I've already seen a similar discussion in this forum (How to compute the strain rate tensor in non-Euclidean coordinates), but I still have problems. The strain tensor is defined in Euclidean/Cartesian coordinates as: $$ \epsilon_{ij} = \frac{1}{2}\left(\frac{\partial u_i}{\partial x_k} + \frac{\partial u_k}{\partial x_i}\right) $$ where $u = u(x,t)$ is the solid displacement function.
I'd like to get the components of this object in cylindrical coordinates. For example the $ \epsilon_{\theta \theta}$ is known to be: $$ \epsilon_{\theta \theta} = \frac{1}{r} \frac{\partial u_{\theta}}{\partial \theta} + \frac{1}{r} u_{r}.$$
In Euclidean/Cartesian coordinates the metric tensor is $ g_{uv} = diag(1,1,1) $ and the strain tensor can be rewritten as half of the Lie derivative of the metric along the solid displacement function u: $$ \epsilon_{uv}= \frac{1}{2}\left(Lg\right)_{uv} = \frac{1}{2}\left(u^{\alpha} \partial_{\alpha}g_{uv} + g_{u\alpha} \partial_{v}u^{\alpha} + g_{\alpha v} \partial_{u}u^{\alpha}\right) .$$
Since this object transforms as a tensor we can get the components of $\epsilon$ in every coordinate system. In cylindrical coordinates the new metric is $ g_{uv} = diag(1,r^2,1)$ and we can now get the theta-theta component:
$$ \epsilon_{\theta \theta} = \frac{1}{2} \left(u^r \partial_r\left(g_{\theta \theta}\right) + 2g_{\theta \theta}\partial_{\theta}u^{\theta}\right) = r u^r + r^2\frac{\partial u^{\theta}}{\partial \theta}. $$
This is completely different from what I've expected. What am I doing wrong?
