I am reading a paper, and see someone decompose a divergence operator as follows, could someone judge and see if it is correct?

$$\nabla \cdot {\bf{v}} = \left( {{\bf{n}} \cdot \nabla } \right){v_n} + {\nabla _\parallel } \cdot {{\bf{v}}_t}$$

$\bf{v}$ is velocity vector field, and $n$ means normal, $t$ is tangent component.

And I do not understand why ${\bf{v}}_t$ is a vector and $v_n$ is not?

  • $\begingroup$ This would be more appropriate for math forum, but someone might want to try to answer here as well $\endgroup$
    – Pygmalion
    May 25, 2012 at 19:25
  • $\begingroup$ Comment to the question(v2): It seems the paper is implicitly assuming that the unit normal ${\bf n}$ is a constant vector field. If ${\bf n}$ is non-constant, there will be more terms on the right-hand side. $\endgroup$
    – Qmechanic
    May 25, 2012 at 19:50
  • $\begingroup$ More terms? Can you elaborate? I dont think $\bf{n}$ is constant, it is surface normal unit vector, and this is a CFD paper, the surface is a bluff body, it is flow around a bluff body. $\endgroup$
    – Daniel
    May 25, 2012 at 19:55
  • $\begingroup$ Well, I was referring to the case where ${\bf n}$ is a non-constant bulk vector field. We were not informed that ${\bf n}$ only lives on a surface. $\endgroup$
    – Qmechanic
    May 25, 2012 at 20:05

1 Answer 1


This might help. Divergence operator is by definition

$$\nabla = \frac{\partial}{\partial x} \vec{i} + \frac{\partial}{\partial y} \vec{j} + \frac{\partial}{\partial z} \vec{k}.$$

and velocity can be written as

$$\vec{v} = v_x \vec{i} + v_y \vec{j} + v_z \vec{k},$$


$$\nabla \cdot \vec{v} = (\frac{\partial}{\partial x} \vec{i} + \frac{\partial}{\partial y} \vec{j}) \cdot (v_x \vec{i} + v_y \vec{j}) + \frac{\partial}{\partial z} \vec{k} \cdot v_z \vec{k} = \nabla_{xy} \cdot \vec{v}_{xy} + \frac{\partial }{\partial z} v_z.$$

Now, let's just put Cartesian coordinate system in a way that $z$ is along normal component...

  • $\begingroup$ No.1 Why did you use $\nabla \vec{v}$ instead of $\nabla \cdot \vec{v}$? $\endgroup$
    – Daniel
    May 25, 2012 at 19:47
  • $\begingroup$ No.2 ${{\vec v}_{xy}}$ just has two component, is it okay to call it vector? And then use a bold print? $\endgroup$
    – Daniel
    May 25, 2012 at 19:49
  • $\begingroup$ Yes, it is scalar product, so dot is appropriate. I'll correct. It is OK and it must be called vector. The result must be scalar. $\endgroup$
    – Pygmalion
    May 25, 2012 at 20:07

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