# Taking derivatives in index notation

So I'm working out some calculus of variations problems however one of them involves a fair bit of index notation. I'm familiar with the algebra of these but not exactly sure how to perform derivatives etc. Here's the specific problem. I'm given $$\mathcal{L}[\mathbb{\eta}] = \frac{1}{2}\rho\dot{\eta_i}\dot{\eta_i} - \frac{1}{2}e_{ij}c_{ijkl}e_{kl}.$$ Repeated indices are summed over from 1 to 3, and each $\eta_i$ is function such that $\eta_i = \eta_i(t,x_1,x_2,x_3)$. $\dot{\eta_i}$ here is $\frac{\partial{\eta_i}}{\partial{t}}$, and $$e_{ij} = \frac{1}{2}(\frac{\partial{\eta_j}}{\partial{x_i}} + \frac{\partial{\eta_i}}{\partial{x_j}}) = \frac{1}{2}(\eta_{j,i} + \eta_{i,j}).$$ $c_{ijkl}$ is a tensor with the following symmetries $$c_{ijkl} = c_{klij} = c_{jikl} = c_{ijlk}.$$ I'm asked to show that upon using the equation $$\frac{\partial{\mathcal{L}}}{\partial{\eta_i}} = \frac{\partial}{\partial{t}}(\frac{\partial{\mathcal{L}}}{\partial{(\dot{\eta_i})}})+ \frac{\partial}{\partial{x_j}}(\frac{\partial{\mathcal{L}}}{\partial{(\eta_{i,j}})})$$ I should get $$\rho\frac{\partial^2{\eta_i}}{\partial{t^2}} - \frac{\partial}{\partial{x_j}}(c_{jikl}e_{kl}) = 0 .$$ Of course I could brute force it but that approach seems like it would be extremely unnecessary. So I would appreciate it if you could give me some examples of rules to follow to do such manipulations correctly and to get me started towards getting the correct equation. If anyone was wondering this problem is related to small-amplitude elastic deformations.

I would also appreciate some references which explain this sort of thing.

-
You should write $\dot{\eta}_i\dot{\eta}_i$, not $\dot{\eta}_i^2$. – enzotib Sep 16 '12 at 17:01

## 1 Answer

You have

\begin{align} \frac{\partial\mathcal L}{\partial\dot{\eta}_h} &=\frac{\partial}{\partial\dot{\eta}_h}\left(\frac{1}{2}\rho\dot\eta_i\dot\eta_i-\frac{1}{2}C_{ijkl}e_{ij}e_{kl}\right)=\\ &=\frac{1}{2}\rho\frac{\partial}{\partial\dot{\eta}_h}\left(\dot\eta_i\dot\eta_i\right)-0=\\ &=\frac{1}{2}\rho\left(\frac{\partial \dot\eta_i}{\partial\dot{\eta}_h}\dot\eta_i+\dot\eta_i\frac{\partial \dot\eta_i}{\partial\dot{\eta}_h}\right)=\\ &=\frac{1}{2}\rho\left(\delta_{ih}\dot\eta_i+\dot\eta_i\delta_{ih}\right)=\\ &=\frac{1}{2}\rho2\dot{\eta}_h=\\ &=\rho\dot{\eta}_h \end{align}

and similar calculation for other derivatives.

-