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The left part of Navier-Stokes equation is: $\dfrac{D\vec{v}}{D t}= \dfrac{\partial\vec{v}}{\partial t}+ \vec{v} \cdot\nabla \vec{v}$

Let's take $\vec{v}$ as a two dimentional vector: $(u,v)$. Then: $\dfrac{\partial\vec{v}}{\partial t}$ is $(\dfrac{\partial u}{\partial t}, \dfrac{\partial v}{\partial t})$, which is a vector. However, $\vec{v} \cdot\nabla \vec{v}$ will be $(u,v)\cdot(\dfrac{\partial u}{\partial x},\dfrac{\partial v}{\partial y}) = u\dfrac{\partial u}{\partial x}+v\dfrac{\partial v}{\partial y}$, which is a scalar. How a "vector" can be summed with a "scalar"?

Of course my thought is wrong somewhere. Please help me.

T X
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  • directional derivative – timur Dec 10 '18 at 07:40
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    $\nabla\vec{v}$ is not $(\dfrac{\partial u}{\partial x},\dfrac{\partial v}{\partial y})$, it is the rank-2 tensor with entries $\dfrac{\partial u}{\partial x},\dfrac{\partial u}{\partial y},\dfrac{\partial v}{\partial x},\dfrac{\partial v}{\partial y}$ – user10354138 Dec 10 '18 at 07:46
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    @user10354138 -- Then the dot must be interpreted as tensor contraction. I guess that produces the same results, but it's probably not what was meant by the equation. – mr_e_man Dec 10 '18 at 07:51
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    @mr_e_man Yes it is what is meant by the equation, in the definition of $\frac{D}{Dt}=\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla$ as the material derivative. – user10354138 Dec 10 '18 at 08:16
  • @user10354138 -- I mean it's probably supposed to be $(v\cdot\nabla)v$ instead of $v\cdot(\nabla v)$. – mr_e_man Dec 10 '18 at 19:18
  • @mr_e_man Aren't they equivalent? – Dylan Dec 11 '18 at 07:47
  • @Dylan -- Yes, but the first only uses vector operations, while the second uses tensors. The author could have written it without knowing, or expecting the reader to know, about tensors or matrices. – mr_e_man Dec 15 '18 at 00:14

2 Answers2

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As stated in the comments, this is the material derivative. If $\mathbf{v} = (v_1,v_2)$ then

$$ (\mathbf{v}\cdot \nabla) \mathbf{v} = (\mathbf{v}\cdot \nabla v_1, \mathbf{v}\cdot \nabla v_2) = \left(v_1 \frac{\partial v_1}{\partial x} + v_2\frac{\partial v_1}{\partial y}, v_1\frac{\partial v_2}{\partial x} + v_2\frac{\partial v_2}{\partial y}\right) $$

You can also intepret it as a matrix product

$$ \mathbf{v}\cdot \nabla \mathbf{v} = \begin{bmatrix} v_1 & v_2 \end{bmatrix}\begin{bmatrix} \dfrac{\partial v_1}{\partial x} & \dfrac{\partial v_2}{\partial x} \\ \dfrac{\partial v_1}{\partial y} & \dfrac{\partial v_2}{\partial y} \end{bmatrix} $$

where $\nabla \mathbf{v}$ represents a rank-2 tensor

Dylan
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You're right in that scalars and vectors don't add, but have unfortunately - and understandably - got muddled up with how the operators are put together. If you put brackets around (v dot del) then it will make sense. This will be a scalar operator just like partial_t.

Paul Childs
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