Elasticity/Tensors

Let ${\displaystyle {\boldsymbol {A}}\,}$ and ${\displaystyle {\boldsymbol {B}}\,}$ be two tensors. Then the sum ${\displaystyle ({\boldsymbol {A}}+{\boldsymbol {B}})\,}$ is another tensor ${\displaystyle {\boldsymbol {C}}\,}$ defined by

${\displaystyle {\boldsymbol {C}}={\boldsymbol {A}}+{\boldsymbol {B}}\implies {\boldsymbol {C}}\bullet \mathbf {v} =({\boldsymbol {A}}+{\boldsymbol {B}})\bullet \mathbf {v} ={\boldsymbol {A}}\bullet \mathbf {v} +{\boldsymbol {B}}\bullet \mathbf {v} ~.}$

Let ${\displaystyle {\boldsymbol {A}}\,}$ be a tensor and let ${\displaystyle \lambda \,}$ be a scalar. Then the product ${\displaystyle {\boldsymbol {C}}=\lambda {\boldsymbol {A}}\,}$ is a tensor defined by

${\displaystyle {\boldsymbol {C}}=\lambda {\boldsymbol {A}}\implies {\boldsymbol {C}}\bullet \mathbf {v} =(\lambda {\boldsymbol {A}})\bullet \mathbf {v} =\lambda ({\boldsymbol {A}}\bullet \mathbf {v} )~.}$

The zero tensor ${\displaystyle {\boldsymbol {\mathit {0}}}\,}$ is the tensor which maps every vector ${\displaystyle \mathbf {v} \,}$ into the zero vector.

${\displaystyle {\boldsymbol {\mathit {0}}}\bullet \mathbf {v} =\mathbf {0} ~.}$

The identity tensor ${\displaystyle {\boldsymbol {\mathit {I}}}\,}$ takes every vector ${\displaystyle \mathbf {v} \,}$ into itself.

${\displaystyle {\boldsymbol {\mathit {I}}}\bullet \mathbf {v} =\mathbf {v} ~.}$

The identity tensor is also often written as ${\displaystyle {\boldsymbol {\mathit {1}}}\,}$.

Let ${\displaystyle {\boldsymbol {A}}\,}$ and ${\displaystyle {\boldsymbol {B}}\,}$ be two tensors. Then the product ${\displaystyle {\boldsymbol {C}}={\boldsymbol {A}}\bullet {\boldsymbol {B}}}$ is the tensor that is defined by

${\displaystyle {\boldsymbol {C}}={\boldsymbol {A}}\bullet {\boldsymbol {B}}\implies {\boldsymbol {C}}\bullet \mathbf {v} =({\boldsymbol {A}}\bullet {\boldsymbol {B}})\bullet {\mathbf {v} }={\boldsymbol {A}}\bullet ({\boldsymbol {B}}\bullet {\mathbf {v} })~.}$

In general ${\displaystyle {\boldsymbol {A}}\bullet {\boldsymbol {B}}\neq {\boldsymbol {B}}\bullet {\boldsymbol {A}}}$.

The transpose of a tensor ${\displaystyle {\boldsymbol {A}}\,}$ is the unique tensor ${\displaystyle {\boldsymbol {A}}^{T}\,}$ defined by

${\displaystyle ({\boldsymbol {A}}\bullet \mathbf {u} )\bullet \mathbf {v} =\mathbf {u} \bullet ({\boldsymbol {A}}^{T}\bullet \mathbf {v} )~.}$

The following identities follow from the above definition:

${\displaystyle {\begin{aligned}({\boldsymbol {A}}+{\boldsymbol {B}})^{T}&={\boldsymbol {A}}^{T}+{\boldsymbol {B}}^{T}~,\\({\boldsymbol {A}}\bullet {\boldsymbol {B}})^{T}&={\boldsymbol {B}}^{T}\bullet {\boldsymbol {A}}^{T}~,\\({\boldsymbol {A}}^{T})^{T}&={\boldsymbol {A}}~.\end{aligned}}}$

A tensor ${\displaystyle {\boldsymbol {A}}\,}$ is symmetric if

${\displaystyle {\boldsymbol {A}}={\boldsymbol {A}}^{T}~.}$

A tensor ${\displaystyle {\boldsymbol {A}}\,}$ is skew if

${\displaystyle {\boldsymbol {A}}=-{\boldsymbol {A}}^{T}~.}$

Every tensor ${\displaystyle {\boldsymbol {A}}\,}$ can be expressed uniquely as the sum of a symmetric tensor ${\displaystyle {\boldsymbol {E}}\,}$ (the symmetric part of ${\displaystyle {\boldsymbol {A}}\,}$) and a skew tensor ${\displaystyle {\boldsymbol {W}}\,}$ (the skew part of ${\displaystyle {\boldsymbol {A}}\,}$).

${\displaystyle {\boldsymbol {A}}={\boldsymbol {E}}+{\boldsymbol {W}}~;~~{\boldsymbol {E}}={\cfrac {{\boldsymbol {A}}+{\boldsymbol {A}}^{T}}{2}}~;~~{\boldsymbol {W}}={\cfrac {{\boldsymbol {A}}-{\boldsymbol {A}}^{T}}{2}}~.}$

The tensor (or dyadic) product ${\displaystyle \mathbf {a} \mathbf {b} \,}$ (also written ${\displaystyle \mathbf {a} \otimes \mathbf {b} \,}$) of two vectors ${\displaystyle \mathbf {a} \,}$ and ${\displaystyle \mathbf {b} \,}$ is a tensor that assigns to each vector ${\displaystyle \mathbf {v} \,}$ the vector ${\displaystyle (\mathbf {b} \bullet \mathbf {v} )\mathbf {a} }$.

${\displaystyle (\mathbf {a} \mathbf {b} )\bullet \mathbf {v} =(\mathbf {a} \otimes \mathbf {b} )\bullet \mathbf {v} =(\mathbf {b} \bullet \mathbf {v} )\mathbf {a} ~.}$

Notice that all the above operations on tensors are remarkably similar to matrix operations.

Let ${\displaystyle {\boldsymbol {S}}}$ be a second order tensor. Let ${\displaystyle \lambda }$ be a scalar and ${\displaystyle \mathbf {n} }$ be a vector such that

${\displaystyle {\boldsymbol {S}}\cdot \mathbf {n} =\lambda ~\mathbf {n} }$

Then ${\displaystyle \lambda }$ is called an eigenvalue of ${\displaystyle {\boldsymbol {S}}}$ and ${\displaystyle \mathbf {n} }$ is an eigenvector .

A second order tensor has three eigenvalues and three eigenvectors, since the space is three-dimensional. Some of the eigenvalues might be repeated. The number of times an eigenvalue is repeated is called multiplicity.

In mechanics, many second order tensors are symmetric and positive definite. Note the following important properties of such tensors:

1. If ${\displaystyle {\boldsymbol {S}}}$ is positive definite, then ${\displaystyle \lambda >0}$.
2. If ${\displaystyle {\boldsymbol {S}}}$ is symmetric, the eigenvectors ${\displaystyle \mathbf {n} }$ are mutually orthogonal.

For more on eigenvalues and eigenvectors see Applied linear operators and spectral methods.

To make the above relations more compact, we introduce the Kronecker delta symbol

${\displaystyle {\delta _{ij}={\begin{cases}1&~{\rm {{if}~i=j~.}}\\0&~{\rm {{if}~i\neq j~.}}\end{cases}}}}$

Then, instead of the nine equations in (1) we can write (in index notation)

${\displaystyle \mathbf {e} _{i}\bullet \mathbf {e} _{j}=\delta _{ij}~.}$

Recall that the vector ${\displaystyle \mathbf {u} \,}$ can be written as

${\displaystyle {\text{(2)}}\qquad \mathbf {u} =u_{1}\mathbf {e} _{1}+u_{2}\mathbf {e} _{2}+u_{3}\mathbf {e} _{3}=\sum _{i=1}^{3}u_{i}\mathbf {e} _{i}~.}$

In index notation, equation (2) can be written as

${\displaystyle {\mathbf {u} =u_{i}\mathbf {e} _{i}~.}}$

This convention is called the Einstein summation convention. If indices are repeated, we understand that to mean that there is a sum over the indices.

We can write the Cartesian components of a vector ${\displaystyle \mathbf {u} \,}$ in the basis ${\displaystyle (\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3})\,}$ as

${\displaystyle u_{i}=\mathbf {e} _{i}\bullet \mathbf {u} ~,~~~i=1,2,3~.}$

Similarly, the components ${\displaystyle A_{ij}\,}$ of a tensor ${\displaystyle {\boldsymbol {A}}\,}$ are defined by

${\displaystyle {A_{ij}=\mathbf {e} _{i}\bullet ({\boldsymbol {A}}\bullet \mathbf {e} _{j})~.}}$

Using the definition of the tensor product, we can also write

${\displaystyle {\boldsymbol {A}}=\sum _{i,j=1}^{3}A_{ij}\mathbf {e} _{i}\mathbf {e} _{j}\equiv \sum _{i,j=1}^{3}A_{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j}~.}$

Using the summation convention,

${\displaystyle {{\boldsymbol {A}}=A_{ij}\mathbf {e} _{i}\mathbf {e} _{j}\equiv A_{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j}~.}}$

In this case, the bases of the tensor are ${\displaystyle \{\mathbf {e} _{i}\otimes \mathbf {e} _{j}\}}$ and the components are ${\displaystyle A_{ij}\,}$.

From the definition of the components of tensor ${\displaystyle {\boldsymbol {A}}\,}$, we can also see that (using the summation convention)

${\displaystyle {\mathbf {v} ={\boldsymbol {A}}\bullet \mathbf {u} ~~~\equiv ~~~v_{i}=A_{ij}u_{j}~.}}$

Similarly, the dyadic product can be expressed as

${\displaystyle {(\mathbf {a} \mathbf {b} )_{ij}\equiv (\mathbf {a} \otimes \mathbf {b} )_{ij}=a_{i}b_{j}~.}}$

We can also write a tensor ${\displaystyle {\boldsymbol {A}}}$ in matrix notation as

${\displaystyle {\boldsymbol {A}}=A_{ij}\mathbf {e} _{i}\mathbf {e} _{j}=A_{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\implies \mathbf {A} ={\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}}~.}$

Note that the Kronecker delta represents the components of the identity tensor in a Cartesian basis. Therefore, we can write

${\displaystyle {\boldsymbol {I}}=\delta _{ij}\mathbf {e} _{i}\mathbf {e} _{j}=\delta _{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\implies \mathbf {I} ={\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}~.}$

The inner product ${\displaystyle {\boldsymbol {A}}:{\boldsymbol {B}}\,}$ of two tensors ${\displaystyle {\boldsymbol {A}}\,}$ and ${\displaystyle {\boldsymbol {B}}\,}$ is an operation that generates a scalar. We define (summation implied)

${\displaystyle {{\boldsymbol {A}}:{\boldsymbol {B}}=A_{ij}B_{ij}~.}}$

The inner product can also be expressed using the trace :

${\displaystyle {{\boldsymbol {A}}:{\boldsymbol {B}}=Tr({\boldsymbol {A^{T}}}\bullet {\boldsymbol {B}})~.}}$

Proof using the definition of the trace below :

${\displaystyle {Tr({\boldsymbol {A^{T}}}\bullet {\boldsymbol {B}})={\boldsymbol {I}}:({\boldsymbol {A^{T}}}\bullet {\boldsymbol {B}})=\delta _{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j}:(A_{lk}\mathbf {e} _{k}\otimes \mathbf {e} _{l}\bullet B_{mn}\mathbf {e} _{m}\otimes \mathbf {e} _{n})=\delta _{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j}:(A_{mk}B_{mn}\mathbf {e} _{k}\otimes \mathbf {e} _{n})=}}$

${\displaystyle {A_{mk}B_{mn}\delta _{ij}\delta _{in}\delta _{jk}=A_{mk}B_{mi}\delta _{ij}\delta _{jk}=A_{mk}B_{mj}\delta _{jk}=A_{mj}B_{mj}=A:B}}$

The trace of a tensor is the scalar given by

${\displaystyle {\text{Tr}}({\boldsymbol {A}})={\boldsymbol {I}}:{\boldsymbol {A}}=\delta _{ij}\mathbf {e} _{i}\otimes \mathbf {e} _{j}:A_{mn}\mathbf {e} _{m}\otimes \mathbf {e} _{n}=\delta _{ij}\delta _{im}\delta _{jn}A_{mn}=A_{ii}}$

The trace of an N x N-matrix is the sum of the components on the downward-sloping diagonal.

The magnitude of a tensor ${\displaystyle {\boldsymbol {A}}\,}$ is defined by

${\displaystyle \Vert {\boldsymbol {A}}\Vert ={\sqrt {{\boldsymbol {A}}:{\boldsymbol {A}}}}\equiv {\sqrt {A_{ij}A_{ij}}}~.}$

Another tensor operation that is often seen is the tensor product of a tensor with a vector. Let ${\displaystyle {\boldsymbol {A}}\,}$ be a tensor and let ${\displaystyle \mathbf {v} \,}$ be a vector. Then the tensor cross product gives a tensor ${\displaystyle {\boldsymbol {C}}\,}$ defined by

${\displaystyle {{\boldsymbol {C}}={\boldsymbol {A}}\times \mathbf {v} \implies C_{ij}=e_{klj}A_{ik}v_{l}~.}}$

The permutation symbol ${\displaystyle e_{ijk}\,}$ is defined as

${\displaystyle {e_{ijk}={\begin{cases}1&~{\text{if}}~ijk=123,231,~{\text{or}}~312\\-1&~{\text{if}}~ijk=321,132,~{\text{or}}~213\\0&~{\text{if any two indices are alike}}\end{cases}}}}$

The dyadic product ${\displaystyle {\boldsymbol {\nabla }}{\mathbf {v} }\,}$ (or ${\displaystyle {\boldsymbol {\nabla }}{}\otimes \mathbf {v} }$) is called the gradient of the vector field ${\displaystyle \mathbf {v} \,}$. Therefore, the quantity ${\displaystyle {\boldsymbol {\nabla }}{\mathbf {v} }}$ is a tensor given by

${\displaystyle {{\boldsymbol {\nabla }}{\mathbf {v} }=\sum _{i}\sum _{j}{\cfrac {\partial v_{j}}{\partial x_{i}}}\mathbf {e} _{i}\mathbf {e} _{j}\equiv v_{j,i}\mathbf {e} _{i}\mathbf {e} _{j}~.}}$

In the alternative dyadic notation,

${\displaystyle {{\boldsymbol {\nabla }}{\mathbf {v} }\equiv {\boldsymbol {\nabla }}{}\otimes \mathbf {v} =\sum _{i}\sum _{j}{\cfrac {\partial v_{j}}{\partial x_{i}}}\mathbf {e} _{i}\otimes \mathbf {e} _{j}\equiv v_{j,i}\mathbf {e} _{i}\otimes \mathbf {e} _{j}~.}}$

'Warning: Some authors define the ${\displaystyle ij}$ component of ${\displaystyle {\boldsymbol {\nabla }}{\mathbf {v} }}$ as ${\displaystyle \partial v_{i}/\partial x_{j}=v_{i,j}}$.

Let ${\displaystyle {\boldsymbol {A}}\,}$ be a tensor field. Then the divergence of the tensor field is a vector ${\displaystyle {\boldsymbol {\nabla }}\bullet {\boldsymbol {A}}}$ given by

${\displaystyle {{\boldsymbol {\nabla }}\bullet {\boldsymbol {A}}=\sum _{j}\left[\sum _{i}{\cfrac {\partial A_{ij}}{\partial x_{i}}}\right]\mathbf {e} _{j}\equiv {\cfrac {\partial A_{ij}}{\partial x_{i}}}\mathbf {e} _{j}=A_{ij,i}\mathbf {e} _{j}~.}}$

To fix the definition of divergence of a general tensor field (possibly of higher order than 2), we use the relation

${\displaystyle ({\boldsymbol {\nabla }}\bullet {\boldsymbol {A}})\bullet \mathbf {c} ={\boldsymbol {\nabla }}\bullet ({\boldsymbol {A}}\bullet \mathbf {c} )}$

where ${\displaystyle \mathbf {c} }$ is an arbitrary constant vector.

The Laplacian of a vector field is given by

${\displaystyle {\nabla ^{2}{\mathbf {v} }={\boldsymbol {\nabla }}\bullet {{\boldsymbol {\nabla }}{\mathbf {v} }}=\sum _{j}\left[\sum _{i}{\cfrac {\partial ^{2}v_{j}}{\partial x_{i}^{2}}}\right]\mathbf {e} _{j}\equiv v_{j,ii}\mathbf {e} _{j}~.}}$

If ${\displaystyle \Omega }$ is a region in space enclosed by a surface ${\displaystyle \Gamma \,}$ and ${\displaystyle {\boldsymbol {A}}\,}$ is a tensor field, then

${\displaystyle {\int _{\Omega }{\boldsymbol {\nabla }}\bullet {\boldsymbol {A}}~dV=\int _{\Gamma }\mathbf {n} \bullet {\boldsymbol {A}}~dA}}$

where ${\displaystyle \mathbf {n} \,}$ is the unit outward normal to the surface.

If ${\displaystyle \Gamma \,}$ is a surface bounded by a closed curve ${\displaystyle {\mathcal {C}}}$, then

${\displaystyle \int _{\Gamma }\mathbf {n} \bullet ({\boldsymbol {\nabla }}\times {{\boldsymbol {A}})}~dA=\oint _{\mathcal {C}}\mathbf {t} \bullet {\boldsymbol {A}}~ds}$

where ${\displaystyle {\boldsymbol {A}}\,}$ is a tensor field, ${\displaystyle \mathbf {n} \,}$ is the unit normal vector to ${\displaystyle \Gamma \,}$ in the direction of a right-handed screw motion along ${\displaystyle {\mathcal {C}}}$, and ${\displaystyle \mathbf {t} \,}$ is a unit tangential vector in the direction of integration along ${\displaystyle {\mathcal {C}}}$.

Let ${\displaystyle \Omega }$ be a closed moving region of space enclosed by a surface ${\displaystyle \Gamma \,}$. Let the velocity of any surface element be ${\displaystyle \mathbf {v} \,}$. Then if ${\displaystyle {\boldsymbol {A}}(\mathbf {x} ,t)\,}$ is a tensor function of position and time,

${\displaystyle {\cfrac {\partial }{\partial t}}\int _{\Omega }{\boldsymbol {A}}~dV=\int _{\Omega }{\cfrac {\partial {\boldsymbol {A}}}{\partial t}}~dV+\int _{\Gamma }{\boldsymbol {A}}(\mathbf {v} \bullet \mathbf {n} )~dA}$

where ${\displaystyle \mathbf {n} \,}$ is the outward unit normal to the surface ${\displaystyle \Gamma \,}$.

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

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

for all vectors ${\displaystyle \mathbf {u} }$.

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

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

for all vectors ${\displaystyle \mathbf {u} }$.

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

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

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

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

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 ${\displaystyle {\frac {\partial {\boldsymbol {F}}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}={\frac {\partial {\boldsymbol {F}}_{1}}{\partial {\boldsymbol {F}}_{2}}}:\left({\frac {\partial {\boldsymbol {F}}_{2}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {T}}\right)}$

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