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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

    lurflurf

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

    Tom Mattson

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    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

    lanedance

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    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
    http://en.wikipedia.org/wiki/Del
     
  6. Mar 3, 2009 #5
    Tom,

    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

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    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

    lurflurf

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    ∇ 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

    lurflurf

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    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
    ∇()
    ∇.()
    ∇x()
    as three different operators
    then there are things like
    a.∇()
    (ax∇).()
    (ax∇)x()
    (ax∇)()
     
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