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Meaning of (A dot nabla)B

  1. Mar 21, 2012 #1
    What does this mean?

    [itex](\vec A\cdot\nabla)\vec B[/itex]

    I read somewhere that

    [itex](\vec A\cdot\nabla)\vec B=\vec A\cdot\nabla\vec B[/itex]

    but this must be nonsense since you can't take the gradient of a vector.
  2. jcsd
  3. Mar 21, 2012 #2

    Char. Limit

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    Well, you could try extending it out by components. Assuming three-space, we have as such:

    [tex]\left(a_x \frac{\partial}{\partial x} + a_y \frac{\partial}{\partial y} + a_z \frac{\partial}{\partial z}\right) \left(b_x \hat{i} + b_y \hat{j} + b_z \hat{k}\right)[/tex]

    Which would be the interpretation I'd go with. What that means, on the other hand, is harder to determine.

    EDIT: Should probably say that I assumed [itex]\vec{A} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}[/itex], and same with B.
  4. Mar 22, 2012 #3
    What you're talking about is called the Advective or Convective operator and describes the change in a property due to flow of continuous media (in Fluid Mechanics anyway).

    Char limit is correct. Don't worry about whether it makes sense or not the way it's written because it is essentially taking notation abuse to the next level (similarly to the way curl and divergence are written as cross and dot products of the gradient with the vector!). This is what the thing actually looks like:

    (\vec A\cdot\nabla)\vec B=\vec A\cdot\nabla\vec B=\left(a_x \frac{\partial}{\partial x} + a_y \frac{\partial}{\partial y} + a_z \frac{\partial}{\partial z}\right) \vec B=a_x \frac{\partial\vec B}{\partial x} + a_y \frac{\partial\vec B}{\partial y} + a_z \frac{\partial\vec B}{\partial z}[/tex]
    [tex] \vec A=(a_x,a_y,a_z) [/tex]

    Does that help? Can you see how it can be written both ways? You can think of the gradient of a vector as a vector of vectors, but as I said, don't worry about doing that. You should use the formula you wrote initially only as a sort of mnemonic device for what it actually is (the final formula I wrote). You should note, however, that when you're actually doing the calculations (I'm assuming you're seeing this in the Navier-Stoke's Equations), you won't do this as one big vector, but rather you'll have separate equations for the x, y, and z components of B.
    Last edited: Mar 22, 2012
  5. Mar 22, 2012 #4
    Yes, thanks a bunch.
  6. Mar 22, 2012 #5


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    You SURE can take the gradient of a vector!
    What you end up with is a matrix, and your left-hand side in your identity is a scalar operator on a vector, yielding a vector C, whereas your right hand side is a vector-matrix product, yielding a vector D..
    The equality sign means that C=D
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