# Index notation of double contraction with second order tensor derivative

I'm trying to wrap my head around an equation that involves the derivative of a second order tensor valued function of a second order tensor, then double-dot producted with another second order tensor: $$\frac{\partial\mathbf{E}}{\partial\mathbf{F}}:\delta\mathbf{F}$$ The tensor $\mathbf{F}$ is a gradient (so contra- vs. co-variant comes into this somehow, probably), and $\delta\mathbf{F}$ is an infinitesimal change in $\mathbf{F}$. I'm trying to reconcile this with what I've been learning about index notation, especially with respect to upper vs. lower indices, and which indices are summed over. E.g. an almost-definitely-wrong guess: $$\frac{\partial E_{i}^{.j}}{\partial F_{k}^{.l}}\delta F_{j}^{.l}$$

For some background, this equation is from Finite Element Method Simulation of 3D Deformable Solids, Sifakis & Barbic, 2015. $\mathbf{F}$ is the deformation gradient tensor of continuum mechanics, and the full equation reads $\mathbf{E}=\frac{1}{2}(\mathbf{F}^T\mathbf{F} - \mathbf{I})$, so that $\frac{\partial\mathbf{E}}{\partial\mathbf{F}}:\delta\mathbf{F} = \frac{1}{2}(\delta\mathbf{F}^T\mathbf{F} + \mathbf{F}^T\delta\mathbf{F})$. I figure I may understand how one gets from the lhs to the rhs if I can play with the equation in index form.

• Here is the equation for the differential of the strain in both notations $$dE=\frac{\partial E}{\partial F}:dF\,\,\implies\,\,dE_i^{.j}=\frac{\partial E_i^{.j}}{\partial F_p^{.q}}\,\,dF_p^{.q}$$
– greg
Apr 30, 2018 at 6:12
• That's great, thanks! Could you perhaps turn that into an answer, and even better, explain how you chose the indices to sum over and which are upper and lower?
– Dave
Apr 30, 2018 at 19:18

## 1 Answer

Consider the vectors
\eqalign{ e &= {\rm vec}(E) \cr f &= {\rm vec}(F) \cr } Because of our familiarity with matrices, writing the differential in each notation is obvious \eqalign{ de &= \frac{\partial e}{\partial f}\cdot df \cr de_i &= \frac{\partial e_i}{\partial f_p}\,df_p \cr } Note that the independent variable $f$ appears in the denominator of the gradient and in the differential on the RHS. And the contraction occurs between those two terms. The index $(i)$ on the dependent variable remains free.

If we replace the vectors by their underlying matrices, each index becomes two indices, $(i\rightarrow i,j)$ and $(p\rightarrow p,q),\,$ but now $(i,j)$ are free and $(p,q)$ are contracted. \eqalign{ dE &= \frac{\partial E}{\partial F}: dF \cr dE_{ij} &= \frac{\partial E_{ij}}{\partial F_{pq}}\,dF_{pq} \cr } And if you're working in a domain where covariant-vs-contravariant has any consequences, you can raise the appropriate indices.

• Excellent, thanks, just what I needed!
– Dave
May 2, 2018 at 11:58