Rigid observer calculus and objective strain rate

Let $I\subseteq \mathbb{R}$ be a nonempty open interval. For $r=1,2$, $C^r$ means that all coordinate partial derivatives through total order $r$ exist and are continuous. Let $U\subseteq {\mathbb{R}}^3$ be open and let $\chi \in C^2(I\times U,{\mathbb{R}}^3)$ be such that every slice ${\chi }_t:U\to {\chi }_t(U)$ is a bijection with $C^2$ inverse and $J(t,a):=\det D_a\chi (t,a)>0$, where $D_a\chi$ is the matrix of reference-coordinate partial derivatives. Write $H(t,a):=(t,\chi (t,a))$ and $\mathrm{\Omega }:=\{(t,x):t\in I,x\in {\chi }_t(U)\}$. The map $H$ is injective and its derivative is block triangular with determinant $J>0$, so the inverse function theorem gives local $C^1$ inverses; injectivity makes them agree on overlaps. Hence $\mathrm{\Omega }$ is open and the joint inverse $A(t,x):={\chi }_t^{-1}(x)$ is $C^1$. Define $u(t,x):={\mathrm{\partial }}_t\chi (t,A(t,x))$, which is $C^1$. For any $C^1$ vector field $f$ on $\mathrm{\Omega }$, let $D_{x}f$ be its matrix of spatial coordinate partial derivatives and define $D_{t}f:={\mathrm{\partial }}_{t}f+(D_{x}f)u$. Set $L:=D_{x}u$, $D:=(L+L^{\mathsf{T}})\mathrm{/}2$, $W:=(L-L^{\mathsf{T}})\mathrm{/}2$, $\theta :=\mathrm{tr}L$, and $a_m:=D_{t}u$. Here $\mathrm{tr}M:={\sum }_i{M}_{ii}$ for a square matrix $M$, $I_3$ denotes the $3\times 3$ identity matrix, and

$SO(3):=\{Q\in {\mathbb{R}}^{3\times 3}:Q^{\mathsf{T}}Q=I_3,\det Q=1\}\mathrm{.}$

Let $Q\in C^2(I,SO(3))$ and $c\in C^2(I,{\mathbb{R}}^3)$, hold time fixed, and change spatial coordinates by $x^{\ast }=Q(t)x+c(t)$. This map preserves Euclidean distances and orientation because $Q^{\mathsf{T}}Q=I_3$ and $\det Q=1$; the coordinate change does not assert that the observer is inertial. Set ${\chi }^{\ast }(t,a):=Q(t)\chi (t,a)+c(t)$ and ${\mathrm{\Omega }}^{\ast }:=\{(t,x^{\ast }):t\in I,x^{\ast }={\chi }^{\ast }(t,a)\text{ for some }a\in U\}$. Define the transformed velocity, material derivative, rate tensors and acceleration from ${\chi }^{\ast }$ by the same formulas on ${\mathrm{\Omega }}^{\ast }$; in particular, $D_t^{\ast }f={\mathrm{\partial }}_{t}f+(D_{x^{\ast }}f)u^{\ast }$ and $a_m^{\ast }:=D_t^{\ast }{u}^{\ast }$. For corresponding points $x=\chi (t,a)$ and $x^{\ast }={\chi }^{\ast }(t,a)$, the transformed fields satisfy

$u^{\ast }(t,x^{\ast })=Q^{\mathrm{\prime }}(t)x+Q(t)u(t,x)+c^{\mathrm{\prime }}(t),L^{\ast }(t,x^{\ast })=Q(t)L(t,x)Q(t{)}^{\mathsf{T}}+Q^{\mathrm{\prime }}(t)Q(t{)}^{\mathsf{T}},$

$D^{\ast }(t,x^{\ast })=Q(t)D(t,x)Q(t{)}^{\mathsf{T}},W^{\ast }(t,x^{\ast })=Q(t)W(t,x)Q(t{)}^{\mathsf{T}}+Q^{\mathrm{\prime }}(t)Q(t{)}^{\mathsf{T}},{\theta }^{\ast }(t,x^{\ast })=\theta (t,x)\mathrm{.}$

Thus $D$ transforms objectively, meaning its transformed tensor components obey $D^{\ast }=QD{Q}^{\mathsf{T}}$ under every such $Q,c$; $W$ acquires the observer spin $Q^{\mathrm{\prime }}{Q}^{\mathsf{T}}$ and is not objective under general time-dependent rotations. If $a_m:=D_{t}u$ is the material acceleration, then

$a_m^{\ast }(t,x^{\ast })=Q(t)a_m(t,x)+2Q^{\mathrm{\prime }}(t)u(t,x)+Q^{\mathrm{\prime }\mathrm{\prime }}(t)x+c^{\mathrm{\prime }\mathrm{\prime }}(t)\mathrm{.}$

When $Q^{\mathrm{\prime }}=0$ and $c^{\mathrm{\prime }\mathrm{\prime }}=0$, the acceleration formula reduces to $a_m^{\ast }=Q{a}_m$. These are mathematical kinematic identities and assert no force law or balance equation.

R. Abeyaratne, Continuum Mechanics, Volume II, §3.8 (printed pp. 89–93; equations (3.96)–(3.102))