How to understand the equality about $(\mathbf{q}\cdot \mathrm{grad})\mathbf{q} $? For the relation, 
$$(\mathbf{q}\cdot \mathrm{grad})\mathbf{q} = (\mathrm{curl}\mathbf{q})\times\mathbf{q}+\frac{1}{2}\mathrm{grad}|\mathbf{q}|^2,$$
is there any physics, geometry, or basic intuitive explanation? Thanks.
 A: You obtain the above relation by putting $\vec A=\vec B =\vec v$ in the following relation:
$$\nabla(\vec A\cdot \vec B)=(\vec A\cdot \nabla)\vec B+(\vec B\cdot \nabla)\vec A+\vec B \times (\nabla \times \vec A)+\vec A \times (\nabla \times \vec B)$$
Now as you can see, the physical/geometrical/intuititve explanation for your relation is the same as that for this relation.
And this can be easily done as L.H.S. denotes the gradient of the scalar field "the projection of $\vec A$ on $\vec B$" and hence on. 
A: to begin with this is down to algebra. if we let the components of the vector field $q$ be $(f,g,h)$ and use subscripts for differentiation then, looking at the $x$-component
$$
((\nabla\times  q) \times q)_x = ((\nabla\times  q)_yh-(\nabla\times  q)_zg \\
=(f_z-h_x)h-(g_x-f_y)g \\
= (f_yg+f_zh) - (g_xg+h_xh)
$$
if we insert an extra term $f_xf$ in each parenthesis the value of the whole is not changed and we have:
$$
(f_xf+f_yg+f_zh) - (f_xf+g_xg+h_xh) \\
= q\cdot \nabla f -\frac12 \frac{\partial}{\partial x}(f^2+g^2+h^2) \\
= \left(q\cdot  \nabla q -\frac12\nabla |q|^2 \right)_x
$$
A: In fluid mechanics, $\vec{q}$ is the velocity of the fluid. The term $\vec{q}\cdot\nabla\vec{q}$ is the advective part of the total time derivative of $\vec{q}$. It describes the change in the velocity of the fluid as one travels along a fluid parcel. Such a change can happen because of two reasons:
(1) The magnitude of $\vec{q}$ changes. It is described by the $\nabla(q^2)/2$ term or (2) The direction of $\vec{q}$ changes. This happens because of the fluid's vorticity. If the vorticity $\vec{\omega} = \nabla\times\vec{q}$ has a component normal to $\vec{q}$ then it induces a change in direction.
