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Suppose I have a steady flow and I want to find the rate of change of pressure of a bit of fluid. This depends on the velocity of the fluid and the pressure gradient,

$$\frac{\mathrm{d}P}{\mathrm{d} t} = \mathbf{v}\cdot (\nabla P)$$

The equation is simply saying that the pressure changes because I'm moving to a new place in the flow, so I take how quickly I'm moving there and multiply by how much the pressure is changing in that direction.

If I want to find the acceleration of a bit of fluid, conceptually it's the same thing - multiply the velocity by the velocity gradient in the direction of motion, so one could write

$$\mathbf{a} = \mathbf{v}\cdot (\nabla\mathbf{v})$$

(or leave out parentheses entirely). However, I more commonly see

$$\mathbf{a} = (\mathbf{v} \cdot \nabla) \mathbf{v}$$

In tensor notation it's the same thing, but it parses differently. $\nabla v$ has some intuitive physical meaning to me (a tensor for the gradient of the velocity field) whereas $(v\cdot \nabla)$ doesn't seem to parse into anything particularly meaningful on its own.

This $(v\cdot \nabla)$ notation avoids having to figure out what the gradient of a vector field is; is that why it's written that way? Or is there some other reason, such as that there's a good intuition for $(v\cdot \nabla)$ other than "the rate of change due to spatial variation operator", or am I overthinking a trivial bit of notation, or what?

example of $(v\cdot \nabla)v$ notation: https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations#Incompressible_flow

example of $v\cdot (\nabla v)$ notation: https://en.wikipedia.org/wiki/Material_derivative

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    $\begingroup$ It's done for the reason you said. For beginners who are not yet accustomed to tensors, the v dot del version is easier to relate to, and they don't have to deal with the gradient of a vector field (tensor). $\endgroup$ – Chet Miller May 21 '17 at 11:06
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    $\begingroup$ Related, if not a dupe: physics.stackexchange.com/q/160229/25301 $\endgroup$ – Kyle Kanos May 21 '17 at 11:53
  • $\begingroup$ This is definitely not a duplicate. The question here is about notational misunderstanding while the question linked is about whether the two notations are actually the same, which OP openly admits to knowing is true. $\endgroup$ – Bob Knighton May 21 '17 at 21:40
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The intuition for this statement is actually rather simple. While both statements are notationally the same, I personally think that $\left(\textbf{v}\cdot\nabla\right)\textbf{v}$ is a lot more intuitive and holds more physical meaning.

Let's consider viewing the world from the perspective of a particle in your flow. Then if I define a function $f(\textbf{x})$ (which can be any field -- scalar, vector, or tensor), I will measure that the tendor field changes in time with (assuming that $\partial_tf(\textbf{x})=0$)

$$\frac{\mathrm{d}f}{\mathrm{d}t}=\left(\textbf{v}\cdot\nabla\right)f(\textbf{x})$$

This is simply the chain rule. More intuitively, we see that the change of the function along the particle flow is just the directional derivative along $\textbf{v}$. Similarly, if we want to define the acceleration of a body along the flow lines, we take the effective time derivative of $\textbf{v}$, which gives us the result you claimed seemed unmeaningful. In fact, this operator is probably the most physically meaningful differential operator in the problem.

I hope this helped!

PS: Today I learned that writing $\LaTeX$ on your phone is incredibly slow and frustrating.

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The reason is that $\mathbf v\cdot (\nabla \mathbf v)$ is ambiguous: $$ \text{it could mean}\ \ v_i\frac{\partial v_j}{\partial x_i}, \ \ \text{but it could also mean}\ \ v_i\frac{\partial v_i}{\partial x_j}, $$ i.e. $\nabla\mathbf v$ is an asymmetric tensor and the notation $\mathbf v\cdot (\nabla \mathbf v)$ does not specify which of the two tensor sectors gets addressed by the contraction. Placing the contracting tensor on the left makes it a good bet that it contracts with the derivative (and this can be done so long as you specify your notation up front), but it's not completely unambiguous the way that $(\mathbf v\cdot\nabla)\mathbf v$ is.

Similarly, the way I see it, the notation $(\mathbf v\cdot\nabla)$ makes immediately obvious a message along the lines of "this isn't really a full gradient, it's just the directional derivative along $\mathbf v$ that's in play here", which is often precisely the message that needs to be transmitted.

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