Is it true that $(\mathbf{v}\cdot\nabla)\mathbf{v} = \mathbf{v} \cdot (\nabla \mathbf{v})$?

Question

I have seen that sometimes the Navier-Stokes equations are written with the term $(\mathbf{v}\cdot\nabla)\mathbf{v}$ expressed as $\mathbf{v} \cdot \nabla \mathbf{v}$. However, is it true in general the following equality for any vector field $\mathbf{v}$?

$$(\mathbf{v}\cdot\nabla)\mathbf{v} \stackrel{?}{=} \mathbf{v} \cdot (\nabla \mathbf{v})$$

A couple of examples: in Batchelor's An Introduction to Fluid Dynamics, equation (2.1.2) defines the mass derivative of the velocity field (which is the LHS of the NS equation) as

$$\frac{\partial \mathbf{u}}{\partial t}+\mathbf{u} \cdot \nabla \mathbf{u}$$

Whereas, in Landau-Lifschitz Fluid Mechanics, 2nd edition, the Navier-Stokes equation is written in equation (15.7) as

$$\frac{\partial \mathbf{v}}{\partial t}+(\mathbf{v} \cdot \nabla) \mathbf{v}=-\frac{1}{\rho}\nabla p+\frac{\eta}{\rho} \Delta \mathbf{v}$$

On the other hand, in Wikipedia, it appears written both ways in the same article of the Navier-Stokes equations.

Answer 1

The equality is true because both sides in index notation are $$ v^\mu\nabla_\mu v^\nu\,. $$ On the level of components this "multiplication" is associative. Then we sum over the repeated index $\mu$. What this describes is the directional derivative of the vector field $v$ into the direction of $v$.

Answer 2

According to the NRL Plasma Formulary,

$$ \nabla\cdot(\mathbf{AB})=(\nabla\cdot\mathbf{A})\mathbf{B}+(\mathbf{A}\cdot \nabla)\mathbf{B} $$

For $\mathbf{A=B=v}$, we have

$$ \nabla\cdot(\mathbf{vv})=(\nabla\cdot\mathbf{v})\mathbf{v}+(\mathbf{v}\cdot \nabla)\mathbf{v} $$

I don't think I've ever seen the Navier-Stokes equations expressed as other than $\mathbf{v\cdot\nabla v}$, and clearly, from the above, this is not equal to $(\mathbf{v}\cdot \nabla)\mathbf{v}$.