From 8044008e45f87b95d7a8c9f0fce1847ceedfb09a Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 3 Apr 2021 16:04:40 +0200
Subject: Expand knowledge base

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+---
+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.
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