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4 files changed, 364 insertions, 5 deletions

diff --git a/content/know/concept/cauchy-strain-tensor/index.pdc b/content/know/concept/cauchy-strain-tensor/index.pdc
index 2994674..f150723 100644
--- a/content/know/concept/cauchy-strain-tensor/index.pdc
+++ b/content/know/concept/cauchy-strain-tensor/index.pdc
@@ -54,7 +54,7 @@ $$\begin{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, these points respectively had these positions,
+Before the displacement, those points respectively had these positions,
 where we define $\va{A}$ as the "old" version of $\va{a}$:
 
 $$\begin{aligned}
@@ -159,7 +159,7 @@ $$\begin{aligned}
     = 2 \va{a} \cdot \hat{u} \cdot \va{b}
 \end{aligned}$$
 
-The Cauchy strain tensor $\hat{u}$ is a second-order tensor,
+The Cauchy strain tensor $\hat{u}$ is a second-rank tensor,
 and can alternatively be expressed like so:
 
 $$\begin{aligned}
@@ -320,7 +320,7 @@ we remove it, and isolate the rest for $\delta(\dd{\va{S}})$:
 $$\begin{aligned}
     \boxed{
         \delta(\dd{\va{S}})
-        = \big( (\nabla \cdot \va{u}) \va{1} - \nabla \va{u} \big) \cdot \dd{\va{S}}
+        = \big( (\nabla \cdot \va{u}) \hat{1} - \nabla \va{u} \big) \cdot \dd{\va{S}}
     }
 \end{aligned}$$
 
diff --git a/content/know/concept/cauchy-stress-tensor/index.pdc b/content/know/concept/cauchy-stress-tensor/index.pdc
index a26e2a8..080254d 100644
--- a/content/know/concept/cauchy-stress-tensor/index.pdc
+++ b/content/know/concept/cauchy-stress-tensor/index.pdc
@@ -108,7 +108,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The stress components $\sigma_{ij}$ can be written as a second-order tensor
+The stress components $\sigma_{ij}$ can be written as a second-rank tensor
 (i.e. a matrix that transforms in a certain way),
 called the **Cauchy stress tensor** $\hat{\sigma}$:
 
@@ -177,7 +177,6 @@ $$\begin{aligned}
     F_{s, i}
     = \oint_S \sum_j \sigma_{ij} \dd{S_j}
     = \int_V \sum_{j} \nabla_{\!j} \sigma_{ij} \dd{V}
-    = \int_V \nabla \cdot \vec{\sigma}_i \dd{V}
 \end{aligned}$$
 
 In any case, the total force $\va{F}$ can then be expressed
diff --git a/content/know/concept/hookes-law/index.pdc b/content/know/concept/hookes-law/index.pdc
new file mode 100644
index 0000000..94ceb1b
--- /dev/null
+++ b/content/know/concept/hookes-law/index.pdc
@@ -0,0 +1,245 @@
+---
+title: "Hooke's law"
+firstLetter: "H"
+publishDate: 2021-04-02
+categories:
+- Physics
+- Continuum physics
+
+date: 2021-03-09T17:31:37+01:00
+draft: false
+markup: pandoc
+---
+
+# Hooke's law
+
+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/) $\hat{u}$.
+
+Importantly, all forms of Hooke's law are only valid for small deformations,
+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 $x$:
+
+$$\begin{aligned}
+    \boxed{
+        F
+        = k x
+    }
+\end{aligned}$$
+
+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 $A$.
+
+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 $L$.
+
+$$\begin{aligned}
+    \frac{F}{A}
+    = \Big( k \frac{L}{A} \Big) \frac{x}{L}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \boxed{
+        \sigma_{xx}
+        = E u_{xx}
+    }
+\end{aligned}$$
+
+Where the proportionality constant $E$,
+known as the **elastic modulus** or **Young's modulus**,
+is the general material parameter that we wanted:
+
+$$\begin{aligned}
+    E
+    = k \frac{L}{A}
+\end{aligned}$$
+
+Due to the microscopic structure of some (usually crystalline) materials,
+$E$ 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.
+
+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** $\nu$:
+
+$$\begin{aligned}
+    \boxed{
+        \nu
+        \equiv - \frac{u_{yy}}{u_{xx}}
+    }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \hat{\sigma} =
+    \begin{bmatrix}
+        T & 0 & 0 \\
+        0 & 0 & 0 \\
+        0 & 0 & 0
+    \end{bmatrix}
+    \qquad
+    \hat{u} =
+    \begin{bmatrix}
+        T/E & 0 & 0 \\
+        0 & -\nu T/E & 0 \\
+        0 & 0 & -\nu T/E
+    \end{bmatrix}
+\end{aligned}$$
+
+
+## General isotropic form
+
+The general form of Hooke's law is a linear relationship
+between the stress and strain tensors:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\sigma}
+        = 2 \mu \: \hat{u} + \lambda \Tr\!(\hat{u}) \: \hat{1}
+    }
+\end{aligned}$$
+
+Where $\Tr{}$ is the trace.
+This is often written in index notation,
+with the Kronecker delta $\delta_{ij}$:
+
+$$\begin{aligned}
+    \boxed{
+        \sigma_{ij}
+        = 2 \mu u_{ij} + \lambda \delta_{ij} \sum_{k} u_{kk}
+    }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    T
+    = \sigma_{xx}
+    &= 2 \mu u_{xx} + \lambda (u_{xx} + u_{yy} + u_{zz})
+    \\
+    &= \frac{2 \mu}{E} T + \frac{\lambda}{E} T - \frac{\nu \lambda}{E} (T + T)
+    \\
+    &= \frac{T}{E} \Big( 2 \mu + \lambda (1 - 2 \nu) \Big)
+\end{aligned}$$
+
+Meanwhile, the other diagonal stresses $\sigma_{yy} = \sigma_{zz}$
+are expressed in terms of the strain like so:
+
+$$\begin{aligned}
+    0
+    = \sigma_{yy}
+    &= 2 \mu u_{yy} + \lambda (u_{xx} + u_{yy} + u_{zz})
+    \\
+    &= - \frac{2 \nu \mu}{E} T + \frac{\lambda}{E} T - \frac{\nu \lambda}{E} (T + T)
+    \\
+    &= \frac{E}{T} \Big( \!-\! 2 \nu \mu + \lambda (1 - 2 \nu) \Big)
+\end{aligned}$$
+
+After dividing out superfluous factors from the two preceding equations,
+we arrive at:
+
+$$\begin{aligned}
+    E
+    = 2 \mu + \lambda (1 - 2 \nu)
+    \qquad \quad
+    2 \nu \mu
+    = \lambda (1 - 2 \nu)
+\end{aligned}$$
+
+Solving this system of equations for the Lamé coefficients
+yields the following result:
+
+$$\begin{aligned}
+    \boxed{
+        \lambda
+        = \frac{E \nu}{(1 - 2 \nu)(1 + \nu)}
+        \qquad \quad
+        \mu
+        = \frac{E}{2 (1 + \nu)}
+    }
+\end{aligned}$$
+
+Which can straightforwardly be inverted
+to express $E$ and $\nu$ as a function of $\mu$ and $\lambda$:
+
+$$\begin{aligned}
+    \boxed{
+        E
+        = \mu \frac{3 \lambda + 2 \mu}{\lambda + \mu}
+        \qquad \quad
+        \nu
+        = \frac{\lambda}{2 (\lambda + \mu)}
+    }
+\end{aligned}$$
+
+Hooke's law itself can also be inverted,
+i.e. we can express the strain as a function of stress.
+First, observe that the trace of the stress tensor satisfies:
+
+$$\begin{aligned}
+    \Tr\!(\hat{\sigma})
+    = \sum_{i} \sigma_{ii}
+    = 2 \mu \sum_{i} u_{ii} + \lambda \sum_{i} \sum_{k} u_{kk}
+    = (2 \mu + 3 \lambda) \sum_{i} u_{ii}
+\end{aligned}$$
+
+Inserting this into Hooke's law
+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 $u_{ij}$,
+leading us to Hooke's inverted law:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            u_{ij}
+            &= \frac{\sigma_{ij}}{2 \mu} - \frac{\lambda}{2 \mu (3 \lambda + 2 \mu)} \delta_{ij} \sum_{k} \sigma_{kk}
+            \\
+            &= \frac{1 + \nu}{E} \sigma_{ij} - \frac{\nu}{E} \delta_{ij} \sum_{k} \sigma_{kk}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
diff --git a/content/know/concept/navier-cauchy-equation/index.pdc b/content/know/concept/navier-cauchy-equation/index.pdc
new file mode 100644
index 0000000..d3802d3
--- /dev/null
+++ b/content/know/concept/navier-cauchy-equation/index.pdc
@@ -0,0 +1,115 @@
+---
+title: "Navier-Cauchy equation"
+firstLetter: "N"
+publishDate: 2021-04-02
+categories:
+- Physics
+- Continuum physics
+
+date: 2021-04-02T15:04:55+02:00
+draft: false
+markup: pandoc
+---
+
+# Navier-Cauchy equation
+
+The **Navier-Cauchy equation** describes **elastodynamics**:
+the movements inside an elastic solid
+in response to external forces and/or internal stresses.
+
+For a particle of the solid, whose position is given by the displacement field $\va{u}$,
+Newton's second law is as follows,
+where $\dd{m}$ and $\dd{V}$ are the particle's mass and volume, respectively:
+
+$$\begin{aligned}
+    \va{f^*} \dd{V}
+    = \pdv[2]{\va{u}}{t} \dd{m}
+    = \rho \pdv[2]{\va{u}}{t} \dd{V}
+\end{aligned}$$
+
+Where $\rho$ is the mass density,
+and $\va{f^*}$ is the effective force density,
+defined from the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$
+like so, with $\va{f}$ being an external body force, e.g. from gravity:
+
+$$\begin{aligned}
+    \va{f^*}
+    = \va{f} + \nabla \cdot \hat{\sigma}^\top
+\end{aligned}$$
+
+We can therefore write Newton's second law as follows,
+while switching to index notation,
+where $\nabla_j = \pdv*{x_j}$ is the partial derivative
+with respect to the $j$th coordinate:
+
+$$\begin{aligned}
+    f_i + \sum_{j} \nabla_j \sigma_{ij}
+    = \rho \pdv[2]{u_i}{t}
+\end{aligned}$$
+
+The components $\sigma_{ij}$ of the Cauchy stress tensor
+are given by [Hooke's law](/know/concept/hookes-law/),
+where $\mu$ and $\lambda$ are the Lamé coefficients,
+which describe the material:
+
+$$\begin{aligned}
+    \sigma_{ij}
+    = 2 \mu u_{ij} + \lambda \delta_{ij} \sum_{k} u_{kk}
+\end{aligned}$$
+
+In turn, the components $u_{ij}$ of the
+[Cauchy strain tensor](/know/concept/cauchy-strain-tensor/)
+are defined as follows,
+where $u_i$ are once again the components of the displacement vector $\va{u}$:
+
+$$\begin{aligned}
+    u_{ij}
+    = \frac{1}{2} \big( \nabla_i u_j + \nabla_j u_i \big)
+\end{aligned}$$
+
+To derive the Navier-Cauchy equation,
+we start by inserting Hooke's law into Newton's law:
+
+$$\begin{aligned}
+    \rho \pdv[2]{u_i}{t}
+    %= f_i + \sum_{j} \nabla_j \sigma_{ij}
+    &= f_i + 2 \mu \sum_{j} \nabla_j u_{ij} + \lambda \sum_{j} \nabla_j \bigg( \delta_{ij} \sum_{k} u_{kk} \bigg)
+    \\
+    &= f_i + 2 \mu \sum_{j} \nabla_j u_{ij} + \lambda \nabla_i \sum_{j} u_{jj}
+\end{aligned}$$
+
+And then into this we insert the definition of the strain components $u_{ij}$, yielding:
+
+$$\begin{aligned}
+    \rho \pdv[2]{u_i}{t}
+    &= f_i + \mu \sum_{j} \nabla_j \big( \nabla_i u_j + \nabla_j u_i \big) + \lambda \nabla_i \sum_{j} \nabla_j u_{j}
+\end{aligned}$$
+
+Rearranging this a bit leads us to the Navier-Cauchy equation written in index notation:
+
+$$\begin{aligned}
+    \boxed{
+        \rho \pdv[2]{u_i}{t}
+        = f_i + \mu \sum_{j} \nabla_j^2 u_i + (\mu + \lambda) \nabla_i \sum_{j} \nabla_j u_j
+    }
+\end{aligned}$$
+
+Traditionally, it is written in vector notation instead,
+in which case it looks like this:
+
+$$\begin{aligned}
+    \boxed{
+        \rho \pdv[2]{\va{u}}{t}
+        = \va{f} + \mu \nabla^2 \va{u} + (\mu + \lambda) \nabla (\nabla \cdot \va{u})
+    }
+\end{aligned}$$
+
+A special case is the **Navier-Cauchy equilibrium equation**,
+where the left-hand side is just zero.
+That version describes **elastostatics**: the deformation of a solid at rest.
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.