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

  • 2
    $\begingroup$ 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). $\endgroup$ Commented May 21, 2017 at 11:06
  • 2
    $\begingroup$ Related, if not a dupe: physics.stackexchange.com/q/160229/25301 $\endgroup$
    – Kyle Kanos
    Commented May 21, 2017 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. $\endgroup$ Commented May 21, 2017 at 21:40
  • $\begingroup$ For me personally, this answer was much more helpful than the existing answers. I wasn't wondering about the physical meaning though, just how exactly the notation works. physics.stackexchange.com/a/576977/71668 $\endgroup$
    – MaxD
    Commented Feb 9, 2021 at 19:57

2 Answers 2


The reason is that $\mathbf v\cdot (\nabla \mathbf v)$ is ambiguous:

  • it could mean $\displaystyle v_i\frac{\partial v_j}{\partial x_i}$
  • ...but it could also mean $\displaystyle v_i\frac{\partial v_i}{\partial x_j}$.

In other words, $\nabla\mathbf v$ 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.


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 tensor field changes in time with (assuming that $\partial_tf(\textbf{x})=0$)


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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.