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+---
+title: "Hooke's law"
+date: 2021-04-02
+categories:
+- Physics
+- Continuum physics
+layout: "concept"
+---
+
+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.
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