---
title: "Cauchy strain tensor"
firstLetter: "C"
publishDate: 2021-03-31
categories:
- Physics
- Continuum physics

date: 2021-03-31T09:43:40+02:00
draft: false
markup: pandoc
---

# Cauchy strain tensor

**Strain** quantifies the deformation of a solid object.
If the body has been deformed, e.g. by pulling or bending,
its constituent particles have moved a bit.
Let $\va{X}$ be the original location of a particle,
and $\va{x}$ its new location after the deformation.
We can thus define the **displacement field** $\va{u}$:

$$\begin{aligned}
    \va{u}
    \equiv \va{x} - \va{X}
\end{aligned}$$

We restrict ourselves to **infinitesimal strain**,
where $\va{u}$ is so tiny that the material's properties are unchanged,
and a **slowly-varying strain**,
where the particle's neighbourhood has been distorted,
but not completely changed.

A key challenge when quantifying deformation
is that we need to somehow exclude movements of the *entire* body:
for example, you can bend a twig in your hands while walking or dancing,
but we are only interested in the twig's shape change,
not in your movements.
The above definition of $\vu{u}$ includes both,
so we should be careful how we extract the strain from it.


## Definition

We use the **Eulerian description** of deformation,
where the new position $\va{x}$ is the reference,
and the old position $\va{X}$ is expressed as a function of $\va{x}$:

$$\begin{aligned}
    \va{u}(\va{x})
    \equiv \va{x} - \va{X}(\va{x})
\end{aligned}$$

Let us choose two nearby points in the deformed solid,
and call them $\va{x}$ and $\va{x} + \va{a}$,
where $\va{a}$ is a tiny vector pointing from one to the other.
Before the displacement, those points respectively had these positions,
where we define $\va{A}$ as the "old" version of $\va{a}$:

$$\begin{aligned}
    \va{X} = \va{X}(\va{x})
    \qquad
    \va{X} + \va{A} = \va{X}(\va{x} + \va{a})
\end{aligned}$$

Because the new positions $\va{x}$ are our reference,
we would like to write $\va{A}$ without $\va{X}$.
To do so, we use the definition of $\va{u}(\va{x})$, yielding:

$$\begin{aligned}
    \va{A}
    &= \va{X}(\va{x} + \va{a}) - \va{X}(\va{x})
    \\
    &= \big( \va{x} + \va{a} - \va{u}(\va{x} + \va{a}) \big) - \big( \va{x} - \va{u}(\va{x}) \big)
    \\
    &= \va{a} - \va{u}(\va{x} + \va{a}) - \va{u}(\va{x})
\end{aligned}$$

Using the fact that $\va{a}$ is tiny by definition,
we expand the middle term to first order in $\va{a}$:

$$\begin{aligned}
    \va{u}(\va{x} + \va{a})
    \approx \va{u}(\va{x}) + a_x \pdv{\va{u}}{x} + a_y \pdv{\va{u}}{y} + a_z \pdv{\va{u}}{z}
    = \va{u}(\va{x}) + (\va{a} \cdot \nabla) \va{u}(\va{x})
\end{aligned}$$

With this, we can now define the "shift" $\delta\va{a}$
as the difference between $\va{a}$ and $\va{A}$ like so:

$$\begin{aligned}
    \delta{\va{a}}
    \equiv \va{a} - \va{A}
    = (\va{a} \cdot \nabla) \va{u}(\va{x})
\end{aligned}$$

In index notation, we write this expression as follows,
with $\nabla_j = \pdv*{x_j}$ simply being the partial derivative
with respect to the $j$th coordinate:

$$\begin{aligned}
    \delta a_i
    = \sum_{j} a_j \nabla_j u_i
\end{aligned}$$

Where $\nabla_j u_i$ are called the **displacement gradients**,
and are just one step away from the desired definition of strain.
Note that these gradients are dimensionless,
so we can more formally define a *slowly-varying* displacement $\va{u}(\va{x})$
as one where $|\nabla_j u_i| \ll 1$.

