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Vector Potentials

  1. Nov 23, 2012 #1
    It's not hard to show that the function:

    [tex]g = \frac{1}{2} (c \times r)[/tex]

    is a "vector potential" function for the constant vector "c". That is, that:

    [tex]\nabla \times g = c[/tex]

    The calculation is straightforward to carry out in Cartesian coordinates, and I won't reproduce it here.

    However, my question concerns the following. It is also a standard result in vector algebra that we have:

    [tex]a \times (b \times c) = (a \cdot c) b - (a \cdot b) c[/tex]

    My question is, when taking the "curl" of the vector potential above, why can't I just treat the "del" operator as though it were a vector, and write something like the following:

    [tex]\nabla \times (c \times r) = (\nabla \cdot r) c - (\nabla \cdot c) r [/tex]

    [tex]\nabla \times (c \times r) = 3c [/tex]

    where the last equality follows because (a) the divergence of a constant vector is 0; and (b) the divergence of the radial vector (i.e., the vector [x,y,z]) is 3.

    However, when factoring back in the factor of (1/2) to obtain the final "answer", the "answer" obtained using this method is (3/2)c, rather than the "correct" answer of c.

    Why is this?

    Is it because you can't treat the "del" operator as though it's a "vector" in this case?

    Or am I doing something else wrong?
  2. jcsd
  3. Nov 24, 2012 #2
    In general, you cannot treat the del operator as a vector because the "elements" of [itex]\vec{\nabla}[/itex] are not numbers, but partial derivative operators. As far as I know, the problem with this is that partial derivative operators don't follow commutativity. With real numbers, we don't usually think twice about replacing [itex]a b[/itex] with [itex]b a[/itex]. However, [itex]\frac{\partial}{\partial x} f(x)[/itex] is NOT the same as [itex]f(x) \frac{\partial}{\partial x} [/itex]. The first one is the derivative of f(x), while the second one is not even a function, it's like a weighted operator.

    So basically what tends to happen is that in proving an identity for vectors, you end up with something like [itex]a b[/itex] and you replace it with [itex]b a[/itex] without even thinking about it. That's fine, but you now can't use that identity for the del operator, because if [itex]a[/itex] is a partial derivative and [itex]b[/itex] is a function, that move was a big no-no. That's why you can't just plug the del operator into identities that have been derived for real numbers. What makes this so tricky is that a lot of the vector identities do actually work for the del operator, so it's very tempting to assume that they all do.

    I would guess that in the derivation of the vector identity you stated, you have to apply commutativity ([itex]ab = ba[/itex]) at some point.
  4. Nov 24, 2012 #3


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    hi psholtz! :smile:
    you can

    but (a.∇) is defined as the operator ax∂/∂x + ay∂/∂y + az∂/∂z …

    try it that way :wink:
  5. Nov 24, 2012 #4
    You still can't just plug in the del operator, though. The actual vector calculus identity takes a similar form to the vector identity, but it has four terms instead of just two:

    [tex] \nabla \times (\mathbf{A} \times \mathbf{B}) = \mathbf{A} (\nabla \cdot \mathbf{B}) - \mathbf{B} (\nabla \cdot \mathbf{A}) + (\mathbf{B} \cdot \nabla) \mathbf{A} - (\mathbf{A} \cdot \nabla) \mathbf{B} [/tex]

    (I stole the LaTeX code directly from the wikipedia "Vector Calculus Identities" page)

    Notice that even if you replace [itex]\nabla[/itex] with an actual vector like [itex]\mathbf{C}[/itex], you don't end up with the same identity written in the original post. You end up with an extra factor of two.

    I think the best policy is just to treat vector calculus identities (involving the div operator) and "standard" vector identities as completely separate. Certainly there are some enticing parallels, but there are also things that just don't work.
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