# Continuum mechanics/Time derivatives and rates: Wikis

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# Study guide

Up to date as of January 14, 2010

## 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 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 $\dot{J} = 0$.