I recently come across a paper in which the notation of some equation confuses me a lot. Let's say, if I have an expression represented by delta $\delta_{jk},\delta_{jl}$, infinitesimal strain tensor $e_{jk}, e_{jl}$, and some components of a vector ($x^0_{k}, x^0_l$):


And if I differentiate this with respect to one component of strain, say $e_{\alpha \beta}$, it gives me:

$\frac{\partial s}{\partial e_{\alpha \beta}}=2(\delta_{\alpha k}+e_{\alpha k})x^0_kx^0_\beta$-----------------------(2)

How can I get (2) based on (1)? could anybody give me some details? Thank you!



$A(k)_j = (\delta_{jk}+e_{jk})x_k $

Now differentiate w.r.t $e_{ab}$:

$A(k)'_j= \delta_{ja}x_b $



so product rule says:

$s'=A(k)A'(l) + A'(k)A(l)$

Now plug and chug:

$s=(\delta_{jk}+e_{jk})x_k\delta_{ja}x_b + \delta_{ja}x_b (\delta_{jl}+e_{jl})x_l$


$s'=\delta_{aj}(\delta_{jk}+e_{jk})x_kx_b + \delta_{aj}(\delta_{jl}+e_{jl})x_lx_b$

Note that dummy contractions on $k$ and $l$ are separate, so you can combine the terms:


Now contract Kronecker deltas:

$s'=2(\delta_{ak} + e_{ak})x_kx_b$

Add superscripts and convert latin (a, b) to greek, as needed.

  • $\begingroup$ Thanks! My solution is also attached below, I miss the fact that $\frac{\partial e_{ij}}{\partial e_{lk}}=\delta_{li}\delta_{jk}$ $\endgroup$ – Sizhe Nov 2 '17 at 2:41

Using the fact that $\frac{\partial}{\partial e_{\alpha \beta}}(e_{ij})=\delta_{i\alpha}\delta_{j\beta}$:

$$\begin{align}\frac{\partial}{\partial e_{\alpha \beta}}(s)&=\frac{\partial}{\partial e_{\alpha \beta}}(\delta_{jk}x_k^0e_{jl}x_l^0)+\frac{\partial}{\partial e_{\alpha \beta}}(e_{jk}x_k^0\delta_{jl}x_l^0)+\frac{\partial}{\partial e_{\alpha \beta}}(e_{jk}x_k^0e_{jl}x_l^0)\\&=\delta_{jk}x_k^0\delta_{\alpha j}\delta_{\beta l}x_l^0+\delta_{j\alpha}\delta_{\beta k}x_k^0\delta_{jl}x_l^0+2\delta_{\alpha j}\delta_{\beta k}x_k^0e_{jl}x_l^0\\&=x_k^0\delta_{\alpha k}x_{\beta}^0+x_\beta^0\delta_{\alpha k}x_k^0+2x_\beta^0e_{\alpha k}x_k^0\\&=2(\delta_{\alpha k}+e_{\alpha k})x_{\beta}^0x_k^0\end{align}$$


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