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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: \begin{equation} \frac{\partial\mathbf{E}}{\partial\mathbf{F}}:\delta\mathbf{F} \end{equation} 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: \begin{equation} \frac{\partial E_{i}^{.j}}{\partial F_{k}^{.l}}\delta F_{j}^{.l} \end{equation}

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.

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    $\begingroup$ Here is the equation for the differential of the strain in both notations $$dE=\frac{\partial E}{\partial F}:dF\,\,\implies\,\,dE_i^{.j}=\frac{\partial E_i^{.j}}{\partial F_p^{.q}}\,\,dF_p^{.q}$$ $\endgroup$
    – greg
    Apr 30, 2018 at 6:12
  • $\begingroup$ 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? $\endgroup$
    – Dave
    Apr 30, 2018 at 19:18

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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.

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

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