e} _{r}+{\frac {\partial S_{z\theta }}{\partial z}}~\mathbf {e} _{\theta }+{\frac {\partial S_{zz}}{\partial z}}~\mathbf {e} _{z}\end{aligned}}}$$

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

$${\displaystyle \mathbf {v} \times \mathbf {c} =\varepsilon _{ijk}~v_{j}~c_{k}~\mathbf {e} _{i}}$$

where ${\displaystyle \varepsilon _{ijk}}$ is the permutation symbol, otherwise known as the Levi-Civita symbol. Then,

$${\displaystyle {\boldsymbol {\nabla }}\cdot (\mathbf {v} \times \mathbf {c} )=\varepsilon _{ijk}~v_{j,i}~c_{k}=(\varepsilon _{ijk}~v_{j,i}~\mathbf {e} _{k})\cdot \mathbf {c} =({\boldsymbol {\nabla }}\times \mathbf {v} )\cdot \mathbf {c} }$$

Therefore,

$${\displaystyle {\boldsymbol {\nabla }}\times \mathbf {v} =\varepsilon _{ijk}~v_{j,i}~\mathbf {e} _{k}}$$

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

$${\displaystyle \mathbf {c} \cdot {\boldsymbol {S}}=c_{m}~S_{mj}~\mathbf {e} _{j}}$$

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

$${\displaystyle {\boldsymbol {\nabla }}\times (\mathbf {c} \cdot {\boldsymbol {S}})=\varepsilon _{ijk}~c_{m}~S_{mj,i}~\mathbf {e} _{k}=(\varepsilon _{ijk}~S_{mj,i}~\mathbf {e} _{k}\otimes \mathbf {e} _{m})\cdot \mathbf {c} =({\boldsymbol {\nabla }}\times {\boldsymbol {S}})\cdot \mathbf {c} }$$

Therefore, we have

$${\displaystyle {\boldsymbol {\nabla }}\times {\boldsymbol {S}}=\varepsilon _{ijk}~S_{mj,i}~\mathbf {e} _{k}\otimes \mathbf {e} _{m}}$$

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

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

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

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