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I am studying advanced fluid mechanics and sometimes you see equations written in index notation like $$ Dv_i= \partial_t v_i +v_j\partial_jv_i$$ but sometimes you find this arrow/vector notation (what is the name of this notation?)

$$ D \vec{v} = \partial_t \vec{v} + (\vec{v}\cdot \vec{\nabla})\vec{v} $$ which is what I used when I learn vector calculus.

My problem arises when you have tensors, like the stress tensor, let us call it $\bar{\sigma}$. For example, one of Navier-Stokes equations (stationary flow) reads $$ (\vec{v}\cdot \vec{\nabla})\vec{v} = -\vec{\nabla}\cdot\bar{\sigma} = -\partial_j\sigma_{ij} $$ Is there a reasoning on why the divergence on 2-rank tensors acts on the right (as if $\nabla$ is a vector column) ? How does one apply this to higher order tensors?

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    $\begingroup$ I think the stress tensor is symmetric, but, indeed, one can have „divergence to the right” and „divergence to the left”. $\endgroup$
    – DanielC
    Mar 8, 2019 at 20:28
  • $\begingroup$ @DanielC how would you denote that? $\endgroup$
    – Mauricio
    Mar 8, 2019 at 23:47
  • $\begingroup$ @Mauricio with difficulty. $\endgroup$ Jul 26, 2019 at 23:09
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    $\begingroup$ stress tensor is not allways symetric: "However, in the presence of couple-stresses, i.e. moments per unit volume, the stress tensor is non-symmetric." - this situation appear e.g. for polar-fluind in electro-magnetic fields, for polymers - non-Newtonian fluids... $\endgroup$ Dec 31, 2019 at 10:01

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I think that the question was answered in the comments, but your main concern seems to be "how would you denote these in vector notation?".


My answer to this is either (1) you don't, or (2) if you must then you have the freedom to denote it any way you like. The reason for the fact that there is no standard agreement on a "vector" notation is because with tensors with rank greater than 1 it becomes much more confusing than it's worth.

For that reason I recommend option (1)


Example: Suppose you want to take the derivative w.r.t the second index of a tensor. Then you can either write

$$ \partial_{i_2} T^{i_1i_2 \cdots} \qquad or \qquad\vec{\mathcal{D}}\ \cdot \stackrel{\leftrightarrow}{T} $$

In my mind the second equation is essentially useless and above all confusing. The problem with the one on the right is that you are trying to package way too much information into a vector notation. That works find if you have a single index but loses this allure in proportion to how many indices your tensor has.

If you make any attempt to salvage the "vector" notation on the right, you will most likely invent the notation on the left, as it is superior in every single way.

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In this answer I use $x=x_1, y=x_2, z=x_3$ and Einstein notation. Lets take tensor A

$$A = \begin{bmatrix} a_{11} && a_{12} && a_{13} \\ a_{21} && a_{22} && a_{23} \\ a_{31} && a_{32} && a_{33} \\ \end{bmatrix} $$

On wikipedia in this article I found following information (in article they use S instead A) for cartesian coordinate system:

$$ \nabla\cdot A = \cfrac{\partial A_{ki}}{\partial x_k}~\mathbf{e}_i = A_{ki,k}~\mathbf{e}_i = \begin{bmatrix} \frac{\partial a_{11}}{\partial x} + \frac{\partial a_{21}}{\partial y} + \frac{\partial a_{31}}{\partial z} \\ \frac{\partial a_{12}}{\partial x} + \frac{\partial a_{22}}{\partial y} + \frac{\partial a_{32}}{\partial z} \\ \frac{\partial a_{13}}{\partial x} + \frac{\partial a_{23}}{\partial y} + \frac{\partial a_{33}}{\partial z} \\ \end{bmatrix} $$

The result is contravariant (column) vector. But in this article is mention that $\mathrm{div}(A) \neq \nabla\cdot A$ and

$$ \mathrm{div}(A) = \nabla\cdot A^T = \cfrac{\partial A_{ik}}{\partial x_k}~\mathbf{e}_i = A_{ik,k}~\mathbf{e}_i = \begin{bmatrix} \frac{\partial a_{11}}{\partial x} + \frac{\partial a_{12}}{\partial y} + \frac{\partial a_{13}}{\partial z} \\ \frac{\partial a_{21}}{\partial x} + \frac{\partial a_{22}}{\partial y} + \frac{\partial a_{23}}{\partial z} \\ \frac{\partial a_{31}}{\partial x} + \frac{\partial a_{32}}{\partial y} + \frac{\partial a_{33}}{\partial z} \\ \end{bmatrix} $$

When A is symetric: $a_{ij}=a_{ji}$ then $\mathrm{div}(A) = \nabla\cdot A$

Wiki also mention that some authors use alternative definition: $\nabla\cdot A = \cfrac{\partial A_{ik}}{\partial x_k}~\mathbf{e}_i $ probably only for case when A is symmetric (for which that alternative definition is equal to original). However alternative definition is NOT compatible with general curvilinear definition which I found on wiki too:

$$ \nabla\cdot A = \left(\cfrac{\partial A_{ki}}{\partial x_k}- A_{li}~\Gamma_{kk}^l - A_{kl}~\Gamma_{ki}^l\right)~\mathbf{g}^i $$

Currently I don't know what is exact definition of: 𝐠𝑖 - probably it will give something like 𝐞𝑖

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