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How does DEL(dot)V differ from V(dot)DEL?

  1. Jun 13, 2012 #1
    I came across this in Griffith's E&M book and can't quite figure it out. After going through the vector product rules with DEL. I noticed sometimes the terms are written as V[itex]\cdot[/itex]∇ and other times it's ∇[itex]\cdot[/itex]V. I searched more and I found things like convective operators and operators in general but it made things more confusing for me.

    If it is V[itex]\cdot[/itex]∇ then does that mean it's:

    [itex]V_x\frac{d}{dx} + V_y\frac{d}{dy} + V_z \frac{d}{dz}[/itex] ?

    I'm not sure how to interpret that expression if it correct..
  2. jcsd
  3. Jun 13, 2012 #2
    It is [STRIKE]less of an expression and more of[/STRIKE] an operator, if that helps.

    [itex]V\cdot\nabla[/itex] is ready to act on a function f(x,y,z). In other words, [itex](V\cdot\nabla)f[/itex] is the same as [itex]V\cdot(\nabla f)[/itex]. In the end, (after operating on a function f) you get another "scalar field", rather than a vector field.

    I'm just saying scalar field to confuse, it's another way of saying (single-valued) function (e.g. over ℝ3).
    Last edited: Jun 13, 2012
  4. Jun 13, 2012 #3
    I saw a homework problem in this or a related class, where one was asked something about ∇∇f. Or something like that (something bad), and you had to prove some identity. But at that level, there was some ambiguity as to interpret, and you could easily interpret it incorrectly. The only way to resolve the ambiguity, was to sift through some very abstract and confusing... I think it was tensor analysis, and lo I found which way I was to interpret this notation.

    The moral is, in Griffith's, you will encounter some strange notations, but I think you'll find that while you might never have seen them in any courses, you can use what you've already learned before to interpret them, without expecting any ambiguity.
  5. Jun 14, 2012 #4


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    If [itex]\vec{f}[/itex] is a vector valued function then [itex]\nabla\cdot \vec{f}[/itex] is a scalar function while [itex]\vec{f}\cdot\nabla[/itex] is an operator that will map a vector valued function to a scalar.
  6. Jun 14, 2012 #5
    There are some frustrating abuses of notation in Griffiths. Another thing you must be careful of is when you show Maxwell's equations lead to the wave equation, and you want to show some solution for the E-field satisfies ∇[itex]^{2}[/itex][itex]\vec{E}[/itex]=[itex]\frac{1}{c^{2}}[/itex][itex]\frac{\partial^2}{\partial t^2}[/itex][itex]\vec{E}[/itex]. Make sure you are careful about using the vector laplacian rather than the scalar laplacian. (Remember, vectors equal vectors!) This may seem trivial in cartesian coordinates, but it is often messed up in other coordinate systems.
  7. Jun 14, 2012 #6
    True, finding a digestible treatment of all the derivatives (grad, curl, div, laplacian) in various coordinates is hard to find at this level, and so the inside cover is your friend. Notice how they act on the spherically symmetric V=k/r and [itex]\overline{E}=\frac{k}{r^2}\widehat{r}[/itex].

    Boas in Mathematical Methods in the Physical Sciences gives a reasonable treatment in Chapter 10, Tensor analysis, section 9 Vector operations in curvilinear coordinates. I found difficulties in getting the notation to agree with what I had learned from Lee's Introduction to Smooth Manifolds, where we work more with the exterior derivative, which when given a metric can be related to grad div and curl (laplacian is just composition of div and grad?). Boas would get you there much faster; Lee with better care.

    (My challenge to those who have used Boas and Lee: how to interpret the following in Lee's framework, equation 8.6 in ch 10 of Boas:
    [tex]d\textbf{s}=\textbf{e}_rdr+\textbf{e}_\theta rd\theta+\textbf{e}_zdz[/tex]
  8. Jun 14, 2012 #7
    True, but the inside cover doesn't give, for example, the vector laplacian in spherical coordinates, which is a PITA to do. :D
  9. Jun 14, 2012 #8
    Thanks for all the replies.

    I'm catching on to the idea of the math but I'm having trouble interpreting V⋅∇ physically. In those product rules this comes up:

    (V⋅∇)F = [itex]V_x\frac{dF}{dx} + V_y\frac{dF}{dy} + V_z \frac{dF}{dz}[/itex]

    The divergence of F multiplied by V, what is the physical meaning of that?
  10. Jun 15, 2012 #9
    Not everything needs to be physically interpreted, though it's a good brain exercise often. (So is trying to solve all the exercises in the book, ugh.) But if you insist, my best clue off the top of my head would be to dig through the book and find where Griffith's uses it, maybe in a proof. Maybe you'll find that it's buried in a long proof of some other fact. Nevertheless, maybe we can say that V dot del measure how much a function grows in the direction of V. Oh, isn't that the directional derivative, ha!!

    EDIT: multiplied by the length of V, since directional derivative uses a unit length vector usually (I've seen otherwise, can't remember where, makes no sense).
  11. Jun 15, 2012 #10


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    The word "multiplied" is a bit vague. You need to be clear about the difference between multiplying a vector by a scalar, and the dot-product of two vectors.

    ∇F is the gradient of a function, i.e. the direction of the vector ∇F is the direction of the maximum rate of change of F, and the magnitude of ∇F tells you how big or small the rate of change is.

    V . ∇F is like any other scalar product. It depends on the magnitudes of V and ∇F, and the angle between them. You can interpret it either as "the amount of ∇F in the direction of V" or "the amount of V in the direction of ∇F".

    A particular case is when V . ∇F = 0, whcih means F doesn't change if you move in the direction of V. In other words, V points in a direction that lies in the plane perpendicular to ∇F.
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