Now, to solve the problem of macroscopic movements,
we take another tiny vector $\va{b}$ starting in the same point $\va{x}$ as $\va{a}$.
Here is the trick: if the whole body is uniformly translated or rotated,
the scalar product $\va{a} \cdot \va{b}$ is unchanged,
but if there is a non-uniform distortion, it changes.
We thus define the scalar product's difference like so:

$$\begin{aligned}
    \delta(\va{a} \cdot \va{b})
    \equiv \va{a} \cdot \va{b} - \va{A} \cdot \va{B}
\end{aligned}$$

Where $\va{B}$ is the old version of $\va{b}$.
Since these vectors are all tiny, we apply the product rule:

$$\begin{aligned}
    \delta(\va{a} \cdot \va{b})
    &= \delta\va{a} \cdot \va{b} + \va{a} \cdot \delta\va{b}
\end{aligned}$$

It is more informative to switch to index notation here.
Inserting $\delta\va{a}$ and $\delta\va{b}$ yields:

$$\begin{aligned}
    \delta(\va{a} \cdot \va{b})
    &= \sum_{i} \delta{a}_i \: b_i + \sum_{i} \delta{b}_i \: a_i
    \\
    &= \sum_{ij} \nabla_j u_i \: a_j b_i + \sum_{ij} \nabla_j u_i \: a_i b_j
    \\
    &= \sum_{ij} \big( \nabla_i u_j + \nabla_j u_i \big) \: a_i b_j
\end{aligned}$$

At last, we define the **Cauchy infinitesimal strain tensor** $\hat{u}$
such that it has $u_{ij}$ as components:

$$\begin{aligned}
    \boxed{
        u_{ij}
        \equiv \frac{1}{2} \big( \nabla_j u_i + \nabla_i u_j \big)
    }
\end{aligned}$$

Which allows us to rewrite the shift of the scalar product in the following compact way:

$$\begin{aligned}
    \delta(\va{a} \cdot \va{b})
    &= 2 \sum_{ij} u_{ij} a_i b_j
    = 2 \va{a} \cdot \hat{u} \cdot \va{b}
\end{aligned}$$

The Cauchy strain tensor $\hat{u}$ is a second-rank tensor,
and can alternatively be expressed like so:

$$\begin{aligned}
    \boxed{
        \hat{u}
        \equiv \frac{1}{2} \big( \nabla \va{u} + (\nabla \va{u})^\top \big)
    }
\end{aligned}$$

Where $\top$ is the transpose. Being defined from the scalar product,
all macroscopic movements of the body are removed from the tensor,
which turns out to make it symmetric, i.e. $u_{ij} = u_{ji}$.


## Geometry

So far we have used Cartesian coordinates,
but we can choose any three vectors $\va{a}$, $\va{b}$ and $\va{c}$,
and **project** $\hat{u}$ onto this basis.
For example, the component $u_{ab}$ then becomes:

$$\begin{aligned}
    \boxed{
        u_{ab}
        = \frac{\va{a} \cdot \hat{u} \cdot \va{b}}{\big|\va{a}\big| \big|\va{b}\big|}
    }
\end{aligned}$$

And so forth, for the other eight components.
The basis in which $\hat{u}$ is diagonal is the one formed by its eigenvectors,
and their directions are the **principal axes of strain**
at that point in the solid.
Because $\hat{u}$ is symmetric, such a basis always exists.

Given a vector $\va{a}$, its relative length change
due to the deformation is simply given by:

$$\begin{aligned}
    \boxed{
        \frac{\delta|\va{a}|}{|\va{a}|}
        = u_{aa}
    }
\end{aligned}$$

To find the angle change $\delta\theta$
between two vectors $\va{a}$ and $\va{b}$,
we start with the product rule:

$$\begin{aligned}
    \delta(\va{a} \cdot \va{b})
    = \delta(\big|\va{a}\big| \big|\va{b}\big| \cos\theta)
    = \delta\big|\va{a}\big| \big|\va{b}\big| \cos\theta
    + \big|\va{a}\big| \delta\big|\va{b}\big| \cos\theta
    - \big|\va{a}\big| \big|\va{b}\big| \sin\theta \: \delta\theta
\end{aligned}$$

We isolate this for $\delta\theta$, using the fact that
$\delta(\va{a} \cdot \va{b}) = 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab}$
thanks to the projection $u_{ab}$:

$$\begin{aligned}
    \delta\theta
    = \frac{\delta\big|\va{a}\big| \big|\va{b}\big| \cos\theta
    + \big|\va{a}\big| \delta\big|\va{b}\big| \cos\theta
    - 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab}}
    {\big|\va{a}\big| \big|\va{b}\big| \sin\theta}
\end{aligned}$$

