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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:

A:

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

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

B:

$\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.

Thanks!

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Regarding the notation:

.... am I correct?

No.

$(\epsilon\cdot \mathbf{z})$ 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.

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

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