Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

1 files changed, 53 insertions, 53 deletions

diff --git a/source/know/concept/cauchy-strain-tensor/index.md b/source/know/concept/cauchy-strain-tensor/index.md
index a628820..bae7bb8 100644
--- a/source/know/concept/cauchy-strain-tensor/index.md
+++ b/source/know/concept/cauchy-strain-tensor/index.md
@@ -11,9 +11,9 @@ 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** $\va{u}$:
+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** $$\va{u}$$:
 
 $$\begin{aligned}
     \va{u}
@@ -21,7 +21,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We restrict ourselves to **infinitesimal strain**,
-where $\va{u}$ is so tiny that the material's properties are unchanged,
+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.
@@ -31,15 +31,15 @@ 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,
+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 $\va{x}$:
+where the new position $$\va{x}$$ is the reference,
+and the old position $$\va{X}$$ is expressed as a function of $$\va{x}$$:
 
 $$\begin{aligned}
     \va{u}(\va{x})
@@ -47,10 +47,10 @@ $$\begin{aligned}
 \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.
+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 $\va{a}$:
+where we define $$\va{A}$$ as the "old" version of $$\va{a}$$:
 
 $$\begin{aligned}
     \va{X} = \va{X}(\va{x})
@@ -58,9 +58,9 @@ $$\begin{aligned}
     \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:
+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:
 
 $$\begin{aligned}
     \va{A}
@@ -71,8 +71,8 @@ $$\begin{aligned}
     &= \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 $\va{a}$:
+Using the fact that $$\va{a}$$ is tiny by definition,
+we expand the middle term to first order in $$\va{a}$$:
 
 $$\begin{aligned}
     \va{u}(\va{x} + \va{a})
@@ -80,8 +80,8 @@ $$\begin{aligned}
     = \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:
+With this, we can now define the "shift" $$\delta\va{a}$$
+as the difference between $$\va{a}$$ and $$\va{A}$$ like so:
 
 $$\begin{aligned}
     \delta{\va{a}}
@@ -90,24 +90,24 @@ $$\begin{aligned}
 \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:
+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**,
+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 $|\nabla_j u_i| \ll 1$.
+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 $\va{a}$.
+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,
+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:
 
@@ -116,7 +116,7 @@ $$\begin{aligned}
     \equiv \va{a} \cdot \va{b} - \va{A} \cdot \va{B}
 \end{aligned}$$
 
-Where $\va{B}$ is the old version of $\va{b}$.
+Where $$\va{B}$$ is the old version of $$\va{b}$$.
 Since these vectors are all tiny, we apply the product rule:
 
 $$\begin{aligned}
@@ -125,7 +125,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 It is more informative to switch to index notation here.
-Inserting $\delta\va{a}$ and $\delta\va{b}$ yields:
+Inserting $$\delta\va{a}$$ and $$\delta\va{b}$$ yields:
 
 $$\begin{aligned}
     \delta(\va{a} \cdot \va{b})
@@ -136,8 +136,8 @@ $$\begin{aligned}
     &= \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:
+At last, we define the **Cauchy infinitesimal strain tensor** $$\hat{u}$$
+such that it has $$u_{ij}$$ as components:
 
 $$\begin{aligned}
     \boxed{
@@ -154,7 +154,7 @@ $$\begin{aligned}
     = 2 \va{a} \cdot \hat{u} \cdot \va{b}
 \end{aligned}$$
 
-The Cauchy strain tensor $\hat{u}$ is a second-rank tensor,
+The Cauchy strain tensor $$\hat{u}$$ is a second-rank tensor,
 and can alternatively be expressed like so:
 
 $$\begin{aligned}
@@ -164,17 +164,17 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where $\top$ is the transpose. Being defined from the scalar product,
+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. $u_{ij} = u_{ji}$.
+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:
+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:
 
 $$\begin{aligned}
     \boxed{
@@ -184,12 +184,12 @@ $$\begin{aligned}
 \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,
+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.
+Because $$\hat{u}$$ is symmetric, such a basis always exists.
 
-Given a vector $\va{a}$, its relative length change
+Given a vector $$\va{a}$$, its relative length change
 due to the deformation is simply given by:
 
 $$\begin{aligned}
@@ -199,8 +199,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-To find the angle change $\delta\theta$
-between two vectors $\va{a}$ and $\va{b}$,
+To find the angle change $$\delta\theta$$
+between two vectors $$\va{a}$$ and $$\va{b}$$,
 we start with the product rule:
 
 $$\begin{aligned}
@@ -211,9 +211,9 @@ $$\begin{aligned}
     - \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 $u_{ab}$:
+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 $$u_{ab}$$:
 
 $$\begin{aligned}
     \delta\theta
@@ -223,7 +223,7 @@ $$\begin{aligned}
     {\big|\va{a}\big| \big|\va{b}\big| \sin\theta}
 \end{aligned}$$
 
-By recognizing the length change $\delta|\va{a}|/|\va{a}| = u_{aa}$,
+By recognizing the length change $$\delta|\va{a}|/|\va{a}| = u_{aa}$$,
 we arrive at the following expression:
 
 $$\begin{aligned}
@@ -234,8 +234,8 @@ $$\begin{aligned}
 \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 $\va{u}$:
+so the change of the line element $$\dd{\va{l}}$$
+is easy to express using the displacement field $$\va{u}$$:
 
 $$\begin{aligned}
     \boxed{
@@ -244,9 +244,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Next, we calculate the change of the differential volume element $\dd{V}$
+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 $\va{c}$:
+spanned by $$\va{a}$$, $$\va{b}$$ and $$\va{c}$$:
 
 $$\begin{aligned}
     \delta(\dd{V})
@@ -280,7 +280,7 @@ $$\begin{aligned}
     &= (\va{a} \cross \va{b} \cdot \va{c}) (\nabla \cdot \va{u})
 \end{aligned}$$
 
-Here, we recognize the definition of $\dd{V}$,
+Here, we recognize the definition of $$\dd{V}$$,
 leading to the following infinitesimal volume change:
 
 $$\begin{aligned}
@@ -290,8 +290,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Finally, for the surface element $\dd{\va{S}} = \va{a} \cross \va{b}$,
-we use that the volume element $\dd{V} = \va{c} \cdot \dd{\va{S}}$:
+Finally, for the surface element $$\dd{\va{S}} = \va{a} \cross \va{b}$$,
+we use that the volume element $$\dd{V} = \va{c} \cdot \dd{\va{S}}$$:
 
 $$\begin{aligned}
     \delta(\dd{V})
@@ -300,7 +300,7 @@ $$\begin{aligned}
     = (\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 $\delta(\dd{V})$,
+By comparing this to the previous result for $$\delta(\dd{V})$$,
 we arrive at the following equation:
 
 $$\begin{aligned}
@@ -308,8 +308,8 @@ $$\begin{aligned}
     = (\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 $\delta(\dd{\va{S}})$:
+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{