By recognizing the length change $\delta|\va{a}|/|\va{a}| = u_{aa}$,
we arrive at the following expression:

$$\begin{aligned}
    \boxed{
        \delta\theta
        = \frac{(u_{aa} + u_{bb}) \cos\theta - u_{ab}}{\sin\theta}
    }
\end{aligned}$$

Now, everything so far has been about tiny vectors,
so the change of the line element $\dd{\va{l}}$
is easy to express using the displacement field $\va{u}$:

$$\begin{aligned}
    \boxed{
        \delta(\dd{\va{l}})
        = (\dd{\va{l}} \cdot \nabla) \va{u}
        %= (\nabla \vec{u})^\top \cdot \dd{\va{l}}
    }
\end{aligned}$$

Next, we calculate the change of the differential volume element $\dd{V}$
by treating it as the volume of a tiny parallelepiped
spanned by $\va{a}$, $\va{b}$ and $\va{c}$:

$$\begin{aligned}
    \delta(\dd{V})
    = \delta(\va{a} \cross \va{b} \cdot \va{c})
    &= \delta\va{a} \cross \va{b} \cdot \va{c} + \va{a} \cross \delta\va{b} \cdot \va{c} + \va{a} \cross \va{b} \cdot \delta\va{c}
    \\
    &= (\va{a} \cdot \nabla) \va{u} \cross \va{b} \cdot \va{c}
    + \va{a} \cross (\va{b} \cdot \nabla )\va{u} \cdot \va{c}
    + \va{a} \cross \va{b} \cdot (\va{c} \cdot \nabla) \va{u}
\end{aligned}$$

We can reorder the factors like so
(write it out in index notation if you are not convinced):

$$\begin{aligned}
    \delta(\dd{V})
    &= (\va{a} \cdot \nabla) \va{u} \cross \va{b} \cdot \va{c}
    + (\va{b} \cdot \nabla) \va{a} \cross \va{u} \cdot \va{c}
    + (\va{c} \cdot \nabla) \va{a} \cross \va{b} \cdot \va{u}
\end{aligned}$$

By applying a couple of vector identities,
we can rewrite this more compactly as follows:

$$\begin{aligned}
    \delta(\dd{V})
    &= \Big( \va{b} \cross \va{c} (\va{a} \cdot \nabla) \cross \va{b}
    + \va{c} \cross \va{a} (\va{b} \cdot \nabla)
    + \va{a} \cross \va{b} (\va{c} \cdot \nabla) \Big) \cdot \va{u}
    \\
    &= (\va{a} \cross \va{b} \cdot \va{c}) (\nabla \cdot \va{u})
\end{aligned}$$

Here, we recognize the definition of $\dd{V}$,
leading to the following infinitesimal volume change:

$$\begin{aligned}
    \boxed{
        \delta(\dd{V})
        = \nabla \cdot \va{u} \dd{V}
    }
\end{aligned}$$

Finally, for the surface element $\dd{\va{S}} = \va{a} \cross \va{b}$,
we use that the volume element $\dd{V} = \va{c} \cdot \dd{\va{S}}$:

$$\begin{aligned}
    \delta(\dd{V})
    = \delta(\va{c} \cdot \dd{\va{S}})
    = \delta\va{c} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
    = (\va{c} \cdot \nabla) \va{u} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
\end{aligned}$$

By comparing this to the previous result for $\delta(\dd{V})$,
we arrive at the following equation:

$$\begin{aligned}
    \nabla \cdot \va{u} (\va{c} \cdot \dd{\va{S}})
    = (\va{c} \cdot \nabla) \va{u} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
\end{aligned}$$

Since $\va{c}$ is dot-multiplied at the front of each term,
we remove it, and isolate the rest for $\delta(\dd{\va{S}})$:

$$\begin{aligned}
    \boxed{
        \delta(\dd{\va{S}})
        = \big( (\nabla \cdot \va{u}) \hat{1} - \nabla \va{u} \big) \cdot \dd{\va{S}}
    }
\end{aligned}$$



## References
1.  B. Lautrup,
    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
    CRC Press.