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Consider the following expression for order-2 tensor $T$ and vector $v$, in Cartesian coordinates, where $\nabla$ is the gradient operator, and $e_i$ is a unit vector in the $i$ direction.

$$T:\nabla v = T_{ij} e_i e_j : \frac{\partial v_m}{\partial x_n} e_n e_m = T_{ij}\frac{\partial v_m}{\partial x_n} e_i\cdot e_m e_j \cdot e_n=T_{ij}\frac{\partial v_m}{\partial x_n}\delta_{im}\delta_{jn}=T_{ij}\frac{\partial v_i}{\partial x_j}$$

However, I am told that that the above expression should yield $$T_{ij}\frac{\partial v_j}{\partial x_i}$$

Where am I going wrong here?

For reference, $T = T_{ij} e_i e_j, v = v_k e_k$, and so $\nabla v = \frac{\partial v_m}{\partial x_n} e_n e_m$.

megamence
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2 Answers2

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It seems to me that you should use the uniform approach to the tensors index order. If $T:= T_{ij} e_i e_j$, in this case $\nabla v = A_{nm}e_ne_m=\frac{\partial v_m}{\partial x_n} e_n e_m$ $T:\nabla v = T_{ij} e_i e_j : \frac{\partial v_m}{\partial x_n} e_n e_m = T_{ij}\frac{\partial v_m}{\partial x_n} (e_i\circ e_j),(e_n \circ e_m)=T_{ij}\frac{\partial v_m}{\partial x_n} (e_i,e_n)\circ(e_j, e_m)=T_{ij}\frac{\partial v_m}{\partial x_n}\delta_{in}\delta_{jm}=T_{ij}\frac{\partial v_j}{\partial x_i}$

Svyatoslav
  • 20,502
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I already addressed in a comment below my answer to your other question:$$e_i:e_j=e_i^Te_j=\delta_{ij},\,e_ie_j:e_ne_m=(e_ie_j)^Te_ne_m=e_j^Te_i^Te_ne_m=e_j^T\delta_{in}e_m=\delta_{in}\delta_{jm},$$in accordance with @Svyatoslav's answer, which uses a different, equivalent method.

J.G.
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