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I know that $\frac {dv}{dt}=a$ is acceleration, but:

  1. What is convective acceleration of a flow velocity?

  2. What is difference between $(v\cdot \nabla) v$ and $v\cdot (\nabla v)$?

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How about making an attempt and showing people where you are stuck? It is much easier to help you if you do this. :-) – Magpie May 5 '13 at 1:09

Let $\mathbf v(\mathbf x,t)$ denote the velocity of the fluid at position $\mathbf x$ and time $t$.

Suppose that we imagine traveling on a path $\mathbf x(t)$ through the fluid, then we can ask ourselves

If we travel along the curve $\mathbf x(t)$ through the fluid, then what will be the rate of change of the flow velocity of each point in the fluid that we pass through as we move through the fluid?

The answer is called the convective acceleration with respect to the curve $\mathbf x(t)$ and is obtained as follows using the chain rule:

$$ \mathbf a_{\mathrm{cv}, \mathbf x}(t) = \frac{d}{dt} \mathbf v (\mathbf x(t),t) = \dot{\mathbf x}(t)\cdot \nabla\mathbf v(\mathbf x(t),t) + \frac{\partial \mathbf v}{\partial t}(\mathbf x(t), t) $$

Now, suppose that $\mathbf x(t)$ represents a curve that is moving along with the fluid itself $$ \dot{\mathbf x}(t) = \mathbf v(\mathbf x(t), t) $$ then, we get the following expression for the convective acceleration: $$ \mathbf a_{\mathrm{cv}}(t) = \mathbf v(\mathbf x(t),t)\cdot \nabla\mathbf v(\mathbf x(t),t) + \frac{\partial \mathbf v}{\partial t}(\mathbf x(t), t) $$ Which is usually abbreviated as $$ \mathbf a_\mathrm{cv}(t) = \mathbf v\cdot\nabla \mathbf v + \frac{\partial\mathbf v}{\partial t} $$ As for your second question, there is no mathematical difference between the results of $(\mathbf v\cdot\nabla) \mathbf v$ and $\mathbf v\cdot(\nabla \mathbf v)$, but the difference in the order of operations here means that different sub-parts of the expressions mean different things.

In the first case, we first form the differential operator $$ \mathbf v\cdot\nabla = \sum_i v_i\frac{\partial}{\partial x_i} $$ and then we apply this operator to each component of the velocity vector field to obtain $$ (\mathbf v\cdot\nabla)\mathbf v = \left(\sum_i v_i\frac{\partial v_1}{\partial x_i}, \sum_i v_i\frac{\partial v_2}{\partial x_i}, \sum_i v_i\frac{\partial v_3}{\partial x_i}\right) $$ In the second case, we first component-wise gradients to obtain $$ \nabla\mathbf v = (\nabla v_1, \nabla v_2, \nabla v_3) $$ and then we dot the result with $\mathbf v$ to obtain \begin{align} \mathbf v\cdot(\nabla\mathbf v) &= (\mathbf v\cdot\nabla v_1, \mathbf v\cdot\nabla v_2, \mathbf v\cdot\nabla v_3) \\ &= \left(\sum_i v_i\frac{\partial v_1}{\partial x_i}, \sum_i v_i\frac{\partial v_2}{\partial x_i}, \sum_i v_i\frac{\partial v_3}{\partial x_i}\right) \end{align} which is the same result as the first case.

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I do not like the above answer because it does not explain well why $X(t)=V(x(t),t)$. I like the start, it actually led me to my version: $$\frac{df(x(t),t)}{dt}=\frac{d(f(x),t)}{dx}\times\frac{dx}{dt} + \frac{df(x(t),t)}{dt}$$ $$\frac{dx}{dt}=v$$ I think it is called Maxwell's partial differentials. Please do comment. – user71640 Jan 26 '15 at 14:32
@ArthurMabentsela There is a dot above the $\mathbf{x}$ which denotes a time derivative. The equation $\dot{\mathbf x}(t) = \mathbf v(\mathbf x(t), t)$ is simply the statement that if the particle is moving with the fluid, then its velocity matches that of the fluid. – joshphysics Jan 26 '15 at 20:25

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