# Navier-Stokes : divergence or covariant derivative of a tensor : 1 vector result?

I don't understand very well the following definition concerning Navier-Stokes equation :

where

$$\vec{u}\otimes\vec{v}$$ is a tensor (2,0), isn't it ?

This is not scalar since $$\vec{u}\,\vec{v}^{T}$$ is not a scalar product and there are also vectors in the others terms.

So, should I consider $$\nabla \cdot(\rho \vec{u}\otimes\vec{u})$$ like the divergence of a tensor (2,0) (I don't grasp this operation) or the covariant derivative of a tensor (2,0) : but in this case, I get also a tensor (2,0), and this is not a vector

If someone coud epxlain me, this would be fine, Regards

• \cdot is how you produce a centered dot by the way – Triatticus May 8 '19 at 23:00

It is indeed the divergence, as you can tell by the dot product. You can think of it as simply the matrix product of the row vector

$$\nabla = (\partial_x, \partial_y, \partial_z)$$

with the matrix $$\rho\, \mathbf{u} \otimes \mathbf{u}$$. In component notation, we would write this as $$\sum_j \partial_j (\rho u_j u_i)$$, which you can check leads to the same result. I am assuming for simplicity that we are using Cartesian coordinates in Euclidean space, so no covariant derivatives.

Note that there is a bit of a potential ambiguity here: if we want to take the divergence of some 2-tensor $$\mathbf{T}$$, do we multiply by $$\nabla$$ on the left or on the right? In component language: do we contract with the first index (giving $$\partial_j T_{ji}$$) or with the second (giving $$\partial_j T_{ij}$$)? Well, there is no difference if $$T_{ij}$$ is symmetric, which is the case here: $$\mathbf{u}\otimes\mathbf{u}$$ is a symmetric tensor, so $$\nabla \cdot (\rho\, \mathbf{u} \otimes \mathbf{u})$$ can only mean one thing. But if the tensor is not symmetric, the dot-product notation can be dangerous.

• thanks for your answer. Just a last precision, In english mathematical domain, is the scalar product is the same signification as the dot product ? – youpilat13 May 9 '19 at 9:51
• @faya13 Yes, they mean the same thing. – Javier May 9 '19 at 11:05

Indeed, it is the divergence of a tensor. You must interpret this as pure notation: it is just a short way to write a sum of many derivative terms.

The mnemonic rule is just to write what you see. Interpret "nabla" as an operator:

$$\nabla=\left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)$$

So that

$$\nabla \cdot \left(\begin{array}{ccc} a & b & c \\d & e & f \\g & h & i \end{array}\right) =\left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)\cdot \left(\begin{array}{ccc} a & b & c \\d & e & f \\g & h & i \end{array}\right)$$

And then you would perform a matrix product normally. The only change is that a multiplication of "derivative times letter" is actually "derivative of the letter".

$$\left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)\cdot \left(\begin{array}{ccc} a & b & c \\d & e & f \\g & h & i \end{array}\right)=\left( \frac{\partial a}{\partial x}+\frac{\partial d}{\partial y}+\frac{\partial g}{\partial z}, \qquad \frac{\partial b}{\partial x}+\frac{\partial e}{\partial y}+\frac{\partial h}{\partial z}, \qquad \frac{\partial c}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial i}{\partial z} \right)$$

So you've got a sum of 3 derivatives per component, which you can group in a single vector.

Note. This comes from the fact that, considering an Eulerian formulation of the fluid, the rate of change of variables varies not just locally, but also due to the displacement of the flux itself. This is the Reynodls transport theorem, application of the chain rule:

$$\frac{d \phi}{dt}=\frac{\partial\phi}{\partial t}+ \vec{v}\cdot(\vec{\nabla}\phi); \qquad \forall\phi$$

And the above $$\vec{v}\otimes \vec{v}$$ term is just one way of rewriting this (in the case of $$\phi=\rho\vec{v}$$, momentum)

• thanks, it's clearer for me, regards – youpilat13 May 9 '19 at 9:55
• Sorry to bore you again but I just realized that in Reynolds transport theorem (equation juste above), the gradient appears only just on $\Phi = \rho \vec{v}$ and not like $\nabla \cdot(\rho \vec{u}\otimes\vec{u})$ like in my original question at the beginning of my post : could you clarify or explain please this difference ? Regards – youpilat13 Nov 20 '19 at 17:33
• Sure, no problem. For momentum conservation, each component has a substantial derivative. For example: $\partial_t (\rho v_x) + \partial_x (\rho v_x^2) + \partial_y (\rho v_x v_y) + \partial_z (\rho v_x v_z )$ This is equivalent to the $\otimes$ term, when you join the 3 equations. – FGSUZ Nov 20 '19 at 21:29
• ok, thanks but how to prove that $\vec{v}\cdot(\vec{\nabla}\phi) = \nabla \cdot(\phi \otimes\vec{v}) \quad \forall\phi=\rho\vec{v}$ ? – youpilat13 Nov 20 '19 at 22:51
• @youpilat13 Since this will only be used for momentum equation, you can prove it manually, checking that both forms are the same. Another option is using index notation. Hope this helps. If not, you can ask another question, where I (and other people) can give longer answers haha. – FGSUZ Nov 21 '19 at 0:20