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+---
+title: "Cauchy strain tensor"
+date: 2021-03-31
+categories:
+- Physics
+- Continuum physics
+layout: "concept"
+---
+
+**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 $\nabla_j \equiv \ipdv{}{x_j}$ 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 $\va{u}$:
+
+$$\begin{aligned}
+    \boxed{
+        \delta(\dd{\va{l}})
+        = (\dd{\va{l}} \cdot \nabla) \va{u}
+    }
+\end{aligned}$$
+
+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.