1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Confusion about properties of the del operator

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data
    I was reading some notes about the del operator, and they make the statement

    ∇ x (Ua) = U(∇ x a) + (∇U) x a.

    However, I disagree with this because it seems to me that in the right hand side of the equation for the second term, the ∇ is operating on a, since a appears after it.

    I also worked on some other examples to see if understood the properties of del. For instance, for an arbitrary vector function A and position vector r,

    (A dot ∇)r = A and

    (A x ∇) dot r = 0.

    Both of these examples seem to confirm that ∇ operates on what appears after it (crossed, multiplied, or dotted, but not added or subtracted).

    If I expand the right hand side of the above equation, I don't get the left hand side.

    I think that the statement should read
    ∇ x (Ua) = U(∇ x a) + -a x (∇U)

    Could somebody help clear my confusion? Thanks.

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 3, 2009 #2


    User Avatar
    Homework Helper

    I would probably write it the way you do, but botth forms are correct.
  4. Mar 3, 2009 #3

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No, this goes back to the order of operations that you learned in arithmetic. Compute what's in the parentheses first, which is the vector [itex]\nabla U[/itex]. Then cross that vector with the vector [itex]\vec{a}[/itex]. In any case it would make no sense for [itex]\nabla[/itex] to act on [itex]\vec{a}[/itex]. Gradients act on scalars, not vectors.

    It does act on [itex]\vec{r}[/itex]. That's because [itex]\vec{A}\cdot\nabla[/itex] and [itex](\vec{A}\times\nabla )\cdot[/itex] are operators, not vectors.

    I do get it to work out. So perhaps you should post what you've done so we can examine it.
  5. Mar 3, 2009 #4


    User Avatar
    Homework Helper

    The del operator is a differential operator, compare the first equation you gave with the ordinary product rule with a(x,y,z) and U(x,y,z) as U is scalar it can be taken to either side of the cross product without affecting the product.

    So the equations you have written at the top & bottom are actually equivalent as you have changed the order of the cross product

    have a look at this
  6. Mar 3, 2009 #5

    Thinking back to (A x ∇) dot r = 0.
    (A x ∇) is a vector isn't it?

    and ∇ is a vector and an operator.

    Could you please clarify what you mean? Thanks.

    When I expand (∇U) x a, I am doing it like this
    (U daz/dy + az dU/dz - U day/dz - ay dU/dz)i + ...
    Last edited: Mar 3, 2009
  7. Mar 3, 2009 #6

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I wouldn't call [itex]\nabla[/itex] a vector. When I say "vector" (in normal 3d space), I mean an element of [itex]\mathbb{R}^3[/itex], which [itex]\nabla[/itex] certainly is not. Rather, [itex]\nabla[/itex] is an operator on [itex]\mathbb{R}^3[/itex]. And so is [itex](\vec{A}\times\nabla )\cdot[/itex].
  8. Mar 3, 2009 #7


    User Avatar
    Homework Helper

    ∇ is a vector operator
    it is not a vector
    treating it as if it is a vector will lead to much confusion
    for example b.(axb)=0
    b.(∇xb) need not be 0
  9. Mar 3, 2009 #8
    I think I'm starting to understand what you are saying. ∇U is the operation on a scalar function which yields a vector function (gradient), so that really can't operate on anything. However, when you dot or cross ∇ with another vector, that does operate on the terms after it. Right?
  10. Mar 3, 2009 #9


    User Avatar
    Homework Helper

    It is true that the usual ∇ is a right acting operator. Another symbor should be used otherwise. It is very dangerous (as in wrong) to think of say crossing ∇ with a vector, it is much better to think of
    as three different operators
    then there are things like
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook