# Nonlinear finite elements/Kinematics - time derivatives and rates

## Time derivatives and rate quantities

### Material time derivatives

Material time derivatives are needed for many updated Lagrangian formulations of finite element analysis.

Recall that the motion can be expressed as

$\mathbf {x} ={\boldsymbol {\varphi }}(\mathbf {X} ,t)\quad {\text{or}}\qquad \mathbf {X} ={\boldsymbol {\varphi }}^{-1}(\mathbf {x} ,t)$

If we keep $\mathbf {X}$  fixed, then the velocity is given by

$\mathbf {V} (\mathbf {X} ,t)={\frac {\partial {\boldsymbol {\varphi }}}{\partial t}}(\mathbf {X} ,t)$

This is the material time derivative expressed in terms of $\mathbf {X}$ .

The spatial version of the velocity is

$\mathbf {v} (\mathbf {x} ,t)=\mathbf {V} ({\boldsymbol {\varphi }}^{-1}(\mathbf {x} ,t),t)$

We will use the symbol $\mathbf {v}$  for velocity from now on by slightly abusing the notation.

We usually think of quantities such as velocity and acceleration as spatial quantities which are functions of $\mathbf {x}$  (rather than material quantities which are functions of $\mathbf {X}$ ).

Given the spatial velocity $\mathbf {v} (\mathbf {x} ,t)$ , if we want to find the acceleration we will have to consider the fact that $\mathbf {x} \equiv \mathbf {x} (\mathbf {X} ,t)$ , i.e., the position also changes with time. We do this by using the chain rule. Thus

${\cfrac {D\mathbf {v} (\mathbf {x} ,t)}{Dt}}=\mathbf {a} (\mathbf {x} ,t)={\frac {\partial \mathbf {v} (\mathbf {x} ,t)}{\partial t}}+{\frac {\partial \mathbf {v} (\mathbf {x} ,t)}{\partial \mathbf {x} }}\cdot {\frac {\partial {\boldsymbol {\varphi }}(\mathbf {X} ,t)}{\partial t}}={\frac {\partial \mathbf {v} }{\partial t}}+{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {V} ={\frac {\partial \mathbf {v} }{\partial t}}+{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v}$

Such a derivative is called the material time derivative expressed in terms of $\mathbf {x}$ . The second term in the expression is called the convective derivative..

Let the velocity be expressed in spatial form, i.e., $\mathbf {v} (\mathbf {x} ,t)$ . The spatial velocity gradient tensor is given by

${\boldsymbol {l}}:={\frac {\partial \mathbf {v} (\mathbf {x} ,t)}{\partial \mathbf {x} }}={\boldsymbol {\nabla }}\mathbf {v}$

The velocity gradient ${\boldsymbol {l}}$  is a second order tensor which can expressed as

${\boldsymbol {l}}=l_{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}={\frac {\partial v_{i}}{\partial x_{j}}}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}$

The velocity gradient is a measure of the relative velocity of two points in the current configuration.

### Time derivative of the deformation gradient

Recall that the deformation gradient is given by

${\boldsymbol {F}}={\frac {\partial {\boldsymbol {\varphi }}}{\partial \mathbf {X} }}$

The time derivative of ${\boldsymbol {F}}$  (keeping $\mathbf {X}$  fixed) is

${\dot {\boldsymbol {F}}}={\frac {\partial }{\partial t}}~\left({\frac {\partial {\boldsymbol {\varphi }}}{\partial \mathbf {X} }}\right)={\frac {\partial }{\partial \mathbf {X} }}~\left({\frac {\partial {\boldsymbol {\varphi }}}{\partial t}}\right)={\frac {\partial \mathbf {v} }{\partial \mathbf {X} }}={\boldsymbol {\nabla }}_{\circ }\mathbf {v}$

Using the chain rule

${\dot {\boldsymbol {F}}}={\frac {\partial \mathbf {v} }{\partial \mathbf {x} }}\cdot {\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}={\frac {\partial \mathbf {v} }{\partial \mathbf {x} }}\cdot {\frac {\partial {\boldsymbol {\varphi }}}{\partial \mathbf {X} }}={\boldsymbol {l}}\cdot {\boldsymbol {F}}$

Form this we get the important relation

${\boldsymbol {l}}={\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}~.$

### Time derivative of strain

Let $d\mathbf {X} _{1}$  and $d\mathbf {X} _{2}$  be two infinitesimal material line segments in a body. Then

$d\mathbf {x} _{1}={\boldsymbol {F}}\cdot d\mathbf {X} _{1}~;~~d\mathbf {x} _{2}={\boldsymbol {F}}\cdot d\mathbf {X} _{2}$

Hence,

$d\mathbf {x} _{1}\cdot d\mathbf {x} _{2}=({\boldsymbol {F}}\cdot d\mathbf {X} _{1})\cdot ({\boldsymbol {F}}\cdot d\mathbf {X} _{2})=d\mathbf {X} _{1}\cdot ({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}})\cdot d\mathbf {X} _{2}=d\mathbf {X} _{1}\cdot {\boldsymbol {C}}\cdot d\mathbf {X} _{2}=d\mathbf {X} _{1}\cdot (2~{\boldsymbol {E}}+{\boldsymbol {\mathit {1}}})\cdot d\mathbf {X} _{2}$

Taking the derivative with respect to $t$  gives us

${\frac {\partial }{\partial t}}(d\mathbf {x} _{1}\cdot d\mathbf {x} _{2})=d\mathbf {X} _{1}\cdot {\frac {\partial {\boldsymbol {C}}}{\partial t}}\cdot d\mathbf {X} _{2}=2~d\mathbf {X} _{1}\cdot {\frac {\partial {\boldsymbol {E}}}{\partial t}}\cdot d\mathbf {X} _{2}$

The material strain rate tensor is defined as

${\dot {\boldsymbol {E}}}={\frac {\partial {\boldsymbol {E}}}{\partial t}}={\frac {1}{2}}~{\frac {\partial {\boldsymbol {C}}}{\partial t}}={\frac {1}{2}}~{\dot {\boldsymbol {C}}}$

Clearly,

${\dot {\boldsymbol {E}}}={\frac {1}{2}}~{\frac {\partial }{\partial t}}({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}})={\frac {1}{2}}~({\dot {\boldsymbol {F}}}^{T}\cdot {\boldsymbol {F}}+{\boldsymbol {F}}^{T}\cdot {\dot {\boldsymbol {F}}})~.$

Also,

${\frac {1}{2}}~{\frac {\partial }{\partial t}}(d\mathbf {x} _{1}\cdot d\mathbf {x} _{2})=d\mathbf {X} _{1}\cdot {\dot {\boldsymbol {E}}}\cdot d\mathbf {X} _{2}=({\boldsymbol {F}}^{-1}\cdot d\mathbf {x} _{1})\cdot {\dot {\boldsymbol {E}}}\cdot ({\boldsymbol {F}}^{-1}\cdot d\mathbf {x} _{2})=d\mathbf {x} _{1}\cdot ({\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {E}}}\cdot {\boldsymbol {F}}^{-1})\cdot d\mathbf {x} _{2}$

The spatial rate of deformation tensor or stretching tensor is defined as

${\boldsymbol {d}}={\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {E}}}\cdot {\boldsymbol {F}}^{-1}={\frac {1}{2}}~{\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {C}}}\cdot {\boldsymbol {F}}^{-1}$

In fact, we can show that ${\boldsymbol {d}}$  is the symmetric part of the velocity gradient, i.e.,

${\boldsymbol {d}}={\frac {1}{2}}~({\boldsymbol {l}}+{\boldsymbol {l}}^{T})$

For rigid body motions we get ${\boldsymbol {d}}={\boldsymbol {\mathit {0}}}$ .

#### Lie derivatives

Most of the operations above can be interpreted as push-forward and pull-back operations. Also, time derivatives of these tensors can be interpreted as Lie derivatives.

Recall that the push-forward of the strain tensor from the material configuration to the spatial configuration is given by

${\boldsymbol {e}}=\phi _{*}[{\boldsymbol {E}}]={\boldsymbol {F}}^{-T}\cdot {\boldsymbol {E}}\cdot {\boldsymbol {F}}^{-1}$

The pull-back of the spatial strain tensor to the material configuration is given by

${\boldsymbol {E}}=\phi ^{*}[{\boldsymbol {e}}]={\boldsymbol {F}}^{T}\cdot {\boldsymbol {e}}\cdot {\boldsymbol {F}}$

Therefore, the rate of deformation tensor is a push-forward of the material strain rate tensor, i.e.,

${\boldsymbol {d}}={\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {E}}}\cdot {\boldsymbol {F}}^{-1}=\phi _{*}[{\dot {\boldsymbol {E}}}]$

Similarly, the material strain rate tensor is a pull-back of the rate of deformation tensor to the material configuration, i.e.,

${\dot {\boldsymbol {E}}}={\boldsymbol {F}}^{T}\cdot {\boldsymbol {d}}\cdot {\boldsymbol {F}}=\phi ^{*}[{\boldsymbol {d}}]$

Now,

${\boldsymbol {E}}=\phi ^{*}[{\boldsymbol {e}}]\quad \implies \quad {\dot {\boldsymbol {E}}}={\frac {\partial }{\partial t}}\left(\phi ^{*}[{\boldsymbol {e}}]\right)$

Also,

${\boldsymbol {d}}=\phi _{*}[{\dot {\boldsymbol {E}}}]=\phi _{*}\left[{\frac {\partial }{\partial t}}\left(\phi ^{*}[{\boldsymbol {e}}]\right)\right]$

Therefore the rate of deformation tensor can be obtained by first pulling back ${\boldsymbol {e}}$  to the reference configuration, taking a material time derivative in that configuration, and then pushing forward the result to the current configuration.

Such an operation is called a Lie derivative. In general, the Lie derivative of a spatial tensor $\mathbf {g}$  is defined as

${\mathcal {L}}_{\phi }[{\boldsymbol {g}}]:=\phi _{*}\left[{\frac {\partial }{\partial t}}\left(\phi ^{*}[{\boldsymbol {g}}]\right)\right]~.$

### Spin tensor

The velocity gradient tensor can be additively decomposed into a symmetric part and a skew part:

${\boldsymbol {l}}={\frac {1}{2}}~({\boldsymbol {l}}+{\boldsymbol {l}}^{T})+{\frac {1}{2}}({\boldsymbol {l}}-{\boldsymbol {l}}^{T})={\boldsymbol {d}}+{\boldsymbol {w}}$

We have seen that ${\boldsymbol {d}}$  is the rate of deformation tensor. The quantity ${\boldsymbol {w}}$  is called the spin tensor.

Note that ${\boldsymbol {d}}$  is symmetric while ${\boldsymbol {w}}$  is skew symmetric, i.e.,

${\boldsymbol {d}}={\boldsymbol {d}}^{T}~;~~{\boldsymbol {w}}=-{\boldsymbol {w}}^{T}~.$

So see why ${\boldsymbol {w}}$  is called a "spin", recall that

${\boldsymbol {l}}={\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}$

Therefore,

${\boldsymbol {w}}={\frac {1}{2}}({\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}-{\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {F}}}^{T})$

Also,

${\boldsymbol {F}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}\quad \implies \quad {\dot {\boldsymbol {F}}}={\dot {\boldsymbol {R}}}\cdot {\boldsymbol {U}}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {U}}}$

Therefore,

${\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}=({\dot {\boldsymbol {R}}}\cdot {\boldsymbol {U}}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {U}}})\cdot ({\boldsymbol {U}}^{-1}\cdot {\boldsymbol {R}}^{T})={\dot {\boldsymbol {R}}}\cdot {\boldsymbol {R}}^{T}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {U}}}\cdot {\boldsymbol {U}}^{-1}\cdot {\boldsymbol {R}}^{T}$

and

${\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {F}}}^{T}=({\boldsymbol {R}}\cdot {\boldsymbol {U}}^{-1})\cdot ({\boldsymbol {U}}\cdot {\dot {\boldsymbol {R}}}^{T}+{\dot {\boldsymbol {U}}}\cdot {\boldsymbol {R}}^{T})={\boldsymbol {R}}\cdot {\dot {\boldsymbol {R}}}^{T}+{\boldsymbol {R}}\cdot {\boldsymbol {U}}^{-1}\cdot {\dot {\boldsymbol {U}}}\cdot {\boldsymbol {R}}^{T}$

So we have

${\boldsymbol {w}}={\frac {1}{2}}~({\dot {\boldsymbol {R}}}\cdot {\boldsymbol {R}}^{T}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {U}}}\cdot {\boldsymbol {U}}^{-1}\cdot {\boldsymbol {R}}^{T}-{\boldsymbol {R}}\cdot {\dot {\boldsymbol {R}}}^{T}-{\boldsymbol {R}}\cdot {\boldsymbol {U}}^{-1}\cdot {\dot {\boldsymbol {U}}}\cdot {\boldsymbol {R}}^{T})$

Now

${\boldsymbol {R}}\cdot {\boldsymbol {R}}^{T}={\boldsymbol {\mathit {1}}}\quad \implies \quad {\dot {\boldsymbol {R}}}\cdot {\boldsymbol {R}}^{T}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {R}}}^{T}={\boldsymbol {\mathit {0}}}$

Therefore

${\boldsymbol {w}}={\dot {\boldsymbol {R}}}\cdot {\boldsymbol {R}}^{T}+{\frac {1}{2}}~{\boldsymbol {R}}\cdot ({\dot {\boldsymbol {U}}}\cdot {\boldsymbol {U}}^{-1}-{\boldsymbol {U}}^{-1}\cdot {\dot {\boldsymbol {U}}})\cdot {\boldsymbol {R}}^{T}$

The second term above is invariant for rigid body motions and zero for an uniaxial stretch. Hence, we are left with just a rotation term. This is why the quantity ${\boldsymbol {w}}$  is called a spin.

The spin tensor is a skew-symmetric tensor and has an associated axial vector ${\boldsymbol {\omega }}$  (also called the angular velocity vector) whose components are given by

${\boldsymbol {\omega }}={\begin{bmatrix}w_{1}\\w_{2}\\w_{3}\end{bmatrix}}$

where

$\mathbf {w} ={\begin{bmatrix}0&-w_{3}&w_{2}\\w_{3}&0&-w_{1}\\-w_{2}&w_{1}&0\end{bmatrix}}$

The spin tensor and its associated axial vector appear in a number of modern numerical algorithms.

### Rate of change of volume

Recall that

$dv=J~dV\qquad {\text{where}}~J=\det {\boldsymbol {F}}$

Therefore, taking the material time derivative of $dv$  (keeping $\mathbf {X}$  fixed), we have

${\cfrac {d}{dt}}(dv)={\dot {J}}~dV={\cfrac {\dot {J}}{J}}~dv$

At this stage we invoke the following result from tensor calculus:

If ${\boldsymbol {A}}$  is an invertible tensor which depends on $t$  then

${\cfrac {d}{dt}}(\det {\boldsymbol {A}})=(\det {\boldsymbol {A}})~{\text{tr}}\left({\cfrac {d{\boldsymbol {A}}}{dt}}\cdot {\boldsymbol {A}}^{-1}\right)$

In the case where ${\boldsymbol {A}}={\boldsymbol {F}},~J=\det {\boldsymbol {F}}$  we have

${\cfrac {d}{dt}}(J)=J~{\text{tr}}\left({\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}\right)$

or,

${\dot {J}}=J~{\text{tr}}({\boldsymbol {l}})=J~{\text{tr}}(\mathbf {d} )$

Therefore,

${\cfrac {d}{dt}}(dv)={\text{tr}}(\mathbf {d} )~dv$

Alternatively, we can also write

${\dot {J}}={\frac {1}{2}}~J~{\boldsymbol {C}}^{-1}:{\dot {\boldsymbol {C}}}$

These relations are of immense use in numerical algorithms - particularly those which involved incompressible behavior, i.e., when ${\dot {J}}=0$ .