# “Vector add to scalar” in left part of Navier-Stokes equation

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.

• directional derivative – timur Dec 10 '18 at 7:40
• $\nabla\vec{v}$ 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 7:46
• @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 7:51
• @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 8:16
• @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

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

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.

• Scalars and vectors can be added in geometric algebra, though that's not happening here. – mr_e_man Dec 10 '18 at 7:46
• That's interesting. It occured to me after I said it that there was likely some abstract spaces where mixing could occur. – Paul Childs Dec 10 '18 at 7:51