3

I'm having a hard time finding a source with an actual derivation on this. I understand the divergence of a vector field in cylindrical coordinates. However, for a tensor, how do I go from this

$$\nabla \cdot \mathbf{S} = \left( \hat{r} \frac{\partial}{\partial r} + \hat{\theta} \frac{1}{r} \frac{\partial}{\partial \theta} + \hat{z} \frac{\partial}{\partial z}\right) \cdot \begin{pmatrix} S_{rr} \: \hat{r} \otimes \hat{r} + S_{r \theta} \: \hat{r} \otimes \hat{\theta} + S_{rz} \: \hat{r} \otimes \hat{z}\\ + S_{\theta r} \: \hat{\theta} \otimes \hat{r} + S_{\theta \theta} \: \hat{\theta} \otimes \hat{\theta} + S_{\theta z} \: \hat{\theta} \otimes \hat{z}\\ + S_{z r} \: \hat{z} \otimes \hat{r} + S_{z \theta} \: \hat{z} \otimes \hat{\theta} + S_{z z} \: \hat{z} \otimes \hat{z}\end{pmatrix}$$

to this

$$\nabla \cdot \mathbf{S} = \begin{bmatrix} \frac{\partial S_{rr}}{\partial r} + \frac{S_{rr}}{r} + \frac{1}{r} \frac{\partial S_{\theta r}}{\partial \theta} + \frac{\partial S_{zr}}{\partial z} - \frac{S_{\theta \theta}}{r} \\ \frac{1}{r} \frac{\partial S_{\theta \theta}}{\partial \theta} + \frac{\partial S_{r \theta}}{\partial r} + \frac{S_{r \theta}}{r} + \frac{S_{\theta r}}{r} + \frac{\partial S_{z \theta}}{\partial z}\\ \frac{\partial S_{z z}}{\partial z} + \frac{\partial S_{r z}}{\partial r} + \frac{S_{rz}}{r} + \frac{1}{r}\frac{\partial S_{\theta z}}{\partial \theta}\end{bmatrix}$$

?

I understand why the term $- \frac{S_{\theta \theta}}{r}$ appears (because the unit vectors are differentiated, which does not give zero), but I can not find the way to execute this calculation. I probably am missing some tensor algebra rules. How exactly is it done?

I care especially for the top row and the $- \frac{S_{\theta \theta}}{r}$ term. $\hat{r}$, $\hat{\theta}$ and $\hat{z}$ are supposed to be unit vectors, obviously.

Felix
  • 33
  • Please use mathJax. – Kurt G. Mar 23 '23 at 19:53
  • @KurtG. Now nicely formatted. – Felix Mar 23 '23 at 20:08
  • https://math.stackexchange.com/questions/833179/covariant-derivative-in-cylindrical-coordinates/4643576#4643576 – K.defaoite Mar 23 '23 at 20:37
  • @K.defaoite are there other ways to explain it? I am an engineer ... – Felix Mar 23 '23 at 21:09
  • Is what you're looking for the definition of dot product here? It ought to be $u\cdot(v\otimes w) = (u\cdot v)w$. The derivatives can be applied before hand with the product rule because the product rule applies to any bilinear product like $\otimes$. So for instance: – Nicholas Todoroff Mar 23 '23 at 22:02
  • $$(\hat r\partial_r)\cdot(S_{r\theta}\hat r\otimes\hat\theta) = \hat r\cdot((\partial_r S_{r\theta})\hat r\otimes\hat\theta + S_{r\theta}(\partial_r\hat r)\otimes\hat\theta + S_{r\theta}\hat r\otimes(\partial_r\hat\theta)) = (\partial_r S_{r\theta})(\hat r\cdot\hat r)\hat\theta + S_{r\theta}(\hat r\cdot\partial_r\hat r)\hat\theta + S_{r\theta}(\hat r\cdot\hat r)\partial_r\hat\theta.$$ – Nicholas Todoroff Mar 23 '23 at 22:02
  • And in this case the second and third term drop out (derivative is zero) and in the first the unit vectors become $\left(\hat{r} \cdot \hat{r} \right) \hat{\theta} = 1, \hat{\theta}$ therefore it ends up in the second row? – Felix Mar 23 '23 at 22:16
  • @Felix Yes, correct. This is precisely the second term you see in the second row. – Nicholas Todoroff Mar 24 '23 at 01:51
  • @NicholasTodoroff You are a saint. I might seem like a simpleton to you now, but I find this amazing and it works! Can you please provide your comment as a normal answer so I can accept it? I am sure there will be more students who will find this question useful. – Felix Mar 24 '23 at 12:59
  • @NicholasTodoroff I'll wait a few more days and then answer the question myself, referring to your solution in the comments. – Felix Mar 26 '23 at 11:01

1 Answers1

2

The dot product, as best as I can guess, is meant to be a left tensor contraction so that $$ u\cdot(v\otimes w) = (u\cdot v)w. $$ Because the tensor product is bilinear the product rule for the derivatives $\partial_r, \partial_\theta, \partial_z$ is valid (where I've used the abbreviated notation $\partial_a = \partial/\partial a$). So for instance we can compute $$\begin{aligned}{} [(\hat\theta/r)\partial_\theta]\cdot[S_{\theta\theta}\hat\theta\otimes\hat\theta] &= \frac1r\hat\theta\cdot\Bigl[(\partial_\theta S_{\theta\theta})\hat\theta\otimes\hat\theta + S_{\theta\theta}(\partial_\theta\hat\theta)\otimes\hat\theta + S_{\theta\theta}\hat\theta\otimes(\partial_\theta\hat\theta)\Bigr] \\ &= \frac1r\Bigl[(\partial_\theta S_{\theta\theta})(\hat\theta\cdot\hat\theta)\hat\theta + 0 - S_{\theta\theta}(\hat\theta\cdot\hat\theta)\hat r\Bigr] \\ &= \frac{\partial_\theta S_{\theta\theta}}r\hat\theta - \frac{S_{\theta\theta}}r\hat r. \end{aligned}$$ This is the origin of the particular term $-S_{\theta\theta}/r$.

Nicholas Todoroff
  • 6,543