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

1 files changed, 29 insertions, 29 deletions

diff --git a/source/know/concept/hookes-law/index.md b/source/know/concept/hookes-law/index.md
index 87e04de..292b735 100644
--- a/source/know/concept/hookes-law/index.md
+++ b/source/know/concept/hookes-law/index.md
@@ -12,8 +12,8 @@ In its simplest form, **Hooke's law** dictates that
 changing the length of an elastic object requires
 a force that is proportional the desired length difference.
 In its most general form, it gives a linear relationship
-between the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$
-to the [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) $\hat{u}$.
+between the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{\sigma}$$
+to the [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) $$\hat{u}$$.
 
 Importantly, all forms of Hooke's law are only valid for small deformations,
 since the stress-strain relationship becomes nonlinear otherwise.
@@ -22,8 +22,8 @@ since the stress-strain relationship becomes nonlinear otherwise.
 ## Simple form
 
 The simple form of the law is traditionally quoted for springs,
-since they have a spring constant $k$ giving the ratio
-between the force $F$ and extension $x$:
+since they have a spring constant $$k$$ giving the ratio
+between the force $$F$$ and extension $$x$$:
 
 $$\begin{aligned}
     \boxed{
@@ -35,23 +35,23 @@ $$\begin{aligned}
 In general, all solids are elastic for small extensions,
 and therefore also obey Hooke's law.
 In light of this fact, we replace the traditional spring
-with a rod of length $L$ and cross-section $A$.
+with a rod of length $$L$$ and cross-section $$A$$.
 
-The constant $k$ depends on, among several things,
-the spring's length $L$ and cross-section $A$,
+The constant $$k$$ depends on, among several things,
+the spring's length $$L$$ and cross-section $$A$$,
 so for our generalization, we want a new parameter
 to describe the proportionality independently of the rod's dimensions.
-To achieve this, we realize that the force $F$ is spread across $A$,
-and that the extension $x$ should be take relative to $L$.
+To achieve this, we realize that the force $$F$$ is spread across $$A$$,
+and that the extension $$x$$ should be take relative to $$L$$.
 
 $$\begin{aligned}
     \frac{F}{A}
     = \Big( k \frac{L}{A} \Big) \frac{x}{L}
 \end{aligned}$$
 
-The force-per-area $F/A$ on a solid is the definition of **stress**,
-and the relative elongation $x/L$ is the defintion of **strain**.
-If $F$ acts along the $x$-axis, we can then write:
+The force-per-area $$F/A$$ on a solid is the definition of **stress**,
+and the relative elongation $$x/L$$ is the defintion of **strain**.
+If $$F$$ acts along the $$x$$-axis, we can then write:
 
 $$\begin{aligned}
     \boxed{
@@ -60,7 +60,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where the proportionality constant $E$,
+Where the proportionality constant $$E$$,
 known as the **elastic modulus** or **Young's modulus**,
 is the general material parameter that we wanted:
 
@@ -70,7 +70,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Due to the microscopic structure of some (usually crystalline) materials,
-$E$ might be dependent on the direction of the force $F$.
+$$E$$ might be dependent on the direction of the force $$F$$.
 For simplicity, we only consider **isotropic** materials,
 which have the same properties measured from any direction.
 
@@ -78,8 +78,8 @@ However, we are still missing something.
 When a spring is pulled,
 it becomes narrower as its coils move apart,
 and this effect is also seen when stretching solids in general:
-if we pull our rod along the $x$-axis, we expect it to deform in $y$ and $z$ as well.
-This is described by **Poisson's ratio** $\nu$:
+if we pull our rod along the $$x$$-axis, we expect it to deform in $$y$$ and $$z$$ as well.
+This is described by **Poisson's ratio** $$\nu$$:
 
 $$\begin{aligned}
     \boxed{
@@ -88,12 +88,12 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Note that $u_{yy} = u_{zz}$ because the material is assumed to be isotropic.
+Note that $$u_{yy} = u_{zz}$$ because the material is assumed to be isotropic.
 Intuitively, you may expect that the volume of the object is conserved,
 but for most materials that is not accurate.
 
-In summary, for our example case with a force $F = T A$ pulling at the rod
-along the $x$-axis, the full stress and strain tensors are given by:
+In summary, for our example case with a force $$F = T A$$ pulling at the rod
+along the $$x$$-axis, the full stress and strain tensors are given by:
 
 $$\begin{aligned}
     \hat{\sigma} =
@@ -124,9 +124,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where $\Tr{}$ is the trace.
+Where $$\Tr{}$$ is the trace.
 This is often written in index notation,
-with the Kronecker delta $\delta_{ij}$:
+with the Kronecker delta $$\delta_{ij}$$:
 
 $$\begin{aligned}
     \boxed{
@@ -135,10 +135,10 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The constants $\mu$ and $\lambda$ are called the **Lamé coefficients**,
-and are related to $E$ and $\nu$ in a way we can derive
-by returning to the example with a tension $T = F/A$ along $x$.
-For $\sigma_{xx}$, we have:
+The constants $$\mu$$ and $$\lambda$$ are called the **Lamé coefficients**,
+and are related to $$E$$ and $$\nu$$ in a way we can derive
+by returning to the example with a tension $$T = F/A$$ along $$x$$.
+For $$\sigma_{xx}$$, we have:
 
 $$\begin{aligned}
     T
@@ -150,7 +150,7 @@ $$\begin{aligned}
     &= \frac{T}{E} \Big( 2 \mu + \lambda (1 - 2 \nu) \Big)
 \end{aligned}$$
 
-Meanwhile, the other diagonal stresses $\sigma_{yy} = \sigma_{zz}$
+Meanwhile, the other diagonal stresses $$\sigma_{yy} = \sigma_{zz}$$
 are expressed in terms of the strain like so:
 
 $$\begin{aligned}
@@ -188,7 +188,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Which can straightforwardly be inverted
-to express $E$ and $\nu$ as a function of $\mu$ and $\lambda$:
+to express $$E$$ and $$\nu$$ as a function of $$\mu$$ and $$\lambda$$:
 
 $$\begin{aligned}
     \boxed{
@@ -212,14 +212,14 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Inserting this into Hooke's law
-yields an equation that only contains one strain component $u_{ij}$:
+yields an equation that only contains one strain component $$u_{ij}$$:
 
 $$\begin{aligned}
     \sigma_{ij}
     = 2 \mu u_{ij} + \frac{\lambda}{2 \mu + 3 \lambda} \delta_{ij} \sum_{k} \sigma_{kk}
 \end{aligned}$$
 
-Which is therefore trivial to isolate for $u_{ij}$,
+Which is therefore trivial to isolate for $$u_{ij}$$,
 leading us to Hooke's inverted law:
 
 $$\begin{aligned}