# Why does the velocity gradient tensor have additive decomposition but the deformation gradient does not?

In describing the decomposition of the deformation gradient tensor given by $$\mathbf{F}=F_{iJ}=\partial x_i/\partial X_J$$, Malvern (1969, Introduction to the mechanics of a continuous medium) mentions that due to the finite nature of the displacements implied in the definition of $$F_{iJ}$$, it does not have an additive decomposition as the infinitesimal displacement gradient tensor ($$e_{ij}$$) does, given by $$e_{ij}=\partial u_i/\partial x_j=\epsilon_{ij}+\omega_{ij}\\\epsilon_{ij} = (e_{ij}+e_{ji})/2\\\omega_{ij} = (e_{ij}-e_{ji})/2.$$$$F_{iJ}$$ and $$e_{ij}$$ are defined in the reference coordinate system. Instead $$F_{iJ}$$ has a polar decomposition given by $$\mathbf{F}=\mathbf{R}\cdot\mathbf{U}$$, where $$\mathbf{R}$$ is the rotation tensor and $$\mathbf{U}$$ is the right stretch tensor. However, in between the decomposition proofs of $$\mathbf{F}$$ and $$e_{ij}$$, Malvern (1969) does an additive decomposition of the velocity gradient tensor $$v_{ij}$$, which is defined in spatial coordinates under finite deformation, by $$v_{ij}=\partial v_i/\partial x_j=d_{ij}+w_{ij}\\d_{ij} = (v_{ij}+v_{ji})/2\\w_{ij} = (v_{ij}-v_{ji})/2,$$ where $$d_{ij}$$ is the rate-of-deformation tensor and $$w_{ij}$$ is the vorticity tensor. My question is, if under finite deformation an additive decomposition is not possible, why does $$v_{ij}$$ have an additive decomposition instead of polar decomposition? This is done similarly in other textbooks. I am not a physicist so a thorough explanation is appreciated.

• I suspect it's simply because $v_j = dx_j/dt$, and $dx_j$ is basically the same thing as $u_j$, whose gradient tensor does have a linear decomposition. But I don't know how to state this with more rigor (at least not off the top of my head.) – Michael Seifert Aug 4 '20 at 20:20
• That crossed my head but if it were the case there would be no physical basis to relate the velocity gradient $v_{ij}$ and the deformation gradient $\mathbf{F}$ as is done in $v_{ij}=\mathbf{\dot{F}}\cdot\mathbf{F}^{-1}$. – Rodrigues Aug 4 '20 at 20:34

## 1 Answer

In the non-infinitesimal case, we have \begin{align} E_{ij} &= C_{ij}-\delta_{ij}= F^T_{ik}F_{kj}-\delta_{ij}= \frac{\partial x_k}{\partial X_i}\frac{\partial x_k}{\partial X_j}-\delta_{ij}=\\ &= \frac{\partial}{\partial X_i}(X_k+u_k)\frac{\partial}{\partial X_j}(X_k+u_k)-\delta_{ij}=\\ &= \left(\delta_{ik}+\frac{\partial u_k}{\partial X_i}\right)\left(\delta_{jk}+\frac{\partial u_k}{\partial X_j}\right)-\delta_{ij}= \\ &= \frac{\partial u_k}{\partial X_i}+\frac{\partial u_k}{\partial X_j}+\frac{\partial u_k}{\partial X_i}\frac{\partial u_k}{\partial X_j} \end{align} In the infinitesimal case $$X_i\approx x_i$$ and the product term is small, the relation reduces to \begin{align} E_{ij}\approx \frac{\partial u_k}{\partial x_i}+\frac{\partial u_k}{\partial x_j}=e_{ij} \end{align}