I'm trying to figure out how to carry a tensorial operation. I am not sure if I implemented correctly, as the results I get are not what I was expecting. I found these two representations in different documents:


$\mathbf{H}=(\mathbf{\epsilon}\cdot \mathbf{ z})\mathbf{ z} $

$\mathbf{h}=\mathbf{H}: \mathbf{\dot\gamma} $


$\mathbf{h}=\mathbf{\epsilon}_{ijl} \mathbf{ z}_l\mathbf{ z}_k \mathbf{\dot\gamma}_{jk} $

Where $\epsilon$ is the third order permutation tensor, $\mathbf z$ and $\mathbf h$ are vectors and $\dot \gamma $ and $\mathbf H $ second order tensors. I implemented B as:

$ \mathbf{h}_{i} = \sum_j \sum _l \sum _k \mathbf{\epsilon}_{ijl} \mathbf{ z}_l\mathbf{ z}_k \mathbf{\dot\gamma}_{jk} $

A does not make much sense to me, because the dot product within the brackets would yield a second order tensor, and I dont know how you can get a second order tensor from a second order tensor and a vector. Furthermore, the double inner product between two order tensors should yield an scalar and not a vector. Am I correct?

If you are wondering, the context is fluid mechanics, particularly to find the hydrodynamic torque on an elongated particle, it is the contribution to the torque from the rate of strain ($\dot \gamma$) which tries to align the particle along its principal direction. $\mathbf{z}$ is the normalized particle orientation. I see it as the projection of the dyadic product of the orientation vector to the rate of strain tensor.



Regarding the notation:

.... am I correct?


$(\epsilon\cdot \mathbf{z})$ is as you said, a second order tensor. But then $(\epsilon\cdot\mathbf{z})\mathbf{z}$ is a third order tensor: you are taking the tensor product between a second order one with a first order one (a vector) and so end up with a third order one (2+1 = 3). In particular, the assertion that $\mathbf{H}$ is a second order tensor is wrong.

Then you can take $\mathbf{H}:\dot{\gamma}$, where the $:$ operation means inner product on two indices, and finally get a vector as desired.

  • $\begingroup$ That makes sense. Thanks. $\endgroup$ – jcperezma Aug 11 '16 at 1:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.